text:
Why is chloride oxidised instead of water in copper chloride electrolysis?
For the Electrolysis of Copper Chloride:
Cathode: $\ce{Cu^{2+} + 2e- <=> Cu}$
Anode: $\ce{2Cl <=> Cl2 + 2e- }$
I am confused about the reaction taking place at the anode. Wouldn't $\ce{H2O}$ rather undergo oxidation since it has an increased reducing ability? eg. It is above $\ce{2Cl-}$ on the Table of Standard Reduction Potentials.
Am I missing something? Or do all halogens discharge more readily than water?

The standard redox potential of chlorine is $E^\circ = +1.358\ \mathrm{V}$; the actual potential depends on the concentration of $\ce{Cl-}$:
$$\ce{Cl2 + 2e- <=> 2 Cl- }\quad\quad E^\circ = +1.358\ \mathrm{V}$$
$$\begin{aligned}
E&=E^\circ+\frac{RT}{2F}\cdot\ln\frac{1}{\left[\ce{Cl-}\right]^2}\\
&=1.358\ \mathrm{V}-0.05916\ \mathrm{V}\times\log{\left[\ce{Cl-}\right]}
\end{aligned}$$
The standard redox potential of oxygen is $E^\circ = +1.229\ \mathrm{V}$; the actual potential depends on $\mathrm{pH}$:
$$\ce{O2 + 4H+ + 4e- <=> 2H2O}\quad\quad E^\circ = +1.229\ \mathrm{V}$$
$$\begin{aligned}
E&=E^\circ+\frac{RT}{4F}\cdot\ln\left[\ce{H+}\right]^4\\
&= 1.229\ \mathrm{V}-0.05916\ \mathrm{V}\times\mathrm{pH}
\end{aligned}$$
The redox potentials $(E_{\ce{O2}} < E_{\ce{Cl2}})$ suggest that oxidation of $\ce{H2O}$ to $\ce{O2}$ at the anode should be preferred over oxidation of $\ce{Cl-}$ to $\ce{Cl2}$.
However, depending on the material and shape of the anode, the overpotential of oxygen at the anode can be very large. Therefore, the oxidation of $\ce{Cl-}$ to $\ce{Cl2}$ at the anode is feasible.
Nevertheless, at high $\mathrm{pH}$ and at a low concentration of $\ce{Cl-}$, the oxidation of $\ce{H2O}$ to $\ce{O2}$ is still preferred over oxidation of $\ce{Cl-}$ to $\ce{Cl2}$. Hence, if also $\ce{H2}$ and $\ce{OH-}$ are produced during the electrolysis of aqueous solutions (this is less relevant for the reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ which is considered in the question, but for example important for the chloralkali process)
$$\ce{2H2O + 2e- -> H2 + 2OH-}$$
the electrolysis of chloride cannot be made complete in a single step since $\mathrm{pH}$ is increased and the concentration of $\ce{Cl-}$ is decreased during the electrolysis:
$$\ce{2 H2O + 2 Cl- -> H2 + 2 OH- + Cl2}$$

It depends upon the concentration of chloride.  If a dilute NaCl solution is electrolysed, $\ce{O2}$ is produced at first, along with $\ce{H2}$. As the volume of the solution decreases $\ce{[NaCl]}$ increases, and eventually $\ce{Cl2}$ is formed. You can use the Nernst equation to find out the chloride concentration at which $\ce{Cl2}$ is formed over $\ce{O2}$.

Another contribution is overpotential

Yes water has a higher reduction potential. But what is happening at your anode isn't reduction, it's oxidation. So a lower reduction potential for $\ce{Cl-}$means that it's more readily oxidised than water.
For additional information, see my other answer on this recent question on electrolysis and $\ce{Cl2}$ production.

Really? I thought water is more likely to undergo oxidation than Cl−. My chem teacher cleared me up and said that the halogens will rather oxidise than water (Cl- is an exception in the table due to its position on the Standard Reduction Potentials Table)

Water does oxidise some of the time due to its abundance in the cell (it's completely surrounding the electrode, right?) however Cl- preferentially oxidises over water due to having a lower reduction potential (therefore a tendency to oxidise more than water). It's a thermodynamic argument: Cl- oxidation is preferred on thermodynamic grounds, however due to rapid kinetics water will also oxidise.

As for "My chem teacher cleared me up and said that the halogens will rather oxidise than water" you need to be careful what you're stating, by halogen do you mean anion of halogen, or diatomic halogens etc.? If you're referring to Cl-, there are no exceptions here. Cl- prefers to oxidise over water.


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How can I measure carcinogenic risks of an M-ENDO-Agar LES substrate after heat destruction?
I'm about to conduct a study of possible toxic traces from the substance M-ENDO-Agar LES in substrate after destruction with heat (I think it is 160℃) for 180 minutes. The substance is carcinogenic.
I want to find a way to measure if the substrate is still toxic in any way after the heat treatment. How could I conduct such a study?

Most assays of carcinogenicity are designed to measure DNA damage. The 2 most common of those assays are the Ames test and the Comet Assay.
The Ames test uses a special set of strains of bacteria that have known phenotypes. You expose the bacteria to the suspected carcinogen and then look for changes in phenotype. Different strains will be susceptible to different types of mutations, deletions, insertions, frameshifts, etc.
The comet assay uses mammalian cell lines. You have to suspend the cells in an agar and break them up. This will create little pockets of cell debris in the agar. You fluorescently stain the DNA and run an electric current through it. If the DNA is damaged and broken up, it will migrate through the gel, creating tails, hence the "comet". If the DNA is intact, it will not migrate.
I've never done either assay, just heard about them in class. Wikipedia has more detail than I do.
http://en.wikipedia.org/wiki/Ames_test
http://en.wikipedia.org/wiki/Comet_assay

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Nitration of Aromatic Compounds
I am told by Dr. Klein that in the nitration of benzene, hydrogen sulfate anion acts as the base and deprotonates the sigma complex.

My question is why would the hydrogen sulfate anion be doing the deprotonation? We've just established that sulfuric acid is a stronger acid than nitric acid; hence sulfuric acid is able to protonate nitric acid to some extent. Now are we expected to believe that the conjugate base of a stronger acid is stronger than the conjugate base of a weaker acid? Why not have nitrate anion do the deprotonation? It is a stronger base than hydrogen sulfate anion.

There's usually a lot of water present with these acids, why couldn't water remove the proton?

Would water removing the proton be the kinetically favored product? And thermodynamically favored product now that I think of it ... water is the strongest base of all three options.

Sorry, I don't understand "be the kinetically favored product."

As in there's more water than either hydrogen sulfate ion and nitrate ion, so water abstracting a proton is more likely than either of the other two doing the same.

Yes, removal of the proton by water would be kinetically and thermodynamically favored.

It's not advisable to consider these types of reaction schemes in a slavishly literal fashion. In reality, there are multiple proton transfer equilibria occurring simultaneously, and the reactions will generally be quite rapid. While it's logical and technically correct to depict the most basic of the molecules (among those present in sufficiently large concentration) performing the proton abstraction, it's also of literal practical interest or importance in many cases. I suspect the reason that $\ce{HSO4-}$ was chosen is to emphasize that the sulfuric acid is actually catalytic, so it should be regenerated in the course of the depicted reaction mechanism. Of course, water is produced as a byproduct, so the sulfuric acid will not exist primarily in its neutral molecular form.

"It's not advisable to consider these types of reaction schemes in a slavishly literal fashion." It (HSO4- doing the deprotonation) was stated in the author's commentary, not implied through a mechanism.

Also what do you mean water is produced as a byproduct?

@Dissenter, well, the preliminary reaction is protonation of nitric acid to give $\ce{H2NO3+}$, which dehydrates to give $\ce{NO2+ + H2O}$.

Oh okay. Is it true that pure sulfuric acid is rarely used as a reagent? I remember Martin saying something to that effect. Something about sulfuric acid dehydrating organic material.

@Dissenter, I've used sulfuric acid at very high (around $95%$, if I remember correctly) concentration, as well as fuming sulfuric acid, but in general it's true that it's an extremely aggressive dehydrating agent and relatively hazardous in those forms. Furthermore, w/r/t this reaction, my understanding is that anhydrous nitric acid is very difficult to prepare and doesn't have a great shelf-life due to its tendency to decompose. I've never used anything more concentrated than the standard azeotropic $68%$.

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text:
Amount of Phenolphthalein required
We just had chemistry practical few days a go. The topic was strong acid-base titration, in which we had to find the normality and concentration of $\ce{HCl}$.
My teacher instructed me to add 2 drops of Phenolphthalein  in 10 ml $\ce{HCl}$, and then add $\ce{NaOH}$ from buret and take a reading.
We needed to repeat the process 4 times and then find the constant reading, but in that I never used to take 2 drops of Phenolphthalein, instead I used to take 5-6 drops and then also used to get almost same reading as of pilot reading.
So my question is whether it is compulsory to take 2 drops of Phenolphthalein or it was just a instruction so students don't waste it?
In this same instruction given. (Procedure point 2)

Using a little extra phenolphthalein won't matter. You just need enough so that the color is visible once the pH becomes basic enough. If you were titrating a large volume, you might need more than a couple of drops, but 2 drops is enough to be sure that the color will be visible in a typical lab-scale titration vessel. One drop would probably even work. The reason they say "two drops" and not "at least two drops" is that they don't want you to be wasteful.

Actually, using the whole bottle should give a significant difference. But nobody would be that naive.

I slightly disagree. There are several reasons, why one should not use too much indicator. Indicators are usually acids and bases and the measurement will be affected when dealing with significantly diluted systems. Also for a colour change you need approximately a 1:10 ratio of protonated vs unprotonated indicator. Using too much indicator will result in adding too much acid and base. The human eye is not the most reliable measuring instrument so the difference in tiny quantities should not matter too much. Accurate titrations can be done by conductometric measurements.

So we can conclude from answer and comments that I should not use more than 2 drops

@LDC3 - never underestimate the capacity of people to do stupid things!

@Martin - yes, this is true. I was assuming we were talking about a range of 1-5 drops in maybe 30-100 mL of solution, which is typical for undergraduate labs.

@Freddy - yes, stick with 2. But if you accidentally use 3, it shouldn't affect your results by  measurable amount.

@thomij i Have tried with 5 drops but no measurable change :)

Phenolphtalein is slightly acidic. It does affect titration because if you add too much, then it will alter the concentration of whichever solution you are adding it in. If you added it to the acid, it will cause that acid to be more acidic than you may have calculated. The effect of that on the titration is that afterwards when you are adding the base in the acid to find the end-point is that the end point will not be the true end-point since the pH of the acid is lower because of the indicator. The amount of base needed to add to the acid will be more so it can neutralize it and this while ruin your entire experiment since it will not be accurate. It will not have a large difference but it's still different.

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text:
Sulfur Trioxide - Ionic Character
I am told that the sulfur trioxide molecule exhibits charge separation because of poor p-orbital overlap. Sulfur's 3p orbitals are much bigger than oxygen's 3p orbitals, and thus the $\ce{S=O}$ bond is best described as $\ce{S+ - O-}$.

So is the author's argument suggesting that sulfur trioxide is unhybridized, and that it uses pure s orbitals to form sigma bonds and pure p orbitals to form pi bonds? Because that's the only way I can see sulfur as having three empty p orbitals to utilize. If we were to assign hybridization through counting the number of sigma pairs of electrons (3) we would arrive at sp2 hybridization, and there aren't enough p-orbitals to go around.
Also while we're on this topic, Wikipedia says that sulfur trioxide is best represented as having one $\ce{S=O}$ bond and two $\ce{S+ - O-}$ bonds. Now, does bond length differ between an  $\ce{S+ - O-}$ and an  $\ce{S=O}$ bond? Wikipedia suggests not, as only one bond length is given. If bond length is equivalent between the two, why?

Sulfur trioxide is planar with 120 degree O-S-O bond angles.  If the molecule were unhybridized and we used the 3 orthogonal p-orbitals for our sigma bonds, then we would expect the O-S-O angle to be 90 degrees.  Further, in the unhybridized scheme, there are not enough s orbitals to form 3 sigma bonds that are pure s. Just considering s and p orbital involvement, we could say that the sulfur is $\ce{sp^2}$ hybridized.  This would produce 3 $\ce{sp^2}$ orbitals for sigma bonds from sulfur to oxygen and leave one p orbital for one pi bond.  We can draw 3 resonance structures for this arrangement, so we would expect all bond lengths to be equivalent.

Right, but the author almost explicitly states that three p-orbitals are used. So where can the s-orbitals for the sigma bonds come from? The author's argument struck me as off.

I don't read it that way.  Perhaps the author is saying that there are 3 S=O double bonds, but they are not such great double bonds - in fact each one is really about 1/3 of a double bond (as the resonance structures suggest). There is no way to create a planar molecule with 120 degree angles using 3 p orbitals for each of 3 S=O pi bonds.

@Dissenter, ron: For a full orbital scheme of sulfur trioxide, see this answer of me. The bond order of about 1.4 is consistent with the resonance picture. If you do not want to employ the concept of hybridisation here, you could describe the MO with symmetry adapted ligand group orbitals. This of course would not change a thing, since these descriptions are mathematically equal.


text:
How does sulfuric acid dehydrate sugars?
I know sugars have $\ce{-OH}$ groups, and that sulfuric acid is a strong acid. What I'm failing to see is how exactly sulfuric acid dehydrates sugars. Does it protonate the hydroxyl groups, making them good leaving groups? What about the case of sucrose?

For example, how are the oxygens "taken care of" by the sulfuric acid? Are they doubly protonated? Is there some established mechanism for the dehydration of sugars by sulfuric acid, because I can't find any - I just find general equations and I've looked on Google and through textbooks.
After some further research I learned that the gas, sulfur dioxide, is evolved. This suggests to me that this reaction is also redox in nature. So is this reaction both an acid-base reaction and a redox reaction? Sulfur in sulfuric acid has an oxidation state of +6; sulfur in sulfur dioxide has an oxidation state of +4. Plus the elemental carbon formed has an oxidation state of 0 while in sucrose, carbon has various oxidation states.

I doubt sugar dehydrates into pure carbon. Afaik both glukose and fructose can be relatively easy dehydrated into furan derivatives, that are known for their ability to polymerize under sever conditions and condense with pretty much any carbonyl compound, including furfural derivatives, else produced in this conditions. So, in sulfuric acid the product is a complex resin with significant ( 10+%) of non-carbon atoms. Most I wrote is pretty easily googlable, so feel free to educate yourself.

When I googled I found several reaction equations implying that "C" was formed, or pure carbon.

Pure carbon is actually quite hard to obtain. Deep carbonization usually occurs around 700-1500 Celsium and higher . Charcoal, usually assumed being pure carbon actually may contains up to ~10%  non-carbon atoms, and given that sugar-sulfuric acid reaction occurs at temperatures up to 200-300 C I can safely conclude the product contains a good measure of non-carbon atoms.

@Dissenter, the acid catalyzed dehydration of alcohols is usually and E1 reaction. This paper is one of the top Google results for "dehydration of glucose" and shows 5-hydroxymethylfurfural as one of the major products, among others: http://www.sciencedirect.com/science/article/pii/S0896844606002117

Nice catch @jerepierre; corrected in OP.

@permeakra Here is how carbon is made from sugar. https://www.youtube.com/watch?v=nqDHwd9rG0s

@Dissenter How much dehydration do you wish? All the way to elemental carbon or just the removal of the first hydroxyl groups?

All the way to elemental carbon.

Some sites claim that some of the produced carbon is entirely oxidized to CO2.  I'm doubtful;  I suspect the aldehyde moiety of reducing sugars or of the resultant furfurals is oxidized to an alcohol, which if further dehydrated would result in C-H bonds in the elemental carbon.  In accordance with this, the C:H ratio in sugar carbonized with H2SO4 was found to be ~14:1 mol:mol in this paper.

$\ce{H2SO4}$ protonates the alcohol to make $\ce{R-OH2+}$ and $\ce{HSO4-}$, the $\ce{HSO4-}$ then takes a hydrogen from the carbon atom adjacent to the alcohol, yielding an alkene and water.  Alcohol dehydration reactions using acid are just elimination reactions.
Additionally, at temperatures above 340 degrees Celsius, sulfur trioxide exists in equilibrium with sulfuric acid in significant concentration.  The presence of sulfur trioxide explains the presence of sulfur dioxide  (via dissociation), or some redox chemistry occurring with the sugar.

They are E1 reactions. Not E2. There is no base.

HSO4- is the base.

Not nearly strong enough for E2.

Depends on substrate, works for primary.

In strong mineral acid conditions, sucrose, splitted to glucose/fructose, undergoes dehydration (elimination of water by OH groups protonation), mainly to 5-(hydroxymethyl)furan-2-carbaldehyde (or HMF, 5-hydroxymethylfurfural).[1]

(Stereochemistry omitted. Glucose is more stable than fructose, is reported to follow the path by isomerization to fructose (via acyclic endiol) in this conditions, or a via another mechanism proposed:[2] C-2 (in glucose numbering, i.e. C-3 in pyrane numbering) with protonated OH group is attacked by pyrane oxygen, forming intermediate oxiranium ring…)
(Pentose carbohydrates, like xylose, form furfural itself.)
HMF then polymerizes (possibly together with other byproducts like levulinic acid) to form rather insoluble humin, whose fragment may look like:[3]

I don't know if it's possible for such structure to be further dehydrated to carbon, under non-pyrolytic conditions.
References:

Feather, M. S.; Harris, J. F. Dehydration Reactions of Carbohydrates. Advances in Carbohydrate Chemistry and Biochemistry 1973, 28, 161–224.
Dee, S. J.; Bell, A. T. A Study of the Acid-Catalyzed Hydrolysis of Cellulose Dissolved in Ionic Liquids and the Factors Influencing the Dehydration of Glucose and the Formation of Humins. ChemSusChem 2011, 4 (8), 1166–1173.
van Zandvoort, I.; Wang, Y.; Rasrendra, C. B.; van Eck, E. R. H.; Bruijnincx, P. C. A.; Heeres, H. J.; Weckhuysen, B. M. Formation, Molecular Structure, and Morphology of Humins in Biomass Conversion: Influence of Feedstock and Processing Conditions. ChemSusChem 2013, 6 (9), 1745–1758.

I think sulfuric acid is hygroscopic. When you put sugar in it, it will take out water of crystallization from it and if not, most probably it will dehydrate it by taking $\ce{-OH}$ groups from it.

The sulfuric acid brakes up the bonds of the sugar and absorbs the water from the crystallization

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text:
Biological Consequences of Asteroid Mining—Death by Isotope?
It's been documented that NASA hope to capture an asteroid in 2025, and have subsequent aims to mine that asteroid. If if this is successful, we would expect other asteroids to be mined in the future.
A consequence of this is that relative atomic masses of elements mined—those with two or more stable isotopes—will no longer be faithful to our current periodic table. Ruthenium alone, a high-value rare earth, has five stable isotopes alone, ranging from $^{98}\ce{Ru}$ to $^{102}\ce{Ru}$. The relative abundances of these isotopes are bound to differ in other places other than Earth, and so would the weighted average.
Say after mining our ruthenium, we use it in an electronic device and consequently that device is thrown away. For argument's sake, also say that this ruthenium—of an isotopic distribution never before seen—leaches into the environment and later comes into contact with biological systems. Could it be possible that compounds hitherto considered non-toxic on Earth become toxic to biological systems by virtue of a newly realised isotopic discrimination?
As an example, $^{13}\ce{CO2}$  is effectively discriminated against in uptake by plants in comparison to $^{12}\ce{CO2}$  (due to it being a diffusion limited reaction). Ruthenium is purported to be carcinogenic, but the most abundant isotope on Earth of Ruthenium is $^{102}\ce{Ru}$ which could be effectively non-toxic by virtue of the fact that  compounds containing these atoms takes so long to diffuse, they don't diffuse across biological membranes at all.
If a sample of Ruthenium from the moon found its most abundant isotope in $^{98}\ce{Ru}$, then this would be expected to diffuse much faster than its heavier counterparts. Therefore could it be possible that a compound previously thought of as non-toxic could become toxic via this effect. Does this sound plausible?
NB: I've deliberately missed out other routes of exposure such as ingestion to only consider diffusion controlled reactions.

Can you provide a source for plant's not using C13? Because I know they use at least some C14, that's how radiocarbon dating works. And by your argument, C14 should diffuse even more slowly than C13.

I didn't say that they don't use it, but that there is an active discrimination. And yes C14 will diffuse even more slowly. Radiocarbon dating relies on absolutely minute quantities of C14. As for evidence, the δ13C isotope signature is well known in biogeochemistry for example.

Also, to allude to C3 and C4 plants, carbon dioxide uptake is different in the different species. In C4 it occurs via the Hatch-Slack pathway which is less diffusion controlled than C3's Calvin-Benson.

I like the question, but my guess is that every answer would boil down to "we don't know, here are some proposed (or: performed) tests about it [refs 1, 2, 15 and 42]" ...

@tschoppi, I realised it's more of a discussion than a question with a definite answer, although those references might be what I'm looking for!

The largest biological consequences of isotopes are found, of course, with hydrogen.  Both deuterium and tritium are toxic.  Since many biological processes rely on resonance effects, too much deuterium messes up metabolism.  Tritium, while having the same effect as deuterium, adds the beta decay radiation dose.

This is an interesting question and you raise a number of points, let's step through them.

A consequence of this is that relative atomic masses of elements
  mined—those with two or more stable isotopes—will no longer be
  faithful to our current periodic table.

But this is already happening.  $\ce{^235U}$ constitutes 0.72% of uranium found on earth and decays to the stable isotope $\ce{^207Pb}$, which is found in a natural abundance of 22.1%.  Before we get to the asteroid the abundance of various isotopes and their weighted average mass is already shifting.

98 Ru , then this would be expected to diffuse much faster than its
  heavier counterparts

Diffuse much faster?  First off, we're talking about diffusion, not chemical reaction.  Elements or compounds usually enter the body by ingestion of one sort or another (eating, breathing) rather than diffusion, and these ingestion pathways wouldn't involve isotopic discrimination.  That is, if we breath air that contains isotopes in a certain ratio, then that's the ratio that will initially appear in our lungs.  Nonetheless if diffusion were found to be an issue, my guess is that isotopic discrimination by diffusion would be a very small effect.  For a gas the maximum separation of two isotopes is given by
$$\mathrm{\sqrt{\frac{[MW~of~compound ~with ~isotope~1]}{[MW~of~compound ~with ~isotope~2]}}}$$
In the case of separating $\ce{^235UF6/^238UF6}$ this amounts to a fractionation ratio of 1.0043 after 1 pass.  Ruthenium is lighter, so the effect would be larger, about 1.02 ($\ce{^102Ru/^98Ru}$) after 1 pass, even less if it is in a molecule.
Chemical reaction within our body of ingested isotopic compounds would also show  discrimination due to primary kinetic isotope effects.  The maximum primary kinetic isotope effect is proportional to the reduced mass as follows:
$$k~ \thicksim ~\sqrt{\frac{m_1 + m_2}{m_1 m_2}}$$
Applying this to $\ce{^102Ru}$ and $\ce{^98Ru}$ yields a primary kinetic isotope effect of 1.020, again, even less if the element is incorporated into a compound.
To me, the effects look small and since shifts in natural abundance have been occurring for a long time here on earth, with no one raising a flag, my guess (and that's all it is, a guess) is that it's not something to worry about.  Still, I'd feel better if a NASA toxicologist looked at the question and said "no problem"

For the second point I realised that talking about diffusion relating to human health is probably a bit daft, although there's no reason why it couldn't happen in plants such that the effect is passed down a food chain. If we apply your formula for gaseous separation to carbon dioxide as mentioned, then this amounts to a fractionation ratio of 1.011, which is less than atomic ruthenium!

If gas centrifuges can sort out U238 and U235, which have a tiny percent difference, could they be used to remove CO2 from air? CO2 is about 40g/mol, and O2 seems like the next heaviest at 32g/mol, so the percent difference is larger and separation should be more efficient. But how much energy would it take to run the centrifuges?

@user137 Interesting thought.  Sounds like it should work, I know that mean free path and membrane porosity are also design factors, and as you point out, energy considerations.

According to https://www.ncbi.nlm.nih.gov/pmc/articles/PMC396263/, the discrimination against uptake of $\ce{^{14}CO_2(g)}$ relative to $\ce{^{12}CO_2(g)}$  by plants during photosynthesis is due to diffusion.  They conclude this because  the strength of the effect is 2%, which corresponds nicely to the Graham's diffusion equation (which you showed in your post:  $\sqrt{\frac{44}{46}}=0.98$).

Whatever the isotopes are for asteroidal material (and they are mostly close to those seen down here on Earth), they are contained in the 5 to 100 tons of meteoritic material that falls onto the Earth's atmosphere (and thus filters down to us on the surface) every day. It will be a long time before the cumulative pollution from asteroid (or lunar) mining can match the meteoritic rate.

Also, if you follow what @ron said, for large isotope effect you generally need lighter elements. So the isotope effect for deuterium and hydrogen can be large. But even in these cases biological effects can be mostly observed only when one use high concentrations of D2O.
I don't think that diffusion control would play any role in the biology of metal ions. If we are talking about transition metals with biological role, nature is generally relies on transport enzymes and similar to carry around them. Isotope effects are most pronounced in chemical reactions with relatively high activation barrier and strong vibrational coupling with the given atom. In practice it means enzymatic reactions with active centers where there is a significant role of the given metal.
Do we even know any enzymes in which's active center Ru plays important role? You can ask this for any elements that we expect to mine, In practice, isotope effects hardly can seen even for iron or cobalt atoms, which are pretty common active centers compared to most other metals. If you study natural iron enzymes, the isotopic ratio of 56/57 iron is not anomalous. So I wouldn't expect much biological effect to anything that heavy or heavier.
If we are going beyond biological function, and talking about metals and toxicity, most of the time the complex formation between e.g. proteins and the given metal ions is the important question to study. I don't think is it much influenced by isotope effects, and I definitely would not expect that a toxic heavy metal became harmless or similar drastic change.


text:
Using non-food grade NaOH for pH control
I would like to use non-food grade $\ce{NaOH}$ for $\ce{pH}$ water treatment. Any problem with this mixing with water that contains iron?
The $\ce{NaOH}$ will be pumped into a well. As a $25\%$ solution. Then sent through an ozone filter and a water softener.  Eventually, it will be used for pet and human consumption.

In addition to what thomij has already pointed out, note that $\ce{NaOH}$ etches glass and will degrade various polymers (PET and polycarbonate, among others).

What about passing the water through calcium carbonate or magnesium hydroxide to raise pH?

Non-food grade sodium-hydroxide can (and most often do) contains heavy metals. Also, it is pretty corrosive, so you may yourself generate different carcinogens or poisonous side-products inside the well during this process.  If you want to use this for human / pet consumption, you may want to consider other solutions, e.g. @brinnb's suggestion.

Iron hydroxide is insoluble, so it will precipitate out. If there are other metals that aren't  alkali metals or alkaline earth metals heavier than calcium, they will also precipitate out. Most of these reactions will release some heat, but if the concentration of the metal ions is low, that shouldn't matter too much.
If you plan to drink the water, using non-food grade $\ce{NaOH}$ might cause a problem, because you won't know what else is in there.
If you add more details, we can probably tell if there are any other things to be concerned about.

In addition to the other answers, I'd like to add a comment about the risks:
A 25% solution of NaOH in large quantities is very very dangerous. On drop in the eye and you will be permanently blind.
It is much worse than acid. Acid causes something akin to a burn that keeps the acid away from the unburned tissue below. NaOH on the other hand dissolves the tissue and penetrates very deeply. The drop in your eye will liquefy your eye.
When you pump it you'll have a lot of pressure, the solution might corrode the plastic and suddenly you have a leak that sprays corrosive lye all over you.
When you mix NaOH with water, it will heat up. Maybe enough to make it boil explosively. Hot NaOH solution is even more corrosive.
I don't know much about well treatment, if you really need to use NaOH, use at least both a full face shield and safety goggles + raingear and have someone stand by with a water hose to wash of any spills quickly. Keep kids and pets away.
I suggest you check if using a carbonate or hydrogen carbonate (e.g. sodium carbonate, sodium hydrogen carbonate) would work. These are much safer.
For the non-food grade, I would be careful. Some industrial stuff is just as clean, some is heavily contaminated. Maybe the people selling it have an analysis of the batch. In any case, make sure you rinse your system very well afterwards.


text:
Detection of magnesium in a buffer?
I have a PolyA tailing kit for putting PolyA tails on the 3' end of RNA. The kit comes with a proprietary 5X buffer and a 25mM MnCl2 solution, and in the standard set up I use 20uL of 5X buffer and 10uL of 25mM MnCl2. When trying to force the enzyme to accept an alternative substrate I did the reaction without the MnCl2 and it worked. Further research revealed that the enzyme also works with Mg++ ions, even though the rate is faster with Mn++, and some old papers on the enzyme mention using both Mg and Mn in the buffer. This leads me to believe that the buffer contains Mg++, but I don't know for sure because the manufacturer won't tell me.
I immediately thought I could use atomic absorption spectroscopy, but our department does not have one. I looked up colorimetric assays online, but the kits cost about $300, which is more than I want to spend on this idea. What I'm asking is if anyone knows a relatively simple method to detect magnesium in a solution? I have access to a UV-Vis spectrophotometer and fluorimeter.
EDIT: After digging for other phone numbers for the manufacturer I found someone who would tell me a little about the buffer, and it does contain magnesium. He couldn't me how much though. Spectro-Phone-otemotry.

Magnesium 2+ forms an insoluble precipitate with hydroxide ion. Perhaps add a source of hydroxide ion? Mg2+ also is detectable by flame test.

Just tried a flame test to see what would happen, it gave me bright orange. The buffer probably contains sodium chloride. Old literature recipes I've seen use anywhere from 100 to 300mM NaCl, so a 5X buffer could have 500 - 1500mM NaCl and overwhelm the MgCl2. I tried a flame test on the MnCl2 solution as well, hoping to see yellowish green, but only saw a few orange sparks, so maybe my ethanol burner isn't good enough for flame testing.

Even trace amounts of Na+ can make the flame go yellow/orange, rendering the test rather useless.

Manganese hydroxide is also insoluble. I don't know if you can distinguish it from magnesium hydroxide.

Thankfully the manganese is kept in a separate vial, so I can choose to add it or not.

$\ce{Mg^2+}$ forms complexes with 4-(2-Pyridylazo)resorcinol that can be detected spectrophotometrically: Simultaneous spectrophotometric determination of calcium and magnesium in water, described by E. Gómez, J.M. Estela, V. Cerdà in Analytica Chim. Acta 1991, 249, 513-518.
(DOI: 10.1016/S0003-2670(00)83027-5)

Will order some from sigma and give it a try. Looks like I can get 1g for about $25. Don't know how soon I can get back with an answer.

There is also the venerable complexometric determination of magnesium with EDTA.  Determining hardness of tap water is a popular lab exercise, your chemistry department's teaching lab might even have EDTA solution and complexometric indicator in stock.  It's worth a shot.

Can you get an indicator that is specific for Mg but not Mn?

Most probably you will have other metal ions in the solution, eg Ca++, which can make this measurement useless.


text:
Rate-determining step and steady state approximation failure
1) When can't the rate-determining step be applied, and why? How do you recognize these cases?
2) When can't the steady state approximation be applied, and why?

When can't the rate-determining step be applied, and why? How to recognize those mechanisms?

The rate determining step (often abbreviated as "rds") of a reaction is the slowest step in a reaction.  Every reaction has a "slowest step", therefore, every reaction has an rds.  The slowest step will be the step where we pass over a transition state with the highest energy, e.g. the highest energy point along the reaction coordinate. Look at the figure below, the reaction coordinate shown in the top left diagram

illustrates a reaction with a transition state, but no intermediate.  There is only one step in this reaction, passing over that transition state (marked by the double-headed arrow) is the rds for that reaction.  The drawing in middle and right side of the top row show a reaction with an intermediate, and consequently, two transition states.  In each case the double-headed arrow identifies the rds - the highest energy point along the reaction coordinate.  The 3 drawings in the bottom row illustrate the case of a reaction with 2 intermediates (and therefore 3 transition states) along the reaction coordinate - in each case the arrow identifies the transition state with the highest energy, the rds.

When can't the steady state approximation be applied, and why?

Consider the following reaction
$$\ce{A <=>C[{k_1}]\   [B] <=>C[{k_2}]\ C}$$
$\ce{A}$ is the starting material, $\ce{[B]}$ is an intermediate and $\ce{C}$ represents the product.  Either the drawing in the middle or right in the top row represent the reaction coordinate for this reaction.  The steady state assumption applies to reactions that have one or more intermediates along the reaction coordinate.  The steady state assumption states that the rate of change of the concentration of the intermediate is approximately zero.  In other words, the intermediate is being created at about the same rate at which it is
$$\ce{\frac{d[B]}{dt} = 0}$$
being destroyed.  The steady state assumption is best applied to situations where the intermediate is present in low concentration (e.g. $\ce{k_2 >> k_1}$).  A general rule of thumb for the assumption to be valid is for
$$\ce{\frac{k_2}{k_1} > 10 }$$
Looking back at the figure, we now see that this condition is met in the top row, center drawing.  Here is a Wikipedia article on the steady state assumption, the section on "Validity" is nice and clear if you'd like to read further on the subject.

All in all, the SSA can be applied when the rate of consumption of the intermediate is more than ten times faster than the rate of its formation. Thanks.

I know this topic is old but why do you want the K2 to be higher then k1 for the steady state? This will actually produce more P + E so it will deviate from a constant ES concentration?@ron

@RickBeeloo  k2>>k1 guarantees a low concentration of the intermediate - a requirement for the Steady State Assumption to be valid.


text:
What are the priorities for substituent groups on a cyclohexane?
I was wondering about this specific scenario:
If you have a cyclohexane, for example, and the cyclohexane has two consecutive substituent groups, for example a methyl group and an ethyl group, how do you decide where the numbering starts?
Would it be 1-ethyl-2-methylcyclohexane or would it be 2-ethyl-1-methylcyclohexane?
Whoever answers this, could you just give some sort of general rules on what to do in this situation (e.g. does it follow alphabetical order, etc.)?

If the numerical sequence of locants is equivalent, the numbering should be assigned to correspond with alphabetical order. Hence, in your example, correct would be 1-ethyl-2-methylcyclohexane. Quoting the IUPAC Blue Book, Rule A-2.4:

If two or more side chains are in equivalent positions, the one to be assigned the lower number is that cited first in the name.

Please note that the link http://www.acdlabs.com/iupac/nomenclature/79/r79_36.htm leads to obsolete IUPAC recommendations of 1979.


text:
Coefficients and Parameters for contracted Gaussian basis sets
I'm trying to write a program to calculate fixed-point Hartree-Fock level energies of molecules (for my amusement) and everything makes sense but this. I've been agonizing over this for almost 3 hours now. I've tried pretty much every Google search I can think of (mostly returning results that were either too vague or without enough detail) and looked in almost every book/ebook I own (pretty much always too vague). Any help whatsoever would be extremely appreciated.
As far as I understand, an STO-NG contracted Gaussian basis function has the following form:
$$\phi_{\mu}^{\textrm{CGF}}(\vec{r}) = \sum_{p}^{N_{\textrm{PGF}}} d_{p\mu} (x-X_A)^{i_{p\mu}} (y-Y_A)^{j_{p\mu}} (z-Z_A)^{k_{p\mu}} e^{-\alpha_{p\mu}|\vec{r}-\vec{R_{A}}|^{2}}$$
where the contraction coefficients, $d_{k\mu}$ and exponents, $\alpha_{k\mu}$ are chosen such that $\phi_{\mu}$ provides the 'best fit' to a Slater-type orbital having a Slater exponent $\zeta$, $\vec{R_{A}}=(X_A,Y_A,Z_A)$ is a fixed reference centre (usually a nucleus) and $l_{p\mu} = i_{p\mu} + j_{p\mu} + k_{p\mu}$ defines the angular momentum of the primitive Gaussian function:
$$\phi_{p\mu}^{\textrm{PGF}} = (x-X_A)^{i_{p\mu}} (y-Y_A)^{j_{p\mu}} (z-Z_A)^{k_{p\mu}} e^{-\alpha_{p\mu}|\vec{r}-\vec{R_{A}}|^{2}}.$$
I'm having problems interpreting data files documenting contraction coefficients and exponents for various basis sets.
For example, please consider the following snippet from the Gaussian 94 STO-3G basis set file documenting contraction parameters for Carbon:
****
C     0
S   3   1.00
     71.6168370              0.15432897
     13.0450960              0.53532814
      3.5305122              0.44463454
SP   3   1.00
      2.9412494             -0.09996723             0.15591627
      0.6834831              0.39951283             0.60768372
      0.2222899              0.70011547             0.39195739
****

I have read in various places that there are enough coefficients here to describe 5 contracted functions, but I see only enough for 3 maximum (if the exponents are the same, but the coefficients change between 2S and 2P). I'd be eternally grateful if someone could explain for me, precisely how one would determine the following information from such a basis set data file:
1) Which EXACT basis functions appear within each contracted function?
2) How many contracted Gaussian basis functions there are?
If possible, would you be able to do the same for the following snippet, again for Carbon, Gaussian 94, but this time for the more complicated 6-31++G** basis set?:
****
C     0
S   6   1.00
   3047.5249000              0.0018347
    457.3695100              0.0140373
    103.9486900              0.0688426
     29.2101550              0.2321844
      9.2866630              0.4679413
      3.1639270              0.3623120
SP   3   1.00
      7.8682724             -0.1193324              0.0689991
      1.8812885             -0.1608542              0.3164240
      0.5442493              1.1434564              0.7443083
SP   1   1.00
      0.1687144              1.0000000              1.0000000
SP   1   1.00
      0.0438000              1.0000000              1.0000000
D   1   1.00
      0.8000000              1.0000000
****

Have you had a look at the original publication? Maybe it is easier to find the coefficients described in there. For the STO-nG family it should be described by Pople et. al.. Little information is gained on wikipedia.

Thank-you for your reply, Martin.
The thing is, I can't access the paper (I graduated last year) and I've checked out the Wikipedia article, but again, it's really quite qualitative and very vague with the specifics.

Thank-you for the paper! I've wasted so much time googling around in the hopes that someone had uploaded it.
A quick flick through seems to suggest that the researchers were dealing in spherical Gaussians, but I will enjoy the read nonetheless and I'm sure all of the information will be useful.

@SeanMacfoy We're all friendly here, and this definitely wasn't "stupid."

I agree, a paper like this should be open access. I hope it helps you - all implementations were based on this one, so it must be possible somehow.

There is so-known plane-wave basis set approach that may be more interesting in educational purposes 2)I think it would be much easier to dig sources of some existing package for the work. In particular, sources for GAMESS (fortran) and mpqc (c++) are available for educational and personal use.

Have you already had a look at the manual of gaussian, specifically the gen keyword, where they explain how to input a basis set manually? http://www.gaussian.com/g_tech/g_ur/k_gen.htm

In case someone will come by this question, here's a nice short write up on this topic http://www.tau.ac.il/~ephraim/Appendix_1.pdf

Take a look at the carbon STO-3G first. It means for each AO, you form it with three Gaussians. For carbon, you have a 1S, 2S and (three) 2P. 5 functions when finished.
The top line says you are defining the basis for carbon; no big deal.
Next line says you are creating an S shell, with three primitives, with a 1x scale factor.
Then you list your three rows of primitives.
The first column is the Gaussian exponent. The next column is the S expansion coefficient. These are the three Gaussians that make up your 1S "Slater" atomic orbital.
In the next section, you define SP. What this is, and I quote from the Gaussian website now: "Note that 2SP requests the best least-squares fit simultaneously to S and P slater orbitals and is not equivalent to separately specifying the best S and the best P expansions."
S and P (and all higher angular momenta, actually) can be related by a recursion relationship. It's an efficient way of computing integrals. I believe the Obara-Saika recursion takes advantage of this, though in Gaussian the integrals are formed using the PRISM algorithm.
The first column is the exponent for your Gaussian, the second column is the S expansion coefficient, and the third column is the P expansion coefficient. They just share the same exponent. Thus you have 4 + 1 (from before) = 5 AOs total formed with three primitives each.
To jump ahead to your last example, the pattern simply continues. The first section gives six primitives to form one 1S (the 6 in 6-311++G**). The second section gives three Gaussians for you valence S and P, the third section is one Gaussian for valence SP. This is the 31 in 6-31++G**. You have a split valence description, one with three primitives, one with a single primitive.
The next section is the diffuse function, hence the really small exponent in the first column. This is the ++ part of the 6-31++G**. The small exponent gives a really wide spreading Gaussian. Helpful for describing electronic activity far from the nucleus. (It's helpful to compare with the huge exponents in the first section, which gives really tight functions that describe the core electrons).
Finally, you have the D functions which are polarization functions. Carbon doesn't have a D valence shell, but we include them to describe polarization of electronic density in molecules.
Note that depending on your program, you either have 5D AOs or 6D AOs. This is the difference between spherical and cartesian representations, and is more of a computational trick than anything else. Gaussian uses 5D by default, except in some cases.

The tab sections Input, Examples and Basis Function Overview of Gen in Gaussian's online manual present, in tandem, a summary guide to what the various parameters listed in the question are and what the way they are ordered implies.
Under the Input section, there is helpful text next to Defining a shell., which also has a useful pseudo-code block listing the order of the variables for defining a custom basis set:

IType   NGauss   Sc    Shell descriptor line: shell type,
                        number primitive gaussians, and scale factor.
α1   d1μ                Primitive gaussian specification:
α2   d2μ                  exponent and contraction coefficient.

 …

αN   dNμ                There are a total of NGauss >primitive gaussian lines.

I wanted to add the above as a comment as I don't think it is a direct answer to the question, but rather something useful that supplements the answer by jjgoings above.

That would be too long for a comment anyway.


text:
What properties does a solvent with a high dielectric constant have?
My professor commented that water has a higher dielectric constant than alcohol.
Okay, but on what basis?
He didn't elaborate; he asked us which one seemed "more electric" and most answered water. I guess from intuition.
But why exactly is water "more electric"? I think it has to do with the relatively high extent of auto-ionization in water, correct? Alcohols are generally less polar than water and thus cannot stabilize ions as effectively as water can. And I've even seem some computers submerged completely in oil. I'm guessing that only works because oil doesn't ionize much and thus isn't able to conduct a lot of electricity - or enough to short out any circuits.

It's basically a measure of polarity, not ionization. It doesn't have to do with the ability to conduct electricity but with the ability to permit an electric field. Try researching permittivity.

Asking what substance seems more electric sounds odd to me ;)

Before I start, please note that I am not a chemist so be sure to verify what I say...
The electric conductivity of water depends on what is dissolved in it. For example, salt water will be far more conductive than distilled water. So if you're professor was talking about sea water, the answer could actually be different from purified water.
Due to the dissolved salt ions, sea water is a better conductor of electricity and so by definition it will be a poor dielectric, since according to this Wikipedia article: to be a good dielectric a substance must be both a good insulator and able to be polarized by an electric field.
Purified water on the other hand will be a better dielectric because it does not contain any significant amount of ions and is therefor not very conductive (i.e. is a better insulator than sea water).
I think (but don't quote me on an exam) that the reason alcohol is a poor dielectric is because it is not nearly as polar as water is, and according to the above, the second requirement for a dielectric is that it could be polarized by an electric field.
(This might also have to do with the size of the molecules, since water molecules are generally smaller than many kinds of alcohols, and can therefore re-orient themselves more easily in the presence of an electric field, but this is a guess.)
The Solvent Classifications section of this Wikipedia page also mentions that the polarity of a solution can be roughly gauged by the solution's dielectric constant. So this too seems to support the idea that polarity and dielectric properties are related.
If anyone has any improvements or corrections, please let me know in the comments and I'll update the answer accordingly.

What do you mean that salt water is more conductive than distilled water but is less dielectric?

The salt ions in sea water make it a better conductor of electricity than purified water. The reason this makes salt water a poor dielectric is because by definition: a dielectric "is an electrical insulator that can be polarized by an applied electric field". Therefor, since sea water is a better conductor, it would be a poorer insulator and hence a poorer dielectric. I guess one could say that the conductivity and dielectricity of a substance are inversely related, with the added requirement that to be dielectric it must also be polar.

@Dissenter, I updated my answer with a few more details to make it a little clearer because upon rereading it, I found that I had some confusing wording in it. :)

The ability to conduct electricity and the dielectric constant are not particularly or directly related. The answer confuses the two concepts and doesn't really address the intent of the question.

Thanks @matt_black! Could you provide some more details and clarification so I can point readers to your response?

The dielectric's is effectively a measure of the solvents ability to hold a charge and the Ɛr is a measure of its permittivity.
From a chemical perspective the degree of this is inversely proportional to the born energy of an ion (or self ionization energy) so a solvent that is better at holding an electric charge like water will lower the energy requirements for a species to become an ion.
Look up the Born equation, there are a few key relations in there but the main one would be the E=1/Ɛr relation, bigger dielectric, lower energy requirements.
Polarity and dielectric's are not the same thing, polarity relates to separation of charges over space, dielectric's relate to permeability and strengths of electric fields, but yes they do share some strong similarities and trends and a lot of the time you can make assumptions like "this solvent has a higher dielectric constant, so therefore will be more conductive.
When talking about salts, 'dissolution' and 'solvation' are two different processes, dissolution happens first when the solvents electric field goes into the crystal lattice and lowers the energy requires for ionization, then the ions are solvated by solvents. First step is all about the dielectrics and is about ionizing the species in the lattice, the second step is all about the polarity and becomes some annoying surface physics issues where even geometries can muddle things.
But I feel like the main issue with computers in water is that there are dissolved minerals in water that will cause all sorts of redoxs to happen.


text:
What is the length of a unit cell of CuCl assuming that it is fcc?
The density of $\ce{CuCl}$ is given – $x\ \mathrm{g/cm^3}$
The crystal structure is assumed to be fcc.
My teacher is claiming the we can apply the formula
$$\rho=\frac{Z \cdot M}{a^3 \cdot N_\mathrm A}$$
And he took $M = 35.5\ \mathrm{g\ mol^{-1}}+63.5\ \mathrm{g\ mol^{-1}} = 99\ \mathrm{g\ mol^{-1}}$,
$Z = 4$ as it is fcc.
But my question is how can we take $M = 99\ \mathrm{g\ mol^{-1}}$? It means every particle in the fcc has mass $99\ \mathrm u$ doesn't it?

Well, yes, depending on the detail you want to go into. You can find a representation of the crystal structure of CuCl at, for example, http://www.webelements.com/compounds/copper/copper_chloride.html - you will see that the Cu atoms are indeed a simple fcc lattice.  The Cl atoms are on sites that are an interpenetrating fcc lattice.  The end result is the zinc blende (ZnS) lattice prototype, very familiar to semiconductor folks as the crystal structure for GaAs an associated III-V compounds. If all sites were occupied by the same atom, this is known as diamond cubic, commonly seen in diamond, Si, and Ge (amongst others).
But, yes, one way to consider the crystal (particularly for your purposes here) is an fcc crystal with a Cu-Cl basis. Just don't expect the x-ray diffraction to correspond to fcc.


text:
Is superglue a good thermal insulator?
How good of a thermal insulator is superglue compared to other materials like ceramics, plastics, and metals?
I know this might not seem like a chemistry question, but I was hoping someone here could provide a more meaningful answer than what I found online, which was basically along the lines of, "It's not a good heat conductor so don't use it in cooling applications." I was hoping for a little more substance than that, so that's why I wanted to ask it on here.
Ideally, I would love for the answer to include one of the following things:

An approximate numeric comparison between superglue (liquid or gel) and other materials (for example plastics, metals, and ceramics).
Or a qualitative answer, like the example I gave above, but with a short explanation based on the chemical characteristics.
Or perhaps a chart of thermal conductivities of various superglues relative to other common materials.

If a similar question already exists -- or there is a free resource available online that can answer it -- then a link with a short explanation would be a perfectly good answer as well.

EDIT 1:
I know it sounds like I'm asking for a very big answer, but the suggested answers above were just suggestions. Really, the most important thing (for me at least) is to get a feel for where superglue falls in relation to other common materials.
EDIT 2:
In response to Greg's comment: I would have liked to share my finding on the subject, but unfortunately I have not been able to locate any formal research on the matter. The best I was able to find so far were forums and blog articles, which were usually either vague or offered little-to-no supporting evidence of their answers. Perhaps I just don't know the right places to look for information like this, and if this really is the case then I hope you all can give me some direction as to where I can find some research on this matter.

While it is not a homework type of question, it would help if you show how far you got on your own, what problems you faced. On one hand your question is very broad ("compare me all the materials! metals, ceramics, everything!") on the other hand you suggest you ask for great many details ("compare me all different superglues").

Practically it makes no difference whether its conductivity is high or low as its so thin when applied and damn difficult to measure as its so thin. It is practically a plastic.

@Greg Yes, you have a valid point. I don't really need an extremely in depth comparisons of everything.;) I just wanted to provide the y'all with a fair amount of flexibility in answering this question. Becasue almost any answer will be better than what I found online: "Superglue isn't a great thermal conductor" etc. Basically I just wanted to have an answer with some context and comparison in it.

I don't think it is at least before it's cured; superglue is pretty volatile before curing.

@user2617804 Hmmm great point. So does that mean that superglue is chemically/thermodynamically comprable to plastic? Because for example if it had ceramic-like properties, then even a thin layer could have a significant effect on heat transfer.

@Dissenter Yes, I completely agree that it's probabbly a fairly good heat conductor in its liquid state, but perhaps this might be more difficult than we imagine due to viscosity and layer thickness effects -- since very thin and viscus glue will not have as much molecular motion as thick or less-visid glue. In my case however, I would like to find out about only solid superglue since this is its most common state.

I just spent 30 minutes trying to find some number for the heat transfer coefficient of polycyanoacrylates. No luck there. It might be impossible to give an answer to this question without doing original research.

Thermally conductive transfer tapes by 3M have a heat transfer coefficient of 0.6 W/(K m).

@tschoppi Thanks for checking into that! And yes I completely agree with you, it seems there is no published research on the subject (probably because there are very few applications where this question would merit the expense of rigorous study).

@tschoppi And it funny that you mention the 3M tape because I actually ended up using that recently in one of the DIY projects I was working on :) haha

@tschoppi Speaking of polycyanoacrylates, after a lot of digging around I did manage to find a paper that spoke briefly on the topic but it was part of a larger discussion about something else that I don't quite remember now and the topic diverted rather quickly.


text:
What is C2H5-O-CH=CH2?
Someone painted an original graffiti:

source
I wonder what kind of reaction was depicted; what's that product substance?
The graffiti reads:
$\ce{C2H2 + C2H5OH -> C_2H_5-O-CH=CH_2}$
Does it have any more or less common uses?

The particular reaction is of acetylene ($\ce{C2H2}$) and ethanol ($\ce{C2H5OH}$) to yield ethyl vinyl ether (IUPAC name ethoxyethene). It's one example of Reppe chemistry, a fairly diverse set of reactions, often operating on alkenes and alkynes (though certainly not exclusively). The specific reaction in the question is a vinylization reaction, though it was Reppe's carbonylation reactions which were probably of greater practical (read: economic) import historically, as they were famously used in the Monsanto process and in production of various acrylates and other acrylic acid derivatives.
The vinylization reactions could still be synthetically useful, as they provide an extremely simple alternative route to vinyl ethers without having to proceed via elimination, substitution, coupling, olefination, or substrate-specific oxidation reactions.
I'm not aware of any specific uses for ethyl vinyl ether, however it could certainly be used as a building block in any number of organic syntheses.

Both divinyl ether and diethyl ether are anasthetics. Wikipedia article on divinyl ether lists differences between the two. One can suspect ethyl vinyl ether's properties will sit squat in the middle between them.

@SF., yes, you're right. A targeted search immediately found the following article. I have to suspect it's been retired for that purpose, like most non-halogenated ethers.


text:
How to calculate the equilibrium composition after a change in partial pressure of one component?

Question
  Ammonium carbamate dissociates as shown below.
  $$\ce{NH2COONH4 (s) -> 2NH3 (g) + CO2 (g)}$$
  In a closed vessel containing ammonium carbamate in equilibrium with ammonia and carbon dioxide, ammonia is added such that partial pressure of $\ce{NH3}$ now equals to the original total pressure. What is the ratio of the new total pressure to the original pressure?

I tried solving the problem and I know I am quite close to the answer which is $\frac{31}{27}$.
Firstly I took out $K_p$ for equilibrium assuming total pressure to be $p$.
So,
$K_\mathrm{p} = \left(\frac{2p}{3}\right)^2 \times \frac{p}{3} = \frac{4p^3}{27}$
After this the par pressure of ammonia becomes $p$ so let's say that of $\ce{CO2}$ is $\frac{p}{3} + x$.
Now since $K_\mathrm{p}$ doesn't change so:
$\frac{p}{3} + x = \frac{4p^2}{27}$ (cancelling one of the $p$)
But solving further and finding new pressure and dividing by the old one doesn't give the answer

As follows:
Just to simplify the math a little bit, assume pressure of $\ce{CO2_{(g)}}$ is p. Thus, pressure of $\ce{NH3_{(g)}}$ is 2p at equilibrium. Total pressure in this case is 3p.
Let 3p equal new pressure of $\ce{NH3_{(g)}}$.
Added pressure of $\ce{NH3_{(g)}}$ would shift the equilibrium to the left, so say we lose 2x pressure of $\ce{NH3_{(g)}}$ and x pressure of $\ce{CO2_{(g)}}$. New equilibrium pressures will be 3p-2x for $\ce{NH3_{(g)}}$ and p-x for $\ce{CO2_{(g)}}$. Thus, ratio of new to old is $\frac{4p-3x}{3p}$. You can solve for x in terms of p using $K_p$.
Solve for x and find the ratio, which should be $\frac{31}{27}$

how do i solve further?please explain. multiplying 3p-2x and p-x and equating it to 4p^3/27 and solving for x is a crazy thing to do. please explain to me.

what will be the initial pressure for co2 in the 2nd case wont it be p minus something?

After equilibrium shifts to the left due to addition of NO2, the CO2 pressure will be p minus something or I denote as x


text:
How do people check the type of defect in a crystal?
There are many types of defects like

Vacancy
Interstitial
Schottky
Frenkel
Non-Stoichiometric
Alien Impurities

But given a sample, how do scientists know which kind of defect is present in it [practically]?

As you said there are many different kind of defects in crystals. There are basically two different methods to find the Crystallographic defect.

Mathematical classification methods
Computer simulation methods

Mathematical classification methods:
A successful mathematical classification method for physical lattice defects, which works not only with the theory of dislocations and other defects in crystals but also, e.g., for disclinations in liquid crystals and for excitations in superfluid $\ce{^{3}He}$, is the topological homotopy theory.
Computer simulation methods:
Density-functional theory, classical molecular dynamics and kinetic Monte Carlo simulations are widely used to study the properties of defects in solids with computer simulations. Simulating jamming of hard spheres of different sizes and/or in containers with non-commeasurable sizes using the Lubachevsky-Stillinger algorithm can be an effective technique for demonstrating some types of crystallographic defects.

Source: Wikipedia

Given amount of atoms in typical macroscopic crystal, there is no need to check if any type of defect is present - all of them are presence because of sheer numbers and the law of big numbers. For estimating rough amounts different spectroscopic methods may be used. However, it is an indirect technique, which requires somehow obtained idea of the spectrum of the defect in question. Unless a very rough idea is enough, like detecting impurities using different nuclear spectroscopy methods, the only universal way is usage of quantum chemistry methods, and they are often frustratingly inaccurate when spectrum is a property of interest.
For nanosized structures, surface,near-surface and surface-influenced defects may be detected using hi-definition microscopy techniques, like scanning tunneling microscopy, which may provide subatomic resolution. However, in this case a different problem arises. Humanity usually does not need specific knowledge about some particular crystal, more a typical picture for some kind of systems is of interest. In such a case the only way is to make a lot of observations and somehow reduce it to typical values, and such a procedure is usually debatable, as samples for atomic-resolved spectroscopy must be prepared in very specific ways (see method descriptions for details), and so are very different from what should be present in conditions of interest.
Was this answer of any use?

Yeah...Thanks! But the fact that every macro crystal contains EVERY defect ... isn't it a bit of a stretch (or am i wrong ? !)


text:
How do scientists check which kind of unit cell is present in a crystal?
There are many types of unit cells like:

Tetragonal        Monoclinic      Orthorhombic        Rhombohedral
Hexagonal         Triclinic       Cubic

But how do scientists identify the type when a sample of the compound is given to them?

X-ray diffraction. Usually computer will tell you automatically nowadays but in the old days you had to check whether the values you got for the miller planes were possible with the symmetry of different unit cells. They discarded the ones that didnt match until the would find the one unit cell which matched the miller planes from XRD


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Determining boiling point on a created temperature scale

1.Assume that you construct a thermometer using gallium as the fluid instead of mercury, and that you define the melting point of gallium as 0 °G and the boiling point of gallium as 1000 °G. What is the melting
  point of sodium chloride (801 °C) on the gallium scale?

First we determine the ratio of $^{\circ}\mathrm G/^{\circ}\mathrm C$ by dividing the melting points. This states that $0.4599\ ^{\circ}\mathrm G=1\ ^{\circ}\mathrm C$.
$$\dfrac{1000\ ^{\circ}\mathrm G-0\ ^{\circ}\mathrm G}{2204\ ^{\circ}\mathrm C-29.78\ ^{\circ}\mathrm C}=\dfrac{1000\ ^{\circ}\mathrm G}{2174.22\ ^{\circ}\mathrm C}=0.4599\ ^{\circ}\mathrm G/^{\circ}\mathrm C$$
To calculate the boiling point, this is done: $$T=0.4599\ ^{\circ}\mathrm G/^{\circ}\mathrm C\times\left(801\ ^{\circ}\mathrm C-29.78\ ^{\circ}\mathrm C\right)=355\ ^{\circ}\mathrm G$$
My question is, why are we subtracting by $29.78\ ^{\circ}\mathrm C$?

Unlike an absolute temperature scale like degrees Kelvin, Celsius and your hypothetical °G are relative scales, where the zero points are set to the melting points of water and gallium, respectively. In your first step, you've accounted for the difference in the size of one degree Celsius and one °G, but not the difference in the zero points.
Instead of sodium chloride, think about what the melting point of gallium is on each scale. You've defined it as 0 °G, but you know it's not 0 °C. What do you have to do to correct for this to get a value in Celsius?
Incidentally, this is also why the conversion between Celsius and Fahrenheit is not just multiplying by a constant.

It zeroes things out.  It is analogous to the use of "32" in the equation to convert Fahrenheit to Celsius, where "32" represents the melting point of ice on the Fahrenheit scale.  In your G scale you must subtract the melting point of gallium in the Centigrade scale in order to get 0G at the melting point of gallium
$$\ce{^{o}C = (^{o}F-32)*(5/9)}$$
$$\ce{^{o}G = (^{o}C-29$.$78)*($0.4599$)}$$


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Residual stress in polymer coatings
A paper talks about the residual stress when coating polymers.
It says that residual stress is the sum of the "heat stress" (due to the difference in thermal expansion coefficients) and the "shrinkage stress" (due to the solvent evaporation and changes in volume). It then states that heat stress usually increases with temperature while shrinkage stress decreases with temperature. Therefore the shape of the graph of residual stress vs temperature would be a parabola with the minimum changing depending on the polymer/system system.
It then goes to find the starting point of the shrinkage stress and states that it is dependant on the glass transition temperature.
From what I understand, that would happen when the glass transition temperature is between room temperature and the drying temperature.
However, if the glass transition temperature is below both room temperature and drying temperature, what would happen? How this would affect the shrinkage stress? Or did I get the whole connection between glass transition temperature and shrinkage stress wrong?
Thank you

EDIT:
I read the paper linked in your comment, and see that my initial guess was incorrect (see below) - for this system, it is as you say - for temperatures above the bulk pure polymer glass transition temperature, you would not expect to see shrinkage stress due to evaporation.
END EDIT
I don't have access to that paper - so this is just an educated guess, but usually physicists use the glass transition temperature to characterize polymer behavior as a function of temperature because things roughly (very roughly) scale with $T_g$. In other words, if I know how the behavior of one linear homopolymer depends on temperature, and I know a certain response curve scales with $T_g$, then I can scale the temperature curve of another linear homopolymer using the $T_g$ and sometimes get accurate results.
They could be linking it to shrinkage stress because $T_g$ also characterizes the dynamic behavior of polymers (how they "move" in response to thermal fluctuations and external stress). This is essentially the principle of time-temperature superpostion - or the idea that as temperature increases, polymers move the same way, only faster. So heating up is like fast-forwarding time, and cooling down is like slowing time down.
The glass transition temperature is a discontinuity in this behavior - at a certain low temperature, the polymers freeze in place (on large length scales). This phenomenon is not well understood, but we have found empirically that the temperature at which it occurs can be used to scale response curves in certain situations, and since it is fairly easy to measure, it gets used a lot in these types of models.
So in summary, and again this is an educated guess, I would imagine that the glass transition temperature would be used in this model regardless of the temperature you were actually operating at - it is probably a fundamental parameter of the curve that is used to put the "reality" (the type of polymer) into the model.

Im not sure as it says that the shrinkage stress arises when the temperature equals the glass transition temperature so that stress shouldnt arise if those temperatures dont cross right? This is the paper https://www.mediafire.com/?kflzazi435xf3x4

@Daniel - I read the paper and have updated my answer - it looks like you are correct, and for systems above the glass transition temperature, you would not expect to see a shrinkage stress due to evaporation.


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Problems with thermodynamic derivatives
How is the following true?
$$\left ( \frac{∂S}{∂P} \right )_{T}=\frac{-C_{P}}{T}\left ( \frac{∂T}{∂P} \right )_{S}$$

I know that
$$C_{P}=T\left ( \frac{∂S}{∂T} \right )_{P}$$
which means the right-hand side of the first equation becomes
$$-\left ( \frac{∂S}{∂T} \right )_{P}\left ( \frac{∂T}{∂P} \right )_{S}$$
but how is that equal to $\left ( \frac{∂S}{∂P} \right )_{T}$ ?

There's an excellent book by Dill and Bromberg called Molecular Driving Forces. It derives most of these relationships from first principles, and is excellent even if you're weak on the maths side! It's a fantastic introduction to thermodynamics.

It's a consequence of the triple product rule for partial derivatives of thermodynamic state variables.
In summary, if you have three interdependent (not independent) variables, then this is true:
$$
\left ( \frac{\partial{X}}{\partial{Y}} \right )_Z \left ( \frac{\partial{Z}}{\partial{X}} \right )_Y = - \left ( \frac{\partial{Z}}{\partial{Y}} \right )_X
$$
Substituting in your variables, if we let $X = T$, $Y = P$, and $Z = S$, we get:
$$
- \left ( \frac{\partial{S}}{\partial{T}} \right )_P \left ( \frac{\partial{T}}{\partial{P}} \right )_S =  \left ( \frac{\partial{S}}{\partial{P}} \right )_T
$$
Note that the order of multiplication of the two partial derivatives doesn't matter.
Now, why this rule holds for thermodynamic variables is a little bit more complicated - but I think the wikipedia article that I linked to has a pretty good derivation if you are interested.


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How to calculate the heat absorbed by a reversible process?
Given a nonideal substance inside a box with diathermal walls, the pressure is increased reversibly. How do you calculate the heat absorbed?

I assume we use
dE = dq + dw = dq - pdV
=> dq = dE + pdV
But I'm not sure how to deal with the "dE" term.

I assume you mean the first law as
$$dU = \delta q + \delta w$$
where $dU$ is the (state) change in internal energy, $\delta q$ is the (path dependent) work, and $\delta w$ is the (path dependent) work.
Saying the walls are diathermal says something about temperature (as you should confirm). Then, you should find a relationship between temperature change and internal energy change for the substance in your system.


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How can I detect a specific molecule on a surface?
Based on the comments and answers, this question wasn't very clear, so I have completely overhauled this question to make it much clearer in another post:
How to create the equivalent of luminol for a compound?

Original Post/Question:
I'd like to be able to detect trace amounts of the chemical 'monomorine I' (5-methyl-3-butyloctahydroindolizine) on a surface. (It is an ant pheromone)
Is there an easy way to determine if the molecule is visible in IR or UV? (before I spend the money on this equipment)
Is there a strategy for causing a visible reaction? or even a reaction visible by IR or UV?
Advice on where to even begin would be a huge help.
(I do not have a chemistry lab with a Mass Spectrometer at my disposal)

Perhaps a fluorescently labelled monoclonal antibody?

In http://dx.doi.org/10.1007%2FBF01926645 they obtained the structure from spectral analysis: MS, IR and NMR. You can find more about this compound here, quite possibly also a full spectrum. From your question I really do not understand what you are trying to accomplish. Will you wash the substance of the surface, is ich chemically or physically adsorbed to it? All in all, I guess GCMS would be your best bet to figure out trace amounts...

Simple UV spectroscopy is not an option.
You won't find anything at all above 254 nm - probably just a a weak band in the 220 nm range. Indolizidines are pretty dead in UV.
Derivatisation towards a fluorescent compound would require clevage of a $\ce{C-N}$ bond, formally a dealkylation of the tertiary bridgehead amine.
I have no idea if a reagent like 2-naphthyl chloroformate, which does form fluorescent carbamates upon heating with tertiary amines will work here and honestly, I wouldn't bother to try.
As pointed out in Martin's comment, GC/MS after washing/extraction probably is the way to go.

Is it possible? Sure. Could you, considering practical issues, use it here? No.
There is definitely one spectroscopic technique but it's fairly difficult to do and requires the use of ultrafast lasers. Such lasers require vibration-damped tables, constant (and low) relative humidity as well as a constant temperature—this is not therefore a field approach. Furthermore, the whole set up (optical parametric amplifiers etc.) costs hundreds of thousands of pounds.
For those interested anyway, Sum Frequency Generation spectroscopy only works for non-centro-symmetric systems. So if you have a bulk solution, on average the system is symmetric and no spectrum is obtained. At surfaces however this symmetry is broken and molecules even along with their orientations can be observed. This works for both UV/vis and IR. Because the lasers are ultrafast, transient species can often be tracked as a given chemical transformation is occurring, whatever that may be (as long as there is a given chemical signature to lock onto).
A short review of this technique can be found here.


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Causes of reversible reaction
It is generally said that reactants react so that they can achieve a lower energy state. Then why does a reversible reaction occur in the first place? Since turning reactants into products increases their energy - an undesirable effect from the point of view of reacting molecules.

related:  https://chemistry.stackexchange.com/a/124602

First, let's tackle what an irreversible reaction is. Baking a cake, for example, is commonly considered "irreversible." The eggs, flour, milk, etc. which you started with are no longer present in their original forms after the cake has been baked.
Now, can you reconstitute the eggs, flour, milk, etc.? Of course you could take the molecules of the cake and piece them together to make eggs, flour, and milk, but doing so would be a very formidable task. A lot of energy will have to be expended to reconstitute the egg, for example, because eggs are made of protein and with heat (baking) proteins are denatured (lose their original 3D conformation and possibly their linear makeup or amino acid sequence as well).
Another example of an irreversible reaction is combustion of hydrocarbons - i.e. burning gasoline in your car. This reaction is definitely reversible in the long run - it just takes a few million years, a lot of heat, and a lot of pressure for the hydrocarbons to be formed again. But for semantic's sake we say that combustion is irreversible because it's not reversible on a really relevant time scale.
So in general, the difference between irreversible and reversible isn't exactly well defined - i.e. it might be hard to reverse a reaction (turn a cake back into eggs, etc.) or it might take a long time to reverse a reaction, but what's hard? What's long? Tellinging, the IUPAC (International Union of Pure and Applied Chemistry) does not try to define irreversible vs. reversible reaction.
A deeper treatment would involve discussion of equilibrium constants (symbolized by capital K) but I'll let another poster get into this if necessary.

It is generally said that reactants react so that they can achieve a
  lower energy state. Then why does a reversible reaction occur in the
  first place?

Good question. Remember that we can always add energy to make an unfavorable reaction proceed. For example, the sodium ion, which is isoelectronic with neon, is stable with a full octet of electrons. However, we can still take away more electrons. It just takes a rather sizable application of energy.

Thanks for the compliment. So what it really means that in reversible reactions( those reactions which are easily reversed ),the energy released and other factors like temperature, pressure etc. themselves favour the backward reaction. Also am I wrong in thinking that backward reaction takes place easily since the difference between two energy states is small enough that in same conditions both energy states can be achieved?

It is not the case that in all reversible reactions the backwards reaction is favored. Reversible reactions are fundamentally equilibria.

It is generally said that reactants react so that they can achieve a lower energy state.

Here is a slightly more accurate statement: In the absence of non-PV work, the system (closed, at constant pressure) will achieve lower Gibbs energy when a reaction occurs. The less convoluted way of saying this is the second law of thermodynamics applies to chemical reactions.

Then why does a reversible reaction occur in the first place?

The term "reversible reaction" is not exactly what it sounds like. A reaction is called reversible when - at the particular level - both forward and reverse reactions take place. It does not mean that concentrations change in one direction and then suddenly change in the other direction. In fact, the reaction (again, in the absence of non-PV work), always goes toward equilibrium, not away from it. Equilibrium always implies that all reactant and products are present. If you start without product, the reaction will go forward; if you start without reactant, the reaction will go in reverse. If you start with all species present, it depends on the relationship of the reaction quotient Q and the equilibrium constant K.

There are several reasons for the reverse reaction to be favored; one is temperature. In the following reaction:
$$\ce{2NO_2<=> N_2O_4 + energy}$$
The right is the favored product, unless the temperature is increased. With enough heat, then the reactant is favored.

There are two factors that determine if a process will be spontaneous or not:
Enthalpy- Reactions that give off energy tend to be spontaneous.
Entropy- Entropy is a measure of the randomness or disorder in a system. Generally reactions that increase the randomness of the system are spontaneous.
The Gibbs free energy is a way of combining the entropy and enthalpy of a system into one value.
This might be able to help explain the question.

It is very simple! A reversible reaction is a reaction that reaches equilibrium. The conditions are: 1. Reactants and products must be present in the concentrations or activities that satisfy the equilibrium constant. 2. they must be properly mixed. 3. there must be sufficient energy present [temperature and source] for both forward and reverse reactions to have palpable rates. 4. there must be a mechanism for the reaction.  In this situation the process can be perturbed in either direction. If only products or reactants are present and the other requirements are met the reaction will proceed in either direction to equilibrium.
At equilibrium Delta G = zero = Delta H - T Delta S; Delta H/Delta S = T.  This means for an equilibrium Delta H and Delta S have the same sign and that the reaction in one direction is energy driven and in the other direction entropy driven. If this is not true, the reaction will not reach equilibrium and is not reversible.
A nonreversible reaction can be thermodynamically reversible by being broken into a series of reversible steps the sum of which gives the original state. This requires a low entropy energy input; an example being the relationship between respiration and photosynthesis; Every day, if you are lucky, there is fresh food on the table.


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Can a multi-species system oscillate around equilibrium?
In reading about chemical oscillations such as those that occur in the Belousov–Zhabotinsky reaction (BZ), it's often reported that these reactions were initially not taken seriously, because of a result that a homogeneous system cannot oscillate around its equilibrium point. It is usually claimed that the assumption of detailed balance guarantees this. I want to understand this result, but I can't find any specific references to it.
I understand very well why this argument doesn't apply to the BZ reaction, since the latter doesn't oscillate around its equilibrium point. However, I'm interested in understanding what this argument is, and whether it does really apply to all oscillations around the equilibrium point.
Kondepudi and Prigogine (2002) imply that the argument is as follows: for a single reaction, one cannot move from one side of the equilibrium to the other without passing through the equilibrium. Once the system reaches equilibrium it can't then keep on reacting, as this would contradict the second law, and therefore an oscillatory approach to equilibrium is impossible in such a system.
This of course makes sense for a system that only involves one reaction. However, for a system in which multiple reactions can take place it doesn't seem to be enough. For example, let's imagine that I have reactions $\ce{A <=> B}$, $\ce{B <=> C}$ and $\ce{A <=> C}$, whose kinetics are determined by some complicated set of catalytic effects. Let's also suppose that the Gibbs energies of A, B and C are the same, so that the equilibrium contains all three reagents in a 1:1:1 ratio.
Now it doesn't seem impossible that I could start off with a high concentration of A and watch the following sequence of events:

most of the A is converted into B, leaving a little bit of A unconverted;
most of the B is converted into C, leaving a little bit of B unconverted;
most of the C is converted into A, leaving a little bit of C unconverted;
this cycle repeats, with the unconverted material building up over time until the system converges to its equilibrium concentrations.

If I were to plot the concentration of the three reagents against one another as a 3D plot, it would look something like this (with apologies for my shaky drawing)

The black dot is the equilibrium point to which the concentrations eventually converge. The path is confined to a plane, because the total concentration of A+B+C cannot change. The free energy decreases as the path approaches the equilibrium point, so the second law by itself doesn't seem to rule out this behaviour.
If a system behaved this way, it would be an example of oscillations around equilibrium in a multi-reaction system. My question is, is there a result that says this cannot happen, if we assume the kinetics obey detailed balance at the equilibrium point?
If so, what is it called, how is it derived, and where can I read about it? If not, is there a known example of a chemical system that does approach equilibrium through oscillations in this way? (A 'toy' system will do, as long as the kinetics obey detailed balance.) Note that the BZ reaction is again not an example of this, since its oscillations do not take place around the equilibrium point.
Perhaps part of the argument is that the reasoning described by Kondepudi and Prigogine must hold for each reaction, independently. For the three-reagent system I described, this seems to be enough to rule out oscillations around equilibrium: the $\ce{A <=> B}$ reaction cannot pass through its equilibrium point, regardless of what happens to the concentration of C. But for a four-species system this no longer seems true: then the $\ce{A <=> B}$ reaction can move away from its equilibrium, as long as it's coupled to the $\ce{C <=> D}$ reaction. If oscillations around equilibrium really are impossible in every conceivable system this seems quite non-trivial, and it would be nice to see a proof.
Reference: Dilip Kondepudi and Ilya Prigogine (2002) Modern Thermodynamics: From Heat Engines to Dissipative Structures. Wiley. 2nd edition.

Though I do not know of an answer to this, one thing I can say is that the reactions in the example will not proceed the way you state them. Since all have the same or a similar $\Delta{}G$, the reaction does not complete towards B leaving some A. So it might be a ln 'allowed' path, in any system with enough molecules is simply very unlikely to follow it, i.e. it does not happen.

"Every conceivable system" is for this question "every conceivable kinetics"?

@JohnH.K. yes - or to be more precise, every conceivable kinetics that obeys detailed balance, i.e. $k_f/k_r = e^{\Delta G^0/RT}$. (If you ignore detailed balance then it's easy enough to come up with "toy" systems that do oscillate around an equilibrium value.)

The comments have already mentioned one problem with the question: the trajectory of the example system includes both exergonic (OK) and endergonic (not OK for closed systems) stretches.  Think about the total Gibbs free energy of the example with time.  The regions where the trajectory is moving away from the equilibrium point are endergonic (i.e. free energy is rising) and would not be allowed unless work is being done to the system somehow.
The other problem is ambiguity in what is meant by "around" equilibrium.  Passing through the equilibrium point isn't allowed -- almost by the definition of equilibrium.  Gibbs free energy would have to rise (i.e. an endergonic process would be required) after reaching equilibrium.
Orbits "around" equilibrium that always spiral closer to it would potentially be OK.  But it depends on exactly what variables are in the orbit.  Obviously, Gibbs energy must always decrease in an equilibrating system.  Concentrations don't necessarily have to.  A good graph is this one: case A is disallowed by the second law, but case B isn't.

(source: pojman.com)
The page where I found that graph does a good job of explaining the topic in much more detail: see here.  They present chemically feasible reaction mechanisms that lead to oscillations.

I'm afraid you've rather misunderstood the question. The trajectory I sketched has no endergonic sections and is always moving toward the equilibrium. Your plot at the end is the point of the question. I'm asking for a proof of it for a general multispecies system.

Apologies for misconstruing your question Nathaniel.  I still don't really understand what you are asking for.  Are the Brusselator, the BZ reaction, etc., not examples of oscillating "around" equilibrium?  Can you give a formula or mathematical definition of what you mean by "around equilibrium"?

Also, apologies if I misread your graph.  Perhaps it's just a limitation of 3D plotting on a 2D screen, but to my eye there are stretches were the red line first gets closer to the equilibrium point, then moves away from it.

Yeah, the Brusselator etc. don't oscillate around equilibrium. In my plot the red line is supposed to be constrained to a plane defined by the grey lines, with the equilibrium in the centre. It might get closer according to Euclidean distance, but (my inexact drawing aside) it should always be going downhill according to the free energy, which is a nonlinear function of position.

In terms of a mathematical definition of oscillating around equilibrium, this would basically be that if I linearise the dynamics around the equilibrium point, the leading eigenvalues of the resulting Jacobian matrix are complex. But basically it's the right-hand side of the figure you posted - the point is that in a multispecies system the concentrations could look like the top-right plot, while the free energy still looks like the bottom-left one.

This actually happens in physics. For example, consider a damped swinging pendulum. The "free energy" here is kinetic plus potential energy, which always decreases (turned into heat by friction), but the position and velocity of the pendulum both oscillate between being greater and less than their equilibrium values.

Thanks for the new comments.  I can't give a mathematical proof, but in systems at the thermodynamic limit (i.e. with very large numbers of particles where the laws of statistical mechanics and equilibrium thermodynamics apply and temperature is defined), passing through the equilibrium point is not possible.  To map your example onto chemistry, consider an adiabatic pendulum and include T as a state variable. The equilibrium point is not passed as the pendulum swings, because the friction heats the system; the equilibrium temperature is higher than the initial.


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Under what circumstances is working in the chemistry lab hazardous to health in the long term?
I would like to know when working in the Chemistry lab would be hazardous to health in the long term. Because as everyone who worked in the chemistry lab knows that they would be handling many compounds that are harmful to human health. Many of those reagents for example especially organic ones are carcinogenic. While some are toxic to humans. Even with safety precaution, we would inevitably be exposed to those chemicals when working with them such as inhaling them. So actually I wonder whether it is safe to work in the chemistry lab especially with so many carcinogens. Im afraid of getting cancer due to this.
(I'm already halfway through my undergrad chemistry course.)

There has probably been a serious study on this, but I'll give you some anecdotal evidence. I work in a group that has chemists that have been in the lab for 35, 35, 40, and 50 years carrying out organic synthesis (and some radiosynthesis). None of them have any health problems that can be attributed to chemical exposure. I am more concerned with lab accidents (fire, etc). With proper precautions, chemistry can be performed safely in both the short and long term.

unless you have some specific conditions, say allergy, and have no plans to work in dangerous field, you are safe. Granted, there are many hazardous substances in some labs because of their field of work, but most have toxic dose that cannot accidentally do damage in case of all safety rules are followed. There are, however, some things to avoid, so consider taking toxicology course =).

Just be careful and follow all safety procedures. Labs do contain harmful substances, and mishandling can lead to fires, explosions, or contamination. Accidents can and have lead to deaths before, as illustrated by this UCLA incident http://latimesblogs.latimes.com/lanow/2011/12/charges-in-students-death.html  and even if you're following protocol, extremely toxic compounds can still kill you, as illustrated here: https://en.wikipedia.org/wiki/Karen_Wetterhahn  If done right, laboratory work is safe, but you must take responsibility for your safety and follow the protocols.

It highly depends on a lot of factors given that no two labs are necessarily identical.  As a computational chemist, the only dangerous thing I run into is distraction.  Two monitor screens can lead to a serious case of distraction.  I should probably go see a doctor.

I have (hopefully) narrowed the question and was wondering if it can be reopened in its current form.

May I start with a disclaimer saying that whatever long term studies can be done, this will always be a subjective question. All I can do here is offer a few comparisons and examples to show that you shouldn't (in my opinion) be worried.
There answer to the question "Am I more likely to come into contact with carcinogenic/hazardous substances than a regular person?" is quite likely yes. At undergraduate level you will be using a range of strong, harmful, poisonous etc. substances that you could not get outside of a lab environment or at school.
However, the answer to the question "Am I more likely to suffer poor health as a result of working in a lab?" is more difficult.
As you correctly state in your question, there are always many safety procedures in place when you undertake experiments which have been carefully thought, require the scientists to have some minimum level of training in the techniques (perhaps learning with less dangerous substances) out and always err on the side of safe rather than sorry. I myself have been working with a new nanotube based material recently and though unlikely to be carcinogenic, all experiments were to be carried out using gloves in a fume cupboard with extraction as the material hasn't been fully classified or tested yet (again: "safe rather than sorry").
Of course, there are unfortunate cases (for example Marie Curie), but in this day and age the risk assessments are so extensive and always erring on the side of caution that the chance of something going terribly wrong are no worse, I feel, than something catastrophic happening in many other walks of life (e.g. mountaineering accident, plane crash, shark attack).
One perhaps reassuring and rather amusing case is that you are far less likely to succumb to terminal cancer if you work in the nuclear industry by virtue of the extensive risk assessments, precautions and health checks. There is so much safety in place at a reactor, that exposure is minimal (indeed, the background radiation from radon in Cornwall is higher than that next to a nuclear power plant!!) and the industry provides so many health checks for workers that and cancer (radiation induced or natural) is often picked up at a very early, curable stage. Earlier than if you simply went to a doctor when you began to feel ill as a regular person.
Therefore depending on the policy of your employer, you could in fact be at a lower risk of dying from cancer, and I see no need for you to worry for your health in the lab!!
As mentioned when I started, this is a subjective issue, but I really see no need for worry.
Best,
Jack


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Most negative and most positive value for proton chemical shifts
What are the most negative and the most positive values for proton chemical shifts recorded till present?

I suspect the most positive values can be found in superacidic media such as $\ce{HF - SbF5}$, which contain extremely poorly solvated, essentially bare $\ce{H^+}$ ions.

What are the values?

It's been ages, but I stumbled on this article at random, showing a number of very closely related compounds with proton chemical shifts that vary from -58 ppm to +140 ppm! There are also some amusing carbon-13 shifts to be had.

I don't know if the following two examples present the "largest" upfield and downfield proton-nmr chemical shifts, but I suspect they're in the running.  The dihydropyrene dianion example has 16 pi electrons

delocalized around the periphery of the pyrene frame. It fits the 4n rule with n=4, so it is antiaromatic.  The [16]-annulene dianion has 18 pi electrons and fits the 4n+2 rule with n=4, so it is aromatic.  Note how the direction of the ring current reverses between antiaromatic (paramagnetic current) and aromatic (diamagnetic current) systems.  More intesting proton chemical shifts can be found in this compilation.
Edit
Guilty of thinking organic.  If we're including inorganic proton shifts, then how about IrHCl2(PMe(t-Bu)2)2 which has a chemical shift of -50.5!

Thanks. I examined the link you posted. I see the rhodium-pentacyano-hydride has around -10 ppm and that theoretically, a naked proton would have a chemical shift of 40 ppm (which is probably the maximum possible). So I will take the most negative shift to be -10 ppm, and the most positive "measured" to be 21.24 ppm. If anyone finds some even more extreme values, I would like to see them.