text:
Why are the majority of cleaning solutions basic?
When testing for pH for common household cleaning products, such as dishwasher fluid, I noticed that majority of them tend to be basic? Why is that? Do acid make good cleaning solutions as well?

As far as I know many household cleaning products contain anionic surfactants which are supposed to be mild on skin. I don't have any reference, though.

Dishwashing agents are usually anionic detergents—i.e. they are alkylbenzenesulfonates.
They consist of two parts:

Sulfonate part which is hydrophilic(water loving i.e. dissolves in water)
Alkylbenzene part which is lipophilic(fat loving i.e dissolves in fats and oils)

Thus, when in water, they dissociate to give $\ce{RSO3-}$ and $\ce{Na+}$ .  $\ce{Na+}$ is from a strong base (making it a weak conjugate acid), while $\ce{RSO3-}$ is from a weak acid (making it a weak conjugate base), thus the resulting solution is basic.
The sulphonate ion bonds to water, while alkylbenzene bonds to oil, protein, or fat . Thus fats, oils, proteins, and grease dissolve in water
The following is the mechnism of a dishwashing agent:

These are the ingreadients mentioned on a dishwashing bottle in my house:

As you can see, sodium dodecylbenzenesulphonate (an alkylbenzenesulfonate) is one of the main ingredients.

Do acid make good cleaning solutions as well?

Yes they can make a good cleaning solution, but not as a dishwashing agent.
According to Wikipedia ,

Acidic washing agents are mainly used for removal of inorganic deposits like scaling. The active ingredients are normally strong mineral acids and chelants. Often, there are added surfactants and corrosion inhibitors.

is that one correct: Sulfonate + Alkylbenzene = surfactant?

 Basic dishwashing agents?

Some dishwashing agents contain substantial amounts of NaOH, see below.
Sometimes NaClO solutions are used as disinfectant for dishwashing.
$\require{mhchem}$
As $\ce{Cl- + ClO- + 2 H+ <=>> Cl2 ^ + H2O}$, they are kept basic (and $\ce{Cl-}$free, but $\ce{3 ClO- -> 2 Cl- + ClO3-}$ and $\ce{2 ClO^- -> 2 Cl^- + O2 ^}$) for stabilization.

@Ashu asked whether I'm really sure, so here's an example: with 15 - 30 % KOH + 15 - 30 % NaOCl, pH 14 "Liquid intensive cleaner for dishwashing machines" (technical data, saftety sheet both German, though)
Soaps (chemical meaning) are salts of fatty acids. Na- and K-soaps are good for cleaning and as salts of weak acids and strong bases they are basic.

 Dishwashing liquid chemistry?
If we approximate the food rests you want to clean away during the dishwashing as mixture of carbohydrate, proteins and lipids, then your dishwashing liquid needs that take care of those 3 substance classes in water.

carbohydrates are hydrophilic - no problem here.
amphiphils are needed to clean lipophilic substances like lipids, oils, greases as Ashu explained already. They can take care of quite a bit of protein as well.
basic solutions usually cause a faster hydrolysis of the amide bonds in proteins than acids. Usually quite a bit faster: NaOH-solutions immediately give a "soapy" feeling while e.g. HCl doesn't.
As far as I know here (Germany), basic liquids for household dishwashing machines are uncommon/seldom/not used at all (I'm not so sure about the tablets). The actually used solutions are less corrosive (for both the machine and your fingers if you more or less accidentally decide to use it for dishwashing by hand).
This page from a New Zealand poison centre talks about the "dishwasher [...] powder or tablets [...] are often highly alkaline"
For dishwashing liquids where you are thought/supposed to put your hands in, the pH is usually neutral to slightly acidic, so you need to scrub a bit more, but don't dissolve your skin.

 Acids for cleaning?
Acids (usually acetic acid or citric acid, maybe HCl) are used in "household chemistry" to clean limescale (CaCO$_3$). But that usually isn't the problem of your dirty dishes.
Dishwashing machines use ion exchangers to soften the water (or phosphate in the dishwashing agent), so you don't get limescale or lime soaps inside the machine.

Some dishwashing agents contain substantial amounts of NaOH .. are you sure?

@Ashu: yes. I updated the answer with some technical and safety data for a gastronomic product (Germany) and a New Zealand Poison Central speaking about household dishwaser tablets and powder being "highly alkaline". If I remember (unlikely), I'll look at the ingredients of such tabs the next time I'm in the supermarket.


text:
Why are there no edge dislocations in a face-centered cubic (FCC) lattice?
In crystallography, we know that the Burgers vector is parallel to the dislocation direction in a screw dislocation, and perpendicular in a edge dislocation.
Furthermore, I have in my lecture notes "FCC cannot have edge dislocations" but there is no explanation of why, aside from "in FCC, the Burger's vector belongs to the family of the dislocation direction vectors"
I don't understand what this means.

I have to stretch my memory to remember how this goes (and read a bit from here).
First, how do you define burgers vector? You take the end point of the dislocation and make a circle around it as it were a perfect lattice, the extra (or missing step) from the full loop is the burgers vector.
Now, is the burgers vector also a lattice vector? If yes, then it is (edge)dislocation. If no, then it is only a partial (edge)dislocation (partial dislocations have their own name).
Now, how does this work on a fcc-lattice case. On quick read for link I pasted, it seems that fcc-lattice cannot have edge dislocations, because the burgers vector is not a lattice vector.
You may want to go this through by yourself and check that I didn't make any mistakes.

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text:
How are $\sigma$ and $\sigma^\pm$ determined in Hammett plots?
The Hammett plot is commonly invoked in organic chemistry to reason about the plausibility (or implausibility) of various reaction mechanisms. The vertical axis is essentially the logarithm of an equilibrium constant (or rate constant) measured relative to a hydrogen functional group, which I understand. However, I am at a complete loss for understanding what is being plotted on the horizontal axis, which conventionally are denoted $\sigma$ and $\sigma^\pm$. What are these parameters and how are their values determined for various functional groups? In practice it seems that people look them up in tables, but where do the values in these tables come from?

Hammett wanted to find a way to quantify the effects electron-withdrawing (EWG) and - donating (EDG) groups have on the transition state or intermediate during the course of a reaction. Initially he took the $pK_{\mathrm{a}}$-values of benzoic acids carrying the respective functional group in para or meta position (ortho acids and aliphatic acids weren't used because steric effects would overlay with the electronic effects) as a guide and plotted them against the logarithm of the reaction rates.
So, he based his scale on the following reaction reaction:

Later Hammett decided not to use the $pK_{\mathrm{a}}$s themselves for his correlation but defined a new parameter, which he called $\sigma$.
This $\sigma$ shows how electron-donating or -withdrawing a group is relative to $\ce{H}$ as a ratio of
the $\log K_a$s or the difference of the $pK_{\mathrm{a}}$
between the substituted benzoate and benzoic acid itself.
If the acid required to determine $\sigma$ for a new substituent was not available, $\sigma$ could be determined by correlation with other reactions.
A different value of $\sigma$ for any given substituent was needed for the meta and para positions (sometimes called $\sigma_{\mathrm{m}}$ and $\sigma_{\mathrm{p}}$, respectively).
The equation for $\sigma$ is
\begin{equation}
\sigma_{\ce{X}} = \log \left( \frac{ K_{\mathrm{a}}(\ce{X-C6H4COOH}) }{ K_{\mathrm{a}}(\ce{H-C6H4COOH}) } \right) = pK_{\mathrm{a}}(\ce{X-C6H4COOH}) - pK_{\mathrm{a}}(\ce{H-C6H4COOH})
\end{equation}
or in short notation
\begin{equation}
\sigma_{\ce{X}} = \log \left( \frac{ K_{a} (\ce{X}) }{ K_{a} (\ce{H}) } \right) = pK_{\mathrm{a}}(\ce{X}) - pK_{\mathrm{a}}(\ce{H})
\end{equation}
where $\ce{X}$ is the functional group, i.e. the EWG or EDG, whose effect on the reaction rate shall be evaluated.
It is also possible to determine the $\sigma$ values via the logarithm of the ratio of the rate constants for the aforementioned reaction with substituent $\ce{X}$ (in short notation: $k_{\ce{X}}$) and with $\ce{X}=\ce{H}$ (in short notation: $k_{\ce{H}}$)
\begin{equation}
\sigma_{\ce{X}} = \log \left( \frac{ k_{\ce{X}} }{ k_{\ce{H}} } \right)
\end{equation}
Update
AcidFlask's comment reminded me that I forgot to say something about $\sigma^{+}$ and $\sigma^{-}$ values.
As mentioned above the "normal" $\sigma$ values are based upon the difference of the $pK_{\mathrm{a}}$ between a substituted benzoate and benzoic acid itself.
But for benzoates one cannot draw resonance structures that delocalize the negative charge onto the benzene ring via the $\pi$ electron system.
Yet, many reations of interest create negative or positive charges that can be stabilized by delocalization via resonance with the substituent.
For these reactions, one finds that Hammett plots using $\sigma$ values have considerable scatter.
Therefore, two new substituent effect scales were produced, one for groups that stabilize negative charges via resonance ($\sigma^{-}$), and one for groups that stabilized positive charges via resonance ($\sigma^{+}$).
The $\sigma^{-}$ scale is based upon the ionization of para-substituted phenols (I've found different accounts for this. According to the German Wikipedia the $\sigma^{-}$ scale is based upon the ionization of para-substituted anilins.),

for which groups like nitro can stabilize the negative charge via resonance.
The $\sigma^{+}$ scale is based upon the heterolysis ($\mathrm{S}_{\mathrm{N}}1$) reaction of para-substituted cumyl chlorides (phenyldimethyl chloromethanes),

for which groups like amino can stabilize the positive charge via resonance.
The $\sigma^{\pm}$ values can then be evaluated by the aforementioned equation
\begin{equation}
\sigma_{\ce{X}}^{\pm} = \log \left( \frac{ K_{a}^{\pm} (\ce{X}) }{ K_{a}^{\pm} (\ce{H}) } \right) = \log \left( \frac{ k_{\ce{X}}^{\pm} }{ k_{\ce{H}}^{\pm} } \right)
\end{equation}
but with the rate constants $k_{\ce{X}}^{\pm}$ and $k_{\ce{H}}^{\pm}$ (or the equilibrium constants $K_{a}^{\pm} (\ce{X})$ and $K_{a}^{\pm} (\ce{H})$) measured for their respective defining reactions.

Do the $\sigma$ values depend on the specific reaction or the specific starting material?

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text:
Has QTPIE ever been applied to 1D metals such as polyacetylene?
QTPIE is a fluctuating charge model that substantially improves modeling of polarization and charge transfer. I was curious whether it has ever been applied to 1D metals such as polyacetylene to provide new perspectives on topics such as Peierls' condensation and phase solitons?

Not answering your question, but related: all I could find is  this e-print discussing the validity of QTPIE to describe the acetylene molecule.

How flattering that someone asked this question ;)
The short answer is yes, I've tried QTPIE on series of conjugated molecules like the oligoacetylenes and oligoacenes. The results, however, are fairly dismal, as can be seen in the preprint. The problem is that in these empirical charge models, the polarizability grows faster than the size of the molecule, which means that these models predict infinite polarizabilities and charge transport behavior for bulk materials, which is obviously nonsense.
The problem is quite intimately related to that of self-interaction error in approximate density functional theories, which also exhibit similar problems.

For para. 2, see ref. 44 of the above preprint. For SIE in DFT, see ref. 11 ibid.

AcidFlask, thanks. Flattery, maybe, but the interest is genuine: I did a deep dive on 1D metal many years ago, and came away mostly unimpressed. At least two thirds of the papers were pretty much publish-now crud with no real experimental implications, and precious little insight on how to model either the chains or the solitons more accurately. I never could tell for sure if anyone had any meaningful lab evidence of the nominal fractional charges, which were usually hidden by pairing. The long-ago "almost quark finding" of 1/3 charges on levitating lumps may have been related... possibly.

After some review: A genuinely interesting paper, both in its frank discussion of the fluctuation problems and in your strategy of distinguishing bare and effective electronegativities. I will be curious to see where you take such ideas as you move forward with your research.


text:
How is silica bead bonded into polystyrene?
What kind of bonds silica bead forms with polystyrene? Silica is now as ~100 $\mu$m spherical particles in bulk polystyrene that is made by melting.
How can I bond the silica particles

more tightly
more loosely

to the polystyrene?
Can this be generalized to any nano/micro particle and plastics? I am thinking now metal oxides and other thermoplastics.

The main chemical groups on the surface of silica beads are silanol groups –SiOH. The main control one over their surface chemistry is thus the functionalization of these silanol groups.

First, You should define what You mean with "silica beads"! Some kinds of (the dozens of materials called silica) have maybe silanol groups, but most important Polystyrene will not form bonds to them. Why do You think there are bonds?

Polymers do hydrogen bonding (van der Waals, etc.) with  each other, right? Then silanol groups should bond with these? Lets assume that these particles are basic hardware stuff that is used for sand blasting (vwr part number 201-0481 [EN]GLASS BEADS SILIBEADS S D0,04-0,07MM). I would assume that they are very inert. And the question is, how can I activate or deactivate (functionalize/defunctionalize) these bonds to react with polystyrene?

You can functionalize silica to add polymer "brushes" to the surface. You can also buy pre-made functionalized silica, although the downside to that is it isn't always easy to get the manufacturer to tell you exactly what is on the surface.
This isn't really going to increase the attractive interactions between the silica and the polystyrene, however. Polystyrene has a bulky aromatic ring hanging off of each monomer, and no sites that can participate in hydrogen bonding. When you add hydrophobic brushes to the surface of the silica, what you are doing is limiting the ability of the silica to aggregate due to hydrophilic interactions with itself. As a result, given enough kinetic energy (temperature for melt casting, or sonication for solvent casting) you can get the particles to disperse. Then, if you solidify the polymer quickly enough, they will be kinetically trapped in a dispersed state.
Depending on what your goal is, this could be enough for your application. However, if you want to get the particles to strongly attract to the polystyrene, you aren't going to be able to get much farther than van der Waals interactions. That's a limitation of the polystyrene, not the silica.
Other particles (gold, or carbon nanotubes, for example) will be more hydrophobic and easier to disperse without functionalizing, but you still are not going to get very strong attractive forces between the particles and the polymer.


text:
Predict XPS Spectrum of Organic Monolayers on a Semiconductor
X-ray photoelectron spectroscopy (XPS) is a particularly useful technique for examining the surfaces of all sorts of materials. Organic monolayers are particularly difficult to characterize completely via this technique due to the fact that many carbon peaks do not differ much by chemical shift. Traditionally quantitative information is obtained via a lot of guess-work and forced peak fitting. It would be better to predict peaks shapes and peak positions beforehand to aid in the interpretation and prevent a terribly mis-guided interpretation. How well can theory help the non-specialist with this? I wouldn't have the time to develop a program to do this either. Preferably one could point me to the right direction in terms of a research paper/software.

Some density functional theory packages, such as Quantum Espresso, can calculate XPS core level shifts using the pseudopotential formalism. See Phys. Rev. B 74, 045430 (2006).


text:
How to compute molecular formula?
I have the following question to solve:

Tungsten, $\ce{W}$, and chlorine, $\ce{Cl}$, form a series of compounds with the following compositions:
  \begin{array}{rr}
\text{Mass % W}  & \text{Mass % Cl}\\
\hline
    72.17     &  27.83\\
    56.45     &  43.55\\
    50.91     &  49.09\\
    46.36     &  53.64\\
\end{array}
  If a molecule of each compound contains only one tungsten atom, what are the formulas for the four compounds?

My answer is as follows:
For one gram of tungsten, chlorine has mass
$$
\begin{align}
\frac{27.83}{72.17} &= 0.3302~\mathrm{g}\\[3pt]
\frac{43.55}{56.45} &= 0.7715~\mathrm{g}\\[3pt]
\frac{49.09}{50.91} &= 0.9643~\mathrm{g}\\[3pt]
\frac{53.64}{46.36} &= 1.157~\mathrm{g}.\\
\end{align}
$$
Since $\frac{0.7715}{0.3302}\approx\frac{7}{3}$, $\frac{0.9643}{0.3302}\approx3$, and $\frac{1.157}{0.3302}\approx\frac{7}{2}$, the number of atoms of chlorine for a given mass of tungsten are respectively in the ratio $6:14:18:21$.
So if a molecule of each compound contains only one tungsten atom, the formulas are $\ce{WCl_{6}}$, $\ce{WCl_{14}}$, $\ce{WCl_{18}}$, and $\ce{WCl_{21}}$.
Is this correct?

No. Look up "empirical formula" in a chemistry book or Google it.

This question served as a bad example for the homework policy. (It might be quite a good fit nowadays though.)

We don't have a homework policy yet, but I'm quite sure that it will be against providing full answers.

Hints:

You're supposed to take into account molar masses. The ratios in a formula are molar ratios, so you need to use those, not mass ratios.
Read up a bit on emperical formulae, as Janice suggested.

This answer is preventing the question from being deleted no?

Just my opinion, the question should be kept since the homework policy uses it as an example of a bad question. That's especially true since the other two bad examples there have already been deleted.

It seems that we may have become more lax on homework questions because if OP posted this question today, I think this would be considered pretty good.


text:
Is iron in the brown ring compound in a +1 oxidation state?
In the standard brown ring test for the nitrate ion, the brown ring complex is:
$$\ce{[Fe(H2O)5(NO)]^{2+}}$$
In this compound, the nitrosyl ligand is positively charged, and iron is in a $+1$ oxidation state.
Now, iron has stable oxidation states +2 and +3. Nitrosyl, as a ligand, comes in many flavours, of which a negatively charged nitrosyl is one.
I see no reason why the iron doesn't spontaneously oxidise to +3 and reduce the $\ce{NO}$ to −1 to gain stability. But I don't know how to analyse this situation anyway. I think that there may be some nifty backbonding increasing the stability, but I'm not sure.
So, why is iron in +1 here when we can have a seemingly stable situation with iron in +3?

Textbooks say it is FeII and NO . FeII is rather obvious, because it is used in big excess in the reaction. And NO- is unknown to textbooks.

It's a matter of assignment. I guess it comes from stablility argument, that NO$^+$ is isoelectron to N$_2$, which is a stable triple bond ion. On the other hand, Fe$^{2+}$ or Fe$^{3+}$ will produce spin polarization around NO molecule, which may be detected from EPR spectroscopy(?)

Your basic assumption is incorrect: the iron in $\ce{[Fe(H2O)5NO]^{2+}}$ is Fe(III), and the ligand is $\ce{NO-}$.

According Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of
Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction
Revisited
The correct structure is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$
For many years it was thought that iron was reduced to $\ce{Fe^{I}}$ and $\ce{NO}$ oxidized to $\ce{NO+}$, based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin $\ce{Fe^{III}}$ ($S=5/2$) antiferromagnetically couples with $\ce{NO-}$ ($S=1$) for an observed spin of $S=3/2$.

This is a very popular question in twelfth grade in India. National level exams have asked this question with the correct answer there being that Fe is in +1 oxidation state. Feels weird to learn the wrong fact.

Lots of wrong facts get learned.  The whole point of science is to dig out the wrong facts on our way to the truth.

Oxidation State of "Fe" in brown ring complex depends on the binding mode of NO ligand to Iron (whether is it "bent" or "linear"). Unless we specify the IR frequencies of NO ligand for different modes, we can not say the oxidation state of "Fe". Depending on the IR stretching  frequencies of NO ligand which is bound to the Iron, it can be NO+, NO- or simply NO.

This doesn’t really answer the question.

In the brown ring complex, $\ce{[Fe(H2O)5(NO)]^{2+}}$ five water molecules are present, they are harder to stabilise the $+3$ oxidation state of $\ce{Fe}$ and hence here $\ce{NO}$ present as $-1$.
If we consider that $\ce{NO}$ is $+1$, then $\ce{Fe}$ is $+1$, which is very unstable in the co-ligand region like $\ce{H2O}$.
i.e, in brown ring complex oxidation state of Iron is $+3$.

Also, since another answer also gives the oxidation states, can you be specific about what you want to add to the discussion? Thanks!

Here iron becomes +1 due to presence of the ligand $\ce{NO}$ +1. The compound $\ce{[Fe(H2O)6]^{2+}}$ is formed when $\ce{FeSO4}$ dissolves in water during the brown ring test.
Now, when we add an aqueous solution of nitrate ions in $\ce{FeSO4}$ solution $\ce{Fe^{(II)}}$ are converted to $\ce{Fe^{(III)}}$ and $\ce{NO3-}$ are converted to $\ce{NO}$.
 These $\ce{NO}$ being an odd electron compound can use  three of its electrons among which it donates two electrons to $\ce{Fe^{(II)}}$ while one it shares with incompletely filled $\ce{d}$ orbital and displace one water ligand bonded with secondary valency to $\ce{Fe}$.
This way $\ce{Fe^{(II)}}$ is converted to $\ce{Fe^{(I)}}$ and not $\ce{Fe^{(III)}}$ In these regard we get the colour brown instead of yellow for $\ce{Fe^{(III)}}$ or green for $\ce{Fe^{(II)}}$

Are you sure? Wikipedia says it's Fe+3, see above answer. Also, I find it rather hard to follow your arguments, please be a little clearer :)

@ManishEarth The wikipedia page looks absurd. It says it has Fe in +3 and NO+ which actually dont add up to the +2 charge of the complex, but to +4.

@ManishEarth Looks like the wikipedia page has been edited from its original by a single user which strikes controversy. And then changed again.

The $\ce{Fe(NO)2}$ fragment has strongly delocalized electrons, it is considered as covalently bound. The semiempirical calculations show that the electron density is polarized in a manner that allows to ascribe the delta+ charge to $\ce{NO}$ groups. Therefore formally Fe is considered +1. Remember this formal assignment is somewhat simplified...
If interested, read e.g.
http://pubs.rsc.org/en/content/articlehtml/2011/dt/c0dt01244k
and references.

Welcome to Chemistry.SE! Could you summarize the contents of that article you linked. It is behind a paywall for some folks.

Since the reference is about dinitrosyl iron complexes it doesn't seem relevant to the single nitrosyl complex in the question.

The other answers are all wrong. Draw Molecular orbital diagram of $\ce{NO}$ and you will see.
That electron in antibonding orbital, it is so easy to lose, hence it easily goes to $\ce{NO^+}$ and $\ce{Fe^{(II)}}$ to $\ce{Fe^{(I)}}$.
Remember if you read something extraordinary which claims something out of daily routine (especially in chem) then it must be true.
Link : http://en.wikipedia.org/wiki/Nitrate_test
Wikipedia also says Iron is in $+1$

http://pubs.acs.org/doi/pdf/10.1021/ic010628q

The oxidation state of Fe is +1 and not +3 as most answers have stated
NO+ is a strong ligand from crystal field theory(not so sure but please let me know if it is wrong ) hence Fe is in +1 Oxidation state
And as far as@Davephd answer is considered even in +1 state the spin of the compound is 3/2 as NO+ is a strong ligand
Also NO is a paramagnetic substance .so the unpaired electron it can give up to Fe to reduce it from Fe +2 to +1 and No becomes NO+
Fe+ in presence of NO+ the electonic configuration changes to 3d(2,2,1,1,1)(i dont know how to edit it so please forgive me) and so in the complex sp3d2 hybridisation takes place
And so the spin still remains 3/2 here

$\ce{NO+}$ is a strong field ligand, but not a strongly binding ligand per se. It cannot bind well to cationic metals due to its cationic nature.

@Jan https://en.wikipedia.org/wiki/Metal_nitrosyl_complex

and what is that supposed to tell me? All it does is reinforce my arguments as far as I skimmed it …

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text:
Is the Springborg 6D phase space model used in modern molecular orbital modeling?
In various papers written in the 1980s, Michael Springborg explored an interpretation of the Wigner phase space function as an electron density in a six-dimensional $(q,p)$ phase space. He applied it with some success to several simple compounds.
Is the Springborg 6D phase model used in modern molecular orbital modeling? If not, are there specific weaknesses that make it unsuitable for current computational approaches?

ADDENDUM 2012-05-14:
In a world dominated by Hilbert spaces, the orbitals of chemistry are interesting because they are one of the few quantum models that retain the 3D spatial image of the system. (My best recollection is that Schrödinger liked such low-dimensionality approaches since they produced comprehensible waves.)
Alas, the problem is that you need both position $q$ and momentum $p$ to get an accurate picture of the overall electron state. Using a space-only 3D $q$ representation forces the analysis of molecules to flip back and forth at somewhat arbitrary thresholds between localized (position space $q$) and delocalized (momentum space $p$) views of the electrons in the molecules or metal crystals in question.
That bothers me, both from physics and computational integrity perspectives.
The problem is that despite the striking dynamic differences of the $p$ and $q$ spaces, Pauli exclusion works the same way in both. There is nothing intuitive about why having a half-unit of spin should induce the same exclusion behavior in both representations. One can say, "that's just the way it works," but accurate modeling of orbitals of diverse sizes likely requires accurate capture of this persistent feature.
If that's true, this older work by Springborg (he still heads a research group, BTW) strikes me as a possible opportunity for taming some models, making them smoother and saner over a much broader range of molecular sizes. That's because Springborg's 6D electron density functions -- if they work OK in broader contexts -- might help eliminate any need for abrupt switches between "mostly position" or $q$ 3D views (classic "electron clouds," whatever those mean) and "mostly momentum" or $p$ 3D views.
A good place to test whether 6D $(q,p)$ electron distributions might provide higher computational stability would be modeling a range of lengths for a long-chain carbon polymer such as polyacetylene. (I always pick polyacetylene for its simplicity, but there are other good choices.) Springborg appears not to have explored that domain since his papers addressed mostly simpler bonds, as best I can tell. (I think; I don't have all of them.) The instabilities of approximate density functional theories may (or may not) be related to $(q,p)$ transition instabilities. That's not a question I can or should try to address, as folks out there know that domain orders of magnitude better than I ever will.

ADDENDUM 2012-05-16
I mentioned "algorithmic opportunity" and wondered what I meant since, as @AcidFlask reminds us, the $q$ and $p$ basis sets are fully interchangeable in representing any stationary wavefunction.
The issue, I think, is one of accumulating uncertainty. Picture doing some calculation in the $q$ representation. The estimate should give low error rates for a highly localized electron, but higher ones for a delocalized electron in, say, a conductive polymer
Next, Fourier-transform the result into $p$, and do the next iteration of whatever you are trying to calculate. The nature of the errors should change, but in the case of a delocalized electron, it's likely to work out a bit more accurately.
You reverse-transform that and repeat to converge to a solution that looks pretty stable in both spaces. That, I would argue, is the "algorithmic opportunity" part of the 6D argument: By incorporating both views, you can come to an overall solution that diverges less from reality when applied to a broad range of molecular orbital sizes.
Are there problems with that analysis? Oh my yes. You'd have to be nuts to do complete Fourier transforms twice per full iteration of your convergence, and even then, the devil is always in the details about whether it even works.
But the operative word was "opportunity." Such an analysis hints that a dual perspective may lead to stabler approximation methods. The new algorithm would necessarily "live" closer to a continuous 6D representation with constraints than a per-iteration transform back and forth. Can that be done? I don't know; as I said, folks like AcidFlux know the details orders of magnitude better than I ever will.
And finally, for any non-mathematical readers (AcidFlux, shoo, go away now!) who may have wandered in, the section below provides a simple, easy-to-imagine visual analogy to explain why location and momentum both important play a significant role at the level of molecules.
Imagine a very flexible membrane with many embedded holes (but the holes don't stretch). Label one side of the membrane "ordinary space" (abbreviated $q$, don't ask why), and the other side "momentum space" (labeled $p$, and I told you not to ask). Next, put water balloons with fixed quantities of water through each hole, glued to the holes.
The $q$ side of each balloon now represents the "real space" representation of the water balloon. If you squeeze that part very tightly, it shrinks until it looks more like a particle than a balloon. You think you have "captured" that balloon n a small space, but beware! On the other side, the $p$ side of the membrane, the balloon has gotten comparatively huge! So your point-like "capture" of the balloon on the $q$ side was more illusion than reality since in the broader view that includes both sides, all you did was shift a lot of the balloon from the $q$ world into the hidden $p$ side.
Note also that you don't need to see both sides at once because each side of the balloon fully determines the size of the other side. From the amount of water on one side, you can always calculate precisely how much water is in the balloon on the other, hidden side. And the opposite is also true! If you know the size of the balloon on the $p$ side, you also automatically know its size on the $q$ side.
That kind of equivalence between two views that appear quite different at first glance is an example of a "basis set." A basis set has enough "pieces" in its tool kit to allow you to represent something completely, in the sense that you have all of the information needed to predict how the system evolves.
Alas, complete is not always the same as convenient. For example, if you try to stack the balloons together as they appear on the $q$ side, you quickly discover that the hidden $p$ side of the balloons can make the stacking difficult. The balloons won't pack together as readily as their forms on the $q$ side suggest. The problem, of course, is that you have those vast $p$ sides of the balloons interacting in a very different way over in the $p$ or "momentum space" side of the membrane.
It's not that any information is missing! After all, since the $q$ side provides a complete basis set for describing the balloons, you know that when a balloon is small, it must necessarily be large in $p$. The trick, then, is to make sure that you don't have to recalculate the size of the balloons on the other side every time you try to do something with them on the $q$ side. That's inefficient at best and could quickly turn into a significant source of errors.
So, it's simpler and a lot less calculation to keep simultaneous, redundant images of both sides of the membrane and change them only when necessary, e.g., after squeezing some of the balloons on either (or both) the $q$ and $p$ sides. That is inherently redundant since you can always get away with changing and maintaining just one side.
However, if you've ever done much algorithm design, you know that "caching" mostly persistent results can be a great way to reduce calculation times, sometimes hugely. More subtly, it can also help stabilize the problem if it turns out that you have complementary error modes across your different "views" (basis sets) of the final result.
So, that's what I'm suggesting may also be valid for the physics of electrons. Like those water balloons, electrons, as in Planck's "Theory of Slosh" (and yes, of course, I just now made that up!), like to spend some of their time point-like in one space (called $q$ or position space; ordinary, real space), or in another more abstract space (called $p$ or momentum space). Interestingly, just as with the balloons, they bump into and push each other in both of these two spaces. Particles that do that are called fermions.
So why are fermions so turf-protective in both of these quite different spaces? I don't think anyone knows, at least not with the kind of depth that is as satisfying as when chemistry started coming together and making deep sense. It's just one of those little mysteries of how the universe works, one (there are several, actually) that's still sitting there thumbing its nose at us and saying, "haha, gotcha!"
And that's cool! After all, what's a universe without a few mysteries still hanging around?

Alas, the problem of course is that you need both position q and momentum p to get an accurate picture of the overall electron state.

This is not true in quantum mechanics; it is sufficient to characterize the wavefunction $\left\langle x |\psi\right\rangle$ in position space or $\left\langle k|\psi\right\rangle$ in momentum space. Which is chosen is entirely a matter of convenience: most molecules have the most compact description in position space, while most crystals have compact descriptions in momentum space. It can be proven that position alone, or momentum alone, form a complete basis separately. In fact, position-momentum uncertainly says that an electron state cannot be characterized completely in both position and momentum simultaneously. (This, by the way, has nothing to do with fermionic symmetries and Pauli exclusion, which is another quantum-mechanical axiom unto itself. The same would have held for looking at bosonic particles also.)
It is, however, true that a probability density in phase space $(q,p)$ is needed to completely characterize a classical system. Wigner's function was introduced to address this quantum-classical correspondence and is regarded as the quantum analogue of a classical probability distribution in phase space. However, it is also well known that it can take on negative values and therefore cannot be interpreted in the usual way as a probability distribution. This affects all methods using the Wigner function: in the Springborg paper in the OP, there are regions of negative "probability density" in the lithium hydride molecule being analyzed.
Suffice to say, the question of how to deal with negative probabilities is an extremely controversial subject.
Having said that, the Wigner function has found uses in ab initio molecular dynamics when a classical phase space density is required to sample dynamical trajectories from the potential energy surface generated by quantum electronic states. The regions of negative probability are usually neglected in such procedures with seemingly no loss of accuracy (to my knowledge). I am not aware of other common usages in electronic structure theory. (The other problem brought up in the OP about the inaccuracies in DFT are conventionally attributed to incorrect modeling of electron correlation effects, which is a matter of incorrect treatments of fermion symmetries and its consequences when projected down into coordinate or momentum space.)

Thanks! A very readable and nicely pointed answer. And yes, I do know that either $q$ or $p$ provides a complete basis set, bridged by the Fourier transform. I was thinking of how ratty each gets in certain domains, and that molecules span both extremes. But me implying both are needed is just wrong. Hmm. I had no idea negative probabilities were involved; that's an interesting surprise.

@AcidFlux: What I said was badly phrased, but my intent was this: Because Pauli exclusion applies in both $q$ and $p$, you need all six dimensions to model accurately how sets of fermions will reach an energy minimum. I really did mean it only for fermions, for just that reason. I am essentially viewing this a a topological algorithm in a 6D space, one in which multiple fermions interact until they reach a good minimum. The resulting forms will of course be fully equivalent whether represented in $q$ or $p$, but I am postulating (right or wrong) that 6D has interesting algorithmic potential.

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


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What does the charge distribution around a Stone-Wales defect look like?
I'm reading a book on carbon nanotubes (P.J.F. Harris, Carbon Nanotube Science, to be specific) at the moment and puzzling about Stone-Wales or 5775 defects. This is where two adjacent carbons undergo a 90-degree rotation about their mutual centroid and generate a defect consisting of two five membered rings and two seven membered rings, as I've illustrated below (This is only supposed to be qualitative, and is on a finite PAH fragment rather than a nanotube, but you should get the idea):

I note that tropylium ($\ce{[C7H7]^+}$) is aromatic as a cation, and the ever famous cyclopentadienide ($\ce{[C5H5]^-}$) is aromatic as an anion, so what I'm wondering is: do the 5 and 7-membered rings in a Stone-Wales defect pick up either formal or partial negative and positive charges to maintain aromaticity in a fashion consistent with the rest of the structure?
P.S.
Big props to F'x, who attempted electronic structure calculations on the PAHs depicted, regrettably without SCF convergence.

I've just noticed your use of equilibrium arrows - is this really an equilibrium?

FWIW, here are some Hückel calculations on the 5775 PAH.
I'm showing the first 7 energy levels. Each molecular graph presents the contribution of each atom to the electron density ($\mathrm{Abs}[c_i]^2$). This is not normalized. I do not use wavefunctions, only the coefficients multiplying them, assuming
$$\Psi_{\mathrm{mol}} = \sum_i^n c_i\psi_i$$
In this approximation (HMO), the first 7 energy levels will be occupied (14 pi electrons).
I will first give you pictures of each energy level, and then a linear combination of each. Take it for what it is, as there is much information lost in the process.
To me, this simply suggests that indeed, the system will behave as if composed of tropylium and cyclopentadienyl ions. But this is not publication quality calculation either =)

Here are compared the linear combination of the first 7 energy levels of the 5775 and 6666 systems. It is clear that electron distribution is not the same in both PAHs, which hints that my calculations are not so far fetched.
5775

6666


text:
Why does gum get harder to chew after a while?
My Sister asked me this:

"Why is it that when we chew gum,  it is soft and mushy to begin with , but slowly gets firmer and firmer like after 20 mins of chewing?"

I think  it is because when we initially chew gum it has plenty  of sugar causing rapid salivation and hence mushiness and the rubber polymer is very flexible to begin with and slowly looses its elasticity hence becomes firmer.
Is this the reason?
What is the chemistry involved ?

@CHM what are plasticizers used for BTW

Also something I've noticed is that when I have a pack of gum sitting in my pocket for days at a time it becomes really soft and mushy, almost to the point where you can't chew it, but then after chewing it for a while it seems to stiffen up. I was curious as to if this is due to the moisture, heat, or something different

I would be surprised if there was reticulation of the gum's polymers upon contact with your saliva. According to this manufacturer, plasticizers are used, and the plasticizers appear to be fats. This would mean that, upon chewing, the plasticizers would be removed from the gum, rendering it unpleasantly harder.
Plasticizers are used to render a polymer more pliable, more elastic, than it naturally is. This could get its own question, imo.


text:
Reaction between silver nitrate and aluminum chloride
I googled a bit about what $\ce{AgNO3 + AlCl3}$ will yield and found out the following:
$\ce{3AgNO3 + AlCl3 -> Al(NO3)3 + 3AgCl}$
Why is that? I know that $\ce{Ag}$ is higher up in the reactivity series than $\ce{Al}$ but that does not make sense to me in this problem. Is it because $\ce{Cl}$ is just one when $\ce{NO3}$ is multiple?

Le Châtelier's principle only states that a system previously at equilibrium will want to stay at equilibrium - that is, if we perturb it, it will try to go back to equilibrium. This system is initially not at equilibrium, and therefore we don't need his principle.
I think the reaction proceeds because the formation of AgCl(s) is enthalpically favored. Molecular cohesion (enthalpy) is greater in AgCl than in AgNO3, thus promoting its formation. How do I know that? Well, it seems AgCl is less ionic than AgNO3 since it is insoluble in water.
This reaction is a type of metathesis reaction. It can be called double displacement reaction or salt metathesis reaction, depending on who you're talking with.

EDIT
The other answers make an error, I think. Both suggest that the solubility product of AgCl is of importance in determining why the reaction takes place. My opinion is that the argument, as presented, is circular, in the following manner:
The solubility product of AgCl in water is very low, therefore we observe formation of a precipitate.
vs
AgCl is insoluble in water, therefore its solubility product is low.
It's easy to see that the fact AgCl is insoluble in water implies that it has a low Ksp in water. If someone can predict and quantify the solubility product of AgCl in water from first principles, then that's another story, but the explanations offered only rely on observation (experiment) to quantify the Ksp.
My argument goes down the ladder. If you can answer why AgCl is insoluble in water, then you can definitely predict that Ag+ and Cl- ions from different sources (i.e. in different solutions) put into contact will spontaneously form a precipitate.
Now you maybe thinking this argument is also circular, but it's not.
We have access to experiment, which clearly informs us that AgCl is insoluble in water. We can confidently assume (but we might be wrong!) that the AgCl bond is less ionic than the AgNO3 bond, since it practically does not dissociate/has a low Ksp in water. A bond with a greater covalent character is more stable, since the electrons are shared and not transfered.
Perhaps the crystal structure of AgCl makes it especially stable to solubilization from H2O. I just think it's obvious that the enthalpy of formation of AgCl is favorable, since it has to counter an unfavorable entropy loss in the formation of an ordered crystal lattice.

As you can see, I cannot provide you with a thorough answer, it would be extremely time consuming (for me) and I'd have to conduct many many calculations. I really hope someone can clarify the why. Maybe you can do your own research from the answers provided here and come back with a more "elaborate" answer.

I talked to my teacher today and she said it exactly as @CHM did. AgCl(s) is favored due to it is has a more covalent fashion than NO3

Ag is higher up in the Reactivity serie than Al but that do not make sense to me in this problem

indeed this is no redox problem, oxidation states do not change.

instead it is a phase transition equilibrium
$$\ce{Ag+ + Cl- <=> AgCl v}$$

As you marked this question as homework:

Edit the question and put the oxidation numbers to confirm that it is not a redox reaction.

Look up the solubility constant of AgCl. Together with the total initial concentration of $\ce{Ag+}$ and $\ce{Cl-}$, you can then calculate the equilibrium concentrations.

You'll see that the precipitation moves the reaction towards equilibrium (unless the concentrations in the aqeous solution are not extremely low).

Here are two experiments that you can do to check that it is really a dynamic equilibrium (though much on the side of the precipitate)
$$\ce{Ag+_(aq) + Cl(aq)- <=>> AgCl_(s)}$$

If you disturb the equilibrium by removing either $\ce{Ag+}$ or $\ce{Cl-}$, the AgCl will dissolve again.

Here are some ideas:

Complex (mask) the $\ce{Ag+}$, e.g. by $\ce{NH3}$ or $\ce{CN-}$have you ever done this? If not, you should really try. It is really fast.
reduce the $\ce{Ag+}$
oxidize the Cl-, so that $\ce{Cl2 ^}$ is removed out of the system.

(you could also remove the AgCl, but that isn't practical, as you'd never be sure whether you observe formation of more precipitate or whether you didn't remove all of the AgCl)

Crystal growth in saturated solutions. You may have to wait quite a while for this one, though...

F'x pointed us to a paper Paradigms and Paradoxes: The Solubility of AgCl in Water. As it is behind a paywall, I'll try to summarize a few points out of it.
First of all, it's actually kind of an overdue homework question:

It is well established in the pedagogical literature that AgCl is insoluble in water while NaCl and
KCl are soluble [...] What is usually left unsaid, however, is why AgCl is so much less soluble than NaCl or KCl.

Here are some interesting points for the question:

Liebman concludes from thermodynamic data (and contrary to HSAB expectations) that $\ce{Ag+}$ is better solvated than $\ce{Na+}$ or $\ce{K+}$

Lattice energy for AgCl ($\ce{Ag_{(g)}^+ + Cl_{(g)}^- -> AgCl_{(s)}}$) is more negative than for NaCl and KCl.

The paper estimates that the enthalpy of $\ce{AgCl_{(g)} -> AgCl_{(s)}}$ is comparable to that for the alkali chlorides.

Liebman concludes from that, that the AgCl crystal is nothing special but is rather similar to NaCl and KCl

So the point that is different is that the formation of diatomic AgCl is much more favorable than formation of diatomic NaCl or KCl.

My personal conclusions: It's a really tough question. I'll stick to the experiment with measurable properties such as equilibrium constants. ;-)
That is, insolubility of AgCl is no surprise to me: after all, I'm used to the fact for years. And given the insolubility of AgBr and AgI, I'd guesstimate unsoluble even though AgF is soluble (after all $\ce{F-}$ is usually different from the other 3). But I find the particular situation far away from any textbook kind of reasoning that should introduce general chemical concepts to students. Unless the message is supposed to be that reality is complicated... (which OTOH allows for the beauty of reality and makes it interesting).

@cbeleites:  The second part of your answer would work as an answer here as well, you may want to try answering: http://chemistry.stackexchange.com/questions/410/why-is-ceagcl-less-soluble-than-ceagno3

To analyse such situations, you first must realize that nearly everything is present as ions. For example, if you took the reaction $\ce{NaCl +KBr->NaBr +KCl}$, the "direction" it goes in has no real meaning since $\ce{Na+}$, $\ce{Cl-}$, $\ce{K+}$, $\ce{Br-}$ are present as ions. There is no actual reaction going on here, just a mixing of ions.
In this case, we have something slightly different. $\ce{AgCl}$ precipitates here:
$$\ce{3AgNO3_{(aq)} + AlCl3_{(aq)} -> Al(NO3)3_{(aq)} + 3AgCl v}$$
Similar to the $\ce{NaCl}$ case above, $\ce{NO3-}$ and $\ce{Al^{3+}}$ are present as ions on both sides--so nothing to worry about there. We can safely remove them from the equation. This reduces the reaction to:
$$\ce{3Ag+ + 3Cl- -> 3AgCl v}$$
Now, we have reduced the question to "why does $\ce{AgCl}$ precipitate?". The easiest way to decide is by looking at thermodynamic stability--solid silver chloride is more stable. We can experimentally prove this, or use standard enthalpies.
Reactions proceed in the direction of more stability, so the above reaction goes forward.

Of course, I sort of "assumed" that $\ce{AgCl}$ precipitates in the beginning--this was so that I wouldn't have to complicate stuff. Nevertheless, we can, in a similar way, experimentally prove that the others are found in ionized form and generally do not precipitate. There are some nice arguments why $\ce{AgCl}$ is less soluble than $\ce{AgNO3}$  in this post.


text:
How to understand this form of writing the solution: (some salt • n H₂O)?
For example — "high purity chloride salt of Zinc $(\ce{ZnCl2.2H2O})$" or "various concentrations of $\ce{FeCl2.4H2O}$". What does the number before $\ce{H2O}$ mean?

The salt's crystal lattice's repeating unit is constituted of n molecules of salt and m molecules of water. Such salts are called hydrates. Wikipedia even has a nice picture of a hydrated vs non hydrated salt.
You can have anhydrous ferrous chloride, as well as ferrous chloride tetrahydrate. The number indicates how many water molecules are present in the crystal lattice's unit cell.
EDIT
This has just crossed my mind: be careful when preparing solutions of metal complexes to make sure you know what compound you're working with. One gram of anhydrous salt contains more equivalents than a gram of hydrated salt. This information should be obvious on the bottle. If not, ask a lab tech if they can identify the hydrated/anhydrous salt by memory.

Note that the older notation for coordination compounds used a dot as well, before coord compounds were properly understood.

You should look up water of crystallization.
These H2O molecules that are written after that salt are actually the water of crystallization, i.e., the no. Of water molecules that attach to one formula unit of a salt.
You may know about the fact that hydrated CuSO4 salt (blue vitriol) is blue in color. This salt has these same water of crystallization. Its formula is CuSO4.5H2O. But upon heating, it loses water of crystallization due to evaporation and simply reverts to its anyhydrous state CuSO4.


text:
Rotation energy barrier for carboxypyrazolate
In this molecule, which is 3,5-dimethyl-4-carboxypyrazolate (unless I got it wrong):

I do not think there is free rotation around the bond indicated in red, because one can write a resonance structure with significant reasonable weight where a double bond would be located between the two groups.
However, how can I go further than that and qualitatively describe the energy barrier for the rotation. I have found (here and there) that in the case of benzoates, the barrier is in the range 3–6 kcal/mol. Should I expect it to be higher or lower in the above molecule?

According to ChemDraw, the name is 4-carboxylato-3,5-dimethylpyrazol-1-ide

@F'x - can't you transit the carboxylate dihedral angle?

@RichardTerrett I could, but I really would like to have some qualitative argument about which way it is expected to be

One of way to answer this question is to put it in some QM software. The system is small enough and has higher symmetry so computations shouldn't take to long, even on an older computer.
For co-planar geometry the energy calculated at B3LYP/6-31G(d,p) is –492.132206748 a.u.
The geometry with dihedral angle of 90° between $\ce{CO2}$ and $\ce{C3N2}$ is –492.116719625 a.u. That makes roughly 41 kJ/mol (10 kcal/mol).
Now if you really want to be sure if that's the right energy you should make frequency calculations on both structures to find out if they are minima or transition states. This can take a bit longer (I haven't done it) and I would expect that the better structure (with lower energy) should have no imaginary frequencies and the higher one should have one (TS). If by some accident both of them are minima than TS has probably a dihedral angel somewhere around 45°.
Of course this is only computation. It gives as clues but not the exact answer. You can go for higher level of theory, add solvent modelling and... or you can take your compound and try to make temperature depending NMR spectra. With a bit of luck you can freeze your TS and find out how high is your barrier.
p.s.
It would be good if you mention in your post what tools (in this case computational) are available to you. I really would prefer to explain how to get the number instead of giving just dry answer.

I know how I can calculate the numbers, but I was wondering if one could intuitively argue the energy barrier (carboxypyrazolate vs. benzoate) one way or another. Thanks anyway for the calculations…


text:
Temperature of pure oxygen combustion
What is the temperature of pure oxygen being combusted? I was wondering about it.

If you're curious what the "hottest" fuel is, it's dicyanoacetylene (C4N2).  The center of the flame at room temperature is roughly 9010 degrees Fahrenheit.

Oxygen is not a fuel; it is an oxidant, as others have noted. However, ozone is a fuel, and ozone flames exist.

There are flames burning in Chlorine, Fluorine etc. Big amounts of the chlorine from salt electrolysis is burned with hydrogen when demand for lye is bigger than for chlorine. Diborane and F2 is said to be one of the most energetic flames.

Other answers have emphasized the fact that oxygen is not a fuel, but an oxidizing agent. I'll try to underline another point: there is no uniquely defined “combustion temperature” for a chemical compound. This is unlike phase transitions like melting and ebullition, which happen at fixed temperature for a given pressure.
The temperature of a combustion process depends on many parameters, including the heat of combustion, the ratio of air to fuel, the heat capacity of both and the initial temperature. The adiabatic flame temperature is one of such combustion temperatures under certain conditions. The autoignition temperature and flash point, on the other hand, are characteristics of a given chemical compound; they are not, however, combustion temperatures.

I was going to mention this, but I wasn't too sure of it--lots of flames are said to burn at "xyz degrees", so I neglected to. Any idea what those temperatures are?

Oxygen is not a fuel by itself, it's only the oxidant. Combustion involves a fuel (acetylene, for example) and an oxidizer (here, O2). So if you have no fuel, but all oxidizer, then what's left to burn?
Combustion is just a name we give to certain reactions. The point to get here is that the burning and its associated temperature is dependent on the reactants' chemical nature and stoichiometry. When a flame has a higher temperature than another, it means that both reactions give products that have different enthalpies of formation.
The greater the difference between the enthalpy of the products and reactants, the greater the flame temperature.

Pure oxygen does not burn. Oxygen is a supporter of combustion, not a combustible material(fuel) itself.
Combustion reactions are as follows:
$$\text{combustible stuff} \ce{+ O2 -> } \text{various oxides} + \text{heat}$$
Note that nitrous oxide ($\ce{N2O}$) is also a supporter of combustion, which means that you can use it to carry out a combustion reaction without oxygen--but this is only because $\ce{N2O}$ generates oxygen itself.
Since combustion is a reaction with oxygen, what's left for oxygen to oxidise? Itself? Not likely. You need a fuel to release heat for a combustion reaction.


text:
Why are strong acids and bases not suitable as primary standards?
This is one of the questions with which I have puzzled over, and can arrive at no definite conclusion. Why are strong acids or bases, such as $\ce{H2SO4}$,$\ce{ HNO3}$, $\ce{HCl}$, and $\ce{NaOH}$ not suitable primary standards?
How will their higher or lower pH affect the accuracy of the results of a titration?

strong Bases in general are not suitable as they will react with $\ce{CO2}$ from the air. This is actually more problematic than hygroscopicity, as it means that you also cannot store diluted $\ce{NaOH}$ solution (openly) as standard.
Instead, $\ce{Na2CO3}$ or $\ce{KHCO3}$ are much better alternatives (for strong acids).
The acids are either a gas ($\ce{HCl}$) or are produced from gas. I don't see how you can produce a conentration reliably enough for use as primary standard.
even if you would manage to produce saturated $\ce{HCl}$, it's much more difficult and less reliable than known alternatives (see below).
but that wouldn't work e.g. for $\ce{H2SO4}$ as $\ce{SO3}$ is soluble in $\ce{H2SO4}$ (Oleum).
$\ce{HNO3}$: redox reactions are not only not particularly useful, but actually harmful as they use up acid equivalents:
$\ce{NO3- + 2e- + 2 H+ -> NO2- + H2O}$
also here, a solid alternative (KH-Phthalate) is available. If you don't have that, you may get away with oxalic acid which is still better than the acids suggested here.

There's nothing particular about strong acids or strong bases that disqualifies them from being a primary standard, but most of them are too unstable or difficult to work with to be useful as a primary standard.  A primary standard should have these qualities:

High purity
Stability in presence of air
Absence of any water of hydration which might vary with changing humidity and temperature.
Cheap
Dissolves readily to produce stable solutions in solvent of choice
A larger rather than smaller molar mass

Now for the examples you gave:
$\ce{H2SO4}$ is hygroscopic, so it's concentration will change quite readily in air.
$\ce{HNO3}$ is very hygroscopic as well.  The $\ce{NO3-}$ anions can also start redoxing other things which isn't particularly useful for titrations.
$\ce{HCl}$ is a gas, so it will be difficult to determine the concentration of the solutions (you'd have to compare it against another primary standard), and high-concentration solutions are unstable.
$\ce{NaOH}$ is hygroscopic as well.
As Georg pointed out below, one generally needs high-purity solid compounds so they can be weighed accurately (and in addition to the cons above, these substances are difficult to find/weigh in solid form).  I hope this helps.

The only correct point is for NaOH. For HCl the reasoning is silly! diluted HCl solutions are absolutely stable, if only You avoid vaporizing the water in them. The HCl is not volatile at all in diluted solution. For all other the main point misses: one needs to weigh the primary titration standards reliably! This calls for stable solid compounds.

@Georg: if you could be sure of the concentration in the first place you could also weigh a liquid...

I edited to clarify the instability of HCl.  Also, good point on the weighing--I added it to the answer.


text:
What's the difference between precipitate and turbidity?
In many tests for various radicals, a "<color> precipitate" or "<color> turbidity" is mentioned. For example, lead acetate with chromate ions gives a yellow precipitate, but hydrogen sulphide with potassium dichromate gives a yellow(or white) "turbidity".
In the reaction itself, the turbidity is noted as a precipitate with a down arrow, but I'm wondering if there is any difference between the two. It looks like turbidity refers to a colloid/suspension being formed, but I'm not sure.
I'd like to know:

The physical differences/distinguishing features between turbidity and precipitate, if any.
The observable differences/distinguishing features between the two, if any.

My understanding is that turbidity is an adjective describing the presence or degree of particulate matter suspended in a liquid (see turbidimetry), whereas a precipitate is a particulate that can result in turbidity.
The turbidity generated by a precipitate is a function of its ability to form a stable suspension. If the precipitate simply drops to the bottom of a container or readily flocculates, the resultant turbidity will be low. The turbidity of a suspension is also going to be a function of the solvent and the nature and concentration of dissolved ions - a dramatic example of this is the rapid clarification of muddy freshwater at river deltas mixing with salt water.
If the word turbidity is being used as a noun, it is probably referring to a light-occluding suspension.


text:
Why is slush hydrogen at the triple point?
Slush hydrogen is a mixture of liquid and solid hydrogen at the triple point considered as a possible vehicle fuel. What is the need of having it at the triple point? Couldn't any other set of thermodynamic conditions consistent with liquid–solid equilibrium work as well?

I don't know the right answer but I would guess that the higher density of slush hydrogen compared to liquid hydrogen might play a role.

@Philipp yes, but you don't need the triple point for that, just liquid–solid equilibrium (which happens on a line in the phase diagram, not only at the triple point)…

This is just a shot in the dark, but wouldn't rocket fuel (or other slush hydrogen applications) require hydrogen as a gas?  If one stores it at the triple point, that might allow you to have high density and use it as a gaseous fuel.

I'll have a stab at this one, despite the fact I've forgotten almost the totality of engineering thermodynamics, as I do recall seeing these papers before.
One of the references1 in the Wikipedia article describes the production of slush hydrogen through the auger technique, contrasting it with the freeze-thaw method. Specifically, the auger method can be performed above the triple point pressure of hydrogen, whereas the freeze-thaw method uses the latent heat of vaporisation to freeze some of the hydrogen. In this technique the vaporisation of some of the hydrogen (prompted by the application of a vacuum) at the triple point causes some of the liquid hydrogen to freeze, generating a crust of solid hydrogen which is then broken up to make slush.
As far as I can tell, the triple point conditions are only required to make the freeze-thaw process go, and are not required for storage.
The auger technique, also described in the report1, seems to work on the basis of shaving solid hydrogen which is refrigerated with liquid helium. Voth compares the technique to a conventional ice shaver (albeit much, much colder!). In this instance there's no requirement for triple point conditions, which is given as one of the advantages of the technique.
Finally, Haberbusch and McNelis2 give us this sad quote, in reference to a 1991 paper:
"According to Hans, projected annual demand for slush hydrogen in the year 2010 is 584 million pounds to support a fleet of 20 space-planes"
One day, maybe...

(1) Voth, R.O.; Producing Liquid-Solid Mixtures of Hydrogen in an Auger; National Bureau of Standards Department of Commerce; 1978, pp. 1-16
(2) Haberbusch, M.S. and McNelis, N.B.; Comparison of the Continuous Freeze Slush Hydrogen Production Technique to the Freeze-Thaw Technique; NASA Technical Memorandum 107324, 1996, p. 1

Ten years on, and... well, one can still hope.

I have studied triple-point systems, though not principally in hydrogen.  I have worked with phase equilibrium in dynamic (moving) water systems, looking at changes in density and surface tension.  Because of this, I have studied the phenomenon more broadly.
While a system of mass and energy to remains (is maintained) in phase-equilibrium (i.e. the triple point), the introduction of new forces – be they pressure, e-field, magnetic polarization, thermo, etc – can lead to an ongoing reaction cycle.
I won't speculate too deeply about what that does in this H system, but will leave this paragraph for another specialist to answer.
The phase-equilibrium point can make it easier to push a reaction toward a specific kind of change / characterization of the media.  Particularly, the closeness to the gas phase means the volatility of [part of] the media is being utilized to

bond in external media (doping)
perform redox or some other electronic reaction
liberate some media in a thermo exchange that crystalizes others
change the crystallization morphology

Pragmatically, I've read this is applied in reactor\feedstock scenarios, where the 'reaction' portion would in this case be held at the triple-point, whilst the feed/flow of media causes the outlet portion to take on a certain character, as the inlet media is prepped.  The principle to understand here is that the reaction portion act more homogeneously, rather than local regions of differentiated materials.


text:
Solution based measurement of Solvent-Accessible Surface Area of macromolecules
The Solvent-Accessible Surface Area (SASA) is a valuable metric for looking at protein folding and protein-protein interactions. However, this measurement is typically done by calculating the SASA from a solved (and generally static) structure.
Chemical probes like diazirine and hydroxyl radicals show some bias regarding where they tend to bind. I'm realizing while I'm writing this question that NMR is a perfectly valid method to determine a solution based structure and then calculate the SASA. Similar strategies has also been used to examine structured RNAs. I'm curious about the variety of these methods and how accurate they are.

Also asked at http://biology.stackexchange.com/q/1909/389

As you noted, the solvent-accessible surface area is a geometric concept, and as such is most easily calculated using computational geometry tools from a given molecular (or crystalline) structure. However you obtain this atomistic structure (X-ray, neutrons, molecular simulation, NMR, …) is of course of significant importance to the resulting area, but the calculation method will be the same.
There are, however, a few studies which have attempted to link the solvent-accessible surface area to directly measurable properties, and in particular to spectroscopic properties. A quick search of the literature reveals three examples:

Amide H/2H Exchange MALDI-TOF Mass Spectrometry
UV Raman directly on the exposed residues
Using diazirine as a probe


text:
How long does it take to run an MS/MS experiment
I am not planning to do these types of experiments but I would like to know the order of magnitude of the time it takes to run an LC-MS/MS experiment. Let's assume that I have the sample prep (protein) finished and I just need to load the sample into the LC. How long would I need to wait before I would get MS/MS data?
I know that for a typical HPLC experiment, the sample takes around 20 minutes to go through the column.

The LC is the rate-limiting step. That can take anywhere from a couple of minutes to hours, depending on what you're eluting. The MS-MS part should be a breeze.

The MSMS part is the detection of the LC in this case (nb. you don't need to do LC-MSMS, you can do direct infusion quite happily if you don't need to separate anything), so the actual experiment wouldn't need to take much longer than any other LC (as CHM says).
However, calibration + set-up of these things, and extensive cleaning afterwards, can make it a day's job to run just a few samples (even direct infusion). If you don't have much information on the system already, you may also find that you need to do a 'sighting' run, to work out which ions you need to examine more closely (MSMS). And then there is data processing.....
All in all: quick, if you know exactly what you want, and someone is running everything for you. If you want to argue that it is an easy job, you can, but be aware that this is not always gong to be the case.

You can adjust your run times to your preference, based on your conditions.  For example, I am doing LC-MS/MS on steroids (low molecular weight compounds) extracted from human urine samples through a liquid-liquid extraction. It only takes me 10 min per sample from the time the run starts until the very end of it.

May you share details on your column and eluent? This would be a nice addition to show how different elution conditions can affect separation time.


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what kind of suspension liquid should be used with ferrofluid (so it does not stain the glass)
Please take a look at the following video:

I am working on a new project, and I need to find whats the best liquid to hold ferrofluid inside the glass (or maybe even plastic) container, so that ferrofluid (that easily stains everything) does not stain or stick to glass?
Also, is there any special preparation for the glass needed?

What is the ferrofluid you plan on using?

It is, FerroFluid EFH-1 by FerroTec Corp
this one - https://www.amazingmagnets.com/show-decimal-ferrofluid-0250.aspx

I played with ferrofluids when I was younger. I don't recall them being particularly staining on pyrex? I don't know why borosilicate would be any worse off the top of my head. Are you sure a water mixture with this ferrofluid wouldn't work? It would stain the glass presumably?

I tried it, it stains the glass and leaves a trace.. if you go over it with a magnet few times quickly you might be lucky and remove the stain but if you leave it for some time its... painted :)

Definitely don't use plastic. I tried to make some ferrofluid back in high school, and it ended up melting the plastic and making a big mess.

hm... this one doesnt melt plastic, it LOVES plastic :)))...stains like hell

I don't have a full answer, but a hint at how to choose your liquid. You want the ferrofluid not to mix with it, so your fluid has to be immiscible with the ferrofluid solvent (or carrier fluid; ferrofluids are colloidal suspensions). In the case of your “FerroFluid EFH-1”, the solvent is a light mineral oil, so you should go with a polar solvent.
Secondly, you want the ferrofluid solvent not to touch the glass, to avoid staining. In order to do so, you need your liquid to wet the bottle more than the mineral oil, i.e. to have a smaller contact angle. You can play on both the bottle material and nature of the fluid to achieve this. I suggest protic polar solvents, which should wet regular glass well: water, ethanol, isopropanol, acetic acid, …

Hey thanks a lot for all this, it will really help in the search for the right one. I will narrow down the list of all clear liquids i can find with these filters you gave me and see for each how it goes. So far even alchohol didnt prevent it 100%, the only 100% that i've seen is hydrogen peroxide (tried it few hours ago) - though the ferrofluid acts weird.. completely different than in alchohol :)

@DamjanStankovic I know I'm late to the party, but just wondering what was your final solution?

I had success with the following:
1) Glass bottle, ideally with flat sides.  If you have a round bottle you may not be able to see the ferrofluid due to diffraction.
2) A solution of ~25% distilled water (a.k.a. deionized water) and ~75% of store bought isopropyl alcohol at 91% concentration.  That means that ~70% of my solution was actually isopropyl alcohol.  I eyeballed the ratio between the water and the isopropyl alcohol, those numbers are likely to not be exactly correct.
Procedure that I followed:
Add some isopropyl alcohol to the glass container.  Then add some (or all) of your ferrofluid to the container.  Then add some water to bring your isopropyl alcohol concentration to about 70%.  Next, use a magnet to settle the ferrofluid to the bottom and then remove the magnet.  If your ferrofluid has little bits that are floating, then you need to add more water.  If your ferrofluid looks too stringy then I would add more alcohol.  When I was done the ferrofluid had zero staining, like you see in the videos online, except for in a few spots.  I think for better results I would need a higher quality glass container.
notes: Don't put too much ferrofluid in your container, otherwise you will have an uninteresting blob.  I recommend using gloves.  To clean the ferrofluid stuff off my hands I used warm water, laundry detergent, and lots of scrubbing with a towel.  If you mess up, you can use a magnet to hold your ferrofluid in place while you pour off some of your suspension fluid.  In addition, you can use a straw and your thumb to pipet some of the ferrofluid, however some will get stuck in the straw, so I wouldn't do that unless you have a bunch of ferrofluid.  Finally, I only finished like 30 minutes ago, so I hope it will all look good tomorrow.  If something bad happens I will make another post.
notes, part 2: If all you want is a cool desk toy, then I recommend just buying one of the pre-made ferrofluid toys online. (I saw some for ~$30).  You will get a lot more ferrofluid for your money if you buy just the ferrofluid, however making the toy is a bit of work.
GOOD LUCK!
edit:  no combinations of the following worked for me: acetic acid, hydrogen peroxide, pep plastic.

No biggie.  For aqueous ferrofluid, silanize the glass with Rain-X automotive windshield treatment.  If you want both a hydrophobic and oleophobic glass surface, fluoroalkylsilane surface coupling
http://www.gelest.com/gelest/forms/GeneralPages/Applications/hydrophobic_materials.aspx
GLASSCLAD 18
AQUAPHILE AQ
AQUAPHOBE CF
AQUAPHOBE CM

What about an ammonia-based fluid mix? I saw one guy on youtube that said he made it with Windex (the liquid was a clear greenish color) and another guy said he made it with Windolene(he's in the UK) but it went cloudy after a while. Meanwhile doing some other research I found someone who says their product (they sell ferrofluid desk toys) should be kept at room temp and will go cloudy if kept too cold. My guess is the guy in the UK had it accidentally get too cold.. because it's the UK, lol.
Those are my ideas, hope they help. :)

Do you have the links to the "guy on youtube"?  It's always helpful to have answers that are supported by references, giving the OP additional information to address their question.


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Why is silver chloride less soluble than silver nitrate?

Related: Reaction between silver nitrate and aluminum chloride

Experimentally, $\ce{AgCl}$ is insoluble in water, but $\ce{AgNO3}$ is soluble. They're pretty common in a lab (well, $\ce{AgCl}$ is a common precipitate)--so I think most of us know this.
By Fajan's rules, on the other hand, larger anion $\implies$ more polarization/covalent character $\implies$ less solubility. But, $\ce{NO3-}$ is the larger anion, yet $\ce{AgNO3}$ is more soluble.
Is there any theoretical reason for this?

In the comment to my previous answer, you asked for a theoretical reason for the solubilities, not considering energy data. Since I know from energy considerations that the issue is not the solvation of the anions, I can present a reason based on the strength of the ionic bond in the two compounds. This reference (as well as others) states the bonding in $\ce{AgCl}$ has an unusually high covalent character which makes it a tighter bond. The $\ce{Ag+}$ ion and the $\ce{Cl-}$ ion are close to the same size (with the silver ion being smaller), so they can approach each other quite closely. In silver nitrate, the $\ce{NO3-}$ ion is larger and does not allow as close an approach as the chloride ion, so the bond is weaker, easier to break up, and the salt is more soluble.

I'm afraid this is rather a non-answer (or why is it so difficult to answer this)...

Pretty much all nitrates are soluble. This is often explained by the exceptionally good delocalization of the negative charge.
On the other hand, silver salts in general aren't well soluble (I recall only fluoride, nitrate and perchlorate as soluble. Sulfate, carbonate, oxide, sulfide (of course), even the stochiometric cyanide (if I remember correctly) isn't soluble.
So in the case of the nitrate, usually good solubility of nitrates seem to win over the usually not so good solubility of silver salts.
For the Chloride, it is maybe similar: the Chloride stands between the good solubility of AgF and the low solubility of AgBr. Just here, "unsoluble" wins.
(There's also - related to Fajan I think - the HSAB concept which tries to extend also to hard-hard unsoluble salts [e.g. $\ce{CaF2}$] -  However, for the case here it is  about equally unhelpful as Fajan. I think those two concepts work better in a row of similar ions, like Ag-halogenides)
I think the difficulty in explaining/predicting from theory may come from the fact that there are contrary but strong effects to be considered, so it is hard to predict what exactly the result will be (loss of significance: subtracting two almost equal large numbers, you may not even be able to be sure about the sign of the result).

substance     solubility / (mol/l)
AgF               1.4 ⋅ 10¹   = 1.4e1
AgNO3         1 - 5   ⋅ 10⁰   =   1e0 - 5e0
AgCl              5   ⋅ 10⁻⁴  =   5e-4
AgBr              7   ⋅ 10⁻⁷  =   5e-7
AgI               9   ⋅ 10⁻⁹  =   9e-9
Ag2S              1.7 ⋅ 10⁻¹⁷ = 1.7e-17

The solubility between different halogenides varies 10 orders of magnitude between fluoride and iodide. That corresponds to a 10fold change in $\Delta_RG^0$. A factor 10 in energy is like bond strength covalent vs. H-bond or like rotational vs. vibrational vs. electronic transition of a molecule.

Have a look at the paper that I linked from the other question. It is actually focused on $\ce{AgCl}$.

I cannot see the exponents in the solubility table with my device.  Tried using mathjax and blue sparks came flying off the screen.  (OK, actually it just didn't work.)  Can you modify your syntax so the exponents fo not need special characters?

@OscarLanzi: There you are.

In order to dissolve a salt, you have to break apart the ions and hydrate them in solution. You can use the enthalpies of hydration of the ions, and the crystal lattice energy of the solid, to predict which compounds will dissolve.
I found data that the crystal lattice energy of $\ce{AgCl}$ is -916.3 kJ/mol (experimental), while the lattice energy of $\ce{AgNO_3}$ is -820 kJ/mol. (Since lattice energy is defined as the energy released when the ions join to form a solid, it is always negative. Reverse the sign for the amount of energy you need to put in to break the crystal apart into separate ions in the gas phase.) From this data you can see that it takes quite a bit more energy to break apart the $\ce{Ag+}$ and $\ce{Cl-}$ ions than it does the $\ce{Ag+}$ and $\ce{NO3-}$ ions, presumably because the chloride ion is smaller and more tightly held. If the enthalpy of hydration value is about the same for the two cases (I couldn't find actual values), then it may not be enough to overcome the extra energy needed to break up the ions.
So, both enthalpy of hydration and crystal lattice energy are important considerations for solubility.There is quite a nice discussion about how to think about solubilities of salts, and what factors affect solubility using a Born-Haber cycle, in this reference.
Added this paragraph to the answer after putting it in the comments below first.
Calculation from enthalpy of hydration data (-850.7 for $\ce{AgCl}$ and -794.4 for $\ce{AgNO3}$ says that $\ce{Cl−}$ is the more soluble ion, by -61.3 kJ/mol (since the contribution from solvation of $\ce{Ag+}$ would be the same in each case). Nitrate would be the less soluble ion (from energy concerns) in agreement with Fajan's rules. The enthalpy of hydration data was obtained from problems in a chemistry book.

It's still experimental--you've taken the energy data.I'm looking for a theoretical reason..It's the exact opposite of what Fajan's rules predict.

Fajan's rules are talking about polarizability of the ions which would affect how well they're solvated, but solvation is only one aspect of solubility with the other part depending on crystal lattice energy. You can't look at the solubility of the $\ce NO_3^-$ ion in isolation, you have to consider the compound. Your question might end up being "why do nitrate salts have a lower lattice energy" as opposed to looking at hydration energy. For these two, the difference in the lattice energy is larger than the difference in the hydration energy (calc. -850.7 for AgCl and -794. 4 for $\ce AgNO_3$.

Sorry, one more comment after thinking. Calculation from enthalpy of hydration data above says that $\ce Cl^-$ is the more soluble ion, by -61.3 kJ/mol (since the contribution from solvation of $\ce Ag^+$ would be the same in each case). Nitrate would be the less soluble ion (from energy concerns) in agreement with Fajan's rules.

Look at the crystalline structure of $\ce{AgCl}$ vs $\ce{AgNO3}$.
Chlorides usually form face centred cubic crystals while nitrates form trigonal planar crystals. Nitrate is a triangular molecule with a positive charge on the nitrogen and a (2–) charge on the oxygens.
In FCC each ion interacts with six ions of opposing charge in an octahedral arrangement.
But with a nitrate the nitrate ions are only interacting with 2 ions of opposing charge above and below the 2/3 negative charge on the oxygens point to neighboring positively charged nitrogen. This arrangment is far more unstable and in silver nitrate ions are far more likely to be whisked away by ion-dipole interactions.

There is not a positive "charge" on the nitrogen, nor a negative "charge" on the oxygen. The nitrate ion is held together with covalent bonds and the single negative charge is delocalized over the ion. Are you talking about oxidation numbers? The nitrogen-oxygens bonds would be polar with more electron density on the oxygens, but the electron density would be evenly spread over the 3 oxygens.

Can you point me to a reference for nitrates forming trigonal planar crystals? I'm not familiar with that. The shape of the nitrate ion (held together with covalent bonds) would be trigonal planar in VSEPR theory. A paper on the shape of silver nitrate crystals is described here pubs.rsc.org/en/Content/ArticleLanding/1966/J1/j19660000123

I'm not much sure about this. I too was much curious about precipitates and this is similar to what I had read somewhere on the net (can't find the site, got it from Wikipedia)

Solubility occurs under dynamic equilibrium, which means that solubility results from the simultaneous and opposing processes of dissolution and phase joining (e.g., precipitation of solids). The solubility equilibrium occurs when the two processes proceed at a constant rate.

So according to this the process of phase joining is very very more than dissolution in $\ce{AgCl}$ thus it is insoluble in water, but in $\ce{AgNO3}$ it is just the opposite.

In $\ce{AgCl}$ and $\ce{AgNO3}$ the cation is same, i.e. silver ion; now compare the anion. Larger the anion means less is the effective nuclear charge on the valence shell. Hence more is the deformation of the anion due to the cation. More deformation means more covalent nature of the bond. More covalency means less solubility.

PRETTY SIMPLE: bonding between AgCl is more covalent in nature and the size of the both the ions are almost simmilar in size and thus it is insoluble in water.


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What is it about the relationship between the Fe²⁺ and Fe³⁺ in magnetite that makes it diamagnetic?
Hematite is composed of $\ce{Fe2O3}$, and is paramagnetic, whereas magnetite is $\ce{Fe3O4}$ and is diamagnetic.  Magnetite's nature is due to the presence of both $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ (Wikipedia even goes so far as to call it $\ce{FeO}\cdot \ce{Fe2O3}$, but explains that this is not a "solid solution").
I can surmise that in the crystalline solid, there are divalent and trivalent cations dispersed within.  I can also assume that there might be some sharing of electrons between the oxygens and each of these cations.  One would assume this also happens within the hematite crystal, so what is it about "tossing" the divalent cation into the mix of the crystalline structure that makes the magnetite diamagnetic?
Does the charge differential between the two cations cause a permanent dipole, if so, why don't all of the small dipoles simply cancel each other out?

Magnetite has a spinel structure with two types of $\ce{Fe}$ sites: octahedral and tetrahedral, respectively in green and orange below.

(in this structure, oxygen atoms are the small red balls).
This structure is called a spinel, with $\ce{Fe^{2+}}$ ion in tetrahedral coordination and $\ce{Fe^{3+}}$> ions in octahedral sites. The coupling between atoms from these two lattices are superexchange interactions, which result in an antiparallel alignment of spins. However, because the magnetic moments of the two lattices are not equal, their addition results in a net magnetic moment, even though they are of opposition alignment. This is called ferrimagnetism, and in practice most of its macroscopic properties are similar to ferromagnetism, though they stem from a completely different microscopic magnetic ordering.

Nice with the spinning diagram. Thanks for clarifying, I've obviously got to read more about it!

Magnetite is not diamagnetic. It is ferrimagnetic
Ferrimagnetism requires layers of unequal but opposite magnetic moments in the crystal lattice. The two iron ions have different magnetic moments, and they probably are arranged nicely in the lattice, so the ferrimagnetic property comes from there.

Thanks I had no idea that was a classification.  Could you explain a bit more about "arranged nicely"?

@jonsca read the Wikipedia pagr. Or F'xs awesome answer ;-)


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Is soap the natural form of Sodium Lauryl Sulfate?
I was told that "Sodium Tallowate, sodium cocoate and sodium palm kernelate are the 'natural' forms of SLS".  Is this true?  I thought "natural" SLS would be SLS derived from a natural product such as coconut oil.

Hi jrennie, welcome to Chemistry StackExchange! Could you describe a bit what you mean by “the natural form of SLS”? Wikipedia has a nice page on SLS, which you linked to. It includes details on its production, which appears to be from coconut or palm kernel oil.

Conventional soaps are the sodium salts of long chain carboxylic acids (aka fatty acids). These are formed by the hydrolysis of triglycerides, a process known as saponification. This process breaks the triglyceride at its ester linkages and yields three fatty acid molecules and glycerol molecule.
As the aforementioned wikipedia article indicates, sodium lauryl sulfate is formed by refunctionalising the fatty acid lauric acid. In this case, the $\ce{R-OOH}$ (carboxylic acid) head group has been replaced with an $\ce{R-OH}$ (alcohol) group, which is in turn replaced by $\ce{R-OSO3^{-}}$ (sulfate). Here I am using $\ce{R}$ to refer to the long alkyl chain of the molecule.
I'm not sure whether sodium lauryl sufate occurs naturally, however the other molecules you mention are different in the sense that they have carboxylate ($\ce{R-COO^{-}}$) head groups whereas sodium lauryl sulfate has the sulfate head group ($\ce{R-OSO3^{-}}$). Both classes of molecule will act as surfactants as they have polar head groups which are hydrophilic, and nonpolar alkyl tails which have an affinity for fats and oils.
Lauric acid derivatives spotter's guide
(These were drawn using the nifty ChemFig package for (La)TeX.)

The last two (sodium laurate and sodium lauryl sulfate) are sodium salts. Both carboxylic acids ($\ce{R-COOH}$) and hydrogen alkyl sulfates ($\ce{R-OSO3H}$) are fairly willing to ionise to give the corresponding carboxylate ($\ce{R-COO^{-}}$) or sulfate ($\ce{R-OSO3^{-}}$). Sodium ions happily associate with these anions, balancing the negative charge.

The original discussion re: SLS and the compounds used to derive it can be found here: http://parenting.stackexchange.com/questions/5047/how-to-test-whether-my-toddler-is-allergic-to-detergent/5049#comment7644_5049
Sodium Tallowate, sodium cocoate and sodium palm kernelate are derived from coconut oil, palm kernel oil and tallow. Sodium cocoate is the sodium salt of fatty acids from coconut oil; lauric acid is the primary component of this oil. Sodium tallowate and palm kernelate likewise contain the fatty acids used for SLS formulation.


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Side reactions of N-hydroxysuccinimide esters with nucleophiles
N-Hydroxysuccinimide (NHS) chemistry is commonly used in chemical biology to react with primary amines like lysine and the N-terminus of proteins. I was curious how labile NHS esters are to other nucleophiles like secondary amines and sulfhydryls. This knowledge will be helpful to know the extent of side products when I do a bioconjugation experiment.

The reactivity of NHS activated esters has mostly been desribed quite anecdotically. They are susceptible to hydrolysis in water, the kinetics of which is dependent on the nature of the NHS activated group (ester vs carnonate vs carbamate) as well as on the pH of the reaction buffer1, which is one way to show that primary amines are not the only nucleophile which can react with NHS-activated esters.
As for the chemical groups that have been observed to react with NHS activated esters, I did find a couple of examples upon looking up this issue. For instance, the Zenobi group have attempted a quite systematic study on the groups that can react with doubly NHS activated cross-linkers. Citing the abstract2 :

As soon as additional cross-linkers were attached or loops were formed, other amino acids were also involved in the reaction. In addition to the primary amino groups, serine, threonine and tyrosine showed significant reactivity due to the effect of neighboring amino acids by intermediate or permanent Type-1 cross-link formation. The reactivity is highly dependent on the pH and on adjacent amino acids.

In an earlier work, the reactivity of a biotinylation NHS-ester based reagent was assessed on a single peptide,  [D-Lys(6)]gonadotropin releasing hormone. A similar reactivity of -OH groups was observed, with in addition that of the arginine guanidinium group3:

In addition to the O-acylation of Ser(4) and Tyr(5) in this peptide, we have also identified a novel biotinylation of the Arg(8) side chain.

Other groups have also observed some reactivity with cysteins (sulfhydryl groups), but in a situation where the reagent is included within a ligand of the target protein. One can assume, as do the authors, that specific positionning of the ligand within the binding site can lead to binding with a cystein group4,5.
From these earlier work, one cas conclude, in answer to your question that although primary amines are the most reactive group for NHS-ester derivatization, other groups present in peptide side-chains (-OH for tyrosine, serine, threonine, guanidinium for arginine and sulfhydryls for cysteins).
The last issue raised in the question remains that of the secondary amine groups. It is widely accepted that pH influences strongly the kinetics of NHS derivatization, mostly through protonation of the amine groups which cannot act as nucleophiles in protonated state. This has been put to use in controlling the buffer pH in order to favor N-terminal reaction vs lysine side chain reaction. Although, in my experience, it is not at all a straightforward experiment and it is highly dependent on the nature of the derivatization group. One could hypothesize that since secondary amines are more basic then primary amines, the reactivity of secondary amines towards NHS should be reduced compared to primary amines, but I have no experimental or litterature proof for this part of the answer.

please check out our paper we published already in 2008:
Chances and pitfalls of chemical cross-linking with amine-reactive N-hydroxysuccinimide esters. Kalkhof S, Sinz A. Anal Bioanal Chem. 2008 Sep;392(1-2):305-12. doi: 10.1007/s00216-008-2231-5
Another paper on NHS ester reactivity is this:
Stefanie Mädler, Claudia Bich, David Touboul, Renato Zenobi, Chemical cross-linking with NHS esters: a systematic study on amino acid reactivities,J Mass Spectrom.2009 May;44(5):694-706. doi: 10.1002/jms.1544.
In a community-wide cross-linking/MS study, one of the outcomes was that NHS esters react with OH groups in ca. 30% of all cases.
Best regards,
Andrea Sinz.

Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference.


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By what reaction does (–)-sparteine decompose in ambient conditions?
In its pure form (–)-sparteine is an only-slightly-yellow clear, viscous liquid. Yet, after only a week of being kept in dry, normal atmosphere (in a round bottom flask covered by a septa), some of it reacts with something to form a sticky, orange-brown substance. I've had to distill the sparteine I have twice due to whatever reaction is occurring, and I would like to know exactly what the reaction is so I don't have to distill every time I need it. The literature I found mentioned that it readily absorbs carbon dioxide in a 1-1 molar ratio, and I know that tertiary amines do sometimes react with CO2, but I haven't been able to piece it together as something that could happen in ambient conditions.

Besides absorbing/reacting with carbon dioxide, amines also slowly react with oxygen to form the N-oxide. After you distil the sparteine, store it under nitrogen or argon, or store it under vacuum, and store it cold. Doing even one of these will extend the shelflife.

Amines left out in the atmosphere will air-oxidize to give N-oxides, which typically have a yellow color. These N-oxides can decompose to give further byproducts. I would store your sparteine under nitrogen or argon atmosphere. The reaction proceeds through a radical chain mechanism with molecular oxygen [1].
From Wikipedia article for triethylenetetramine:

This oily liquid is colourless but, like many amines, assumes a yellowish color due to impurities resulting from air-oxidation.

From [2, p. 99] (I realize that sparteine is not an aromatic amine, but the result is the same):

It can be observed that arylamines develop a yellow or black color if left exposed to air for prolonged periods.

If you have access to triethylamine or other amines in your lab, do a control experiment: leave one vial open to air, and leave the other under identical conditions except sealed under an inert atmosphere. A few days worth of observation will convince you of the role in oxygen in the yellowing of amines.
References

Aleksandrov, A. L. Oxidation of Amines by Molecular Oxygen: 12. Reactions of Peroxide Radicals of Triethylamine and N,N-Dimethylcyclohexylamine. Russ Chem Bull 1980, 29 (11), 1740–1744. DOI: 10.1007/BF00949211.
Lawrence, S. A. Amines: Synthesis, Properties and Applications; Cambridge University Press, 2004. ISBN 978-0-521-78284-5.

James, could you give some examples of the oxidation reactions, with a mechanism if it's known?  As of now, I'm not sure that this answers the question, as it doesn't give much more information than what what already available in the comments.

Tertiary amines are widely used as electron donors in photoinduced electron transfer (PET) reactions. Electron transfer to an electronically excited acceptor yields the radical anion of the acceptor and the radical cation of the amine. When the stereoelectronic requirements are met, the radical cation deprotonates to give an α-aminoalkyl radical.
Note that the deprotonation is observed for acyclic and thus flexible trialkylamines, but not for DABCO ([2.2.2]-diazabicyclooctane, lacking a sufficient orbital overlap between an α-C-H bond and the (half-filled) orbital on the nitrogen. I'd say that the situation in sparteine is pretty similar to that in DABCO.
With molecular oxygen, the formation of an α-aminoalkyl radical as the first intermediate is nevertheless conceivable. The mechanism, however, is different.
Taking the triplet state character of ground state oxygen into account, i'd say the most likely step is the abstraction of an axial hydrogen atom at one of the α-bridgehead carbon atoms next to a nitrogen atom.