text:
Is the S–O bond strength in sulfur trioxide or sulfite anion larger?
Can you please explain the difference between the bond enthalpies in sulfurtrioxide $\ce{SO3}$ and the sulfite anion $\ce{SO3^{2-}}$?

Due to symmetry constraints ($D_\mathrm{3h}$) in $\ce{SO3}$ there 6 electrons in $\pi$ type orbitals. In a wider sense of the term this molecule is Y-aromatic, but the HOMO actually represents two electrons in in-plane lone pair orbitals of oxygen. The LUMO is an antibonding $\pi$ orbital with respect to all $\ce{S-O}$ bonds.
The following valence molecular orbitals were derived at BP86/cc-pVTZ, Energy level decreases from top to bottom (enlarge).

The bonding picture is usually trivialised as each $\ce{S-O}$ bond being a double bond. But this is actually far away from the truth, as it does not respect the charge of $q=+2$ at the sulfur atom and the charges at the oxygens with $q=-\frac23$. This is due to the fact, that the sulfur atom actually only contributes to one of the three $\pi$ bonding orbitals, resulting in electron density deficiency.
However, the $\ce{S-O}$ bond in sulfurtrioxide is still stronger and shorter, than in the sulfite anion. (BO means bond order)
\begin{array}{lrr}
              & \mathbf{d}(\ce{S-O}) & \text{BO}(\ce{S-O})\\\hline
\ce{SO3}      & 1.458                & 1.4                \\
\ce{SO3^{2-}} & 1.578                & 1.1                \\\hline
\end{array}
The main reason for this is, that upon reduction of sulfurtrioxide to sulfite, $$\ce{SO3 + 2e- <=> SO3^{2-}},$$ you will fill the LUMO. As already stated, this orbital is antibonding with respect to the $\ce{S-O}$ bonds. This is energetically unfavourable for the molecule, therefore the molecule distorts to get rid of the symmetry constraints, i.e. it adapts the lower $C_\mathrm{3v}$ point group. This results in pyramidalization and a different bonding pattern.
You will now no longer find any $\pi$ type bonding orbitals. More accurately this molecule is best described consisting of three equal $\ce{S-O}$ covalent single bonds, and three lone pairs at each oxygen. This results in a charge at each oxygen of $q=-1$ and at the sulfur of $q=1$ (which does not fit the traditional Lewis type bonding description.)
This is also reflected in the molecular orbitals (BP86/cc-pVTZ, energy decreases from top to bottom, enlarge).

It is also very important to understand that the latter molecule would not be stable in gas phase, since it is too highly charged. Without compensation of solvent and/ or counter ions, this molecule will most certainly fragmentate.

@santimirandarp The blue colour of the sky can be explained as the reflection of the blue ocean; unfortunately that doesn't mean this is correct. The octet rule is an observation, it should not be used without the proper explanation when it is possible to simplify a concept with it, and when not. The partial charge of the central sulfur is approximately +2; that is not represented by the totally double-bonded structure.

@santimirandarp I don't understand what you were trying to say with 'where is the hole'. Models might be useful if they are not wrong, but one still has to know their limitations. In that case the limitation of the octet rule is 'up to period 2' (strictly); sulfur isn't in there, which makes it wrong to use. In any case, you need resonance to satisfy symmetry and the charge separated structure; or you cannot satisfy the octet rule... the models fail here; don't twist your brain to make them fit, be more open towards seemingly more complicated explanations.

My hypothesis is that the bond enthalpy of the S-O bond in the sulfate anion is smaller than that of the S-O bond enthalpy in sulfur trioxide. I would say that the S-O bond strength lies with atomic size.
In sulfate anion, we have the electronegative oxygen atoms stabilizing the bulk of the negative 2 formal charge.
Bearing in mind that anions are generally bigger than their respective neutral counterparts, and that cations are smaller than their respective neutral counterparts. Why? This lies with electron-electron shielding and $\ce{Z_{eff}}$, or effective nuclear charge.
Having more electrons in a valence, such as in the case of the oxygens in the sulfate anion as opposed to the oxygens in sulfur dioxide, decreases $\ce{Z_{eff}}$. This in turn allows the electrons to be further from the nucleus, and this obviously increases size.
The increased size of the oxygens in the sulfate anion in turn lengthen the S-O bonds, and this in turn decreases the bond enthalpy of the S-O bonds in the sulfate anion.

Sulfur Trioxide is typically represented with Sulfur double bonded to each of three oxygens.  In this context, the oxygens have full octets, while the Sulfur has an 'expansion'.
Because the oxygens are full (and sulfur is pretty close to 'full'), to add electrons to this molecule requires the placement of electrons into 'antibonding orbitals'.  These orbitals would be pi-antibonding orbitals, weakening the S-O bonds somewhat.
Please be advised, it is difficult to speak about the enthalpy of either bond(sigma or pi) within a double bond.  Typically the enthalpy of a double bond is compared to a single bond and the difference is assigned to the 'pi-bond'.


text:
Why is the C-Br bond considered polar?
Why is a C-Br bond considered polar? From electronegativity considerations, both carbon and bromine have very similar electronegativities - 2.5 and 2.8 - respectively.
Nonetheless, I am told that such a bond would be polar. We generally classify bond between atoms with EN differences < 0.5 as non-polar; why would C-Br be an exception?

What is your source for this view? Who thinks this bond is polar?

I poked around on the internet and found a lot of sites use the Pauling electronegativity scale (link 1, link 2).  For carbon the value is 2.55 and for bromine 2.96.  Not quite a 0.5 difference, but closer - and this "rule" is somewhat arbitrary to begin with.  Also the dipole moment of bromobenzene is around 1.7 D (link 3), certainly not insignificant.  I too would have said bromobenzene is polar.  It is not strongly polar, whether it is "a little" polar, "mildly" polar, "somewhat" polar, etc. seems to be the question - but that's just semantics.  Maybe the reason many people consider the $\ce{C-Br}$ bond to be polar is because it has a very observable effect on electrophilic aromatic substitution reaction rates.  Specifically, why should the relative rate for electrophilic aromatic substitution at the meta position in bromobenzene be slower that the rate in benzene itself unless you invoke the inductive effects caused by a polar $\ce{C-Br}$ bond?

I think that's what the person who made this claim had in mind, and it makes sense, thank you!

Would a C-Br bond show up on IR spectroscopy?

Yes, see  http://www.chem.ucla.edu/~bacher/General/30BL/IR/ir.html

Cool, does that also lend credence to the idea that the C-Br bond is polar?

Not sure, but I don't think so.  A requirement for ir absorption is for a vibration to produce a net change in the molecular dipole moment.  Couldn't stretching a non-polar bond in a polar molecule produce a change in the molecular dipole moment?

What about looking at NMR shifts, @ron? Would that give us a hint at the polarity of a C-Br bond?

That is possible, particularly with 13C-NMR.  You'd need the right model compounds, but there is a substantial  shift of around 180-200 ppm per unit charge in 13C-NMR spectroscopy.

The only truly non-polar bonds are between the same elements in the same environment. Everything else is polar. In the $\ce{C-Br}$ bond already the different masses and the larger radii would induce polarity. Whoever proposed this rule made a very crude approximation. @Dissenter there is therefore no reason, why this bond should be IR inactive. It has a dipole moment to begin with, so every displacement of the atoms will induce a change of the dipole moment. The remaining question is only, how intense this signal will be compared to the other signals of a molecule.

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sort: votes


text:
2p Orbital Occupancy

What does the question mean by how many electron pairs are in 2p orbitals? How does one tell? Given that both the oxygen and the nitrogen are $\ce{sp^2}$ hybridized, wouldn't there be two 2p orbitals?

In the resonance structure drawn at the top of your image, the $\ce{p}$ orbital on carbon and the $\ce{p}$ orbital on oxygen share a bonding pair of electrons, a lone pair of electrons reside in a nitrogen $\ce{p}$ orbital, two lone pairs reside in two oxygen $\ce{sp^2}$ orbitals.  So for this case, the most commonly drawn representation, there is one lone pair in a $\ce{p}$ orbital.  In your middle resonance structure, there is a lone pair in an oxygen $\ce{p}$ orbital, and another lone pair in the nitrogen $\ce{p}$ orbital; hence, 2 lone pairs in two different $\ce{p}$ orbitals.  In your bottom resonance structure, there is a lone pair in an oxygen $\ce{p}$ orbital and a shared bonding electron pair in the $\ce{C-N}$ $\ce{pi}$ bond; therefor, one lone pair in a $\ce{p}$ orbital for this case.
This seems to be a poorly worded question.  Since the real molecule is represented by a combination of the 3 resonance structures, the correct answer would be "somewhere between one and two lone pairs in $\ce{p}$ orbitals."  Adding the detail contained in the above paragraph would probably help the grader understand your explanation.

Thank you. How do pi electrons affect molecular geometry?

I'm not quite following, could you be more specific?

Usually when assigning geometries we only consider sigma bonds; boron trifluoride is sp2 just as a carboxylic acid group's carbon is sp2. Do pi electrons affect geometry; if so, how?

I still might be missing your point.  I'd say pi electrons affect geometry very much.  The bond length in ethylene is shorter than in ethane. The single bond in 1,3-butadiene is shorter than in ethane.  All of the C-C bonds in benzene have the same length, no alternation. That said, this could probably be a separate question.

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text:
Does any element have nutritional (caloric) value?
Is there any single element (e.g. possibly carbon) that can be eaten, which has a nutritional value such that it provides calories?
I found on Wikipedia (dietary element) that there are certain elements that our bodies require.  I suppose oxygen could count as such a nutritional element, although we breathe it instead of eat it.

A bit tricky question since most elements consumed are converted into their ionic form (as chloride, per the acid in our stomach) or, as one has noted, per the action of bacteria in our gut.

It is not the substance but the chemical reaction that has caloric value. So if elemental carbon can be oxidized by elemental oxygen in the body and the products can be removed, it has caloric value.

While your question refers to human metabolism, it should be noted that bacteria can use elemental hydrogen, elemental sulfur, and elemental iron as energy sources:
See Wikipedia's page on Microbial metabolism, particularly the section on chemolithotrophy.
If any of these bacteria are found in the human gut, and their metabolic processes support our own (as is the case for symbiotic gut bacteria), then effectively we can use these elements for energy as well.

I don't follow your logic. If the bacteria ozidize these fuels, they will use the energy generated to synthesize more biomass or to power their flagella or any number of other metabolic processes, none of which provide energy to the human host. That extra biomass just means more cells go out with our excreted waste.

@Andrew  Much of the energy we produce goes towards supporting homeostasis -- basic maintenance of our biology.  Many gut bacteria are symbotic, thus contributing to this task. The idea is that while our main metabolic engines are powered by carbs/proteins/fats, we are also supported by auxiliary engines (within the bacteria) that run on the fuel sources they use.   And if they use, say, elemental sulfur to run their engines, and their engines help to power/support us, then effectively we are partly being powered by metab. of sulfur as well. I've added a phrase in my answer to reflect this.

Interesting question! You may get a better answer, but I believe the answer is no.
Calories are units of heat and represent energy available to the body. The form of fuels that can be (biochemically) "burned" by the body are hydrocarbons; i.e. molecules of carbon and water, like sugars. Carbon alone I do not believe can work.
The only other single element fuel is hydrogen, and again I do not think this is biologically available to the body. By the way, oxygen is required as the oxidizer in any combustion reaction, and we definitely need oxygen (from breathing) to convert the food we eat to energy. But oxygen is not a fuel: it is the oxidizer that is reduced during combustion.

Caloric value is basically stored energy.  Plant-based foods create this stored energy through the absorption of sunlight, and in some extreme situations, I imagine there may be plants that generate energy from thermal vents and such.  They then store that energy in a molecular combination that you can think of as compressing a spring, giving it potential energy.
The same goes for animals, whereas rather than sunlight that energy is created from eating either plant or other animal life.
Mass inherently has energy only at the mass-energy equivalent level, and biologically we are not capable of accessing that.
From the conservation of energy, if the element you are putting in your body is not part of a molecular "energy spring" as per the analogy above, it will not spontaneously posses energy just because you are ingesting it.
As far as Carbon, this is one of the great misunderstandings about fuel.  We burn Hydrocarbons as a fuel source.  Wood originally, then coal, and now we have jet fuel.  They are basically all the same thing, with decreasing Carbon to hydrogen ratios.  The energy stored is actually being released from the hydrogen bonds.  The ideal goal is for us to be using carbon-less hydrocarbons as fuel sources as the only burned waste is water as it combusts with the air, no CO2 providing the highest weight to fuel ratio (no carbon along for the ride reduces the weight) and doesn't pollute.

Does this mean that there are no single elements that provide nutritional value (e.g. it has to be a molecule to store chemical energy that can be used by the body?)

To the best of my knowledge, yes.  Then again, there is always someone who knows more than you.

@Jonathan I'm positive that there is no single atom which can provide energy to the body. Even if one of the elements were able to function as fuel alone (and everything I know about digestion tells me that none does), it would be a molecule of that element, as in H2. The body processes food chemically. To extract energy from a single atom, you'll need a nuclear reaction. So, quite literally, it has to be a molecule.

Most metals consumed (albeit in only tiny amounts for select elements) will be directly converted to ions by the HCl in our stomach.
Then, as the question includes the word 'nutritional', commonly cited elements (in their ionic form) of particular import include: potassium/sodium (regulates fluid balance, muscle contractions, and nerve signals), iron (linked to hemoglobin and the transportation of oxygen in the blood), calcium/magnesium (maintains strong bones), iodine (required for creation of thyroid hormones and regulating the body's metabolism), chromium (important in the breakdown of fats and carbohydrates) and copper (instrumental in formation of red blood cells, maintenance of nerve cells and the immune system), to mention a few.
Of the mentioned elements, those having a caloric impact include chromium and iodine.
Note, elemental carbon, especially in the form of charcoal pills or activated carbon, should be avoided (not nutritional or even caloric), see discussion here.
Further, I would argue, that as the human body generates a chemical-based electric current (see this discussion), it is probably not wise to introduce a very good cathode (namely carbon). Unwanted solvated electrons and radicals created therefrom, most likely do not promote good health, and in ones' body, may supply an electrode to promote electrochemical (or battery) cell activity.

I clarified my consumption of elements are there are converted to their ions form, in our stomach, except for elemental carbon, which I discussed separately. I do not cite oxygen. See added comment above.

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text:
Crystallization of Glycerin
Is it possible to crystallize vegetable glycerin?
I saw a myth about it online and was wondering if it is possible.

Hi Chris, welcome to Chemistry SE. I suggest you put in a link to the page you read this myth on to help us answer your question better.

I guess this link explains a lot : http://www.gamefaqs.com/boards/961351-nine-hours-nine-persons-nine-doors/57582553

yes it is possible, for example by adding a seed crystal under the right conditions.
See the reference for further information: http://www.google.com/patents/US2233606


text:
Why is it ACETone?
Acet* indicates Ethyl, but does not offer two Carbon atoms, but three.
Acetaldehyde is Ethanal, Acetic acid is Ethanoic acid, but Aceton is Propanon (yes, I'm aware that there is no Ethanon).
Why is Acetone called Acetone?

Here's a link to the first page of a book entitled, "The History of Acetone, 1600-1850" by Mel Gorman.  The author points out that acetone was known in the Middle Ages and was frequently produced by heating dry lead acetate.  I suspect that the "acetate" (or whatever the Latin, French or German term was) root stuck and then it was just modified a bit - to acetone - to make it distinct.

But heating lead acetate produces acetic acid, not acetone...

see http://www.jstor.org/discover/10.2307/27757220?uid=3739568&uid=2129&uid=2&uid=70&uid=4&uid=3739256&sid=21103817468101

Could you update your question with quotes and source?

Updated as you requested.

According to Fownes' Elementary Chemistry 12th Edition (pp. 676ff), ketones are obtained from the dry distillation of calcium salts of the fatty acids. So calcium acetate yields acetone, calcium propionate yields "propione" (diethyl ketone), etc. Lead salts also are stated to work.

The naming is done on the basis of the number of carbons present in the parent chain.The first member of each group is "form-". However we do not have a "formone" as ketonic groups are non-terminal.The $\ce{CH3CO-}$ groups are called "acetyl" groups(as they already contain 2 atoms of carbon) and thus the very first compound that it can form with a methyl group is propanone or acetone.
However these are old-fashioned names and do not really need much intervention.The IUPAC nomenclature has logic and rightly names the compound "propanone".


text:
(+)-Bicuculline, what does (+) mean?
I was looking up some ion channel blockers. What does (+) or (-) below mean?
(+)-Bicuculline
(-)-Bicuculline methiodide

Bicuculline is a chiral molecule.  That means that its mirror images are non-superimposable.  These non-superimposable images are termed "enantiomers."  When you see a "(+)" or "(-)" in front of a chemical name it tells you that only one the enantiomers is present (e.g. it is not a racemic mixture of the enantiomers).  Chiral molecules have the ability to rotate the plane of a polarized light beam that passes through it. Enantiomers will rotate the plane of polarization in exactly equal amounts (same magnitude) but in opposite directions.
Dextrorotary designated as (+), indicates clockwise rotation (to the right)
Levorotary designated as (-), indicates anti-clockwise rotation (to the left)

You should also mention that dextrorotary is designated as D (D-Glucose) and levorotary as L (L-tyrosine). But there are some exceptions. L-alanine and a few more L-amino acids are dextrorotary and some D-sugars are levorotary.

@ron could you please comment why http://www.chemistry.mcmaster.ca/~chem2ob3/nhw_temp/old_old_labmanual/expt1/2ob3-exp1b.html page shows (R)-(-)-Carvone? Wouldn't R (to the right) match with a (+) ?

@K-Feldspar No, R and S are nomenclature terms that describe the spatial configuration at a chiral carbon using the Cahn-Ingold-Prelog priority system. (+) and (-) describe the direction of rotation of polarized light when it passes through the sample.    There is no relationship between R and S and (+) and (-).


text:
Which ion of iron is produced in a reaction between iron and copper(II) sulfate?
Using these equations:
\begin{aligned}
\text{(Fe2):}&&\ce{ Fe + CuSO4 &-> FeSO4 + Cu}\\
\text{(Fe3):}&&\ce{ 2Fe + 3CuSO4 &-> Fe2(SO4)3 + 3Cu}
\end{aligned}
Beginning with $\pu{0.78 g}$ of iron, the theoretical yields of copper are $\pu{0.89 g}$ (with $\ce{Fe^{2+}}$) and $\pu{1.33 g}$ (with $\ce{Fe^{3+}}$), and the actual yield was $\pu{0.78 g}$ of copper. I know to find of the ratio of the moles of iron used to the moles of copper formed (88% for $\ce{Fe^{2+}}$, 58% for $\ce{Fe^{3+}}$), and so the clear choice would be that this reaction formed $\ce{Fe^{2+}}$ ions.
But my teacher said this would form $\ce{Fe^{3+}}$ ions, but why is that? Shouldn't it have formed $\ce{Fe^{2+}}$? The reaction formed $\pu{0.12 mol}$ of copper, compared to the $\pu{0.14 mol}$ in the theoretical yield, and the other $\pu{0.02 mol}$ could have just been lost somewhere along the reaction process.

Judging from the standard potentials of the reactions involved, equation (Fe2) should be correct: http://chemistry.stackexchange.com/questions/13569/which-ion-of-iron-to-use/14793#14793

When "doing" science, usually you try to (in-)validate a model of the workings of the world. In the case of chemists, this world is mostly confined to the scale of molecules.
That said, let's proceed with
A Theoretical Model
Listed below are some standard reduction potentials I've gathered from the Handbook of Chemistry and Physics, 95th ed. (probably paywalled for you):
$$ \begin{align}
&&E^0_\text{red} / \mathrm{V}\\
\hline
\ce{Fe^{2+} + 2e-  & <=> Fe} & -0.447 \\
\ce{Fe^{3+} + 3e-  & <=> Fe} & -0.037 \\
\ce{Cu^{2+} + 2e-  & <=> Cu} & 0.342 \\
\end{align} $$
Now we can calculate the potential (and the resulting electromotive force EMF) for each of the reactions:
$$ \begin{align}
E^0_\text{(Fe2)} = E^{0}_\text{red}(\text{red}) - E^0_\text{red}(\text{ox}) &= 0.342~\mathrm{V} - (-0.447~\mathrm{V}) = 0.789~\mathrm{V} \\
E^0_\text{(Fe3)} &= 0.342~\mathrm{V} - (-0.037~\mathrm{V}) = 0.379~\mathrm{V}
\end{align} $$
It is clear that the reaction (Fe2) releases more energy and thus should be preferred over the other reaction. This means that we expect a yield of 100% Cu and not 150% Cu.
Now that we have gathered some theoretical understanding of the matter, let's proceed to the interesting part, which is
Confirmation With Experimental Results
You have correctly calculated the theoretical yields for copper to be 0.89 g and 1.33 g, respectively. The corresponding amounts of copper are 0.014 mol and 0.021 mol.
You have measured the amount of copper produced, and it clocks in at 0.78 g. This is already a strong indication that we should side with (Fe2) on this one. But just to really be sure, let's look at the error that we would have if it was (Fe3):
$$ \varepsilon = \frac{\Delta x}{x} = 41.4\% $$
This is a very high error, especially when compared to the error for (Fe2), which is only 12.1%. Assuming that you did not perform the reaction with pure Fe but a mixture of iron with some iron oxides as contaminants, did not use an analytical scale and performed the experiment in fairly uncontrolled conditions (are you sure the reaction was complete when you aborted?) this error is not as bad as it seems.
With the experimental results in our pocket, we can now move on to the final part:
The Conclusion
We have shown, using theoretical predictions and experimental confirmation, that the reaction observed was the oxidation of iron to iron(II) under reduction of copper(II) to copper.
By the way, most of the calculations I did with an iPython Notebook, which you can view here. I might have messed something up, you never know.

Your analysis misses a subtle point.  If you consider just the ocidationnof Fe(II) to Fe(III), that has a standard potential of +0.783 V.  Therefore Fe(II) is more stable at the potential required to reduce copper, and in fact Fe(III) would oxidize copper metal back to Cu(II) (which then reacts with more elemental iron).  This reaction cycle tends to have ugly consequences in pickling of steel strip.

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text:
Drawing (R)-1,1,2-trimethylcyclohexane
I tried drawing out (R)-1,1,2-trimethylcyclohexane but I can't seem to figure it out. I think I"m getting hung on assigning priorities.
I've labeled the chiral carbon with an asterisk. I've also drawn out all the groups it is attached to. I gave the hydrogen the lowest priority (4) and I'm giving the methyl carbon a priority of 3. I am giving the other carbons priorities of 1 and 2 (as can be seen in the picture). However, this doesn't seem to make any sense; I get a molecule that I can neither assign S nor R to since the numbers are out of order.

Also is this really an image of (R)-1,1,2-trimethylcyclohexane?

According to ACD ChemSketch, the last figure is correct. As for your figure on the paper, you can rearrange the groups so that group 1 and 2 are in the top and bottom position. You just need to ensure you get groups 3 and 4 in the correct positions.

You've correctly identified the chiral center in your molecule (the carbon with the asterisk), that is the first step.  When doing work with chiral centers, it is very important to use solid lines to denote bonds in the plane of the screen, bold wedges to denote bonds coming from the plane of the screen towards the viewer and dashed wedges to denote bonds from the plane of the screen away from the viewer.  Drawing all 4 bonds as simple solid lines, as you did, will often lead to confusion.  Look at the figures below, both enantiomers of 1,1,2-trimethylcyclohexane are shown.  The next step in assigning R/S configuration is to rank the 4 groups around the chiral carbon by molecular weight, with "1" being assigned to the highest molecular weight group, and "4" to the lowest (you also did this correctly).  Now, grab the "4" group in your hand so that groups "1" through "3" form a circular pattern between your eye and your hand.  If moving from 1->2->3 involves clockwise movement, then this is the "R" configuration.  If the movement is counterclockwise, then it is the "S" compound you are observing.

Take a look at this link for a nice picture of what I'm describing.
You cannot determine the R/S configuration of the chiral carbon in the compound you've drawn on paper at the top of your post because it is unclear which group 3 or 4 is projecting towards the viewer and which is projecting away from the viewer.  Finally, note that my "R" drawing matches the "R" drawing at the bottom of your post.


text:
CuI, CuSCN Solubility in Ammonia Solution
I am trying to dissolve copper(I) iodide and copper(I) thiocyanate in ammonia.  I am quite sure that they both dissolve in liquid ammonia, and are insoluble in ethanol.
If I were to purchase a 2.0M solution of ammonia in ethanol, like the one here, would the CuI and CuSCN still dissolve in the ammonia solution?
Also, are there any other solvents that you know of for these materials?

AFAIK, $\ce{CuI}$ is soluble in pyridine and aqueous potassium/sodium iodides.

The reason these salts are soluble in ammonia (including aqueous ammonia -- so why buy ammonia in EtOH?) is that they form complexes with ammonia:
$$
\ce{CuI}\text{(s)}+2\ce{NH3}\text{(aq)}\to\ce{Cu(NH3)_2^+}\text{(aq)}+\ce{I-}\text{(aq)}.$$
Have fun!


text:
Is there any emerged master principle in chemistry, like the theory of evolution in biology?
The question is about emergence and reductionism.
Except for a few exceptions, such as dark matter, most phenomena in Nature could be reproduced from the standard model and general relativity with great accuracy. Up to a Planck scale, there is no inconsistency between general relativity and quantum theory by means of effective theory. However, at different level of complexity, a new principle may emerge. One example is the theory of evolution in Biology.
Is there any similar emerged master theory in Chemistry that comparable with evolution in Biology?
Here, the emerged master theory means:

It is not just a brute-force ab initio computation (emergence). Here I have excluded explicitly correlated Gaussian.
The theory is universally valid. Not like 8- and/or 18-electron principle and/or VSEPR and/or periodic table, we know there are many exceptions.
The theory should generate experimentally testable predictions, hopefully quantitative. Here I have excluded Bader's QTAIM at its current form. Atom in molecule and types of chemical bonds are anyway not observables. One may see some correlation between Laplacian at bond critical point with other molecular properties, and do some fitting or hand-waving argument. Such fitted relation could be useful. However, it is a fitting, not a derivation from its principle. I do not regard it as a prediction.
The theory should not be discovered in condensed matter physics first. Here I have excluded spontaneous symmetry breaking.
I do not expect the statement like, there is no master principle in chemistry is the master principle in chemistry.

Chemistry is an extension of physics.  Seeing as how you have excluded theory that is so rigorously defined as to allow quantitative computation, and also ignored the fact that Bader's QTAIM is derived from Schwinger's fundamental representation of quantum mechanics, I really don't see what is left to form a master theory.  At least you discarded the octet rule, and hopefully also hypervalency and d-orbital participation.

QTAIM is concerned, among many things, with the topological characteristics of the electron density, especially those that involve energy-lowering (bonding!) interactions between atoms (things that can be identifiable because they have the same boundary conditions that are consistent with QM, and one can practically execute this identification).  The best place to see it in action is in crystallography, where x-ray diffraction can actually see the electron density to high enough resolution to identify things like critical points.

what are you trying to do, divine the future?  If you have a question, evaluate system A (predict its properties), then evaluate system B (predict its properties), and then you have A-B (the predicted change).  Whether A or B have actually been observed is pretty much immaterial, due to the predictive power of quantum mechanics.

But what are you trying to predict? Prediction of a chemical product/mechanism due to conservation of state symmetry? Well no doubt. Recall that the MO picture is formed once the state symmetry is already known.  Hardly a prediction.  Also, I can write many wrong (though valid!) qualitative MO pictures, the right one is found through observation and/or computation.  How is this a prediction?

Great, we have come to agreement that prediction is encoded in the computation, which is of course couched in the gory details of quantum mechanics.  I put forward quantum mechanics as the "master principle."  On the other issue, I would separate the question of utility/usefulness and verity/truthfulness.  Though Hoffman may question the former, he certainly has never questioned the latter. (I have discussed with R.H. and R.F.W.B. to good length.)

How about contradiction?  I can point to a bazillion contradictions of qualitative MO theory, which you seem to be holding up as the paragon of prediction.  Can you name a contradiction of QTAIM?

(The problem is that I can trace the lineage of the rhetoric of insisting on an unspecified "prediction" back to Cornell.)  Prediction can be had from difference of descriptions determined from rigorously defined and realizable scientific concepts. Quantum Mechanics and Statistical Mechanics are the "master theories" of chemistry.

QTAIM is not an energy minimizing principle (though it belongs to the same $H\Psi=E\Psi$ theory so it could be round-about)--it can be applied posthoc. One of its predictions(?) is that two atoms, which can be easily discerned at infinite separation, can still be discerned when they are brought into proximity. This is in contrast to another view where atoms are no longer identifiable, just one large supersystem with an overall boundary condition at infinity. QTAIM is great news to chemists who think and operate with functional groups (small collections of atoms).

I feel our conversation has taken a turn for the better. :-) Perhaps a "what good is QTAIM" and "what are its experimentally testable" properties would make a good question separate from the current one (and probably of broad interest as well.)

@EricBrown It would be great if you could ask a self-answered question to capture some of this discussion, if the OP is not interested in doing so.

In the future, it's better to have these kind of extended discussions in chat.

I find interesting that you excluded QTAIM, because "Atom in molecule and types of chemical bonds are anyway not observables" whereas you mention general relativity and standard model, which are based in unobservables such as curved spacetime, quantum fields, and bare particles. Also interesting you mention that "Up to a Planck scale, there is no inconsistency between general relativity and quantum theory by means of effective theory" when special relativity and quantum mechanics are mutually inconsistent (quantum field theory avoids and hide the problems).

You may be asking for something quite a bit deeper, and it's rather hard to strictly adhere to the requirements in the question, but after Eric's interesting edit showing some outside-the-box thinking, I went along those lines and came up with some options. I'll let you decide which (if any) make the cut.
Two good candidates for master principles in chemistry may be acid-base theory and reduction-oxidation theory. I think ultimately they're just somewhat coarse measures of the relative positions between HOMOs and LUMOs for different species, but after a couple centuries of research and some tabulation, we've come up with general top-level rules for understanding the chemical behaviour of large classes of compounds, with significant predictive power, without necessarily resorting to complicated calculations.
The periodic table itself could also work; different elements possess certain chemical similarities which allows them to be grouped with one another. Mendeleev famously used this to predict a handful of new elements, such as germanium (though he did also predict some other elements which never came to be).  A related principle which is simple but of great importance is that elements are entirely characterized by the number of protons in their nuclei. Every single atom in the Universe with 79 protons in its nucleus is a gold atom, no matter how many neutrons and electrons are present.
Also, I wonder if you'd count materialism as a master principle which emerged in chemistry; every material in the Universe is made of atoms (not counting dark matter). If two materials are the same, then they must be made of the same types and proportions of atoms in the same configurations. If two materials are different, then either the atoms are different, or they are present in different proportions, or they are present in different configurations.
Finally, I myself would actually have included the octet rule as a master principle. It has many holes and limited applicability, but it is uncanny that such a simple concept works at all. I find it a surprising way to effectively skip tons of gritty quantum mechanical calculations, helping chemistry education to begin in school around ages 14-15.

Quantum Mechanics and Statistical Mechanics are the "master theories" of chemistry. Prediction can be had from difference of descriptions determined from these rigorously defined and realizable scientific concepts.
Relativistic contributions can be seen especially in heavier elements.
Edit: In the spirit of your question, perhaps you are looking for something that has evolved from chemistry, that may not have been directly apparent in the physics community.  I would propose that the concept of a functional group--a rather predictably-behaving collection of atoms--is a master principle.

The scope of chemistry is beyond Quantum Mechanics and Statistical Mechanics, so they cannot be the "master theories" of chemistry. To get a proper foundation for Chemistry it is needed to generalize both Quantum Mechanics and Statistical Mechanics. https://doi.org/10.1111/j.1749-6632.2003.tb06091.x

My question is, is there any similar emerged master theory in Chemistry that comparable with evolution in Biology?

I would answer classical electrostatics. Sure, you definitely can't use electrostatics to rationalize all of chemistry, but you can use it to a great extent in early and introductory chemistry.
Think about organic chemistry. The organizing principle of introductory organic chemistry is "nucleophile attacks electrophile." This will help you correctly draw most of the arrows you need to draw in organic chemistry.
There are a few classes of reactions that sort of fall out of this "motif," however:
Exceptions to motif: nucleophile attacks electrophile

Main topic of chemistry is transformation of substances.
Theory describing the transformation is the Transition state theory. It allows us to discern favourable processes from less favourable, show how the speed of reaction changes with temperature.
Once spiced with more advanced concepts, like Free energy relationship, it allows us to predict how the reactivity will change upon substitution.


text:
Suggestions for research into different fluids in heat pipes
I'm doing some research on heat pipes (check here for wiki). One of the things I'm planning to do is try some different fluids in the pipes. Water (commonly used) has a freezing point that's too high, so I'm looking into fluids that don't freeze up in winter and disrupt the equilibrium in the pipe.
I've randomly picked ethanol so far, but I'm looking for suggestions from someone more knowledgeable then me.
I'm mainly interested in vertical heat pipes that don't need a wick like here on wikipedia, but rely on gravity to get the drops of condensed fluid back to the warmer area.
Potential things that could matter:

Viscosity of the fluid
Bubble forming at boiling point (bubbles can form an isolating barrier perhaps?), is there a fluid that boils in vacuum with very small bubbles?
Surface tension

All suggestions are very welcome, please explain why you think a certain fluid is a good choice.
Thanks in advance,
Johan Rensink


text:
R,S Configuration
I don't understand why this molecule is assigned S instead of R.

The H is assigned a priority of 4; the methyl group on the chiral carbon gets 3; the left carbon gets 1 and the right carbon gets 2. The rotation should therefore be clockwise.
I guess what I have an issue with is assigning priorities in cases like these: what does moving away from the chirality center mean? Do I choose the H? The C? The other C?
Do we just ignore the attached hydrogens? It seems like the hydrogen on the carbon left of the chirality center is ignored.

And in the above case, why are we comparing an O with a C? The first atom to the left of the chirality center is a C, not an O.

And in the last picture I give my configuration for the molecule in the first picture. Why doesn't the upper right carbon have a priority of 2? Double bonded carbon counts twice; the rest are hydrogens.

The right carbon gets a "1", so the compound is "S".  Attached hydrogens are not ignored.  Here is a link to a nice set of rules.  See rule #3 for handling double\triple bonds, that's where you made an error. It says, "Atoms participating in double/triple bonds are considered to be bonded to an equivalent number of similar “phantom” atoms by single bonds. Note: “phantom” atoms are bonded to no other atoms".  See the page for a clearer picture.

In the case of the diol you show near the bottom of your post, the 2 carbons attached to the chiral center are equal, so we next look at, and compare, all of the substituents attached directly to them.

Thank you. So the upper right carbon, by virtue of being part of a double bond, is considered by the R,S system to have two single-bonded carbons bonded to it? And the other carbon in the C=C double bond similarly has two single-bonded carbons attached?

The carbon in the double bond closest to the chiral center is considered to have 3 carbons and a hydrogen attached to it, while the terminal double bond carbon (the one that is the "tie-breaker on this side of the moelcule") is considered to have 2 carbons and 2 hydrogens attached to it.

As you move away from the chiral center along two chains you are comparing, for each atom in the chain consider the highest priority atom it is bonded to first.
See step 1, parts 2 and 3 in this reference: http://www.chem.sc.edu/faculty/shimizu/333/Chem_333/5a.vii.html
For double and triple bonds you have to add ficticous atoms before you evaluate.
See step 1, part 4 in the above reference.


text:
Lewis Acidity and Bacteria
How does Lewis acidity correlate with bactericidal activity? I read on Wikipedia that the two are positively correlated but no explanation was given.
Does it have to do with the movement of electrons - i.e. a current? I'm guessing that having a current passed through you isn't good for you.
I suppose also that the interactions (i.e. coordination bonds) of metal (ions) with bacterial components can interfere with cellular processes.

Do you have a link to the article?

Yes. " Antimicrobial efficacies generally followed Lewis acidity values of the various metals." http://en.wikipedia.org/wiki/Antimicrobial_copper-alloy_touch_surfaces

Perhaps this would get more attention on biology SE. You might try reading this (if you haven't already): http://en.wikipedia.org/wiki/Antimicrobial_properties_of_copper#Mechanisms_of_antibacterial_action_of_copper I would imagine that a stronger Lewis acid would be more likely to participate in those mechanisms because it is more able to accept electrons.

A review on $\ce{Ag+}$ quotes the following situations, which may be provoked by the Lewis acidic behavior of a metal cation.

Silver treated cells exhibited electron-light regions in
  their cytoplasm thus forcing DNA to condense. Authors state that
  condensed DNA molecules lose their ability to replicate.

I presume that $\ce{Ag+}$, acting as a Lewis acid, will reduce the negative charge on DNA, thus inducing DNA aggregation.

Electron-dense region were also observed around the cell wall inside
  the cytoplasm, and the cytoplasmic membrane experienced  shrinkage
  and  detachment  from  the  cell wall.

There was also indication  that  silver  ions  interact  with  thiol
  groups leading to the deactivation of enzymatic proteins.

Here one sees the action of $\ce{Ag+}$ on thiols which may be interpreted as an acid-base interaction of the Lewis type.

Also the effect of $\ce{Ag+}$ on the respiratory electron chain of E. coli was stressed, since $\ce{Ag+}$ is toxic to microorganisms due to inhibition of the
  respiratory chain at multiple sites.

It appears that in many cases, one can link the antibacterial function of metal ions to their acidic properties. And certainly, those discussed above are not the only ones.

Metal ions are by far some of the most common biological Lewis acids, and are vital to the function of many different cell activities. Excess of a particular species of metal ion, however, can result in cytotoxicity both by catalyzing certain reactions, and inhibiting others by coordinating with unbonded electron pairs present on atoms in many macromolecules, particularly sulfur. In the examples I am presenting, copper, zinc, and silver ions all exhibit their cytotoxicity by forming coordination complexes with the molecules within the cell, hindering their function.
Note: I have emboldened the exact reasons for the cytotoxicity within the text for ease of reading.
Copper$^{[1]}$

Recent evidence suggests a copper toxicity mechanism in which the reduced $\ce{Cu+}$ ion is instrumental. Multiple investigators note that copper toxicity to bacteria is sustained or even enhanced in anoxic conditions where peroxide formation is minimal. $\ce{Cu+}$ toxicity in the E. coli cytosol can be explained by its intense thiophilicity, which is sufficient to competitively disrupt key cytoplasmic iron-sulfur enzymes both in vitro and in vivo. Indeed, other “soft” thiophilic metal ions have been found to exert comparable toxicity. Together, these data provide compelling evidence linking copper toxicity to iron displacement from solvent-exposed dehydratase iron-sulfur clusters, resulting in metabolic disruption and branched chain amino acid auxotrophy.

Zinc$^{[2]}$

Although nearly 30% of all proteins contain metal ions, an excess of certain transition metal ions can exert significant toxicity. The potential roles of such metals in resisting bacterial invasion are increasingly being recognized. Notably the importance of copper in macrophages for bactericidal activity has recently been demonstrated. Our present study provides the first direct evidence that extracellular $\ce{Zn(II)}$ can exert its toxicity towards S. pneumoniae by competing with $\ce{Mn(II)}$ for binding to the PsaA (Pneumococcal surface adhesin A) solute binding protein, thereby preventing the acquisition of $\ce{Mn(II)}$ via the Psa permease.

Silver$^{[3]}$

It is thought that silver atoms bind to thiol groups ($\ce{-SH}$) in enzymes and subsequently cause the deactivation of enzymes. Silver forms stable $\ce{S-Ag}$ bonds with thiol-containing compounds in the cell membrane that are involved in transmembrane energy generation and ion transport. It is also believed that silver can take part in catalytic oxidation reactions that result in the formation of disulfide bonds ($\ce{R-S-S-R}$). Silver does this by catalyzing the reaction between oxygen molecules in the cell and hydrogen atoms of thiol groups: water is released as a product and two thiol groups become covalently bonded to one another through a disulfide bond. The silver-catalyzed formation of disulfide bonds could possibly change the shape of cellular enzymes and subsequently affect their function.
The silver-catalyzed formation of disulfide bonds can lead to changes in protein structure and the inactivation of key enzymes, such as those needed for cellular respiration. 30S ribosomal subunit protein, succinyl coenzyme A synthetase, maltose transporter (MalK), and fructose bisphosphate adolase were identified with high probability as proteins with decreased expression once cells are treated with a $\mathrm{900\:ppb\:\ce{Ag+}}$ solution. It is hypothesized that silver ions bind to the 30S ribosomal subunit, deactivating the ribosome complex and preventing translation of proteins. The proteins that were found to be downregulated upon treatment with $\ce{Ag+}$ serve important functions to the cell: succinyl-coenzyme A synthetase, an enzyme involved in the TCA cycle, catalyzes the conversion of succinyl-CoA to succinate while phosphorylating ADP to produce ATP; fructose bisphosphate adolase is an enzyme involved in glycolysis that catalyzes the breakdown of fructose-1,6-bisphosphate into glyceraldehydes 3-phosphate and dihydroxyacetone phosphate; MalK is a cytoplasmic membrane-associated protein involved in the transport of maltose. In one way or another, all of these proteins play a role in energy and ATP production for the cell, so the decreased expression of any one of these proteins could lead to cell death.
Another one of the suggested mechanisms of the antimicrobial activity of silver was that $\ce{Ag+}$ enters the cell and intercalates between the purine and pyrimidine base pairs disrupting the hydrogen bonding between the two anti-parallel strands and denaturing the DNA molecule. Although this has yet to be proved, it has been shown that silver ions do associate with DNA once they enter the cell.

[1] Chaturvedi, K. S.; Henderson, J. P. Pathogenic Adaptations to Host-Derived Antibacterial Copper. Front. Cell. Infect. Microbiol. Frontiers in Cellular and Infection Microbiology. 2014, 4.
[2] Mcdevitt, C. A.; Ogunniyi, A. D.; Valkov, E.; Lawrence, M. C.; Kobe, B.; Mcewan, A. G.; Paton, J. C. A Molecular Mechanism For Bacterial Susceptibility to Zinc. PLoS Pathog PLoS Pathogens. 2011, 7.
[3] https://microbewiki.kenyon.edu/index.php/Silver_as_an_Antimicrobial_Agent#Mechanism_of_action


text:
Configuration, chirality
I'm having trouble assigning priorities for this molecule:
I tried assigning priorities with the rules for doubly bonded atoms in mind. I know that the lowest priority group should be going into the page (not on a wedge) but I read that I can simply assign a priority with the fourth priority group on a wedge and then take the opposite of whatever I assigned as the true configuration of the molecule. Problem is, I get R. Answer key says S.

Your drawing is correct and yes, "I read that I can simply assign a priority with the fourth priority group on a wedge and then take the opposite of whatever I assigned" this is true too.  In your drawing 1->2->3 is a clockwise rotation (therefore "R" but), but the lowest priority item (4) is coming out towards you.  Therefor, the R/S assignment needs to be reversed (from "R") to "S".

Thank you! I think I need to not only purchase a molecular model kit but also an analog clock.

I once had a friend who had an analog wall clock very near to the corner of a room.  On the adjoining wall, very near to the corner was a mirror...don't do that.

The clock method should be ignored as this example demonstrates. Point your thumb in the direction of the lowest priority group (4). Then your fingers will point in descending order (1 -> 2 -> 3). If not, you are using the wrong hand. Right hand = R; left (sinister) hand = S.


text:
What other nuclei besides hydrogen and carbon 13 could be detected using NMR and is H1-NMR synonymus to NMR?
From my understanding NMR spectroscopy works by analyzing the resonance frequency of each structure around a hydrogen or carbon. Since every structure is unique and has unique resonance frequency it is easy to identify them. Is this correct?
What other nuclei besides hydrogen and carbon 13 could be detected using NMR?
Is H1 NMR the same as saying NMR?

Have you used a common web-based search engine to research this question? Try reading this wikipedia page.

The term $\ce{^1H}$ NMR is more specific than the term "NMR" in that it is referring only to NMR experiments involving protons.  It is usually relatively straight forward to identify which atom in a molecule is responsible for a certain NMR signal.  There is also a lot of other information that can be extracted from an NMR signal by doing secondary NMR experiments that can help with peak assignment.  That said, some molecules are so large and complex that signal assignment can become problematical. As to which other nuclei can be detected by NMR, any isotope with an odd number of protons and/or neutrons will have a magnetic moment and be detectable by the NMR experiment.  Here is a listing of some nuclei with non-zero spin that have been studied by NMR (see this Wikipedia article):
$\ce{^1H}$, $\ce{^2H}$, $\ce{^6Li}$, $\ce{^10B}$, $\ce{^11B}$, $\ce{^13C}$, $\ce{^14N}$, $\ce{^15N}$, $\ce{^17O}$, $\ce{^19F}$, $\ce{^23Na}$, $\ce{^29Si}$, $\ce{^31P}$, $\ce{^35Cl}$, $\ce{^113Cd}$, $\ce{^129Xe}$,  $\ce{^195Pt}$.

It's not always easy to interpret NMR spectra. Different nuclei can show the same chemical shift, leading to overlapping signals. NMR signals are also not infinitely sharp, overlapping signals are rather common.
But there are also more advanced experiments you can use to solve these issues, if you have enough of your sample and enough time on a spectrometer.
1H NMR is the most common type because 1H is almost 100% abundant, and hydrogens are present in most organic compounds. 13C and 15N are also used often because they are common in organic chemistry and especially in biology. Those isotopes are rather rare in nature though, this makes experiments involving them much, much less sensitive than proton NMR. If you want to perform more complicated experiments with these nuclei you have to isotope-label your sample, which is routinely done for proteins, but extremely expensive or even impossible for small organic molecules.
19F is another isotope that works well for NMR, it is also around 100% abundant and therefore very sensitive.
In general, you can do NMR on any spin 1/2 nucleus. Nuclei with higher spins is also possible, but that gets more difficult.

Since every structure is unique and has unique resonance frequency it is easy to identify them.

Firstly, in NMR we do not deal with structures, but instead nuclei. Scroll all the way to the bottom to see a list of some NMR active nuclei on the website of Hans J. Reich (University of Wisconsin):
Summary of Nuclear Properties.
You'll note that some are listed as having low sensitivity which is related to certain properties which we won't go into.
The good thing about NMR, from an organic chemistry point of view at least, is that carbon has only one NMR active nuclei, $\ce{^{13}C}$, that additionally has very low abundance $(\approx1\%)$.
Hence, when you perform a $\ce{^1H}$ NMR experiment on an organic molecule like ethanol, for example, you are unlikely to see carbon nuclei interacting and causing splitting.
You may ask how we obtain $\ce{^{13}C}$ NMR spectra? Well, it is possible, just not easy. Beyond having very low abundance it also possesses a long relaxation time which essentially means that when you excite a $\ce{^{13}C}$ nuclei there is a long period of time that elapses before it relaxes back to the ground state and is capable of being excited again to obtain a clean spectra.


text:
Why is crystalline aspirin washed with chilled water in the purification process?
Take as example this
or even better this:

Wash the crystals with 2-3 ml of chilled water.

Why does the aspirin need to be cooled down? And why do we wash the aspirin with chilled water?

Because aspirin is less soluble in cold water. In fact, aspirin is not very soluble in water at all, which is why you are supposed to take it with lots of water. You are not "chilling the aspirin", you are preventing it from dissolving too much so that you have a better yield of your product. You wash it to remove impurities that do dissolve in water.


text:
Effect of temperature and Le Chatelier's principle

The question
  $$\ce{2NO(g) + H_2(g) <=> N_{2}O(g) + H_{2}O} + \pu{351kJ}$$
  If the temperature is increased in the reaction above, which of the following occurs?
A. The concentration of $\ce{H_{2}O}$ increases
  B. The concentration of $\ce{N_{2}O}$ decreases

My teacher said the answer is B;
however, I thought that since the reverse reaction is endothermic, adding more heat would produce more $\ce{2NO}$ and $\ce{H_2}$ which will cause an increase in pressure. But in order to maintain the same pressure, the system needs to decrease the number of moles so it favors the forward reaction because there are 3 moles of the reactants and 2 moles of the products.
Did my teacher make a mistake? If he did not can you please explain why he is correct.

Your teacher is right, let's see why. First off, here's our reaction $$\ce{2 NO + H2 -> N2O + H2O}\qquad \Delta H = -351\ \mathrm{kJ/mol}$$
So the reaction to the right is exothermic, it gives off heat.  That tells us the products are more stable than the reactants and at equilibrium $\mathrm{\frac{[Products]}{[Reactants]}~>1}$.  Here we could guess that if we heat the reaction (put heat back into the reaction), we will tend to level out the equilibrium thereby pushing the reaction back to the left.
To examine this guess let's work an example.  You were given an enthalpy of reaction, but for a "back of the envelope" calculation, let's run a comparison and assume that the change in entropy will be roughly the same for the two conditions (temperatures) we're going to compare.  With this assumption, we can let $$\Delta G= -RT\cdot\ln K_\text{eq}= -351\ \mathrm{kJ/mol}$$  We know $\Delta G$, $R=8.31\times10^{-3}$ and let's say the data is given for room temperature, so $T=298\ \mathrm K$.  Solving for $K_\text{eq}$ we find $K_\text{eq}=3.6\times10^{61}$.  Now let's let our reaction equilibrate $100^\circ$ above room temperature (now $T=398\ \mathrm K$).  $\Delta G$ and $R$ are fixed, so lets resolve the equation and we now find $K_\text{eq}=1.2\times10^{46}$, indeed heat has pushed our reaction to the left.  The teacher was right.

So basically since $\ce{N2O (g) + H2O + 351~kilojoules -> 2NO(g) + H2(g)} $  adding more energy will be causing the endothermic reaction to occur  more?

Yes, at very low temperatures an equilibrium will favor the thermodynamically most stable side of the equilibrium the most.  As you increase the temperature the relative difference in stability between reactants and products becomes less significant and the equilibrium will begin to approach 1

which will cause an increase in pressure. But in order to maintain the same pressure, the system needs to decrease the number of moles so it favors the forward reaction because there are 3 moles of the reactants and 2 moles of the products.

This is a frequent misconception about Le Chatelier's principle.  Pressure only changes the equillbrium if pressure changes in a way that changes concentrations, for example by changing volume or adding more of a reactant or product.  If pressure changes in a way that doesn't change concentrations, for example by adding a noble gas to the mixture, or in this case changing temperature, equillibrium is unchanged.
When temperature changes, the equillibrium constant itself changes, according to the Van't Hoff Equation.  For exothermic reactions, increasing temperature favors the reverse reaction.  For endothermic reactions, increasing temperature favors the forward reaction.

So the equilibrium constant is governed by partial pressures, not total pressure.

yes, the equillibrium constant is defined as a function of partial pressures or concentrations for idealized situations, fugacity or activity for real situations

I'm not sure if you could actually predict what would happen if you raise the temperature of the reaction system. That would change the equilibrium constant K for this equation.
Given that you don't know what the concentration/pressure of each system component is, there is the possibly that K could be adjusted so that it equals Q when you raise the temperature for the "reaction." That would make both answers incorrect since at equilibrium there is no net change in system composition.
Tell your teacher that. He's being ambiguous by not specifying if we're raising the temperature of the reaction system, or whether we're simply heating up the reactants or what.
This example illustrates why your teacher is ambiguous:
$\ce{flour~+~milk~+~eggs ->cake}$
If you heat up the entire reaction system, then you'd have a warm cake, and a warm bag of flour, some warm milk, and some warm eggs. Or maybe you'd have a warm cake. It depends on how much heat you add. Here's one ambiguity.
If you add heat to the reactants only - i.e. you bake the combination of reactants - then you'd get cake.
If you add heat to the product only, then you'd get a warm cake.

Regardless of however crudely your teacher might have phrased the question, you've thought too much into the question. True, adding heat to the product side will encourage the reverse reaction, but you don't know to what extent and it's best not to speculate what might happen also as a result.

In a system at equilibrium, how can you heat up just the reactants OR just the products? How can you partition where the energy goes into a system? In your example, no matter how much energy you put in or take out of the system, you will never retrieve your cup of milk from your cake.

For an exothermic reaction, $\frac{\Delta G^0(T)}{T}$ increases with increasing T. Since $\ln K=-\frac{\Delta G^0}{RT}$, K decreases with increasing T for an exothermic reaction at all temperatures (not just at the standard temperature of 298 K).


text:
What products will the following reactions yield?
Assume these 2 reactions. I must write the Lewis structures of the products.

$$\ce{C6H5COOH + CH3CH2OH ->[\ce{H+}] ?}$$
(comment: A benzene ring + $\ce{CO_2H}$ on one end, I don't know its name)
$$\ce{CH3CH2OH + (CH3CO)2O ->[\ce{H+}]?}$$ (comment: acetic anhydride)

My approach:
For the first reaction I will get acetic anhydride (source)

For the second reaction I will get aspirin and acetic acid.
Questions:

Is the above correct?
Is there a connection between the two reactions? Is it an intermediate step of some sort? What is the importance between the two reactions?
Why is $\ce{H+}$ added?
What is the benzene ring and the $\ce{CO2H}$ called?

No, I hesitate to give the correct products since this is homework, but maybe some of what's below will help
Yes, the two reactions are both esterifications; converting an acid, an acid anhydride, an acid chloride, etc., to the corresponding ester
The proton is a catalyst, it is not consumed in the reaction.  It protonates the carbonyl oxygen, making the carbonyl carbon more susceptible to nucleophilic attack
benzoic acid

Well even if it is homework how will I learn if I don't see the right answer at least once?

Is this the only product of the first reaction? http://en.wikipedia.org/wiki/Ethyl_benzoate

For the second reaction will I get CH3C(O)OC6H5 + CH3COO-?

Yes, you got the first one and I think you got the second one, but your alpha-structure is a little ambiguous.  You can find the second structure here  http://www.sigmaaldrich.com/content/dam/sigma-aldrich/structure3/016/mfcd00009109.eps/_jcr_content/renditions/medium.png

@ron In an esterification of un-symmetrical anhydride,  how to decide which part of anhydride gets added to form ester? Please reply.


text:
R, S Configuration and Branching - Which Path?
Which path do I take when we have a branched chiral center? None of the textbooks I've consulted adequately covers this or even provides an example problem in which there was branching.

Consider the above:
1) Chiral carbon with the following attachments H, an carboxylic acid group, a methyl group, and then the ring.
2) H gets the lowest priority.
3) The methyl group gets the second lowest priority since it has just one carbon and then hydrogens; the other paths have either more carbons or oxygens!
4) Between the carboxylic acid group and the path along the ring, we have an oxygen and a hydrogen going for the carboxylic acid group. But which path do I take? After comparing the carbon in the carboxylic acid group to the carbon on the ring, do I compare the double-bonded oxygen to the next carbon on the ring or the single-bonded oxygen? How does one choose in cases of branching such as these?

No,  you have a carbon figuratively attached to 3 oxygen atoms vs.  a carbon figuratively attached to 3 carbon atoms; the former has higher priority.

Even if it would be only one Oxygen compared to a branched Carbon, the former will still have higher priority.

As @ron mentioned $\ce{-CO_{2}H}$ has higher priority so the numeration given by @ADG in the fischer projection is incorrect because he assumes that the ring has higher priority than $\ce{-CO_{2}H}$. You can also think in the ring priority as $(C,C,C)$ and $\ce{-CO_{2}H}$ priority as $(C,O,O)$. As you may notice$(C,O,O)$ has higher priority than $(C,C,C)$ (the total atomic number of the substituent group in the first case is higher).
See the following figure:

(Source)
In this case I don't think you need to continue the path, but if that's the case you can see how it works in the figure.

The $\ce{CO2H}$ moiety should be thought of as $\ce{(C,O,O,O)}$ because the carbon is bonded via three bonds to oxygen.

convert it to fischer projection:

you can see it is (S)
Edit: Added Explanation
Going according to atomic numbers.

Atomic no.1

1.$\ce{H}$- lowest priority.

Atomic no. 6

2.$\ce{CH3}$ - next priority.Has $\ce{H}[1]$
3.$\ce{C6H4..}$ - next priority. Has $\ce{C}[6]$
4.$\ce{COOH}$ - next priority.Has $\ce{O}[8]$


text:
How do I find out the number of bonds an atom must link to?
I've always been told that carbon bonds with four hydrogens and that nitrogen bonds with three hydrogens, like in benzene and hexazine, respectively.
What's the name of this type of bonding and how does one come up with the number?

It's named a covalent bond. The amount of hydrogen atoms that can be bond (or any other atom) can be  calculated most of the time using the octet rule, that states that any atom will try to get the electron configuration of a noble gas (i.e. 8 electrons in the outer shell). The electrons from the covalent bond count for both atoms that take part in the bonding.
As an example consider $\ce{N}$ that has five electrons in its outer shell, and $\ce{H}$ that has just one. $\ce{N}$ wants three more electrons and can get them from covalent bond with hydrogen. $\ce{H}$ want one more electron (this is an exception) in its outer shell to fill it up, so will kindly receive them from nitrogen.

Atoms, especially those in the first 2 or 3 rows of the Periodic Table, try to attain a filled outer shell of electrons.  This gives them an inert gas electron configuration and a special stability.  They can achieve this filled outer shell of electrons in one of two ways, they can donate or accept electrons (ionic bonding) or share electrons (covalent bonding).  An example of ionic bonding would be sodium with an electronic configuration of $\ce{1s^2 2s^2 2p^6 3s^1}$.  By donating (losing) one electron it attains the inert gas electronic configuration of neon ($\ce{1s^2 2s^2 2p^6}$).  On the other hand, atoms like carbon ($\ce{1s^2 2s^2 2p^2}$) and nitrogen ($\ce{1s^2 2s^2 2p^3}$) would have to gain or lose too many electrons to achieve an inert gas configuration, so they typically form covalent bonds by sharing electrons.  In order for carbon to achieve the neon configuration it would need to share 4 more electrons.  That is why carbon typically forms 4 covalent bonds.  In the case of methane, $\ce{CH4}$, each hydrogen atom brings one electron to share with the carbon atom and in turn each hydrogen shares one electron from carbon.  By sharing electrons like this, each hydrogen achieves a stable helium configuration and carbon achieves a stable neon electron configuration.  In the case of nitrogen, it only needs to share 3 more electrons in order to achieve the neon configuration.  It can do this by forming 3 covalent bonds.  If we look at ammonia, $\ce{NH3}$, we can apply the same reasoning we applied to methane.  By sharing electrons with 3 hydrogen atoms, both the nitrogen and hydrogen atoms can achieve inert gas electronic configurations.  Simply stated, atoms try to form as many bonds as necessary, be they ionic or covalent, in order to achieve an inert gas configuration.  While there are many exceptions to this approach, especially for heavier atoms, it does seem to work reasonably well for the lighter elements.


text:
How does polarimetry account for the orientation of the molecules?
If you have some sample in solution (of a pure R substance), wouldn't the individual molecules all have a random orientation in space? Or do the photons emitted into the solution cause all of the molecules to align? Or is it just that, statistically, a given percentage of the particles will have an orientation which corresponds to the polarized light source.

Short answer: The light beam interacts with all of the randomly aligned molecules in its path.  There are roughly $\ce{6 x 10^{23}}$ molecules in a mole of material. If we were to dissolve a micro mole of this compound in 10 ml (1 deciliter is a typical polarimetry sample size) of solvent we would have $\ce{6 x 10^{17}}$ molecules in our sample cell.  Since the molecules are randomly oriented, there are enough molecules so that the polarized light beam will sample all possible orientations of the molecule with respect to the light beam many, many times over.  This allows for measurements to be made in an accurate and statistically reproducible manner.
Detailed answer: A beam of light has associated with it an electric and magnetic field.  In simple polarimetry experiments it is the electric field (or vector) associated with the light beam that is of interest.  Let's say the beam is propagating in the Z direction, then the propagating electric vector lies in an infinite number of planes that contain the Z axis.  When we plane-polarize the electric vector associated with the light beam, we remove all of the planes that the electric vector resides in except for one.
The next important point is to realize that a plane polarized electric vector associated with a beam of light can also be represented by (is equivalent to) two circularly polarized beams (or vectors), rotating in opposite directions (clockwise and counter-clockwise), each one-half the intensity of the plane polarized electric vector.  What we wind up with is two helices, one right-handed and one left-handed, traveling through the sample in our polarimeter.
Let's say our polarimeter sample contains a significant concentration of chiral molecules in solution.  Although the chiral molecules are randomly aligned with respect to the light beam, each chiral molecule will interact with the right-handed or clockwise rotating helical electric vector differently than it will interact with the left-handed or counter-clockwise rotating helical electric vector.  Our sample must contain a large number of molecules so that, despite all of the different molecular alignments between our molecules and the beam, when averaged over the entire sample, the difference in interaction of the right- and left-handed helices with our chiral sample will be repeatable.  It is not hard to have a significant concentration of molecules because a reasonably long sample path length is used.
The unequal (dissymmetric) interaction between our chiral sample and the right- and left-handed helices results in a rotation of the plane of polarization of the emerging light beam from its original position - and this angle of rotation is what the polarimeter measures.


text:
Is there a good sol gel method for carbon nanotube thin film deposition?
I was searching for a sol gel method for deposition of carbon nanotube thin films and there aren't too many articles on that.  Does anybody know the reason?  Is it because Carbon nanotubes are difficult to disperse in many solvents or?  Does anyone know if it is possible to effectively use a sol gel method to make a CNT thin film?

I think it may depend on what you mean by "effectively." I imagine that it would be difficult to get a thin film of oriented CNTs using a sol-gel method, since the gel would necessarily have randomly oriented tubes.
The other potential issue is making the sol-gel in the first place. Unless they are functionalized, carbon nanotubes won't gel in hydrophilic solvents. It might be possible to create gels in hydrophobic solvents, but a quick search didn't turn anything like that up.
This link has a brief description of a typical process - you can see that the idea is to make the sol-gel with something that will gel in aqueous solution first (silica, for example), and then add the nanotubes.
So, if you want pure CNT thin films, that would probably be a lot more difficult.


text:
Monomers and polymers
At Wikipedia a monomer is defined to be

a molecule that may bind chemically to other molecules to form a
  polymer.

A polymer is defined to be

a molecule composed of many monomers.

These definitions are obviously circular.
Are there non-circular definitions of monomers and polymers?

Circular definitions for a circular concept.  These seem pretty good. I might embellish the definition with "a molecule or collection of atoms (monomer) that may bind with other like molecules to make a new molecule (polymer) that is comprised of their sequence."

Let's have a look at what IUPAC say; they would be a reliable source to get definitions of chemical terminology from:
Monomer:

"A substance composed of monomer molecules."

Monomer molecules:

"A molecule which can undergo polymerization thereby contributing
  constitutional units to the essential structure of a macromolecule. "

Polymer:

"A substance composed of macromolecules."

Macromolecule:

"A molecule of high relative molecular mass, the structure of which
  essentially comprises the multiple repetition of units derived,
  actually or conceptually, from molecules of low relative molecular
  mass. "

These definitions are consistent with the latest IUPAC Compendium of Polymer Terminology and Nomenclature, IUPAC Recommendations 2008
Note that, historically, the definition of a polymer has changed somewhat. Up until about 80 years ago, polymers were seen as molecules comprised of similar empirical formula - benzene, for example, was considered a polymer of acetylene. It was Staudinger's work in the 1920's (contributing to his Nobel Prize) that defined polymers as macromolecules.

The term "monomer" only has meaning in the context of discussing polymerization.
A polymer is a molecular chain where the links of the chain are repeated identical or a least similar units.  Examples where the units are idential would be polypropylene, polyvinylchloride and polysiloxane.  Example where the units are similar but not necessarily identical would be DNA and proteins.
A monomer is a molecule that can react to form a single unit of a polymer.
For example, propylene is a monomer of poylpropylene and amino acids are monomers of a protein.


text:
How does formylation of a nitrogen change the hydrophobicity of a target protein?
How does formylation, of the sort shown here:

(from this Wikipedia page) affect the hydrophobicity of the target protein?  Can we say anything general about adding a formyl group to a nitrogen?

Formylmethionine is used as the start amino acid for a protein chain in certain organisms (the prokaryotes.)
As the alpha nitrogen is formylated, it cannot be in the middle of the chain. This is never seen in real proteins, partly because there would be two carbonyl groups attached to the nitrogen, which would have a very strong electron withdrawing effect and be quite unstable.
http://en.wikipedia.org/wiki/N-Formylmethionine
Other organisms (the eukaryotes, including the animals and humans) use methionine as the start amino acid of the chain.
http://en.wikipedia.org/wiki/Genetic_code
As there can only ever be one N-formylated methionine in a real protein (and it is often removed after protein synthesis in a step called post-translational modification) the effect on hydrophobicity will not be very much.
If on the other hand, you were to take an existing protein and formylate the -(CH2)4-NH2 side chains on all the lysine amino acid residues, I would expect a slight increase in hydrophobicity in neutral solution. In acid solution I would expect the difference to be more marked, as the formylated NH2 groups would lose their ability to become ionised by accepting a proton from the acid.


text:
How do I go about calculating protein concentration to make a set of protein standards for protein quantification via dot blot?
I have been given a protein of unknown concentration and have been asked to make a set of standards using a protein with a known concentration so we can quantify roughly via dot blot.
I need to know how to work out the calculation part. I would like to learn both mass and molar.For example, the information I have for my standard protein is it is at a concentration of 30mg/ml and 550uM. I want to put, for example, 1ul dots of this protein on the membrane for my blot. I want several dilutions of my standard, for example ranging from 20pg to 100ng. How do I work it out to get varied concentrations from the stock to my 1ul volumes on my blot?


text:
Effect of electronegative groups and length of alkyl chain on inductive effect among carboxylic acids
According to my teacher, the inductive effect in $\ce{CH3COOH}$ exists because of the electron deficiency on $\ce{C}$ (partial positive charge $\delta +$) as a result of the electronegativity of the double bonded $\ce{O}$. Although the single-bonded $\ce{O}$ is more electronegative than $\ce{C}$, the electron density of the single bonded $O$ moves to $C$ in order to “neutralize” the $\delta+,$ leaving $\ce{H}$ less attached to the molecule:

Why do we consider that double-bonded $\ce{O}$ establishes that $\delta+$ on our $\ce{C}$ (and creates an electron withdrawing inductive effect −I) and not the single-bonded $\ce{O}$? Is this because of the double π-bond? Why?
Now supose we have the following carboxylic acids:
$$
\begin{array}{rr}
\hline
\text{Acid} & \mathrm{p}K_\mathrm{a} \\
\hline
\ce{CH3COOH} & 4.80 \\
\ce{CH2FCOOH} & 2.66 \\
\ce{^-OOCCH2COOH} & 5.69 \\
\ce{^-OOC(CH2)4COOH} & 5.41 \\
\hline
\end{array}
$$
For me it makes sense that $\ce{CH2FCOOH}$ is more acidic than $\ce{CH3COOH}$ because $\ce{F}$ is an electronegative element that creates a partial negative charge attracting the electron density and leaving $\ce{H}$ in the carboxyl group less attached and easier to remove.
On the other hand, the fact $\ce{^-OOCCH2COOH}$ is less acidic than $\ce{^-OOC(CH2)4COOH}$ seems less intuitive. My teacher's presentation states that $\ce{O2^-}$ linked to an alkyl chain will create an electron donating inductive effect +I. Why? Isn't there a withdrawing resonance effect −R?
If $\ce{O}$ is an electronegative element, then why doesn't it create an −I effect instead of +I? Because if that was the case adding carbons $\ce{(CH2)4}$ in the chain would reduce the the −I effect and the $\ce{^-OOC(CH2)4COOH}$ would be less acidic than $\ce{^-OOCCH2COOH}$, and that isn't the case.

I think both oxygens will have an inductive effect on the bonded carbon, but since the single bonded oxygen is also attached to a hydrogen, its inductive effect is decreased relative to that of the double-bonded oxygen. You can think of both oxygens as having an inductive effect, but the net inductive effect is definitely away from the carbon and in the direction of the double-bonded oxygen. Also did you mean to write negatively charged diatomic oxygen? I think you meant to write $\ce{O^{2-}}$.

The last pka is for the negative ion? Isn't it obvious, that in that case leaving hydrogen cation has to overcome attraction from two negative charges instead of one and have troubles with it?

You're right, I was thinking in another compound, but the next quetions still remains, just forget the sentence "On the other hand the fact O−2CCH2COOH is less acidic than CH3COOH for me it's less intuitive". I'm going to edit the question and show you the other compound

But still why the electron density from the single bonded $O$ tends to move towards the $C$? Because of eletrostatic interactions? Yes I meant to write the negatively charged diatomic oxygen, where you have one single bonded $O$ and a double bonded $O$ therefore existing withdrawing Resonance effect $-R$ I think

I would be very careful with making any conclusions using the second compound. It has chain long enough to form internal cycle, requiring much more accurate estimation of effects other them +/- M +/- I. And the difference is low.

Why do we consider that double bonded O establishes that δ+ on our C
  (and creates an electron withdrawing inductive effect −I) and not the
  single bonded O?

Look at the resonance structures for acetic acid in the top line of the following drawing.  We can see that the middle resonance structure takes the pi electrons involved in the $\ce{C=O}$ double bond and moves them to oxygen, creating a positive charge at the carbonyl carbon.  In the third, or left-most resonance structure we see that the "single-bonded oxygen" is also involved in resonance delocalization, but it does not create any positive charge at the carbonyl carbon.  Aside from resonance factors, both oxygens will create partial positive charge at the carbonyl carbon through inductive effects.

To answer the question about the relative $\ce{pK_{a}s}$ of these acetic acid derivatives let's proceed as follows.  First, let's consider resonance effects of the substituent ($\ce{X}$) to be negligible.  This is because 1) the substituent is not directly attached to the carbonyl group or the resulting carboxylate anion, 2) it turns out there is a node at $\ce{C2}$ in the carboxylate anion (zeroth order approximation), so the $\ce{X}$ group is disconnected in a resonance sense.  So we need only consider the inductive effect of $\ce{X}$ upon both the reactant and the product and this should allow us to estimate how the equilibrium involving acid ionization will shift.  $\ce{X=H}$ is our baseline and we will compare the other two compounds against it.  When $\ce{X=F}$ the starting acid ($\ce{A}$) will be inductively destabilized due to the electron withdrawing inductive effect of fluorine.  On the other hand, the inductive effect of fluorine will serve to stabilize the resultant carboxylate anion ($\ce{B}$).  These two effects (destabilization of the reactant and stabilization of the product) will act in concert to push our equilibrium to the right, making fluoroacetic acid a stronger acid than acetic acid.  In the case of $\ce{X=CO2^-}$, the inductive electron donating effect of the $\ce{-CH2CO2^-}$ group should be stronger than the $\ce{CH2-H}$ group in acetic acid, so our reactant should be more stabilized.  On the other hand, the inductively electron donating $\ce{-CH2CO2^-}$ should destabilize the already electron rich carboxylate anion ($\ce{B}$).  Here both inductive effects again work in concert, but now to stabilize the reactant and destabilize the product making this derivative less acidic than acetic acid.  In addition to this inductive effect there is also an electrostatic effect.  When $\ce{X=CO2^-}$, we will have two negative charges in $\ce{B}$.  This electrostatic effect further destabilizes $\ce{B}$.  The combination of inductive and electrostatic effects in this last case will make this acid much less acidic than acetic acid.

I still have one last doubt. I edited my question and introduced $O_{2}^{-}C(CH_{2}){4}COOH$. It's about the electron donor effect of $CO{2}^{-}$. Can you check again my question and explain why this group acts like an electron donor instead of electron withdrawing like $F$?

In this case the $\ce{CO2^-}$ group is too far away to have an inductive effect.  The inductive effect of the $\ce{(CH2)_{4}}$ group should be just a bit more electron donating than a methyl group (acetic acid) which should make this acid slightly less acidic than acetic acid.  But we that this compound is much less acidic than acetic acid.  Again the destabilizing electrostatic effect of having two negative charges in the same molecule must be the major factor making this acid so much less acidic than acetic acid.


text:
Conjugate Theory and Redox
I'm trying to apply conjugate theory, which I can apply very well to acids/bases, to redox. Can you verify my logic, which I have broken down below?
1) I know that solid sodium is a strong reducer.
2) I know this because solid sodium when placed in water reacts violently, forming, among other products, sodium ion.
3) Reducers are oxidized. Reducers lose electrons. Correct; $\ce{Na^{+}}$ is the product of placing solid sodium in water.
4) So in this half-reaction: $\ce{Na<=>Na^{+} +e^-}$, sodium is the reducer. $\ce{Na^{+}}$ is the conjugate oxidizer.
5) Oxidizers are reduced. Oxidizers gain electrons.
6) Because sodium is a strong reducer, which means it has a strong tendency to be oxidized, or a strong tendency to lose electrons, its conjugate oxidizer must be weak. Oxidizers are reduced; oxidizers gain electrons. The potential for $\ce{Na^{+}}$ to gain the electron that $\ce{Na}$ just lost must be small, or $\ce{Na}$ would not be a good reducer.
7) Therefore, a strong reducer's conjugate oxidizer must be weak.
8) Likewise, a strong oxidizer's conjugate reducer must be weak.
9) Likewise, a weak reducer's conjugate oxidizer must be strong.
10) Likewise, a weak oxidizer's conjugate reducer must be strong.
Also, if anyone could point me to a good reference on gaining insight into redox, tha would be great!
And while we're on the topic of redox, what's the mechanism for this reaction? Are both bonds cleaved homolytically? I see how the nucleophilic chlorines attack the electrophilic hydrogens but how's the bond broken?
$\ce{H_2 + Cl_2 ->2HCl}$

Your question covers two different topics:

Redox properties can be well described with redox potentials attributed to a given reaction. It is essentially the same concept that you are talking about: we assign a number (conveniently it is an actual measurable potential) to an electrode reaction which includes an oxidized and reduced form of the species we are talking about. This number, the redox potential, can show you how strong oxidizer/reducer is the corresponding corresponding form, and just as you guess, strong oxidizers generally have a weak reducer pair.
The $\ce{HCl}$ formation reaction is generally a radical chain reaction, triggered e.g. by light, and not a redox reaction. The key here is to split the $\ce{Cl2}$ (e.g photo-chemically), then the obtained free radicals induce a chain-reaction with elementary steps like
$\ce{Cl + H2 -> HCl + H}$ or
$\ce{H + Cl2 -> HCl + Cl}$
The two gas mixed is more or less inert in total darkness.

I didn't know that about the HCl reaction. It was a picture in the redox chapter of an introductory level text I consulted. Thank you though for clarifying. Also by radical formation I assume that you mean the Cl2 must be split homolytically then!

The formation of Hydrogenchloride from the elements is still a redox reaction, even if its mechanism is a radical one. In one of the reactions $$\ce{H. +Cl2 <=> [HCl2]. <=> HCl + Cl.}$$ The oxidation state of hydrogen is increased while the oxidation state of chlorine is decreased.

You are right. Thank you for the clarification and cleaning up the answer!