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Increasing elasticity of microcrystalline wax
I'm doing sculpture and I've been playing around with some microcrystalline wax. The pliability of it once warm is great, but I'm running into the problem of it tearing / ripping when I stretch it. The same issue causes problems when smoothing it out, as the surface "gives way" under my finger if I press hard enough.
I've tried mixing in petrolatum and mineral oil in various different quantities but with no success; both just make the wax stickier and softer with no elasticity benefits. Is there another method which will increase the stretchiness?

The wax (presumably a mixture of mostly n-alkanes?) constantly re-crystallises under strain, making it brittle in the same direction. Petroleum acts as a plasticiser, easing the flow of the amorphous phase, but also easing the recrystallisation even more. In effect it doesn't really help with your problem.
You could try to find another, harder wax with longer alkyl chains, and try softening that by mixing in your present material.
And generally take your time when working it. When the material becomes less opaque, i.e. whiter, then it is already hardening. Stay below that threshold. High strain rates let the crystallites grow, which are stiff, so at the edges of the crystallites, the strain rates are even higher.

Natural bee wax is most certainly not a mixture of n-alkanes. Other than that, you are mostly right.

I was thinking the OP meant paraffin. Beeswax is rather not microcrystalline, is it?

@IvanNeretin Why "mostly"? ;-)

I use microcrystalline wax all the time but mine is now soft and tacky and very bendable when I pour sheet wax.  However, with you issue, it is best not to BEND the wax while it is still warm, even though this is tempting, it will break and tear.  It is best once you have your form or sheet, to leave the wax go completely cold before attempting to twist it or bend it.. works well and better.. don't try bending or manipulating when it's just completed it's cooling process.. I'd pour a few waxes and leave them for a day until they are well cured and cold.


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Why are lactate and lactic acid used synonymously in biochemistry?
In science there's always a misuse of the two terms. I know that lactate is the conjugate base of lactic acid.
Why are they used like is the same thing? For exemple: in lactic acid bacteria (LAB) fermentation it's often said that anaerobic glycolysis produces lactic acid, but if you actually look at the reaction it produces lactate.
Is this relevant? Because lactate is a negative compound while lactic acid is a neutral one.

Minor point, but in the case of lactic acid bacteria, the pH of the growth environment is often low enough that significant amounts of lactic acid are present. Furthermore, secretion of the product into the medium is often effected by lactate+proton symport, so the actual product of the fermentation is lactic acid, not lactate.

If you were to produce any lactic acid in the body, it would pretty much immediately get deprotonated, so by right it's always going to be lactate. The $\mathrm pK_\mathrm{a}$ of a typical carboxylic acid is at most 5, and at physiological pH levels (~ 7, give or take) the ratio of $\ce{HA}:\ce{A-}$ is on the order of $1:100$. (As getafix mentioned, lactic acid is somewhat more acidic than usual, because of the electron-withdrawing hydroxyl group, so the ratio would be expected to be even larger. But these are just meant to be very rough numbers, for illustration.)
You could say that it's sloppiness on the part of the biologists, but past a certain point you don't really care whether it's $\ce{HA}$ or $\ce{A-}$, and that's why the distinction isn't really enforced. From a biochemical perspective the more important thing is the oxidation state as well as other byproducts (ATP, carbon dioxide, etc.) Likewise the citric acid cycle is called that despite citrate being the actual compound in the cycle.
And perhaps the most egregious case of sloppiness: DNA stands for deoxyribonucleic acid, but in the body the phosphate groups are deprotonated, so chemically speaking it really should be called the conjugate base of DNA: "deoxyribonucleate". But again, nobody really cares - as long as we know that it has a negative charge, we can call it by a strictly incorrect name and it's fine.
There are probably tons of physicists out there who hate us chemists for using non-rigorous terminology, too!

I was once given this "rule of thumb" in a biochem class:

If the $\mathrm{pH}$ of a solution is one or more units below the $\mathrm{p}K_\mathrm{a}$ of a group, then said group is protonated and if it is one or more unit above, then the group is de-protonated"

If you are anything like me, then you probably don't like rule of thumbs without some form of derivation/justification, then read the discussion here
According to this table, the $\mathrm{p}K_\mathrm{a}$ of lactic acid is $3.86$. Assuming, physiological $\mathrm{pH}$ hovers around ca. $7$, we notice that since $\mathrm{pH}>\mathrm{p}K_\mathrm{a}$, lactic acid would be found in its depronated form, i.e as lactate.
I would say using the terms interchangeably is sloppy use of terminology on the biochemists' part, and nothing more.

Is this relevant?

I would say relevance would depend on context. However, I would assume most biochemists/biologists are aware of the distinction, yet (for some reason unknown to me) continue to be sloppy in their use of terminology.

Lactic acid and lactate are part of an equilibrium reaction in aqueous solution. Even when the pH is far away from the pKa, there is still a small fraction of the minor species, and protonation/deprotonation reactions occur. If lactate were a bunny and the hydrogen ions in solutions were hats, lactic acid would just be the bunny after putting on the hat. If you were to do some analytical chemistry with a biological sample, you might have the lactic acid/lactate in a different solvent or at different pH, so you might talk about it as lactic acid or lactate depending on circumstances.
The animation conceptually shows how the major species changes with pH. It does not show the minor species.

Source: https://commons.wikimedia.org/wiki/File:Low_pH_favors_protonation_while_high_pH_favors_deprotonation.gif

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Electrolysis of water with aluminum at the negative pole
I am doing electrolysis of water to collect the hydrogen and I'm running into a problem: on the positive side I'm using graphite from a pencil (I cannot use any other material because it corrodes too fast) and on the negative side I was using copper. Copper was working really fine, the water was clear and no problem. But I don't have copper with a big surface area so I decided using aluminum (from coca cans) and my solution started to get dirty because of the aluminum and the aluminum is slowly disappearing.
Why? Any other eletrode that I used for the negative pole was not corroded (only on the positive pole everything corrodes). Why is aluminum making my water dirty? Do you think if I use iron this problem will not happen?

Aluminum is pretty active. Once the passivation layer is breached, it would react with water quite readily.

Please don’t use shorthands like $-$ or $+$ for negative and positive poles/cathode and anode, respectively. It took me quite a while to understand that this question is actually valid. Also, please don’t resort to colloquialisms such as cause and try not to forget any apostrophes. Thanks.

You'll probably need to add more information in order to get any useful answers. What, if anything, did you add to your water? What potential are you using? Could you take a picture of the aluminum cathode and the water after a test where you've observed the aluminum to dissolve?

If you are using salt to make the solution more conductive the process will cause the creation of Sodium Hydroxide, which will react with aluminium.

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Which resources teach spectrum analysis of NMR and IR spectra?
I’m looking for a textbook that can teach me NMR and IR spectrum reading and analysis.
I really don't understand how to read the spectra or how to analyse them.

Try Oxford Chemsitry Primers, specifically for NMR I recommend Peter J Hore's book, called Nuclear Magnetic Resonance (I have the second edition)

Depends on what you want. PJ Hore's primer focuses mainly on the physical principles of NMR. For application of NMR, IR, MS etc. to organic molecule identification I would go with Tim Claridge's primer Introduction to Organic Spectroscopy.

If this question is off-topic, this one must be too. (However, this could be a dupe of that.)

@Jan I agree. The main resource thread certainly doesn't have spectroscopy recommendations, though. I will edit some in when I'm free then close as a dupe.

Hesse, Meier, Zeeh, Spektroskopische Methoden in der OC. Very good German book, covers all NMR, IR, MS that the ordinary organic chemist needs. There is an english translation available.

This question doesn't make a lot of sense without knowing your background. Student, professional? Biology, engineering, physics? I guess you're not a chemistry student?

I found Oxford Chemistry Primers to be quite good. There are many topics and the books themselves are very thin, but provide an adequate amount of information for undergraduate level.

[Introduction to spectroscopy]
Donald L Pavia; author.   How about this book??  is it good???


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For water, 1 mole = 18 g= 22.4 liter. How to explain?
For water, 1 mole = 18 g= 22.4 liter. But practically, how 22.4 liter is equal to 18 g?

22.4 liters of liquid water does not weigh 18 g. Pay attention to the phase.

For water, 1 gram is approximately 1 mL.
22.4 L/mol is the molar volume of an ideal gas at STP.
Since water is not a gas at STP, you can't apply that.
If you had one mole substance that is a gas at STP (hydrogen, nitrogen, carbon dioxide, etc.) you could say that they occupy a volume of approximately 22.4 L.

22.4 L/mol is for the "old" STP. The current definition of STP is 0 C and 100 kPa which has a volume of 22.7 L/mol.

Water can be a gas if it's water vapor.  22.4 L is the volume of 18 g of water vapor at 1 atm and 0 C.

Water can't exist in the gaseous state at these conditions.


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Transformation of fats in presence of alcohol
I have always observed when cooking that if I have some amount of fat like oil or butter in a hot frying pan and I pour some alcohol like wine on it, then there is a reaction and the quantity of matter which is left is strongly reduced because it evaporates. What is the name of this reaction and what are its products? Why is it used in cooking?
Maybe the question has been asked somewhere in a more erudite way, however I know really little about chemistry so I couldn't find it on neither Google nor SE.

There is no reaction. Alcohol (and water, for that matter) quickly evaporates and vanishes; that's it.

So what remains is really the fats and the alcohol ? How can you prove it ?

I can't prove a thing, if my words come across to you with the opposite meaning. Did I say anything about the alcohol? Did I say that it will remain? What did I say, really?

really not much you are right about that ! Elaborate your answer then, please

you actuall said there is no reaction. So both should remain if they do not completey evaporate. It is true, if my frying pan is very hot then everything will be evaporated and solid, black and stinky and I won't eat. And that's bad for me so I avoid it in general, therefore the temperature is sufficiently low in the problem considered so that I have some stuff remaining. The question is : what is this stuff ??

Let's just not forget that we're dealing with cooking, i.e., ridiculously complex chemical composition.
The main factor here is temperature.

Alcohol (Ethyl Ethanol) evaporates, as Ivan said. Although, since you're using a beverage (wine, beer, etc.) and not pure alcohol, so you're going to have some remains from the beverage, such as sugars, tannins, complex acids, etc.
If you have a very high temperature, you can have your organic compounds - the aforementioned sugars, tannins, the fats that you already put there, etc. - degrading and reacting with each other. You DO NOT want that happening, because the products can taste or smell really bad (for example, rancid smells are the product of fat degradation)

Speaking of cooking (and I do that a lot), you shouldn't add an alcoholic beverage or ANY kind of liquid directly to heated fat - butter or oil, not only because of the safety problem but because it is actually counterproductive. When you fry something - like meat or onions - and then add alcohol to form a reduction you're using the solvent properties of a alcohol + water solution to maximize de-scaling of the remnants in your frying pan (which are the products of a reaction called Maillard's reaction, very tasty ;)) and have maximum flavor in your sauce by adding the flavor from the beverage.
Hope it helped!


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Problems with Daniell cell
In Zemansky and Dittman's book on thermodynamics, pages 41–43, the authors discuss the Daniell cell as a thermodynamic system.
Their cell consists of two saturated $\ce{CuSO4}$ and $\ce{ZnSO4}$ solutions separated by a porous wall, with a copper strip dipped in the first and a zinc in the second, and with salt crystals in the bottom of each solution.

There are two problems:
The first is that the authors say that the cell is rechargeable, whereas this Wikipedia page says it not, so who is right?
The second is that the book states that in discharging mode:

If the electrolytes are saturated solutions, a current driven by the chemical reaction at constant temperature and pressure will not alter the concentrations of the electrolytes.

Well I can't understand how this happens. I can imagine that the $\ce{ZnSO4}$ solution will be over-saturated and the extra amount of it will deposit on the crystals,  but how can the other solution keep a constant concentration?
Any help is appreciated.

Why would it change? It is a saturated solution. If some CuSO4 vanishes, more will dissolve and keep it saturated.

@IvanNeretin so $\ce{CuSO4}$ dissolves spontaneously till it reaches saturation ?

Of course it does, like all compounds do.

@IvanNeretin forgive my ignorance, i clearly know nothing on this, but would the deposit dissolve even if the solution wasn't saturated at the beginning ?

Of course it would. That's the meaning of the word "saturated". Unless the solution is saturated, more compound would dissolve.


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Where on the backbone does a peptide get protonated at low pH?
As a particular example, how does the backbone of a peptide look at $\mathrm{pH}~2$?  Where does the backbone get protonated in such an acidic solution?
I know that at low $\mathrm{pH}$ the free amino acids have $\ce{-NH3+}$ and $\ce{-COOH}$ groups, but when the residues are linked together into peptide bonds, where do the extra protons go?

In the backbone of a polypeptide, you would find amide bonds formed from the amino (NH2) and carboxyl (COOH) groups of two amino acids. Water is lost in this condensation process:

The peptide backbone is considered to simply be the repeating Cα–C–N units. In the peptide shown below (a hexapeptide), the backbone is bolded. The side chains of each amino acid residue (R1, R2, ...) are not part of this backbone.

Amides are only very weakly acidic and very weakly basic, and do not undergo appreciable protonation or deprotonation under aqueous conditions.
Therefore, you only consider the terminal amino and carboxyl groups. So, at pH 2, both the N- and C-termini will be protonated:

Of course, if any of your residues contain a group that could become protonated, you need to consider its pKa as well.

I suggest to visualize the question on your own using the Avogadro molecule editor.
I'm not adding screenshots from my own computer since I run Avogadro with a localized German menu.

Generate some tripeptides
From the menu chose Build --> Insert --> Insert peptide
Chose three amino acids. Knowing the three-letters codes is an advantage ;-)
You might want to try this with amino acids that contain additional $\ce{-NH2}$ groups, such as Lys and those having additional $\ce{-COOH}$, such as Glu.
Adjust structure for pH
From the menu chose Build --> Add Hydrogens for pH...
Select a pH of 2 and have a look for changes.

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Are fossil fuels and petroleum the same thing?
I've been reading articles about renewable energy, and in comments people talk about fossil fuels and petroleum as examples of non-renewable types of fuel. Sometimes they are used interchangeably, or even mentioned as different things. I couldn't find explicit explanation for which is which, and if one is a subset of the other. Can someone clarify this for me?

Fossil fuels are those that originate from ‘fossilised’, anaerobically partially degraded prehistoric living species (mostly plants and animals) and that take a few million years to form hence are not considered renewable.
Depending on how old the deposit is and which conditions it encountered across Earth’s history, the deposit will either turn into something gaseous, liquid or solid — and these categories are typically labelled ‘natural gas’, ‘crude oil’/‘petroleum’ and ‘coal’. Thus, petroleum is only one type of fossil fuel: the one that presently is in the liquid aggregate state.
However, equating petroleum and fossil fuels isn’t entirely unwarranted, since petroleum is both the most important fuel and the scarcest today.

I would also count coal, lignite, and natural gas as fossil fuels. Like petroleum, they are non-renewable.


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Reaction of 2-ethyl-5-methyl-3-phenylfuran under acidic conditions

I was asked:

Does protonation occur to the electronegative oxygen?
What is the function of the heat?

I think that there will be protonation at C-3 which will break the aromaticity and the ring will open to form a diketone product.

Your answer looks perfectly reasonable. Acid hydrolysis of furans is a well established route to 1,4 diketones. The heat is required to overcome the resonance stabilisation of the furan ring.


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Can fluorescent & phosphorescent molecules still be fluorescent or phosphorescent as ions and not as salts?
The way my pchem professor explained it is that fluorescence /phosphorescence is caused by collision excitation. During this collision the spin of the promoted electron is flipped to a parallel spin of the orbit it left. The Pauli Exclusion Principle prevents relaxation/emission until the electron is some how able to "flip" its spin.
Most of the phosphorescent compounds I've found online have been salts which makes a bit of sense, (I imagine that the oscillation of the ionic bond is what causes the collision excitation). However, not all of them are salts. In lab previously our solution of sodium fluorescein exhibited a small degree of phosphorescence even when in basic conditions. I'm curious how a molecule can promote to a triplet spin multiplicity internally.
My professor says that the ionic component is not necessary and these compounds are typically ionic for the purpose of stability and longevity.
If the ionic bond isn't contributing to the triplet spin multiplicity then what is causing it?

Fluorescence derives from an electronically excited state after the molecule has been excited by absorbing a photon, usually in the visible or uv parts of the spectrum. Almost all organic molecules fluoresce to a greater of lesser extent, most common are aromatics for example, benzene and its derivatives as well as most dye molecules such rhodamines or coumarins. They can be ions or neutral.
When a molecule is electronically excited an electron is transferred from the highest occupied molecular orbital (HOMO) into an anti-bonding orbital usually the lowest one (LUMO). From here the molecule, which now has excess energy, can emit a photon and return to the ground state. The lifetime of the singlet excited state  varies between different types of molecules but is generally speaking around a few nanoseconds and not longer than microseconds.
By definition the excited state produced has the same spin as the ground state, which is usually means spin paired electrons $\ce{^ (v)}$. They remain paired even in the excited state (called a singlet excited state). Fluorescence is emission from a molecule with spin paired electrons, the transition is called 'allowed' and happens rapidly.
Occasionally the molecule can suffer a perturbation which causes one of the spins to be flipped to produce 'parallel' spins $\ce{^ ^}$ (heavy atoms or paramagnetic species can cause this). In this case a triplet state is produced and now the transition back to the ground state is far more difficult as the spin-flip has to be undone. The emission from such a state is called phosphorescence, it is long lived vs fluorescence, microseconds and longer, and as the triplet state is lower in energy than the singlet phosphorescence occurs at a longer wavelength in the same molecule, as fluorescence.
The figure  shows the energy levels. Different process compete with one another, thus the singlet can loose energy by fluorescence (with rate constant $k_f$) or by non-radiatively crossing to a triplet state (intersystem crossing) or non-radiatively to the ground state (internal conversion).

continued after question below.
The breaking of formally ‘ forbidden’ transitions is very common indeed, possibly the best example is in the observation of phosphorescence from almost any aromatic or dye molecule. Absorption in these molecules is generally a spin allowed transition, (single to singlet), however, triplet states, from which phosphorescence is observed,  can only be produced by a spin forbidden process, singlet to triplet, and just as in phosphorescence, is a spin forbidden process and becomes allowed due to spin-orbit coupling. In the case of the manganese ions and in other transition metal the forbidden transition is in absorption.
The general treatment starts with allowing the two states to interact via some  operator. In the case of photons (absorption / emission) it is an electric dipole operator, i.e. dipole moment operator. If the operator is $H_0$, the rate of the transition from one state to another is
$$W_{if}^0 \approx \int \psi_f H_0 \psi_i d\tau = \int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau $$
where $\sigma$ represents the spin wavefunction and $\psi$ the spatial one. As the operator does not depend on spin then
$$\int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau = \int \psi_f H_0 \psi _i d\tau\int\sigma _f \sigma_i dtau = 0$$
and the  result is zero because the spin wavefunctions belonging to i and f are orthogonal (as singlet & triplet) and so the last integral is zero.
If we suppose that it is possible to mix a little of the initial state wavefunction into the final one and generate a new state with wavefunction $\psi_m$ then $$\psi_m=\psi_f + b\psi_i$$ where b is some mixing constant we can determine later on. Now using this wavefunction as the final state and repeating the calculation above and separating out terms then as orthogonal spin terms are zero the only term left is
$$W_{if} \approx \int \psi_m H_0 \psi_i d\tau = b\int \psi_i \sigma_i H_0 \psi_i \sigma _i d\tau$$
in which case the integral is no longer zero and evaluates to a number.(Note that all the indices are the same unlike the equation for $W_{if}^0$)  This equation shows that there is a finite rate for the process with operator $H_0$. In the case in point, spin-orbit coupling, the operator $\Omega$ depends on the mass of the atoms involved and the product of orbital and spin angular momentum. This is why it is also called a ‘heavy atom effect’ when applied to emission processes.
The constant b is deribed using perturbation theory and turns out to have the form
$$b=\frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i}$$
where $\Omega$ is the spin-orbit operator. Thus the rate for the process finally becomes
$$ W_{if} \approx \frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i} $$
and we can use symmetry to decide if this integral is zero or is finite. The dependence on energy gap between states is clear, rapid rate constants for small energy gap. (The degenerate case when the energy gap is zero needs special and a more complex treatment)

Thank you for the response, but these were things I already knew.  I am most interested in what causes the intersystem crossing and why some molecules can stay excited for many hours while others can only remain phosphorescent for a few minutes.

some details are in answer to this question http://chemistry.stackexchange.com/questions/28883/non-radiative-transitions-or-what-features-of-rhodamine-result-in-it-being-so-hi/55287#55287, but in essence its due to spin-orbit coupling, which is the interaction between spin angular momentum and orbital angular momentum. It allows some singlet character to be incorporated in the triplet. The coupling is enhanced by paramagnetic species and heavy atoms, I, Xe etc.  Weak coupling means long lived triplet states, strong short lived. Molecular symmetry can also play a part in 'forbiddenness'


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Development of experiments to optimize HPLC-MS procedure using 2D central composite full factorial design
I have recently been studying development of experiments and how to statistically plan/optimize an experiment for x number of variables based on y number of responses.
To begin with, I am just going to use a 2D central composite full factorial design (CCD) to make sure I don't screw up bigger designs later. My two factors are 'load pH' and 'load conductivity'. My min/max values will be 6.5 - 7.5 and 5.5 - 6.5 respectively. My alpha value = sqrt(2) and I have scaled my model so far to (from min to max):
$$
-\sqrt{2}, -1, 0, +1, \sqrt{2}
$$
In a 2D CCD where 9 experiments are needed (depending on number of centre points), what relevance are the -1 and +1 values? To clarify, if my 2D array is effectively a 3x3 grid then the x and y axis will respectively read $-\sqrt{2}, 0, +\sqrt{2}$. What is the purpose of -1/+1?

The best source of experimental design pedagogy for beginners (and beyond) is Kevin Dunn's course, http://learnche.org/pid/ ... the more appropriate question is do you really need the root two values?  That is, do you need the two purple squares on your diagram above x1 above at this stage?  Have a look at http://learnche.org/pid/design-analysis-experiments/full-factorial-designs/using-two-levels-for-two-or-more-factors from Kevin Dunn.

I appreciate the swift response and thank you for linking me some reading material.

I have just realized what's happened. My design was for a face centred composite (FCC) design. In this situation only 3 levels are needed, -1, 0 and 1.
I've confused myself by constructing five levels for a central composite design (CCD) in which the root 2 values are used to extend outside the characterised square space and detect curvature better. Thus, I either need to switch to CCD or ignore the root 2 values and continue with an FCC design.
I hope that makes sense.

I'm voting to close this question as off-topic because it is a better fit for Cross Validated, the Stack Exchange site for statistical questions.

The question is too old to migrate, so the preferable course of action is to re-ask it at [stats.se] and delete it here.

FCC is a variant of a CCD design. It's usual to run your 2 factorial design first and establish which factors are relevant.  In your case, it won't matter so much but if you have 5 or more factors, it does.  The 2 factorial establishes your range with an upper and lower limit (+1 and -1). When you move to CCD,  you now have options. Using alpha of root 2 gives your "response surface design" orthogonality and rotatability, meaning, loosely, it is easier to analyse for variance in your model. The reason for varying alpha usually comes down to operational reasons e.g if a negative time or negative mass would result.  This is when a FCC might be used instead. If you ignored your coded +1 and -1, in effect your root 2 values would become your new +1 and - 1. The rationale for this is when a formula is generated to model each response,  the one using coded factors (set at +1 or -1) will indicate the relative influence of each factor, which the actual factor values will not do. The formula using coded values will not base it on your alpha value.


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Chromyl chloride test side products
I have read the test for chloride ions (chromyl chloride test). The reaction is:
$$\ce{K2Cr2O7 + 4NaCl + 6H2SO4 -> 2CrO2Cl2 + 2KHSO4 + 4NaHSO4 +3H2O}$$
My doubt here is if chlorine gas is evolved in a side reaction. Is it evolved in detectable amount? The side reaction is:
$$\ce{6Cl- + (Cr2O7)^{2-} + 14 H+  → 3Cl2 + 2Cr^{3+} + 7H2O}$$
Can someone please confirm this?

Yes, in a very good old book about preparative chemistry it is stated that chlorine is a side product in the synthesis of chromyl chloride.

Can you give the reference of the book? I am interested to check it out, assuming it's still in print.

Sure, I can give you the reference, but it is in Polish...

In short, potassium dichromate (VI) is not strong enough to oxidse chloride ions. However, if potassium permanganate (VII) is used, the chloride ions will be oxidse because potassium permanganate (VII) is strong enough to oxidse it. For more explanation, I write my explanation in the answer section below.

Chlorine gas will be evolved if there is water in the reaction mixture. In completely dry condition, the amount of chlorine evolved will probably be very small.

If you write the ionic equation, it seems that chlorine gas will produce. However, we see their half equation and standard potential.
$$\ce{ Cr2O7^2- + 14H+  + 6e-  ->  2Cr^3+  + 7H2O}$$
Standard potential: $E^\circ = \pu{+1.33V}$
$$\ce{ 6Cl-  ->  3Cl2 + 6e-}$$
Standard potential: $E^\circ = \pu{-1.36V}$
Thus, the standard potential cell for this ionic you say is equal to  $$\pu{+1.33V} + (\pu{-1.36V}) = \pu{-0.03V}$$
Because the standard potential cell is negative, so, it is call not energetically feasible or this reaction will not happen.
P.S.: The explanation only work under standard conditions.

As you noted the difference in potentials is very small, therefore some of the chloride can get oxidized as practice shows (some sources give 1.36 V value for both chlorine and Cr(VI) reduction).

This is not correct. Potassium dichromate does oxidise chloride ion in solution in presence of acid. Unless someone specifically measures the reactants, they are not in standard condition( 1 molar) so, the general method of adding potentials does not work.


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Why do amino acids hydrogen bond rather than ionic bond when they are forming secondary structures in proteins?
When peptide bonds are formed between amino acids, electron delocalisation causes the N to be more positive and the O to be more negative. As a result, why does 'hydrogen bonding' occur to form secondary structures such as alpha helices and beta pleated sheets, rather than 'ionic bonding'?
Is this simply a case of semantics, or is it the lack of transfer of an electron that makes the distinction between the two?

This looks like a homework question. Perhaps you should look at the primary structure of polypetides before looking into the secondary structure.

@D.J.Lawson please consider adding any relevant information to your question (which also shows that you did research to your question). As of now your query is likely to be closed/downvoted.

Why do you think there are ions here?

For a proper ionic bond, you would need ions. However, the backbone of your peptide chain does not have ionic groups. While the resonance structure you drew is correct, note that it is only a resonance structure. It does a good job at explaining the lesser reactivity of an amide bond with respect to an ester or an acid chloride. And it also shows the partial negative charge of the oxygen. But nitrogen is not positively charged in an amide bond (it is slightly negatively charged). Thus, there is no positive counterion that your negative carbonyl oxygen could bond to.
A hydrogen bond is a very viable possibility, though. You may know that it requires three things:

a hydrogen atom (check)
bonded to an electronegative atom (typically $\ce{O, N, F}$ — check)
and another electronegative atom that can receive (typically the same atoms — check)

So a hydrogen bond is possible where an ionic interaction is not.

Note that ionic interactions do happen in peptides. The side chains aspartate and glutamate are deprotonated (negatively charged) at ambient $\mathrm{pH}$ and the residues of lysine and arginine are positively charged. These can now form actual ionic interactions — although typically there will also be a significant hydrogen bonding component in these tertiary structure-creating bonds, too.

So although the resonance structure depicts a positive charge on N, it retains its electronegative character and still has a partial negative charge?

@D.J.Lawson Yes.

(this should be a comment)
I think you are not quite understanding the concept of partial charged or charged. The N atom in the polypeptide chain (so the N in the peptide bond) is partially charged! to be in an ionic state it should have 4 bonds instead of 3.
 The peptide bond which is formed(above) just contains a N atom which is bounded to three other atoms. The oxygen atom is also partially charged but not charged! read this about H-bonding and partial charge


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Formation of a pH gradient using carrier ampholites
In a technique called isoelectric focusing, the crucial part is the formation of a pH gradient using carrier ampholytes. What is the exact principle by which the gradient is formed? How do the ampholytes "know" exactly where to migrate, since they are the only protonizable chemical species in the solution?


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Can all nine essential amino acids be produced synthetically?
I would like to know if all nine of the essential amino acids can be produced synthetically with high purity without amines or other impurities through microbial activity during the production process.

If you're using microbes, it's not synthetic. You can use the Strecker reaction otherwise: https://en.wikipedia.org/wiki/Strecker_amino_acid_synthesis But you'll need a resolution for the desired enantiomer.


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Are hydrogen atoms in acetone molecule acidic?
My chemistry book says that acetone is aprotic, but I think that, since the conjugate base of acetone has possible  resonating structures, the hydrogen atom must be free to move out as $\ce{H+}$ (as the resulting structure is more stable than the previous one). Where am I going wrong?

What is the conjugate base of acetone?

What is the pKa of a hydrogen $\mathrm{\alpha}$ to a ketone?

@DHMO conjugate base of hydrogen is the molecule we get on removing a H+ unit from the molecule, as per my knowledge .

@zhe sorry, I haven't yet studied the concept of pKa .

You literally said that "the conjugate base of acetone has possible resonating structures". I'm asking you what "the conjugate base of acetone" is.

There are levels upon levels of wholly different meanings to the word "acidic". Yes, acetone is acidic (in a way), but not like ordinary acids. And yes, it is aprotic.

Generally, we refer to protic as meaning "can hydrogen bond." This may be a bit of an oversimplification, but it might help you understand what the problem is here.

@Ivan neretin, will u please explain what aprotic really means and/or why this compound is aprotic ?

@Dhmo, it's CH3COCH2-

You can't remove a hydrogen like this...

Why, you can, but it is pretty hard.

@Dhmo , that's what the question is . Why can't we remove the H when the resultant molecule is resonance stabilized . We can't remove it means it's aprotic

@ Ivan neretin , what makes it hard?

@PyroRecorcinol You'd better wait till you get familiar with the concept of pKa. Before that, think of it this way: there are strong acids, like H2SO4 and HCl, and there are weak acids, like H2S. Then there are even weaker acids, which are so weak that they don't work as acids in water. And then there are other acids, which are many orders of magnitude weaker than that.

The point is that after removing one proton, the resulting structure is in fact not resonance stabilised. The terminal double bond of the enolate is not very favourable. If you have a tertiary C-atom there, then the enolate is sterically less hindered and has an advantage. If you have another double bond at the beta C-Atom, then you get resonance stabililsation.

When the alpha hydrogen moves out, it's the first step of tautomerism.. given the small percentage of enol form present, you can safely say it's not much acidic

With respect to a large number of other molecules, deprotonating acetone is remarkably simple. Also, I reject Ivan’s opinion that acetone be aprotic; it can very well protonate given appropriate bases.

You’re not going wrong anywhere. Acetone can indeed be deprotonated. The resulting structure is called an enolate.
$$\ce{H3C-C(=O)-CH3 <<=> H3C-C(-O^-)=CH2}$$
When compared with truly acidic compounds such as $\ce{HCl}$ and even $\ce{NH4+}$, acetone is a very, very weak acid, though. That does not mean it should be considered aprotic. Given a strong enough base — e.g. $\ce{HMDS}$ ($\ce{(Me3Si)2N-}$) — it can be deprotonated. Weaker bases will allow for partial (but probably hardly verifiable) deprotonations.

What's the threshold for something being aprotic, then? (Where does, for example, toluene lie?)

@orthocresol Very good question! Somewhere between acetone and DMSO would be my guess so a $\mathrm{p}K_\mathrm{a}$ value of $25$~$35$. Toluene has ~$40$ so it’s definitely out.


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Does anyone know what this piece of glassware is?

I recently found this piece of lab glass with a jacket and spout. It says eubilicmeter but I cannot find any information about it online and I do not know what it is supposed to do.  It is hollow in the middle, sort of bell shaped. What is its name and what is it for?

I believe this is an ebulliometer, from the Latin ebullientem (to boil).
It's a used to measure boiling points. Clearly, there are some spelling issues with the sharpie label here.

Yes seems like that

Actually, this is a fun question to punt over to the Indoeuropean SE. :)

There might be fewer spelling errors that it first appears.  I'd say that the sharpie could be "eobiliometer" (allowing for handwriting, especially on a slick surface, and where it looks like some ink might have rubbed off), which only mistakes an "i" for a "ul".

It certainly was supposed to read  'ebulliometer', but the question is if it really is an ebulliometer? It does not look like one, maybe it's just a part of it. It is a vessel with double walls with vacuum in it (see that sealed tube at bottom right?). The central volume is thermally insulated - an ebulliometer must have a part that is heated, I don't see anything like that here.


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What if solute and solvent exist in equal quantities?
A solution is made of solute and solvent, like salt in water, however, what if the solute and solvent exist in equal quantities?
Which one is the solute and which is the solvent? If one of them is water I think it will definitely be the solvent, but what if neither of the two parts are water?

related discussion http://chemistry.stackexchange.com/questions/23519/can-there-be-more-solvent-in-a-solute

As far as I can recollect, I think in such a situation you can arbitrarily select one of them as solute and one as solvent. However, according to convention, it is better to assign the solvent as a substance which is usually used as a solvent in other situations.


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Particle in a box models for polyenes and quantum dots
I'm kind of new to physical chemistry and trying to understand how, if at all, the correspondence principle (in the limit of large quantum numbers, quantum mechanics reproduces classical behavior of a system) relates to transition energies for electronic excitement in conjugated polyenes and quantum dots.
Our experiment used UV/vis spec to measure absorption and calculated theoretical values of energy with the particle in a box/sphere models. So why does the difference between experimental and theoretical energy get larger as length gets smaller?
Someone proposed the correspondence principle here, but I don't see how that has to do with going from the ground state (HOMO or valence band) state to the first excited state (LUMO or conduction band) that we're dealing with in the experiment. What am I not understanding here? Thank you!

I think that the agreement is pretty accidental. The particle in a box is a very crude model of a molecule but it does show basic trends, bigger box (i.e. more conjugation) gives  lower energy levels with smaller gaps between them. Thats it; don't read any more into it than that and use it as a 'rule of thumb' sort of thing.


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Sublimation and recrystallization of caffeine
I'm looking to do an extraction of caffeine from energy drinks. I want to use both sublimation and recrystallization in my experiment for advanced purity.
My question is which should happen first, recrystallization or sublimation? Does it even matter which happens first?

You might want to break this question into 2 separate ones: one about sublimation and recrystallization; and another about solvents for extraction. As it stands, the query is rather broad and might be closed as such by the community.

Why? This sort of inquiry makes me think that this is about some sort of quantitative analysis. Is that the case?

This sounds like a nice project.
For the purification of caffeine, sublimation in vacuum and recrystallization from ethanol are both conceivable.
But first you have to separate it from the other ingredients, such as carbohydrates.
According to the manufacturer, 100 mL of Red Bull, a popular brand, contain 11 g sugars and 32 mg caffeine.
In order to isolate the caffeine, two methods are possible:

extraction with $\ce{CH2Cl2}$
solid phase extraction on a C18 column

The latter method might be interesting. Imagine that your pour the energy drink through a preconditioned column of this adsorbent. The caffeine will  be retained, whereas everything else passes through.
In a second step, you wash the caffeine off the adsorbent using ethyl acetate. Then, dry the organic solvent with sodium sulfate, filter off, evaporate the ethyl acetate, dry the crude material and check the melting point. If necessary, use any of the above two methods for purification.
If you'd like to read this in more detail, have a look at a procedure provided at Open BU, an open access initiative of the Boston University.

UPDATE
I'd probably use just one of the methods for purification, depending on the equipment available. But if you insist on using both, I'd recrystallize first.

Hello, thank you so much for your reply I genuinely appreciate it. I fully intend to use Dichloromethane for the separation of caffeine intially from the rest of the products but since I have 40 hours of lab hours available I may also be able to take a look into the second methods.

My only issue is in terms of purification does it matter whether the recrystallization is done before the sublimation ? or vice versa?


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Reason for negative Joule Thomson coefficient of Helium and Hydrogen at NTP conditions
Recently, while reading my textbook I came to know that Helium, Hydrogen and Neon are the only gases which have negative Joule Thomson coefficient at NTP conditions, i.e heating effect is observed when these gases expand from a region of high pressure to a region of low pressure. I was wandering about possible explanations for this strange behaviour (since almost all gases display cooling effect during free expansion at NTP conditions) but nothing came to my mind. Can anyone give me a reason why these gases behave differently at room temperature and 1 atm pressure?
One thing I noticed is that all the three gases in question have compressibility factor(Z) always greater than 1, i.e. negligible intermolecular forces of attractions, but how can this explain their abnormal behaviour?

The JT coefficient changes sign at the inversion temperature, this is true for all gases.

And of course you can relate it to Z, I shall write up an answer later.

Note: You can skip section I, and go straight to section II and/or the end of section III (specifically the conclusions subsection), if you are already familiar with the basic mathematical machinery/definitions
I. Preliminaries

Feel free to skip all of this if you are familiar with it
First, we have an adiabatic expansion, i.e $\mathrm{q} = 0 $.
Second, our process is carried out as follows: a gas at pressure $p_i$ flows into a region of lower pressure $p_f$ without significant change in kinetic energy. This is called the Joule–Thomson expansion.
Obviously, this process is inherently irreversible. Additionally, one can show that this expansion isenthalpic (i.e $H_i = H_f$). Unlike a free expansion, work is done, causing a change in internal energy; the exact specifics of this change will depend on the initial and final conditions and also on the nature of the working fluid.
What we measure experimentally is a change in temperature with respect to pressure at constant $H$, and we call it $\mu_{\mathrm{JT}}$ (Joule-Thomson Coefficient).
$$ \mu_{\mathrm{JT}} = \left(\frac{\partial T}{\partial P} \right)_H = \frac{V(T \alpha-1)}{C_p} \tag{1}$$
where $\alpha$ is the coefficient of thermal expansion $$\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p$$
All real gases have an inversion point at which the value of  $\mu _{\mathrm  {JT}}$ changes sign. The Joule–Thomson inversion temperature, depends on the pressure of the gas before expansion.
$$\begin{array}{cc}
\hline
\text{Temperature} & \text{sign of}\ \partial P & \text{sign of}\ \mu _{\mathrm  {JT}} & \text{thus sign of}\ \partial T \\ \hline
\text{below  inversion temp.} & - & + & - \\
\text{above inversion } & - & - & + \\
 \hline
\end{array}$$
The final pressure is always lower than the initial pressure and thus the change is always negative. Depending on whether you are above or below the inversion temperature, your change in temperature can either be positive (warming) or negative (cooling). I would like to stress that the inversion temperature depends both on the temperature and pressure.

II. Qualitative Analysis
The arguments given below might seem a little "hand-wavy"/crude, but bear with me.
$$ H = U + pV$$
$$ \Delta H = \Delta U + \Delta (pV)$$
During a Joule-Thomson expansion (isenthalpic, adiabatic), the change in $pV$ represents the work done by the fluid.
If $PV$ increases,at constant enthalpy, then U must decrease as a result of the fluid doing work on its surroundings; the working material is using its own internal energy to do work. This produces a decrease in temperature and results in a positive Joule-Thomson coefficient.
Decrease in$pV$ means that work is done on the fluid and the internal energy increases. This can thus lead to an increase in temperature and the Joule-Thomson coefficient will be negative.
For an ideal gas, $pV$ does not change during a Joule-Thomson expansion.
As a result, there is no change in internal energy; also, there no change in temperature; and the joule-thomson coefficient is zero.
In real gases, $pV$ does change. The deviation from ideality (in general) is contained in $Z$, the compressibility factor. Thus, it would make sense if we could make some connection to it.
The ratio of the value of "real" $pV$ to that expected for an ideal gas at the same temperature is called the compressibility factor,
$$Z := \frac{pV}{nRT}$$
For a gas, this is typically less than unity at low temperature (attractive forces dominant) and greater than unity at high temperature.
At low pressure, the value of $Z$ always moves towards unity as a gas expands (tend towards ideality).
In general at low temperature, $Z$ and $PV$ will increase as the gas expands, resulting in a positive Joule-Thomson coefficient.
At high temperature, $Z$ and $pV$ decrease as the gas expands; if the decrease is large enough, the Joule-Thomson coefficient will be negative.
It follows, that for the $Z>1$, Joule-Thomson coefficient will be negative and the gas gets warmer. This is the case for the gases in question at the given conditions.
A graph from wikipedia showing how the Joule-Thomson Coefficient changes sign (for nitrogen):

Note: The question originally asked about gases, but I'll make a small point about condensed phases and supercritical fluids (since you can see them in the graph). At high pressures, these phases can't respond to an increase in pressure by reducing their volume, and $pV$ increases as pressure increases. The molecules are being forced together, and such the Joule-Thomson coefficient is negative.
Of course intermolecular forces, attractive and repulsive, and the limits in which they are dominant are key in explaining the changes of $Z$, and consequently of $\mu_{\mathrm{JT}}$. This point is further developed in the next section.

III. A slightly more quantitative, but ultimately Qualitative Analysis
To see how intermolecular interactions come into play, let's look at two model systems: an ideal gas (no intermolecular interactions) and a Van der Waals gas (both attractive and repulsive interactions).
Using equation (1) and the definition of $\alpha$ we can write
$$ \mu_{\mathrm{JT}} = \frac{V(T \alpha-1)}{C_p} = \frac{T \left(\left(\frac{\partial V}{\partial T}\right)_p -V\right)}{C_p} \tag{2}$$
For a perfect gast $\left(\frac{\partial V}{\partial T}\right)_p = \frac{V}{T}$ and so, $ \mu_{\mathrm{JT}} = 0 $
Since the Van der Waals gas equation is cubic in $V$, I will use the first few terms of the virial expansion and solve for $V$ to obtain $\left(\frac{\partial V}{\partial T}\right)_p$. I am omitting explicit computations, you may do them as an exercise.
$$Z = 1 + \frac{1}{RT} \left(b- \frac{a}{RT} \right)p + .... \tag{3}$$
Since $Z = \frac{p\tilde{V}}{RT}$ where $\tilde{V}$ is the molar volume of the gas; I am doing this for convenience's sake, I don't want to lug around a $n$ all over the place.
$$\tilde{V} = \frac{RT}{p} +  \left(b- \frac{a}{RT} \right) + .... \tag{4}$$
After differentiating we obtain
$$\left(\frac{\partial V}{\partial T}\right)_p = \frac{R}{p} +  \left(- \frac{a}{RT^2} \right) + .... \tag{5} $$
Substituting (4) and (5) into (2), and dropping all of the higher order terms.
$$ \mu_{\mathrm{JT}} \approx \frac{\frac{2a}{RT}-b}{C_p}$$
Now, consider the low temperature limit of this expression. The $\frac{2a}{RT}$ becomes large, and $b$ is negligibly small compared to it so
$$ \mu_{\mathrm{JT}} \approx \frac{2a}{RTC_p}$$
All the quantities on the right hand side of this equation are positive. so $\mu$ is positive. The attractive forces are dominant, since only the paramter $a$ appears in our expression.
Similarly, the high temperature limit is
$$ \mu_{\mathrm{JT}} \approx \frac{-b}{C_p}$$
Here, $\mu$ is clearly negative. The repulsive forces are dominant, since only the parameter $b$ appears in our expression.
Conclusions

The existence of an inversion temperature becomes obvious in a gas with both attractive and repulsive forces, since $ \mu_{\mathrm{JT}}$ changes sign from positive to negative.
The connection to attractive and repulsive forces also becomes more explicit.
In the low temperature limit, attractive forces are dominant and the gas needs to break free from these interactions as it expands. This requires energy, which has to be supplied by $U$ the internal energy of the gas. Thus the gas cools.
At high temperatures, the repulsions are dominant. As the gas expands, the over potential is lowered, and energy is released. Thus, the gas warms up.
Obviously, the exact magnitude of the high and low temperature limit depends on the nature of working fluid, since the parameters $a$ and $b$ are different for each gas. The gases you are talking about, have barely an attractive interactions.

Caveats

Our analysis based on the Van der Waals gas seems to suggest that the inversion temperature and $\mu_\mathrm{JT}$ is a function determined by temperature alone. This has experimentally proven to be false. This discrepancy arises because we dropped the higher order terms in our expansion which contained the pressure dependance.
This model is ultimately qualitative. It will give you the wrong numbers when you try to compute something, but it explains the general trends well.

Thanks for the detailed answer!! I think the bottom line is that.....due to decrease in interatomic/intermolecular potential energy of the gases in question.....their kinetic energy increases which results in increase in temperature.....since K+P=U and U should be constant overall (kinetic + potential). Am I correct?

@SirJMP I'm not sure I follow your argument.

Since joule-thomson effect is isoenthalpic process, so everything is done at the cost of internal energy.
Taking $a$ as the term involving intermolecular forces of attraction and $b$ as the excluded volume, we get:
$$T_i = \frac{2a}{Rb}$$
As per above equation inversion temperature is the function of intermolecular forces of attraction. When the temperature is high and above $T_i$ intermolecular forces of attraction are weak and kinetic energy is high for particles. When in such condition gas expands the kinetic energy of particles increases rapidly and small amount of energy is spent to break forces of attraction among particles of gas and hence increase in kinetic energy is dominant over energy spent. Since temperature is the kinetic manifestation of kinetic energy($K =\frac{1}{2}mv^2$ and $v =\sqrt{\frac{3}{2}RT}$). Hence temperature of the gas increases.
However if gas is below inversion temperature the forces of attraction are huge compared to kinetic energy and expansion of such a gas involves spending much energy for breaking intermolecular forces compared to rise in kinetic energy of particles as a result of expansion. Hence use of internal energy for breaking intermolecular forces dominates over increase in kinetic energy and thus gas shows cooling.
Thus from the above arguments it can be understood that at inversion temperature both factors balance and hence gas shows neither cooling nor warming under conditions of expansion.
Taking case of $\ce{H}$, $\ce{He}$ these particles are very small and intermolecular forces of attraction are very weak and hence the inversion temperature is below room temperature (forces of attraction are stroger below room temperature). Thus when such systems are allowed to expand from region of high pressure to low pressure these show heating effect rather cooling(explained above).


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Why is an alkyne more acidic than an alkene? Or, why is acetylene more acidic than ethylene?
pKa acetylene ~ 25
pKA ethylene ~ 44
I'm not looking for the "because alkyenes are sp hybridized and more s character means their electrons are held tighter" answer because I'm not the biggest fan of hybrid atomic orbital theory.
Is there an alternative answer that looks at this question from a Molecular Orbital (MO) Theory perspective instead? Or maybe a different conceptual analysis as to why the electron pair is more stable on the alkyene carbon than an alkene carbon? Does magnetism (from the ring of charge in the triple bond) somehow have an effect?
Or is it really as simple as the fact that the because the pi electrons are higher energy and farther away from the nucleus, the remaining electrons participating in sigma bonding must counteract the strong Coulombic pull from the carbon nucleus by moving closer to it?

hybrid orbital explanation: http://chemistry.stackexchange.com/questions/32341/relative-acidities-of-alkanes-alkenes-and-alkynes

Three of the carbons electron pairs hang on one side, so it is pretty positive on the other side, and pulls the CH electron pair closer. But blunt, that explanation, but what's wrong with it? Whatever you come up with, it is never more that a rationale. Nature does not answer "why" questions. ;-)


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Is there a way to tell what the coordination of a complex ion will be without knowing the ligand?
Do complex ions tend to form a certain number of coordinate bonds, or does it completely depend on the ligand molecule?

There is no way you can predict this a priori. For example, cobalt(III) forms an octahedral hexaaquacomplex $\ce{[Co(H2O)6]^3+}$. If you add chloride to the solution, you will form a tetrahedral tetrachloridocomplex $\ce{[CoCl4]-}$. This was a pure ligand exchange reaction as shown in the equation below.
$$\ce{[Co(H2O)6]^3+ + 4 Cl- <=>> [CoCl4]- + 6 H2O}$$


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Ranking bond lengths of dienes

For the bond lengths (ranking  them longest to shortest) I think that it would be A>C>B
My thought is that bond-B is a double bond (so the shortest), bond-A is the longest because it is only connected to one double bond when C is connected to two double bonds.
Does that logic work out for these problem types?

This is a bit tricky. B and C are very similar...

I'd say that $C$ is not really a single bond; it has a significant $\pi$ component to it. So yes, it is $A>C>B$

looking at this as 1, 4-dimethyl butadiene then I agree with your reasoning; A is longest (single bond) , B shortest and C slightly shorter than A but much longer than B as it has slight double bond character due to delocalisation of the butadiene.

Your answer is correct, but that's not the exact train of thought I would go along. "Connected to $n$ double bonds" doesn't make much sense unless you relate it to either hybridisation or bond order (via resonance forms).


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Hydrogenation side product: potential mechanisms for C-C bond making
In the analysis of the product mixture of a new catalytic hydrogenation  reaction I'm trying out I am getting this side product molecule (in low quantities; less than 1% w/w).

Is this a likely / explainable side product? What would be a potential mechanism?
I'm skeptical because the C8 increases to a C9. I assumed C-C bond making wasn't a very common occurrence in this sort of reaction? Or is it?
The reaction conditions are fairly mild: 10 barg & 80 C. Somehow I associated this sort of hydrogenation with functional group identity changes & not Carbon-backbone changes.
Or am I wrong?
More info. Solvent is Methanol. pH is alkaline by small qty. of NaHCO3 initially added. Catalyst is Pd on C.
Edit: Supposing the MeOH solvent reacts with the starting material, would this be the reaction to expect?

Any mechanism to go from this alcohol-ether to the impurity I'm seeing?

With no conditions disclosed (catalyst, solvent, acidity etc) this is not a reasonably easy question. However, I would check purity of the reagent, since styrene used to make it could contain some 2-phenylpropylene, especially if it was technical grade. I would also check that the reaction mixture didn't have any sources of nucleophilic methyl synthone.

Thanks @permeakra. Sorry, I forgot to add those. Solvent is Methanol. pH is alkaline by small qty. of NaHCO3 initially added. Catalyst is Pd on C.

@permeakra Does that info. help answer?

Your methanol adduct seems most likely to me. How are you characterizing the impurity? Mass spec? NMR? Other?

@jerepierre Thanks! Yes, by mass spec. But how does the adduct transform itself into the product I'm seeing? No doubt the skeletons match but how does the ether change into a C-C bond? Would hydrogenation do the trick?

@curious_cat I wouldn't expect hydrogenation to do that. You could test it by making the ether you suggest intentionally, and then subjecting it to your hydrogenation conditions.

First of all, I would check for impurities in styreneoxide. There is a good chance it was prepared from styrene having some 2-phenylpropylene in it, and this could account for your impurity.
Other than that, I would think among lines that hydrogenation is usually reversible, so above strong hydrogenation catalys phenylethanol is in equilibrium with corresponding aldehyde and methanol is in equilibrium with formaldehyde. Crotone condensation of said carbonyl compounds with subsequent hydrogenation may account for the impurity you see here.

Is alphamethyl styrene the same as 2-phenylpropylene? Then it is present as an impurity in styrene. Have to figure out what molecule it gets converted to at the styrene oxide stage.

@curious_cat yes, it is. The you are dealing with impurity produced from another impurity.


text:
Term symbol for atomic nitrogen: why can't S be -1/2?
The spectroscopic notation for an element is given by $^{2S+1}L_{J}$. I want to use this notation for nitrogen with configuration
$$\mathrm{[1s^22s^22p^3]}$$
This is what I did. $l=1$ for p orbital. There are three electrons in p orbital, two electrons: spin up and spin down, and the remaining electron in spin up position, so that my $$S = \frac 12 + \left(-\frac{1}{2}\right) + \frac12 = \frac12$$ which gives
$$J=l+s=1+1/2=3/2$$
My query is can't the remaining electron in p orbital be in spin down position so that my $S$ becomes $-1/2$ and so that $J=l+s=1/2$.

To get you started have a look at my answer to this question http://chemistry.stackexchange.com/questions/37038/how-to-find-the-number-of-atomic-microstates-for-a-given-electronic-configuratio/54333#54333

Why not go one step further and make all the electrons have spin down? Then you should get $S = -3/2$, no?
In fact, what's happening is: you are confusing the projection of the spin quantum number, $M_S$, with its magnitude, represented by the spin quantum number $S$.
The $\pm1/2$ that you are summing are the individual projections of the electron spins, $m_s$. When you add up projections, you get a projection $M_S$. This projection can indeed be $-1/2$, and it can indeed be $-3/2$ as well, but the term symbol does not depend on $M_S$; it depends on $S$.

To obtain the total spin quantum number $S$ you have to use the Clebsch-Gordan series to couple different sources of angular momenta together. In this case we would first couple the spins of two electrons
$$S_{12} = s_1 + s_2, s_1 + s_2 - 1, \cdots, |s_1 - s_2|$$
Note that for an electron, it has spin $s = 1/2$. Again, the projection of the spin can be $+1/2$ or $-1/2$, but that is not what we are interested in. We are interested in the magnitude of its spin, which is only $1/2$, not $-1/2$.1 If we plug $s_1 = s_2 = 1/2$ into that equation above we get
$$S_{12} = 1, 0$$
and now if we couple this to the spin of the third electron we get
$$S = S_{12} + s_3, S_{12} + s_3 - 1, \cdots |S_{12} - s_3|$$
Again, $s_3 = 1/2$ only. But $S_{12}$ can take on two values so we will have to evaluate this sum two times, one time for each possibility:
$$\begin{align}
S_{12} = 1 &\Longrightarrow S = \frac 32, \frac 12 \\
S_{12} = 0 &\Longrightarrow S = \frac 12
\end{align}$$
Note that the absolute value in the Clebsch-Gordan series forbids the possibility that $S < 0$. This makes sense since a magnitude of a spin cannot be negative.
In the end your two possibilities are $S = 3/2$ or $S = 1/2$. Due to Hund's first rule the $S = 3/2$ term is the ground state and the $S = 1/2$ terms are excited states.

Next, the $L$ in the term symbol is not equal to 1 just because the p orbitals have $l = 1$. Note that one is a small $l$ and the other is a big $L$. The big $L$ in the term symbol is obtained again by coupling the individual values of $l$ for each orbital. In this case all three p orbitals have $l =1$ and you would have to work through the Clebsch-Gordan series again.
$$L_{12} = l_1 + l_2 , l_1 + l_2 - 1, \cdots , |l_1 - l_2|$$
$$L = L_{12} + l_3, L_{12} + l_3 - 1, \cdots, |L_{12} - l_3|$$
and if you set $l_1 = l_2 = l_3$, should reach the conclusion that $L$ can take on the values of $3,2,1,0$. However you have to be careful about this, because the $L = 3$ value is forbidden by the Pauli principle, which is not taken into account by the Clebsch-Gordan series. A term with $L = 3$ would necessitate the existence of a microstate with $m_L = +3$, which would imply that all three electrons are in the p orbital with $m_l = +1$, which is not possible.

1 Technically, the spin is $\displaystyle \sqrt{\frac{1}{2}\left(\frac{1}{2} + 1\right)}\hbar$, but whatever. The correct phrasing is that the quantum number representing the magnitude of the spin can only be $1/2$.

I'm sorry, but I got a little confused by your statement A term with L=3 would necessitate the existence of a microstate with mL=+3, which would imply that all three electrons are in the p orbital with ml=+1, which is not possible.. Could you, please, explain it in a little more detail? I don't see the connection to $m_l$ state...

@Eenoku (1) a term with $L = 3$ necessarily encompasses several states with $M_L$ ranging from $-L$ to $+L$. So, if that term is possible, it would imply the existence of states with $M_L = +3, +2, +1, 0, -1, -2, -3$. (2) $M_L$ is the direct sum of individual values of $m_l$. For $M_L = +3$ and three unpaired electrons in p-type orbitals (which have $m_l = +1, 0, -1$), the only way to accomplish this would be with three electrons in the $m_l = +1$ orbital, which is forbidden. Thus, the $L=3$ term cannot exist, which is why there is no $F$ term symbol for a $2p^3$ configuration.

An algorithmic approach to writing down term symbols for an electron configuration can be found here. If you know group theory, that can help.
Anyway, if you follow the procedure described in the linked wikipedia article, then you will arrive at the following terms. $^4S,\,  ^2D,\ \text{and } ^2P $.
Now, the rule's for determining the ground state term are:

The lowest energy term is that which has the greatest spin multiplicity.
  and
  For terms with the same spin multiplicity the term with the greatest orbital angular momentum is lower in energy

So in this case, the ground state term would be $^4S$. Now, J can take on values $J = |L+S|, |L+S-1|....|L-S|$
As an example, for the ground state term, $L = 0$ and $S = \frac{3}{2}$
So, $J = \frac{3}{2}$
The situation you describe, where one electron is spin up and the other two are spin paired is possible, and they give rise to the other two terms, and aren't the ground state term, which is what you were interested in? (I am guessing)

It can be spin down, but that would give you a doublet state, compared to a quartet if all electrons were spin up. The ground state is the one with the largest multiplicity.


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Computational Chemistry: calculation of sterical effects
Is there a way to calculate sterical forces between atoms/groups of atoms (preferably with free and open source tools)? I have heard that this is possible within the framework of NBO. I am using ORCA, nwchem and Gaussian09 mostly. Sterics are obviously one of the most important aspects of organic chemistry but still I could find close to no resources on the computational chemistry aspects of it. It would be nice to have a way of quantifying sterics and also visualizing it somehow.
Does anyone have some experience on that?

Just ran a cursory search and ran into http://dx.doi.org/10.1063/1.3555760 on calculations of the anomeric effect. Not sure if the reference points at what you had in mind.

What about using Bader's QTAIM? There is some literature in that regard.

I think that you're searching for non-covalent interactions; if so, here's the usual Wikipedia link, https://en.wikipedia.org/wiki/Non-covalent_interactions
But more importantly here's the paper describing the software, http://pubs.acs.org/doi/abs/10.1021/ja100936w and here's Prof Rzepa's blog with 'non-covalent interaction' searched, http://www.ch.imperial.ac.uk/rzepa/blog/?s=non-covalent+interaction
That's the start, post again when you want to know how to implement it.

Useful links thanks for that. : ) But I don't really mean the broader non-covalent interactions I mean steric repulsion. And I have seen some papers describing "new algorithms" how to accomplish that but I am searching for an already existing software.

It looks like a software package called NBO 5.9 will do it.  Again from Prof Rzepa's blog, http://www.ch.imperial.ac.uk/rzepa/blog/?p=1887#comment-4793 ... hopefully this time.

NBO 5.9 is propietary though (and probably also very expensive), I know there is the open source alternative JANLA but I could not find information about if this includes sterics as well. Anybody used JANLA?

The software is JANPA - I did not use it so far.


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Valences of atoms in covalent compound
1 does the valency of an atom in a covalent bond  equals the number of bonds it has with  the other atoms in that compound?
2 If "yes" then typically to bonds  chlorine has in NaCl04
It has a valency of 7 ?? I know it has an oxidation number of 7+ at that case.
And what I know too that oxidation number / state = valency

the things you are mixing are oxidation state and valency.there is a lot of  difference between them.

Yes I know that valency and oxidation state have diffrent defintions but the same numeric value.Am I right?

No you are wrong.In carbon monoxide the oxidation state is +2 while valency of carbon is 4.

@THELONEWOLF. The oxidation state of what?

Valency is defined as the maximum number of univalent (e.g. hydrogen) atoms that an atom can bond to. In NaClO4, Cl is indeed has 7 bonds, but this is beyond the scope of the definition of valency and is referred to as "hypervalency" or "octet expansion" and basically means that the atom in question has more than 8 electrons in its valence shell.
The idea that oxidation state = valency does not always apply. The numbers may be the same in some cases, but not always, especially not for carbons in organic compounds. The oxidation number shows you the how electron rich/deficient is a given atom in a compound (single atoms have o.s. of 0).
Example: the C atom is tetravalent in all organic compounds. Take methane CH4 and methanol CH3OH. In both molecules the carbon is connected to 4 other atoms, however, in CH4 its o.s.= -4 and in CH3OH it's -2. Another example would be formaldehyde HCHO, where the carbon has an oxidation state of 0.


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For isomers, what does the term "connectivity" mean?
I would like an explanation for the term "connectivity" in the question "Do the compounds have the same connectivity? I do not understand what "connectivity", a vague term, means.

Same atoms are connected to each other (but in different positions).

Connectivity describes how the atoms are connected to each other. Consider a daisy chain (image courtesy of Wikipedia).

All the daisies are similar enough much like carbon atoms are similar enough between them; we cannot distinguish. Counting ten daisies in that chain, I could also represent it as (with the ends of the chain reconnecting):
$$\ce{-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-{}}$$
This compound would by cyclodecane.
Now imagine you replaced two daisies with violets. Of course, violets are immediately distinguishable from daisies like oxygen atoms are from carbon atoms. We now have a number of different options how to replace daisies with violets. They could be next to each other, separated by one, two three or four daisies. That would give rise to the following structures:

1-peroxecane
$\displaystyle\ce{-O-O-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-{}}$

1,3-dioxecane
$\displaystyle\ce{-O-CH2-O-CH2-CH2-CH2-CH2-CH2-CH2-CH2-{}}$

1,4-dioxecane
$\displaystyle\ce{-O-CH2-CH2-O-CH2-CH2-CH2-CH2-CH2-CH2-{}}$

1,5-dioxecane
$\displaystyle\ce{-O-CH2-CH2-CH2-O-CH2-CH2-CH2-CH2-CH2-{}}$

1,6-dioxecane
$\displaystyle\ce{-O-CH2-CH2-CH2-CH2-O-CH2-CH2-CH2-CH2-{}}$

In connectivity terms, in the first case and the first case only the two violets are connected to each other. In the second case and the second case only, there is a single daisy connected to two violets. These should be the obvious cases of different connectivities, although each different violet placement gives rise to a new connectivity.
These compounds are different; especially the first (which is a peroxide) and the second (which is the formyl acetal of a dialcohol) have very different chemical (and physical) properties from the others (which are simply cyclic diethers).
Added difficulty arises for issues that I have purposefully ignored in my daisy chain analogy: it does not consider three-dimensionality. Of course, atoms in reality are connected in a three-dimensional manner. You want to be sure that the connectivity is the same in all three dimensions (but not necessarily in the same orientations). Taking butan-2-ol as an example, we speak about identical connectivity whether we are dealing with (R) or (S)-butan-2-ol; because the same carbons atom are still connected to the same other atoms (e.g. carbon 2: Connected to a methyl group, an ethyl group, a hydrogen and a hydroxy group). But a different connectivity would be butan-1-ol, where carbon 2 is suddenly connected to a hydroxymethyl group and a second hydrogen (in addition to the ethyl group and the hydrogen which remained unchanged).

Connectivity refers to how the atoms are attached to one another.
Constitutional isomers have the same molecular formula, just the way the atoms are connected together, or connectivity, differs.


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Does any chirality-dependent inorganic chemical reaction exist?
In the particle physics, there are non mirror-symmetric reactions.
In the organic chemistry, particularly in complex proteins, there are reactions producing only a single chirality of molecules.
But what is in the intermediate regime? Does any inorganic chemical reaction exist with an asymmetric chirality behavior?
I suspect, maybe it would be possible with highly asymmetric electron clouds, probably with a significant EM octopole moment.

Note: as generally in the case of the "does ... exist" questions, giving a "no" answer is harder, but it is still possible.

I have the impression that my question is probably not very HQ in the grammar/spelling/terminology sense, so any improvements are welcomed.

Interesting idea! I really can't answer on inorganic chemistry in magnetic fields, but I remember some failed experiments (by a colleague who shall remain unnamed) on asymmetric resolution of organic compounds in an ESR magnet. On the more serious and maybe less related side, there's been a lot of research on photogenerated radical (ion) pairs in magnetic fields.

Here's a ref: https://www.iupac.org/publications/pac/2009/pdf/8101x0019.pdf but as I mentioned, this isn't about inorganic chemistry. You might find it helpful anyway :)

There are no reactions producing only a single chirality, unless they start from a single chirality. Besides, there are some reactions producing a slight excess of one enantiomer over the other as a result of some asymmetric physical influence (circularly polarized light, etc). Inorganics is no different from organics in this regard.

@IvanNeretin That's why the experiment which I mentioned miserably failed :D Achiral molecules can crystallize in chiral space groups. Sometimes, a reproducible preference for one enatiomer in one lab was observed. The reason typically isn't magnetic magic, but chiral contamination carried in by air.

@IvanNeretin With organic things I've tried to address the significantly chirality-dependent reactions in the biochemistry. Could you explain, why? I think, some asymmetry could be inserted in the system externally, for example by electric or magnetic field, or by both.

There are various chiral complexes so I don't see a problem.

@peterh What's to explain? That's right, some chirality must be introduced into the system from beyond, otherwise you'll never have any. Chemistry is perfectly symmetric, the way physics was before the discovery of CP violation.

@IvanNeretin That's not entirely correct. If you couple spontaneous symmetry breaking and a process that enhances selectivity (e.g., recrystallization), you can achieve enantiomeric naturally, though you can't control which one. So this means that you wouldn't be able to predict the selectivity or reproduce it, but it could occur.

@Zhe Maybe the direction of the spontaneous symmetry breaking could be controlled by some initiator, for example by an already chiral crystal.

Yes, that's right; the resulting enantiomer would be determined by pure chance, hence if you run the experiment many times, it would turn out "right" and "left" about equally often. Of course you might use a seed crystal, and probably with great results; that's the mentioned "chirality from beyond".

I think the answer to your question is that there could be some predictable symmetry breaking element (e.g., polarized light) that initiates a predictable symmetry breaking in a chemical system, but the problem is that this is no longer the realm of chemistry. I think this is also exactly what @IvanNeretin is saying.

@Zhe I think it could be also a chiral molecule.

@peterh No because that chiral molecule had to come from some kind of symmetry breaking. You're looking at a chicken-egg situation here.

See http://oldwww.iucr.org/iucr-top/comm/cteach/pamphlets/21/node7.html and http://pubs.acs.org/doi/abs/10.1021/ja907535g for some near-inorganic examples.

There are plenty of compounds that on crystallization from a racemic mixture, provide two sets of distinct crystals, with each crystal type being nearly enantiomerically pure.  Of course, there are equal amounts of both crystal types, but I think even a hypothetical non-chiral observer could sort such macroscopic crystals into two separate bins, a la Pasteur and tartaric acid.

There is a lot of speculation here about complex mechanisms and so on. But doesn't the fact that some inorganic compounds crystallise into single enantiomer, chiral crystals (quartz, for example) basically prove the point? Inorganic chirality isn't fundamentally any different to organic chirality.

Maybe the reaction rates between two chiral compounds could also differ if they differ also in their relative chirality. If there is compound1 and compund2, both chiral, then the reaction rate of L-compound1+L-compound2 is the same as R-compound1+R-compound2. But L-compound1+R-compund2 may have a different rate (or even different reactions).

@matt_black I think it means, yes they exist, but none of them is important?

A crystalline form of silicon dioxide known as α-quartz is optically active and can spontaneously crystallize in one of the two optically active form. However, which one of the two you get is defined purely by chance.
This, however, is different from "asymmetry" in particle physics, see here for details.

Here is a possible answer from inorganic chemistry centered on the $\ce{ClOClO}$ dimer.
Apparently, the $\ce{(ClO)2}$ dimer also exists as $\ce{(OCl)2}$ per the following important reaction leading up to chlorate (problematic from an environmental perspective) formation:
$$\ce{OClOCl + ClOClO -> OClO + Cl2 + OClO}$$
where $\ce{OClO}$ is chlorine dioxide ($\ce{ClO2}$).
Reference

Kinetics and mechanism of chlorate-chloride reaction by Rafaela T. P. Sant'Anna; Cristina M. P. Santos; Guilherme P. Silva; Ricardo J. R. Ferreira; André P. Oliveira; Carlos E. S. Côrtes; Roberto B. Faria, J. Braz. Chem. Soc. vol.23 no.8 São Paulo Aug. 2012  Epub, DOI:10.1590/S0103-50532012005000017 (PDF)

Ok, but it seems to me only an isomer. $(ClO)_2$ doesn't even seem to me like a chiral compound.

More research suggests a characteristic property associated with chiral compounds being a response to "circularly polarized light (CPL) "  (see  http://che.sjtu.edu.cn/EN/Research/tiao_mu/2014/7/31_Construction_and_Mechanism_Research_of_Chiral-Mesoscopic_Inorganic_Materials.html ). The example in that reference uses amino acid doped CuO nano particles.  I believe with that criteria, the list of pure inorganic compounds is limited and includes some esoteric complexes.

Per a source (see http://che.sjtu.edu.cn/EN/Research/tiao_mu/2014/7/31_Construction_and_Mechanism_Research_of_Chiral-Mesoscopic_Inorganic_Materials.html) that appears to directly address the organic/inorganic divide to quote:
"(3) Formation mechanism of chiral inorganic materials
Electronic transition-based optical activity (ETOA) is ubiquitous in organic materials. However, chiral inorganic materials exhibiting optical activity at their characteristic absorption bands attributed to electronic transitions are rare. Our group has reported several work in the field as follows. All of them, including of TiO2 nanotubes, CuO nanoflowers, chiral carbonaceous nanotubes, N-doped chiral carbon nanotubes et al exhibited optical response to circularly polarized light and their antipodal materials showed the mirror-imaged circular dichroism (CD) signals, which could be attributed to the electronic transitions from the lower energy to a higher energy under a dissymmetric electric field."


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Internal Energy Definition, Why is dU zero when dT is zero?
According to my Equation sheet, $$\delta U=\left(\frac{\delta U}{\delta T}\right)_V\,\delta T+\left(\frac{\delta U}{\delta V}\right)_T\,\delta V=C_V\,\delta T-C_V\left(\frac{\delta T}{\delta V}\right)_U\,\delta V$$ Where $\delta U$ is change in internal energy.
So one of the homework solutions said that for a system with a constant Temperature, $\delta T=0$ $\Delta U=0$. And I remember one of the TAs saying that $\Delta U=C_V\,\delta T$ and since $\delta T=0$ the value goes to zero, which makes sense. But she didn't even include the second term: $-C_V\left(\frac{\delta T}{\delta V}\right)_U\,\delta V$.
I was trying to make sense of the derivative but I'm really having trouble with the second term. The small letter $U$ outside the parenthesis is supposed to mean constant $U$.
Why am I allowed to ignore the second term?
The equation for enthalpy is very similar, and I have the same problem:
$$\delta H=\left(\frac{\delta H}{\delta T}\right)_p\,\delta T+\left(\frac {\delta H}{\delta H}\right)_T\,\delta H=C_p\,\delta T-C_p\left(\frac{\delta T}{\delta p}\right)_H\,\delta p$$

$\Delta T = 0$ only implies $\Delta U = 0$ under certain conditions (the same conditions apply to the differentials $\mathrm{d}T$ and $\mathrm{d}U$). The single most important condition is that the system must consist of only an ideal gas, in which case the partial derivative $(\partial T/\partial V)_U$ is equal to zero. I wrote quite a bit about this particular topic here.

Thanks for the link. it is very helpful. I will learn to derive the thermo equations! The problem did specify that the gas was ideal.

The second term in each of your two equations can be expressed solely in terms of the equation of state for the solid, liquid, or gas:$$\left(\frac{\partial U}{\partial V}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]$$$$\left(\frac{\partial H}{\partial P}\right)_T=\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]$$
In the case of an ideal gas, the right hand sides of both these equations are equal to zero.


text:
How to balance the equation for the reaction of aluminum bicarbonate and hydrochloric acid?
Here's what I have to balance:

Solid aluminum bicarbonate and hydrochloric  acid combine to form an aluminum chloride solution and carbonic acid.

Or in an equation form:
$$\ce{\text{__ } Al(HCO3)3 + \text{__ } HCl -> \text{__ } AlCl3 + \text{__ } H2CO3}$$
Side Note:
$\ce{Al}$ has a charge of $3+$ and $\ce{(HCO3)}$ has a charge of $1-$. To balance the ionic compound we have to change it to: $\ce{Al(HCO3)3}$.
The underscores before each compound in the chemical equation are just there for number values we can input in order to balance the equation.
My question is: what would be the reactants and products in this chemical equation? Would $\ce{HCO3}$ as a whole compound be a reactant itself?

The properly balanced equation is:
$\ce{\text{}Al(HCO_3)_3 + \text{3 }HCl \bond{->} \text{ }AlCl3 + \text{3 }H_{2}CO_{3} } $

This is a homework question. Please avoid providing the answer directly.

My question is... what would be the reactants and products in this chemical equation?

You give the equation, you just need to balance it.

Would HCO3 as a whole compound be a reactant itself?

No, $\ce{HCO3^-}$ is an anion.


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Cyanide's Lewis structure - electronegativity or octet rule?
The formula for the cyanide ion is $\ce{CN-}$. That gives us a total of ten valence electrons to work with. There are two obvious ways to build the Lewis structure.

$\ce{:^-C#N:}$
$\ce{:C=N^-::}$

(I hope that was clear. I'm not sure how to make ASCII Lewis structures. In that second example, the nitrogen has two lone pairs.)
In the first example, each atom has a complete octet, but the carbon has a formal charge of -1. In the second example, the carbon atom does not have a complete octet, and the nitrogen (the more electronegative atom) has a formal charge of -1.
My understanding is that when formal charge cannot be avoided, negative formal charge should reside on the most electronegative atom.
Which rule takes precedence over the other? Using cyanide as an example, should I:

Follow the octet rule and go with the first example?
Follow the electronegativity rule and go with the second example?
Treat them as resonance structures?

Octet rule should always take priority. I do not know of any counterexamples to this. (In fact, they are resonance structures. The same rules - octet, electronegativity - allow you to judge which is the greater resonance contributor. In this case the first resonance form contributes more.)

Related and related.

Which rule takes precedence over the other? Using cyanide as an example, should I:

Follow the octet rule and go with the first example?
Follow the electronegativity rule and go with the second example?
Treat them as resonance structures?

You should treat them as resonance structures (both are valid Lewis structures, they have the same connectivity of atoms, and they are related by one of the usual electron-pushing operation, i.e. double bond turning into single bond and lone pair).
Then, you can look at the other rules to make a guess which resonance structure should have higher weight. The octet rule takes preference over the formal charge rule:

The structure on the left follows the octet rule for both atoms, whereas the one on the right has only six electrons associated with the carbon atoms. So if you only want to write a single Lewis structure, the one on the left is better. If you have space, write the pair of resonance structures with a double arrow ($\ce{<->}$) between them.
Combining the cyanide ion with a hydrogen ion, there are two resulting molecules that exist, hydrogen cyanide (HCN) and hydrogen isocyanide (HNC). Hydrogen cyanide, for which you can write a Lewis structure that follows the octet rule, is more stable than hydrogen isocyanide (for which you can't). Once these two molecules get deprotonated, they yield the same ion, cyanide, described by the two resonance structures.


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How to calculate the pH of a propanoate buffer after addition of sodium hydroxide?
I've calculated the pH of the buffer solution, as seen.
The question I'm referring to is part d), which asks: 'Calculate the pH of the solution after 0.01 mols of NaOH are added to  500$cm^3$ of the solution'.

The addition of the NaOH will "use up" 0.01 mol of propanoic acid and "create" 0.01 mol of propionate. To make things easier, just work in molarity. Thus 0.02 M NaOH.
The pH of the final solution can be found by adjusting the equation you used in part b,
$pH = -log \left( 1.26 \times 10^{-5}\frac{0.1-0.02}{0.05+0.02} \right)=4.84$
The result agrees with our expectations. Since the solution is a buffer, the pH is not expected to drastically change with the addition of a strong base. It is buffered.


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Synthesis of imine
This synthesis problem is giving me a lot of trouble. The main problem is that the four steps I have down I think are the only ones I need but there is another blank. Anyone have any thoughts?

With so many options, I wouldn't be surprised if there was another equally valid path you were expected to find. Nevertheless, if you insist on finding something "wrong" with your perfectly valid path, I would say that you might want to have a H2O step after the LiAlH4 reduction. (Usually when writing chemical schemes, aqueous workup after every step is implied so you are not supposed to have to write H2O after every step.)

That worked, thank you. I'm just so frustrated because how was I supposed to know that? I can't find a strict h2o step anywhere near this section of my notes

In the lab, there are quite a few reactions where you could probably get away with not performing an aqueous workup. However, with the LiAlH4 step, it is crucial: 1) You need to protonate the alkoxide 2) You need to destroy any remaining LiAlH4 in solution. So, I figured that if you had to add a H2O step, it was probably after the LiAlH4.

And you need to remove all of the lithium salts.

How are you supposed to know that?  I understand your frustration.  Most reactions have a 'workup' step which for the most part is about neutralizing strong, reactive reagents.  In this case, the strong reagent being neutralized is the reasonably strong base alkoxide.  This step is normally considered 'trivial' by organic chemists, and is often not dwelled on in lecture discussion. The only comfort I can offer is this will get easier with exposure to more synthetic pathways.


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Identity a metal based on its properties

A metal X can be cut with a knife. When placed in water, metal X floats and melts. What is X? The options are calcium, copper, magnesium and potassium.

I think it can't be potassium because it will react vigorously with water, So I believe it's calcium. But the only problem is that calcium is denser than water so it will sink.

Come to think of it, "reacts vigorously" is pretty much the same as "floats and melts".

Copper is much denser than water and it does not react with it. Magnesium is also denser than water and reacts only very slowly. Calcium will also sink in water, although it will react with it producing hydrogen, so small pieces of the element may float on the surface. However, it would definitely not melt. The correct answer is potassium - it is a very soft metal, less dense than water and reacts with it producing enough heat to melt it (potassium melts at only $\pu{63.5^\circ C}$).


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What is the difference between hexanes and petroleum ether
I'm trying to use hexanes as the nonsolubilizing recrystallization solvent in a multisolvent recrystallization (benzene is the solubilizing solvent) that requires an oxygen free atmosphere and hexanes is present in the glovebox. I am using it in place of light petroleum ether (not in glovebox) and was wondering what the differences between hexanes and light petroleum ether were to determine if I can adequately substitute the two.

My understanding is that they're fairly similar, but hexanes is purified to be mostly hexanes, whereas for petroleum ether, you get a mixture of hydrocarbons that boil in a given range. For recrystallization purposes, the substitution is probably fine. I guess it depends on how finicky your recrystallization is.

Petroleum ether:
This is a mixture of a lot of different compounds (aliphatic hydrocarbons in this case) that as a mixture have a boiling point within a specified range. The contents are dealt with as having low amounts of aromatic compounds and high amounts of this and that, not one specific formula. It's like a blackbox to me and can be used for crystallisations that are usually easy to do.
Hexane:
In the synthesis laboratory I worked in, hexane was dealt with as we would deal with petroleum ether, since it was a mixture the many possible constitutional isomers and not pure n-hexane. We bought the technical grade hexane in our chemical store and purified it with a solvent purification system to have it dry and compatible for glove box use. Since they appeared so clean in our Schlenk flasks I thought in the beginning that it was pure n-hexane but a simple 1H NMR spectrum taught me otherwise.
The hexanes mixture was used for medium difficulty crystallisations.
Pentane:
The pentane we bought and purified, however, was n-pentane and was used for difficult crystallisations that were not easy to reproduce. Already among organometallic compounds there are examples in which obtaining crystals at all is an enormous struggle and having single crystal quality is simply a matter of hope.
Therefore, by going from petroleum ether to hexane to n-hexane (or n-pentane) is a matter of going from a wide mixture to a compound to a pure chemical.

Any rationale for this finding, that stuff crystallises better from pentane than hexane or petrol ether? Btw., the use of n-hexane is frowned upon, because it is supposed to be rather more toxic than e.g. heptane. A mixture named "isohexane" seems to be widely used as substitute, where cyclohexane is regarded as too expensive.

Ah, thank you! I knew about benzene being frowned upon but not n-hexane.

The difficulty in crystallisation comes from highly symmetric clusters, large solvent spaces in the crystal lattice, or sometimes the compound crystallises only once and cannot be recrystallised because it decomposes. To find the optimal crystallisation conditions, it's best to rely on solvents with high purity and always the same properties. This cannot be found in petroleum ether. Petroleum ether is different in laboratories all over the world and pure compounds have a higher chance of leading to a reproducible experimental procedure.


text:
Why is the Debye Hückel Limiting Law called the "limiting law"?
I could not find any source which explains this point well.
I looked up Atkin's physical chemistry and Castellan's physical chemistry book but could not quite understand the logic bhind it being called the limiting law.

It is only valid in the limit of low concentration. You see significant deviations from it at higher concentrations

For a 1:1 electrolyte, a  limit of $10^{-3} \pu{mol dm^{-3}}$ is typically put on the concentration. There are similar equations that can be used at higher concentrations but involve parameters that have to be fitted to the data.

The Huckel limiting law allows us to determine the activity coefficient of an ion in a dilute solution of a known ionic strength.
Here's an appealing answer which makes perfect sense to me that I found on EveryScience

The name 'limiting law' is applied because in the limit of arbitrarily low molalities (ie as the concentration of the solute gets
  closer and closer to zero), all solutions are expected to behave in a
  manner consistent with the expression.


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Are there art supplies that create a new color by mixing through a chemical reaction?
When two elements or compounds react together, they form new compounds, which more often than not assume new colors. If the two elements or compounds don't react, they form uniform or non-uniform mixtures, in which case the new color which results from the mixing is the result of the direct combination of the light of emitted by the original elements or compounds. These colors affect the red, green, and blue light receptors inside human eyes, producing a perceptual sensation of color.
I want to know, are there any safe colored substances or color media, which can be used together by an artist, notably a typical artist working from home on paper, cardboard, or canvas materials, and which when mixed together to form a new color, as artists do, actually generates a color resulting from a chemical reaction rather than substance mixing, and, above all, the chemical reaction is safe in terms of not causing damage to the eyes or skin.

Can you please reword the first paragraph? Admittedly, I might not have fully understood what you asked. Anyway: I am not aware of a system where different dyes are reacted with each other. This doesn't mean that there isn't any ;-) Chemical reactions to play a role when reactive dyes are (irreversibly) anchored to a fibre or when the dye is first applied in its reduced form and then reacted (with oxygen) to yield the final colour. This is pretty much what happens during vat dying with indigo or purple.

Interesting answer @KlausWarzecha. What about the colors wood, brick, plastic, metal dyes for houses, furniture and cars, as well aspens, pencils, pastels, oil paint, and eyeliner, and lipstick, nail polish are made of. Are there any chemical reactions taking place there, before these dyes are used, or are those just mixtures as well, and if so, then of what?

As for dyes, as chemical compounds are scarce, both from a sustainability point of view, but abstractly, in terms of number of combinatorial combinations, I would guess that dyes must result from mixing different combinations of substances, otherwise we could not have that many colors. I wonder where these substances are taken from.

What about hair dyes. Do they actually just blend, or do they react upon entering the hair. What are they made of, and what makes them safe and skin friendly versus harmful to both the skin and the environment?

Ceramic glazes often change colors during firing - this is typically just an oxidation reaction.

Unlikely, because the components of this reaction would likely be susceptible to reactions with air or water. Definitely it must be a special combination, which is hardly usable as paint individually. Real paint also is usually not coloured with dye, but pigments, i.e. inorganic particles.  You can sure make something that works, but it probably won't be harmless.


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What equipment is used for two-dimensional IR spectroscopy and how can the molecular structure of a compound be resolved?
On the Wikipedia page about two-dimensional IR spectroscopy it is said that 2D IR spectroscopy can be used to study the

molecular dynamics with femtosecond time resolution.

According to Wikipedia it

has for example been used successfully to study hydrogen bond forming and breaking and to determine the transition state geometry of a structural rearrangement in an iron carbonyl compound.

Now I'm curious. What equipment is necessary to perform 2D IR experiments? Does somebody have experience with 2D IR spectroscopy and could explain how the geometry of the system is inferred just from the (2D) IR spectrum.

As I understand, you have one broadband light source to do simple (non-FT) absorption spectroscopy, and another one to excite the system beforehand. Now you vary the time distance between excitation & measurement, and FT every point in the measured spectra along that development time axis.

You can find a schematic drawing of the box-CARS 2D IR setup in the publication
Enhancing signal detection and completely eliminating scattering using quasi-phase-cycling in 2D IR experiments, Robbert Bloem, Sean Garrett-Roe, Halina Strzalka, Peter Hamm, Paul Donaldson
https://pdfs.semanticscholar.org/e134/cdbb98a8d899a17848f8242c8160d4913f0f.pdf

Usually, we would prefer the answer doesn't boil down to just giving a link. Can you summarize some key points from the paper or maybe include the image in this answer?