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Oxidation number of phosphorus in phosphine
Electronegativity of $\ce{H}$ is greater than $\ce{P}$, but according to my book in $\ce{PH3}$, $\ce{P}$ has oxidation number of -3. How is it possible? Is any of the data wrong?

Here is a way to get the answer. If you place a reagent with oxidation number of H = -1 (hydride) it would produce dihydrogen gas. For example: $\ce{KH + H2O-> KOH + H2}$.
If you place a molecule with oxidation number of $\ce{H} = +1$ dihydrogen gas is NOT formed. $\ce{NH3 + H2O -> NH4+ + OH-}$. Phosphine doesn't form dihydrogen gas on contact with water so $\ce{H}$ is in oxidation state +1 and $\ce{P}$ is in oxidation state -3. Reaction with strong acids leads to $\ce{PH4+}$ which has similar structure to the ammonium ion. For these reasons even for $\ce{As}$ in $\ce{AsH3}$ the oxidation state of $\ce{As = -3}$.
The problem with electronegativity is that we are trying to assign a single number, while we actually need a function. We take average of electronegativity of $\ce{H}$ in $\ce{HF, NaH, H2O, CH4, NH4+}$ and come up with a single number. This works well if you need to guesstimate the behavior, but it is not accurate. It doesn't even account for softness/hardness of a partner.
Analysis "will $\ce{PH3 + H2O}$ form $\ce{H2}$?" (no) vs "will $\ce{PH3 + P2O3}$ form $\ce{P + H2O}$?" (yes) is more accurate way to assign formal charges.

But H is more electronegative than P...how do you explain that?

@ZOZ please see updated version.

@ZOZ As a rule of thumb, H will pretty much always be in oxidation state +1. You can work your way through many redox reactions starting with this assumption! Only in some rarer cases like metal hydrides (the KH example above) will H be have the oxidation number -1.

Electronegativity is a mere indication of what is happening and you will find many different values for the electronegativity of H and P (and all the other elements).
If you have a look at the C-H bond, electronegativities are 2.55 for carbon and 2.2 for hydrogen, i.e. the difference is much larger than for phosphorus (2.19) and hydrogen (2.2). However, C-H bonds are generally regarded as non-polar.
What really makes phosphorus different in its reactivity is that it is extremely oxophilic, i.e. it will try to form P-O or P=O bonds on every occasion. This is why PH3 is pyrophoric, for example.

Oxidation numbers are generally used to determine if a molecule acts as a reducer or oxidiser in a given reaction. The same molecule can be a reducer in one reaction and oxidiser in another if it is an intermediate oxidation state. Knowing the electronegativity (however defined) does not help in this instance.
If the oxidation number of a particular atom is not known it may be deduced from the known oxidation numbers of other atoms associated with it in a molecule. For example, the oxidation state of Pu in PuO$_2^{2+}$ is not known. It is calculated as
charge on Pu in PuO$_2^{2+}$ + charges on two O's (or 4-) = charge on PuO$_2^{2+}$ or 2+
thus Pu has an oxidation number of 6+.
In some cases (S$_4$O$_6^{2-}$, Fe$_3$O$_4$) the oxidation number can be fractional. Thus oxidation numbers can be negative, positive, zero or fractional. Of course the same atom can have different oxidation numbers in different molecules. An example is shown below for nitrogen
-3 in NH$_3$, 0 in N$_2$ , +1 in N$_2$O, +2 in NO, +3 in HNO$_2$, +4 in NO$_2$, +5 in HNO$_3$
where NH$_3$ is only a reducer and HNO$_3$ only an oxidiser, but the other compounds can be reducers or oxidisers. Note that it is zero in N$_2$.
A molecule is an oxidiser of its oxidation state decreases, for example iodine oxidises the strong reducer H$_2$SO$_3$ e.g.
I$_2$ + H$_2$SO$_3$ +H$_2$O $\rightarrow$ 2HI+H$_2$SO$_4$,
the iodine oxidation state changes from 0 to -1. In a reduction the oxidation state increases.
You can see that in your compound the phosphorous' oxidation state is -3.


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Why aren't the amine groups in the nucleic acid bases protonated?
Nucleobases G, C and A carry an amine group ($\ce{-NH2}$). In amino acids, the amine groups are considered protonated at physiological pH. Why isn't it the case for the amine groups of nucleobases?
I found some references discussing the protonation of nucleobases, but I am not enough expert in either DFT (http://www.ncbi.nlm.nih.gov/pubmed/19049279) or Mass Spectrometry (http://www.ncbi.nlm.nih.gov/pubmed/17920289/) to  get a rational from this. Plus, neither they nor the references they cite and that I have briefly examined actually consider protonation of the amine groups.

Amines are basic because of the lone pair on the nitrogen. In the nucleobases, the amine is adjacent to a double bond, which allows the lone pair to be delocalized. This reduces the basicity of the amine.

There are 2 answers:

As f' answered, the amino (NH2) group in a nucleobase is conjugated
with the heterocycle. Therefore, its pKa is closer to the one of
aniline. In other words, it is much less basic than an alkylamine.
Any drawing of a molecule in chemistry or biology is only a mere representation (agreed on by scientists) of the reality. For example, amino-acids are most usually drawn as H2N-R-COOH. We all know that it does not reflect reality but we all use that representation because it is much more readable than +H3N-R-COO- (which itself is also wrong), and because the real form of the amino acid will greatly depend on the experiment conditions. So you must always consider in what medium the molecule is to determine, more or less, its major form.

Ah! I had indeed overlooked the delocalization of the lone pairs. Thanks a lot!

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Which dicarboxylic acid has the most acidic hydrogen?

Which of the following acids (maleic, fumaric, succinic, or malonic) has the most acidic hydrogen?

I think that malonic acid should have the most acidic hydrogen due to the presence of an active methylene group.
The extraction of hydrogen atom from malonic acid would lead to high resonance stabilization, which is not the case in other acids.
However the answer key says that maleic acid has the most acidic hydrogen. Can someone please explain it to me why?

Related: Difference in acid strength of oxalic acid and malonic acid

These aren't esters but acids - you should care about H atoms in carboxylic groups.

The $\mathrm pK_\mathrm a$ of the $\ce{CH2}$ group in a dicarbonyl compound is roughly $10$, whereas the $\mathrm pK_\mathrm a$ of a carboxylic acid is roughly $5$. For example, dimethyl malonate has a $\mathrm pK_\mathrm a$ of $13$ whereas acetic acid has a $\mathrm pK_\mathrm a$ of $4.7$. So, the C–H bond acidity is not likely to be under consideration in any of your compounds.
Let's draw up some $\mathrm pK_\mathrm a$ data (only first ionisations, all data from Wikipedia):
$$\begin{array}{cc} \hline
\text{Carboxylic acid} & \mathrm pK_\mathrm a \\ \hline
\text{Maleic} & 1.9  \\
\text{Fumaric} & 3.03 \\
\text{Succinic} & 4.2 \\
\text{Malonic} & 2.83 \\ \hline
\end{array}$$
Clearly, all of these relate to ionisation of the $\ce{CO2H}$ group; if the acidity of dimethyl malonate is anything to go by, the $\ce{CH2}$ group in malonic acid is less acidic than the $\ce{CO2H}$ group by a factor of $10^{10}$.
What remains is to analyse this trend. Why is maleic acid the most acidic? The answer is – at least partly – intramolecular hydrogen bonding. When the first carboxylic acid group is deprotonated, the resultant monoanion is stabilised as such:

You are probably familiar with the mantra that a more stable conjugate base implies a more acidic compound. This is precisely what happens here.
Obviously, for fumaric acid, this is not possible because the double bond holds the two carboxyl groups apart. One could, however, argue that similar stabilisation could be derived for malonic and succinic acids. Why is it not a factor, then? It has to do with conformation: for the malonate or succinate anions to enjoy the benefits of intramolecular hydrogen bonding, they have to twist themselves into one specific conformation where the carboxyl groups are close to each other. This results in a loss of configurational entropy (the molecule is not as "free" as it would like to be to explore different conformations). For maleic acid, though, this is not a problem because the (Z)-double bond plonks those two carboxyl groups next to each other, and they don't have a choice about it. This bears some similarity to the concept of preorganisation, where two or more groups are "organised" in a fashion which alters their properties.

That's not the only reason why maleic acid is especially acidic, however. Part of the reason is also because an $\mathrm{sp^2}$ carbon is more electronegative than an $\mathrm{sp^3}$ carbon. An $\mathrm{sp^2}$ orbital contains 33% s-character, compared to an $\mathrm{sp^3}$ orbital which has 25% s-character. s-Orbitals are closer to the nuclei than p-orbitals, and therefore, an electron in an $\mathrm{sp^2}$ orbital experiences a greater effective nuclear charge from carbon; this translates into a greater electronegativity. (For similar reasons, alkenes are more acidic than alkanes.)
So, relative to the succinate anion, both the maleate and fumarate anions are stabilised by electron withdrawal via the inductive effect. This explains why fumaric acid is more acidic than succinic acid; the intramolecular hydrogen bonding explains why maleic acid is more acidic than fumaric acid.
Lastly, of course, malonic acid is more acidic than succinic acid because the electron-withdrawing $\ce{CO2H}$ substituent is fewer bonds away. In fact, succinic acid is barely more acidic than acetic acid, which doesn't have an electron-withdrawing substituent on it.

Additionally, I was wondering does the extended, end-to-end conjugated pi system also increase the stabilisation of the conjugate base for fumaric and maleic acid?

@TanYongBoon You can't actually draw a resonance structure to delocalise the charge across the double bond, so I wouldn't say that it's particularly important here.

I agree that resonance structures can't really show that delocalisation. However, there is, in fact, a set of consecutive parallel p orbitals capable of interacting with each other. Hence, there is still conjugation?

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Redox reactions
First semester chem student and I am wondering about redox reactions. For example, when copper is oxidized by silver ions in the reaction:
Cu + 2Ag+(aq)----->Cu2+(aq) + 2Ag(s) what is causing the copper to transfer 2 electrons (oxidize) to the silver ? At first I thought this was an ionic bond but learning that metals dont combine, that is not true. Does the transfer of electrons always imply a bond? If not, what causes this oxidation reaction, oxygen? And if so why isnt it written in the reaction itself? Fundamental I know, but I am trying to understand this.

There is no bond between them. If you dip Cu wire in AgNO3 solution then silver just deposits on top of Cu. The standard explanation involves the "activity of metals" and the redox potentials. In a sense it is similar to a question "why reaction 2NaI + Br2 -> 2NaBr + I2 goes in forward direction, but not the other way around?"
Please read a chapter on redox potential. Please ask a new question if something remains unclear.

Hi @Sixtytrees Thanks so much for the answer! Weirdly I can't find anything further about redox potential in my book (chemistry the central science 13e.d Brown). It simply talks about redox reactions but nothing about the actual redox potential or the science that explains it in what factors affect the potential. Again im in 1st semester chemistry perhaps this is an advanced topic? Secondly, in my original question does the aqueous solution (specifically the oxygen) have play any role in redox reactions that causes the copper to transfer electrons to the silver. Thanks in advance!


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Paper Data Replication
I read a paper recently - Here. I found the concept of Electride materials very interesting and wanted to mess around with them myself. I tried to replicate the partial electron density plots shown in the paper (Bottom right of picture):

I followed their methods as exactly as I could on Ca2N (It seems to be better researched than Na2Cl as I've seen it in many other papers on the topic, but I didn't want to put their data out that's not public), but I'm not sure how they created that graph/visualization. I did a geometry opt, then SC, found the band structure which matched theirs almost exactly, then did a band-decomposed run (I tried around the fermi level they said the electron resided in, I tried individual bands, etc) and VESTA doesn't pull up anything near what they have when I graph the PARCHG's.
Is there another step to visualize it the same way they have that I am missing, or is it likely to be something else I'm doing wrong?

I'd send an e-mail to the authors. They should be able to help you with their methods if it's not clear from the paper.


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Liquids that keeps the lowest temperature for the longer time after being frozen
I am working on a personal project, however I know nothing about chemistry and I would like to ask you guys for a quick question. I need to use a liquid for my project, which after being put into a freezer will freeze, and will be kept frozen for the most time when taken out of the freezer into room temperatures.
Ideally I would like to know if there is a liquid which can freeze after a couple of hours, and when taken out from the refrigerator it can sustain a temperature of -2 -3 degrees for the longer period, in normal room temperatures. And most importantly it can me used again and again.
Thank you so much guys, I will be looking forward to an answer on this one.

Try salt water.

To maintain a stable temperature while heat is entering a system implies some phase change: that is, the substance is either melting or boiling, keeping the temperature the same, while some part is changed form solid to liquid or form liquid to gas.
Pure water freezes at 0 °C, but if another substance is added, the freezing point (f.p.) may be lowered. Ordinary table salt, $\ce{NaCl}$, can lower the f.p. to -20 °C. Calcium chloride, $\ce{CaCl2}$, is even more effective. Both are used for melting ice from roads, with $\ce{CaCl2}$ preferred in cold weather.
You might test various concentration to see what best meets your needs. Ideally, the solution f.p. should be a bit above the temperature of your freezer.

No liquid can sustain a lower temperature than its direct surroundings for an extended amount of time; that is against the fundamental laws of thermodynamics (don’t ask me whether the first or second law are more relevant here). All you can do is have a liquid that warms up as slow as possible, i.e. has a rather high specific heat capacity.
There is one liquid with an exceedingly high specific heat capacity — at least when compared to all other common liquids. This liquid is water. It will take the more energy to heat water from $0~\mathrm{^\circ C}$ to room temperature than it takes to heat any other common liquid within the same boundaries.


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Standardization of sodium thiosulfate using potassium dichromate
In this process first we need to liberate iodine to react with sodium thiosulfate.
First, sodium bicarbonate is added to a iodate-free solution of potassium iodide. The mixture is shaken until the salts dissolve, then conc. hydrochloric acid is added. The solution is shaken again, then covered and keept in a dark place. I have two questions here:

What is the role of sodium bicarbonate here?
Can conc. sulfuric acid be used to provide the acidic media instead of conc. hydrochloric acid?

I think your procedure is incorrect. Typically you would mix $\ce{KIO3}$ (limiting reagent) with access of $\ce{KI}$. Then you add $\ce{HCl}$ to have a reaction:
$$\ce{KIO3 + 5KI + 6HCl -> 6KCl + 3I2 + 3H2O}$$
This is a way to generate a known amount of $\ce{I2}$.
You would then dissolve $\ce{Na2S2O3}$ in a buffer (sodium bicarbonate). This is done because $\ce{Na2S2O3}$ is not stable under acidic conditions. Then you can titrate $\ce{I2}$ in presence of starch with $\ce{Na2S2O3/NaHCO3}$. $\ce{NaI}$ and $\ce{Na2S4O6}$ are formed in this reaction.
In any case you cannot replace $\ce{HCl}$ with other acids. $\ce{H2SO4}$ and $\ce{HNO3}$ are oxidizers strong enough to turn some more $\ce{KI}$ into $\ce{I2}$ and mess up your measurement.

no I am pretty sure that's the process.because later we add potussium dichromate solution.this is the process of producing iodine gas from potussium iodide only with potussium dichromate.


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What are the molecular orbitals shaped like in diberyllium (mono)cation?
I think one way of describing the result is that the bond in $\ce{Be^+_2}$? cation is a resonance hybrid of two possible pi bonds. Each pi bond has a bond order of 1 so the resonance hybrid structure has two partial pi bonds. But then can it further be in a resonance hybrid of different rotations and so be an even cylinder?
I guess if you wanted to be silly you could say that that the electron configuration for each beryllium ion is like:
$$ \underset{2s}{[\uparrow \downarrow]} \underset{(2p)^2}{[\uparrow \vert \;]} \underset{2p}{[\;]} $$

The resonance hybrid of the two pi orientations does produce an even cylinder. You don't need to include other rotations.

@f" Suppose I had two resonant pi orientations such as in carbon dioxide. Can I freely rotate the parts around each other then?

What do you mean by "freely rotate the parts around each other"? The electron density of the molecule has cylindrical symmetry, if that's what you're asking.

Short answer: the molecular orbitals will be $\sigma$ and $\sigma^*$ and thus cylindrically symmetric.
First off, your electron configuration is wrong. $\ce{Be2+}$ will have 3 electrons, so the $\ce{2p}$ orbitals need not be considered.
Here's a quick MO diagram:

So there aren't an $\pi$ bonds at all. There's a $\sigma$ and a $\sigma^*$ orbital, for a bond order of 0.5.
If I generate orbitals (using Avogadro) then I see something like this for the $\sigma$

And this is the $\sigma^*$


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How does the London force work at absolute zero?
Consider a molecule such as pentane which is mostly attracted to other molecules by dispersion forces. If there is no heat and no fluctuations in electron positions how would a liquid or solid hold together?
I am not sure that zero point energy would work to keep this together.

Your question is unclear to me

"no fluctuations in electron positions" - that's not true. Cooling doesn't change anything in this matter.

As this is 'on hold' i can only comment. The dispersion energy is proportional to the square of the polarisability and to the ionisation energy and to inverse distance to  6th power. The polarisability is a measure of how easily a nearly (transient) dipole can distort the electrons in an atom or molecule. Iodine large effect, pentane small one. Thus the interaction energy has nothing to do with temperature.

Firstly, kinetic energy of electrons is greatest in the ground state of an atom or molecule.
But regarding your actual question, according to quantum electrodynamics, the vacuum consists of fluctuating electric fields.  The electric field polarizes the atom or molecule, even in the ground state, and there is attraction between the polarized atoms or molecules.http://george.ph.utexas.edu/~dsteck/reading/casimir/milonni94.pdf

You assume that absolute zero can be reached - but it is not
possible.
You assume that at absolute zero the molecules would have
any kinetic energy to  drift apart - but they would not have.
You  assume that absolute zero temperature means no fluctuation -  I am
not sure about that either.

In summary, you picked a non-physical state and you are trying to apply your physical intuition on it.


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Why are they called ‘dispersion forces’ if they are an attractice force?
From the name, I thought dispersion forces would be repelling or dispersive forces between atoms, and not attractive forces. What is the dispersion part of the name actually referring to? Is it the electrons moving around?

Although Danny Rodriguez has already excellently exposed what the dispersion force is in simple terms, the word dispersion still demands a better explanation in my opinion.
According to Wikipedia:

The London theory has much similarity to the quantum mechanical theory of light dispersion, which is why London coined the phrase "dispersion effect."
  In physics, the term "dispersion" describes the variation of a quantity with frequency, which is the fluctuation of the electrons in the case of the London dispersion.

So the word "dispersion" here is a mere analogy.
The original work is this one (in German).

Interesting -- so dispersion could be quantified as $\frac{\partial n}{\partial \omega}$, maybe, as a function of the electron coordinates..

@khaverim If your $n$ is electron count, then no. The number of electrons doesn't fluctuate during dispersive interactions, only the spatial part of the electron density does.

Right, density, i.e. instantaneous $n$ at a given $x,y,z$, not an integer-count of electrons in the atom.

More formally $\rho = n/V$ I suppose

It refers to the electrons that are distributed (or dispersed) in the electron cloud which then ensues an attraction between atoms or molecules.
Electrons can be dispersed in any number of different ways in an electron cloud. One way, however, is where all the electrons appear in one side of the electron cloud. When this polarization happens, the atom has a negative charge on one side (where the electrons have gathered) and a positive charge on the other side.
This induces neighboring atoms to polarize the same and therefore attract to each other in a negative-positive-negative-positive chain.
A good example would be helium.


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interaction of grape juice or wine with marble
I have a marble cup which I drink wine from. If I leave a little bit of wine in the cup and wait maybe 30 minutes, the wine becomes discolored and appears black (photos attached). My conjecture is that this is the acid from the wine interacting with the calcium carbonate of the marble to produce calcium choloride? If so, is it toxic to continue drinking wine from the cup? Thanks.

No. You're fine. People have been drinking wine from vessels made from calcium carbonate to glass to various metals since, well, the dawn of wine. Your precipitate or concentrate is not hazardous to your health but might be unpleasant to taste. I recommend diluting it with more wine and continuing on as you were.

Romans had their drinking water arrive to them in lead pipes. As a result they lived shorter than they could have.

Todd - thanks for your comments. Do you happen to have a source for the precipitate or concentrate being non-hazardous? I am curious why I couldn't find any mention of this anywhere on the Internet.

@AdamKapelner - Not offhand, but it could also be something from the wine, too, which I would expect to be non-hazardous. Not all of them are filtered to the same degree and you might be seeing some leftovers from the wine. If you want to get a little more quantitative, I suggest filling the cup with water and testing both the water and the water that has sat in the cup - there are cheap water testing kits such as this one which might put your mind at ease.

@ToddMinehardt - tested it with the product which comes with bacteria, pesticides and lead tests. No lead... Any other toxic compounds I should be worried about? Can you think of any other byproducts of wine + marble I should be concerned about?

@AdamKapelner - In my opinion, you've covered the bases pretty well. There is an off-chance that some mineral was metamorphosed along with the other (main) constituents of the marble, but that's getting into some pretty fine-grained details that would entail (at least) looking at the composition of the marble itself (via XRD at minimum). If I was in your shoes, I would feel comfortable drinking wine from the vessel BUT it is ultimately your choice.

CaCl2 and magnesium are not toxic at these levels. In fact NIH suggests around 1 g calcium per day: https://medlineplus.gov/magazine/issues/winter11/articles/winter11pg12.html
Marble itself is predominantly calcite or dolomite, that is carbonates of calcium and magnesium. Marble also contains other elements. We can't say if these other chemicals are safe just by a photo because composition of marble varies.
When you bought this cup - did they say it is safe to drink from it? Supplier should know if the material is safe or not. If you didn't get assurance that it is safe then you should assume it is not.

Hi sixtytrees, I bought the cup in India and I have no information about it. So I should assume it's not safe?


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Transparency and partially deuterated polymers
I've been skimming over the literature and have come into some conflicting information about the optical properties of partially deuterated polymers.  From what I've seen, in general deuterated polymers have less attenuation of visible and IR than their non-deuterated equivalents.  But I've seen some papers remarking that partially deuterated polymers also have improved optical properties while others reporting that partial deuteration increases opacity versus non-deuterated equivalents.
Is there some sort of general rule, or does it depend on the polymer in question?  The underlying mechanisms governing transparency in polymers, beyond the basics such as crystallinity, are something I haven't studied in depth.

Deuteration lowers the vibrational frequencies C-H to C-D, but does not affect C-C etc vibrations.  Thus some lowering of vibrational frequencies will make the ir region slightly more transparent by bunching more frequencies together. In the UV deuteration hardy affects the spectrum and so here the polymer should be pretty much the same. Unless there is something in the processing of polymers that is improved by deuteration I can't think of anyhthing else.
EDIT
Having read your comments & looked at the data and thought more about it I now think that its the Franck-Condon factors that are responsible. These measure the strength of a transition and for electronic transitions in visible and uv are all proportional to $exp(-\alpha (\Delta R^2/2))$ where $\alpha = \sqrt(\mu k)/\hbar$ where $k$ is the force constant, (same for CH and CD bonds) and $\mu$  is reduced mass 1.7u for CD and 0.9u for CH so the deuterated bonds will be less intense as the exponential will be smaller for CD than CH. ($\Delta R$ is the displacement between levels in ground state and excited state). In the ir the intensity is proportional to the transition frequency multiplied by the square of $ \int \psi_i x \psi _{i+1} dx$ where x is the displacement from equilibrium bond length and $\psi$ a vibrational wavefunction with i quanta. Evaluating the integral shows that a transition is proportional to $\nu ^3$ where $\nu $ is the transition frequency between levels with i to i+1 quanta. The intensity is thus reduced by substituting D for H $via$ $\nu = \frac {1}{2\pi}\sqrt{\frac{k}{\mu}}$.
As CH/CD vibrations are of higher frequency than CC and the effect should be most pronounced at higher ir frequencies and proportional to the amount of deuterated species. Typically deuterated transition frequencies are 0.715 of the protonated values compared to the expected $1/\sqrt(2)$ =0.707, considered to be due to effects off cubic and quartic terms in the potential. This means that the protonated transitions are 2.7 times more intense than deuterated at approx 2900 cm$^{-1}$, the CH frequency in methane.

Thanks for giving an answer.  :)  However, it's not just a slightly higher transparency (vis. and IR), it's a quite massive difference.  For example, page 10, figure 2 here for attenuation rates in PMMA fiber:  http://tinyurl.com/j9sgmun - d8 is fully deuterated, while d5 is partially deuterated.  I'm having trouble finding the reference that I read earlier that talked about reduced transparency in some cases, unfortunately... I think it had to do with changing the crystal size.  Or maybe I misread it...

I'm just trying to get a handle on what one can "expect" with X% deuteration of a transparent, high degree of crystallinity plastic or X% deuteration of a transparent, amorphous plastic, where X is, say, 0%, 2%, 20%, or 100%.

Excellent answer.  But I think you're still missing one aspect (I'll be choosing your answer regardless, I'm just trying to get a grip on the complete picture  :)  ).  A major factor concerning transparency in polymers is the degree and nature of its crystallinity.  Low crystallinity polymers tend to be transparent while high crystallinity polymers tend to be opaque, as light scatters at the crystal/amorphous boundaries. But crystal size matters, too, as the scattering (Bragg scattering, right?) is limited by wavelength. How might deuteration alter crystallinity (size and degree)?

Scattering in general  is due to the different refractive index of one medium compared to another, e.g. polymer beads are visible in a beaker, but fill with water, or a solvent with a closer refractive index to the beads and they 'disappear'.  So if the crystalline defects have a different refractive index than the bulk polymer then you expect scattering. I'm speculating now as I've no experience in polymer fabrication, but the deuterated polymers may be better for a simple reason; they are expensive to make so the starting material may be purer and more care may be taken over its manufacture.

Diffraction really does require some order in the material, such as a $periodic$ change in refractive index.  Light is not diffracted from crystals as  are x-rays. Only if a sufficient number of the crystals were ordered in a periodic way in your polymer would diffraction occur. A random order and orientation would just give light scattering.

Aha, I think I found the key as to why a paper I'd read earlier reported worse transparency with partially deuterated than non or fully deuterated! tinyurl.com/jrawleo - in PE, for example, deuterated melts at 135C but protonated melts at 129C, so the melt fractionalizes as it cools if not rapidly quenched. Okay, I think I've got all of the pieces now.  Thanks!

Just to be clear - the melt issue only occurs with mixtures of deuterated and non-deuterated molecules, not when all of the molecules have roughly the same D/H ratio. Also, re. the earlier comment: polymer crystals are usually denser than amorphous (ex. PE crystals = 1,01g/cm^3, amorphous = 0,85), and density affects refractive index. Also there is (kind of) periodic patterning within crystals (birefringence too) - although I was talking about general scattering, not diffraction.

glad you sorted it out :)


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How does metallic bonding in alkaline earth metals work?
Metallic bonds are said to be delocalised.
Lithium crystals have each lithium atom in direct contact with eight other lithium atoms. Each atom could therefore be in a hybrid of eight chemical bonds with each other.
But a similar explanation does not work for atoms like beryllium.
Some possible arrangements of electrons that beryllium could be a hybrid of configurations of might be:
$$\ce{Be^{+} + Be^{-}}$$
$$\ce{Be^{2+}_2 + Be^{2-}_2}$$
$$\ce{Be^+ + e^-}$$
$$\ce{Be^{+}_2 + e^-_}$$
$$\ce{Be^{2+}_2 + 2e^{2-}_}$$
$$\ce{Be^{2+}_2 + e^{2-}_2}$$
$$\ce{Be^{2+}_2 + e^{2-}_2}$$
But beryllium is diamagnetic and all these suggest paramagnetic properties or a $e^{2-}_2$ pair which doesn't seem right to me although I'm not fully sure how the free electron hybrids would work.
Another possibility is that the electrons are dragged up into the p orbitals.
For example, the atoms would have valence electron configurations of:
$$\underset{sp}{[\uparrow \vert \uparrow]} \underset{p}{[\; \vert \; ]}$$
which would result in hybrids of $\ce{Be2}$ or $\ce{Be_4}$ I think.
However, the atoms would have to be dragged up really high so that they aren't paramagnetic.
Let me list out the molecular orbital theory bonding and antibonding orbitals for diberyllium.
$$ \underset{\sigma}{[\uparrow \downarrow]} \underset{\sigma^*}{[\uparrow \downarrow]} \underset{\pi}{[\; \vert \;]} \underset{\sigma}{[\; ]} \underset{\pi^*}{[\; \vert \;]} \underset{\sigma^*}{[\;]}$$
If I wanted a bonding material that was not paramagnetic I'd have to lift up the energy to something like:
$$ \underset{\sigma}{[\uparrow \downarrow]} \underset{\sigma^*}{[\;]} \underset{\pi}{[\; \vert \;]} \underset{\sigma}{[ \uparrow \downarrow]} \underset{\pi^*}{[\; \vert \;]} \underset{\sigma^*}{[\;]}$$
which is probably too high energy to matter for any purposes.

Where did you get the information that "Lithium crystals have each lithium atom in direct contact with four other lithium atoms."? Everything I find says that its crystal structure at room temperature is body-centered cubic, where each atom has eight neighbors.

That was a typo I meant 8

Metallic bonding is essentially delocalisation of covalent bonds to the logical extreme and cannot be considered as "two-centre, two-electron". The metallic lattice is a network of cations immersed in a "sea" of electrons. As the highest energy orbital is the spherical s, there is no directional influence and hence why beryllium has the efficient hexagonal close packed structure.

But then why is beryllium diamagnetic? If beryllium is like $\ce{Be^+}$ in a sea of $\ce{e^-}$ then it should be paramagnetic. If it is like $\ce{Be^{2+} + 2e^{-}}$ shouldn't it still be paramagnetic?

The two electrons have opposite spins, so there is no overall magnetic moment.

So it is $\ce{Be^{2+} + e{2-}_2}$! But I thought pairs only happen in superconductors with Cooper pairs?

Most lanthanoids have $Ln^{3+}$ + 3$e^-$.


text:
Which is hardest: iron, brass or bone?
I was hopping around random wikipedia articles when I came across the article for the Behemoth. In the description for the beast it says:

His bones are as strong pieces of brass; his bones are like bars of
  iron

So it got me thinking, which of these three substances is hardest: iron, brass or bone?
(I had a quick look at the Mohs scale, which lists iron as 4, but could not find anything for brass or bone.)

Please note that the general meaning of "hardness" is much broader than the Mohs scale, which basically measures "resistance to scratches". Hardness includes  ductility, elastic stiffness, plasticity, strain, strength, toughness, viscoelasticity, and viscosity (list copied from Wikipedia), all of which have specific meanings and little to do with Mohs hardness.

There is quite a difference between a strong material and a hard material.

https://en.wikipedia.org/wiki/Strength_of_materials

@GuntramBlohm I think if I asked for all those things the question would be closed as too broad.

@Johannes_B It's probably a bit late in the day to change the question, so I'll make do with this now. It's clear not that many people are interested anyway.

Thought this was interesting: Modulus of rigidity.  Bronze is higher than iron which raises a question.  Why did iron tools replace bronze tools.  Suspicion: iron is more flexible (less brittle?).  (Some sources saying that the problem with bronze is the need to make an alloy, but no advantage in the edge, hardness or moduli.

These two sources both put bone at a hardness of 5:
http://www.chacha.com/question/how-hard-is-bone-according-to-moh's-hardness-scale
https://answers.yahoo.com/question/index?qid=20110310200841AABwtMj
Whether they are trustworthy is questionable though, so take it as you will.
This source put brass at 3 and iron at 4.5:
http://www.jewelrynotes.com/the-mohs-scale-of-hardness-for-metals-why-it-is-important/
and this image puts brass at 4 and iron at 4-5 (Similar to 4.5):
http://patentimages.storage.googleapis.com/WO2001048807A1/imgf000009_0001.png
While these different sources seem to have conflicting data, I think it would be safe to assume that Brass is the softest of these three materials, Iron comes second, and Bone is the hardest.
Edit: In the description of that monster, the adjective used is 'strong'. You may want to consider how much force each of these materials can withstand instead of how hard they each are :)

Bone is primarily composed of hydroxyapatite. Note that the apatite minerals define the value 5 on the Mohs scale.

Right; the composite material (bone) is much better for load bearing vs weight, and scratching it is not an issue.


text:
Is pentamine or pentylamine the correct name?
For $\ce{CH3CH2CH2CH2CH2NH2}$, is the correct name pentylamine or pentamine?
From http://antoine.frostburg.edu/chem/senese/101/organic/faq/yl-ending.shtml

"The -yl ending means "a fragment of an alkane formed by removing a hydrogen".

How does the above definition apply to this problem?
Does it mean, if the N wasn't there it would be a carbon and 3 hydrogens would attach to the carbon; but with the nitrogen there there is only 2 hydrogens attached to a the nitrogen?

Thank you

Related: Nomenclature of Amines and Can there be two IUPAC names of a compound?

The following IUPAC guidelines suggest the correct systematic nomenclature for simple primary amines.

P-62.2.1.2
Primary amines are systematically named in the following ways:
(1) by adding the suffix ‘amine’ to the name of the parent hydride;
(2) by adding the name of the substituent group R - to the parent hydride ‘azane;
(3) by adding the name of the substituent group R - to the term ‘amine’ used as a preselected parent hydride name for NH3; this method is used only with monoamines.

Method (1) gives the preferred IUPAC name (PIN). So in your example this would give pentan-1-amine.
Method (3) gives what is often the more commonly used name, in your example this would give pentylamine.
Both names are 'fine' and in reality method (3) is the one that you will most often see used/written on bottles of reagent etc. But for the correct IUPAC identifier method (1) should be used.
You should also note that pentamine is a wholly different compound:

Other example:

By method (1). Methanamine
By method (2). Methylamine

The last image shows up as a completely black square for me.

@f''. I do indeed mean method (3).  There was a discrepancy between my initial post and the 2013 IUPAC recommendations (thats to Loong for pointing it out), just forgot to update the link.        Image looks fine to me? Will try re-adding it anway

Thanks for the response. FYI It is showing me this black square @NotNicolaou http://i.stack.imgur.com/SsZ5ks.png

A wikipedia search of each chemical confirmed that it is pentylamine.
The name pentylamine has two parts. The first part, pentyl, means that there are 5 carbons with 11 hydrogens attached to them. This is a pentyl group https://en.wikipedia.org/wiki/Pentyl
The second part, the amine, is the NH2 part. This is when an ammonia (NH3) molecule loses one hydrogen and instead bonds with the organic group, in this case the pentyl group. There are only 2 hydrogens attached to the nitrogen because nitrogen has a charge of -3, whereas carbon has a charge of -4 (in this case, not in all cases).
Pentamine wikipedia page:
https://en.wikipedia.org/wiki/Pentamine
Pentylamine wikipedia page:
https://en.wikipedia.org/wiki/Pentylamine

To the name pentamine, which you asked for (and probably meant pentanamine). Others have already mentioned, that it's a trivial name of $\ce{MeN(CH2CH2N^+Me2Et)2}$. (I'm not sure that it's correct, some other source sources claim “Pentamin” trade name, for its bromide salt.)
(There's also cyclopentamine, which is a trivial, INN, name of say methamphetamine analogue, with cyclopentyl instead of phenyl.)
Please note that it's also a systematic name of a group of compounds (or a part of systematic names), created from penta-, -amine, with elision of ‘a’.
E.g.

cyclopentane-1,2,3,4,5-pentamine.

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text:
Electronegativity of noble gases
SO, I stumbled upon a periodic table of electronegativity-
http://www.thecatalyst.org/electabl.html
Now I don't understand why the first 3 noble gases have blank spaces instead of the electronegativities.
If it is because their octet is complete then why doesn't the noble gases Kr, Xe and Rn have blank spaces under them?

They don't have values there because they aren't on the pauling scale of electronegativity, as they don't form any compounds with other elements. However, argon and neon can technically form compounds with other elements; it is just extremely unlikely. For example there is a compound called Argon Fluorohydride, HArF, that can exist, but is only stable at around 27K (-246C). I don't think Helium can form any compounds other than with itself.
The other noble gases also can react with fluorine (As well as other elements), and make slightly more stable compounds. Hope that helps :)

then why does  Krypton, Radon and Xenom have an electronegativity value? They are noble gases too so they shouldn't be electronegative.

Noble gases have their outer s and p orbitals filled with electrons (just s in case of He). Beginning with Argon, they also have unfilled inner (d, f) orbitals, which allows them to accept electrons, even if it's unlikely they do.

@GuntramBlohm you said beginning with argon. Then why doesn't argon have an electronegativity?

@adi maybe because with argon the outer d orbitals are not as good electron acceptors as their counterparts in the heavier atoms? We do know of at least ine argon compound (https://en.wikipedia.org/wiki/Noble_gas_compound), but argon lacks the broader empirical base in which we can establish a well defined electronegativity.  Such a database has been developed with compounds of krypton and heavier ng's.


text:
What is activating effect?
In the book "Organic Chemistry" by Morrison and Boyd,It is written that

Sodium borohydride cannot reduce weaker electrophiles like esters though Lithium aluminum  hydride can reduce esters to alcohols due to stronger activating effect of Li+ ion than Na+

What is activating effect?
(Google search did not give appropriate results.)

Consider a carbonyl group. We know that, under acidic conditions, $\ce{H+}$ can protonate on carbonyl $\ce{O}$ and hence activate the carbonyl group to nucleophilic attack at $\ce{C}$, and we call such an effect an activating effect. The metal ions $\ce{Li^+}$ and $\ce{Na^+}$ serve the same purpose as $\ce{H+}$, though $\ce{Li^+}$ is a stronger activator due to its high charge density and hence stronger attraction to $\ce{O}$.


text:
What chemicals cause the softer, less-raw taste of purified water sold in bottles?
I've seen the post, "What defines an element's taste?", but it doesn't answer my question.
I took water that is supplied by the municipality corporation and put it through a gravity based water filter. The taste of the water didn't change. It still had a raw, tap-water kind of taste.
However, if I buy one of the 20-liter cans (they charge x+y for the water+can and when you return the empty can, they return y amount of money), the water tastes softer and less tap water-ish.
A 2-liter bottle of mineral water costs x. Which is the same as the 20 liter can cost. So I'm wondering if they really purify it or whether they just add a few chemicals to it to fool people into thinking the water is purified. They might just be supplying us with ordinary tap water or borewell water.
Are there any chemicals that can change the raw taste of water? Maybe the raw taste caused by chlorine.

Water issues tend to be very specific to a certain location, so it is not possible to say exactly for sure what is happening in your situation.
Pure water has no taste.  The taste comes from the impurities.  Impurities, which are almost always present, can partially be removed by the filter shown.  That filter looks like a carbon filter which is effective on chlorine, organic particles, and sediment.  This filter is ineffective on minerals and inorganic compounds.
Minerals give tap water its hard taste. This hard taste is harmless (and may actually serve as a dietary mineral supplement.) Removing the hard taste requires a process like deionization or distillation which is costly. The cost of the water can go up significantly with more mineral removal.
Chlorine, which is used to kill pathogens, can be neutralized with citric acid.  Squeezing a lemon or lime in the water can cut down on a chlorine taste.
Hard water can be chemically softened with something like baking soda; however, this is generally trading one ion for another, so the water will only taste softer.
Try adding a reverse-osmosis filter after the filter shown. The filter shown will treat the chlorine while the reverse-osmosis filter can treat the minerals. (Chlorine may damage a reverse osmosis filter, so it is best to get it out first.)

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text:
Which is stronger, PET or PVC?
I have been given the task of designing a new wheel rim (the metal part of the wheel) but instead of metal, it has to be made out of a polymer.
I'm thinking of using either PVC or PET but I don't know if either of those plastics will be able to support the weight of a car and am not sure how to find that out.
If anyone can think of a different polymer that I could use instead that would be better suited feel free to include it as an answer. This is only theoretical so cost/difficulty to manufacture is not a big hassle.

Weight is an extrinsic variable, stress is intrinsic.  I suggest editing the question.  The polymer's ability to support the car is dependent on the weight of the car and the dimensions of the rim.  The Question you have asked is not answerable.

You need to define what you mean by "stronger".  Are you talking about resisting deflection (and thus rolling losses)?  Yield?  Break?  Impact?  Abrasion?  Lifespan?  Ability to withstand heating from its deflection as the vehicle rolls?  Also, any material can "withstand the weight of a car" if you use enough of it, do you know what sort of geometry you're talking about?
Lastly, the material properties of polymers depend a lot on the manufacturing processes and the details of the raw materials (chain lengths, branching, monomer ratios, etc) and additives.  A good example is PET.  Check out the difference in tensile strength between simple thermoformed PET and BoPET (Mylar), for example; they're vastly higher for BoPET.  And the reason is because it's elongated to near breaking at lower temperatures - thus orienting the polymer - and then given a heat treatment to relax it into its new shape.
I think you need to define your problem statement better.  Think about what properties you want, your geometry, and how you plan to manufacture it.  Also, what are the parameters you're trying to optimize?

Use PVC as it has a greater value of young's modulus:   490,000 psi
PET has a young's modulus of: approximately: approx 430,000 psi
Higher the young's modulus the greater the resistance to deformation.


text:
It is possible to make an homogeneous solution of water and oil?
I wondered if in certain condition like very high pressure(so that water and oil remain liquids) and very high temperature(so that the variation of the Gibbs energy is negative--> ∆G = ∆H-T*∆S) we can make an homogeneous solution of water and oil.
Or, alternatively, mixing water and oil meanwhile applying an intense electric field so that the oil molecules get a charge such that they become soluble in water.
Do you think these methods could work? Is there any other way to mix them?

In fact if you apply an electric field it will form a colloidal emulsion and not a true solution. If you want an emulsion then its even easier, just add a little soap to it and mix. However I don't know weather you would consider an emulsion to be homogeneous or not.

I didn't intend an emulsion. I was thinking about a solution like salt and water for example.

Generally polar and non polar liquids do not dissolve in one another to any great extent. Practically, you would have to find a temperature composition phase diagram for the oil you want to use. This will tell you if there is a region in which the water and oil are mixed as a homogeneous solution. Alternatively if you add surfactant then a ternary phase diagram (triangle plot) is what you need to look for since than micellar or liquid crystal phases may form.
As you mention the free energy I have tried to summarise what happens on dissolving a non-polar molecule in water.
The enthalpy $\Delta H$ change on placing, alkane, into water is small, slightly positive or negative depending on the compound, -9kJ/mol for ethane, +1.6 for toluene for example.  This can be thought of as the cost  making a void in the water to hold the molecule.
Water is unusual in having a strong and highly directional hydrogen bonding nature. Thus adding a non-polar molecule into water is determined by entropic effects. This generally means that $T\Delta S$ is negative and so $\Delta G$ is positive, for cyclohexane into water $\Delta G$ = +28 kJ/mol.
The entropy of water decreases on contact with non-polar molecules since these molecules cannot form hydrogen bonds. The water has to bridge around the non-polar molecules, as this is a lowering of entropy compared to not bridging. This bridging partly restricts the water molecules compared to bulk water. This option is preferable as the alternative of breaking hydrogen bonds, and leaving H atoms pointing at the non-polar molecules,is too energetically costly. Bridging makes it possible to form some H bonds at the expense of entropy loss. The figure shows this (from Southall et al. J. Phys. Chem. p521, 106, 2002 )

As the temperature increases the enthalpy increases as does T$\Delta S$, but $\Delta G$ remains largely constant and remains positive, see sketch below. This suggest that  water and oil will not easily mix.


text:
Cyclopentane conformations
What is the energy minimum of cyclopentane conformational diagram: the halfchair or envelope? How does cyclopentane flip?

As you mention, there are two main conformations that cyclopentane may adopt.

In the absence of substituents around the ring, there is actually only a $\pu{0.5 kcal/mol}$ energy difference between envelope and half-chair (envelope slightly lower in energy) meaning cyclopentane itself (computationally at least) is in rapid equilibration with no 'major' conformation as is observed with cyclohexane. (Note that the energy of the totally planar conformation is about $\pu{5 kcal/mol}$ higher than that of the both bent conformations, and so is not observed to any great degree).
With a single substituent on the ring however, such as a methyl group, the envelope conformation becomes slightly more favourable (to the tune of around $\pu{3.5 kcal/mol}$) giving a barrier to the ring flipping and meaning there is a dominant conformation.

"In the absence of substituents around the ring, there is actually only a 0.5 kcal/mol energy difference between envelop and half-chair meaning cyclopentane itself" In favor of which one?

Are in the envelope conformation the C atoms in a plane, except the rightmost one?

Yes. So in the envelope conformation just one carbon is bent out of the plan

Why do you show two C$_s$ envelope conformations? Is not the left one the same as the right after a rotation.  As all the 5 possible envelope-half-chairs are equally probable surely only one envelope and half chair is needed ? As thermal energy is approx 2.4 kJ/mol  interconversion should be very rapid, I'm guessing but $10^8$ s$^{-1}$ would not be unreasonable.

The left envelope is equal to the right. I think the diagram was just intending to show that to get between two different envelope conformations you have to go via the half chair, which, for unsubstituted cyclopentane has little barrier is fine.

Which one is the most stable: envelope or halfchair?

Envelope, but only by 0.5 kcal/mol (computationally). The take home message is that in the absence of substituents there isn't really a 'most stable' conformation, its in totally rapid equilibration.


text:
Reaction between RCH2=CH2 and CCl4 -- Why does it form RCHClCH2CH(R)CH2CCl3 along with RCHClCH2CCl3?
I am reading Organic Chemistry from Morrison & Boyd and there is this problem in it:

Problem 6.8 From the addition of $\ce{CCl4}$ to alkenes (in presence of peroxides), $\ce{RCH2=CH2}$, there is obtained not only $\ce{RCHClCH2CCl3}$ but also $\ce{RCHClCH2CH(R)CH2CCl3}.$ Using only the kind of reactions you have already encountered, suggest a mechanism for formation of this second product.

I know that the addition of CCl4 to alkenes in presence of peroxides is a free radical reaction. So what I have thought is, while the reaction is going on, there are the following species present: RCH=CH2 (reactant), RC^HCH2CCl3 (^ represents the electron on the free radical) and the product $\ce{RCHClCH2CCl3}$.
When the free radical reacts with $\ce{RCH=CH2}$ it gives the main product $\ce{RCHClCH2CCl3}$ and $\ce{.CCl3}$. So probably the second product is formed by reaction of $\ce{.CH(R)CH2CCl3}$ with $\ce{RCHClCH2CCl3}$?
$$\ce{RCHClCH2-CCl3 + .CH(R)CH2CCl3 ->[???] RCHClCH2CH(R)CH2CCl3 + .CCl3}$$
However this reaction I have never encountered while the question explicitly states that I have to use only the kind of reactions I have already encountered. I don't think anyway that this reaction is possible.
There must be something big that I am missing here ... Can someone give me a little hint?

Is the chain mechanism not given directly above the question on the same page?   Using that you should be able to rationalise where an issue of selectivity might arise to give the various product.

@NotNicolaou Yes, it is given just above that and I have read it but still I don't get it.

google://radical+polymerization

@permeakra I've searched google but nowhere is this product of the reaction mentioned. If you know the answer why don't you post it?

@Kartik Read about the mechanism of radical polymerisation and apply it in your particular case. I don't do it, because it will teach you nothing and I'm lazy.

@permeakra I know the mechanism already. However I dont get how it leads to formation of this product. I have tried a lot to think about it myself but I couldn't. If you just give me a hint probably I will be able to do it.

@Kartik hint: alkene does not care much, which radical will attack it, as long as it is active enough. Alkyl radicals in general are quite active.

The propagation series starts with the addition of the trichloromethyl radical to the alkene to give the more stable secondary radical. This radical can abstract a chloride from carbon tetrachloride to give the expected product. Alternatively, this radical can add to the alkene to give an a new carbon-centered radical with an extended chain. Abstraction of chloride gives the 'unexpected' product.

The key realization here, as permeakra suggested in the comments, is that any radical in solution can theoretically add to the alkene. This means that the penultimate radical intermediate could add to the alkene as well extending the chain. This is the basis of a radical polymerization.

Thank you. I had found the answer myself a few days ago but I didnt have internet at that time so I couldnt reply.


text:
Does any known substance ignite on cooling?
As the title says, I'm interested in knowing if there is any substance — or combination of substances — that ignites (or even increases its chance of spontaneous ignition) when cooled.
I've never heard of such a thing, nor can I find it in Atkins' Physical Chemistry, but I might easily have overlooked it, since I don't know what to call it. Googling gives me information on substances that lower the freezing point or ignition point of explosives/fuels.
It seems that this should obviously be forbidden on thermodynamic grounds, but I can't quite rule out a phase change allowing such behaviour, for instance.

Ignition is about kinetics, not thermodynamics. You can't cheat thermodynamics, but there may be a chance with kinetics. True, most reactions slow down upon cooling, but this is not an absolute law.

I think a lot of substances would ignite if you cool them down from a temperature initially so hot that molecular bonds can't form, but I'm not entirely sure the process would be classified as ignition.

Actually... yes! Iron(II) oxide is thermodynamically unstable below $848~\mathrm K$. As it cools down to room temperature (it has to do it slowly) it disproportionates to iron(II,III) oxide and iron:
$$
\ce{4FeO -> Fe + Fe3O4}.
$$
The iron is in a form of a fine powder, which is pyrophoric (it may catch a fire when exposed to air). You can see it in action here.

For those of us coming from the hot questions list, 848 K is 574.85 C or 1066.73 F. =)

When actually cooled all the way down to room temperature and left alone, $\ce{FeO}$ will stay like that till kingdom come. I guess the disproportionation is possible in a relatively narrow range of temperatures.

@IvanNeretin That's true, you have to cool it slowly. Like all chemical reactions, the rate of the disproportionation drops at lower temperatures.

I doubt ignition could ever be favored by cooling since, as for any combustion you need heat. Also the compound must emit enough vapors for ignition to take place and cooling, of course, goes in the opposite direction.
Some explosives, however, could well detonate (not ignite) if the temperature causes them to crystallize. Diazonium chloride salts, for example, are usually not isolated in their crystalline and dry form because they can cause explosions in such a state, while they are safe in aqueous solution.

Phosphorus vapor
Red phosphorus heated at the atmospheric pressure up to approx. $\pu{400 °C}$ sublimates, and upon cooling it builds up white phosphorus:
$$\ce{P_n ->[Δ] n/4 P4}$$
which spontaneously ignites in air at the room temperature being deposited as a film on the surface (bulk sample ignites at $\pu{50 °C}$):
$$\ce{P4 + 5 O2 → P4O10}$$
which is sometimes observed in a form of glowing light (chemiluminescence), as seen, for example, on the famous painting "The Alchemist Discovering Phosphorus" by J. Wright:

A published procedure for the reference [1]:

PREPARATION
  Obtain two dry Pyrex test tubes, one 6-inch and one 8-inch. To the 8-inch test tube add 0.25 g red phosphorus. Fill the 6-inch test tube 1/2 full of cold water, dry the outside, and insert it into the larger tube. The hot tip of the smaller test tube will be supported by the
  neck of the larger tube.
DEMONSTRATION
  Heat the red phosphorus in the larger tube until a deposit appears on the cold surface of the inner tube. Allow to cool. Upon removal of the smaller tube the white phosphorus on the bottom mill ignite.

Plutonium
Bulk plutonium ignites only above $\pu{400 °C}$. However, when cooled down, it undergoes a series of phase transitions and its density is increased by approx. $11\%$:

This causes a large hazard issue as plutonium sample cracks while cooled and develops a large surface that swiftly reacts with traces of moist in the air forming plutonium hydrides $\ce{PuH_{2...3}}$ and plutonium(III) oxide $\ce{Pu2O3}$ which are both pyrophoric and ignite spontaneously at ambient temperatures.
References

Brodkin, J. Preparation of White Phosphorus from Red Phosphorus. Journal of Chemical Education 1960, 37 (2), A93. https://doi.org/10.1021/ed037pA93.1.

It depends on how you cool it.  For example, if you try to cool an alkali metal by immersion in water, you can get a rapid exothermic reaction.

But that's more than just cooling, it's a completely new reagent.

@fafl That's true, and it's the new reagent rather than just the temperature drop which causes the reaction.  However, immersion in water is one cooling technique that a lot of people are familiar with, and that specific act of cooling the right/wrong substance would generally cause ignition.

There is also the property of some salts to be cryoluminescent, which means they produce light when they are either precipitated upon cooling down a solution or when the filtered compounds crack upon cooling down. In this case, this is also an electronic transition which produces light, but it is reported to slightly warm up in this process as well. So perhaps you can have an excited state high enough to cause a reaction? Would be the only guess I have.


text:
What is a "lower organic acid"?
From the "Technical guide for the elaboration of monographs" by EDQM, edition 7, 2015, page 25:

Where the counter-ion of an active substance is formed from a lower organic acid, a test for related substances of the organic moiety is usually not considered necessary (for example, magnesium lactate used as a source of magnesium).

What is the meaning of "lower organic acid"? An organic acid of simple composition, with a short alkyl group?

Probably a lower molecular weight organic acid, generally the ones that are soluble in water.

The term doesn't seem to have an official IUPAC definition however several sources agree that the lower organic acids are those that are soluble in water.
In the simplest case this could just refer to the first 5 carboxylic acids (C1-C5): methanoic (formic) acid, ethanoic (acetic) acid, propanoic acid, butanoic acid and pentanoic acid.
What is less clear is whether small water soluble acids such as lactic acid  and tartaric acid are counted. (The context of the quote in your question suggests it may well be, though it's wording isn't overly clear)
Since the phrase has no official definition I think it's probably safe to use it as you see fit, ensuring you define what you plan on calling a lower acid and use it consistently. You will equally see people refer to lower amines or halides, which in the same way isn't a well defined phrase but just lets people vaguely define a group of compounds.


text:
How does ice formation glue things together?
Today, I took two crumpets out of the freezer, which had become stuck to each other where moisture between them had frozen. They were two separate crumpets, the top one stuck fast to the bottom one, so they didn't go into the freezer attached to each other in any way.
And it made me wonder, what causes the crumpets to become stuck? Why don't you just have a crumpet, a thin sheet of ice, and then the other crumpet? What's going on at the chemical level that explains why these two crumpets became stuck to one another?

The crumpets you have frozen together are very porous and convoluted at a microscopic level. One would then imagine that this means that any water vapour condensing on them as they start to freeze will percolate somewhat into the surface of the material and then freeze. The interior will stay warmer longer than the surface so this will help percolation. (Imagine an ink drop spreading into paper and you get the idea of percolation). On freezing water crystals become connected and thus many of these stick the two parts together. Rather like a sort of very fine velcro. However, most all foodstuff already contains a lot of water surrounding lipids, proteins and carbohydrates. Thus hydrogen bonding of condensed water to these intrinsic waters may also be important at the molecular level.


text:
When to use Da or u for mass units
Following up on the question Units of mass on the atomic scale,
Daltons (Da) and unified mass units (u) are equivalent, but which is used seems to depend on context.  For example, when discussing single atoms one typically sees the mass expressed as u, but when discussing large biomolecules one typically sees Da or kDa.
Is there a rule about whether to use Da or u? Or, is this simply a matter of convention or left to the writer's choice?

See also this question: http://chemistry.stackexchange.com/questions/54388/how-to-retrieve-dalton-value-from-m-z-value

It doesn’t matter. Both units are basically identical, much like there is no everyday difference between the units mbar and hPa. There may be situations where one is preferred — e.g. hPa are, in my opinion, rarely used outside weather reports — but they should be understood nonetheless.
If there is anything to be said about the two, it seems using dalton is more popular in biologic contexts. But that wouldn’t be a tendency to base anything on.
Being consistent within one’s lab, within the journal one is publishing in or throughout one’s thesis is far more important.

Your second paragraph is more along the lines of what I was asking.  Roughly, it seems to me that physical chemists tend to use u and biochemists use Da, with everyone else using a mixture.  Does that sound right?

@TedCorcovilos Well, semi. As I wrote, Da is rather popular among biologists. But I wouldn’t want to propose any preference of chemists (outside using molar mass and grams per mole). Then again, I don’t have anything to do with physical chemists, maybe I’m biased.

Dalton (Da) is used to characterize polymers. For example, biochemists and molecular biologists specifically to discuss biopolymers such as nucleic acids or proteins.
Usually Da appears in the context "a 60 kDa protein band was observed on SDS-page" or "10 kDa cut off membrane was used for dialysis".
same applies to synthetic polymers.

This is mostly incorrect.  "Da" appears in a much wider variety of contexts than just in biochemistry and molecular biology.

@Curt F. Why don't you write about it if you know the correct answer?


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What is the difference between the structure of nitrito-O and nitrito-N?
I am preparing for chemistry olympiad and I am stuck nitrito-O and nitrito-N. Can anyone explain difference between the structure of nitrito-O and nitrito-N or point a resource which is suitable for a beginner?

http://chemwiki.ucdavis.edu/Core/Inorganic_Chemistry/Coordination_Chemistry/Isomers/Structural_Isomers_in_Inorganic_Molecules#Linkage_Isomerism or https://en.wikipedia.org/wiki/Linkage_isomerism

Please specify your question. Are you having trouble with resonance structures, o-,m-,p- naming or sigma and pi- electron-withdrawing/donating effects?

The last letter of nitrite-O and nitrite-N depicts the atom which is attached to the other atom. This phenomenon is called linkage isomerism and is an important concept in both organic and inorganic chemistry.
Inorganic chemistry

This complex exhibits linkage isomerism which results in two possible ways of attachment of ligand to the central atom. First one, the $\ce{NO2}$ ligates with (attaches to) the central atom through the nitrogen atom and thus is called pentaamminenitrito-N-cobalt(III) chloride and the second one,  $\ce{NO2}$ ligates with the central atom through the oxygen atom and thus is called pentaamminenitrito-O-cobalt(III) chloride. In this case, $\ce{NO2}$ is called ambident ligand and is represented as nitrito-κ-N and nitrito-κ-O, replacing the old system of trivial names such as nitro and nitroso.
Organic chemistry
Alkyl nitrite forms two isomers: $\ce{RNO2}$ (called alkyl nitrite-N) and $\ce{R-O-N=O}$ (called alkyl nitrite-O).For preparation of these two isomers, see Reaction of an alkyl halide with silver nitrite.


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A cereal sample was analysed for Calcium content as follows:
A cereal sample was analysed for Calcium content as follows: A 5g cereal sample was digested with HCl and the resulting mixture made up to 50ml in a volumetric flask. The absorbance of this solution was found to be 0.351. A 7ppm calcium solution gave an absorbance of 0.254. Assuming linearity between absorbance and concentration, calculate the calcium content of the cereal in mg Ca/100g cereal. (8 marks)
Seems so simple but is confusing me as when I take into account calcium content of the cereal in mg Ca/100g cereal.
A 5g cereal sample was digested with HCl and the resulting mixture made up to 50ml in a volumetric flask.
The absorbance of this solution was found to be 0.351.
A 7ppm calcium solution gave an absorbance of 0.254
7 ppm Ca solution/ absorbance 1 x absorbance 2 =
step 1: 7 ppm/ (0.351 x 0.254) = 78.515 ppm ca in final dilute sample.
step 2: 78.5158/50 ml  (of initial volume sample)= 1.57 ppm ca in the first solution.
Supposing the density of the first solution to be near that of water:
step 3: (1000g) x (1.57 x 10^-6) = 1564 g Ca
step 4: (1564 g ca / 5g) = 312.8 g/100 g
answer: 3.128x10^5 mg/100 g would be the initial answer I receive.
My second attempt looking at it a different way...
feeling that I have done step 1 correctly.  Starting at step 2
78.5158/(50 ml x 2) (multiplying initial volume by 2 to achieve 100 ml)
= .785158 ppm ca in 100 ml.
step 3: (1000g x 7.85x10^-4g)=7846 g ca /100 ml
7846000 mg ca / 100 ml
answer: 7.846 x 10^6 mg ca/ 100 ml
third attempt:
Seeing that the initial sample is 5 g and the question is looking for calcium content in a 100 g sample...
5g sample x 20 = 100 g
therefore other components are multiplied by 20
V= 50ml x 20 = 1000 ml
step 1: step 1: 7 ppm/ (0.351 x 0.254) = 78.515 ppm ca in final dilute sample.
step 2: 78.5158/1000 ml  (of initial volume sample x20)= 0.0785 ppm ca in the first solution.
step 3: (1000g) x (.0785 x 10^-6g) = 72.5 g Ca
step 4: (72.5 g ca / 100g) = .725 g Ca
answer: 725 mg Ca content/ 100 g cerial
I feel that I am missing a step in calculations.  I would prefer to read a resource of working this out myself than finding the answer.
I feel more confident in my third attempt of finding a correct answer.
This question is from a past paper.

The absorbance of this solution was found to be 0.351.
A 7ppm calcium solution gave an absorbance of 0.254
You have slightly (~1.5 times) higher absorbance (.35) than a solution of 7 ppm calcium (0.24)

step 1: 7 ppm/ (0.351 x 0.254) = 78.515 ppm ca in final dilute sample.

This is not correct. You should have proportion
7/0.254 = x/0.351   leading to x = 7*0.351/0.254
You get 10 times higher value. This is the first error. You should have done 7*(0.351/0.254) = 9.67 ppm.
In 50 mL it is A, but when it was in 5 mL it was 50/5 = 10 times higher.
7*(0.351/0.254)*(50/5)=96.7 ppm.
This means you have 96.7 mg calcium per 1 kG of cereal (ppm is one millionth, and thus, mg per kG). Then one tenth of that amount of cereal (0.1 kg/1 kg) contains 1/10 of the calcium. that is 9.67 mg.
Once again, you should have had these steps:
So, you should have these steps:
(1) Since the final sample absorbs .35/.254 ~1.38 times more it is 1.38 times more concentrated than 7 ppm. That is 9.67 ppm.
(2) It was diluted 50/5= 10 times, so it was 10 times more concentrated before. That is 96.7 ppm.
(3) For 100 g of cereal: 100 000 mg * 96.7 /1 000 000 = 96.7 mg /10 = 9.67 mg.

Thank you very much sixtytrees.  I Appreciate it.  Are there any resources that you recommend?

I have made another question.  I really hope i've wrapped my head around it. [link] (http://chemistry.stackexchange.com/questions/55335/a-cereal-sample-was-assayed-for-iron-content-as-follows) @sixtytrees

@sixytrees I'm just wondering where 5 ml is coming from in the above workings.


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Reaction of methacrylic acid with BH3
If methacrylic acid (2-methylprop-2-enoic acid) is reacted with first $\ce{BH_3/THF}$ and second  $\ce{H_2O_2/HO^{-}}$ then $\ce{OH}$ group is attached at less subtituted carbon.
But if it is treated with $\ce{CH_3COOH}$ in the second step, what is the end product?

Are there any other conditions for the second reaction, like heating or acid catalysis? If that were the case then I would suggest an ester as the product, after a reduction to the primary alcohol by borane in the first step.

Last step requires a nucleophile. It can be water, base or acetate (from acetic acid. Nucleophile will attach to terminal CH2 group. In your case it will be acetate.

A nucleophile is required, yes. But to form the boron-ate complex shown in my answer. Also, @sixtytrees notice that the nucleophile in the first step of the question is not water it's hydrogen peroxide with base to give a peroxide ion. This forms an ate complex and there is a migration followed by hydrolysis to give the alcohol. NOT a substitution.

It's simply a protonolysis:

Except in the picture D is used instead of H. Hydroboration based reactions always proceed via this boron-ate complex (negatively charged) then migration and loss of leaving group. Search google for carbonylation, amination, cyanidation using boron.

Can you show me the mechanism of my reaction

It's exactly the same. The double bond is the thing that matters. Perhaps I should have made it clear that there is and initial syn addition of BH3 across the double bond. This is called hydroboration. The reaction in my answer starts with 1-methyl-cyclopent-1-ene.

so what is the role of compound which I have shown in image

It contains the double bond which is what reacts with bh3.

The acid would be reduced by the borane to the primary alcohol first.  Subsequent hydroboration of the product allylic alcohol would probably give you the diol.

Can you please draw the mechanism as i am not able to understand like this proparly

Whilst I think this is probably true, it doesn't answer the second part of the question about what happens with 1. $\ce{BH3/THF}$, 2. $\ce{CH3COOH}$.

@bon yes I also think you are correct


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Role of acetone in the Finkelstein reaction
What is the role  of acetone in the Finkelstein  reaction? It has something to do with Le Chatelier's principle but I am not able to understand why it will not proceed without acetone.

Finkelstein reaction is a halogen exchange reaction between haloalkane and a salt of a different halogenide. $\ce{R-CH2-Cl + KI -> R-CH2-I + KCl}$. The driving force in this reaction is precipitation of $\ce{KCl}$. It is important to find a solvent that would dissolve $\ce{KI}$ better than $\ce{KCl}$. Then solution will always have $[\ce{I-}] > [\ce{Cl-}]$ (more iodide anions than chloride anions). As a result this equilibrium will be driven to $\ce{R-CH2-I + KCl}$. Acetone is exactly this type of a reagent.
For this case Le Chatelier's principle means that reaction will be driven to the direction in which one of the reagents gets outside of the solution. Since $\ce{KCl}$ keeps precipitating the reaction will be driven to the desired product.
This reaction is important because there are many ways to make $\ce{RCH2Cl}$, but making $\ce{R-CH2-I}$ is difficult. Finkelstein reaction allows to convert available $\ce{R-CH2-Cl}$ to $\ce{R-CH2-I}$.


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What type of bonding is there among d-block metals?
Metallic bond strength in $\mathrm{d}$-block elements increases up to the middle and then decreases. Why does pairing decrease the strength of metallic bond? What does pairing have to do with metallic bonding?
Some books [1, p. 1129] say there is covalent bonding, which is related to unpaired electrons, and that is why their inter-atomic interaction strength increases up to the middle:

Hardness of these metals indicates the presence of covalent bonds. The unpaired $d$-electrons may overlap to form covalent bonds. The number of such covalent bonds would be regarded as a measure of overall bond strength and hardness of the metal. In a transition series tire number of unpaired $d$-electrons increases on moving from group IIIB $[(n-1)d^1,ns^2]$ to VIB $[(n-1)d^5,ns^1]$ after which pairing begins.
  ...
  The high melting and boiling points of transition elements show that they arc held by strong forces. This is due to the presence of (a) strong metallic bonding because of the overlapping of $(d-1)d$ orbitals, and (b) covalent bonding by the unpaired $d$-orbital electrons. As Zn, Cd and Hg are having completely filled $(d—1)d$-orbitals, their atoms arc not expected to form covalent bonding amongst themselves; hence they arc having relatively low m.p. than the other $d$-block elements.

Are unpaired electrons taking part in covalent bonding or metallic bonding? Why can't paired electrons take part in metallic bonding?

Gurdeep, R. Advanced Inorganic Chemistry, 3rd ed.; Krishna Prakashan Media, 2008; Vol. 1. ISBN 978-81-87224-03-7.

Some educational reading https://en.wikipedia.org/wiki/Electronic_band_structure

Comments are not for extended discussion; this conversation has been moved to chat.

Unpaired electrons are not the only thing that are used for bonding in d-block elements. The outermost ns electrons are also present for bonding.
     and when you ask what type of bonding is present in d-block elements, it means you mean to ask the type of bonding present only amongst these metals, and not with some other element(like p-block elements, say F, Cl). If that is the case, then only metallic bonds are present in d-block elements. For example, copper atoms are held by metallic bonds in the metal crystal.
The factors affecting the strength of a metallic bond are-

No. Of valence electrons (more number means higher attractive forces, and hence stronger bond)
Distance of ionic core from the valence electrons (lesser distance means a smaller size of the metal ion, and hence higher value of the force)
Presence of vacant d-orbitals (the free valence electrons in metal travel through the metal crystal using these vacant d-orbitals. So, if there are more number of vacant d-orbitals, the free electrons are more easily "accessible" for attraction by the ionic cores)

well data disagrees with your first point specifically for d-block elements.. Strength of metallic bond depends only on unpaired valence electrons.


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In Chemistry, is it possible to use the old representation of the atom as a nucleus surrounded by orbiting particles?
In Physics, Newton's laws are enough for many applications. Sometimes, relativity must be used instead or complementary, and sometimes, Quantum Mechanics.
Is this the same case with Chemistry? Are there areas of Chemistry where the simplification of electrons as little particles orbiting the nucleus at distances proportional to their energy level is possible? Or does all of modern Chemistry must use Quantum Mechanics? At what point Chemistry "breaks" if the particles and orbits model is used?

Well, electrons are "point" (we don't know how small) particles, but they have quite strong wave properties and that fact itself isn't particularly useful.

How accurate can chemical reactions be predicted (heat released, absorbed, steps in the reaction, chemical bonds broken/formed) if we disregard the modern notion of orbital and go back to the Bohr model?

Read https://en.wikipedia.org/wiki/Bohr_model 'cause I have a feeling you don't know exactly about what you are talking.

@Mithoron I really don't. I don't know how much the theory used to model the atom influences applied Chemistry. In the end, I don't know if it makes any difference if we view the atom as a mini solar system or as a nucleus and superimposed probability states clouding around it. How much does this matter in practice when describing the hydration of potassium manganate into potassium permanganate? Is there a point in the study of chemical reactions when Quantum Physics is essential?

There are two distinct approaches to model chemical systems, namely Molecular Mechanics (MM) and Quantum Mechanics (QM).
QM uses quantum physics principles to predict the behavior of molecules. this approach is more accurate and more computationally expensive.
MM models molecules as balls connected with springs. This approach is relatively cheap, but less accurate. It requires calibration (you need to have experimental data for distances, angles, dihedrals for atoms you want to model). It cannot model transition states. But again, it is very cheap.
Molecular Dynamic simulations (MD) studies big molecules over long periods of time (long is just microseconds, but for a molecular simulation it is long). You have to use MM here, because you cannot afford QM.
So, if you want to model a small molecule (like acetone) you should use QM. But if you want to model a large molecule such as protein folding, or docking (of a small molecule to a protein active site for example), then you have to use molecular mechanics.
You can also use a QM-MM approach when you model a small and important part of a protein with an accurate QM method and the rest of the protein with cheap MM method. ONOIM is a common approach to do that.

So, do chemists use non-quantum mechanics representation of atoms and their interactions? Yes they do. Whenever they use MM they ignore quantum effects.
Do they use point particles orbiting nuclei as the model? Not really, they use springs connecting balls in these models.

Bohr model of atom was accurate enough to produce valid predictions. So in that field you can use Bohr's model. But in practice this is rarely needed.

Organic chemists (usually) don't care about the "implementation" of the reaction. For ethanol dehydratation reaction all they use is:
(1) Ethanol is an alcohol with a structure R2CH-CR2 OH (important parts are bold, not important parts are labeled as R) without any cycles (cycles might prevent dehydration sometimes). It doesn't have any other active groups (no esters, amids, thiols, double bonds to worry about).
(2) Alcohols that fit R2CH-CR2 OH template react with H2SO4 in three ways: (a) dehydration (2) formation of ether (3) formation of R-O-SO3H (at low temperature).
(3) Yes, I can dehydrate ethanol using sulfuric acid. I need to run the reaction at higher temperatures. I don't know how to decrease yield of ether. I will google the procedure (or check it in scifinder).
So, organic chemists just accept the rules and combine them to get the desired result. They don't really care what exactly was electron doing when all this stuff was happening.
More advanced chemists use reaction mechanisms, but there they operate such concepts as "acid", "base", "nucleophile" "electrophile", "anion", "cathion", "radical". They ignore the quantum physics behind the process. They just say:  "this is a strong nucleophile, so we should protect it first". If they want to know they can talk to theoretical chemists.

When talking about not so complicated chemical reactions, for example the dehydration of ethanol via sulphuric acid to form ether, does this reaction needed Quantum Mechanics to describe the process of which bonds are broken, the amount of heat needed for this reaction to occur, the steps of the reaction, etc. Can all this be predicted with accuracy without using Quantum Mechanics? Or without Quantum Mechanics is Chemistry limited by empirical observation and experimentation?

@FinnTheHuman for mechanistic studies one uses concepts like "base", "electrophile", etc. But all these predictions are based on "methylamine  (is a base because has amine) reacts with HCl (an acid). Buthylamine (also a base) should also react with HBr (also an acid).  Neither quantum mechanics not Bohr model is needed for this prediction.

I didn't see your edit. That comes really close to answering my question, but you talked about organic chemistry only. In inorganic Chemistry reactions, do Quantum Physics play a vital role, or is it just abstracted to cations and anions, and the rest doesn't matter?

@FinnTheHuman Classical inorganic chemistry is pretty much tabulated by now. What I mean is that almost everything you can find in tables. Real progress is being made in organometallic chemistry. There you would operate such ideas as "HOMO/LUMO", weak/strong field ligands, hard/soft acid/base, etc. For any serious prediction you would use QM just because it is affordable. You probably can use MM, but why use a marginal approach with unpredictable results. Another direction is solid state materials. Allows, High Temperature Superconductors, etc. They prefer QM, just because it is more reliable.

@FinnTheHuman You should understand that most of theoretical chemists don't really solve equations themselves.  Constraints there is not your ability to solve equations, but that would be off-topic. You can ask a separate question if you are interested.

@sixtyrees This was an awesome insight in the misterious world of Chemistry for me, Thank you all very much. I'm not into the avant-garde stuff, because my questions are all directed into making a Minecraft mod (not sure how many of you know what this is. Minecraft is a very popular games among children and teenagers. a Mod is a small piece of software that adds more things for the players to do. I'm working on a Chemistry mod, and I want it to teach real Chemistry, not some hocus-pocus video game stuff. So I'm doing my best to get my material straight.

Publish your mod scenario here - we might find and correct some errors.

I dispute that organic chemists don't care about QM. Molecular orbital theory is extremely important for explaining the reactivity of organic species.

@bon A moment organic chemist starts to care about why photo-reaction and thermo-reaction give different products he starts using theoretical chemistry approach. And nowadays noone can work like a pure organic or pure analytical chemist.

@bon Organic chemists should know molecular orbitals theory. However, when organic chemist decides if the compound is aromatic or antiaromatic he uses Hückel molecular orbital method, rather than Bohr or  Schrodinger models. They work with approximations without solving Schrodinger equation because someone else got approximate solution for a large number of systems.

When the fixed-orbit particle model is taught in general chemistry classes, it is usually presented as Bohr's fixed circular orbit model.
However, this was not the culmination of the deterministic-orbit particle model.  In 1916 Sommerfeld expanded the model to include precessing elliptical orbits, and made the theory relativistic. The varying of the eccenticity of the ellipses permits multiple angular momentum values for a given energy.
The Sommerfeld relativistic model correctly predicts the fine structure of the hydrogen atom.
In this one aspect, the deterministic-orbit particle model is superior to the non-relativistic Schrodinger model.  The Sommerfeld model correctly predicts the fine structure, whereas the Schrodinger model does not.


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Determine the quantitative composition of a solid sample containing Co, Mn, Al, C, and O
I have a solid sample that contains Co, Mn, Al, C, and O. The sample mass is ~200 mg. I need to determine what the composition of this sample is. I've looked at atomic adsorption spectroscopy, but that method is unable to detect C or O. I've heard mention that inductively coupled plasma mass spectrometry (ICP-MS) might work, but I am looking for additional potential solutions. I have access to a large number of instruments (university instrument sharing agreement). It is fine if the sample is destroyed as part of the analysis.

How much sample do you have, what accuracy do you need, which equipment do you have, what is the sample made of (is it organic or inorganic sample? If you have enough sample (100 mg) elemental analysis service would work. C, H, N, O, and metals are quantified using classical techniques. If you have small amounts of the sample you have to use several different

Are those all the elements present? X-ray fluorescense might be an idea, although it can have trouble with light elements. TGA could help you to quantify carbon, or carbonate.Is your sample very very likely to be completely oxidized?

It very much depends on element concentrations and the accuracy required for the total assay, but, I would do the metals by ICP-AES or AA and C by a classical combustion technique. ICP-MS would be unnecessarily expensive and since you are looking at compositional levels rather than low trace levels, it just wouldn't buy you anything.
If you know oxygen is the only other element present at compositional (i.e. percent) levels then don't bother with the finickier oxygen analysis and just determine it by difference. If carbon is present at at least a few percent then you only need a few mg, leaving enough to do multiple digestions for the metals analysis if desired.
If you want to do direct oxygen analysis, it can be done by classical pyrolysis techniques, though this presumably high-metals matrix would require a special technique (Mertz) and this analysis could have relative uncertainties of greater than 10%.


text:
Calculating cell potential

First of all, is the problem even worded correctly? From how the problem is worded, it seems that copper is being oxidized as well as iodide ion. This is what I am taught:

Second of all, I'm not getting 0.24 volts as the cell potential. I'm getting 0.19 volts.
Here is my work. What am I doing wrong?

I think you are almost there: log(.009)~-2

It's important to note that just subtracting the cell potentials only work if everything is at 1M, but in the question you've got slightly different concentrations. So actually all you've missed the last step! Your calculations are all correct. You've stopped at:
$$E_{cell} = 0.19 - \frac{0.0592}{2}\log{0.009}$$
and if you evaluate this you'll find that:
$$E_{cell} = 0.19 + 0.060 = 0.25 V$$
Which is just a little bit off from 0.24V

oh wow! To think that 2 hours of frustration was due to a simple calculator error ...

why would the problem list the second half-reaction backwards though?

In electrochemistry you always write the half reaction with reduced species in the left and oxidized species on the right. This is done for consistency purposes. Species with higher potential will reduce and species with lower potential will be oxidize.
I$_2$ + 2e- ----> 2 I$^-$   +0.54     //will reduce (go from left to right), has positive charge
Cu$^{2+}$ + 2e- ---> Cu  +0.34       //will oxidize(go from right to left), has negative charge
Next, you should use Nernst equation to each half reaction. You will get values E$_{Cu^{2+}/Cu}$ and E$_{I_2/I^{2+}}$. Now subtract one from another to get the voltage.

I thought in cell notation the reactant is on the left.

@Dissenter the oxidized form is on the left. Here is why: Cu/Cu$^{2+}$ is reduced by zinc, but is oxidized by  Ag/Ag$^{+}$. It is impractical to have two tables (one for each case), so scientists agreed to always write the reaction in the direction of reduction.

what do you mean the copper is reduced by zinc and oxidized by silver? There is no zinc or silver.


text:
Looking at the equation of a reaction, how do we know it is endotermic or exotermic
For example, $MnCO_3 → MnO + CO_2$ is endotermic, and occurs at 473 K
the polymerization of ethylene $H_2C=CH_2$ to $n(H_2C=CH_2)$ is exotermic
I suspect it has something to do with the standard enthalpy of formation of the different terms of the equation (right side - left side = $\Delta H$ if negative, the reaction is exotermic and releases $\Delta H$ to the environment as heat, if positive, it is endotermic and needs $\Delta H$ to happen.
Is this correct?
If not, What am I missing?

You are correct. If you have enthalpies of formation of each reagent and each product then you sum up enthalpies of each product multiplied by stoichiometric coefficient and subtract enthalpies of each reagent multiplied by stoichiometric coefficient. If the result is negative then the reaction is exothermic and vice versa.