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What is a rigorous definition of gas volume, and how is the Van der Waals equation derived?
I am confused about the justification for the corrections to the ideal gas law in the Van der Waals equation: $$p=\frac{nRT}{V-nb}-a\left(\frac{n}{V}\right)^2$$
I understand that the equation attempts to correct for attractive and repulsive forces between molecules, and that at high volumes the corrections are negligible, at intermediate volumes the $V-nb$ correction dominates (since attractions are dominant and this shrinks the volume, as should occur for attractions), and at low volumes the $a(n/V)^2$ correction dominates, since repulsions are dominant.  However, the 7th edition of Atkins' and de Paula's Physical Chemistry justifies part of this equation as follows:

The repulsive interactions are taken into account by supposing that they cause the molecules to behave as small but impenetrable spheres.  The nonzero volume of the molecules implies that instead of moving in a volume $V$ they are restricted to a smaller volume $V-nb$.  This argument suggests that the perfect gas law $p=nRT/V$ should be replaced by $$p=\frac{nRT}{V-nb}$$ when repulsions are significant.

I am confused by this argument-- what exactly is the definition of 'volume' here?  They seem to be saying that it is the empty space around the gas molecules, but it seems to me that volume should be defined as the space 'taken up' by the gas.  Why should the space taken up by the particles themselves be subtracted?  This isn't done for solids or liquids as far as I know.
These thoughts have also made me realize that I'm not quite sure what it means for a gas to 'take up' space.  Does anyone have a rigorous definition of gas volume?
EDIT: The exchange with Chris in comments has raised further questions.  It now seems to me that the $V-nb$ correction actually accounts for repulsions rather than attractions.  I think I was incorrect in thinking that the $V-nb$ correction dominated at intermediate volumes and $a(n/V)^2$ at low volumes.  If $V-nb$ is for repulsions, then it should dominate at low volumes, but I can't tell from the equation which correction will actually be dominant.  Also I am now wondering whether it even makes sense to connect one correction with repulsions and one with attractions.
EDIT: Follow-up questions for F'x:
- I thought that 'density' meant mass/volume.  Is the use of it to represent the inverse of molar volume (as you have used it) common?
- Where is the phase-transition in the red van der Waals curve?
- I am still a little unclear on gas volume.  Are the 'volume available to the gas' and 'volume of the gas' the same?  You say it's the same as the shape of the container, but aren't we subtracting the volume of the actual gas particles from that?  The $V$ term represents the volume of the gas, correct?  And $V-nb$ represents the 'volume available to the gas'.

I was considering answering this question, but a more rigorous derivation I know is in terms of statistical thermodynamics, which you clearly have yet to study. $b$ is a sphere some constant larger than the sphere representing your atom of interest (gas molecules move around a lot), and $V$ is the total system volume. When you have $n$ particles, they have effectively 'taken up' $nb$ space. The remaining space they can move is $V- nb$, since $V$ is your system's volume. As $nb$ approaches $V$, i.e. less space, pressure goes up a lot. Does that help any?

Thanks Chris.  Your comment has made me realize that what is expressed by the $V-nb$ correction is not really a reduction in gas volume, but a change in the relationship between pressure and volume when the volume becomes very small.  The volume of the gas includes the space taken up by the molecules, but the gas will only behave approximately ideally for the first $V-nb$ of the total $V$ if it is being compressed.  Once we get volumes in the neighborhood of $nb$, pressure shoots up massively with further volume reductions.

The van der Waals equation can't be derived from first principles.  It is an ad-hoc formula.  There is a "derivation" in statistical mechanics from a partition function that is engineered to give the right answer.  It also cannot be derived from first principles.
A gas is a collection of molecules that do not cohere strongly enough to form a liquid or a solid.  The volume of a gas is, as long as the molecules fit into it, the volume of the container holding the gas.  In other words, the volume of a gas is NOT a property of the gas directly, but a property of the container.

The ad hoc part cannot be sufficiently emphasized.

I am accepting this answer because it succinctly points out that Van der Waals can't be derived from first principles, but readers should look at F'x's answer below as well.

I edited the first van der Waals equations in your question, because it was incorrect.

First, the volume available to the gas is pretty much was you think: it's the space left for it to occupy, i.e. the volume delimited by the container. If you think of a gas tank, it's the interior volume of the tank. For systems of macroscopic dimensions, there is no real ambiguity as to its definition, which can be purely geometrical. For nanoscale systems, such as accessible volume in porous media, the definition is much harder: it is typically done by constructing a Gibbs dividing surface between the two phases considered (here, a solid and a gas).

Second, importance of both terms in a real-life example. Let's write the van der Waals equation in terms of density, rather than volume and quantity:
$$ \left(P+a\rho^2\right)\left(\frac{1}{\rho}-b\right)=RT $$
Let's take a practical example of a gas with a large quadrupolar moment, CO2. We have $a$ and $b$ parameters for it from this source, and we can plot $\rho(P)$ at $T=300$ K:

(black: ideal gas; red: van der Waals gas; green: only $b$; blue: only $a$). We see that excluded-volume correction by itself is not much of a gain, but it's actually crucial once you include the attraction, because otherwise the phase transition (the cusp in the blue curve) occurs at much too low pressure.

Finally, regarding the derivation of the van der Waals equation, this is typically a topic of statistical thermodynamics and there are many such demonstrations. My favorite is the classical one by Hill, but it may not use the notations in modern textbooks. I'm also partial to that in McQuarrie’s book. All statistical physics derivation introduce the excluded volume in a very natural fashion, but they show more variation on the way they arrive at the $a$ term (so of them find the exact analytical form of the equation, while in some derivations you find a series expansion whose first term is the van der Waals equation).

See above in my main post for the followups.  Thank you for pointing me to McQuarries definition.  Also, it would be minor but nice if you labeled the axes on the graph, to make this a clear resource for future.


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Regenerating the imaginary data of a 2D NMR spectrum with a hilbert transform
2D spectrum take up a large amount of hard disk space. At my last lab it was standard practice to delete the imaginary portions of the spectrum with xfb n (We used TopSpin) and if needed later, to regenerate it from the Real part of the data with a Hilbert transform (Related, the TopSpin Commands and References, ~pg P-233 in version 2.1.2)
My first question: If the raw data is still stored in the ser file (TopSpin Commands and References v2.1.2, pg P-8) why can't you just reprocess the raw data? Is all the data not there? (I suspect this somewhat, as those files are much, much smaller then the processed files, and TopSpin processes 2D spectra as it acquires them, unlike 1D spectra)
My second (highly related) question: Is there a downside to regenerating the imaginary data from the real data? Is it a perfect replication of the original data, or is it an approximation? Why do we ever bother to store this data if we can recreate it at any time?

If you have the raw data (the FIDs in the ser file) still available, there is no reason to use a Hilbert transformation, you can just process with xfb instead of xfb n again. xfb n also doesn't really discard the imaginary part, it processes the spectrum from the raw data again, but doesn't save the imaginary processed data.
The ser file contains all the recorded FIDs, that data is sufficient (together with the parameter files) to recreate the full spectrum. That your Topspin processes 2D spectra automatically is just some specific configuration, it doesn't do that by default.
As far as I understand it, if you zero-fill your data at least once, all the information content of the imaginary part is then present in the real part of the processed spectrum. So I'd guess that under those conditions, the Hilbert transformation shouldn't have any downside, but I'm not totally sure about that.

Now this is most useful in regenerating the imaginary part of the 2d data, after doing the reconstruction of a non-uniform sampled 2D (or nD) spectrum. Bruker's Topspin doesn't do this by default, with say a phase sensitive dept-edited HSQC; so you can't phase it with out doing 'xht2'.

Welcome to Chemistry.se Duncan. You night find the [tour] useful to get familiar with the site!

Pardon me but is this an answer to the question? I genuinely can't tell.

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What open-source software can produce 3D PDF representation of molecules?
Most of the existing software I know (like VMD or Pymol) say that they can be used to create PDF files with 3D representations of molecules: actual 3D models, which you can rotate inside the Adobe Acrobat viewer. However, they all require the use of “Adobe Acrobat Pro 9 Extended” for its 3D capture feature. From what I can read, this feature does not exist any more in Acrobat X and was transferred to a separate product, Tetra4D, which doesn't work on Mac OS.
So, how can I produce PDF files including 3D molecular models using free (or better, open source) tools?

What I would look for is something that lets you draw molecules and save as a 3D standard (VML, etc). Then something that takes that standard and saves as 3D pdf. Open source of course. Mash 'em up!

Disclaimer

This is not for the faint-hearted.
If you are not interested in using LaTeX, stop reading now, as this uses the media15 or movie9 packages to embed U3D in a pdfLaTeX generated PDF. This answer assumes basic familiarity with LaTeX. I have tested this with pdfLaTeX and it should work fine in XeLaTeX.
One of the tools  I have used (DAZ Studio) is apparently 'free for a limited time'. It is also not available on Linux. FOSS replacement suggestions are welcome.
I'm of the personal opinion that 3D in PDF is not as good as it sounds.

What doesn't work
VMD offers the ability to output as OBJ, which can then be processed with FOSS MeshLab into a U3D file that can be embedded into a PDF. This is sub-optimal for two reasons:

There is no way to tell MeshLab that the VMD-generated mesh is supposed to be Gouraud-shaded. This means that the hard facets of the mesh will be glaringly obvious in the pdf, even if you use an unwieldy degree of subdivision.
Colours are not preserved. Your model will appear in the PDF with a uniform grey texture.

As such, MeshLab is not suitable at this stage.
Preparing a model
VMD offers a number of decent export options to get our geometry out of .PDB or what have you and into a mesh format. I opted to use the VRML format to output a mesh of a ribbon representation of PDB entry 4DM9. I then imported this into Blender.

Blender has no reliable U3D export capabilities, however I can use it to assign Gouraud shading to all of the shapes and assign colours to them. I then use the blender export function to export the scene as a Wavefront .OBJ file, with attendant .MTL file.
Processing with DAZ Studio
I am indebted to the author of this article (John Nyquist), which indicated that DAZ Studio can convert Wavefront to U3D just fine, with preservation of textures. DAZ is a very specific program that seems to be heavily geared towards posing, clothing and rendering humanoid figures in the spirit of programs like Poser. However all we really need is the U3D export function and we can read in the structure just fine.
Sometimes, this is how I see myself:

Okay in all seriousness, don't put a human figure in or on your molecule without good reason. Export as U3D and then we can embed in pdfLaTeX.
pdfLaTeX and movie9
As a proof of concept, I just followed the minimal example given in the media9 manual for getting a 3D model into a pdfLaTeX-generated PDF. Here it is, being viewed in Adobe Reader X (I know, I know!). Bear in mind that very few readers can actually display 3D PDF, which is one of the reasons why I feel that it is not such a good idea for chemical visualisation.

It starts off zoomed way too far in because I did not set up any options for the import (however you can find out all about that in the media9 or movie15 package manuals) and is a bit laggy on account of having over 1 million faces (I went way overboard with subdivision), however the important thing is that it works.
Improvements welcome!
Obviously, this is a pretty Rube Goldbergesque route to getting a molecule into a PDF. There probably exist easier ways which should definitely be suggested.

If you are ready to invest a lot of time to learn Asymptote you can use it to create 3D PDF documents from pdb files with it. Here is an example. It has the advantage that Asymptote has a good LaTeX integration and you can use the code directly in a LaTeX document. But it would be very messy to get the appearance right and I'd say it will only work for Ball-and-Stick models.
Update:
I found some other nice examples of crystal unit cells being depicted with Asymptote, which can serve as some kind of proof-of-concept on using Asymptote to depict (simple) molecules:
Labels and angles in crystal structure using Asymptote

Drawing 3D crystal lattice with tikz/pstricks

Both are from TeX.StackExchange, so if you want to try asymptote you might find some help there, when you run into problems.

That's pretty neat - how is atomic connectivity obtained, though? It looks like most of the atoms are joined in some kind of linear sequence, probably the order in which they are defined in the pdb?

@RichardTerrett: Sorry, but I don't know how it is done in the example. I accidentally came across Asymptote while searching for a good way to embed 3D molecules in PDF documents but haven't had the time to learn it. But I'm pretty sure one can produce code with it that connects certain atoms with one another according to information from pdb files.

There is online converter from PubChem CID to 3D PDF/U3D
at http://xemistry.com/ (http://85.214.192.197/pdf3d to be exact).
As to Asymptote - go to its forum and ask for help.
3D PDF export is under development, so early adopters (AKA guinea pigs) are welcome.
Among other things OBJ import is in the works.
The apparent lack of demand is one of the reasons FOSS support for 3D PDF is slow to emerge - and at lest asking for it in appropriate place is the minimal effort I can think of.
Also you can inspire the developers of your favorite tools to take code from Asymptote or Intel U3D Sample library to produce 3D PDF directly or develop standalone converters.

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How does adding lemon juice to sugar make better caramel?
It's a common cooking advice: if you want a somewhat softer caramel, add some lemon juice to the sugar:

To help prevent the caramel from crystallizing, you can add an acid to the sugar before you begin: add about half a tablespoon of lemon juice to each cup of sugar and mix it with your hands

But how does it work?

The citric acid in lemon juice hydrolyzes the sucrose molecules into glucose and fructose as you heat the sugar. The water in the lemon juice is also part of this process. This mixture of monosaccharides is much less crystalline in the same way that most mixtures are less crystalline than pure substances. Wikipedia has a good article on this process under the title "Inverted sugar syrup". The syrup is "inverted" in the sense that the sign of its optical rotation has changed because sucrose has a large positive optical rotation while fructose has a large negative optical rotation. Glucose has a positive optical rotation, but of a lower magnitude than sucrose.
The reaction is:
$$\ce{\underset{(sucrose)}{C12H22O11} + H2O -> \underset{(glucose)}{C6H12O6} + \underset{(fructose)}{C6H12O6}}$$
The reaction can occur to a small degree by simply heating sugar with water in absence of acid catalysis. I assume the acid is necessary to get good conversion before you burn your sugar or vaporize all of the water.
Honey is a naturally occurring inverted sugar syrup of this type. Honey is resistant to crystallization because it is a mixture of glucose and fructose. Honey is also slightly acidic.

Aside this "invert-sugar-reaction" some more reactions are catalyzed ba the acid. Most of this reactions are the same as in caramelization without acid. But You achieve caramelization faster and at lower temperatures.

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Why does menthol (e.g. peppermint) feel cool to the tongue?
Especially when drinking water after the fact, mint can give a sharp cold sensation inside one's mouth.  What process causes the sensation to occur?

There is a really excellent, detailed answer to this here (http://boundlessthicket.blogspot.com/2012/04/that-cool-mint-feeling-when-i-was-in.html). If you'd like me to summarize it as a formal answer, I'd be happy to...but it's really written up very well.

On [biology.se]: Why do mints make your mouth feel cold?

Menthol it self gives a cold feeling in the mouth because it is active at the same receptor (an ion channel) on the tongue that cold temperature triggers. Interestingly, although they act at the same receptor, they act at different sites, so that provides the intensified response when eating a mint and then drinking water. This reference gives an excellent detailed answer, with references to the original papers, which I'll  summarize here.
Menthol acts at the TRPM8 protein which forms an ion channel that allows $\ce{Na+}$ and $\ce{Ca^2+}$ ions to flow into cells and this sends a signal saying "cool" to the brain. (As an aside, this protein monitors temperature across the body and not just on the tongue.) Cold temperatures actually change the confirmation of this protein, which allows the ions to flow more freely, and sends the signal to the brain. Menthol, on the other hand, stabilizes the open channel (allowing ions to flow even more freely) and also...

“menthol shifts the voltage dependence of channel activation to more negative values by slowing channel deactivation”.  This is very significant to my question because it supports a claim made by the first web page I visited which stated that menthol acts on the receptors, leaving them sensitized for when the second stimulus is applied (i.e. cold water) resulting in the enhanced sensation.  This mechanism of binding is very clearly different from the mechanism of cold affecting the TRP channels.  This is why the sensation is increased when both stimuli are applied, yet is not affected after addition stimulation from the same stimuli (i.e. eating another mint).

All in all, pretty cool.

"conformation" (typo). A great answer.

Where's this citation from?

@Mithoron I don't think you are going to get an answer. It looks as though Janice hasn't been on the site since 2012.

@Tyberius Just felt it should be pointed out. It is probably traceable so maybe someone (or even  myself) will check it.


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Is the order of orientation of electron box diagrams meaningful or arbitrary?
Here is my interpretation when asked to:

By drawing arrows in the appropriate boxes, complete the outer electron structures for Cu and Cu2+

I had no problem in drawing out the electron structure, but am a little confused by the orientation of my arrows.  I understand that each normal box has to have arrows in opposing directions to represent the opposite spins on the electrons, but the mark scheme defines which way round they want them to go:

Source: OCR GCE Chemistry B F334 June 2011 Markscheme pg. 11
Ordinarily I would have said that whether you go up or down first (or even consistently) is irrelevant and arbitrary, but the comment in the guidance

ALLOW single arrow in either direction

suggests to me that with the paired arrows, the order of directions matter.

Is there any chemical reason for putting the downwards spin arrow first?  I would happily accept by convention as a reason but I just want to be sure that I'm not missing anything chemically relevant.

I dont think so.. Your teacher seems to have subcontracted the correction to non-teaching staff :/. Note that unpaired electrons all have the same direction of spin, that's the only remotely related rule.

I'd say it doesn't make a difference, since in atoms and ions the energy levels for two states with different spins (but all the other quantum numbers the same) are degenerate. Furthermore, the choice of the system's $z$-direction, to which $\uparrow$ and $\downarrow$ refer for the spins refer, is arbitrary in a spherically symmetric system.

Raise hell with the grader. There's no reason whatsoever for this answer to be rejected. A Google book search can substantiate your argument, and the first example from a textbook is actually in your order:

I think the guidance “allow single arrow in either direction” is not exclusive with “allow double arrows in either order”, and maybe only single arrows were explicitly mentionned because it was obvious to the writer that order in double arrows doesn't matter.

That's what I found also.

I just cracked open five textbooks (Silberberg, Housecroft and Sharpe, West, McMurray, Simons) and they all adopt the $\uparrow\downarrow$ convention.
The only conceivable reason to prefer any order (namely, alphabetic ordering of the spin orbitals $\alpha$ and $\beta$) would actually suggest your chosen order. There is no chemically meaningful distinction between $\uparrow\downarrow$ and $\downarrow\uparrow$. I would love to know the marker's rationale here.

The order definitely does not matter.
This sort of diagram is for representing the quantum numbers of the electrons in the system. The 3d boxes mean the quantum numbers n=3, l=2 with ml =-2,-1,0,+1,+2 for the 5 boxes.
If a box has an up and a down arrow in it, this means there are 2 electrons with those quantum numbers, one with s=1/2, and one with 2=-1/2. It is completely meaningless which order your draw these in, the result is still the same two electrons with the same set of quantum numbers. Electrons are indistinguishable from each other. If you had a complex system with an exotic lepton such as a tau or muon then the order might matter, but in this case it certainly does not!
I can only assume that the writer of the mark-scheme assumed that UP DOWN and DOWN UP being identical would be obvious to anyone, and so didn't bother to write it. If the answer to a maths question was x * y, most mark-schemes wouldn't bother to write y * x also acceptable, as its obvious to anyone with any subject knowledge.

This is intended as an extension of RichardTerret's answer.
I personally have used/been taught uparrow first (not _explicitly  taught, just that the teachers did it).
Anyway, I wanted to address the "Allow single arrow in either direction" issue.
I think that you have misinterpreted this. Basically, it's allowing some deviation from Hund's rule
Normally, Hund's rule forbids stuff like $[\uparrow\downarrow][\uparrow][\downarrow] $, since the spins cancel out (Hund's rule valls for maximum total spin). The second two ought to be both up or both down.
In early stages, this aspect of Hund's rule is not necessarily taught--Hund's rule is just portrayed as the thing that prevents pairing (from my experience). So, the paper setter may have decided to allow violations of the other aspect(if it is that) of Hund's rule, and thus the "Allow single arrow in either direction". They mean that the directions of single arrows can be mixed up.

There's only a single “single arrow” in each case, so I don't think that can be the reason why…

@F'x: Hmm, yeah. Though it may be a boilerplate for all such questions... :\


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13C NMR spectrum only showing solvent
This is my first time taking a $\ce{{}^{13}C-NMR}$ spectrum and while I'm aware the signal to noise ratio is much lower compared to $\ce{{}^{1}H-NMR}$, I didn't realize it was quite this bad. I took a spectrum and I only see the $\ce{CDCl3}$ triplet. What concentration range is generally appropriate given say 256 scans (10 minutes on our machines?), or is it rare to see any good results after without drastically increasing the scan time?
Spectrum for reference (taken on a 300 MHz Varian Mercury):

What pulse sequence are you using? Are you sure your delays between scans is sufficient for relaxation? Have you calibrated your pulse power/length to ensure you get 90degree pulses? As its your first 13c spectrum, have you tried using a huge concentration of something else (Nujol, ethanol etc) to check its working?

At 256 scans you should see something if your compound has reasonable concentration. The rule of thumb I was taught as an undergrad was ~10 mg of compound for a $^{13}$C scan in under a mL of solvent (400-500 uL if you are measuring precisely, modified by probe size and whatnot of course)
As for an exact concentration based on MW? I'm not sure, I've never really worked it out, since the rule of thumb works well enough.
Next question: Does your compound have protons on some of its carbons? Quaternary carbons will be much harder to see, so it is possible that you won't see them after 256 scans.
However yes, concentration is probably the problem. I made this mistake yesterday and found that after 256 scans I could see a tiny blot on the baseline.

Thanks, I tried again with 1000 scans and still nothing. I suppose the concentration is far too low for my compound which is very heavy.

With the same sample? Remember that throwing more scans at it increases the resolution on $\sqrt{ns}$ where $ns$ is the number of scans. So you are almost always better off increasing the concentration. So going from 256 to 1024 is only doubling your S:N ratio. More concentration is a far better option. The other thing, is are you sure you are setting things up right? Good shimming, etc? Can you see the 1H spectrum decently?

There are several reasons why $\mathrm{^{13}C}$ experiments are far less sensitive than proton experiments.
The natural abundance for $\mathrm{^{13}C}$ is only 1.1%, while $\mathrm{^{1}H}$ has a natural abundance of 99,9%. This difference alone means that if you don't label your sample with $\mathrm{^{13}C}$, the spectra will be around 90 times less sensitive than proton spectra due to the lower natural abundance alone.
$\mathrm{^{13}C}$ also has a four times lower gyromagnetic ration than $\mathrm{^{1}H}$, which also lowers the sensitivity. But the way the experiment is usually measured, you get magnetization transfer from directly-attached protons to the carbons, which increases the sensitivity.
$\mathrm{^{13}C}$ also relaxes more slowly, which means that the time between experiments is longer and you can measure fewer scans than for proton experiments in the same time.
Another factor that is important is which kind of probe your spectrometer is using. For an inverse probe like e.g. a Bruker TXI, the inner coil is the proton coil, which means it's especially suited for proton experiments. For $\mathrm{^{13}C}$ experiments an observe probe like a BBO is more sensitive, as the carbon is measured on the inner coil. If you have multiple spectrometers, you should ask which one is better suited for $\mathrm{^{13}C}$ experiments.
If you have a very small amount of sample, a Shigemi tube allows you to measure in half the sample volume (200-250 $\mu$l) compared to standard tubes, you can achieve a higher concentration that way. They're pretty expensive though and they have to be matched to the solvent, so you would need a CDCl3 tube.
And there are various problems that could drastically reduce your sensitivity for $\mathrm{^{13}C}$, the most important ones if we disregard hardware problems, are proper tuning and matching for $\mathrm{^{13}C}$, and the correct pulse length for $\mathrm{^{13}C}$.
I've avoided your actual question, as the required concentration depends on quite a few factors. I'd recommend you to ask the person responsible for the spectrometers, as they should be able to give you a better estimate based on the actual hardware you have.

You might start seeing signal if you set lb=1 (or, drastically, lb=2) and then type "wft" It won't give you a decent spectrum but you may at least see the beginnings of peaks.
Also, are you confident that this is a 1-H decoupled spectrum? If not, then your S/N is going to be very poor.


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Why is it so difficult to create mock NMR spectra of macromolecules?
From what I understand, if you have a NMR spectra, you can determine the structure of a macromolecule but it is very difficult to determine what the NMR spectra of an expected structure would look like? Why is this the case?

I think you can sort of determine the qualitative NMR spectrum, but getting the heights right is harder. But I'm not sure.

Macromolecular NMR spectra are difficult to compute because macromolecular electronic structures are computationally expensive to solve with accuracy. Obtaining precise chemical shifts and splitting requires accurate treatment of electron density and geometry, especially for simulation of NOESY, which I understand is important in macromolecular structural elucidation. This requires good methodology and large basis sets.
An interesting case study by Saielli and Bagno demonstrates the use of DFT (with cc-PVTZ basis) in predicting NMR spectra for two candidate structures and the correlating them with experimental data, however the mean error in chemical shifts is nontrivial1 (0.2 ppm in the best match). The same people do $^{1}\ce{H}$ and $\ce{^{13}C}$ NMR predictions of a number of complex molecules2 (with intriguing discrepancies in shifts) however this is a far cry from predicting, say, a large protein $^{1}\ce{H}$ NMR spectrum.
I have seen EPR spectra of transition metal clusters simulated as abstract spin systems (for instance by EasySpin) however I presume that this is very qualitative.

(1) Saielli, G.; Bagno, A.; Can two molecules have the same NMR spectrum? Hexacyclinol revisited.; Org. Lett.; (2009) 11, pp. 1409-1412
(2) Bagno, A.; Rastrelli, F.; Saielli, G.; Toward the complete prediction of the $\ce{^1H}$ and $\ce{^{13}C}$ NMR spectra of complex organic molecules by DFT methods. Application to natural substances.; Chem. Eur. J.; (2006) 12, pp. 5514-5525

In summary: There are a freaking huge number of atoms present, the complexity goes up more then linearly, as you have more relaxation types to consider, tumbling speed becomes an issue, and other things. Net result: Bigger molecules = harder, macromolecules are freaking huge. ^^

What order of magnitude of mass can NMR achieve nowadays?  I vaguely recall 10 kDa being some sort of reasonable limit but I don't know how old that figure is.


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When we heat a borax crystal, why does it first expand and then contract?
In the borax bead test, we take a crystal of borax and then heat it.

A small loop is made in the end of a platinum or nichrome wire (as used in the flame test) and heated in a Bunsen flame until red hot. It is then dipped into powdered borax, and the adhering solid is held in the hottest part of the flame where it swells up as it loses its water of crystallization and then shrinks, forming a colourless, transparent glass-like bead (a mixture of sodium metaborate and boric anhydride)

Whenever I've done the experiment, I've noticed that the process looks like this:

It heats up in a second, in this time there are no visible changes
It starts expanding/blooming. Bubbles appear, and little pseudopodia(closest work I can think of) of borax flail about. This takes approx five seconds. At the end of this, the borax looks like whipped cream that's been...well.. whipped .
The thing suddenly contracts into a tiny bead.

Now, as mentioned in the quote, basically borax loses its water of crystallization, and then decomposes into $\ce{NaBO2 +B2O3}$, the small bead.
The flailing about is probably due to steam escaping.
But, the reaction proceeds in such an interesting manner, I can't help but think that there's something else going on as well.
What are the details of the mechanism of the borax bead preparation reaction?

I've never seen this reaction, but it sounds quite entertaining. And I learned a couple of new words in looking at the literature to explain this phenomenon.

Intumescence: The property of a material to swell when heated. Borax is intumescent.
"Calcination is a thermal treatment process in presence of air applied to ores and other solid materials to bring about a thermal decomposition, phase transition, or removal of a volatile fraction. The calcination process normally takes place at temperatures below the melting point of the product materials."

The mechanism for the calcination of borax quite complex and is covered in some detail in this article. Borax decahydrate (BDH), $\ce{Na2B4O7.10H2O}$ is easily dehydrated to Borax pentahydrate (BPH), $\ce{Na2B4O7.5H2O}$ without much visible change. The actual form of the borate ion in both BDH and BPH is $\ce{B4O5(OH)4^2-}$ with 2 of the water molecules actually incorporated as hydroxyl groups in the ion. (The other 2 hydroxyl groups come for the water hydrogens and 2 oxygens from the $\ce{B4O7^2-}$.) That means that the last two water molecules can only be removed by molecular decomposition. So in BPH, you actually have only 3 "free" (i.e., easy to remove) water molecules.
So here's what happens. The borax decahydrate dehydrates smoothly with minimal density changes (no swelling) until you get to borax pentahydrate. At this point, the BPH dissolves  in its own water of dehydration and the "blossoming" you see is from the solution erupting through the partially dehydrated crystal surface as it's heated. The density of the borax changes dramatically as it goes from the pentahydrate (about $0.85\ \mathrm{g/cm^3}$) to the dihydrate  (about $0.14\ \mathrm{g/cm^3}$) which is why it ends up looking "whipped".
To do the final transformation to a tiny bead, the remaining 2 waters of hydration (those tied up as hydroxide ions) are driven off as the $\ce{B4O5(OH)4^2-}$ decomposes to give metaborate, $\ce{BO2^2-}$ and boric anhydride, $\ce{B2O3}$. $$\ce{B4O5(OH)4^2- -> 2 BO2 + B2O3 + 2H2O}$$
Fascinating.

Nice :) Try it some day, borax is common in a lab. Take a platinum loop and heat a few wet crystals in it. Fun! Any reason why the metaborate clumps together? Probably cohesion, IMO

My thought was that it just melted at that point and changed to a glass...the paper talks about sintering as it dehydrates.

Yeah, possibly. The glass, being molten and sticky, would easily coalesce.


text:
Can a NOESY NMR experiment be halted partway through and give meaningful results?
Some NMR experiments can be halted partway through the acquisition and still give perfectly valid results: The traditional 1H experiment for example. Others, like a Saturation Transfer Difference Spectrum will have errors in the magnitude of the peaks in it if it is halted, due to the fact it uses groups of pulses and takes the difference of between them.
Can the standard proton-proton NOESY pulse sequence be halted halfway through and give a usable spectrum, or will this introduce artifacts?

Very late answer, but  well, maybe it will help someone else who reads this.

halted halfway through

2D experiments are acquired as a series of increments where $t_1$ is changed in order to generate the indirect frequency dimension ($F_1$).
On Bruker spectrometers, 2D pulse programmes are always written so that each increment is saved to disk as soon as it has been recorded. So, if you were running an experiment with 256 increments and you stop acquisition halfway you will still have 128 increments saved to disk; there is no need to use the tr command. Furthermore, since $t_1$ increments are usually acquired in a regular, uniform manner, this truncated data will still be suitable for Fourier transformation.* This is already pointed out in Mad Scientist's answer.
However, it omits one salient point: if you record only half of the increments for a 2D experiment, then the resolution in the indirect dimension will be decreased by a corresponding amount. Depending on the processing being used you may also find that there are truncation artefacts, which appear as 'wiggles' around the peak in the indirect dimension (although processing parameters are often chosen to minimise these anyway).
Whether that is 'usable' enough will vary depending on your use case. If you are looking for NOEs between two peaks that are quite close to one another, it's possible that it's not good enough. On the other hand, HSQC spectra are very sparse and you can often get away with it.
If you're unsure, you can test this out by processing the 2D data during acquisition, i.e. before stopping it. Just do xfb (Fourier transform), and if necessary apk2d (phase correction), plus whatever other processing you would normally do. TopSpin will process the data which have been acquired so far. If the results are good enough, you can stop the experiment. If they aren't good enough, don't stop it.

* Off-hand, I'm not sure what happens with NUS spectra, but I expect that it can still be processed as usual (you'll just end up with a lower NUS sampling percentage than what you initially keyed in). Note that using a very low NUS sampling percentage can—and will—introduce reconstruction artefacts.

There are two big differences between the STD experiment and a NOESY that matter when aborting the experiment early. The STD is usually run as a 1D experiment, and it's using the phase cycle to substract the on- and off-resonance spectra from each other.
When you acquire a 1D spectrum with multiple scans, the scans are not saved individually but directly added up to one FID, at least on Bruker spectrometers. So if you abort a 1D experiment (you need to execute tr to do that in most cases, as the FID is only written to disk after all scans are finished in most pulse sequences), you can't just throw away the last incomplete phasy cycle. And with experiments like the STD, your actual signal is the difference of two larger signals, so one incomplete substraction can ruin your spectrum.
In a 2D spectrum the increments for the indirect dimension are measured and written to disk sequentially, so you can throw away the last incomplete increment if you would need to. In most cases the incomplete increment will not be saved at all to disk, so you'll end up with one less data point in the indirect dimension, but no incomplete ones that were aborted in the middle of the phase cycle. This depends on how exactly the pulse program is written, but I suspect that most standard sequences handle it this way.
I've stopped HSQC and NOESY spectra before they were finished pretty often, and never observed any artifacts that could be traced to the abortion of the experiment.


text:
What is the order of stability of the resonance structures of furan?
I want to know which are the most stable furan resonance structures. I know that structure number 1 is the most stable, but which one comes after?

2/5 and 3/4 are identical

And do you want to know which is more stable in order to predict properties like reactivity? Or is it just a question for it's own sake?

I want to ask about which one is more stable thus less reactive ? also which one has the higher charge separation and how does it affect the stability ?

I think your structures 2–5 have a problem as they are lacking a negative charge to balance the positive charge… furan is a neutral molecule, as indicated in your structure 1.

There are no such things as more or less stable resonance structures. See also What is resonance, and are resonance structures real?

A quick post so you can get started on thinking about this.
Remember, resonance is just a model. The electrons are delocalized all the time; the furan does not morph back and forth among the various structures.
You can do a "formal" evaluation of which resonance forms have the most contribution to the structure by looking at formal charges, and there are rules to help you decide, from the formal charges, which resonance forms are most important.
If you're looking at reactivity, the property you're interested in may not be that the one that is "more stable is less reactive". What you might look at instead, is the consequences of attack at the 2 and 3 position, look at the stability of the intermediates formed...a more stable intermediate would tend to drive the reaction along that path.
After looking at these ideas, feel free to come back with specific questions.

I mean i want to know which is more preferable .. especially between structure no. 2 and 3 ?

Evaluate them by the rules for "formal charges" and that answer that part of the question.

"Remember, resonance is just a model." - +1 for this. More explicitly: we use resonance to deal with the limitations of using Lewis structures to represent molecules that exhibit electron delocalization.


text:
Chemical compounds responsible for the colors in flowers?
I know that anthocyanins are a class of compounds responsible for the purple colors found in flower petals.

Anthocyanins (also anthocyans; from Greek: ἀνθός (anthos) = flower + κυανός (kyanos) = blue) are water-soluble vacuolar pigments that may appear red, purple, or blue depending on the pH. (same reference as above)

The anthocyanins, when extracted from flower petals with water, can be used as acid/base indicators, producing red colors in acid and blue in neutral solutions (in basic solutions they turn green, yellow, and then become colorless). When I've done the extractions, some red petals produce a red color in acid and maintain the red color at higher pH than purple flowers (see plate below), but some red flowers produce colors that seem to follow the normal progression.

Row 2 from the top was a purple flower, row 3 a red flower and row 4 also a red flower. (pH values start at 2 on the left and increase by 1 pH unit as you move right by adding the water extracts to buffer samples in the wells.)
So, are there other classes of compounds responsible for the red and yellow colors found in flowers? Can the red colors that give a "normal" anthocyanin pH response arise from pigments held in an acidic environment in the plant?   (That may be a biology question, but I'm interested in the chemistry side of where the color comes from.)

are there other classes of compounds responsible for the red and yellow colors found in flowers?

Interest in colors, specifically dyes, was a real motivator for serious chemical work. Some of the early efforts were aiming to mimic Tyrian purple, also known as Royal Purple, because in antiquity it was worth its weight in gold.
Two classes of molecules come to mind other than anthocyanins in phytochemistry, such as the closely related flavones (its root means yellow) and the distinct carotenes (its root means carrot, but also responsible for flamingo color).

Can the red colors that give a "normal" anthocyanin pH response arise from pigments held in an acidic environment in the plant?

A real sharp organic chemist by the name of Richard Willstatter proposed this  in the early 1900s. He was working on determining the structure of the pigment responsible for the blue color in cornflower. He extracted the same substance coming from a rose and proposed that all flower colors were due to a different pH in the sap. As your experiment clearly points out, however, alkaline conditions are typically required for the blue anion. A Japanese group argued that alkaline conditions in the sap were unlikely and that instead these different colors can be explained by the formation of metal-organic complexes.
It wasn't until 2005 that a definitive crystal structure was determined from the cornflower pigment called protocyanin. It turns out that the blue color comes from a supermolecule containing one $\ce{Fe^{3+}}$, one $\ce{Mg^{2+}}$, two $\ce{Ca^{2+}}$, six anthocyanin and six flavone molecules packaged together in a complex structure. This color, as I vaguely recall, remains blue from pH 7 until pH 1! This agreed with some earlier evidence which suggested a complex species by the temperature dependence of the pigment.
These are not easy studies to conduct and I'm not familiar with how in-depth red colors have been studied, but it is certainly possible a complex species is actually how the red pigments form in vivo. Keeping in mind that plant vacuoles have a pH on the order of 5, we'll generally notice a loss in red above pH 3 in solution for a pure anthrocyanin in solution as hydration of the cation occurs. Stability of the color, and thus cation, obviously must be obtained either via copigmentation or with metals in a plant.

Good answer (but more on the "reds" would be nice). I'd never run across the information about the protocyanin before: what an interesting mechanism. If you run across more information about the red pigments, please come back and add it to your answer. The flavones and carotenoids make sense, too. They don't change much with pH and are less interesting in the experiment. Thanks for the info.

Janice, apart from the benzopyrylium (anthocyane) dyes you mentioned, there's quite some red to be found among the natural hydroxy-substituted naphthoquinones and anthraquinones, e.g.

Purpurin [81-54-9] (1,2,4-Trihydroxyanthraquinone) from Rubia tinctorum

Carminic acid [1260-17-9] from Dactylopius occus, an insect living on opuntia
Alkannin [517-88-4] from Alkanna tinctoria

I hope that helps a bit.
Ahh, I forgot to mention

Lycopene [502-65-8] from Tomato.

While others have already mentioned the carotenoids in general, lycopene is your favourite if you want it really red :)
If you're interested in lycopene in particular, don't bother to buy it from chemical suppliers. The lycopene concentration in tomato paste is that high (and the paste dirt cheap) that the extraction is the way to go. Colleagues of mine working in photosynthesis research did that all the time.
PS. I'm sorry, you asked for red dyes in flower petals, not natural red dyes in general. Otherwise I would have mentioned a member of another class of compounds we didn't cover so far:

Betanine [7659-95-2] from the red beet

Now, I won't ;)


text:
What do the different grades of chemicals mean?
When purchasing chemicals from Sigma, Fisher, or wherever, there are often -grade's attached to their description like reagent-grade, technical-grade, analytical-grade, or more niche-sounding biotech-grade, HPLC-grade, DNA grade (DNase free perhaps?)
Is there some sort of standard for what these actually mean or are they arbitrary, differing from supplier to supplier or even chemical to chemical?  Beyond the 'niche' grades, is there a common order of purity between the others or do they depend on the types of impurities?

Grades don't mean much. Learn how to purify lesser quality chems yourself. Or pay the HUGE markups of the big-box chem suppliers. Example: Sigma Aldrich HPLC Acetone 99.9%. 1 liter costs $94 plus shipping plus hazmat etc. Instead, we buy somewhat recycled/cleaned acetone from local industry. Several liters can be processed daily by one person with a 3-5L flask, a stillhead, and the other general kit involved. Our cost per liter including KMnO4 and a little silica xerogel is under one US dollar. After a day or two in a sealed bottle over a little dessicant and POW. HPLC suitable acetone, clean

Sigma-Aldrich gives a very useful table outlining what the different purity levels are and suggested applications. I was less successful at finding equal documentation from some of the other suppliers, but the analysis for the purity of the chemicals they sell is in the catalog and on the bottle.
In general, technical grade or laboratory grade are the lowest purity. ACS Reagent grade means that the chemical conforms to specifications defined by the Committee on Analytical Reagents of the American Chemical Society (but Aldrich "ReagentPlus" means >95% pure). So, "ACS Reagent grade" chemicals should be comparable from different suppliers.  Analytical grade is generally the most pure.
For some of the other grades, such as HPLC grade solvents, the issue is less about overall purity than about being free of substances that would interfere with a particular application. I've reproduced a piece of the table below to give you an idea of the information available.

"Reagent grade" is not mentioned in the table. So what's "reagent grade" compared to "ACS reagent"?

I guess the follow up question would be "what/where are the ACS specs"?

@NickT The ACS specs depend on the particular chemical under discussion. Janice did provide a link for more info: http://pubs.acs.org/reagents/index.html

At the company Puritan Products, the reagent grade designation is used to describe high purity chemicals for which no established specifications exist. Apparently, this is often the case with products diluted from ACS grade products, as some of these cannot be ACS reagent grade in dilution by the standards that ACS sets.
Source


text:
How to calculate the equilibration time of two connected chambers of salt solution, initially at different concentrations?
I am trying to build a system of two connected chambers, one of which is also connected to a cuvette, so I can perform a spectrometry experiment with different concentrations of $\ce{KNO3}$ solution. In its simplest form, the cuvette and the first chamber would hold solution at the same concentration, while the second chamber would hold solution of higher concentration. I would perform one experiment, open a valve between the chambers, and wait for the concentration to equilibrate before performing the next experiment.
My advisor suggested this system and said that the equilibration time would likely be on the order of minutes.
I used this equation to calculate equilibration time between two chambers:
$$
t = \frac{\Delta m \cdot  l_{v}}{A_{v} \cdot D \cdot 10^6 \cdot \Delta c}
$$
where $t$ = equilibration time in s
$\Delta m$ = the difference in mass of $\ce{KNO3}$ between the chambers in g
$l_{v}$ = the length of the valve in m
$A_{v}$ = the orifice area of the valve in m^2
$D$ = the mutual diffusion constant of $\ce{KNO3}$ and $\ce{H2O}$, which is on the order of 10^-9 m^2/s
$\Delta c$ = the difference in concentration between the chambers in g/cm^3
the factor of 10^6 is to convert between g/cm^3 and g/m^3
For a cylindrical valve 1 cm long and 2 cm in radius (improbably short and wide) and chamber volumes 95 cm^3 and 400 cm^3, this gives an equilibration time of 4.7 days!
This is much more than I was expecting. I can reduce the sizes of the chambers to decrease the time demand, but I would prefer to keep them as large as I can, to increase the final concentration that can be achieved in the cuvette.
Is there something obvious that I'm missing in this calculation? Is there a better way to do it?
Edit:
The equation above was derived by equating the mass flux $j_{m}$ with its definition from Fick's first law (using $c$ as mass/volume rather than amount/volume):
$$
j_{m}=\frac{\Delta m}{A_{v}\cdot \Delta t}=D\cdot\frac{\Delta c}{\Delta x}
$$
I know that strictly the RHS should contain $\frac{\delta c}{\delta x}$ rather than $\frac{\Delta c}{\Delta x}$. Does this approximation make much difference? For now, I am also making the fairly large assumptions that the volumes of solution in each chamber do not change, and that the density of the solution is independent of concentration. I think I can include these considerations later on.
This is an actual system that I need to build (well, ask our machine shop to build for me) and I need to have a good idea of the best valve sizes to use as they have a significant effect on the overall structure.

I think your estimate time based on diffusion alone is probably quite accurate: diffusion is a fairly slow phenomenon in mixing phenomenons, and mass transfer through diffusion alone is quite slow. This is a main reason for which mechanical stirring is extremely common is chemistry labs.

You could, perhaps, determine the diffusion time experimentally by using a small, highly colored compound (permanganate? or copper (II) sulfate absorbs well), and watching the absorbance change to find the maximum.

What is happening is not just diffusion, there is dispersion and free convection (due to concentration gradient). Your advisor is right in that it should be quick, but the reason is not diffusion. refer to "Diffusion" edition 3 by Ed Cussler, Pg 47.

@picakhu, I have this book and I opened pg 47 which only speaks about free convection in a vertical tube. These are stirred solutions with a narrow tube connecting them. This is quite different from the situation presented here. Her advisor probably thinks she will be stirring the solution. And I don't think I agree with your definition of dispersion.

@Chris, I did not define dispersion in my answer. Just said that it is another factor to think about. This problem is far from trivial. I think convection will be the major component.

@picakhu Dispersion in this case is related to diffusion. Any convection in this case is related to diffusion and I don't see why you think this will be dominant. Unless you're stirring there is no bulk convection to consider in the classical treatment.

In the future please specify where your equation came from, justify its use and specify all of your parameters.
When only a concentration gradient is present, the diffusion coefficient, $D$, is typically described by Fick's first law of diffusion. To determine $D$ in these terms we must know the flux, $j$, and the gradient, $d c / dx $.  $c=$ concentration of our species divided by the volume and $x$ is the distance this occurs over.
Flux, $j$, is the (flow quantity/time)/area or (flow rate)/area. Fluxes depending on potential gradients are molecular, but bulk flow flux is called convective. Here we have molecular flux present and the units of this are kmol/(m$^2 \cdot$ s). (It's true that diffusion creates its own convection but there is no bulk flow so we ignore this diffusion-induced convection.) Properly our flux is the product of a transport property by the gradient of the driving force, $ j = (D) ($Gradient of driving force$)$. Note that $D$ has units of m$^2$/s. For example, $D$ is $10^{-5}$, $10^{-9}$, $10^{-10}$ m$^2$/s for some gases, liquids and solids respectively at STP.
Now let's think back to basic physics. Say we have a feather at $h$ height here on Earth and we drop it. It will experience an acceleration due to gravity, $g$. Remember that gravity has units m/s$^2$. We can estimate the time it takes to travel distance $h$ crudely by dimensional analysis: knowing that $g$ has units of m/s$^2$ and $h$ has units of m then time is simply $\sqrt{\frac{h}{g}}$. In fact the equation that describes this is $h = \frac{1}{2}gt^2$ and so we actually needed to multiply our original answer by $\sqrt{2}$. In the same way we can crudely estimate the time by dimensional analysis on the diffusion coefficient, $D$ with units m$^2$/s and going a distance with units of m. So we need to take the distance$^2$ divided by $D$ to get seconds. Say ( $0.01$m )$^2$ divided by $10^{-9}$ m$^2$/s gives $10^5$ seconds which is about 1.1 days. This could easily be 4 days if  we needed to multiply by 3.7 which is certainly possible if your orifice is small. It is calculating these other things which turns out to be tricky. Although it's obvious the single easiest thing you can do is switch to a gas when considering just diffusion because a gas would take on the order of 10 seconds because $0.01^2/(10^{-5}) = 10$. You can also change the time by introducing another flux, such as by stirring the solution.
A better answer requires differential equations. Are you sure you want to venture down that road?

Hi Chris, thank you for your answer. I have added some extra information in the question above. Switching to a gas is not possible, since this system is intended to study the KNO3 solution specifically. Stirring might be possible, though it raises its own problems as the total space for the solution chambers is quite constrained. I am willing to consider differential equations if it's necessary! :)


text:
Choice of doping elements in semiconductors
In a typical semiconducting material, we start with silicon (group IV) and introduce atoms of group V or group III depending on whether we are constructing an n- or p-type semiconductor respectively.
Why is it optimal to choose doping atoms one group away? Why not introduce atoms from groups VI or II, for instance?

This is a great question, but it would benefit from a chemist. My intuition is that the elements one group away, once one electron is shed, have an orbital structure which is identical to the Si, so they sit in the lattice well, creating minimal defect, allowing their electron to join the conduction band without as much stress in the lattice. If you make a +2/-2 ion the nuclear charge will probably bind an electron in the neighborhood, and the lattice will be distorted more, and perhaps the electrons won't even enter the conduction band, or you'll make a disordered insulator. Maybe, maybe not.

Look at the II-VI options from a chemical perspective.
For silicon, Group III boron p-doping would be replaced by Group II beryllium doping, and Group V phosphorus n-doping would be replaced by Group VI sulfur doping.
The beryllium would tend to act like a Lewis base that donates a pair of electrons to some nearby silicon atom, forming a stronger and more localized bond than is possible with the odd electron count created by boron doping. Similarly, the sulfur would tend to act like a Lewis acid and form a localized sulfide bond with some nearby silicon atom, versus the odd electron count created by phosphorus.
So, the main reason for picking elements immediately "next door" (off by one electron) is to avoid introducing paired electrons, which on average will bond more strongly and more locally via a Lewis base-acid process.
That said, there's enough complexity going on there that it's still possible that you could get II-VI or maybe even I-VII doping to work in the right situations.

Ok, this is a helpful start.  I understand now that, for instance, V/III has ionic properties, and hence probably less desirable than V/IV. But there are many semiconductor materials that are ionic; e.g., IV/VI are in use. Hence there is still a question: why not V/III, given that ionic compounds are in use?

@SimonS, sorry, sort of lost track of this one when it moved. I... don't really understand your question, specifically what you mean by "ionic" in this context. If you mean the Lewis acid-base pair idea, that's not necessarily a strongly ionic bond; it's more like a covalent bond for which both electrons came from the same atom. So I was just saying that if you create essentially a free radical, an incomplete bond with no easy way to get a local partner. But it's clearly not a necessity. Beyond that you'd have to get a lot deeper into the modeling, and I'll bow out for that.

This is a great question, and one that I also have and would like to see a better explanation for the answer, but I can offer what I know (I would comment, but I do not have that usability at the moment).
When you add a dopant to say, Si, you are essentially shortening the distance between the valence and the conduction band by adding either an accessible conduction (empty) band just above the original valence band (in the case of p-type like B) or are making the original conduction band more accessible by adding a higher energy valence band (in the case of n-type like N).
My guessed answer to your question is one of two things:

either the group III/V elements are optimal in shortening this band gap, and that II produces excess conduction too high and VI excess valence too low, or...
the change in valence of the dopant too greatly affects the lattice stabilization of the Si crystal. Essentially by causing too many localized vacancies/defects that the lattice strain negatively affects the overall conductivity of the Si.

Semiconductor elements have half-filled outer-most shells so if we doped group VI or II elements which have 6 and 2 electrons in the outer-most shell and they form covalent bonds with each other just like Si doped to Si and that would result in a no-free-electron insulator because there is no vacancy there .


text:
Why is a metabisulfite / starch / HCl solution unstable?
In one variation of the iodine clock demonstration, you add two colorless solutions together and apparently nothing happens. After about 30 seconds to a minute, you get an "instantaneous" change to a dark blue color.
Solution A is $\pu{0.006 M}$ $\ce{KIO3}$ and is stable over at least a year or two.
Solution B is made with starch, $\pu{0.2 g}$ of sodium metabisulfite ($\ce{Na2S2O5}$), and $\pu{1 mL}$ of $\pu{1 M}$ $\ce{HCL}$ in $\pu{200 mL}$ of water. Starch can be from "reagent grade starch", or starch from a packing peanut, or from laundry starch bought at the grocery store. The concentration of the starch ends up at about 0.05% w/v.
Solution B must be made up no more than 24 hours in advance or the reaction doesn't work: the blue color doesn't form or the timing of the reaction is way off, depending on how old the reagent is.
What happens in solution B that makes it unstable, and is there an easy way to stabilize it so that I could have the chemicals needed for the demo ready on the shelf?
The reactions that happen in the demo are:
\begin{align}
\ce{S2O5^2- + H2O &-> 2 HSO3-} & \tag{1}\\
\ce{IO3- + 3 HSO3- &-> 3 SO4^2- + I- + 3 H+} &\text{(slow)} \tag{2}\\
\ce{IO3- + 8 I- + 6 H+ &-> 3 I3- + 3 H2O} &\text{(slow)} \tag{3}\\
\ce{I3- + HSO3- + H2O &-> 3 I- + SO4^2- + 3 H+} &\text{(fast)} \tag{4}\\
\ce{2 I3- + starch &-> blue complex + I-} &\text{(fast)} \tag{5}
\end{align}
I don't see anything in these that tell me why the metabisulfite / $\ce{HCl}$ / starch reagent is unstable.

I can partially answer the question. I know what is happening to solution B. I don't know of a more stable version.
One potential problem is the acid-catalyzed hydrolysis of the starch into glucose, which does
not form the blue complex.
$$\ce{nH2O + (C6H10O5)}_n \ce{->nC6H12O6}$$
For the timing to be off, the metabisulfite must also be decomposing, either by reducing glucose or by reacting with dissolved oxygen. I'm not sure what the specific process is, nor do I know how to stop it.

Will starch hydrolyze that easily? The HCl concentration is only 0.0005 M. Thanks for the answer, this gives me something to explore.

Ah, I only the saw the 1 M. I missed that it was 1 mL of 1 M HCl diluted to 200 mL. Then I expect that the main problem is the metabisulfite decomposing.


text:
Why does meltwater taste different?
I've noticed that water obtained from melting ice cubes (by keeping them in the open) tastes different from water cooled to the same temperature.
Furthermore, the taste goes away if you keep it in the open for a sufficient duration of time.
I can think of these possible explanations:

Dissolved gases being released
Something to do with the crystal structure of ice
Something to do with dissolved minerals

The Internet is giving me all sorts of varying answers.
Exactly what chemical differences exist between meltwater and cold water?

I might guess the dissolved gases (and specifically the chlorine/chloramine) would be ejected faster from melted ice (and slowly from standing water), but I don't know if that's the taste you're looking for.

@NickT: I'm not looking for the taste, I'm looking for all the physical/chemical differences. The taste is just a point proving the existence of those differences. You can make that into an answer, though I'd prefer something more than a "guess" ;-)

I read it had to do with dissolved gasses, that "normal tasting" water has the dissolved gas but freezing water you lose the gasses.  So I tried this home experiment.  I froze water then let melt and tasted it, it tasted bad with the off flavor.  Then I took out a large freezer bag and only put a small amount of the foul tasting water.  I then shook it violently for some time to mix air back into the water.  Boom the water tasted fine again.  No double blind here but seriously is it needed?

Interesting experiment. Did the normal taste return because new gasses where introduced into the water or did some gases (oxygen?) escape from the water?

Oxygen solubility is 10 mg/L for water, and 14.5 mg/L for ice. Even after the ice melts to room temperature, the oxygen level in the melted ice water remains higher than the water that kept a constant room temperature(~12 mg/L vs 10 mg/L). The extra oxygen present is most likely the cause for the difference in taste.

Complete nonsense. The oxygen solubility in water rises towards 0°C (14.6mg/l, 9mg/l at 20°C), and drops to practically zero upon freezing.


text:
What is antisymmetric exchange? What is J-strain? Where does it come from?
I'm reading a paper1 by Sanakis, et al. that characterises the magnetic coupling in the $\ce{Fe3S4}$ clusters present in bacterial ferredoxin II and beef heart aconitase as arising through something called 'antisymmetric exchange'. This looks like it may be a result of the apparent magnetic frustration of the system (three high-spin $\ce{Fe^{3+}}$ ions, all in essentially identical environments, all antiferromagnetically coupled with respect to their neighbours). The paper says that the clusters have a S=1/2 ground state, which is incompatible with a broken-symmetry model.
The wikipedia article on antisymmetric exchange (all four sentences of it) is not particularly edifying. I'm hoping an expert can explain what antisymmetric exchange is and what causes it as I've not heard of it be before. The paper also notes that the exchange situation in the cluster gives rise to three slightly different $J_{ij}$ values (again, despite the symmetry of the ions) and terms this 'J-strain'. This makes me doubt that, for instance, 'antisymmetric' is being used as a synonym for 'antiferromagnetic'.
[Edit]
Here's an example of the general connectivity of this kind of $\ce{Fe3S4}$ cluster (based on D. Gigas FdII (PDB entry 1F2G)). The cluster comprises a hemicubane. $\ce{S_{Cys}}$ denotes a cysteinyl sulfur.

(1) Sanakis, Y.; Macedo, A.L.; Moura, I.; Moura, J.J.G.; Papaefthymiou, V.; Münck, E.; Evidence for Antisymmetric Exchange in Cuboidal $\ce{[3Fe-4S]^+}$ Clusters; J. Am. Chem. Soc.; 2000, 122, pp. 11855-11863

Maybe OP has already contacted the authors of this paper by now, but this was interesting. I'm no expert but I can Google things and this was too long for a comment.
Antisymmetric exchange:
At first I thought it was simply an exchange interaction where the wave function's sign is changed during exchange, now I don't think it's so simple.
Antisymmetric exchange is also known as DM-interaction (for Dzyaloshinskii-Moriya). I don't know exactly what it is, here is the original paper citation - can't find it anywhere though.
I.E. Dzyaloshinskii, Thermodynamic theory of weak ferromagnetism in antiferromagnetic substances, Sov. Phys. JETP 5(6), 1259-1272 (1957)
A definition
A slideshow, which contains some more info (slides 16-23)
J-Strain: Reading this, "J-strain" isn't the only "strain" that's referenced in this paper. Also referenced within is "G-Strain" which has a more elucidating Wikipedia article.
So, maybe it's safe to say, like G-Strain, J-strain is the broadening of the "distributed exchange coupling constants" (a.k.a. J-values). Inferred from a line in the paper: "Both proteins exhibit distributed J-values (J-strain)".
What did J-strain have to do with the experiment? "...The observed J-strain, which presumably represents a distribution of protein conformations “locked in” upon freezing the samples."
Where does the broadening of the J-values (exchange coupling constants) a.k.a. J-strain come from? Well apparently there's this thing called J-coupling that seems relevant. J-coupling is also known as indirect dipole-dipole coupling, and gives info about molecule interconnectivity. J-coupling relates nuclei to local electrons (All this is from the J-coupling article). The broadening of the lines may come from anisotropic interactions, which affect nuclear spin energy levels. Sure enough, the paper says it can't rule out anisotropy. An Alternative source? Non-homogeneous regions in the sample, which could go back to the frozen/locked conformations.
I still think it'd be great if someone doing physics research in this field could elucidate some more though.
Another J-coupling Webpage (U. Wisc)
Of course I should cite the original paper, "Evidence for Antisymmetric Exchange in Cuboidal [3Fe−4S]+ Clusters". Linked in question.

This is about the spin Hamiltonian of the system, and more like naming an empirical observation then giving an actual physical explanation.
So in the most simple (regular) spin systems a Hamiltonian with H=JS1S2 where J is a scalar handles pretty well the behaviour os spin. Note, in this case there is no orientation preference for the coupled spin compared to any internal axis, e.g. ligand orientation or such, only tells if S=S1+S2 or S=S1-S2 is more stable (i.e. ferro or antiferromagnetic material).
Sometimes there is a preference of orientation of S1, S2.. compared to local molecular axis, eg. S1 prefers to align with the local z axis. You can models this with the zero-field splitting, i.e. you have a DS1^2 .. type of Hamiltonian in addition.
Now asymetric exchange is a contribution when the exchange interaction has a directional preference, i.e. JS1S2 with scalar J is not enough as it would be fully anisotropic. While in your system there is a magnetic frustration, the reason for asymmetric exchange is relativistic effects similar to zero-field splitting. There are low energy excited states that mix into the ground state due to strong spin-orbit coupling. Since iron(III) doesn't have such states, I guess it is coming from the extra electron of iron(II). But it is just an uneducated guess. I think it is nothing to do with the spin-strain (that exist with simple JSS coupling, too), it is just the strain makes it easier to observe.
Note, that the spin-Hamiltonian model is just an empirical parametrisation of the behaviour of the spin (e.g. description of the EPR or in-field Mossbauer spectra). Frankly speaking, it is very rare to see someone making out any physical / chemical information out of it. I would read more of Munck/Bominaar group's later papers, as they often put some theoretical model (DFT and ligand field) behind their measurements.


text:
How can I update a map/mesh and coordinates in a PyMol session file?
I have a PyMol session file I was using to create my figures.  It contains a protein and ligand broken into several groups of atoms with a fair bit of overlap.  There are also a few meshes built from a density map for a few figures.
Unfortunately, I wasn't quite done with refinement, and now Refmac has given me shiny new PDB and MTZ files with subtly different coordinates and terms.  Is there a way to "update" my coordinates (despite the myriad groups, all the atoms still have the same name) with the new PDB?  With all the groups I made that are shown/hidden/colored differently for several scenes to produce all the figures, I'd rather not recreate it all manually.
I'm more pessimistic about the map/mesh, but I created them via a script so I can repeat that.  Should I also script the coordinate to figure stuff in the future if it's not finalized?

As a generale rule, I'd script as much as possible. If you don't do it, it will come back eventually and bite you in the... somewhere. I also think that this is more of a question for the official support channels provided by PyMol.

Is your atomic structure coordinates from xray or NMR? GROMACS has a universal file format that can be read into many including pymol and VMD. I haven't used pymol but VMD at a basic level. Wouldn't SO a better place to post this q?

@bonCodigo x-ray. While there's some coding/scripting involved, this is more about usage of PyMol rather than a coding challenge, so it would probably get closed on SO. I'm a bit confused behind the (very high?) vote-to-view ratio (nearly 1 vote per 16 views), so it's not like there are a ton of people on the interwebgoogles that have this problem, but maybe a few really want an answer? Probably should be made into an issue for PyMol...

@NickT wow! X-ray crystallography! A lot of time, coders/scientists/students like to scoop a solution from someone else's resolutions. Due to time, due to the rush etc. So it looks like a lot in need for a solution - waiting for you to post it to PyMOL. Heard their response time isn't too bad. Just a speculation about votes: some who viewed possibly do not have a SE account, so can't vote and lazy to create one... And each time you view your own question with even 1 hr lap - I noticed the view count does update...(unique IP hmm..). 33 votes means to me, people are in need.

NickT and @bonCodigo, such questions are normally considered off topic on Chemistry SE because it's not about Chemistry but the software or coding itself, see the tour page. However a new Matter Modeling SE has recently launched where it is the focus to answer such computational chemistry software questions.

Until someone identifies an ‘update’ function in PyMol, I think the next best thing is to use scripting. (See the pymol wiki) It is an imperfect solution, but it may work for the situation presented in the original post if the session may be reproduced.
To begin capturing the commands in PyMol, including menu selections to a script file, select:
File -> Log File -> Open -> myscript.pml

When done with creating the display panels, select:
File -> Log File -> Close

The input data files and the script itself may then be updated or replaced. In a fresh PyMol session, execute the script with:
File -> Run Script -> myscript.pml

To test the above, I generated a PyMol session where I captured a script as above. I loaded the atomic coordinates of a small protein, a ligand and a 2mFobs-DFcalc electron density map. Then I displayed some panels along with mesh surface around the compound. The co-structure was then re-refined, thus generating modified atomic coordinates and electron density maps. I replaced the original files with the updates and executed the script in a fresh PyMol session. The display panels were updated accordingly.
I recommend the Advanced Scripting Workshop.


text:
How should I go about picking a functional for DFT calculations?
I've been looking to use time-dependent density functional theory, but reading the literature I keep getting overwhelmed by how many different functionals there are.  How should I go about selecting which functional to use?

@Dan - What properties are you most interested it?

@Richard Terrett: I suppose the "time-dependent" part is the most important; I've heard many functionals lose a lot of accuracy when calculating the response of the molecule to some time-dependent forcing.

@Dan, it really boils down to the literature when it comes to DFT.  I would start by looking papers which benchmark DFT with ab initio methods.  DFT should never be used without some sort of verification from another method or experiment.  You could try finding Christopher Cramer's Computational Chemistry book which has a nice table outlining the various functionals on standard datasets and go from there.

A useful review of many functionals and how they came to be is "Perspective: Fifty years of density-functional theory in chemical physics" by Axel Becke, J. Chem. Phys. 140, 18A301 (2014).

A newer question partially overlaps with this one: https://chemistry.stackexchange.com/a/27418/5017

For Structures and Energies: Answer Hazy, Try Again Later
People can talk on this subject for literally days.
It mostly comes down to computational expense vs accuracy - if you have a larger system, you can't afford the fancier functionals. The absolute highest accuracy state of the art functional-wise would probably be those using the Random-Phase Approximation (e.g.), and also the use of more complex range separation techniques (such as the Coulomb-Attenuated Method mentioned below).
Having said that, a lot of people just use B3LYP for everything. B3LYP is parametrised for low-Z main-group chemistry, and so does reasonably well there. It's a hybrid functional (i.e. it includes the exact Hartree-Fock form exchange operator), so it's not the cheapest. I tend to use the series of PBE, PBE0, TPSS - their only parameters are physical constants, so they should never give surprisingly bad results.
A couple of basic guidelines, though (which should be taken as "as far as I know"s):

If you have a delocalised system, you want exact exchange (so, a hybrid functional like PBE0, B3LYP, M06-2X, etc). Without this, most functionals will over-delocalise the orbitals.
If you have a system in which dispersion forces are important, worry. Then use something like $\omega$B97X-D which includes empirical dispersion corrections. Some packages can also apply Grimme's empirical dispersion corrections to any functional.
If you have a periodic system, you probably want to use screened exchange rather than full exact exchange - something like HSE (which tails off the exact exchange past a certain range) rather than PBE0 should be faster while not significantly less accurate.

For TDDFT: Probably CAM-B3LYP
Fortunately, this is a bit easier.
Firstly, you want something that incorporates exact exchange (a hybrid functional), which quickly rules out many functionals. Pure GGAs and LDAs give qualitatively wrong answers for states involving charge transfer. Secondly, functionals that include long-range exchange corrections, such as CAM-B3LYP, also give better results.
Also, look out for triplet instabilities, which can be a problem with the introduction of HF exchange.
(David Tozer is my go-to author for TDDFT testing, though I would warn that despite the above answer, I've never used TDDFT in "production".)
(Now, bring on the functional flamewar.)

Be very careful when characterizing systems with delocalization and dispersion.  When doing so, I highly recommend checking the literature before supporting the data various DFT methods will give.  Sadly, even using a slew of functionals won't help you determine a proper answer without reverting to ab initio methods.  Here is a great paper illustrating how badly DFT is when describing hydrocarbon/water interactions.

And maybe this paper is of any help. One should try to climb from bottom to top. I usually start with BP86 (very robust), always try to skip B3LYP (it is a pain quouting and a very lazy functional), continue via PBE0 and unfortunately end up at the Minnesota functionals. (BTW, I'd be delighted to enjoy a good flamewar :D)

This is a loaded question.
Some groups focus on precisely cooking a recipe from known ingredients to produce useful functionals, the Minnesota functionals being quite popular.
In general, you will have to benchmark available functionals against your property and known analogues to your target of choice and ensure that you are capturing the relevant physics. This requires that you familiarize yourself with the respective assumptions in functional design, e.g., the homogeneous electron gas approximation.

Hi Deathbreath: there's tonnes of DFT questions in Materials Modeling Stack Exchange I wonder if you can commit so that you can be one of the beta users?


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Bonding in diatomic C2, a carbon-carbon quadruple bond?
Carbon is well known to form single, double, and triple $\ce{C-C}$ bonds in compounds. There is a recent report (2012) that carbon forms a quadruple bond in diatomic carbon, $\ce{C2}$. The excerpt below is taken from that report. The fourth bond seems pretty odd to me.

$\ce{C2}$ and its isoelectronic molecules $\ce{CN+}$, BN and $\ce{CB-}$ (each having eight valence electrons) are bound by a quadruple bond. The bonding comprises not only one σ- and two π-bonds, but also one weak ‘inverted’ bond, which can be characterized by the interaction of electrons in two outwardly pointing sp hybrid orbitals.

According to Shaik, the existence of the fourth bond in $\ce{C2}$ suggests that it is not really diradical...
  If $\ce{C2}$ were a diradical it would immediately form higher clusters. I think the fact that you can isolate $\ce{C2}$ tells you it has a barrier, small as it may be, to prevent that.

Molecular orbital theory for dicarbon, on the other hand, predicts a C-C double bond in $\ce{C2}$ with 2 pairs of electrons in $\pi$ bonding orbitals and a bond order of two. "The bond dissociation energies (BDE) of $\ce{B2, C2}$, and $\ce{N2}$ show increasing BDE consistent with single, double, and triple bonds." (Ref) So this model of the $\ce{C2}$ molecule seems quite reasonable.
My questions, since this is most definitely not my area of expertise:

Is dicarbon found naturally in any quantity and how stable is it? Is it easy to make in the lab? (The Wikipedia article reports it in stellar atmospheres, electric arcs, etc.)
Is there good evidence for the presence of a quadruple bond in $\ce{C2}$ that wouldn't be equally well explained by double bonding?

You may be interested in this blog post by Rzepa on the $\ce{CN+}$ cation, which putatively contains a $\ce{CN}$ quadruple bond and is isoelectronic with $\ce{C2}$

@Richard Terrett Thanks for the reference...it's one I hadn't found. So, the quadruple bond is plausible from a calculation stand point (if I'm reading that right). Is there experimental evidence that could/would support one view or the other? As I said, I'm "a bit" out of my field here.

There is an example that C might have quadruple bonds with U

@JaniceDelMar There is no evidence, and there never will be.  The C2 molecule looks like any other homodiatomic:  two fluffy balls of electron density pushed together.  Where are the four ropes in that picture?

It would not necessarily form higher clusters, because maybe 2 C-C -> C-C-C-C  is an endothermic reaction.  The product, too, is a diradical!  It's a non-explanation.

I have not yet been able to read it, but here is another very recent paper by the same authors about the topic: The Nature of the Fourth Bond in the Ground State of C2: The Quadruple Bond Conundrum

Okay, this is not so much of an answer as it is a summary of my own progress on this topic after giving it some thought. I don't think it's a settled debate in the community yet, so I don't feel so much ashamed about it :)
A few of the things worthy of note are:

The bond energy found by the authors for this fourth bond is $\pu{13.2 kcal/mol}$, i.e. about $\pu{55 kJ/mol}$. This is very weak for a covalent bond. You can compare it to other values here, or to the energies of the first three bonds in triple-bonded carbon, which are respectively $348, 266$, and $\pu{225 kJ/mol}$. This fourth bond is actually even weaker than the strongest of hydrogen bonds ($\ce{F\bond{...}H–F}$, at $\pu{160 kJ/mol}$). Another point of view on this article could thus be: “valence bond necessarily predicts a quadruple bond, and it was now precisely calculated and found to be quite weak.”

The findings of this article are consistent with earlier calculations using other quantum chemistry methods (e.g. the DFT calculations in ref. 48 of the Nature Chemistry paper) which have found a bond order between 3 and 4 for molecular dicarbon.

However, the existence of this quadruple bonds is somewhat at odds with the cohesive energy of gas-phase dicarbon, which according to Wikipedia is $\pu{6.32 eV}$, i.e. $\pu{609 kJ/mol}$. This latter value is much more in line with typical double bonds, reported at an average of $\pu{614 kJ/mol}$. This is still a bit of a misery to me…

The real issue is that no one has ever taken a picture (i.e. electron density) of genuine, unambigious, cases of a single, double, triple, quadruple??? bonds. And they never will, because these concepts are not based on quantum mechanics.
Two atoms reside next to each other, and if they have a favorable electrostatic interaction, then a certain type of topology arises in their electron density.  (q.v. Quantum Theory of Atoms in Molecules)
You might as well say that every "bond" is a single-bond, or, equivalently, and infinity-bond.
These types of articles are bogus, as they can not be confirmed experimentally.  They got lucky with the reviewers, and/or an editor who knows their readership is just dying to hear news of a quadruple bond, having heard for so many years that triple is the highest you can go.
I mean, what are people looking for? Four "ropes" that link between the two carbon atoms?  Where is the unambiguous, unbiased dividing line between bond energies of a single/double/triple/quadruple bond?

Electron density can be observed. 2) Quadruple bonds are pretty obvious in metals complexes 3) Even hextuple bonds are theoretically possible in certain molecules...

Yes, that's why I cited electron density as an example of an observable which might be used to confirm. 2) citation needed 3) example needed. Hexatuple bonds?

https://chemistry.stackexchange.com/questions/714/are-there-any-molecules-with-delta-bonds-in-their-ground-states https://chemistry.stackexchange.com/questions/27532/what-would-follow-in-the-series-sigma-pi-and-delta-bonds

A recent publication by Chen & Manz (2019) on the bond order of diatomic molecules did touch on this controversy in their discussion. In their study, they have employed a novel method of analysis, which they called Bond Order Component Analysis (BOCA), to quantum mechanically calculate the bond orders of 288 diatomic molecules and ions. They discussed the diatomics $\ce {C2}$, $\ce {Mo2}$ and $\ce {O2}$ in greater detail. The following edited text taken from the publication focuses on their discussion on the diatomic bonding in $\ce {C2}$:

Fig. 2A displays the natural spin orbitals, their occupancies, and their bond order components for the carbon dimer, $\ce {C2}$. (Here, we adopted the convention of labeling the first valence orbital (rather than the core orbital) as $ 1\sigma_g$.) The $ 1\sigma_g$, $ 1\sigma_u$, and $ 2\sigma_g$ orbital shapes show strong $\mathrm{s–p}$ mixing. Four natural spin orbitals had significantly positive bond order components. The $ 1\pi_{u,x}$, and $ 1\pi_{u,y}$ components ($0.786$) were larger than the $ 1\sigma_g$ component ($0.667$) which was larger than the $ 1\sigma_u$ component $(0.407)$. The $ 2\sigma_g$ orbital had negligible BO contribution $(0.045)$. These bond order components sum to $2.727$. Our findings are in agreement with a carbon dimer bond order of $2–3$ caused by four bonding components,(i.e., two smaller $\sigma$-bonding components and two larger $\pi$-bonding components) reported by several research groups as discussed by Hermann and Frenking.

So yes, there seems to be indeed, four bonding components contributing to the overall bond order of the dimer but each component contributes to less than a bond. Hence, the overall bond order is much less than the value of four.
Reference
Taoyi Chen, Thomas A. Manz, "Bond orders of the diatomic molecules," RSC Adv. 2019, 9, 17072-17092 (DOI: 10.1039/C9RA00974D).

Carbon 100% could form a quadruple bond in C2. Each and every non-noble gas atom is trying to get an outer shell of 8. Diatomic elements are elements that can bond with themselves (the three most common being Oxygen, Nitrogen, and Hydrogen). An acronym to remember the Diatomic elements is H O Br F I N Cl. As you can see, "C" isn't one of them. But, these elements can only bond with themselves because  if they do, they get a full outer shell. 2 Nitrogens share 3 electrons with each other, and have 2 left over on each one. 3+3=6+2=8. Carbon has 4 valence electrons and needs to gain 4. It is light enough to be able to form a quadruple bond, and if it does, it will share all four of its electrons with the other carbon! So each carbon will have a total of 8 valence electrons.


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The Chemistry behind creating Polylactic Acid (PLA)
I've seen a few videos of persons creating plastics from starch by adding an acid, glycerol, and water, however no explanation of the chemistry.
Does anyone know what $\ce{(C6H10O5)_{n} + C3H8O3 + C2H4O2 + H2O -> (C3H4O2)_{n} + (what)}$?
or does it?
I only have a basic understanding of organic chemistry, so sorry if the equations don't make sense.

Polylactic acid polymers are creating a lot of interest these days because they're biodegradable. On the downside, corn starch is often one of the starting materials...and corn is also used as a food source, so there's some controversy over the wisdom of this approach.
I think I've found a site that takes the approach you're basing your question on...but there are a couple of issues with the equation you've written. So, I'll answer this question in two parts. Plastic from starch is what you saw in the video; PLA from starch is what you asked about.
Plastic from starch. The recipe for the plastics-from-starch uses starch, glycerol, acetic acid (the $\ce{C2H4O2}$ in your equation) and water as a solvent. (They do not add lactic acid, the $\ce{C3H8O3}$ in your equation, but lactic acid can be produced by hydrolysis of the starch.) With the starch-acetic acid-glycerol recipe, you do not actually make a PLA polymer, but a starch polymer that is somewhat like it instead. (Ref)
So, starch is a polymer that looks like this:

In water, it tends to ball up into globular clusters due to hydrogen bonding. By adding acetic acid and heating the starch in water, you "denature" the starch and turn it into "disordered chains" that stretch out. When you let these disordered chains dry, they become entangled and form a flexible film, the plastic. In this recipe, the glycerol is included as a  plasticizer, so it is unchanged chemically in the reaction. The glycerol also helps to keep the starch stretched out by stabilizing it with hydrogen bonding. Plasticizers (as the name would indicate) are additives that increase the plasticity of a plastic and make it more durable or flexible. Changing the amount of glycerol in the recipe can have a big effect on the strength of the plastic film you make. This is the process you're seeing in the videos, I think.
Making PLA from starch. This is going to be a bit of a challenge to explain depending on your understanding of organic chemistry, but hopefully the pictures will help. When you make PLA commercially, you digest starch, usually with a bacteria, to turn it into lactic acid (see picture below for the structure of lactic acid). This lactic acid must be purified to make a high quality product. If you just combine a bunch of lactic acid molecules to make PLA, one water molecule would be lost each time they combine...sort of like this. (This is an acid and alcohol combining, not two acids, but the idea is the same.) The water in the polymerization reaction leads to a poor quality plastic.

So, in commercial synthetic schemes, two purified lactic acid molecules are reacted to form lactide...they come together in a controlled condensation reaction, with the loss of water. (Ref for picture.)

Now you can use the lactide, i.e. "lactic acids with the water removed", to do the polymerization to PLA. You add an initiator to open the lactic acid by breaking the bond between the $\ce{O-C}$ in  $\ce{O-C=O}$ and this gives you 2 lactic acid molecules hooked together...which react with another lactide molecule to add two more lactic acid molecules to the growing chain, and so on. And that's how the PLA forms.
So the reaction sequence is strach-> lactic acid-> lactide + $\ce{H2O}$ -> PLA
Probably more than you wanted to know!

No, actually it's exactly what I wanted to know. How to make PLA, as well as what kind of plastic those start to plastic videos are actually making. It's answers like this that will quickly take chemistry out of beta :)

To all: Here's a nice little pdf about actual PLA, including time and temperature: Synthesis and Characterization of Poly (Lactic Acid) for
Use in Biomedical Field

The link to springerlink.com is broken. I am also not able to find any copy saved on the Wayback Machine.


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What is the behavior of a Zwitterion during ion exchange chromatography?
We're interested in purifying a protein that has an azide moiety using ion exchange chromatography. Due to its unique structure, the moiety is a Zwitterion with a novel chemical behavior. My question is how will the molecule behave when I run it through an ion exchange column? Will it be considered positively charged, negatively charged, or neutral or is it much more complicated?

I would expect it to depend on the pH of the mobile phase. The charge of your stationary phase and whether you want the azide to elute will determine what charge you want on the azide. Given that, you should set the pH of the mobile phase either acidic or basic enough to ensure that the azide is fully charged in the direction that you want.
From Wikipedia:

Proteins have numerous functional groups that can have both positive and negative charges. Ion exchange chromatography separates proteins according to their net charge, which is dependent on the composition of the mobile phase. By adjusting the pH or the ionic concentration of the mobile phase, various protein molecules can be separated. For example, if a protein has a net positive charge at pH 7, then it will bind to a column of  negatively-charged beads, whereas a negatively charged protein would not. By changing the pH so that the net charge on the protein is negative, it too will be eluted.

If changing the pH is problematic, apparently increasing the ionic strength can help shield the azide's charge from the column's charge:

Elution by changing the ionic strength of the mobile phase is a more subtle effect - it works as ions from the mobile phase will interact with the immobilized ions in preference over those on the stationary phase. This "shields" the stationary phase from the protein, (and vice versa) and allows the protein to elute.


text:
Why is nitric acid such a strong oxidizing agent?
In my teaching-lab experiments I've seen that nitric acid solutions are able to facilitate the dissolution of metals such as silver, even though they are more active than hydrogen. Does anyone know why nitric acid is so special? Why is it such a strong oxidizing agent?

Its not that special- concentrated sulfuric acid and all the oxyacids of chlorine are oxidising particularly perchloric HCL04.

Because, unlike in other metal dissolution reactions, the $\ce{H+}$ of $\ce{HNO3}$ isn't reduced—the $\ce{NO3-}$ is. The following data and balanced reactions are taken from Wikipedia:
\begin{align}
  \ce{NO3- + 4 H+ + 3 e- &-> NO  + 2 H2O} &
    E^\circ_\mathrm{red} &= \pu{0.96 V} \\
  \ce{NO3- + 2 H+ +   e- &-> NO2 +   H2O} &
    E^\circ_\mathrm{red} &= \pu{0.79 V} \\
  \ce{Ag+(aq) + e-       &-> Ag(s)} &
    E^\circ_\mathrm{red} &= \pu{0.799 V}
\end{align}
Since the standard reduction potential (SRP) of the $\ce{NO2}$ reaction looks smaller than that of $\ce{Ag+}$ (I may be wrong, but by significant digits it can't be greater than $0.799$), one can conclude that it's the $\ce{NO}$ reaction that's occurring here. And the $\ce{NO}$ reaction has a large enough SRP to oxidise $\ce{Ag+}$.
Usually nitrogen compounds are pretty versatile when it comes to redox reactions, since nitrogen shows many oxidation states. So the simple reason for why $\ce{HNO3}$ is so strong an oxidising agent (with respect to other acids) is that it has a different, better path available to it to get reduced.
Note that the exact reduction path (i.e. final reduction products/oxidation state) depends upon the concentration of nitric acid—so much that copper can be oxidised in three different ways.

what about platinum? the E reduction of platinum is greater than NO3, yet NO3 still oxidized Pt. also the same case for Ag

ManishEarth did a good job explaining already. Perhaps I could point out one more you could think of to understand why nitric acid has such a high reduction potential. The nitrogen centre in nitric acid is in +5 oxidation state, which is also the highest known oxidation state of nitrogen. As a result, the acid exists in high energy state to begin with as it is really unstable.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What is a good method to measure the redox potential of a cellular system?
I know how to measure the concentration of free thiols, which is reflective of the redox-potential of cellular compartment such as the cytoplasm or a lysosomes.

What other methods exist?

I am only writing this as a comment because I don't know whether it is doable... but maybe you can use an electrochemical approach (e.g. amperometry or cyclic voltammetry). You could give an oxydative stress to the system and measure the decay of the signal of oxydable species over time.

@nico Amperometry/(cyclic) voltammetry would be inconvenient, since he'd have to recreate a macroscopic copy of the cell interior. Maybe this or that?

@CHM: I read "cellular system" as a group of cell rather than the interior of a single one.

There are many complexes of d-elements, that can reversibly lose/gain one electron, changing their color in process. They are known as Redox Indicators. Some of them are stable enough to be used in vivo.

Electrochemistry;  using microelectrodes.  So, basically, place one microelectrode inside the cell, and another (the reference electrode) just outside the cell, and then use a potentiostat to measure the potential between them. Reference.