text:
Dependence of the rate of a catalysed reaction on substrate concentration?
In my answer to this question: Master equation for catalyst and substrate binding?. I made the assumption along the following lines:
If we have a substrate concentration $[S]$ then the rate of binding of any substrate molecule with a given enzyme is proportional to $[S]$. Such that $$\text{rate}=\alpha [S]$$
I have however, read that it is rather a hyperbolic relation. So are there any (and if so what) assumptions required so that we can assume that the rate is proportional to $[S]$ rather then it been a hyperbolic relation?

There is always a maximum convertion rate $V_{max}$ for every given enzyme-substrate system. In some cases the substrate itself can inhibit the enzyme's activity at high concentrations, because of the formation of some ESS-complexes, which can not dissociate into product and enzyme.
The frequent correlation is described by the Michaelis-Menten kinetics, which introduces the Michaelis constant $K_M$ as the substrate concentration at which half of all enzyme is bound to substrate $-$ and thus the reaction is running at a speed of $\frac{V_{max}}{2}$.
As you can see, the relation is described by a saturation function. But for loose approximations, where $[S] \leq K_M$ you could say its roughly proportional: $$rate = \dfrac{V_{max}}{2 \times K_M} \times [S]$$

This curve is not called hyperbolic and the model never considers the number of binding sites. So the statement "at which half of the binding sites are taken" is not appropriate. Note that just that mathematical relationship does not talk anything about the mechanism.

Moreover, Vmax is not a property of the Enzyme-Substrate pair;  only kcat is the property of the ES pair. Vmax depends on both the nature of the ES pair and the total enzyme present in the sample.

The curve is not hyperbolic. A hyperbolic curve is a result of an equation $f(x)=\dfrac{b}{a}\sqrt{x^2-a^2}$. Which is not the case here.
The enzyme catalysis (Michaelis-Menten model) can be described by the two step reaction.
$$\ce{E + S<=>[k_f][k_r] ES ->[k_{cat}] E + P}$$
In Michaelis-Menten kinetics you make either of the two approximations:

Equilibrium approximation: assuming that first reaction (i.e. enzyme-substrate binding) is in equilibrium. Therefore $k_r[ES]=k_f[E][S]$
Quasi-steady-state approximation (QSS): assuming that the concentration of $ES$ complex remains constant over time. Therefore $k_f[E][S]=(k_r+k_{cat})[ES]$

For both these cases you can substitute $[ES]$ with $[E_0]-[E]$ where $E_0$ is the total enzyme concentration. You can also represent $[E]$ in terms of $[S]$ using any of the above relationship. Then, you assume that maximum catalytic activity would be when all the available enzyme is complexed with the substrate ($[ES]=[E_0]$). In that case the maximum rate of catalysis $V_{max}$ would be $k_{cat}[E_0]$. When you put everything together then the rate of catalysis would turn out to be:
$$\frac{d[P]}{dt}=V_{max}\frac{[S]}{K_M+[S]}$$
Where:

$K_M=\dfrac{k_r}{k_f}$ for equilibrium approximation
$K_M=\dfrac{k_r +k_{cat}}{k_f}$ for QSS approximation

These kind of curves denote saturation kinetics and can also be seen in case of adsorption.


text:
Instability of heavy hydrides
Question:
Why do hydrogen compounds from the third period down (that are electron deficient or have completed octets) readily decompose?
Thoughts:
Apparently, heavier group 14 elements form hydrides which decompose readily because they have valence orbitals with which $\ce{O_2}$ and $\ce{H_2O}$ can coordinate. But I've tried and tried and cannot think of why this makes $\ce{CH_4}$ more stable than $\ce{SiH_4}$. What is the difference between the two (except for the fact that silane contains a larger central atom)?
As for the remaining groups on the periodic table, I cannot fathom how this trend holds.

What do you mean by 'electron deficient or precise'??

I was referring to the number of electrons required by the central atom. Methane, for instance, is electron precise, because the central carbon has a completed octet. Monomeric borane, on the other hand, is electron deficient, because the central boron only has 6 electrons surrounding it.

Ok I guessed that was the case but precise is a weird word to describe it.

There are two factors at work here. One is that going down within a group of the periodic table, say from C to Si to Ge, gives you more electropositive (metallic) elements, so they behave more like metal hydrides and are easier to hydrolyze. A second factor is that with increasing atomic radius as you go down to heavier elements, the overlap between hydrogen's valence orbital(s) and those of the element will be poorer, and the resulting bonds weaker.

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text:
Wave and particle properties of protons and electrons
Do both electrons and protons have wave and particle properties and if they do, do they show them at the same time?

Remember, they are not real waves but rather probability amplitudes that is probability waves. They are both simultaneously associated with probability waves. Think of the atom where the electrons are described with standing waves.

@user36790 Technically you are espousing the Copenhagen interpretation and there are many other interpretations of quantum mechnics. While the Copenhagen interpretation is the most popular method of interpreting the Schrödinger equation, there are many other explanations that do not interpret $\Psi$ as a probability amplitude and $\Psi^*\Psi$ as probability density.

@Jonathon: You are absolutely correct. By far, however, the best theory is Quantum Field Theory. Nevertheless, I thought it would be better to talk about the conventional one:)

I would say that both wave and particle properties aren't "shown at the same time." Our perception of particle or wave property is the result of the experiment, and isn't a limitation of the  particle.

Actually, everything has wave-particle properties and is given by the deBroglie equation: $${\mathrm p}\lambda = h$$ where $\mathrm{p}$ is momentum (mv), $\lambda$ is wavelength and $h$ is known as the Plank constant ($6.626\cdot 10^{-34}\;\mathrm{J\cdot s}$).
As you can see as the mass of an object increases, the wavelength decreases. So in macroscopic objects that we an see, such as ball, its mass (therefore momentum) is so big compared to $h$ that the wavelength is really, really small, so we can't observe it. For light waves, it is the opposite. Due to their relatively large wavelength, they have a really small mass. However we can do experiments that show the particle nature of light. Such as the photoelectric effect which can only be explained by thinking of light as a particle rather than a wave.
Electrons which have a mass of $9.011\cdot 10^{-31}\;\mathrm{kg}$ are in between the two examples that I have given you. It turns out that their mass is perfect for their wave and particle nature to be observed. Protons are roughly 1840 times heavier than electrons, however we are also able to observe their dual nature.
The answer your second question is that they do exhibit these wave and particle properties at the same time. An electron can't simply decide to be a particle for a certain instant than decide to become a wave in the next moment. All objects exhibit their wave-particle properties at the same time.

The wave nature of neutron is used in diffraction experiments, see https://en.wikipedia.org/wiki/Neutron_diffraction

@ssavec - It goes beyond that; "buckyballs" (C60) and much larger molecules have been shown to interfere with the right apparatus. See this old Nature News story. I've just sort of come to accept that the only reason "I am" where I think I am is that almost all of me can almost always be found where I expect to find me to well within experimental tolerances. (I barely notice the fuzziness around the edges, and, honestly, that's probably just my eyes showing their age.)

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text:
Olfactory Receptors Bonding with Organic Material
How strong is the bond between the olfactory receptors and an organic material, like a perfume?
Is there a chemical that is safe for inhalation that would weaken this bond and prepare the olfactory receptors to receive new scents?
I know that materials like charcoal and zeolite absorb these molecules, but they are carcinogenic when inhaled, so I am searching for an alternative. Any and all advice is greatly appreciated!
Thanks

The so-called "bond" you are talking about is relatively weak. Binding between most odiferous compounds (most of them are organic compounds, but there are some inorganic ones like ammonia and hydrogen sulfide) and olfactory receptors (think proteins with a groove that fits the shape of the odor chemical) is rapid and reversible. There are no covalent bonds formed- just weak intermolecular forces like London forces and maybe hydrogen bonding. You need that rapid reversibility in order to tell which way a smell is coming from  or to track something by smell.
For sure, high doses of odiferous compounds can temporarily overload your olfactory receptors - the result being that your sense of smell is degraded or qualitatively changed for minutes or perhaps hours. I do not know for sure but I suspect that some chemical substances (probably only if they are designed to do so) can irreversibly bind to receptors (similar to so-called "suicide" substrates that inactivate an enzyme) and permanently inactivate them, in which case the olfactory cells would have to produce more receptors (hours to days?) in order to restore your sense of smell.
The question I think you need to consider is: is there a chemical substance that would purge the olfactory receptors of old odor compounds and render them freshly able to detect new odors? I think the answer may be no. Even if you could place a zeolite or activated carbon particle next to an olfactory receptor, the odor chemical that is bound to it would still have to diffuse out (on its own) in order to bind with the particle.
But lets see if others out there have any ideas about this. And give my regards to your grandmother!

Thinking some more about this question, I also want to offer that there may be a way to chemically stimulate the expression of the olfactory receptor genes, so that the olfactory cells in your nose produce more (fresh) receptors. This in turn might possibly (but would not necessarily) increase your sensitivity to certain odors.


text:
Vacuum distillation apparatus selection and configuration
I want to assemble a "hobbyist-level" vacuum distillation apparatus for extracting small volumes of "typical" solvents (e.g., water, alcohols, ketones, diols).
I already have standard lab glassware with 24/40 joints, 10mm barbed inlets.
How do I select a suitable pump?

I would want a pump that can run for extended periods to perform vacuum desiccation.
So far my research suggests that I should be looking at diaphragm pumps, but:
I don't know if there are other features or specifications available in a low-three-figure pump that will make life easier.
What kind of tubing do I want?  I know that larger inner diameter is better, but I don't know if something special is used instead of regular tubing to support the vacuum, or what the standard adapters and sizes are to come off the 10mm glass fittings.

What elements need to be part of such an apparatus, and how should they be configured?  E.g., I have seen mention of:

Filters: What are used, and when are they necessary?
Traps: Does this only refer to "cold traps," which I understand to simply be another condenser in front of the pump?  If so, is it correct that this would only help if I can cool the "trap" significantly below the temperature I'm using for the primary distillate condenser?
Manometers
Capillary bubblers: I can't even find anything on these, other than what's mentioned here.

Also, how do these answers change if I'm doing vacuum desiccation of the named solvents (i.e., where the only difference is I don't care to capture the evaporate)?  E.g., can I just let the system exhaust through the pump?

Don't bother with a powered pump: Get a water aspirator (I recommend the $20 Humboldt H-12020, which uses 10mm or 3/8" inner diameter tubing just like your glass), connect to faucet, and using utility water (1-2 gallons/minute) it can pull over 500 torr, which should be adequate for your applications.  Sure, you're flushing away a lot of water, but you don't have to bother with traps or filters.
(Actually, the H-12020 does have a check valve, but to ensure you don't get even a little water sucked into the apparatus when you shut off the aspirator you might want to add a separate valve or simple trap, but that's easy compared to a cold trap.)
You should buy "thick-walled vacuum tubing" for all connections if you want to run full vacuum.
Finally, for safety you do need some bump countermeasure.  A "capillary bubbler" is just a tube that lets a tiny stream of bubbles into the boiling flask.  Yes, that is reducing the vacuum, but a proper capillary tube admits a trivial amount of gas into the system.  When possible, however, I prefer stirring.  Running a magnetic stirrer in the boiling flask also prevents bumping, and is better than a bubbler because it doesn't introduce potentially contaminating gases into the mix (or require the supply of a neutral gas).
A decent reference for practical distillation is Zubrick's Chapter 20 ("Distillation for Dummies") available online here.

A water venturi (aspirator) will mix the solvent into the wastewater system, for some compounds this may be a pollution hazard depending on the situation. There are also similar devices that are designed for compressed air which could  achieve lower pressure because they aren't limited by the vapor pressure of tap water. It is also possible to connect several in series (water and or gas working fluid) for better vacuum at the cost of less efficiency.


text:
Urban air pollution and the smell of freshly cut lawn grass?
Recently, all the lawns in the apartment complex that I live in have had their lawns cut - giving that pleasant smell of freshly cut lawn.
Now, I am aware that the smell arises from a chemical distress signal from within the plant itself - are the chemicals from freshly cut lawn contributors to urban air pollution?
Related: How to determine source of air pollution

The information page from the Compound Interest page of the same name describes it well (below).
A summary of the main points are that grass already emits small amounts of VOCs (volatile organic compounds) - mostly comprising of 6-carbon atom based aldehydes and alcohols.
When the lawn is cut, the main smell is from (Z)-3-hexenal, at which only 0.25 parts per billion can be sensed (smelled) by humans - this compound is short lived, hence why the smell is a freshly cut smell.
As a little twist, in research presented in the paper Air Pollution and the Smell of Cut Grass, the authors suggest that a significant proportion of aerosol VOC's in urban air pollution actually originate from the release of these chemicals when lawns are cut - and given the sheer number of manicured lawns in an urban centre, this starts to make sense.
Some of these released chemicals are highly photoreactive. The flux of VOCs is seasonal as well in many places, contributing to seasonal variations of VOC-based photochemical smog.
Within the article, VOC flux measurements taken in the urban area of Melbourne Australia indicate that 6% of the flux arises from uncut grass; however, this flux doubles when the lawn is cut. The combined flux is comparable to industrial sources.
It is suggested in the article, that

within an
  urban airshed, during the growing season, about one third of the emissions of
  photochemically reactive VOCs come from grass and cut grass.


text:
What structural property creates stickiness?
Recently, me and my mom parked her car in a spot reserved for customers of a super expensive shoe store while going to get some also fairly expensive drinks at Starbucks. When we came back, an employee at the shoe store put a sticker on my mom's window warning us he could have us towed and ticketed. Clearly they're bitter about having no customers. This sticker left a whole bunch of sticky residue on the window.
I knew from experience that water wouldn't get the job done, so I put some canola oil on a paper towel and sure enough the sticky residue dissolved into the oil and I could see it dissolved in the liquid on the paper towel.
This got me thinking. Is it generally true that a sticky substance will be nonpolar? Furthermore, is there a generalization which can be made about structures which are sticky on a macroscopic level?
My only guess is that sticky substances are usually viscous too which leads me to believe that there must be a lot of VDW/dispersion interactions involved in those materials because I understand that high viscosity is related to strong intermolecular attractions. Although, the viscosity argument doesn't work for sticky solids (unless you say that a sticky solid is actually just an extremely extremely viscous liquid).

Pro tip: Embolden important stuff in your question so we can understand what it's trying to say faster.

Thanks for the edit. Ya, I've used bold writing in the past and it makes it look so nice, but I just wasn't thinking much for this one I guess.

Long story short, no, sticky substances don't have to be nonpolar at all. Pour a little wine or beer on a table, and let it dry almost completely: it will get sticky, but still be water-soluble and hence polar. Also, there are certain water-based glues, which are surely sticky and polar. On top of all that, go read about the hagfish, only make sure you're not eating at the moment, for fear of adverse reaction.
As for the structural properties that make a substance sticky (or viscous, which is basically the same thing), these are not so easy to point out. Many sticky things are polymers, i.e., very long molecules with lots of van der Waals interactions, just like you said, and probably lots of interactions of other sorts, too. But this is hardly a rule, as exceptions both ways are plenty. Polyethylene is a polymer, but is not sticky at all. On the other hand, sugar syrup can be pretty sticky, despite containing no polymers.
As for the sticky solids, I would claim that there are none. Of the possible examples, some are actually highly viscous liquids, some are sticky because they are covered with such liquids, and some would stick to things because of electrostatic effects.


text:
Why does mercury form polycations?
For example, mercury (I) is $\ce{Hg2^2+}$ and not $\ce{Hg+}$. What causes  the stability in covalently bonded $\ce{Hg}$ ions?

It isn't unusual. See: Metal Compounds that bond covalently

Still, most metals normally wouldn't do that, so mercury is indeed somewhat unique.

Well metal-nonmetal covalent bonds are present in most of so called salts. Metal-metal bonds like in mercury (I) aren't so ubiquitous but still often found in multi-center complexes.

It's worth noting that Mercury seems to form a wide variety of polycations with varying geometries, such as linear chain and triangular. As far as I can tell from searching through ACS publications, there doesn't seem to be any analysis of why these are stable, just a few papers that show that they exist.

I don't have an excellent understanding of the topic, but I will try to write something up.
Polycations are found with the Hg(I) oxidation state. The mercury cation, $\ce{Hg+}$, has a $\mathrm{5d^{10}6s^1}$ configuration. The mercurous cation $\ce{Hg2^2+}$ can then be considered to simply be formed by overlap of the two $\mathrm{6s}$ orbitals on two separate $\ce{Hg+}$ ions. In this sense, the bonding is qualitatively very similar to that in $\ce{H2}$. Note that the $\mathrm{5d^{10}6s^0}$ Hg(II) cation has no chance of forming such covalent bonds.
Generally, going down a group, orbitals become larger, more diffuse, and the overlap between them decreases. As a consequence, $\sigma$-bonding tends to become weaker going down the group. A simple example is provided by the alkali metal dimers:
$$\begin{array}{c|ccc}
\text{Element} & \ce{H} & \ce{Li} & \ce{Na} \\  \hline
D_\mathrm{0}(\ce{E-E})\text{ / }\pu{kJ mol-1} & 432 & 99 & 71 \\ \hline
\text{Reference} & 1 & 2 & 3
\end{array}$$
These dissociation energies are extremely small. For the analogous $\ce{Hg2^2+}$ cation, the situation is not quite as bleak. The Group 12 metals in general have more low-lying and contracted ns orbitals than the corresponding alkali metals, due to the presence of ten poorly-shielding (n-1)d electrons. However, given that zinc does not form similar cations, it is odd that mercury with its larger 6s orbitals should do so.
The most obvious explanation is simply that Zn does not commonly exhibit the +1 oxidation state. The existence of the +1 oxidation state for Hg is directly related to the relativistic stabilisation of the 6s orbital, which increases the first two ionisation energies4 of mercury.
$$\begin{array}{c|ccc}
\text{Element} & \ce{Zn} & \ce{Cd} & \ce{Hg} \\ \hline
I_1\text{ / eV} & 9.393 & 8.992 & 10.44 \\ \hline
I_2\text{ / eV} & 17.96 & 16.90 & 18.76 \\
\end{array}$$
However, the question still remains as to why Hg(I) chooses to form polycations instead of existing simply as $\ce{Hg+}$. In fact, relativistic effects have also been implicated in this for a long time: Pyykkö and Desclaux wrote in 1979 that5

The relativistic contraction of the half-filled valence s AO makes $\ce{Au2}$ much more strongly bound than $\ce{Ag2}$. Inasmuch as $\ce{Hg2^2+}$ and $\ce{Cd2^2+}$ are isoelectronic with these two molecules, the remarkable stability of the mercurous ion [compared to the Cd analogue] can also be understood.

Therefore, this may most simply be explained as the relativistic contraction of the 6s AOs leading to stronger-than-expected overlap to form a covalent bond. As suggested by the above paragraph, similar effects are seen in the neutral dimers of Group 11 metals:6
$$\begin{array}{c|ccc}
\text{Element} & \ce{Cu} & \ce{Ag} & \ce{Au} \\  \hline
D_\mathrm{0}(\ce{E-E})\text{ / eV} & 1.95 & 1.65 & 2.29 \\
\end{array}$$
The decrease going from Cu to Ag is expected on the basis of weakened overlap, and the dissociation energy of $\ce{Au2}$ would be even lower in the absence of relativistic effects: calculations indicate that relativity contributes roughly $\pu{1 eV}$ to the dissociation energy of $\ce{Au2}$.6 Presumably, the effect in $\ce{Hg2^2+}$ is of a similar significance.

References

Frenking, G.; Bickelhaupt, F. M. The EDA Perspective of Chemical Bonding. In The Chemical Bond: Fundamental Aspects of Chemical Bonding; Frenking, G., Shaik, S., Eds.; Wiley: Weinheim, Germany, 2014; p 128.
Velasco, R.; Ottinger, Ch.; Zare, R. N. Dissociation Energy of Li2 from Laser‐Excited Fluorescence. J. Chem. Phys. 1969, 51 (12), 5522–5532. DOI: 10.1063/1.1671979. Free version provided by the Zare group here.
Jones, K. M.; Maleki, S.; Bize, S.; Lett, P. D.; Williams, C. J.; Richling, H.; Knöckel, H.; Tiemann, E.; Wang, H.; Gould, P. L.; Stwalley, W. C. Direct measurement of the ground-state dissociation energy of Na2. Phys. Rev. A 54, R1006(R). DOI: 10.1103/PhysRevA.54.R1006.
Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed; Oxford UP: Oxford, U.K., 2014; pp 836–7.
Pyykko, P.; Desclaux, J. P. Relativity and the periodic system of elements. Acc. Chem. Res. 1979, 12 (8), 276–281. DOI: 10.1021/ar50140a002.
Pitzer, K. S. Relativistic effects on chemical properties. Acc. Chem. Res. 1979, 12 (8), 271–276. DOI: 10.1021/ar50140a001.

Oh wow! I was going along this line of thought as well. What a coincidence! However, I would attribute the fact that zinc and cadmium do no behave in the same way to the presence of the f subshell in mercury. These f subshell orbitals are very weakly shielding. This causes the electronegativity of mercury to be much closer to that of hydrogen, on the Pauling scale. Because mercury (I) and hydrogen share similar electronegativities and have similar electron configurations, they exhibit similar bonding behaviour.

Perhaps, this would add more evidence to your theory.

Zinc does form the $+1$ oxidation state (Holleman, Arnold F.; Wiberg, Egon; Wiberg, Nils (1985). "Zink". Lehrbuch der Anorganischen Chemie (in German) (91–100 ed.). Walter de Gruyter. pp. 1034–1041.).  But yes, this state is of limited range compared with heavier Group 12 elements.

@orthocresol  "due to the presence of ten poorly-shielding (n-1)d electrons" ; I think it isn't true from what I have studied that d electrons shield ns electrons somewhat more effectively than outer electrons can shield one another , this being the reason why ionisation enthaply doesn't increases sharply in 3d series

@YashAgrawal The IE may not rise sharply, but it still does, which shows that shielding doesn't compensate for the increase in nuclear charge. If you compare Ca (589.8 kJ/mol) and Zn (906.4 kJ/mol) then it's clear that the Zn 4s orbitals are lower in energy than the Ca 4s orbitals, which is the comparison I was making in the text. (data from https://en.wikipedia.org/wiki/Ionization_energies_of_the_elements_(data_page))

Only a few $\ce{Hg(I)}$ compounds (also called mercurous compounds) are known which include mercury(I) halides $\ce{Hg2X2}$ and mercury(I) nitrate $\ce{Hg2(NO3)2.2H2O}$ . They contain the ion $\ce{Hg2^2+}$ actually $\ce{(Hg -Hg)^2+}$ and not $\ce{Hg+}$. The two $\ce{Hg}$ atoms are bonded together using the 6s orbital. Zinc and cadmium also forms stable dinuclear ions but are unstable and has been only detected spectroscopically in melts of $\ce{Zn/ZnCl2}$  and $\ce{Cd/CdCl2}$. They are made by reducing mercury(II) salts with metal. Mercury(I) nitrate can be prepared by dissolving $\ce{Hg}$ in  $\ce{HNO3}$.
Their standard reduction potential are so close that oxidizing agent like $\ce{HNO3}$ converts $\ce{Hg}$ to $\ce{Hg^2+}$ rather than $\ce{Hg(I)}$.
$\ce{Hg^2+ -> Hg}$ $(E^0 = 0.85\,\rm V)$
$\ce{Hg2^2+ -> Hg}$ $(E^0 = 0.79\,\rm V)$
This shows that $\ce{Hg2^2+}$ is stable to disproportionation by a small margin under standard condition.
$\ce{Hg2^2+ <=> Hg + Hg^2+}$ $(E^0 = 0.13\,\rm V)$
The equilibrium constant can be calculated:-
$$E^0 = \frac{RT}{nF}\cdot\ln K$$
$$K = \frac{[\ce{Hg2^2+}]}{[\ce{Hg^2+}]}=170$$
Thus solution of mercury(I) compounds contains one $\ce{Hg^2+}$ for every $\ce{170}$ $\ce{Hg2^2+}$ and the equilibrium is finely balanced. If any reagent is added to the  solution, disproportionation occur.
There are many evidence of mercury(I) having the structure $\ce{(Hg-Hg)^2+}$ like:-

X-ray diffraction
Crystal structure of various mercury(I) compounds have been determined by X-   ray diffraction. For ex-  It was assumed that  $\ce{HgCl}$ consist of $\ce{Hg+}$ and $\ce{Cl-}$ but X-ray diffraction showed $\ce{Cl-Hg-Hg-Cl}$ arrangement. Other compound was also proved of having the $\ce{Hg-Hg}$ bond.
equilibrium constant

Consider the reaction:-
$$\ce{Hg(NO3)2 + Hg -> 2HgNO3}$$
$$\ce{Hg^2+ + Hg -> 2Hg+}$$
Then by law of mass action
$$\frac{\ce{[Hg+]^2}}{\ce{[Hg^2+]}}={\rm constant}$$
Experiment have shown this to be wrong if however:
$$\ce{Hg^2+ + Hg -> Hg2^2+}$$ then $$\frac{\ce{[Hg2]^2+}}{\ce{[Hg^2+]}}={\rm constant}$$
This shows $\ce{Hg2^2+}$ contains$\ce{(Hg-Hg)^2+}$.

concentration cell

For the cell below, $E^0$ was found to be $0.029\,\rm V$ at $25°\rm C$
Hg/mercury(I) nitrate  (0.005M)//mercury(I) nitrate  (0.05M) /Hg
$$E = \frac{2.303RT}{nF}\cdot\ln \frac{c_2}{c_1}$$
Putting values,
$$0.029 = \frac{0.059}{n}\cdot\ln \frac{0.05}{0.005}$$
$$n = 2$$ where n = no. of charges on ion. This confirms $\ce{Hg2^2+}$

Raman spectra

Spectral lines correspond to $\ce{Hg-Hg}$ bond.

magnetic properties

All mercury(I) compounds are diamagnetic as all electrons are paired in  $\ce{(Hg-Hg)^2+}$.

cryoscopic measurements

Observed depression fits for all mercury(I) compounds containing  $\ce{(Hg-Hg)^2+}$ ion.
Also, compounds like $\ce{Hg3(AlCl4)2}$ and $\ce{Hg4(AsF6)2}$ are also discovered and they contains cation with three and four mercury atoms bonded linearly.

-1 - Technically this is NAA. You are just stating the existence of the cation as well as some (weakly presented) evidence for it but OP has no doubt that it indeed is Hg2(2+). The question is about why it takes this form.

All ready answered, but I will give the distilled version:
Zn and Cd are most commonly found in the +2 OX State (d10) and can not metal-metal bond in this state. +1 OX state is possible for these elements but is rare due to there low ionization energies, but they do form metal-metal bonds in the +1 OX state.
Hg has relatively high ionization energy due to it having a higher effective nuclear charge from pore shedding from the f electrons (lanthanide contraction). The high ionization energy makes the +1 OX state more stable for Hg, where it is capable of M-M bonding.

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text:
What is the balanced equation for the reaction of potassium permanganate, iron sulfate, and sulfuric acid?
I'm trying to balance the following redox equation. I think this is happening in acidic solution, the textbook doesn't specify anything more. Can you please help me understand if I got the half reactions correct?
$$\ce{KMnO4 + FeSO4 + H2SO4 -> K2SO4 + MnSO4 + Fe2(SO4)3 + H2O}$$
I found out that manganese is reduced as the oxidation number goes from $+7$ to $+2$ and iron gets oxidized as the oxidation number goes from $+2$ to $+3$.
I wrote the two half reactions, \eqref{Q:red} for the reduction and \eqref{Q:ox} for the oxidation. Did I get them right?
\begin{align}
\tag{A}\label{Q:red}\ce{KMnO4 + H2SO4 &-> K2SO4 + MnSO4}\\
\tag{B}\label{Q:ox}\ce{FeSO4 &-> Fe2(SO4)3}\\
\end{align}
I started working on the reduction half-equation and this is what I came up with after balancing atoms and charge:
$$\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 -> K2SO4 + 2 MnSO4 + 8 H2O}$$
Now comes the trouble. How do I balance iron and sulfur in the oxidation half-reaction \eqref{Q:ox}?

You may have marked this question as a duplicate but I still didn't understand what is it that I got wrong with the oxidation half-reaction, or why do I need to write the sulfuric acid twice in order to balance both half-reactions and obtain the correct answer. Can anyone care to explain? Please?

Well you can ignore all the spectator ions: $\ce{K+, SO4^2-}$ Your half equations are wrong because they don't balance. I would get rid of all the clutter and then balance the half equations using $\ce{H2O, H+}$ and $\ce{e-}$.

You mean I should re-write the given reaction this way? $$\ce{(MnO4)^- +Fe^{2+} -> Mn^{2+} ,+Fe^{3+}} $$ …and write the reduction and oxidation half-reactions from there on? How would I then get the solution (see below, in a comment to @oryza's advice)?

Your first half reaction, for the reduction, is correct:
$$\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 -> K2SO4 + 2 MnSO4 + 8 H2O}\tag1\label{red}$$
For the second half reaction, the oxidation, start by balancing iron:
\begin{align}
\ce{FeSO4 &-> Fe2(SO4)3 }\tag{2a}\\
\ce{2 FeSO4 &-> Fe2(SO4)3 }\tag{2b}
\end{align}
Add $\ce{H2SO4}$ on the left so you can balance sulfur:
\begin{align}
\ce{2 FeSO4 &-> Fe2(SO4)3 }\tag{2c}\\
\ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 }\tag{2d}
\end{align}
Now balance the protons and electrons:
\begin{align}
\ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 }\tag{2d}\\
\ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 + 2 H+ + 2 e-}\tag{2e}\label{ox}\\
\end{align}
Now add \eqref{red} and five times \eqref{ox} so that the electrons are equal on every side and coincidentally the protons also balance:
\begin{align}
\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 &-> K2SO4 + 2 MnSO4 + 8 H2O}\tag1\\
\ce{5 H2SO4 + 10 FeSO4 &-> 5 Fe2(SO4)3 + 10 H+ + 10 e-}\tag{$5\times$2e}
\end{align}
And the final result is:
\begin{align}
\ce{2 KMnO4 + 8 H2SO4 + 10 FeSO4 &-> K2SO4 + 2 MnSO4 + 8 H2O + 5 Fe2(SO4)3}\tag3
\end{align}

Yes you are correct saying that manganese gets reduced from $+7$ to $+2$ and iron gets oxidised from $+2$ to $+3$.
However, if I were you, since they already gave you the complete redox equation, – it is just unbalanced –, I wouldn't bother with writing the half equations. I would just focus on balancing the equation that they have already given you.
Here is a guide for balancing most redox equations:

Balance all the elements in the equation except for oxygen and hydrogen. For polyatomic ions that are spectator ions, just think as them as one big element.
Check if oxygen is balanced. If it isnt, add water to the side which is missing oxygen.
Check if hydrogen is balanced. If it isn't, add protons $(\ce{H+})$ to the side with doesn't have enough hydrogen.
Now that all the elements have been balanced, make sure that the charges are balanced, too. If they are not, add electrons to the side which is more positive.
Check your equation now to make sure that everything is balanced correctly.
When you have balanced both half-reactions, add them so that electrons cancel on each side.

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


text:
Why doesn't silica gel dissolve in ethyl acetate
Why doesn't silica gel dissolve in ethyl acetate when something like aspirin does. Both silica gel and ethyl acetate are highly polar. I though like dissolved like.

Neither aspirin, nor ethyl acetate are highly polar, but rather moderately.

@Mithoron, add the fact that silica is a network solid, so it does not dissolve in much of anything (except for what reacts with it), and you have good answer.

If "like dissolves like" were the only rule, then your bathtub would get dissolved in water, together with much of your kitchenware. Oh, and your car would be gone completely, because it has both polar and non-polar liquids in it.

but seriously. if that rule is incorrect, why is it taught. i guess its akin to the i before e rule.

Come on. Every empirical rule has its limitations.

It is not incorrect, it is just limited.

Not even methanol or water do a good job of dissolving silica gel (although water is better at it than most other solvents). I haven’t actually tried DMF, which is supposed to be one of the most polar solvents out there, though.

@BenNorris Silicon dioxide does dissolve in hi-pressure hi-temperature water. http://www.minsocam.org/ammin/AM62/AM62_1052.pdf

Because (among other things) silica is not actually polar. Yes, theoretically a molecule of $\ce{SiO2}$ should have some degree of polarity to it. However, silica as a solid does not exist as single molecules.
Silica exists as a network of $\ce{Si}$ cations bonded together by $\ce{O}$ anions. One way of looking at it would be to say that each small bead of silica gel (or any other form of silica, glass or quartz) is one huge single molecule of silica.


text:
What reaction will ensue when sulfur trioxide is added to excess barium hydroxide solution?
Question:

If $\ce{SO3}$ was added to an excess solution of $\ce{Ba(OH)2}$, what would be the resulting reaction?

Thoughts:
Initially, I suppose it would form barium hydrogen sulfate, which slowly precipitates barium sulfate in water as the hydroxide is present in excess.
$$\ce{2SO3 + Ba(OH)2 <=> Ba(HSO4)2 \\ Ba(HSO4)2 + Ba(OH)2 <=> 2BaSO4 + 2H2O}$$
Is this correct?

Looks ok, but I wouldn't like to make this reaction in practice ;)

The word "slowly" does not look OK to me. The precipitate would form pretty much instantly.

Basically yes.
Sulfur trioxide reacts with water to form sulfuric acid (this is extremely dangerous!  do not try it at home) in a strongly exothermic reaction which, unless done under very precise control, produces a fine mist of concentrated sulfuric acid vapor (not nice).
$$\ce{SO3 + H2O -> H2SO4}$$
If you are doing this with an excess of barium hydroxide then the sulfuric acid will be deprotonated in two steps as you suggest.
$$\ce{H2SO4 + OH- <=> HSO4- + H2O}$$ $$\ce{HSO4- + OH- <=> SO4^2- + H2O}$$
Barium sulfate will precipitate from the solution, driving the reaction to completion.
$$\ce{Ba^2+ (aq) + SO4^2- (aq) -> BaSO4 (s)}$$
I couldn't find any mention of barium hydrogen sulfate as a compound anywhere so I take this to mean that it is not stable and therefore will not precipitate.


text:
Quantum and organic chemistry of vitamin D synthesis
I am somewhat confused by certain steps in this mechanism.
To start with, in the conversion of 7-dehydrocholesterol to previtamin D3, how does the UV photon affect the pi-orbital system in such a way as to rearrange the bonds as shown in the diagram? Else, what exactly brings about the change shown?
Similarly, in the thermal isomerisation step from previtamin D3 to Vitamin D3, what quantum effects and other changes cause the rearrangement shown?
Since this is a biological reaction within the human body it also takes place under physiological conditions ($37^\circ C$, 1 atm, pH = 7.4, various intracellular ionic concentrations).

I have not enough time to read these articles, but they could have the answer: Michael F Holick, J Investigative Dermatology, 1981, 77, 51–58; doi:10.1111/1523-1747.ep12479237. Principle investigation: M.F. Holick, J.E. Frommer, S.C. McNeill, N.M. Richtand, J.W. Henley, J.T. Potts Jr., Biochemical and Biophysical Research Communications
Volume 76, Issue 1, 9 May 1977, Pages 107-114

These are pericyclic (i.e. concerted) reactions. They do not need particularly high temperatures, but indeed they can be catalysed by either the right photons or temperature. They are not "reactions" per se, rather just intramolecular rearrangements of electrons over a "chain" of connected atoms.
The first reaction is a photolytically catalysed electrocyclic reaction that results in the opening of the 6-membered ring, they are very common.
The second reaction is a thermally catalysed [1,7] sigmatropic migration of the H. The rate is dependent on the nature of the nucleus that moves, in this case H (D would be slower). Quantum tunneling will indeed enhance the rate of its non-quantum counterpart, but it is not the driving force of the step as you seem to wonder.
As for the reason why these reactions happen in the first place, it's because there is favourable overlap in the transition states between the orbitals of the atoms at the two external ends of the "chain" of atoms interested in each rearrangement.
As you can see below, the HOMO in each case permits favourable overlap (phase-wise and sterics-wise) between the ends of the "chains" in both cases. I couldn't quickly find a picture of a photolytic ring opening but in that case the actual HOMO is a SOMO: one electron from the former HOMO has been promoted (by the photon) to the former LUMO. This will change the overall symmetry properties of the atom chains so that a [1,7] H shift is allowed under photolytic conditions.
Electrocyclic reaction:

Hydride shift:

Reference: https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/pericycl.htm
Suggested reading: "Pericyclic Reactions" by Ian Fleming (OUP, 1998).

Could you please illustrate how this "favourable overlap" works, preferrably with a diagram of sorts?

Also I wasn't referring to quantum tunneling, I was thinking more about how an MO diagram or frontier pi-orbitals could be used to explain this phenomenon.

@James Harrison: I've put pictures of the FMOs to clarify as requested. I wrote that bit about the tunneling as it was the only reference to quantum effects that could be made in this context and you were asking about quantum effects in the question, apologies if that wasn't clear.

Thanks very much for adding those in! But one more thing, is rotation still restricted in the C atoms within the conjugated pi system after the 1,7 H shift, or does it otherwise matter that the double bonds seem to be on the same side of the single bond in your FMO diagram, but on different sides in the mechanism in my question?

I see what you mean. I was always given to understand that there is reasonable flexibility around the alkene bonds (at least during the transition state), so that the ends of the conjugated system will be close during the shift, but then readjust themselves soon afterwards to reach a minimum in potential energy - thereby giving the somewhat "straighter" looking π-system that you have in your original picture. Hope this helps.


text:
How does aluminium amalgam work as a reducing agent?
I can't seem to find much information on this topic but how does aluminium amalgam work as a reducing agent?
Could it be used to deoxygenate a carboxylic acid or amino acid to the corresponding alcohol or alkane?
If not , why not?

Joel, in one place you are saying that you are "Looking for an answer drawing from credible and/or official sources" and in another place "Please don't post any links".   Please explain what citation format you would like the references to official sources to be in if you don't like links.  Do you want patent numbers, journal volume and page, what are looking for?

I just mean an answer from someone that is credible/official so it's an original answer and they can explain it without a need for links . I mean no disrespect and fully appreciate your time , I'm not at all saying your not credible/official , I'm sorry if I worded it incorrectly.  :)

The chemistry theory behind why the amalgam reacts selectively as apposed to other reducing agents towards this style of compound is why the question is interesting.

Please see the [help], especially the entry on bounties. In short: They are non-refundable, i.e. you cannot get back your reputation points. You also cannot extend the period of the bounty. You can set a new bounty if you are not satisfied with the answers that have been given. You also might want to read the guide about flagging.

Could aluminium amalgam be used to deoxygenate a carboxylic acid or amino acid to the corresponding alcohol or alkane?

No. Aluminium amalgam has been used to selectively reduce other groups in molecules containing a carboxylic acid group, while preserving the carboxylic acid group.
See for example in Industrial Organic Nitrogen Compounds at page 139:

Some pyridine-carboxylic acids have been reduced to the corresponding dihydropyridine-carboxylic acids with aluminium amalgam.

See also the patent Polyene carboxylic acids and esters and manufacture thereof where aluminium amalgam is suggested as a mild reducing agent which preserves the carboxylic acid or ester group.

So would it reduce an alcohol or a ketone to the alkane?

The first link you posted doesn't work with my phone unfortunately @DavePhD. Do you have a detailed example of a reaction mechanism so i can see how it used ( in theory!).

Also why does it act to protect the functional groups instead of deoxygenating all the functional groups. How is it selective and why won't it work? I can already find many examples of reactions it's used in.

Sorry  @ DavePhD , just posting links and not providing the theory behind what's happening just isn't the answer I was looking for. I appreciate your help tho :)

@Joel as far as I know, aluminum amalgam can be used to reduce a ketone to an alcohol, and zinc amalgam with HCL can be used to reduce some ketones to alkanes (Clemmensen reduction).  But I'm not aware of aluminum amalgam being used in the same way.  Maybe the HCl would react with the aluminum instead to form hydrogen gas, but I'm just guessing.  Sorry if I didn't give you the info you were hoping for.

@ DavePhD, all good , I appreciate your efforts likewise. I've also found it hard to find info regarding this hence the bounty.  :)


text:
What happens to the complex part of the wave function when doing LCAO?
I understand that in general the wave function can take complex numbers, however when talking about combining atomic orbitals to form molecular orbitals we talk about the phase of the orbital being positive or negative to create constructive or destructive interference. It seems as if we have lost the aspect of the wave function taking on complex phase, what happened to it?

It got very scared because sh*t just got real and squared away ;)

Also, generally chemists prefer working with real wavefunctions, which are obtained by taking appropriate linear combinations of the complex wavefunctions. https://en.wikipedia.org/wiki/Table_of_spherical_harmonics#Real_spherical_harmonics This is allowed because any linear combination of degenerate eigenfunctions is also an eigenfunction of the Hamiltonian with the same eigenvalue. Another benefit is that these real wavefunctions have convenient directional properties e.g. the p orbitals are aligned in the x, y, and z directions. Same goes for the d orbitals.

I understand that in general the wave function can take complex numbers [...]

It is not that it can, rather it should: by the very postulates of quantum mechanics wave function is a complex-valued function. So, both the spin orbitals (that are one-electron wave functions) and the Slater determinant built out of them are in principle complex-valued functions.
However, in solving the HF equations by the SCF procedure it is quite typical to impose some constraints on the spin orbitals, for instance, restrict them to be real-valued functions rather then complex-valued ones. As with any other constraint this restriction might lead to what is called SCF instabilities, i.e. situations when relaxing the constraints lead to a different variational solution of lower energy.
To elaborate a bit more let us mention few other well-known (as well as lesser-known) constraints for the SCF procedure:

The infamous restricted Hartree-Fock (RHF) model with the requirement that spin orbitals come in pairs: two spin orbitals corresponding to two different pure spin states are constructed from the same spatial orbital:
$$
\begin{aligned}
  \psi_{2i-1}(1) &= \phi_{i}(1) \alpha(1) \, , \\
  \psi_{2i}(1)   &= \phi_{i}(1) \beta(1)  \, .
\end{aligned}
$$
In the unrestricted Hartree-Fock (UHF) model the requirement above is relaxed and we exclusively use spatial orbitals from one set to construct $\alpha$ spin orbitals and spatial orbitals from another set to construct $\beta$ spin orbitals:
$$
\begin{aligned}
  \psi_{2i-1}(1) &= \phi_{i}^{\alpha}(1) \alpha(1) \, , \\
  \psi_{2i}(1)   &= \phi_{i}^{\beta}(1) \beta(1)   \, .
\end{aligned}
$$
A bit less known is the fact that the unrestricted Hartree-Fock method is not so unrestricted. In fact, there is a constraint here: each and every spin orbital describes an electron in a pure spin state, either $\alpha$ or $\beta$, while in general (in accordance with the postulates of quantum mechanics) electron can be in superposition of this states:
$$
  \psi_{i}(1) = \phi_{i}^{\alpha}(1) \alpha(1) + \phi_{i}^{\beta}(1) \beta(1)
$$
This most general setting is known as the general Hartree-Fock (GHF) method.

And all these three methods (RHF, UHF, GHF) exist in two variants: the real one, in which spin orbitals additionally required to be real-valued functions, and the more general complex one. All this give rise to six variants of the HF method with many possible instabilities between them discussed it some details in the seminal paper by Schlegel & McDouall1:

There is also a very similar earlier paper by Seeger & Pople.2 which suddenly is not freely available.
Its important to realize that any constraint of the mentioned above can only raise the electronic energy: with a great deal of certainty, we may expect that if any constraint is relaxed, the variational procedure will result in a lower energy due to greater variational freedom. Thus, for instance, for the most usual different real variants of the Hartree-Fock method we have
$$
    E_\mathrm{e}(\mathrm{RGHF}) \leq E_\mathrm{e}(\mathrm{RUHF}) \leq E_\mathrm{e}(\mathrm{RRHF}) \, .
$$
The same, of course, is true for the corresponding complex variants of these methods,
$$
    E_\mathrm{e}(\mathrm{CGHF}) \leq E_\mathrm{e}(\mathrm{CUHF}) \leq E_\mathrm{e}(\mathrm{CRHF}) \, .
$$
And for any of the three formalism
$$
    E_\mathrm{e}(\mathrm{C}x\mathrm{HF}) \leq E_\mathrm{e}(\mathrm{R}x\mathrm{HF}) \, , \quad \text{where} \quad x = \mathrm{G}, \mathrm{U}, \mathrm{R} \, .
$$

RHF/UHF/GHF and correct spin symmetry
The possible rise of electronic energy that accompanies introduction of more and more constraints in the GHF -> UHF -> RHF sequence seems to be quite contrary to the goal of the variation method. However, UHF and RHF constraints are nothing but symmetry constraints: they arise by requiring an approximate electronic wave function to have the same spin symmetry as the exact non-relativistic one, i.e. to be an eigenfunction of spin operators, the total spin-squared operator $\hat{S}^2$ and the $z$-component of the total spin operator $\hat{S}_{z}$.

With respect to $\hat{S}_{z}$ operator it is well known that any Slater determinant built out of spin orbitals corresponding to pure spin states is an eigenfunction of $\hat{S}_{z}$. Thus, RHF and UHF wave functions are eigenfunctions of $\hat{S}_{z}$, but GHF wave function is not.
The situation with $\hat{S}^{2}$ is a bit more complicated, but all in all only in RHF (but not in UHF or in GHF) formalism it is possible to construct an approximate electronic wave function which is an eigenfunction of $\hat{S}^{2}$.3

To conclude, as it was pointed out by Löwdin, we always face with a dilemma here: should we seek a solution that is a true variational minimum or should we seek a solution with the correct spin symmetry?

1) H. B. Schlegel and J. J. W. McDouall, Do You Have SCF Stability and Convergence Problems? in Computational Advances in Organic Chemistry: Molecular Structure and Reactivity, Springer Netherlands, 1991, pp. 167-185. DOI: 10.1007/978-94-011-3262-6_2. Free PDF from wayne.edu.
2) Seeger, R., & Pople, J. A., Self‐consistent molecular orbital methods. XVIII. Constraints and stability in Hartree–Fock theory. The Journal of Chemical Physics, 66(7), 1977, 3045-3050. DOI: 10.1063/1.434318.
3) This can be done even for an open-shell electron configuration, although it would require a linear combination of Slater determinants with constant coefficients, none of which alone is an eigenfunction of $\hat{S}^{2}$.


text:
Why does my blood taste like rust?
I thought it is just me, but when I searched it on Google, it revealed that there are many people who experience this:
Why is it that whenever I taste my own blood, I always think it tastes like rust?  I have never eaten or tasted rust, so how can I relate something to it?

Perhaps iron of haemoglobin makes it taste like rust

Related answer

What you are calling "taste" is actually produced by the olfactory sense – "smell."  True taste, which requires contact of a substance with your tongue, is limited to sensations of sweetness, sourness, saltiness, bitterness, and umami.  The chemistry of taste is certainly quite interesting, but anything beyond those "flavors" is produced by smell (via far more complex physiological processes).  You have certainly smelled rust (or, more accurately, various oxidation states of iron that exist in blood, likely from wet or weathered steel).  So you are, in fact, correctly associating the smell of iron oxides from two different sources: steel and hemoglobin.

There is evidence that metallic taste might be caused by galvanic reactions or even specific receptors in the oral cavity. But I am not sure if this has already been proven in some way.

@AstronAUT: Good point.  I should clarify that my answer does not mean metals have no taste in the formal sense.  (A common trick to try to distinguish taste from smell is to simply hold your nose closed while tasting a substance.  It's not perfect, because it doesn't seal the nasal cavity, but it certainly reduces circulation of the substance past olfactory receptors.)


text:
Nomenclature of compounds by the IUPAC
Are all compounds named according to International Union of Pure and Applied Chemistry (IUPAC) recommendations or is there another system?

Certainly, the most important organization for chemical nomenclature is the International Union of Pure and Applied Chemistry (IUPAC). In most cases, the various IUPAC recommendations represent the best references.
However, not all IUPAC recommendations are universally accepted. For example:

The correct internationally standardized name of element 111 is ‘röntgenium’; IUPAC recommendations spell it ‘roentgenium’, which is explicitly considered wrong by international standards.
According to current IUPAC recommendations, the formula of ions like sulfate is written as $\mathrm{SO_4{}^{2-}}$, which has not yet gained much acceptance. According to ISO and various national standards, the correct internationally standardized notation is $\mathrm{SO_4^{2-}}$.
In the 1970 IUPAC recommendations, a chelating ligand with two coordinating atoms was indicated by the term ‘bidentate’. In the 1990 IUPAC recommendations, the term ‘didentate’ was used rather than ‘bidentate’, for reasons of linguistic consistency. However, the term ‘bidentate’ remained in common usage, which is reflected by the reversion to the previously accepted term ‘bidentate’ in the 2005 IUPAC recommendations.
The @ notation for “fullerenes with metals inside” was proposed in J. Phys. Chem. 1991, 95 (20), 7564–7568 and became generally accepted. The corresponding IUPAC proposal, which was presented in Pure Appl. Chem. 1997, 69 (7), 1411–1434, did not use the @ symbol, but it was not as successful as the first one.

Many compounds can have two or more systematic names in accordance with several methods recommended by IUPAC. In addition, various traditional names, which have been coined and used over the years, are retained for use in IUPAC nomenclature. Therefore, current IUPAC recommendations include the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including the many retained names. Nevertheless, other names that are generated in accordance with IUPAC recommendations may be used in general nomenclature.
Apart from the large and universally active IUPAC, further smaller and specialized organizations exists, for example the International Union of Biochemistry and Molecular Biology (IUBMB). Such organizations may issue their own recommendations concerning their field of expertise. However, many of these recommendations are actually harmonized with IUPAC recommendations.
Another large and universally active organization that prescribes independent naming rules is the Chemical Abstracts Service (CAS). CAS names may differ from IUPAC names. For example, the preferred IUPAC name of Buckminsterfullerene is (C60-Ih)[5,6]fullerene, whereas the corresponding CAS name is [5,6]fullerene-C60-Ih.

If I am not wrong, the PIN concept was introduced in 2013 only. So, by "Many compounds can have two or more systematic names in accordance with several methods recommended by IUPAC. " do you mean to suggest that before 2013, many compounds had more than one acceptable IUPAC name? (because in my entire high school I was taught that, given how varied and complex the organic compounds can get, the IUPAC's role is to assign a unique, unambiguous name to each compound...)

The Chemical Abstracts Service (CAS) Registry is a proprietary system used to ensure that every compound has a unique (but not terribly descriptive) "name" (actually number).
IUPAC's International Chemical Identifier (InChI) is becoming a more popular alternative to CAS because it is not only unique to each compound but also non-proprietary and deterministic (i.e., it can be computed from the compound's structure instead of relying on an arbitrary name/number assignment).


text:
Is it possible to use Vmax as Kcat in a Michaelis-Menten reaction?
I am studying a biological system using ODEs. As part of my work I need to use the '$k_\text{cat}$' version of the Michaelis-Menten reaction, which is:
$$\frac{V_\text{max}\cdot [\ce{S}]}{(K_\mathrm m+[\ce{S}])} = \frac{k_\text{cat} \cdot [\ce{E}_\text{free}] \cdot [\ce{S}]}{(K_\mathrm m+[\ce{S}])}$$
I have found in a publication a value of a $V_\text{max}$ for one of my reactions:
$$V_\text{max}=\mathrm{\frac{11.4\;pmol}{min^{-1}{\;pmol^{-1}\;enzyme}} = \frac{0.19\;nmol}{s^{-1}\;nmol^{-1}\;enzyme}}$$
My question is, since this $V_\text{max}$ value is quoted in terms of 1 unit of enzyme (i.e. 0.19 nmol is the maximum one nmol of enzyme can catalyze in a second), can I use this $V_\text{max}$ value as a $k_\text{cat}$?
here is the abstract I found the information from :

Retinoic acid (RA) is a critical signaling molecule that performs
  multiple functions required to maintain cellular viability. It is also
  used in the treatment of some cancers. Enzymes in the CYP26 family are
  thought to be responsible for the elimination of RA, and CYP26A1
  appears to serve themost critical functions in this family. In spite
  of its importance, CYP26A1 has neither been heterologously expressed
  nor characterized kinetically. We expressed the rCYP26A1 in
  baculovirus-infected insect cells and purified the hexahistidine
  tagged protein to homogeneity. Heme incorporation was determined by
  carbon monoxide difference spectrum and a type 1 spectrum was observed
  with RA binding to CYP26A1.We found that RA is a tight binding ligand
  of CYP26A1 with low nM binding affinity. CYP26A1 oxidized RA
  efficiently (depletion Km
  9.4  3.3 nM and Vmax 11.3  4.3 pmoles min1 pmole P4501 ) when supplemented with P450 oxidoreductase and NADPH but was independent of
  cytochrome b5. 4-Hydroxy-RA (4-OH-RA) was the major metabolite
  produced by rCYP26A1 but two other primary products were also formed.
  4-OH-RA was further metabolized by CYP26A1 to more polar metabolites
  and this sequential metabolism of RA occurred in part without 4-OH-RA
  leaving the active site of CYP26A1. The high efficiency of CYP26A1 in
  eliminating both RA and its potentially active metabolites supports
  the major role of this enzyme in regulating RA clearance in vivo.
  These results provide a biochemical framework for CYP26A1 function and
  offer insight into the role of CYP26A1 as a drug target as well as in
  fetal development and cell cycle regulation.

$V_{max}=k_{cat}[E_0]$ where $[E_0]$ is the total enzyme concentration and not just the free enzyme. See my answer here.
You cannot use the value of $V_{max}$ as $k_{cat}$ because while the latter is the property of the enzyme (actually the enzyme-substrate pair), the former is the property of the chemical sample i.e. it is a system parameter.
Obviously, the $V_{max}$ will be same as $k_{cat}$ for a unit-concentration of total enzyme (unit concentration does not mean units of enzyme. It means, for whatever you are using as the units of concentration in your equation, lets say unit-x, the concentration of the enzyme should be 1 unit-x)

Hi WYSIWYG, would you know of a way to convert my Vmax value into a usable Kcat value then? Thanks

@user3059024 Can you provide a link to that paper? I need to know the context.

Sure, here it is http://ac.els-cdn.com/S0006295208007235/1-s2.0-S0006295208007235-main.pdf?_tid=5829d03c-58a0-11e5-bcb5-00000aab0f27&acdnat=1441988320_53246227d2b3ae3be51c6d4cb11485f4

Technically speaking you need to make a few more assumptions to use $V_{max}=k_{cat} [E_0]$.  If a single enzyme molecule has two active sites that both have equal activity, then you probably would want to use $V_{max} = 2 k_{cat} [E_0]$, for example.  And of course the standard restrictions about substrate concentration being in vast excess relative to enzyme would also apply...

@CurtF. Well yes.. you can always extend the model to include more complex kinetics.. The standard restrictions are the part of the MM model and actually it really doesn't explicitly state that. The model just assumes either equilibrium or QSSA which usually holds when the substrate is in excess.


text:
Why will internal energy decrease in this reaction?
1 mol zinc reacts with 1 mol sulfuric acid in a bomb calorimeter. I know since volume is constant work done will be zero .. but why will the internal energy decrease??
I know that the reaction is not endothermic or exothermic so there is no heat change involved, and as E = w + Q, the internal energy should also have been unaffected. But in an mcq book, the answer clearly states that the internal energy is decreasing.

Internal energy loosely can be seen as the measure of energy which are stored in the substance. Internal energy is equal to $$\Delta E = q + w ~~~~~~~~(where ~~w = -p\Delta V)$$
Since we know that volume is constant, w = 0. Also, now we need to determine the sign of q. Where you got stuck is saying that the reaction is neither endothermic or exothermic. Any reaction between a strong acid with a reactive metal will be exothermic. This means that heat energy is released from the system into the surroundings (water). That is why the water gets hot and forms the foundation of how the bomb calorimeter works.
Therefore, since we know that heat is transferring from the system to the surroundings, q must be negative. Therefore, $\Delta E$ will be negative, meaning that the internal energy decreased.

$\Delta E$ is the change in internal energy.

If the reaction is exothermic, and the system you choose includes both the insulated chamber and its mixture of reactants and products, then the change in internal energy of this system will be zero.  However, if the system you choose includes only the mixture of reactants and products, then heat will be transferred from the mixture to the chamber, and the internal energy of your mixture will decrease.


text:
Formulas for non-bonded interaction energies
If one were to calculate the non-bonded interaction energy between two atoms, this would equate to the sum of the vdW + electrostatic potential energies:
$$
E_{\text{non-bond}} = E_{\text{vdW}} + E_{\text{electrostatic}}
$$
Could anyone please explain how I could calculate these potentials providing a formula?
Perhaps given an example with zinc ($\ce{Zn^2+}$) and oxygen ($\ce{O}$), where $\sigma$ and $\varepsilon$ values (defined by Charmm27) are given as:
\begin{array}{ccc}
\hline
\text{Species} & \sigma & \varepsilon \\
\hline
\ce{Zn} & 1.942 & 0.25 \\
\ce{O}  & 3.029 & 0.12 \\
\hline
\end{array}
Ultimately I just wish to understand how they are calculated.
Simply put, I am trying to calculate non-bonded interaction energy between a ligand and a protein. However, after trying to understand each of these terms I am a little lost.

What are  sigma and epsilon?

You've got the Lennard-Jones potential parameters for zinc and oxygen from a certain version of the CHARMM force field. The commonly-used notation is $\varepsilon$ for potential well depth and $\sigma$ is the distance where the pair potential is zero, and $r_{min}$ is the distance which minimizes the L-J potential energy function. The linked Wikipedia article gives the potential as:
$$E_{\rm L-J}(r) = 4\varepsilon\left[\left({\sigma\over r}\right)^{12} - \left({\sigma\over r}\right)^{6}\right] = \varepsilon\left[\left({r_{min}\over r}\right)^{12} - 2\left({r_{min}\over r}\right)^{6}\right]
$$
At $r = r_{min}$, $E_{\rm L-J} = -\varepsilon$ and we can map $r_{min} = 2^{1/6}\sigma$ to work with either form of the L-J potential function.
Accurate Calculation of Hydration Free Energies using Pair-Specific Lennard-Jones Parameters in the CHARMM Drude Polarizable Force Field. J. Chem. Theory Comput. 6(4) 1181–1198 (2010) describes (among many other things) how the L-J potential term is calculated for 2 interacting species (as in your case). See equations (1), (2), and (3) in that paper, reproduced below (where I have used $r_{min}$ instead of $R_{min}$ for consistency).
In the paper, the Lennard-Jones potential is given as
$$E_{\rm L-J}(r) = \varepsilon\left[\left({r_{min}\over r}\right)^{12} - 2\left({r_{min}\over r}\right)^{6}\right]$$
For a pair of interacting species $i,j$:
$$r_{min} = {r_{min,i} + r_{min,j}\over 2}$$
and
$$\varepsilon = (\varepsilon_{i}\cdot\varepsilon_{j})^{1/2}$$
For your system, we compute $r_{min}$ from the given values of $\sigma$ (using the mean of the two values of $\sigma$, as specified in the manuscript):
$$r_{min} = 2^{1/6}\sigma = 2^{1/6}\cdot {1.942 + 3.029\over 2} = 2.789\;\unicode{xc5}$$
and
$$\varepsilon = (0.25\cdot 0.12)^{1/2} = 0.1732\;\mathrm{kcal\;mol^{-1}}$$
Now, at some distance $r$, you can compute $E_{\rm L-J}(r)$.
In the linked publication, the authors also go through the process of how the L-J parameters are derived/fit, which may be of interest to you.
Finally, see this vintage web page authored by a familiar guy which goes through a procedure for taking parameters for computed hydration free energies of cations and deriving L-J parameters for use in the AMBER force field. The procedure is very similar to that used for CHARMM.

Thank you for your detailed reply.  however, in the case where we have two differing atoms with separate $\sigma$ and $\epsilon$ values am I correct in saying the following: $\sigma_{ij}$ = $\sqrt{(\sigma_i\sigma_j)}$  where in the cited paper we have $\epsilon_{ij}$ = $\sqrt{(\epsilon_i\epsilon_j)}$ ?

You want $\sigma = R_{min}$ and $\epsilon = \varepsilon$ in their notation -  I edited my answer to include the math with your numbers, so it should hopefully be a little clearer.

Thank you so much, You have really helped me understand this. with regard to the -2 in the LJ 6 component, could you kindly explain what this means? It does not appear in other literature which I have read.

Also after further reading and for those who are also interested in this question: the $\sigma_{ij}$ value is dependant on the force field used for instance: CHARMM27 uses the Lorentz-Bertelot method for determining $\sigma$ i.e. $\sigma_{ij}=1/2(\sigma_{ij}+\sigma_{ij})$ while another OPLS uses a Geometrical average such that: $\sigma = (\sigma_{i}\sigma_{j})^{1/2}$

The -2 in the L-J expression is a remnant of whether or not $\sigma$ or $R_{min}$ is used in the analytical expression (see the linked Wikipedia page in my answer, above). That said, it's not clear to me that I mapped your $\sigma$ values to $R_{min}$ properly above - $R_{min} = 2^{1/6}\sigma$ - so let me edit that in above.


text:
Why does pouring soda slowly and smoothly seem to keep more of the carbonation?
So, I caught myself gently pouring a can of soda into a cup of ice (making sure that there was room for air to move through so the stream of soda from the can to the cup was consistent and even) instead of just haphazardly tilting and dumping the entire thing in there as quickly as possible. I was doing this in part to prevent spills at my desk, but also to keep the carbonation present in the soda. This got me wondering: why does pouring soda quickly cause it to fizz more and lose carbonation?

Because your soda is a supersaturated solution of $\ce{CO}_2$, and such solutions are non-equilibrium and hence unstable. Any physical disturbance (shaking, touching a rough surface, etc.) causes it to emit the gas and return to equilibrium state. Throw in some salt, and you will see a reaction even more vigorous.
Supersaturated solutions of solid compounds exhibit similar behavior, but they are not that easy to come by (I mean, starting from the typical household items). Sugar solution is too viscous, and table salt is no good because its solubility does not depends on temperature all that much, so you can't really supersaturate it.
There is, however, another pretty common example of non-equlibrium state, and that is superheated water. If you pour some bottled water into a clean cup, heat it in a microwave for a while, and then touch it not too gently, it may almost literally explode in your face, so beware.


text:
How to refer to this method of mixing solids?
The steps of this process are:

Fill a container with water (or other solvent)
Add mixants (is that the right word?) the container
Stir until all mixants are dissolved/suspended
Boil off the water (or other solvent)

The substances do not end up chemically reacting during this process. A real-world application of this is in small-scale manufacture of Rocket Candy, where this is used to mix KNO3 and Sucrose so the mixture can later be ignited and will burn more smoothly. Another example is making a cake - all the ingredients end up dissolved or suspended in the batter to mix them, and then the water is boiled off during baking. (Yes, there are also a lot of other things that happen during cake-making.)
What is the name of this process, if it has one? If not, what is a succinct way to describe it or state that this is what is used?

I think the closest term would be recrystalisation though this is normally used to dissolve one component (the desired one) to allow the removal of impurities, so I would be hesitant to use it but it is used here to describe the rocket candy example you used: http://www.jamesyawn.net/rcandy/index.htm
With regards to your second example, baking a cake is actually a series of interesting and complex chemical reactions and structural changes. The New Zealand institute of chemistry wrote a nice summary of the chemistry of baking that can be found here: http://nzic.org.nz/ChemProcesses/food/6D.pdf
Hope that helps


text:
Choice of base for malonic ester synthesis
Most reaction mechanisms use $\ce{EtO-}$ as a base to deprotonate the alpha hydrogen in the malonic ester sythesis. But in the equilibrium there are still esters that haven't been deprotonated. Why won't the enolate do a nucleophilic attack on the ester, just like in the Claisen reaction?
Why isnt LDA used instead, which would deprotonate all the ester molecules, preventing Claisen reactions?

I suspect steric hindrance plays a part. The enolate is a fairly bulky nucleophile and would be attacking a bulky ester group so I imagine the kinetic barrier is high. The wikipedia page for malonic ester synthesis indicates that diakylation is a problem so clearly this route is much more favourable than the one you are suggesting.

LDA should also work as expected

It's unnecessary.
In the reaction you are suggesting, the enolate would have to attack the carbonyl double bond from above. The optimum angle of attack is roughly $107^\circ$ but since both the attacking enolate and the diester substrate are reasonably bulky, there I suspect there will be a considerable amount of steric hindrance here, creating a large kinetic barrier to the reaction. If you compare this to attack by an alkyl halide then you can see that the latter reaction is strongly kinetically favoured.

Additionally, the problem may not be as bad as you think. According to Bordwell's $pK_\mathrm{a}$ table, diethylmalonate has a DMSO $pK_\mathrm{a}$ of 16.4. If you compare this to ethanol with a DMSO $pK_\mathrm{a}$ of 29.8, you can see that the malonate ester is actually quite acidic, and so ethoxide is easily a strong enough base to achieve facile deprotonation. Even in a protic solvent such as water (this reaction is probably done in ethanol but the effect is similar) which greatly reduces the ethoxide ion's basicity, Evans' $pK_\mathrm{a}$ table gives a value of 13 for dimethylmalonate and 15.5 for methanol (no data given for the ethyl variants) so the malonate ester is still a couple of orders of magnitude more acidic.

In addition:  using a base like a methanolate (or an ethanolate) yields after the deprotonation methanol (or ethanol) that may be distilled off easily as azeotrope, for example with toluene as a co-solvent, favouring the advancement of the reaction.

Understanding how this reaction is actually performed in the lab is key to answering your question.
Typically in the malonic ester synthesis, a full equivalent (or 2 equivalents if you are carrying out a dialkylation) of base is used, so little, if any, ester remains with the $\ce{\alpha}$-proton still in place.

(image source)
Also, keep in mind that both the alkylation and Claisen reaction have an activation energy. This means that it is usually necessary to heat the solution to bring about reaction (see here for a lab example).
The reaction is carried out by first forming the deprotonated malonate ester at room temperature and then dripping it into a refluxing solution of the alkylating agent ($\ce{RX}$ in the above diagram). This procedure achieves  the following:

concentration of non-deprotonated ester is very low
deprotonated ester is formed at room temperature, so even if any non-deprotonated ester remains there is not enough energy (at room temperature) to make it over the energy barrier and react
the deprotonated ester is dripped into the solution of refluxing alkylating agent; therefore, now that there is enough thermal energy to bring about reaction, the relative concentration of alkylating agent (the entire solution) is much higher than the relative concentration of deprotonated ester (added drop by drop) making the desired alkylation reaction much more probable than reaction of deprotonated ester with protonated ester (which, for reasons explained above, is in very low concentration in any case).


text:
Why should space whisky taste different to Earth whisky?
I recently read an article about a vial of whisky being stored on the International Space Station for 3 years, and having its taste compared to a control sample stored on Earth.
What could cause this difference in taste / chemical composition?
Theories so far
I'd assumed that the effects of (absence of) gravity would have altered the taste by affecting how the liquid's molecules interact with one another (i.e. being free to move in space; whilst being separated into layers on Earth), though I've been told that gravity doesn't affect diffusion within a liquid.
Other possibilities that myself & my father came up with:

Effect of UV if the vial's unprotected
Effect of oxidation (due to increased surface area of the liquid in the zero-g environment, if there's any air in the vial)
Effects of increased G during take off / reentry possibly causing increased interactions between the molecules (though this would be short lived).

That reminds me of Linje Aquavits, which are sent with ships from Norway to Australia and back again before being bottled.

Convection will certainly be different. And there might be effects from the higher radiation levels from solar radiation and cosmic rays experienced in space. Be interesting to see what measurements not just taste buds suggest.

@matt_black - See my amended answer, below, especially the part about Ardbeg scotch. The details of the experiment and the results are pretty neat (and are linked in my edited answer).

Well, cosmic radiation ... I say if it doesn't severely harm astronauts, it's very unlikely to harm wiskey. That being said, it's an old wisdom that spirits, stored at 10°C, in a dark cellar, significantly change in taste, and much for the better (well, smoother), in fifteen years. Chemists gut feeling says the processes that happen there are highly sensitive to variations in O2 partial pressure, temperature, maybe even radicals formed from cosmic rays. The idea that microgravity is responsible for differences is however, imo, ridiculous. Diffusion makes sure all samples stay homogeneous.

Except of course that the liquid in a bottle standing upright has no direct contact with the (metallic, polymeric?) lid. In space, there is no upright.

From the press release from Suntory, the Japanese company which sent 6 vials of their whisky into space, which details their study:

Tokyo, Japan, 30 July -- Suntory Global Innovation Center is about to embark upon space experiments on the “development of mellowness in alcoholic beverage through the use of a microgravity environment.” This research will be conducted in the International Space Station’s Japanese Experiment Module (nicknamed “Kibo”), with the cooperation of the Japan Aerospace Exploration Agency (JAXA).
  H-II Transfer Vehicle No. 5, commonly known as “Kounotori5” or HTV5, is scheduled to be launched from JAXA’s Tanegashima Space Center on August 16 (Sunday) carrying alcohol beverages produced by Suntory to the Japanese Experiment Module aboard the International Space Station, where experiments on the “development of mellowness” will be conducted for a period of about one year in Group 1 and for two or more years (undecided) in Group 2.
With the exception of some items like beer, alcoholic beverages are widely known to develop a mellow flavor when aged for a long time. Although researchers have taken a variety of scientific approaches to elucidating the underlying mechanism, we still do not have a full picture of how this occurs.
Our company has hypothesized that “the formation of high-dimensional molecular structure consisting of water, ethanol, and other ingredients in alcoholic beverages contributes to the development of mellowness,” and we have been conducting collaborative researches on this topic with research groups of Professor Shigenao Maruyama of the Institute of Fluid Science, Tohoku University and Professor Mitsuhiro Shibayama of the Institute for Solid State Physics, the University of Tokyo, the Japan Synchrotron Radiation Research Institute and Suntory Foundation for Life Sciences. The results of these collaborative researches have suggested the probability that mellowness develops by promoted formation of the high-dimensional molecular structure in the alcoholic beverage in environments where liquid convection is suppressed.
On the basis of these results, the space experiments will be conducted to verify the effect of the convection-free state created by a microgravity environment to the mellowing of alcoholic beverage.
The Suntory Group aims to use these experiments to help find a scientific explanation for the “mechanism that makes alcohol mellow.”

It appears that the working hypothesis is that the lack of gravity in space precludes liquid convection and thus alters the characteristics of liquids aged in that environment ${\it vs.}$ that on Earth. As you indicate, it's allegedly one way that promotes the formation of compounds which are otherwise not formed when gravity and convection within the ageing liquid are present.
From the same press release, details about the study:

Title of study
          Elucidating the Mechanism Mellowing Alcoholic Beverage
Experimental period
          Group 1: August 16, 2015 (Sunday) (planned) to September 2016 (planned)
  Group 2: August 16, 2015 (Sunday) (planned) to undecided date two or more years later
Content of experiment
          One set of samples consisting of various alcoholic beverages will be stored in a convection-free state in Japanese Experiment Module “Kibo” on the International Space Station, and another set of the identical samples will be stored in Japan for the same period of time. The following methods will be then used to analyze and compare the two sets of samples.

Measurement of substance diffusion coefficient with the use of a phase shifting interferometer$^{1}$ in cooperation with the Institute of Fluid Science, Tohoku University.
Detection of high-dimensional structure by small angle X-ray scattering$^{2}$ using SPring-8, in cooperation with the Japan Synchrotron Radiation Research Institute and the Institute for Solid State Physics, the University of Tokyo.
Measurement of substance diffusion with the use of the NMR method$^{3}$ in cooperation with Suntory Foundation for Life Sciences.

[1] Phase shifting interferometer: Uses the properties of light (visible light) waves to visualize temperature and concentration distributions, etc. that cannot be seen by the eyes, making it possible to detect slight changes in the ingredients of alcoholic beverages and determine how they are distributed.
[2] Small angle X-ray scattering: Can be used to obtain nanoscale (1 millionth of a meter) structure by irradiating a substance with X-rays and observing the scattered X-rays in an angular region of a few degrees or less.
[3] NMR method: Uses the phenomenon of nuclear magnetic resonance (NMR) to extract information from the organic compounds that are the object of measurement. Can be used to obtain information on the molecular structure of a compound and its dynamic properties.
Samples
          Five types of distilled spirits differing aging periods and 40% ethanol: Total of six samples

Suntory is not alone. Ardbeg scotch made a similar trip into space, and you can read about the experiment and results here. They used GC, HPLC, GC-MS, and organoleptic assessment to analyze their samples.
As stated in their summary:

The key aims of the experiment were as follows:-
(i) To determine if conditions of micro-gravity would have an impact on the range of naturally occurring terpene compounds and on the extraction
  of key flavour-active compounds in oak wood by Scotch malt distillate.
(ii) To assess the impact of maturation in conditions of micro-gravity on the flavour profile of oak matured distillate.
(iii) To investigate the presence of novel compounds and flavours in model maturation systems for Scotch malt distillate, as a result of conditions
  of micro-gravity.
The key findings of the above three aims would then be considered in conjunction with actual maturation conditions on Earth to investigate
  techniques for developing novel flavours in Ardbeg Islay Single Malt Scotch Whisky.

In the Results and Discussion section (I do encourage interested parties to jump to the link, and there are several tables and too much information to post here), they note:

GC analysis of major volatile congeners demonstrated no significant differences between the International Space Station samples and the control
  samples on Earth. Since this group of compounds comprising primarily alcohols, aldehydes, ketones and esters, are produced during fermentation
  and concentrated up, or in some cases, catalysed during distillation, their presence and quantities should have been identical in each set of samples
  to start off with. The conditions of micro-gravity on the ISS have not impacted on their levels, as demonstrated by the similar levels in the Earth
  based samples (Table 1.)
A more detailed investigation of these compounds, along with phenolic compounds, was carried out on our behalf by external whisky experts and
  scientists, using GC-MS. In the resulting chromatograms, the areas of 75 separate peaks were integrated, most of which were identified as alcohols,
  esters, acetals and phenols. Again, no major differences were found between the ISS and Earth samples. From an Ardbeg perspective, this was
  particularly interesting as the conditions on the ISS have not impacted on the absolute quantities of the various phenolic compounds. These are
  the compounds which contribute the powerful, smoky, medicinal flavours to Ardbeg spirit.
The most significant variable between the ISS and Earth samples was discovered when the results of the HPLC analysis of key maturation related
  congeners (wood extractives) were considered. The absolute concentration of these compounds was far higher than would normally be expected
  in standard, barrel-matured spirit, almost certainly as a result of the much higher surface area of woody material that the spirit was exposed to in
  the MixStix. However, when the two samples, ISS and Earth, were compared to each other, the lignin breakdown products - which are the major
  flavour active derivatives from oak wood - were present in higher concentrations in the control samples than in the space samples. This indicates
  that conditions of micro-gravity have inhibited their extraction from oak. On the other hand, ellagic acid and gallic acid, which are tannin breakdown
  products, both appeared to be higher in the ISS samples. The levels of these compounds are detailed in Table 2.
By drilling down further into these results, and subjecting them to some statistical analysis, the relative abundance of the main, easily extractable
  lignin breakdown products (vanillin, coniferaldehyde and sinapaldehyde) expressed as a percentage of the total wood extractives, is broadly
  similar in each sample. However, other extractives, such as the syringyl compounds, are very different in their total abundance, with much lower
  percentages in the ISS samples. This result means that while the overall extraction of wood-derived compounds is negatively impacted by micro-gravity,
  the impact is not equal; the compounds can be subdivided into ‘readily extractable’ and ‘less readily extractable’.
While it is clear that micro-gravity is inhibiting the release of wood extractives, not all compounds are affected equally, as would be expected
  in normal conditions. This has resulted in an unusual ratio between the readily extractable compounds and the less readily extractable ones in
  the ISS samples. This offers the intriguing possibility of using the ratios of such compounds as a potential marker to identify ‘unusual’ maturation
  characteristics, and thus a means of identifying for example, spurious age claims in whiskies. This possibility may be further investigated by the Scotch
  Whisky Research Institute.
For studies such as this, where we are examining the impact of differing conditions on either model systems, or on the full scale in actual barrels,
  the key question is always ‘What does this actually mean for the flavour?’, so the organoleptic analysis was of great interest to us. The results from
  the key volatile congeners, wood extractives and phenols analysis did not lead us to believe that from an organoleptic perspective, there would
  be large differences between the ISS and control samples. However, when we finally nosed and tasted the samples in the sensory laboratory,
  the results were unexpected.
Triangle tests gave a high number of correct responses, with almost all participants picking out a difference between the ISS and control samples.
  When we examined the actual aroma and flavour profiles, the differences between the two were as remarkable as they were unexpected; all tasters
  described the Earth control samples as being of the general Ardbeg whisky style, while the ISS samples displayed a quite different set of flavours.
  These have been summarised in the detailed descriptions below.

The above-mentioned MixStix are described in the write-up:

MixStix™ technology, which provides a convenient platform for small scale sample experimentation, is a trade-marked piece of technology used
  by NanoRacks. Essentially, they are small glass vials which can hold a range of materials; in our case these materials were 6ml of Ardbeg new spirit
  distillate, along with a small quantity of charred oak wood shavings (from a once used American oak Bourbon barrel), c. 1mm x 1mm x 6cm in
  size. The two materials are separated from each other by a glass dividing partition, which is broken by bending the Teflon outside covering of the
  MixStix™ at time of experiment activation, thus allowing the two separate components to mix together.

Hmm, woudn't trust this article "high-dimensional molecular structure" seems pseudoscientific

@Mithoron - I added the rest of the information from the press release. Apparently there's some real analytical chemistry happening here.


text:
Alcoholic KOH mechanism with alkyl halides
Does alcoholic $\ce{KOH}$ only show $E_2$ mechanism with alkyhalides?
I have studied with two organic chemistry teachers, one told me that there is always $E_2$ mechanism, except in the case of tertiary alkylhalide and the other told me it can be any depending on the substrate with $\alpha$ hydrogens more than 4 or if it has resonance it will show $E_1$.
Which one is true?

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Also, please put some more effort into your formatting and spelling. It doesn't have to be perfect but I'm sure you can do better than that.

This is not a homework ques.   I studied for it on net for an hour but couldn't find anything ..

A "homework question" is any question whose value lies in helping you understand the method by which the question can be solved, rather than getting the answer itself. This includes not just questions from actual homework assignments, but also self-study problems, puzzles, etc.

If u don't know about it then simply say it ..why are you abusing....

We are not being abusive in any way. We are enacting the policies of the site in a civil manner. And trust me, there are many people on this site who know the answer to this question and many more, which is why it will profit you to show some thought towards you question.

Then the policy of the site is not up to the mark it shouldn't be like this .......i tell you the proper confusion for this is that i have studied from two organic teachers one told me that there is always e2 mechanism except in the case of tertiary alkyl halide ....And the other told me it can be any depending on the substrate .......with alph hydrogens more than 4 or it is havingresonance  it will show e1..

If you don't like the policies of our site then feel free to go somewhere else. If you stay here though, you must play by our rules. If you edit your thoughts into the question and tell us what is confusing you then the question will be reopened and you will get an answer.

@Abhishek - I can see you have edited in your thoughts - I have voted for reopening. Having said that, it would have been done faster if you had followed the rules of the site, as per bon's advice, without the attitude.

Not sure by what your second teacher said. Can you clarify it by using better grammar or explaining in a more open way?

He told that if there is formation of stable carbocation which is in conjugation or having alpha hydrogen more than 4 than it will show E1 mechanism...

With a strong base $E2$ will always predominate over $E1$.
Both elimination mechanisms have a preference for tertiary substrates over secondary substrates over primary substrates. This behavior is on contrast with the substitution reactions, where steric factors are important for $S_N2$. Steric factors are less important for the proton transfer necessary for $E2$.
Since $E2$ and $E1$ have the same substrate preference order, you may feel that you cannot decide which is more likely. However, $E2$ is usually faster because $E2$ is bimolecular. Consider the rate laws:

$\mathrm{rate}_{E2}=k\mathrm{[substrate][base]}$
$\mathrm{rate}_{E1}=k\mathrm{[substrate]}$

If we make the assumption that the rate constants are of similar magnitude, then increasing the concentration of the base increases the rate of $E2$.

E2 does have a stereochemical requirement though. The leaving group must be antiperiplanar to the hydrogen.

Means it will mainly show E2 mechanism....