Why doesn't a catalyst shift the position of equlibrium, when temperature does?

Your presumption in the first paragraph is incorrect. The fact that more particles collide with enough energy to make the activation energy with an increase in temperature means that the rate of reaction increases as temperature goes up, but that doesn't affect the equilibrium of the reaction.

Instead, the effect on temperature has to do with the comparative free energy of the two sides of the equilibrium. The controlling factor in where the equilibrium lies is the difference in free energy between the reactants and products (the ΔG). In fact, the equilibrium constant has a very simple relationship to the free energy: $\ce{\Delta G = -RT ln K }$ ... or $\ce{K = e^{-\frac{\Delta G}{RT} }}$. The ΔG is made up of two components - the enthalpy and the entropy: $\ce{\Delta G = \Delta H -T\Delta S}$, where T is the (absolute) temperature. The side of the reaction which is favored is the one with the lowest (most negative) free energy.

So when we substitute, we get the following relationship for the temperature dependence of the equilibrium constant, in terms of entropy and enthalpy:

$$ K = e^{-\frac{\Delta H}{RT} } \cdot e^{\frac{\Delta S}{R} } $$

The temperature dependence is only in the enthalpy ($\ce{\Delta H}$). Enthalpy, by the way, is what we normally think of as endothermic/exothermic. Roughly, reactions with a positive $\ce{\Delta H}$ absorb heat and are thus endothermic, and those with a negative $\ce{\Delta H}$ give off heat and are thus exothermic.

So imagine we have an endothermic reaction (positive $\ce{\Delta H}$), which means that the exponent for enthalpy in the above equation is negative. When we increase the temperature, that exponent gets smaller in magnitude (closer to zero), which means it becomes less negative. This causes an increase in $e^x$, which means that K gets bigger, which means the equilibrium shifts in the forward (endothermic) reaction. Likewise, if you have the reaction written in the exothermic direction, $\ce{\Delta H}$ is negative, the exponent is positive, an increase in temperature causes the exponent to become less positive, causing a decrease in the $e^x$, which results in a decrease of K -- the equilibrium now shifts in the reverse (endothermic) direction.

So while the statement that increasing temperatures means that the equilibrium shifts in the endothermic direction is true, it has nothing to do with the activation energy of the reaction. The arguments above are from a purely thermodynamic standpoint, and don't involve any consideration about the mechanism of reaction or the activation energy.


The point of using alcohol to extract things from herbs is that alcohol solubilises things that are not soluble in water. So using grape juice as an alternative is pretty pointless since it is just water with a lot of other water-soluble things in it (like sugars).

So using grape juice is a pretty poor alternative to using alcohol.

Grape juice will ferment under some circumstances, converting some of the sugars into alcohol. But this won't work well unless the conditions are controlled carefully and won't work at all if it is cold (45 F is cold). If you don't do what winemakers do, there is no telling whether it will work or whether it will work without producing nasty results. And pasteurised grape juice won't ferment at all if you don't add yeast and keep it clean. If you don't keep it clean you might well get some nasty strain of natural bug rather than a good wine yeast which will have unpredictable results.

Moreover, the best level of alcohol you are going to get even as an experienced winemaker is 10-15%. This is nothing like as useful as pure or 50% alcohol for solubilising or extracting things that are not water soluble. And grape juice with no alcohol isn't going to be a much better solvent than water (and most likely will be worse as it is already full of other dissolved things).

The old school recipes you are trying to avoid exist because people have tried these experiments before and found what works. Trying to invent new recipes that ignore the "old school" experience is a pretty futile task: what you propose isn't in the old school recipes because it doesn't work.


Why does the Michaelis-Menten constant remain constant in the event of non-competitive enzyme inhibition?

This is a question that is relatively easy to answer mathematically.

To keep the algebra a little simpler, let me use a simplified Michaelis-Menten mechanism, where we assume that the enzyme-substrate complex (ES) and the free enzyme are in equilibrium.  (We could use the less restrictive pseudo-steady state hypothesis on (ES) and obtain the same result for non-competitive inhibition, at the expense of more complicated math.)

$$\ce{E + S <=>[K_M] (ES) ->[k3] E + P}$$

Non-competitive inhibition means that the inhibitor doesn't care at all whether the substrate is bound to the enzyme.  Either on (ES) or on (E), it will bind with affinity $\mathrm K_{\mathrm i}$.  We assume that (ESI) does not react, at all, to form products.  So the only other two reactions we need are:

\begin{align}
\ce{E + I &<=>[K_i] (EI)}\\
\ce{(ES) + I &<=>[K_i] (ESI)}
\end{align}

Now to get the rate equation, we first perform a mass balance on enzyme:

$E_0 = E + ES + EI + ESI$

We use the equilibria we've assumed above to get:

\begin{align}
(ES) &= \frac{(E)(S)}{K_M}\\
(EI) &= \frac{(E)(I)}{K_i}\\
(ESI) &= \frac{(E)(S)(I)}{K_M K_i}
\end{align}

Thus

\begin{align}
E_0 &= E\Big(1 + \frac{(S)}{K_m} + \frac{(I)}{K_i} + \frac{(S)(I)}{K_M K_i} \Big)\Big)\\
E &= \frac{E_0}{\Big(1 + \frac{(S)}{K_m} + \frac{(I)}{K_i} + \frac{(S)(I)}{K_M K_i} \Big)}
\end{align}

The rate equation we are after is just the formation of product, $\frac{dP}{dt}=k_3 (ES)$.

$$k_3 (ES) = k_3 \frac{(E)(S)}{K_M}$$

Now we know $(E)$ from our enzyme mass balance, so

\begin{align}
\frac{\mathrm{d}P}{\mathrm{d}t} &= k_3 E_0 \frac{(S)}{K_M \Big(1 + \frac{(S)}{K_m} + \frac{(I)}{K_i} + \frac{(S)(I)}{K_M K_i} \Big)}\\
&=k_3 E_0 \frac{S}{K_M + S + \frac{I}{K_i}\left(K_M + (S)\right)}\\
&=k_3 E_0 \frac{S}{(K_M + S)\cdot (1 +\frac{(I)}{K_i})}\\
&=\frac{k_3 E_0}{(1 +\frac{I}{K_i})} \frac{S}{(K_M + S)}
\end{align}

The only thing different about this expression is that the "$V_{max}$" part of the expression, which would have been $k_3 E_0$ for a simple reaction with no inhibitor, is now $\frac{k_3 E_0}{1 + I/K_i}$.  In other words, $V_{max}$ is the only parameter changed by the inhibition mechanism we have assumed.

The apparent $K_M$ of the non-competitively inhibited reaction is still $K_M$, because if you plug in $S = K_M$, you get a rate that is half of the maximum (of the inhibited reaction).

  Wikipedia attributes the invariance of  $K_M$ in non-competitive inhibition to the fact that the actual binding of substrate to an active site is unaffected. Now I agree that the actual binding of the substrate is unaffected here...but how does that have any effect on $K_M$?

You can think of $K_M$ as an apparent equilibrium constant for the binding of substrate to free enzyme.  So if something doesn't affect substrate binding, it probably won't affect $K_M$.

  So, when $V_{max}$ decreases during non-competitive inhibition (the only point I and Wikipedia concur on), shouldn't the $K_M$ decrease correspondingly (based on the definition of $K_M$? Is the Wikipedia article incorrect in this regard?

No, Wikipedia is correct.  Enzymologists have adopted very confusing terminology for no good reason, but non-competitive inhibition is not the same uncompetitive inhibition.  In uncompetitive inhibition, both apparent $K_M$ and apparent $V_{max}$ are changed.

If I gave you two vials, A and B, of the same enzyme, and one of them had been pre-mixed with a high concentration of a non-competitive inhibitor, and I asked you "what's the $K_M$ you measure for each of these two different enzyme preparations?,"   then you'd get the same number for each vial.  If 5 mM of substrate is the concentration at which enzyme from vial A reaches half of its maximum rate, then you'll find that 5 mM of substrate is also the concentration where enzyme from vial B reaches half of its maximum rate.


The dotted lines in this structure could be thought of as partial bonds. In other words, none of those three bonds has a bond order of 1 with two electrons attributable to that bond.

Another way to look at this structure is that the dashed lines represent multicenter bonding. A common example of multicenter bonding is diborane $\ce{B2H6}$. The structure of diborane features 3-center-2-electron bonds (often abbreviated 3c2e bonds - in this notation, a typical single sigma bond is a 2c2e bond) with two hydrogen atoms shared by both boron atoms:

These bonds in $\ce{CH4+}$ could be considered a 3-centered-3-electron bond as originally each C-H bond would have had two electrons. The radical cation is one electron short, so there are only three electrons shared between those three atoms.


Why does ClO₄⁻ only have 3 double bonds?

I think the best argument comes from resonance.

Drawing perchlorate with three $\ce{Cl-O}$ double bonds and the negative charge on the oxygen allows for three resonance structures to be drawn (four total). This resonance stabilization  dominates the Lewis structure over simply drawing four double bonds to chlorine.

In other words, energetic benefit of charge spread out over four oxygens > charge localized on one chlorine.


Why does the reaction quotient use the products (multiplications) of reactants and products, rather than their respective sums?

The derivation cited can help you understand it mathematically. But the answer also turns out to be satisfactory intuitively. Let me try to explain how.

The term $RT\ln Q_c$ is derived form the entropy part or $\Delta G=\Delta H-T\Delta S$, and in a slightly abstract way, represents entropy of mixing of the products and the reactants. This is why reactions which are spontaneous initially, as concentrations change, reach to the equilibrium state before complete conversion of reactants into products. If there was no entropy of mixing, a reaction which had negative free energy change(under any set of conditions) would continue to have a negative change (hence, spontaneous) until all of the reactants is converted to products, which will be the state of maximum entropy. But since there is entropy of mixing, the state of maximum entropy is reached before all the reactants are converted into products (because somewhere in between the entropy of mixing due to having a mixture of reactants and products outweighs the negative $\Delta G$ for the reactants to convert into the products) , and the concentrations at which this maximum entropy state can be reached is precisely what the equilibrium constant tells us.

To calculate the entropy, we need to realize the fact that certain components of the reaction may have a greater contribution in deciding the equilibrium constant than others because, at the maximum entropy state, the components are present according to their stoichiometric relations and hence one might be in a well-defined excess over other. The relative contributions of each species to the entropy is logarithmic, that is, the increase in entropy is a logarithmic function of concentration (related slightly in a abstract manner to $S=k_B\ln \Omega$). Therefore, the contribution to the free energy and consequently in deciding the $K_c$ is logarithmic in the concentrations. The stoichiometric coefficientts are multiplied to the logarithmic contributions of each(which appear as the power to which the concentrations are raised when brought inside the logarithm) and the relative contributions are simply added, but since the addition of logarithms means multiplications of its arguments, we get the familiar $RT\ln Q_c$ term which gives the mentioned definitions of reaction quotient and the equilibrium constant.

I have used reaction quotient and equilibrium constant interchangeably but the RQ represents an unfulfilled equilibrium constant and hence represents the same drive towards maximum entropy as $K_c$ (which is the state of maximum entropy itself).


This reaction will happen by both SN1 and SN2 mechanism, apart from this SN1' and Sn2' (allylic mechanism ) will also occur. Thus from SN1 and Sn2 one will get primary cyanide and from SN1' and SN2' one will get secondary cyanide .( read organic chemistry by jerry march for the detailed explanation).
I am attaching a screenshot of clayden which explains this beautifully.


It is often asked "what is waving when a particle is described as a wave?" and the correct response is "existence itself is waving."
The wavefunction is most elegantly described as a description of reality for your system. You then act on the wavefuntion with a hamiltonian to yeild an eigenvalue to quantify the energy.


Cerium electronic configuration change from 0 to +2 oxidation state

The lanthanides are most stable in their 3+ oxidation state. For example, lanthanum is perfectly happy changing from

$$\mathrm{La}^0 = [\ce{Xe}]\mathrm{(5d)^1(6s)^2}$$ to

$$\mathrm{La}^{3+} = [\ce{Xe}]\mathrm{(5d)^0(6s)^0}$$

Then, as you go along the lanthanide series, you're adding more 4f electrons. For example:

$$\mathrm{Ce}^{3+} = [\ce{Xe}]\mathrm{(4f)^1(5d)^0(6s)^0}$$
$$\mathrm{Pr}^{3+} = [\ce{Xe}]\mathrm{(4f)^2(5d)^0(6s)^0}$$
$$\mathrm{Nd}^{3+} = [\ce{Xe}]\mathrm{(4f)^3(5d)^0(6s)^0}$$

etc.

This electronic configuration is extremely stable for all lanthanides, and there are very few exceptions to it.

Ce(II) is extremely unstable, and if there's one way to stabilise it, that would be adoption of the Pr(III) electronic structure. This way Ce has no 5d or 6s electrons, which makes lanthanides happy (apologies for the non-scientific way of framing this).


How about
$\ce{O=O+-S+=O+-O-}$ ? I don't know if it exists but it is a closed shell species and has a delocalized pi bonded structure that might give it some stability.


Like ron already stated, the reason for that arrangement is the possibility to form more intermolecular hydrogen bonds. For oxalic acid there are two modifications which depict this quite clearly.

From charge considerations itself it is not obvious, that the periplanar alignment should be better. However one can see, that there might be a little less repulsion. Given here are natural partial charges from NBO/BP86/cc-pVDZ.

A rotational scan reveals that the rotation barrier is just around $7~\mathrm{kJ/mol}$ and the difference between these are just around $2~\mathrm{kJ/mol}$ (BP86/cc-pVDZ). The dihedral angle of $0^\circ$ refers to the hydroxyl moieties being on the same side.

The barrier results mainly from breaking the partial $\pi$ bond between the two carbon centres.

The $\ce{O...H}$ intramolecular distance is around $230~\mathrm{pm}$ and in gas phase and higher temperatures hydrogen tunneling should be also considered.

In aqueous solution, exchange rates between water and oxalic acid are very fast, so that all oxygen atoms should be equal - there is no difference observable at this time scale.
\begin{aligned}\ce{
(COOH)2 + 2H2O &<=> HOOC-COO- + H3+O + H2O\\
               &<=> (COO^{-})2 + 2H3+O\\
}\end{aligned}


Why do Grignard reagents add to propargyl alcohols?

The addition of Grignards to propargyl alcohols occurs via the alkoxide salt, after deprotonation:[1]

Here:

the reactivity may be explained by the stabilising influence of complexation;
the regioselectivity may be explained by the preference of a five-membered chelate ring over a four-membered one; and
the stereoselectivity arises because syn addition requires a five-membered ring with a trans double bond and is therefore precluded.

In fact, propargylic alcohols also get reduced by $\ce{LiAlH4}$ (Hans Reich's website has a page on it), with the same trans stereochemistry observed.

Reference

(1) Forgione, P.; Fallis, A. G. Magnesium mediated carbometallation of propargyl alcohols: direct routes to dihydroxydienes and enediyne alcohols. Tetrahedron Lett. 2000, 41 (1), 11–15. DOI: 10.1016/S0040-4039(99)01994-2.


Dubowski’s Formula vs Henry's law applied to alcohol sensor 'calibration'

Dubowski's formula is a parameterization of the temperature dependence of the water-vapor partition constant (Henry's law constant). It is based on data accumulated over 63 years (see Ref 1). It is assumed that Henry's law holds, in other words, that the vapour pressure over the solution is proportional to the concentration in the solution as $\frac{p}{c} = K_{a/w}=$ constant.
It can be used as a proxy for the concentration of alcohol in the blood although it is not expected to be entirely accurate, as Dubowski comments (Ref 2):

We believe that the conversion of a breath quantity to a blood concentration of ethanol, for forensic purposes, should be abandoned and that the offense of driving while under the influence of alcohol should be statutorily defined in terms of the concentration of ethanol found in the breath in jurisdictions employing breath analysis.

Irrespectively of the details involving breath-alcohol analysis, Henry's law states that the vapor pressure is linearly proportional to the concentration in solution as $p=K_Hc$, and is applicable in practice when the alcohol concentration in the blood is very low (it is in general applicable as a limiting law).The following figure shows the partial pressure of ethanol in dilute aqueous solution at $25 ^\circ C$ as a function of the mole fraction of alcohol in solution, where the dashed line represents the pressure predicted by Henry's law:

For reference, a blood alcohol concentration (BAC) of 0.1%(w/w) translates into a mole fraction of $\pu{3.9E-4}$.
The Henry's law constant is obtained by extrapolating to zero concentration of the solute (here alcohol). In general, water-ethanol solutions are highly non-ideal due to hydrogen bonding and would not be expected to follow Henry's law except for very dilute concentrations of one component in the other. As soon as interactions between solute molecules become significant the law can be expected to break down.
The difference in the functional form of the temperature-dependence in Dubowski's formula compared to Van't Hoff's can be ascribed to differences in units: one uses T in Kelvin, the other in degrees Celsius. With proper application of Taylor series formulas they can be shown to be equivalent within narrow T ranges.
References

K.M.  Dubowski.  Breath-alcohol simulators: Scientific basis and actual performance. J. Anal ToxicoL 3" 177-82 (1979).

M F Mason, K M Dubowski. Breath-alcohol analysis: uses, methods, and some forensic problems--review and opinion. J. Forensic. Sci. 1976 Jan;21(1):9-41.


How small a peak could one identify by GCMS

You can get GCMS peaks for very small components of the overall mixture.

However, most GCMS patterns are electron impact, and hence probabilistic, so if you don't have enough sample, the fingerprint won't be reliable.

In my experience 0.3% produces spectra which are 'good enough'... as long as you don't inadvertently overload your column.


Difference between "physical quantities" and "units"?

Mass is a physical quantity. Its unit may be gram or kilogram.
Volume is a physical quantity. Its unit may be the liter or the cubic meter


Generation of singlet oxygen by irradiation

Rose bengal (pictured below) is a member of the fluorescein family.

(image source: wikipedia)

There are 3 steps in the process where rose bengal sensitizes the formation of singlet oxygen..

\begin{gather}
\ce{RB ->[$h\nu$]\  ^1RB}\tag1\\
\ce{^1RB ->T[ISC]\  ^3RB}\tag2\\
\ce{^3RB + ^3O2 ->T[sensitization]\  ^1RB + ^1O2}\tag3\\
\end{gather}

In the first step, ground state rose bengal (a ground state singlet) absorbs light producing an excited singlet state.  Because of the heavy iodine atoms in the molecule intersystem crossing (ISC) to the excited triplet state rapidly occurs (heavy atom effect). In the presence of ground state triplet dioxygen, a process called "sensitization" occurs. In this step the excited triplet rose bengal transfers its energy to the dioxygen promoting it to the excited singlet state while the rose bengal is deactivated and returns to its single ground state.  In this sensitization step, 2 triplets are converted to 2 singlets so, importantly, spin is conserved.


Percentage of trans-stilbene traces in cis-stilbene H NMR

You have to focus on the $\ce{^1H}$ of the double bond.
They are singlets because both molecules are completely symmetrical.
As per the tables, the chemical shift of the cis isomer is $5.25+1.38-0.07=6.56$ ppm.
On the other hand, the chemical shift of the trans isomer is
$5.25+1.38+0.36=6.99$ ppm.

These theoretical values correspond to the peaks of 6.60 and 7.11 ppm.
Since their integrals are of 1 and 1.1571 respectively, the proportion of trans is of:

$\frac{0.1571}{0.1571+1}100=14\%$


What decides -1/2 and +1/2 spin values of subatomic particles?

Spin is an intrinsic property of any nucleus, and can be 0, 1/2,1, 3/2, 2........ A nucleus of spin I can adopt 2n+1 stable states or orientations. For 1H, I=1/2, and this means it can adopt two possible stable spin states; +1/2 or -1/2. The energy levels of these two spin states are different (and proportional to the strength of the magnetic field); one is slightly higher than the other, and so there is a population difference between the two states. However, the energy difference is incredibly small, and difference in populations (called the Bolztzmann distribution) is only about one in a million. The individual nuclei can hop back and forth between the two states quite readily, provided there is sufficient energy transfer (into or out of) the system. Indeed, this is the basis of the NMR experiment - we can manipulate spins between the two spin states with relative ease.

For a nucleus with a spin greater than 1/2, there are more than 2 stable spin states that the nucleus can exist in. Deuterium (2H) has I=1, and can adopt one of 3 stable spin states. The distribution of spins amongst these states is approximately equal (less the BD of approx one in a million). Any individual nucleus will switch between these states. Lutetium176 has a spin of 7, and so can adopt 15 different stable spin states.


Is there a term that denotes 'molecules and molecular constituents'?

This is pretty much link-only, but anyway...

When searching for official definitions, nomenclature, or recommended procedures, the organisation providing all these is the IUPAC (International Union of Pure and Applied Chemistry)!

An IUPAC book series, known as the Colour Books, provides the normative definitions and guidelines on various topics. These are the results of what chemists in the different committees in the IUPAC divisions have eventually agreed upon.

To my knowledge, all these Colour Books are available online.

As kindly mentioned by Loong in a comment, the online versions are not necessary the latest ones! The online version of the IUPAC Nomenclature of Organic Chemistry is from 1993. In addition, a 2004 draft copy is legally available.

Organic chemists will often refer to the Blue Book (Nomenclature of Organic Chemistry), inorganic chemists to the Red Book (Nomenclature of Inorganic Chemistry), etc.

As far as your question is concerned, the Gold Book (IUPAC Compendium of Chemical Terminology) provides the definitions for molecule and molecular entity.


Mathematics in inorganic and quantum chemistry

For better understanding of molecular symmetry you indeed need some mathematics, namely, some basics of group theory, but aside from that I doubt that you need some mathematics apart form arithmetic and calculus for general, inorganic, organic, and even physical chemistry.

For quantum mechanics you absolutely need to familiarize yourself with the concept of complex numbers and you have to learn some basic approaches for solving differential equations. For an advanced insight into quantum chemistry you need some linear algebra, by which I mean the theory of vector spaces both finite- and infinite-dimensional, not just finite-dimensional part of it, which, in a sense, reduces to matrix algebra.

P.S. Abstract algebra is nothing but a common name for different areas of mathematics which studies different so-called algebraic structures. Say, group theory studies groups, linear algebra studies vector spaces, etc.


Does a sulfuric acid/nitric acid mixture dissolve PTFE (ie. teflon)?

I can't claim to have any personal experience with that sulfuric acid/nitric acid cleaning mixture specifically, but, as far as I'm aware, PTFE is generally inert towards acids. As a testament to that, it is apparently the material used for storing fluoroantimonic acid. I found a chemical resistance chart confirming its stability against strong acids. It is evidently resistant to concentrated sulfuric and nitric acids, as well as aqua regia (a concentrated mixture of nitric acid and hydrochloric acid). I have used the last of those for cleaning PTFE lab equipment before with no problems.

Having said all that, it would be wise to consult the handling and safety data for your particular labware to be certain.


Why do atoms "want" to have a full outer shell?

You are attaching too much importance to Lewis structures. The 8-electron rule and Lewis structures which are derived from it are only rough guidelines for working out the electronic structure of a compound in very broad strokes. Often these broad strokes are accurate enough to make some meaningful statements about molecular properties but it does not accurately describe the true electron or charge distribution in a compound. Take water for example. As you say, the 8-electron rule would predict that the hydrogen atoms each transfer one electron to the oxygen molecule which would then carry a charge of -2. But experiments show a different result: here you can find a paper that determined

  that a charge of approximately $0.5 e$ is transferred from each
  hydrogen atom to the oxygen

which would amount to a partial charge of $-1 e$ on oxygen and here you can find a similar claim:

  The charge distribution depends significantly on the atomic geometry and the method for its calculation but is likely to be about
  $-0.7e$ on the O-atom (with the equal but opposite positive charge equally divided between the H-atoms) for the isolated molecule.

And there not being a $-2e$ charge on oxygen is quite understandable if you look at the electron affinities of oxygen (e.g. here): You see that the first electron affinity is $-142 \, \mathrm{kJ}/\mathrm{mol}^{-1}$ and thus energy is gained by the process of transferring a free electron to a neutral oxygen atom. But the second electron affinity is $+844 \, \mathrm{kJ}/\mathrm{mol}^{-1}$, so another electron transfer is very costly. The quite extreme situation of a -2-charged oxygen described by the 8-electron rule is very rarely found in real compounds. Even in ionic compounds like $\ce{NaCl}$ the electron is not 100% completely transfered from the electropositive to the electronegative element.


Non stationary effects of molecular orbitals

First off, let's define our terms. The quasistatic approximation:

  More generally, the quasistatic approximation can be understood through the idea that the sources in the problem change sufficiently slowly that the system can be taken to be in equilibrium at all times.

So the question boils down to "are there ever any quantum mechanical behaviors that are non-equilibrium?" Or. more precisely, are there fast changes where the system is not really in equilibrium.

Well, as noted in the comments, there's the Born-Oppenheimer approximation which, at least in my opinion, reflects a quasi-static view of electronic structure. That is, the nuclei move slowly relative to the electrons, so despite vibrations, etc. the electrons can be considered in equilibrium with nuclear positions.

Obviously the Born-Oppenheimer approximation fails.

The poster suggested the question was about other non-static behavior. In that case, you fall into time-dependent quantum mechanics, e.g. time-dependent Hartree Fock (i.e., RPA) or TD-DFT.

  Linear-response TDDFT can be used if the external perturbation is small in the sense that it does not completely destroy the ground-state structure of the system. In this case one can analyze the linear response of the system. This is a great advantage as, to first order, the variation of the system will depend only on the ground-state wave-function.

Are there cases where linear response breaks and the variation in the system doesn't depend only on the ground-state wave-function. Yes. Consider the effects of large perturbations, like large electric fields, nonlinear optical effects (two photo absorption), etc.

As another fairly easy example, consider photo-ionization. That is, remove an electron from a system. In the quasi-static approximation, you'd have something like Koopmans' theorem with frozen ground-state orbitals. In reality, there's orbital relaxation even though the ionization process is very fast.


Hückel's rules are a very concise set of very strict rules. Most aromatic compounds do not comply with these.

  Hückel (4n + 2) rule
  Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. The rule is generally limited to n = 0 – 5. This rule is derived from the Hückel MO calculation on planar monocyclic conjugated hydrocarbons (CH)m where m is an integer equal to or greater than 3 according to which (4n + 2) π-electrons are contained in a closed-shell system. Examples of systems that obey the Hückel rule include:

  cyclopropenyl cation (m = 3, n = 0)
  cyclopentadienyl anion (m = 5, n = 1)
  benzene (m = 6, n = 1)

  Systems containing 4n π-electrons (such as cyclobutadiene and the cyclopentadienyl cation) are 'antiaromatic'.
  From IUPAC Gold Book, DOI: 10.1351/goldbook.H02867

Based on this it is actually quite difficult to assess aromaticity to more complex molecules. Aromatic molecules that do not comply with Hückel's rules are for example naphthalene, thiophene, or pyridine, among many more. A more suitable definition, which also makes quite clear how fuzzy the whole subject is:

  aromatic

  In the traditional sense, 'having a chemistry typified by benzene'.
  A cyclically conjugated molecular entity with a stability (due to delocalization) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1HNMR spectrum.
  See also: Hückel (4n + 2) rule, Möbius aromaticity
  The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

  From IUPAC Gold Book, DOI: 10.1351/goldbook.A00441

While we tend to say a molecule is aromatic, that is hardly ever the case. Think about substituted benzene molecules; while the phenyl ring retains aromatic character, the side chains are not necessarily aromatic. When describing the molecule it is better to refrain from absolutes.

Within this framework it is possible to identify molecular entities that posses aromatic character and refer to these. Strictly speaking the answer that 1,2-dihydronaphthalene is aromatic is therefore incorrect. Better is to say that because of the phenyl subunit, 1,2-dihydronaphthalene possesses some aromatic character. In this particular case you could even argue that because of the additional double bond, the  main reactivity of this molecule is more similar to alkenes than it is to aromatic compounds.
Aromatic character is sometimes like a unicorn: it's hard to describe in unambiguous terms, but when you see it, you'd probably know what it is.

There is another definition of relevance in the IUPAC Gold Book, which actually gives a clearer, more systematic aproach for determining aromatic properties, and can be seen as an extension of the above:

  aromaticity
  The concept of spatial and electronic structure of cyclic molecular systems displaying the effects of cyclic electron delocalization which provide for their enhanced thermodynamic stability (relative to acyclic structural analogues) and tendency to retain the structural type in the course of chemical transformations. A quantitative assessment of the degree of aromaticity is given by the value of the resonance energy. It may also be evaluated by the energies of relevant isodesmic and homodesmotic reactions. Along with energetic criteria of aromaticity, important and complementary are also a structural criterion (the lesser the alternation of bond lengths in the rings, the greater is the aromaticity of the molecule) and a magnetic criterion (existence of the diamagnetic ring current induced in a conjugated cyclic molecule by an external magnetic field and manifested by an exaltation and anisotropy of magnetic susceptibility). Although originally introduced for characterization of peculiar properties of cyclic conjugated hydrocarbons and their ions, the concept of aromaticity has been extended to their homoderivatives (see homoaromaticity), conjugated heterocyclic compounds (heteroaromaticity), saturated cyclic compounds (σ-aromaticity) as well as to three-dimensional organic and organometallic compounds (three-dimensional aromaticity). A common feature of the electronic structure inherent in all aromatic molecules is the close nature of their valence electron shells, i.e., double electron occupation of all bonding MOs with all antibonding and delocalized nonbonding MOs unfilled. The notion of aromaticity is applied also to transition states.
  From IUPAC Gold Book, DOI: 10.1351/goldbook.A00442

One of the key elements is the thermodynamic stability and the tendency of retaining the structural type during the course of a reaction. This certainly reflects the benzene-like aspect of the earlier definition.
The last part comes directly from the original (4n + 2) criterion, but significantly weakens it, as it is not reduced to the very symmetry restricted number Hückel derived from the hydrocarbon cycles.
For simple compounds like thiophene it is easy to construct an approximate MO diagram (if you for further simplification ignore the σ-sub-structure). And the same can be done more complex molecules like 1,2-dihydronaphthalene.
In general I am not a big fan of oversimplifying complicated situations; questions that are asking for absolute statements are always oversimplified. In case of aromaticity it is tempting to reduce it to a simple electron-count, which therefore still prevails in textbooks although horribly incomplete; but it is also dangerous, because it neglects a lot of the underlying chemistry.

Aromaticity is a beast; fuzzy definitions, or criteria, the ongoing debate among scientists only show that you need to be open minded and accept that many things are not as easy as it seems at first glance.
For the time being you can help yourself by identifying molecular entities, or moieties that are derived from (simpler) aromatic compounds and check whether a conjugated π-system was destroyed while creating them. As one of the key features you can be almost certain, that aromaticity is retained in all but the most extreme cases.


Sn + HNO3: When will the metastannic acid H2SnO3 appear?

It is a common belief that tin dissolves in dilute nitric acid to produce tin(II) nitrate. The reactions are addressed in a previous answer of mine. However, employing a more hot and concentrated nitric acid yields an unstable compound called hydrated stannic oxide also known as β-stannic acid. The compound also has a α-form which is later converted to β form. The relevant information can be found here:

  [...]tin dissolves slowly, in very dilute nitric acid to produce
  stannous nitrate (but) when hot and more concentrated acid is
  employed, is very unstable and quickly decomposes, yielding the form
  of hydrated stannic oxide known as β-stannic acid. Probably
  α-stannic acid is first produced [...] which then passes into the
  β-form. The product is, therefore, liable to contain both α- and
  β-forms, and the pure β-acid is obtained by dissolving this product
  in sodium hydroxide solution, and then adding excess of concentrated
  soda, which precipitates sodium p-stannate whilst the α-salt remains
  in solution. Pure β-stannic acid, or β-metastannic acid, is then
  obtained by decomposing the sodium salt with acid. It is also formed
  by the hydrolysis of its sodium salt at 60° C., and by boiling a
  dilute solution of stannic chloride, with or without the addition of
  nitric acid.[...]

  [...]Gelatinous, precipitated β-stannic acid has the empirical
  composition $\ce{SnO2.4H2O}$, when air-dried $\ce{SnO2.2H2O}$, and
  when dried in a vacuum, $\ce{SnO2.H2O}$; [...]


The electronegativity difference using the Pauling scale is not accurate in heavy transition metals. This is because the Pauling electronegativity involves the measure of bond energies, while the bonding of several metals (including "noble" metals like gold, silver, platinum, mercury, etc.) with large halides are diffuse and do not have sigma character . This is because the frontier orbitals of these elements are in the $d$ block, so molecular orbital interactions, if any, are diffuse especially in larger 4d and 5d metals. In reality, compounds like $\ce{HgI_2}$, as well as ones like $\ce{PbI_2}$ and $\ce{BiBr_3}$ (which I work with) are ionic salts.


Law of Definite Proportion: Do the numbers indicating moles count?

The stoichiometric coefficients are important. They make sure you have a valid relationship between reactants and products that obeys the Law of Conservation of Mass.

As Jerry hints at in his comment, there is another fundamental law that you need to consider: the Law of Conservation of Mass. Matter cannot be created or destroyed (outside of nuclear reactions and relativist physics). In chemistry this means you need the same number of atoms of each element on both sides of the equation. You cannot create additional atoms from nothing, nor can extra atoms of something vanish.

Consider your first equation:
$$\ce{Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$$

This equation is balanced - it obeys the law of conservation of mass. Each side of the equation has one $\ce{Fe}$, two $\ce{H}$, two $\ce{C}$, and six $\ce{O}$ atoms. This is a valid chemical equation representing a real reaction relationship.

Your second equation is not balanced.

$$\ce{2Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$$

There are now twice as many atoms of each element on the reactant side of the equation as on the product side. What happens to this extra atoms? They cannot vanish! This is not a valid chemical relationship. Balancing this equation generates a scalar multiple of your first equation, which will yield the same ratio of iron (ii) bicarbonate to iron (ii) carbonate.

$$\ce{2Fe(HCO3)2 -> 2FeCO3 + 2CO2 + 2H2O}$$

The Law of Definite Proportions tells you that chemical compounds always contain the same elemental compositions by mass. I am not sure how that is relevant to the ratio of product to reactant in your stoichiometry problem. In other words, it tells you that iron (ii) carbonate is always $\ce{FeCO3}$ and never $\ce{Fe3(CO3)4}$ or $\ce{Fe2(CO3)3}$ or $\ce{FeCO4}$ or anything else. Those other formulas may refer to real compounds (the first two are real), but they are not iron (ii) carbonate.


From a theoretical viewpoint we would expect the Br to be sp3 hybridized, because in the HBr molecule the Br atom has 4 pairs of electrons around it (the 3 lone pairs plus the bonding pair), and symmetry (plus the repulsion of electrons for each other) tells us to expect that 4 pairs of electrons arrange themselves in a tetrahedron, and as you know we explain a tetrahedral arrangement of electron pairs around an atom by arguing the s and p atomic orbitals form sp3 hybrid orbitals.

It's the same valence shell electron pair repulsion (VSEPR) logic that leads us  to expect the 4 bonding pairs around the C in CH4 should be arranged tetrahedrally, and, therefore, the C atom is sp3 hybridized.  (We would also expect that a more careful and precise determination would give us some deviation from this prediction for HBr, since unlike CH4 the 4 electron pairs around the Br in HBr are not equivalent.)

But you are interested in the question of whether there could be some experimental verification of this expectation.  I don't think the experimental dipole moment would work, because it's too crude a picture of the electron distribution, it just tells us whether the electrons spend extra time near the H or near the Br, and we already know it will be the Br because of its much stronger effective nuclear charge for valence electrons.  What you want to know is something like whether the valence electrons spend significantly extra time on the "far" side of the Br, away from the H-Br bond, as opposed to the "near" side, so we can conclude they're in sp3 hybrid orbitals that point away from the H-Br bond.  Offhand I can't think of an experiment that would unambiguously test this.

You could certainly calculate the electron density in the HBr molecule, with a good quantum chemistry package, and for a molecule this small the results should be quite reliable.  That would tell you whether it makes more sense to consider the Br lone pairs as in sp3 hybrid orbitals or atomic p orbitals.


Why are there no cheap diamond equivalents?

There are four properties that make a diamond look like a diamond. I will compare them with the two best diamond simulants at the moment:

cubic zirconia $\ce{ZrO2}$ - a special ("cubic") form of zirconium oxide. Don't confuse it with zircon, a different mineral with the forumula $\ce{ZrSiO4}$.
moissanite $\ce{SiC}$ - the name for the naturally occurring form of silicon carbide, also used for the synthetic gemstones varieties of it.

So what are the differences and why none of them is an all-round replacement for a diamond?

Refractive index: determines the brilliance. This is what you were talking about when you said "optical density". This basically means the amount of light the diamond reflects to the observer. If you change the refractive index, you also have to slightly change the cut of the diamond to achieve optimal brilliance, and then your "diamond" will not look exactly like a diamond. The refractive index of diamond is 2.42. Moissanite is higher: 2.65 and cubic zirconia is lower: 2.17.
Dispersion: determines the fire. This is what you wanted to say when you mentioned that diamonds are sparkling. Fire is the property of diamonds to break the light into different wavelengths thus creating the characteristic play of colors that you get in diamonds. Here's the interesting thing: The dispersion of diamond is 0.044 and interestingly the dispersion of both cubic zirconia and moissanite is higher: 0.060 and 0.104, respectively. So theoretically, if you consider only fire (the play of colors), then they both look better than diamonds.
Birefringence: (the lack of) determines the clarity. When you look through a diamond at all angles, you can clearly see what's behind. This is also true for cubic zirconia. Both minerals are isotropic. On the other hand, moissanite is birefrigent - it will double the image you see through the mineral, causing it to look blurred and lose the visual clarity.
Hardness: while not directly related to optical properties, a mineral with higher hardness will retain it's sharp cut. If the cut is damaged, the gemstone will not look as good and lose its value. Diamond has a hardness of 10, moissanite 9.25 and cubic zirconia 8.25. Note that this scale is not linear: 10 is much harder than 9 compared to 9 and 8, for example.

As you can see, none of the alternatives fully satisfies the entire set of properties that make a diamond. Moissanite may be slightly better, but it's also more expensive. And remember - one of the most appealing things about diamonds is their rarity and natural occurrence. No synthetic simulant can replace that.


Which of the following has maximum nucleophilicity?

According to me, the answer should be (C) in general cases but this might get changed to A) in special cases. The best way to judge their nucleophilicity order  is to check their basicity. The weak bases are generally better nucleophiles in polar protic solvents . The acidity order of the  conjugate acids are $$\ce{CH_3OH < (CH_3)_3COH  < NH_3 < CH_4}$$So, the strongest acid is $\ce{CH_3OH}$ and weakest acid is $\ce{CH_4}$. So, weakest base is $\ce{CH_3O^-}$  and strongest is $\ce{CH_3^-}$. So, the maximum nucleophilic character should be of $\ce{CH_3O^-}$,not generally $\ce{CH_3^-}$ as basicity is inversely proportional to nucleophilicity in general cases. Coming to D), the inductive effect of three methyl groups actually decreases the acidity of tert-butyl alcohol. So, the tert-butoxide ion is more prone to take $\ce{H^+}$ rather than to nucleophilically attack.  It also creates steric crowding, and preferably takes easily ionisable hydrogen ion if possible from the compound.So,  looking only at increased $\ce{+I}$ effect doesn't tell that its nucleophilicity is increased. About A), this $\ce{CH_3^-}$ is usually seen in Grignard reagents, whose basic activity is much much greater than its nucleophilic activity. If acidic hydrogen is present in the compound, it preferably takes that. But, if a compound doesn't possess any acidic hydrogen (special case), then its nucleophilic character is more as being less electronegative than $\ce{O}$, it is very less reluctant to keep yhe lone pair on itself like the case in $\ce{CH_3O^- or (CH_3)_3CO^-}$. Thus in the absence of any acidic hydrogen $\ce{CH_3^-}$ is better nucleophile. So, from the above discussion, it can be concluded that in many of the cases, where the attacked compound doesn't possess any acidic hydrogen, A) is maximum nucleophillic. But in the opposite  cases, C) is maximum nucleophilic.


How to add two hydroxy groups to 1,5-diphenylpentan-3-ol with as high as possible yield?

A few possibilities for direct hydroxylation:

$\ce{SeO2}$ immediately comes to mind as one potential option; it is well known to selectively oxidize allylic positions to yield alcohols and carbonyl compounds, and there appear to be references suggesting it reacts at the benzylic position as well.
In Advanced Organic Chemistry, March notes that cerium(IV) triflate "converts benzylic arenes to benzylic alcohols, although the major product is the ketone when > 15% of water is present." Ref: Laali, K.K.; Herbert, M.; Cushnyr, B.; Bhatt, A.; Terrano, D. J. Chem. Soc., Perkin Trans. 1 2001, 578.
Intriguingly, March also reports a variety of oxidations accomplished enzymatically. I have no experience and very limited knowledge of these reactions, so I'll just give the relevant primary literature references:

Adam, W.; Lukacs, Z.; Saha-Moller, C.R.; Weckerle, B.; Schreier, P. Eur. J. Org. Chem. 2000, 2923.
Hamada, H.; Tanaka, T.; Furuya, T.; Takahata, H.; Nemoto, H. Tetrahedron Lett. 2001, 42, 909.
Adam, W.; Lukacs, Z.; Harmsen, D.; Saha-Moller, C.R.; Schreier, P. J. Org. Chem. 2000, 65, 878.

If those approaches aren't appealing, there are various less direct routes. Radical bromination (with, e.g., NBS under photochemical conditions), followed either by direct substitution with hydroxide, or alternatively elimination followed by subsequent acid-catalyzed hydration, would likely work well. If you need stereoselectivity, there are are numerous epoxidation reactions (that is, after bromination and subsequent elimination) that would likely be appropriate (Sharpless, Jacobsen, Shi, and certainly numerous others). Conveniently, your molecule would already have the allylic alcohol necessary for Sharpless asymmetric epoxidation, should you want to go that route. In any case, it should be possible to open the epoxides at the less hindered carbon with a suitable hydride donor (e.g., LAH) to yield the alcohols after work-up.


What is the difference between distilled water and rainwater?

Rain water passes through the atmosphere and picks up whatever is present there, including particulates such as dust, trace amounts of atmospheric gases, nitrates and nitrites, and so forth.

Rain water is relatively low in mineral content.  In practice, people drink rainwater from streams and wells, where it has had ample opportunity to pick up minerals from soil and rock.

https://www.google.com/search?q=rain+water+composition


Viscosity generally decreases with increasing temperature so honey will become less viscous at higher temperatures. Regarding the actual values, look at this article. It appears that even at quite high temperatures honey remains a couple of times more viscous than water.

I think the reason for higher viscosity in honey is the high sugar () content - hydroxyl groups in sugars can form strong, hydrogen-bonded interactions with water and each other, and the sugar molecules being quite long, they are difficult to shear.


What does the crystal field splitting diagram for trigonal planar complexes look like?

Say you put a ligand in the trigonal complex (or an equatorial ligand in a trigonal bipyramidal complex) on the $x$-axis, where $z$ is the "axial" axis.  Then that ligand is directly aligned with an $x^2-y^2$ lobe and falls on a "seam" of the $xy$ orbital.  But the other two ligands in the plane are closer to an $xy$ lobe ($15°$ off) than to an $x^2-y^2$ lobe ($30°$ off).  These effects balance so that both in-plane $d$ orbital see the same total metal-ligand intetaction.


Precipitation of water-insoluble materials from shower or laundry greywater

Your ambitious project is not that trivial. Soaps are of several types.
a) If you are talking about salts of fatty acids, the classical soaps, then one can precipitate them out with calcium salts. Many other cations form insoluble salts with soaps. Now if you are trying to purify laundry water, dishwasher water, shampoos, liquid soaps, nobody can remove those surfactants from water by simple means. They (manufacturers) have specifically designed them so that they never precipitate out in water, otherwise your clothes and dishes would always have a layer of scum on them. Potassium ions may form ppts with certain surfactants like sodium dodecylsulfate but forget about ppt'ing out these typical surfactants from water used in ordinary household.

b) Alum can quickly help you in cleaning all the particulate matters in waste water, like magic.

c) Filtration on beach sand beds can remove small amounts of certain surfactants from water.

d) Filtration on activated carbon would be the best to remove surfactants.

Well it turned out to be more complicated than you desired. There is no single cure and certainly this water is not good for mopping floors because it will spread nothing but germs from using laundry water and shower waste. You can think of solar distillation as an alternative. That distilled water would very clean.

Note that in certain countries, household waste water including the sewage water after doing natural bacterial treatments is used for watering gardens and small farms. In one large pit, all wastewater is allowed to stand in open air, far away from residential areas. In the first ou all solids settle down, next pit further purifies via same settling and bacterial oxidation process, finally the supernatant water is transferred to third pit, which is now clean enough that gardens and small farms can be watered with it.  Of course, this water is rich in nitrogen and phosphates (from human and laundry waters) which is good for plants.


Why do the field of study "Chemical biology" often return papers that are in general anything disease related?

Because when we try to sell some new molecule to the chemical biology journals' editors and reviewers we think health and drug like actions sell better than a mere signalling molecule or so ?


In analysing acid strength, bond polarity of the bond to the ionisable hydrogen atom is what should be considered, not the polarity of the molecule (i.e. the dipole moment). In the case of alcohols, the polarity of the $\ce {O-H}$ bond should be considered when assessing strength as an acid.

Why does the bond dipole matter? In the process of ionisation $\ce {ROH + H2O-> RO^{-} + H3O^{+}}$, the $\ce {O-H}$ bond is heterolytically cleaved. It is easy to observe that the greater the partial positive charge on the $\ce {H}$ atom, the more easy it is for it to be abstracted by the water molecule, due to stronger electrostatic attractions. Also, the more polar the bond, the lower the covalent character of the bond as the electron density shifts more towards the oxygen. Such a bond is also more easily cleaved. Recall that the covalent character of a covalent bond is highest when the atomic orbitals are of similar energies and that electron density is shared equally between the two atoms.

In the primary alcohol, there is one electron-donating alkyl group attached to the alpha carbon while in the secondary alcohol, there are two such groups attached to the alpha carbon. Due to the greater electron donation to the oxygen atom in the secondary alcohol, where $\ce {R = CHR'R''}$ , the oxygen atom actually withdraws less electron density from the $\ce {O-H}$ bond. We can think of it this way: The oxygen atom has a fixed demand for electron density. When this demand is met by an increased supply from the alkyl chain, the demand for electron density from the hydrogen atom would decrease. Thus, the $\ce {O-H}$ bond is less polar and less easily heterolytically cleaved.

Simarly, for the primary alcohol, where $\ce {R = CH2R'}$, there is also an increased supply of electron density from the alkyl chain R, as compared to $\ce {R = CH3}$. However, the increased supply from $\ce {R}$ is less so compared to that in the secondary alcohol. Thus, the primary alcohol would be a weaker acid than the methyl alcohol but a stronger acid than the secondary alcohol.

Conventionally, acid strength is assessed using conjugate base stability, rather than bond polarity. A brief overview of assessing conjugate base stability can be found here.


Why is the reactivity of primary alkyl halides with nucleophiles (SN2 mechanism) greater than secondary and tertiary alkyl halides?

The order of reactivity by substitution in these two reactions is difference because they have different mechanisms. The substitution of an alkyl halide by a strong nucleophile in a polar aprotic solvent is an SN2 mechanism. A $\ce{C-Nu}$ bond forms and a $\ce{C-X}$ bond breaks at the same time:

$$\ce{CH3CH2Br + NaOH ->[\text{acetone}] Na}\bigg[\ce{CH3CH2(Br)(OH)}\bigg]^{^\ddagger_{-}}\ce{->CH3CH2OH +NaBr}$$

All else equal, the rates of these reactions are controlled by sterics on the alkyl halide, which control how well the nucleophile can react the $\ce{C-X}\ \sigma^*$ orbital. Thus:

$$\text{Methyl}>1^\circ>2^\circ>>3^\circ$$

The reactions of alcohols with hydrohalic acids follow a different mechanism (mostly). These reactions happen in a protic medium ($\ce{HX}$ is protic) with a poor leaving group ($\ce{OH}$), at least initially. These are conditions that favor the step-wise SN1 mechanism, where $\ce{C-X}$ bond breaking occurs before $\ce{C-Nu}$ bond forming. A carbocation is formed as a key intermediate.

$$\ce{(CH3)3COH + HBr -> [(CH3)3COH2+]Br- -> [(CH3)3C+]Br- + H2O -> (CH3)3CBr + H2O}$$

According to the Hammond postulate, the stability of this carbocation influences the rate of reaction. More stable carbocation intermediates suggest lower energy transition states for the formation of those carbocations, which implies lower activation energies and faster rates. Thus:

$$3^\circ>2^\circ>>1^\circ,\text{Methyl}$$


Are all mechanisms involving carbocations concerted?

Many carbocations are stable, observable entities.  Many large, delocalized carbocations like trityl perchlorate form very stable salts (reference see p. 196), other carbocations are easily observed by nmr.

If free carbocations didn't exist as discrete intermediates how could you explain the stereochemical results from an SN1 reaction where a chiral starting material generates a planar, long-lived carbocation that can be attacked from both sides to produce an achiral product?  Any type of concerted process would be expected to proceed with inversion of configuration (SN2).

(image source)


The pKa2 of sulfuric acid is 1.99 (Table of pka).  This is definitely less acidic than pKa1, but calling this a 'weak' acid may be a distraction.

The relevant question is: does $\ce{KHCO3}$ readily deprotonate $\ce{KHSO4}$?  The pKa1 of carbonic acid is 6.35 (same source), making it a weaker acid than  $\ce{KHSO4}$.  The difference is more than 4 orders of magnitude, so yes, I expect that $\ce{KHSO4}$ will more strongly favor depronotation than carbonic acid, and hence $\ce{K2SO4}$ is the product I expect.

To get at this quantitatively, consider the sub reaction:

$$\ce{KHCO3 + KHSO4 <=> K2SO4 + H2CO3 }$$

Which leads to the expression:
$$\ce{K_{eq} = \frac{[K2SO4][H2CO3]}{[KHCO3][KHSO4]} }$$
which can be rewritten as
$$\ce{K_{eq} = \frac{[H^+][K2SO4][H2CO3]}{[H^+][KHCO3][KHSO4]} }$$
Which can be rewritten as
$$\ce{K_{eq} = \frac{[H2CO3]}{[H^+][KHCO3]}\cdot\frac{[H^+][K2SO4]}{[KHSO4]}}$$
which can be written as
$$\ce{K_{eq} = \frac{$K$_{a2,\ce{KHSO4}}}{$K$_{a1,\ce{H2CO3}}} }$$
Taking the log of both sides
$$\ce{pK_{eq} = p$K$_{a2,\ce{KHSO4}} - p$K$_{a1,\ce{H2CO3}} = } -4.36$$

Which is $$\ce{K_{eq} \approx 23,000}$$ heavily favoring products, and hence, $\ce{K2SO4}$ is the dominant product


What assumptions underlie Hess' law and and its analogues?

The typically unstated assumption in applying Hess's law is that you are going from an initial state of pure reactants (say, in separate containers) to a final  state of pure products (also in separate containers).  In the case of ideal gas reactions, the enthalpy of mixing is zero and enthalpy is independent of pressure.  So, for ideal gas situations, Hess's law can be extended and applied directly to starting and ending states in which the reactants and products are mixed (and even simultaneously present) and the partial pressures are all different from the standard pressure of 1 atm.

Of course, since U, S, and G are functions of state, the strict version of the law, applying only to pure reactants and products, still holds.  However, for ideal gas mixtures, partial molar entropy and partial molar free energy are functions of mole fraction, so this needs to be taken into account for mixtures.


Structural representation of substituted benzene

Yes, this graphical representation of a substituted benzene is in accordance with Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008).

  GR-9.4 Variable attachment location

  In addition to allowing the type of attachment to vary, it may also be convenient to indicate that the attachment’s location is variable as well. This type of notation should be restricted to substituents that are known to be bonded to a specific ring, but at an unspecified or unknown atom of that ring. The substituent will always replace a hydrogen atom on one of the ring atoms, and cannot be bound to any atom that lacks an attached hydrogen atom. Unless explicitly specified otherwise, such diagrams imply that the substituent may be bonded to any ring atom that has an attached hydrogen atom. (…)


Help with derivation on why antibonding orbitals are more antibonding

For the 1s orbital in hydrogen, the wavefunction is

$$\psi_\mathrm{1s} = N\mathrm{e}^{-r}$$

where $N$ is some normalisation constant. The normalisation constant is not unique: if $N$ is a satisfactory normalisation constant such that $\left<\psi_\mathrm{1s}\middle|\psi_\mathrm{1s}\right> = 1$, then for any value of $\phi$, the normalisation constant $N\mathrm{e}^{\mathrm i\phi}$ is also valid.

However you can make the assumption, without any loss of generality, that the wavefunctions of both 1s orbitals in dihydrogen ($\ce{H2}$) are of the same phase (i.e. both normalisation constants have the same value of $N$ and $\phi$). As a simple example we can just let $N$ be a positive real number. With this, the wavefunction of the 1s orbitals always have a positive value, and the overlap integral

$$S_\mathrm{AB} = \left<\psi_\mathrm{1s,A} \middle| \psi_\mathrm{1s,B} \right> = \int \psi_\mathrm{1s,A}\psi_\mathrm{1s,B}\,\mathrm{d}\tau$$

always has a positive integrand (no imaginary components, so I dropped the complex conjugate). The integral is therefore positive.

(Note that even though both 1s orbitals have the same exponential form, $\psi_\mathrm{1s,A}$ is not equal to $\psi_\mathrm{1s,B}$ since the orbitals are centred at different points in space.)

In fact, this choice of phase is required so that you can express the bonding and antibonding MOs as the specific linear combinations presented. The "in-phase" linear combination, i.e. the bonding MO, only has the form $N(\psi_\mathrm{1s,A} + \psi_\mathrm{1s,B})$ if you let the two orbitals have the same phase. If you had a different choice of normalisation constant, the bonding MO would be represented by a different linear combination $N(\psi_\mathrm{1s,A} + z\psi_\mathrm{1s,B})$ where $z$ is some complex number determined by the relative phases of the 1s orbitals.

As such, by virtue of expressing the MOs in that form, your overlap integral is necessarily positive.

Note that the above discussion only really applies to 1s + 1s overlap. With other orbitals, which have both positive and negative regions, you can end up with a negative overlap integral depending on the relative phases that you choose.

For example in $\ce{BeH2}$ if you consider the middle Be 2p orbital to overlap with both hydrogen 1s orbitals, one "overlap" will be formed with the positive lobe of the 2p orbital and the other "overlap" will be formed with the negative lobe of the 2p orbital. Assuming that you phase both 1s orbitals to be positive, then the first overlap integral will be positive (positive 1s times positive 2p gives a positive integrand) and the second will be negative (positive 1s times negative 2p gives a negative integrand).

In general your choice of phase is arbitrary; it does not affect the final physical result, and is simply chosen to make the calculations easier. That is to say, the eigenvalues (MO energies) that you get out of it will still be the same. The eigenfunctions may take on a different form, but that is physically unimportant.


How to determine which reagent to use H2; lindlar catalyst vs Na;NH3(l)?

This is a really interesting question. I'm not sure I agree with the end product, but I will get to that in a second. There does appear to be a piece missing from this equation, which is a second step (something has to be responsible for the new bromine in the product).

First thing is to identify how to get to the final product from an intermediate. You've correctly ruled out $\ce{NaNH2}$ which is just a strong base. It's frequently used to deprotonate a terminal alkyne (which you do not have), and in this example, all it would do would be to remove the hydrogen on the alcohol.

You've also ruled out $\ce{H2, Pt}$. This is a standard hydrogenation and will reduce a triple bond fully to the alkane. If that happens, you won't have the functional groups to do anything like a ring closure.

That gives two possibilities, hydrogenation with Lindlar's catalyst or a sodium ammonia reduction. Both of these reactions reduce the alkyne to an alkene (and then stop). The difference is that concerted hydrogenation will add hydrogen in a syn fashion to afford the cis-alkene, whereas the sodium/ammonia reduction is not concerted and will give the more stable trans-alkene.

The cyclization happens via addition of bromine. This is a standard reaction: halohydrin formation. Except that the intramolecular opening via the attached alcohol will be fast, so the alcohol attacks instead of water to provide an ether. Assuming this is the mechanism, you have to think about how the relative stereochemistry of the alkene is carried forward.

The key intermediate is the bromonium ion (the three member ring with a bromine). The intermediate retains the stereochemistry of the alkene. You might need a model to identify this, but the correct bromonium ion that gives the indicated product has the hydrogens on the same side. That means that they were originally syn across the double bond, so the double bond was cis. Then Lindlar hydrogenation was the appropriate reaction for the first step.

Here's my issue with this question: I don't actually think you'll get this product. The technical point is that the opening can proceed via a 5-exo-tet attack or a 6-endo-tet attack. Generally, in these types of reactions, the exo attack is favored for stereoelectronic reasons. So my money is on obtaining the 5-member ring instead of the 6-member ring.


That the question says: "Without any 'binding' forces, the atom couldn't be possibly trapped inside the fullerene", shows you are misunderstanding fullerenes.

Atoms such as Helium and $H_2$ can be confined within the interior of the fullerene.

There are even reports of extraterrestrial helium trapped in fullerenes during astroid or comet impact.

The helium can move around within the fullerene.  It is sterically confined rather than bonded.

In the case of La, the metal ion does not remain neutral, but loses about 3 electron to the fullerene.  So, in addition to being sterically confined in the cage, there is the interaction between postively and negatively charged species.

As calculated in a later article Computed Structure and Energetics of
La@C60 International Journal of Quantum Chemistry, Vol 104, 272–277 (2005), the La has a charge of somewhere betwen 2+ and 3+.  In the lowest energy state, the La is off-center with respect to the fullerene, located between the center and the center of a hexagon face.  Also, the fullerene cage is distorted from icosahedral by Jahn-Teller distortion.

The La ion is located 2.62 Angstroms from the nearest carbon atom (4.6 Angstroms from the furthest), when at the potential miniumun, but is expected to move within the fullerene unless the temperature is very low.  The reference also says "there is obviously a relatively large negative charge on some carbon atoms of the cage", without specifying how this charge is distributed.

Overall I would say, whereas in the case of helium and dihydrogen being confined in the fullerene is completely because of van der Waal repulsive force of avoiding the walls, in the case of La, there is attractive electrostatic force (ionic interaction) between the La3+ and negatively charged carbon atoms.

See also Theoretical investigation of structural and thermodynamic properties
of lanthanum carbides LaCn (n = 2–6) which finds strong ionic interactions between La and Cn due to charge transfer.


Why are electrons transferred in an ionic bond?

The modern definition of electronegativity is due to Linus Pauling. It is the power of an atom in a molecule to attract electrons to itself.

When the electronegativity difference between the two atoms  is greater than 2.0, the bond is considered ionic; i.e. there is complete transfer (between the two atoms) of the electrons in the bond.
The electronegativity (according to Pauling scale) of chlorine is 3.0, while the electronegativity  of sodium is 0.9.


Does an atom want to give or take electrons when it has a shell half full?

When it comes to elements like carbon and silicon, which have 4 valence shells, the maximum possibility is sharing of electrons to form covalent bonds, like in $\ce{CO2}$ or $\ce{CH4}$. However in carbides, like calcium carbide or tungsten carbide, it may gain electrons.


Yes, you're exactly right - multiple papers in chemistry ML drop the units.

There are even comparisons (usually by statistics, ML or comp. sci. researchers) where models are compared by "averaging" errors down a column like that. Of course that's meaningless, since you can't average electron volts or Hartree (energies), Debye (dipole moments), and volume (polarizabilities).

Worse, energies are often computed as atomization energies - so for large molecules, they can be enormous...

In my opinion, a more relevant and meaningful statistic would be the Mean Absolute Percent Error (MAPE) which is unitless and easier to understand (e.g., 1% error? 0.01% error?)

That said, the original QM9 paper gives the units Table 3:
Scientific Data (2014) 1, art. 140022


Is a dithioic acid.

For example, a benzene with two -CSSH hubstituents in 1, 2 positions is a benzene-1,2-dicarbodithioic acid.


Hardness
TC Total Carbon

TIC Total Inorganic Carbon
TOC Total Organic Carbon
VOC Volatile Organic Carbon

COD Chemical Oxygen Demand
BOD Biochemical Oxygen Demand


Why is Hessian Matrix called 'force-constant' matrix?

You are right that multidimensional PES is not simple quadratic, but close to the PES well minima it can be approximated as harmonic oscillator. Also if you really want to include the real PES you can include anharmonicity in your calculation. In that case you have to include an anharonic constant $x_e\;.$ Your PES can be written as
$$E=\left(n+\frac{1}{2}\right)h\nu-x_e\left(n+\frac{1}{2}\right)^2h\nu$$

Usually in computational chemistry the energy can be expanded into a Taylor series
$$E=E_0+\frac{\mathrm{d}E}{\mathrm{d}x}(x-x_0)+\frac{1}{2}\frac{\mathrm{d}^2E}{\mathrm{d}x^2}(x-x_0)^2+\textrm{higher order terms}\;.$$
And you can usually neglect those first order term and higher order terms for most of the practical problems. And then your PES will follow the trajectory of a harmonic oscillator.


Stereochemistry of cycloalkanes having substituents

For a compound to be optically active it must not be superimposable with its mirror image. It's as simple as that. Also note that stereogenic centres are not always required for chirality. There are other types of chirality such as axial and planar chirality where a compound possesses either an axis or plane of chirality. Sometimes it can be the case that just the sterics of a molecule does not allow for free rotation of bonds and the molecule is locked one configuration. I think in the simplest cases of open chain molecules, you could indeed use the structural formula, but with rings which adopt specific 3D configurations, you must refer to the 3D structure.
Again, as said in 1., to determine if a molecule is chiral, it's better to refer to its 3D structure, especially when with meso, when you are dealing with more than one stereogenic  centre. For a compound to be meso, it must not be chiral, but posses stereogenic centeres (look up the definition).
No, you cannot use the simple structure to determine symmetry elements present if that structure is not really planar. It's not necessary to build a model, but you must use wedged/dashed bonds to show stereochemistry.

If someone thinks differently, please comment and correct me.


How to determine the work function and electron velocities from data on the photoelectric effect?

Consider the difference between the kinetic energy in case A and B:
$$K_A-K_B=(E_A-W)-(E_B-W)=E_A-E_B$$ where the work function cancels out because it is the same metal in each experiment.
The right hand side of this is easy to evaluate using $hc/\lambda$, which gives $$E_A-E_B=hc\left(\frac{1}{\lambda_A}-\frac{1}{\lambda_B}\right)=(\pu{1.9878\times10^{-25}J\cdot m})\cdot(\pu{1.333\times10^6m^{-1}})=\pu{2.65\times10^{-19}J}$$

To solve for the left hand side, we simplify to a single variable by recognizing that $v_a=2v_b$ (as Wildcat said, the right hand side is positive and the only way the left hand side will be is if A has the larger velocity): $$K_A-K_B=\frac{m}{2}(v_A^2-v_B^2)=\frac{m}{2}(4v_B^2-v_B^2)=\frac{3}{2}mv_B^2$$
With an electron mass of $\pu{9.109\times10^{-31}kg}$, this gives $v_B=\pu{4.404\times10^5 \frac{m}{s}}$ and $v_A=\pu{8.809\times10^5 \frac{m}{s}}$. The work function can then be solved by evaluating either $K_A$ or $K_B$ and solving for the difference $E_x-K_x=W$ where $x$ is either A or B.


Your answer is simply missing something. Yes, I is a better leaving group, therefore an Sn reaction of an iodide proceeds faster than the one of the chloride. But your starting material in both cases is the chloride. You first need to exchange that chloride for an iodide and that only works well if the iodide is a good nucleophile.


What is the density of sodium trimethylsilylpropanesulfonate (DSS)?

The reported crystallographic density for sodium 3-(trimethylsilyl)-1-propane-sulfonic acid monohydrate is $\pu{1.32 g cm^-3}$ [1].

The title compound which is used as reference for some NMR spectra in $\ce{D2O},$ was obtained casualy during an attempt to crystallize a cephalosporin of a serie we are interesting for […].
The X-ray analysis was undertaken because the expected cephalosporin and the two independent molecules of the title compound which were in fact present in the unit cell have nearly identical molecular weights.
The compound crystallizes from water in the triclinic space group $P\bar{1}$ with unit cell dimensions $a = 5.914(3),$ $b = 10.816(7),$ $c = \pu{18.869(9) Å},$ $α = 97.96(5),$ $β = 95.46(4),$ $γ = 89.76(5)^\circ,$ $V = \pu{1190(1) Å^3}.$
The calculated density is $\pu{1.32 Mg m^-3}$ for $Z = 4.$

Figure 1. Unit cell of sodium trimethylsilylpropanesulfonate monohydrate (CSD: BEHNIL).
Reference

Tinant, B.; Coene, B.; Declercq, J. P.; Germain, G.; Meerssche, M. van. X-Ray Crystal Structure of 3-(Trimethylsilyl)-1-Propane Sulfonic Acid Sodium Salt Hydrate $\ce{C6H15O3SSi^-Na^+ · H2O}.$ Bulletin des Sociétés Chimiques Belges 1982, 91 (1), 95–96. DOI: 10.1002/bscb.19820910116.


Proton NMR spectrum of aminocaproic acid

The N+1 rule, in its ideal form, assumes that for multiplets arising from $\mathrm{^3J}$ couplings between neighbouring $\ce{CH_2}$ groups,

shifts for the spins within a $\ce{CH_2}$ group are identical
all couplings between spins in two different groups are identical

While condition 1 usually holds for substituted alkanes of the form $\ce{A-(CH_2)_n-B}$, condition 2  does not. This leads to the title "Is the n+1 rule ever obeyed?" in a section of a textbook [1], like a cry of despair. The text elaborates:

  At other times, when there is a large difference between $\mathrm{^3J_{AB}}$ and
  $\mathrm{^3J_{BC}}$, distinct peaks, more than five in number, can be seen.
  Deviations of this type are most common in a chain of the type
  $\ce{X-(CH_2)_3-Y}$, where X and Y are widely different in character.

So, the exception is actually closer to the rule, and that is because true magnetic (unlike chemical) equivalence is relatively rare.

[1] Pavia, Lampman and Kriz, Introduction to Spectroscopy, 2nd Ed., Saunders Publishing


Will washing child seat belt straps with baking soda and hydrogen peroxide make them unsafe to use?

I have seen some companies that refurbish child seats, perhaps a new strap kit would put your friend's mind at ease.

If the straps were not soaked for hours in concentrated H2O2 and I saw no damage and could not cause them to break with my strength (healthy adult) I would have used it with my kids.  The baking powder or soda should have no effect.

We bought a used seat for the next size up and have no idea of how it was washed.  Would do the same again.

Stay at home, spend the coin, take out a loan or live with the risk, those are the choices. There are other risks in life like not being sure that the excavator operator is of sound mind and will not swing the bucket into your car at the stop sign, does this stop you leaving the house?

EDIT:
It is possible the dangers of washing are circulated by the seat refurbishing and manufacturing industries to increase sales and not safety.  A car seatbelt can last for decades and sees a lot more use and has no washing warnings.


Why is the reason for color different in these two cases?

These at two different scenarios and two different phenomena. Flame color is due to atomic emission and complexes are colored because of light absorption in the solid state and in solution.

In a high temperature flame, an atom is considered to be a free and independent entity. With heat, the electrons in the atoms are excited to higher energy levels. However, selection rules determine where an electron can land e.g. s electron goes to p, p electron can go to d, however s electron cannot go to d.

In case of transition metals, Hans Bethe (Nobel Prize) developed a theory called called crystal field splitting. In a free gaseous atom, all five d orbitals orbitals have the same energy (called degenerate orbitals), so electronic transitions do not make any sense. It is also forbidden. However in the presence of ligands, the d- orbitals are no longer of the same energy. Electronic transitions are possible within d orbitals. Search crystal field theory if you are interested...


I guess the diagram of phase changes you have in mind looks like this:

Indeed this says deposition is the right name for a gas-to-solid phase transition and I believe that this is the only correct term.

The only thing I can argue why it is obvious here that condensation is the right answer (at least the closest) is that this is the only answer that gives a transition from fluid to solid. B and D are the other way around, and C is not really a phase change (although it could result in one).


A contradiction regarding the reaction coefficient expression in the Nernst equation

This is why for equilibrium expressions we use activity and fugacity. Both of these values are dimensionless and are generally quite closely related to the concentration and partial pressure, respectively, for reasonable concentrations.


What is the term for passing an eluate back through an adsorption column?

This process is similar to re-cycling chromatography or recycling HPLC. In a more strict sense, it does not do any good, at least theoretically, if you keep on recycling all the eluate and transfer it back to the column.

What makes more sense is that you collect fractions and choose certain fractions for recycling.


Even the tetravalent cations (most of them) are heavily hydrolyzed and barely exist in water solutions. Pentavalent cation is even more of a challenge. Protactinium might be a possibility, but you can't have it anyway.

Speaking of anions, there is not even a trivalent single-atom anion that would not hydrolyze instantly and completely upon touching water.

On the other hand, polyatomic anions (and cations, for that matter) can have pretty much any charge you want.


Aluminium hydroxide isn't a partilarly reactive or soluble compound, so you probably don't want to directly mix it with stearic acid; also, you'd likely obtain a tristearate this way.
According to a procedure described in US3056819A patent, a synthetic path would be to obtain aluminum dihydroxychloride $\ce{Al(OH)2Cl}$ first, and then add stearic acid and sodium hydroxide (please disregard full stops after units, this isn't a standard notation these days):

  (1) 2.7 g. (0.1 mole) of metallic aluminium is dissolved in 50 cc. of 2N-HCl (0.1 mole) under heating to obtain a solution of a basic aluminium chloride whose formula is $\ce{Al(OH)2Cl}.$

  (2) To an aqueous solution of 47.58 g (0.1 mole) of potassium alum $(\ce{AlK(SO4)·12H2O})$ is added a solution of sodium bicarbonate to form aluminium hydroxide.
  The aluminium hydroxide is dissolved in 50 cc. of 2N-HCl with stirring and the solution is heated for 1 hour at 100 °C. to give a solution of a basic aluminium chloride whose formula is $\ce{Al(OH)2Cl}.$

  (3) To a solution of 24.2 g. (0.1 mole) of hydrous aluminium chloride in 100 cc. of water is added gradually 16.8 g. (0.2 mole) of sodium bicarbonate with stirring and the solution is heated for 1 hour at 100° C. on a water-bath.

  Any of the solutions of the basic aluminium chloride produced by the above three methods is diluted with water to 200 cc. and added to a solution of 28.45 g. (0.1 mole) of stearic acid and 4 g. of sodium hydroxide (0.1 mole) in 1000 cc. of water, and the mixture is heated on a water bath until the product precipitates.
  After washing with water to remove chlorine ion, the product is extracted with acetone in Soxhlet's extractor to remove free stearic acid, obtaining 33 g. of aluminium monostearate whose formula is $\ce{Al(OH)2(C17H35COO)}.$

References

Hideo, T. Aluminium Soaps and Their Production. US3056819A, October 2, 1962. (PDF)


Nomenclature of aldehydes with double bonds and stereoisomerism

IUPAC has changed some of the rules of nomenclature as of 2013. In aldehyde 1 both the aldehyde group and double bond are included in the base name (noun) and the methyl group is the substituent (adjective) because both chains are $\ce{C4}$. In structure 2 the methyl group of 1 has been changed to ethyl. Now the longest chain is $\ce{C5}$ containing the aldehyde group and the double bond, as a substituent is the modifier methylene. In structure 3 the branch at $\ce{C3}$ has two $\ce{C3}$ units. The one containing the double bond takes priority as the "noun". In structure 4 the chain containing the double bond is shorter, which becomes the "adjective". In your compound 5, the noun is heptanal (in black) and the adjective, which contains the double bond, is in red. The two entities are numbered independently.


E/Z configuration of the central double bond in a highly branched poly-ene

For simplicity, I shall assign numbers to the double bonds so that it is easier to reference them. This numbering is chosen solely after their position in space and does not correspond to the number of carbons in a PIN or otherwise systematic name.

1                 10
  > 5         8 <
2    \       /    11
       > 7 <
3    /       \    12
  > 6         9 <
4                 13

The terminal bonds, 1–4 and 10–13, are easy since they are all disubstituted: (1Z,2Z,3E,4E,10E,11Z,12Z,13E)

1Z                 10E
   > 5         8 <
2Z    \       /    11Z
        > 7 <
3E    /       \    12Z
   > 6         9 <
4E                 13E

Next up are the intermediate double bonds. If we simply follow the atoms, neither side will ever have priority, thus the somewhat less common rule that Z has priority over E takes effect. This allows us to label the double bonds (8Z) and (9Z).

1Z                  10E
   > 5         8Z <
2Z    \       /     11Z
        > 7 <
3E    /       \     12Z
   > 6         9Z <
4E                  13E

However, we cannot determine a stereodescriptor for bonds 5 and 6: both 1 and 2 have the same geometry (Z) as do both 3 and 4 (E). Thus, neither of these double bonds is asymmetric as they only feature three different substituents with the two identical ones being on the same side.

1Z                   10E
   > 5Ø         8Z <
2Z     \       /     11Z
         > 7 <
3E     /       \     12Z
   > 6Ø         9Z <
4E                   13E

Moving on to the final double bond 7, we face the same problem. The substituents 8 and 9 are identical, both feature an E double bond attache to E and Z. Thus again, we are unable to determine a stereodescriptor since the double bond is not asymmetric.

1Z                    10E
   > 5Ø          8Z <
2Z     \        /     11Z
         > 7Ø <
3E     /        \     12Z
   > 6Ø          9Z <
4E                    13E

Note that we can assign priorities to the left-hand side of double bond 7: the bottom half (3,4,6) has a lower priority than the top half (1,2,5) since the latter features Z double bonds where the former features E. But assigning a descriptor to 5, 6 and 7 is impossible due to their symmetric other side.


Simplest way to balance any chemical equation

You can do this in the 'guided trial-and-error' method that LordStryker showed which is probably quickest for simple reactions, or approach it in a purely mathematical fashion which is the method I will explain. This method works well for arbitrarily difficult reactions.

Your chemical equation contains 3 atomic species: $\ce{C}$, $\ce{H}$ and $\ce{O}$. This means that you need 3 equations to balance. First I write the chemical reaction as follows:
$$\ce{a C12H26 + b O2 -> c CO2 + d H2O}$$
Now I will write the 3 equalities for the 3 atomic species that we have:
\begin{align}
12a &= c \tag{for C}\\
26a &= 2d \tag{for H}\\
2b  &= 2c + d \tag{for O}
\end{align}
The numerical constants in front of $a$, $b$, $c$ and $d$ come from the number of atoms that are in the molecule for which they are included.

As you may have noticed, this system of equations is ill-defined, because we have 4 unknowns ($a$ to $d$) and only 3 equations. The reason is that we can pick any multiple of the equation without it becoming incorrect,1 just against convention. This means that we can simply set one of the unknowns to any value. Let's say $a = 1$ for now and revisit this choice after solving the equations.

With $a = 1$ it immediately follows that $c = 12$ and $d = 13$. From those two it then follows that $b = 18.5$. Our balanced reaction becomes:2
$$\ce{C12H26 + 18.5 O2 -> 12 CO2 + 13 H2O}$$

In principle this is already correct, but convention is to have only integer numbers in the equation (i.e. no decimal points) therefore we need to revisit our choice of $a = 1$ and pick it such that all numbers will become integers. In this case you can easily see that this will happen for $a = 2$ which then results in
$$\ce{2C12H26 + 37 O2 -> 24 CO2 + 26 H2O}$$

If for some reason it is not easy to see which value for $a$ you would need then you can multiply by a large power of 10 to make all the decimals disappear and then check the greatest common divisor of the 4 numbers and divide by that to obtain the same equation.

Notes

Compare $10=10$ to the multiplied version $30=30$, both are correct, but they are just different by a factor $3$.
I'm not sure whether you are familiar with linear algebra, but if you are you will probably have noticed that the set of equations is a linear set so you could solve it through matrix manipulations which makes this method applicable to arbitrarily complex chemical reactions.


If a chemical formula for a mineral contains a comma, are there limits on the amounts of each element?

Actinolite is a mineral that is part of a so-called solid-solution series, which means that you get different but related minerals depending on the amounts of (in this case) iron and magnesium that are present. The magnesium-rich end member is tremolite; the iron-rich end member is ferro-actinolite; and what lies between is actinolite.

According to Wikipedia's article on actinolite, the formula is more precisely:
$$\rm Ca_{2}(Mg_{4.5-2.5}Fe^{2+}_{0.5-2.5})Si_{8}O_{22}(OH)_{2}$$

So, for your question:

  is it possible for the amount of magnesium to be less than iron and still be actinolite, or must magnesium always have the greatest amount?

the answer is no: at best, Mg and Fe can be present in equal amounts for the mineral to still be called actinolite, but note that in that particular case you are on the "border" between actinolite and ferro-actinolite, so perhaps what you choose to call it is up to you. And as mentioned above, if magnesium is dominant, the mineral is tremolite.

In the case where iron is the dominant atom, the mineral is called ferro-actinolite with the formula:
$$\rm Ca_{2}(Mg_{2.5-0.0}Fe^{2+}_{2.5-5.0})Si_{8}O_{22}(OH)_{2}$$


What pressure is created by the electrolysis of water?

The maximum gas pressure you can expect in a Hofmann Voltameter is the atmospheric pressure plus the pressure from the water column in the middle cylinder ($1~\pu{atm} / 10~\pu{m}$).

For a reasonably sized apparatus, you will not reach conditions where you had to consider "back pressure" effects that might stop production of hydrogen or oxygen.

Given that the cross section of all three cylinders is the same, the pressure of hydrogen is slightly higher than oxygen pressure since the water column that adds pressure is higher (as can be seen in the Wikipedia article you've linked.)


Explain the function of this Galvanic Cell in detail

in the case of electrochemical cells/Galvanic cells: Oxidation at the anode (the A and O of electrochemistry, german people will know what I mean), reduction at the cathode (in electrolysis its the other way round)
$EMF = E_{red} - E_{ox} = E_{cat} - E_{an} > 0$ for the reaction to run without external voltage (<0 in the case of electrolysis) "the higher redox potential always oxidizes the lower one"

In your case this means running the cell without applying any voltage gives an $EMF = 1.33 - -0.25 V = 1.58 V$

reduction at cathode:

Cr$_2$O$_7^{2-}$$_{(aq)}$ + 14 H$^+$$_{(aq)}$ + 6 e$^-$ $\rightarrow$ 2Cr$^{3+}$$_{(aq)}$ + 7 H$_2$0$_{(l)}$
(in the electrolyte -6e charges)

oxidation at anode:

3 Ni$_s$ $\rightarrow$  3 Ni$^{2+}_{(aq)}$+ 6e$^-$
(in the electrolyte +6e charges)
(already multiplied by 3 for the total reaction)

total:

Cr$_2$O$_7^{2-}$$_{(aq)}$ + 14 H$^+$$_{(aq)}$ + 3 Ni$_s$  $\rightarrow$ 2Cr$^{3+}$$_{(aq)}$ + 7 H$_2$0$_{(l)}$ + 3 Ni$^{2+}_{(aq)}$

Now there is however a clear symbol in the electric cycle standing for a voltage source "V". This is indeed confusing and I would ask what was the meaning of that if the figure was actually titled "Galvanic Cell" which is the other way round. So if it would't be for the title and we would consider running the cell with external voltage, this would then turn around all reactions which I wrote before.

Other things that might be relevant:

K$_2$SO$_4$ is responsible for the charge transfer between the cells. Since one cell produces positive charges in solution and the other one negative ones the electrolyte has to establish that the overall charge in both cells stays neutral. The salt bridge connects the electric cycle, without it, the cell won't run for long
the Gibb's energy is related to the EMF. $\Delta G = -n\cdot F\cdot EMF$, so again a positive $EMF$ gives a negative $\Delta G$ which means the reaction runs "alone" without external voltage, the Gibb's energy provides the driving force. n is the number of electrons involved in the redox reaction (6) (F is the Faraday constant, a natural constant)


How can I update a map/mesh and coordinates in a PyMol session file?

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

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

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

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


Ring opening of aziridinylmethyl and oxiranylmethyl radicals

The 2-methyl-3-phenyloxiranylcarbinyl radical has been shown EXPERIMENTALLY to cleave the C-O bond kinetically while the C-C bond is cleaved thermodynamically. This means that, in this example, The oxygen-centered radical forms faster than the highly stable benzyl radical. The formation of the oxygen radical is necessarily reversible. See ref. 11 in your ref. 2 (Marples paper)

Ziegler, F. E.; Petersen, A. K. Allyloxy Radicals are Formed Reversibly from Oxiranylcarbinyl Radicals: A Kinetic Study, J. Org. Chem. , 1995, 60, 2666.


The 2nd law is

$d S \ge \displaystyle \frac{d q}{T}$

so the best you can do is zero for a reversible process.   Or if you don't like inequalities,

$d S = \displaystyle \frac{d q}{T} + d S_{irr}$

where $d S_{irr}$ is the entropy production due to the irreversible processes within the system.

You can find a intuitive in depth discussion in Prigogine's book:

Modern Thermodynamics: From Heat Engines to Dissipative Structures

http://books.google.com/books?id=6qFxngEACAAJ&dq=Modern+Thermodynamics:+From+Heat+Engines+to+Dissipative+Structures&hl=en&sa=X&ei=IveUU8HvH8KVqAbDvILoBA&ved=0CCcQ6AEwAA


How to convert sodium aluminate back to sodium hydroxide?

I don't think you will have much luck recovering the hydroxide.  To do so you would need to reverse the reaction that happened when the aluminum dissolved, which was

$$\ce{2 Al + 2 NaOH + 2 H2O -> 2 NaAlO2 + 3 H2}$$

Your idea is essentially to substitute reduction at an electrode for reduction with hydrogen gas.  Unfortunately electrochemical reduction of $\ce{Al(III)}$ is very difficult.  If your idea was feasible, it would revolutionize aluminum manufacture!  Instead of doing electrolysis of molten cryolite at ~1000 degress with sacraficial carbon electrodes, aluminum manufacturers could use your process.  So if you could get it to work, you'd be rich...but I am not hopeful.

Note however that you don't necessarily need to do electrolysis to recover sodium hydroxide.    There is hope!  You could just boil down your solution so that it is supersatured with respect to aluminum hydroxide, and then wait for the aluminum hydroxide to precipitate, which would happen faster if you add seed crystals of aluminum hydroxide $\ce{Al(OH)3}$) to initiate precipitation.

$$\ce{2H2O + NaAlO2_{(aq)} -> Al(OH)3_{(s)} + NaOH_{(aq)}}$$

This step is part of the Bayer process for commercial refining of aluminum ore.


Does a positive electron affinity actually correspond to the release of energy?

It may help to take a look at the two processes you are comparing in terms of reactions. Ionisation follows the following reaction:

$$\ce{A ->[][$\Delta E_\mathrm{i}$] A+ + e-}\tag{1}$$

This process involves tearing charges apart, separating a positive and a negative one. Even if the original atom was somewhat unstable, it will be unfavourable to remove that electron simply due to electrostatic interaction. You write in the question that you were able to deduce this already.

The electron affinity on the other hand describes the process in this following reaction:

$$\ce{A + e- ->[][EA] A-}\tag{2}$$

Here, a charged particle is approaching a neutral particle and we have the same number of charged particles (and the same charges) on both sides of the equation. This means that we don’t have a strong, obvious physical effect that will make all electron affinities sign-equal for very simple reasons. Instead, we need to consider the details more.

The process often involves a release of energy because most atomic configurations have space for extra electrons at little or no extra cost. Except for the noble gases and the second group, atoms don’t have completely populated sub-shells so there should always be a position available in a subshell at an acceptable exothermic energy level. While adding an electron will increase electron-electron repulsion and increase all energy levels by a certain amount, this increase is typically less than the first ionisation energy which corresponds to the energy level of the highest occupied atomic orbital. Thus in electron affinity, we in principle gain the entire ionisation enthalpy minus the destabilising contribution of the new electron. (Caution! This is a very simplified picture and not suitable for generalisation or even strictly for explanation. It gets the basic point across though.)

Therefore, most electron affinities are exothermic with the notable exceptions of group 2, group 18 and nitrogen.