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Does soda-lime glass contain carbon?
Soda-lime glass is made by melting together soda (sodium carbonate), lime (calcium oxide), dolomite (calcium magnesium carbonate), silica (silicon dioxide), alumina (aluminum oxide), and small quantities of other compounds. Wikipedia says this is done in a furnace at "up to" 1675 °C. Many of the carbonates decompose at lower temperatures, 650 °C for dolomite and 851 °C for soda. However, soda-lime glass begins to soften at 700 °C.
Given the high decomposition temperature of carbonates like soda and dolomite, which is close to and sometimes higher than the temperature at which soda-lime glass begins to soften, one might think that there is a potential for some carbonate to remain dissolved in the glass. This is perhaps supported by the fact that silicon and carbon are both Group IV elements and could share spaces in the same lattice structure.
However, I haven't been able to find a quantitative analysis of soda-lime glass which measures oxidized carbon content. Is this because it's difficult to measure, or because no one in the glass industry cares, or because it's actually very close to zero?

https://en.wikipedia.org/wiki/Soda-lime_glass gives contents, so... I don't exactly get why you're asking this.

@Mithoron: "I haven't been able to find a quantitative analysis of soda-lime glass which measures oxidized carbon content"

At the higher temperatures the carbonates are completely decomposed and you are left with the oxides. The carbonate content should be quite negligible in the finished product.

Soda lime is made by mixing a few components, which are added to help with reducing the melting point. And at those temperatures any remaining carbon in form of carbonates would decarbonise, therefore as you mentioned, there is virtually close to zero amount of carbon.

Nature abhors a vacuum, the vacuum in this case being absolutely zero carbon atoms in the glass. But the "decomposition temperatures" of the carbonates do not account for CO2 being displaced by another acid such as silica, so probably the carbon is largely gassed out.


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Protonation of compounds with two or three neighboring nitrogens
I am working with organic compounds. Many of them are similar to this ones:

The R group (which I paste for completeness) is:

I am a bit confused about the protonation state of those amine-compounds at pH=7.4, in aqueous solution.
N-substitued amines are more basic than the other ones. For example methylamine is more basic than ammonia.
But the protonation state of the ring is more complex than that, and I can't understand how to protonate it.
I suppose that in a kind of analogy with the guanidine (product) we could protonate again the double bonded nitrogen (imine).
Will amine be protonated? Will that compound has $+2$ charge?
Guanidine
It will possibly be protonated in the double bounded nitrogen, because it has more negative density charge and resonation is not broken. Any other point of view will be welcome.

These are guanidines and it's imino group which gets protonated - it's not "secondary amine".

This is quite similar to the question that I am trying to reason as well.

" with two or three neighboring nitrogens" There's no two nitrogens right next to each other in these structures...

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Confusion on light absorption by electrons
Say an atom's electron can jump between energy levels n = 2 and n = 3 with x joules of energy. If a light beam of frequency f provides x+y joules (where y is a positive number), that is, more than enough to excite the electron to the next energy state but not enough to reach n = 4. The electron should accept this quantized quantity of energy and then
later emit it. On emission spectra, however, the extra "y" J of energy are not accounted for, only the minimum value x to transition between n = 2 and n = 3 (the energy difference between orbital levels). Where, then, does the extra y energy go? It cannot simply "not be absorbed" because it exists along with the rest of the energy, x, in quantized packets. If it contributes to raiseing the KE of the electron, it must at some point be emitted, but that would give way to an incorrect value on the emission spectra. What am I missing?

"The electron should accept this quantized quantity of energy and then later emit it." Why should it?

Depending on how large the value $y$ is, the premise of your question can be wrong.
On the other hand, if you jump between two electronic energy levels where $y$ is on the order of vibrational states, it is entirely possible this extra energy will be stored in vibrations of the molecule on the excited state. This energy is then typically radiated away quite quickly until you reach the ground state of the higher electronic energy level, and then the emission takes place where you could measure the energy $x$ again but in an emission spectrum.
It is entirely possible, however, that the emission process will happen from above the $v=0$ state on the excited surface. It is also possible that you will emit down to different vibrational energy levels on the ground state electronic surface. That is, $0(v=0)\leftarrow1(v=0)$, $0(v=1)\leftarrow1(v=0)$, $0(v=2)\leftarrow1(v=0)$, etc. are all possible and will be observed with varying intensities.
So, in general if you pump in some amount of energy to excite an electron, you don't need to see that same amount of energy coming back out from electronic transitions. You would need to look quite closely at infrared radiation and other energy to make it add up.
As another point, it is actually possible for this energy $x+y$ to be absorbed even if $x+y$ is very far away from the energy of a state. This is how scattering Raman spectroscopy works. You have to think about it differently, however, because the state you are excited to is not a stationary state, but rather a virtual state. See the question I asked here.
So, the short answer is that energy very frequently gets added to other degrees of freedom besides the electronic ones. The same thing goes for vibrational transitions where the excess energy gets put into rotational states. There are rules for how these transitions (and how they couple) will take place.
The other answer, which is less interesting, is that it's totally possible for the molecule to simply not absorb this $x+y$ precisely because there is nowhere to put this extra $y$.

While jheindel made a lot of interesting points, for most purposes the best answer is just the last sentence

that it's totally possible for the molecule to simply not absorb this $x+y$ precisely because there is nowhere to put this extra $y$.

In fact that is the reason that absorption spectra with clear lines exist in the first place: if any transition with energy $x$ could just swallow any $x+y$ as well then (regardless of what ultimately happened with the $y$) any gas would block basically the entire visual spectrum. But because only the particular transition lines can efficiently be absorbed, most of the light gets through instead even if the gas is dense – most photons simply won't interact with the molecules, instead they pass right through/past them as it were.
One way to derive why this happens is a semiclassical view: the energy levels that we're always considering are Eigenstates of the Hamiltonian; that means they don't change in time. In particular, there is no movement of the charge density. However, the transition between two such states $E_0,E_1$ can be viewed as a superposition between two eigenstates, and such a superposition is itself not and eigenstate but “vibrates” with frequency $\frac{E_1-E_0}{h}$. It thus acts as an “antenna” tuned just so it can interact with photons of energy $E_1-E_0$.

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Why are plastics lipophilic?
My wife prepares lunch for me every morning. After eating it on my desk, I typically wash the lunchbox and utensils in the company's kitchen.
I have noticed that it is very difficult to remove the oiliness from the plastic containers (think Tupperware, but also harder plastics that look like polycarbonate): I need to rub the detergent-soaked sponge extensively and rinse with very hot water, several times. This does not happen with glasses, porcelain dishes or metal utensils like forks.
In other words, all the plastic containers I have come in contact with seem to be quite lipophilic.
Is this because edible oils and plastic are chemically "similar" (i.e., both contain long-ish chains of hydrocarbons), or is there another reason that I am missing?

I suspect it's not that plastics are very lipophilic (maybe a little), but that it's more hydrophobic than metals etc.

So, metals and glass are lipophobic and that makes oil harder to stick (easier to remove)?

Yes @Nicolas Miari. With a lot of differences between them, you can still say common plastics and oils have similar "hydrocarbon like" fragments. Note that what is lipophilic is hydrophobic. Ergo we need a detergent (a molecule with both parts).

I really hate that phenomenon too. I think if the plastic surface could be made more hydrophilic I see a good business... But surely it has been already considered and probably overall it does not make sense for household applications. I want to add that, hands being wetted, I think that we feel the plastic to remain more greasy even when is already washed because of its hydrophobicity.

@Alchimista I don't know about the greasiness of hands; I judge both by touch and look. The still-oily surface looks very different, reflects light according to the oil pattern. When clean, the reflection is even.

First, many plastics are somewhat porous. Since plastics are permeable to water vapor and oxygen, coffee and other food products are sometimes packed in two-layer bags of polyethylene (for strength) laminated to aluminum foil (to reduce permeability).
Second, lipids and other non-polar substances can dissolve in and and migrate through common plastics used in food containers, such as polyethylene and polypropylene, as you and @Alchimista note. This is why carotene, lycopene and other chemicals in tomatoes eventually stain poly containers permanently: they've dissolved into the plastic.
In a similar way, mercury dissolves into gold (e.g. jewelry) in a matter of seconds, leaving a silvery stain. That stain then disappears in a week or two as the mercury migrates throughout the object.
The reverse phenomena also occurs, where substances in plastic packaging can migrate out of the packaging.
As the Bard might have put it, "Will all the water in the ocean wash this food from my cans?"

Thank you; this is the kind of in-depth, additional information I was looking for beyond my superficial observation/guessing.

I'm not sure this actually answers the question which is why plastics are lipophilic.

@matt_black : True, the question contained the answer: the molecules are similar, i.e. non-polar compounds.

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Electron configuration of Y(2+) ion
Why is the electron configuration of $\ce{Y^2+}$ $\mathrm{[Kr]5s^1}$ rather than $\mathrm{[Kr]4d^1}$? It seems as though there are numerous anomalies from the typical electron configuration rules, but can one predict these exceptions (and if so, how)?
What causes one electron to be removed from the $\mathrm{5s}$ orbital and one from the $\mathrm{4d}$ orbital, is it simply due to the $\mathrm{5s}$ energy decreasing below the $\mathrm{4d}$ energy after the first ionization?


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Does the rate of dissociation of PCl5 increase in presence of Cl2
I'm sorry if this is somehow inappropriate to ask but I've found a rather strange question on my text book that asks "Why does the rate of dissociation of $\ce{PCl5}$ increase in the presence of $\ce{Cl2}$ ?".
Somehow I couldn't find any relevant answer to this question, which led me to possibly believe that this might be a printing mistake. But nonetheless I wanted to ask if this particular case is an exception or not.

$PCl_5$ tends to decompose to $PCl_3$ and $Cl_2$. This might help: https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle ;)

According to Le Chatelier's principles the case is exactly opposite to the views of my question.I did figure this out myself.But I'm not sure if there is any particular exception in terms of PCl5 dissociation

"Exactly opposite"? Care to clarify? Additionally, are you sure you understood the idea (Le Chatelier's principle) correctly?

Yikes! I'm sorry but I've accidentally put my own explainations as my question in there.I've corrected them.Sorry for the inconvenience

That edit completely changes your post...perhaps the question was stated incorrectly? Pretty sure the question has been wrongly worded, but you should wait for someone else to put in their opinion here. Also, where did you see this question?

I just think there has to be an error in the question itself, i.e. increase should be decrease.

@air Yup, think so too. I edited the question to include the "erratum" tag (yes, it exists) :D

@Paracetamol I've found this in a chapter that deals with Equilibrium State and Le Chatelier's Principles on my Chemistry text book.I'm an International student and the Author is domestic, hence unrecognizable.

@Saad Meh, just asked. It's advisable to include excerpts, that are pertinent to your question, from whatever material you use. Perhaps a PDF version of the text is available online? In that case, try putting in the link at the end of your post. Cheers!

Please include the citation to the text you are using (no matter how obscure you think it is). If it is a published text book it should have and isbn, if it doesn't, it could just be a random statement off of a stranger in the street and has little more credibility.

I'm sorry but as I've told It's a domestic Author and the language is 'Bangla'

If you think you're up to it, try translating the excerpt to English and post that (alongside the original Bangla) in your question. A lot of users here use books/articles written in their native tongue. Take a page out of our beloved Moderator @Mart's book in this regard ;) [ https://chemistry.stackexchange.com/questions/76959/what-did-actually-happen-when-staudinger-tried-to-synthesise-diamond ]


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What type of chemical reaction is it when sodium bicarbonate is added to citric acid?
I can't tell what type of chemical reaction it is? Is it a combustion chemical reaction? Or Displacement or decomposition or something else?

Reactions don't really have types. But if you need to put a sticker on this one, call it a displacement followed by decomposition.

If the citric acid was vinegar instead what type of chemical reaction would that be?

The same. $\mathstrut$

Are you interested in baking or CO2 or both?

The first step is usually called an acid-base reaction or a proton transfer. Any acid of similar strength to citric or acetic (vinegar) acids will react with bicarbonate this way, and it is very fast in water solution at room temperature. The products are carbonic acid and the sodium salt of the acid used--for example, sodium citrate.
Carbonic acid is energetically unstable relative to carbon dioxide and water, to which it is rapidly converted via what most chemists would call an elimination reaction. I used to think that this occurs instantly, but carbonic acid is a species that actually exists in solution and can be observed and studied.
While there are many named organic reactions, I  agree with Ivan that there is no one universally agreed upon way to name them, so others may wish to weigh in with their opinions about this reaction.


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Why can alkoxides leave in the E1cb mechanism?
Why can alkoxides leave in the $\text{E}_{\text{1cb}}$ mechanism? I refer to the mechanism below. I would think that the anion formed from the first step would be more stable than the alkoxide ion, because the charge on the alkoxide ion is intensified by the electron-donating alkyl group, making it less stable. Correct me if I am wrong.

image source: https://en.wikipedia.org/wiki/E1cB-elimination_reaction

The fact that the elimination of $\ce{EtO-}$ leads to the generation of a carbonyl bond makes this a feasible reaction. Also, the $\ce{EtO-}$ ion generated can then take $\ce{H+}$ from water taking the equilibrium further in the direction of the forward reaction ( the net constant would be ~ $10^2$).


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Separation of Regioselectively Alkylated Ionic Solid
How would one go about separating a solid mixture of regioselectively alkylated solids.  I'll give a specific example.
Take the hypothetical alkylation of methylpentazole with MeI.
There are three (3) different products possible.  How would you go about the separation of these products after recrystallization of the product mixture (with the assumption the reactants and products don't explode).
The problem is that you cannot simply distill off each part of the mixture due to the high boiling point of the solids.

Do they show separation by HPLC?

This is hypothetical, but I would assume that two of the products would not separate, namely the 2- and 3- methylated salts.  And the third salt would probably not form anyways due to loss of aromaticity.

Then I think it would be a real struggle to separate the isomers


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Order of C-O bond length

Find the correct order of increasing $\ce{C-O}$ bond length in $\ce{CO}$, $\ce{CO2}$, $\ce{CO3^2-}$.

My solution: as bond length is inversely proportional to the bond order, the order would be $\ce{CO} < \ce{CO2} < \ce{CO3^2-}$. Am I correct?

https://chemistry.stackexchange.com/questions/80421/how-to-rationalise-the-increasing-bond-length-order-in-the-carbonate-ion-carbon/80422#80422 Same test, maybe?

The order is correct but this is my procedure.
Stronger the bond smaller the length


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What is the shiny structure under a metal coating?
I performed a hot dip aluminizing on steel for 15 seconds. Then, after I bent the steel to remove the coating easier, the steel under was rough, shiny, and metallic, and was different from the original steel.The layer seems to be random and has bumpy structures. I can't mechanically removed the layer. I couldn't have analyze it chemically. Could I conclude that it is an intermetallic layer of Al and steel?

Comment not answer because I am not sure of what is. But I suggest to look for something like thermite reaction. How does it sound? It should loose is shine more or less soon

Based on my readings, thermite reaction involves Al and rust. But before I hot dip aluminize the steel, I cleaned it first with HCl to remove rust, so the metals involve is mostly the steel itself and Al. Would the process still affects the shiny appearance?

No, it is was just a plausible suggestion. I was thinking about a "cleaning" of the steel, but if you treated with HCl then no.

Likely an intermetallic , I think there are a couple different ones. I am more familiar with "spray and diffuse"  or vapor deposition ( Alon). Alonizing is very long ( > 24 hr) but gives a high integrity coating , can't remember the thickness.  As I remember the "diffuse" cycle takes a couple hours. I expect your process produced a thinner coating but still with some intermetallics.

Is these process related if my dipping time is only 15 secs?


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Determine reaction concentration at chemical equilibrium with multiple reactions
I have a problem understanding how to calculate the change/dependence of the concentrations at the chemical equilibrium using multiple reactions.
The concentration of my reactants and products at chemical equilibrium are given.
E.g.: For a simple reaction like $\ce{C + O <=> CO}$ at chemical equilibrium with concentrations $\ce{10C}$, $\ce{5O}$ and $\ce{2CO}$, I calculate the equilibrium constant $K_{eq} = 2/(5*10) = 0.04$. Then I can use the ICE table and calculate $x$ to derive the concentration of $\ce{CO}$ at chemical equilibrium at various input concentrations for the reactants:
e.g., for an input concentration of $\ce{22C}$ and $\ce{7O}$ (and $\ce{0CO}$) I get a concentration of about $\ce{3CO}$.
BUT how do I calculate the change of my products, if I have to consider multiple reactions with the same reactants involved?
E.g., $\ce{C + O <=> CO}$ and $\ce{C + N <=> CN}$ again with given concentrations at chemical equilibrium of $\ce{10C}$, $\ce{5O}$, $\ce{30N}$, $\ce{2CO}$, and $\ce{8CN}$.
These two reactions are obviously coupled, so solving them independently is not possible (without proper decoupling). I'm also not sure if the equilibrium constant for the reactions can be determined. If I just 'ignore' one of the reactions at a time and calculate the equilibrium constant using the remaining concentrations, I would neglect the coupling which can have a strong effect, e.g., an increase in the input concentration of $\ce{C}$ may influence one reaction by ~90% and have almost no influence at the other (depending on a saturation effect).
I'm also not sure if it is NOT allowed to throw all products and reactants together and just consider them as a single reaction like $\ce{10C + 5O + 30N <=> 2CO + 8CN}$, calculate the equilibrium constant and apply my variation of input concentration like in the example above. How would I account for the case N=0 and CO != 0?!
I think that the only reliable approach would be an iterative calculation of the concentration until convergence is achieved. But I don't have a clue how to do this.

EDIT: I have found part of the solution myself. As I mentioned above, I have to use a system of coupled equation and solve it. So here are my thoughts about it. There has to be conservation of mass (or elements), so the following equations have to hold:
$\ce{C + CN + CO -} nC = 0$
$\ce{N + CN -} nN = 0$
$\ce{O + CO -} nO = 0$
where $nX$ represents the concentration of a certain species. Second, the equilibrium constants also need to be specified like the following:
$\ce{CO - K_1*C*O = 0}$
$\ce{CN - K_2*C*N = 0}$
Solving these coupled equations numerically is no problem. I did some testing and can confirm that at least the uncoupled case ($nN = 0$) leads to exact the same results as the method above. Extending this case to include multiple and more complicated reactions should not be a problem. The only remaining question is how to determine the equilibrium constants for these coupled equations?!

EDIT: I used the same approach to solve the following system of equations for the equilibrium constants:
$\ce{K_2*K_1 - \frac{CN}{(C*N)}*\frac{CO}{(C*O)} = 0}$
$\ce{K_2/K_1 - \frac{CN}{CO}*\frac{O}{C}} = 0$
I used the solution for these equilibrium constants, plugged them into the equations above and calculated the concentrations at equilibrium. The solutions are consistent. (Note: Got myself confused and interchanged $K_1$ with $K_2$ within my computations.)

Conclusion To calculate variations of the input concentrations at chemical equilibrium 2 major steps are needed.
1) Calculate the equilibrium constants $K_x$ for every reaction $x$ at given concentrations. Solve that system of $x$ equations (details in 2nd edit) with a numerical solver of your choice.
2) Adjust/variate the input abundance and use another system of $z$ equations (conversation of mass/element and reaction equations - details in 1st edit) to calculate the new concentrations at equilibrium.
Note: Make sure to always have $n$ equations for $n$ variables.

Answer already found by myself. Short conclusion:
To calculate variations of the input concentrations at chemical equilibrium 2 major steps are needed.
1) Calculate the equilibrium constants $K_x$
for every reaction $x$ at given concentrations. Solve that system of x equations (details in 2nd edit) with a numerical solver of your choice.
EDIT: No numerical solver is needed, since the equations simplify to:
$\ce{K_1 - \frac{CN}{C*N} = 0}$
$\ce{K_2 - \frac{CO}{C*O} = 0}$
which is very interesting, since this simplifies the whole problem a lot. It also means that at chemical equilibrium a reaction does not care if a reactant is depleted via another reaction e.g., is bound within another product, or if the (initial) concentration of that reactant is just lowered to achieve the same amount in equilibrium. No reaction describes the connection between $\ce{CO}$ and $\ce{CN}$ e.g., $\ce{CO + N <=> CN + O}$, so basically the concentration of $\ce{CO}$ and $\ce{CN}$ ARE DECOUPLED (looking at the decoupled equilibrium constants), although they seem to be coupled via their common reactant ($\ce{C}$) e.g., changing $\ce{N}$ (or $\ce{O}$), changes the amount of produced $\ce{CN}$ (or $\ce{CO}$), changes the amount of $\ce{C}$ and further the amount of the other product $\ce{CO}$ (or $\ce{CN}$) .
This gives rise to a new question: How would a 'forced coupling' between $\ce{CO}$ and $\ce{CN}$ via $\ce{CO + N <=> CN + O}$ influence the concentration of both products?
Since we are at chemical equilibrium and have already taken 'indirect coupling via $\ce{C}$' into account, I guess that introducing a triangular-like reaction scheme would only introduce a redundant equation and won't give new information. Maybe I will test for that and share my results.
EDIT: Testing showed no difference in using a triangular-like reaction scheme (including the reaction above). As stated above, the 'new' reaction/equation does not contain new information. The 'new' equation can be written as a linear combination of the other 2 reaction equations. Therefore, one of these 3 reaction equations can be canceled/eliminated since 1) we need $n$ equations for $n$ variables and 2) an equation consisting of a linear combination of other equations cancels out (leads to a zero result) when solving the system of equations.
This means we only need to consider (in this case) the 3 equations for mass/element conservation and 2 reaction equations for the 5 variables (concentration of 3 reactants and 2 products).

2) Adjust/variate the input abundance and use another system of $z$
equations (conversation of mass/element and reaction equations - details in the edit of the question) to calculate the new concentrations at equilibrium.
Note: Make sure to always have $n$ equations for $n$ variables and don't mess them up.


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Is flow battery cycle life really as high as the literature claims?
I've seen a few reviews/journal articles now which claim that flow battery cycle life reaches the 10,000 mark and beyond with good capacity retention. However all of the primary sources I've came across only conduct tests for tens or hundreds of cycles, and there is almost always a significant capacity drop.
The general claim is that RFBs have a much better cyclability than solid state batteries such as lithium due to there being no physical change at the electrode, but I can't find any sources to back that up.
Can anyone show me a primary source which tests flow batteries (Preferably vanadium of iron chrome) for extended periods?


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Effect of temperature and pressure on structure of solids
What is the effect of these conditions on the structure of solids? My book has the answer that confuses me. I got nothing about it. Can you please explain? The answer of the book is

With the increase in temperature, coordination number of ion decrease. With the increase in pressure the coordination number of ions increase.

Roughly speaking, that is generally correct. Increasing temperature, which implies increasing atomic vibrations, might make a less dense, more open crystal structure the preferred option. Increasing pressure will favor small volume crystal structures, increasing the number of nearest neighbors.


text:
Good resource for stereochemistry of more than 4 substituents?
I have several classes of problems I'm facing in which I'd like to be able to discuss stereochemistry of non-tetrahedral centers. Specifically, I'd like to be able to check that the absolute configuration around a (for instance) trigonal bipyramidal transition state matches the absolute configuration that I've calculated.
For instance, these (chemically meaningless) trigonal bipyramidal structures are not superimposable:

What's a good resource discussing "higher order" stereochemistry? I'm particularly interested in classification of stereochemistries and in exchange rules for different orderings (analogous to the parity rules for tetrahedral centers). I'm coming from a computational/algorithmic slant.
The best (only) resource I've found so far is the opensmiles specification: http://opensmiles.org/opensmiles.html#_trigonal_bipyramidal_centers. I'm happy to purchase a book, or academic paper, or read program source code.

https://github.com/SiMolecule/centres which is being used in Jmol currently

@gilleain - That appears to work for tetrahedral, cis/trans, and axial. It does not seem to include trigonal bipyramidal or octahedral (as far as I can tell, examining https://github.com/SiMolecule/centres/blob/develop/core/src/main/java/com/simolecule/centres/Descriptor.java#L28)

https://chemistry.stackexchange.com/questions/5977/optical-isomerism-in-co-ordination-compounds

I'm not sure how easy this is to implement algorithmically, but for what it's worth, the official IUPAC recommendations are as follows.
Section P-93.3 in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) describes the guidelines for assigning configuration at non-tetrahedral atoms. The portions of this answer will be quoted from there.
These guidelines, and further details on certain aspects, are also discussed in Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) (link), section IR-9.3.

Stereodescriptors for nontetrahedral configurations are composed of three parts.
(1) A symbol indicating overall geometry called the 'polyhedral symbol';
(2) A symbol called the 'configuration index' that describes the configuration of ligands around the central atom;
(3) A symbol indicating the 'absolute configuration' associated with the central atom called the 'chirality symbol'.

For the first example you provided:

which is a trigonal bipyramid, the appropriate 'polyhedral symbol' is TBPY-5 (P-93.3.2, Blue Book). The numbers in red are the priorities according to the Cahn–Ingold–Prelog system, which will soon be required.
The 'configuration index' is

a series of numbers to identify where each atom or group is located. It is based on the CIP priority order of the atoms attached. The atom or atoms with highest priority are given priority number one '1'; the next priority number two '2', and so on. [...] (P-93.3.3.1)

For a trigonal bipyramid, this "consists of the priority numbers of the two apical atoms (lower number first if different), representing the reference axis of the system" (P-93.3.3.5.1). In this case, the two apical atoms have priority numbers 1 and 2, so the configuration index is simply 12.
The 'chirality symbols' available for a trigonal bipyramidal system are 'A' and 'C'.

The atoms in the plane perpendicular to the reference axis are viewed from the side with the atom or group of highest priority on the reference axis. If the direction from the atom in this plane with highest priority to the atom with the next higher priority is clockwise the chirality symbol is 'C', if anticlockwise 'A' (P-93.3.4.2.1).

Looking along the reference axis (which is the axis containing the axial ligands), from the direction of the higher priority atom (oxygen), the equatorial ligands decrease in priority going in a clockwise order. So, the chirality symbol is C. (Essentially, this is analogous to assigning an R-configuration for an ordinary chiral carbon.)

The three parts of the stereodescriptor are enclosed in parentheses and separated from each other by hyphens and cited at the front of the name separated from it by a hyphen. (P-93.3.1)

The full stereodescriptor is therefore (TBPY-5-12-C).


text:
SO3(2-) to Na2S2O5 ratio
Is there a way of calculating the amount of sodium metabisulfite from the sulfite content?
I found online instructions for the test kit for sulfite that suggests multiplying by $2.37$ to obtain $\pu{mg/l}$ of $\ce{Na2S2O5}$ from the $\pu{mg/l}$ of $\ce{SO3}$ found in a sample.
Where did they get that $2.37$ from?
This is the formula I came up with:
$$\frac{\pu{mg/l} (\ce{SO3}) \times \text{total volume} \times M(\ce{Na2S2O5})}{M(\ce{SO3})}$$
i.e.
$$\frac{\pu{5 mg/l} (\ce{SO3}) \times \pu{100 L} \times \pu{190.107 g/mol}}{\pu{80.063 g/mol} \times 100} = \frac{5 \times 100 \times 190.107}{80.063 \times 100} = \pu{27.306 g} (\ce{Na2S2O5}) \text{ in 100 liters}.$$

Where's the online that indicates 2.37?? What exact ratio do you need? SO3:S2O5?


text:
What does it mean to say that a strong base is only slightly soluble?
I've often seen the hydroxides of Calcium, Strontium and Barium described as "strong bases that are only slightly soluble," or similar.  But strong bases dissociate completely in solution, so I would guess this means that such hydroxides dissolve only a small amount, but the dissolved portion dissociates completely.
But the problem with this interpretation is that these hydroxides are ionic compounds, and my understanding is that "dissolve" is essentially equivalent to "dissociate" for ionic compounds, i.e. ionic compounds dissolve by dissociating.
If there's no distinction between dissolving and dissociating for ionic compounds, what does the description "strong base that is only slightly soluble" mean?  How does it differ from the term "weak base?"
If there is a distinction, and it is possible for ionic compounds to dissolve without dissociating, how does such dissolution work?  I have a model for the distinction between dissolution and dissociation of covalent compounds: first the compound dissolves into separate molecules, then each molecule dissociates into two or more ions.  If ionic compounds can dissolve without dissociating, into what sort of particles do they dissolve? Formula units? Small clusters of ions?  Something else?

And you guess well. What IS dissolved IS fully dissociated. Stop here and you see your puzzle is solved :)  (being fully dissociated does not require that more material, then, has to dissolve)

@Alchimista If an ionic compound can dissolve without dissociating, how do we model the dissolution process?  Into what sort of particles does the compound dissolve?

No one says that! I say that IF it get dissolved, THEN (implies) it is dissociated. If it would be more, it will dissociate. But there is not more! That when it in liquid phase it is dussociated does not imply that it has to be in liquid phase. You are mixing two equilibria here. Between solid hydroxide and its ions in solutions (Kps) AND and the one in acid-base reaction. They are two different things. Of course when your hydroxide does not dissolve anymore, if you add acid you force it to do it (Le Chatellier). This is what links the two equilibria. Look also at answer by a-cyclohexane-mol..

You know what, read about coordination chemistry then aquocomplexes and polycentric clusters and you'll probably get what happens with cations in solutions.

Solubility is different from dissolution, which is itself different from dissociation.
Solubility tells you the extent to which a substance dissolves. Dissolution is the kinetic process of solvation, when solvent molecules interact with solute molecules or ions. Dissociation is, from IUPAC,

The separation of a molecular entity into two or more molecular entities (or any similar separation within a polyatomic molecular entity). Examples include unimolecular heterolysis and homolysis, and the separation of the constituents of an ion pair into free ions.
The separation of the constituents of any aggregate of molecular entities.

Dissociation is the mode of solvation for ionic compounds.

I think your confusion stems from associating solubility and dissolution improperly. ChemGuide directly answers your question here:

Some strong bases like calcium hydroxide aren't very soluble in water. That doesn't matter - what does dissolve is still 100% ionised into calcium ions and hydroxide ions. Calcium hydroxide still counts as a strong base because of that 100% ionisation.

Ok.  But what does dissolution of consist of, if not dissociation?  In the case of sucrose, I have a good model for understanding dissolution and dissociation as separate processes.  Dissolution is the process of breaking down into separate sucrose molecules, whereas dissociation involves the further breakdown of sucrose molecules into ions.  If Ca(OH)₂ can dissolve  without dissociating, what are the particles into which it dissolves?  Formula units?  Small clusters of ions, larger than formula units but still microscopic?  Something else?

*If Ca(OH)2 can dissolve without dissociating : It CANNOT. That is the simple point, and you more or less got it, in the first part of your question

@Chad, that's a fair question. I think my answer is quite misleading as-is, and I'm quite uncertain as to how well $\ce{Ca(OH)2 (aq)}$ holds up under scrutiny, but let me edit my answer and see if it resolves your concerns.

@Alchimista What about a compound like Fe(OH)₂, which is commonly described as a weak base, not a "strong base that is only slightly soluble."  Can Fe(OH)₂ dissolve without dissociating?  If so, how does that process work?  If not, what is the distinction between a weak base and "strong base that is only slightly soluble?"

@Chad, I've removed the misleading analogy with the mysterious $\ce{Ca(OH)2 (aq)}$ and hopefully clarified the distinction between dissolution and dissociation. Can you provide a reference as to where you see $\ce{Fe(OH)2}$ being described as a weak base?

Iron II hydroxide does not dissolve in water. Which brings us to my comment were I mentioned Le Chatelier. The very small amount that goes in solution dissociates, according to the low Kb. Perhaps the fact that there are not molecules in ionic compounds makes you think that there are not small particles of them.... There are of course, down to the smallest size one can ascertain, which is likely not well known. I think while writing this I got what blocked you. Yes you can have aggregates of Fe(OH)2 floating in solutions. Microscopically speaking. It is the trait d'union with your solid excess

Useful request @a-cyclohexane-molecule - I don't see Fe(OH)₂ being described as a weak base!  I thought I had, but I was wrong.  As I look around I do see some references to other transition metal hydroxides described as weak bases.  AgOH, for instance.  See http://herh.ccrsb.ca/staff/FarrellL/Chem12/acids/acid-base-theory-notes.pdf

@Alchimista So would it be correct to say that ALL metal hydroxides are strong bases (in the sense that whatever portion of them dissolves also dissociates), but the transition metal hydroxides like Fe(OH)2 are negligibly soluble?

It seems a quite general statement: OH- which is in both Kb and Kps. Then we likely run into formal hydroxides with covalent bonds. Let us try solve clearly your original question first. To me the answer is that by a cyclohexane mol. Perhaps tomorrow I ll post an answer, at the moment I am getting crazy with the small box in the app. Just let me know if shall I refer to the original question although edited or you came up with a new one or a new generalisation. See you asap

Qualitative descriptors such as "slightly soluble" usually refer to the maximum amount of a substance that can be dissolved in water. The use of the word "slightly" indicates that, for example, less $\ce{Ca(OH)2}$ can dissolve in a given volume of water than a comparable molecule like $\ce{NaOH}$ in the same amount of water.
If you were to continually add small amounts of $\ce{Ca(OH)2}$ to 1L of water (numbers are just picked at random) at some point no more $\ce{Ca(OH)2}$ would dissolve. All the $\ce{Ca(OH)2}$ that dissolved at that point DID completely dissociate, but after the solution is saturated, more $\ce{Ca(OH)2}$ will not continue to dissolve. After this point, the excess $\ce{Ca(OH)2}$ that you added would just sink to the bottom of the beaker and stay there.
Note: My description relates to what is observable in the macro view. The microscopic view shows that an equilibrium  between the solid and ions will be established once the saturation limit is reached. The equilibrium will not cause the average concentration to exceed the saturation limit.

Actually, when I learned my strong bases in high school my teacher told me that those "slightly soluble" Ca(OH)2 etc were not strong bases.
But I think the idea is even though they are only slightly soluble, they are still more soluble than other even weaker bases such as say Fe(OH)3.
All solids dissociate completely until a saturation point; they do not dissolve proportionally like aqueous acids since the concentrations of the solids (concentration is not amount!) do not change.  So, what you call a "strong" base is based on how strong you have to get.
List of Ksp (dissociation constant; generally greater means more dissociation):
Aluminum hydroxide  Al(OH)3 1.3×10–33
Barium hydroxide    Ba(OH)2 5×10–3
Cadmium hydroxide   Cd(OH)2 2.5×10–14
Calcium hydroxide   Ca(OH)2 5.5×10–6
Chromium(II) hydroxide  Cr(OH)2 2×10–16
Chromium(III) hydroxide Cr(OH)3 6.3×10–31
Cobalt(II) hydroxide    Co(OH)2 1.6×10–15
Copper(II) hydroxide    Cu(OH)2 2.2×10–20
Iron(II) hydroxide  Fe(OH)2 8.0×10–16
You can find more at:
http://webcache.googleusercontent.com/search?q=cache:http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm

@Chad. I should have commented below, not here. Sorry. Where is the problem with the equilibria AB(solid) double arrow AB(solv) double arrow A(solv) + B(solv)? In case of full dissociation you end always with higher solubility, as is indeed the case. Yes you have a neutral entity involved, the smaller one or as cluster, within their solvation sphere. This should be experimentally verified. Anyway you are dealing with the Kps value.

I have to admit that jumping between the various parts of the question as well as the editing/comments almost confused me, or at least gave me some discomfort. So I try to clarify the situation although I am fond of reading some of the previous answers, in particular that given by a_cyclohexane_molecule. To this end, I go through your text. Please note that capital is not shouting but an attention driver.
1 - "But strong bases dissociate completely in solution, so I would guess this means that such hydroxides dissolve only a small amount, but the dissolved portion dissociates completely."
Yes, this is the meaning and what happens.
2 - ".... problem with this interpretation is that these hydroxides are ionic compounds, and my understanding is that "dissolve" is essentially equivalent to "dissociate" for ionic compounds, i.e. ionic compounds dissolve by dissociating.
Not completely right. The ACTUAL amount of THESE hydroxides - NaOH and Ca(OH)2 - that goes in solution (solubility) is fully dissociated. It is again point 1), that you inadvertently     generalise to all hydroxides, whatever their base strength is.
Think of Me(OH)n in water. Me ions coexist with Me(OH)n-1, Me(OH)n-2 etc. positively charged ions (according to the stoichiometry). In which sense the dissolved hydroxide is fully dissociated? It is not, even if you postulate that what goes in solution undergoes at least one complete first dissociation process, i.e. no neutral entities can dive in solution (I think this is matter for a separated question. We are in the case "to which extent a bond is ionic?", at the end. Or even as to ask what is the biggest neutral agglomerate of ionic compounds that can be filtered out of a solution. It is not a coincidence that most hydroxides are tricky to handle :) Unfortunately real things rarely fit in absolute definition).
3 - " if there's no distinction between dissolving and dissociating for ionic compounds, what does the description "strong base that is only slightly soluble" mean? How does it differ from the term "weak base?"
Here is where some definitively confusion comes in. First question in the paragraph. We answered already. There is a distinction, as above. 100% dissociation does NOT  imply that the same amount is in solution. Microscopically, let us take for granted that each whatever small portion of the compound that gets dissolved does dissociate. It does ONLY IF gets dissolved, AND IF is not like the case Me(OH)n above.
Second question in the paragraph. Let me stay in the realm of Arrhenius: a strong base definitively gives you a solution with a high pH.
Differently, a very weak base does not generate a considerable amount of OH-, their final concentration it is about the natural one in water (from Kw). (You get see this in B-L or L theories, too, and you need those theories to rationalise the behaviour of ammonia, for instance).
3 - " If there is a distinction, and it is possible for ionic compounds to dissolve without dissociating, how does such dissolution work? I have a model for the distinction between dissolution and dissociation of covalent compounds: first the compound dissolves into separate molecules, then each molecule dissociates into two or more ions. If ionic compounds can dissolve without dissociating, into what sort of particles do they dissolve? Formula units? Small clusters of ions? Something else?"
I think that answer 2 applies here, too.
The point is that if you stick to hydroxides (except alkaline and earth-alkaline), metal and amphoteric hydroxides are weak bases because

are so sparingly soluble in water that by a practical, Arrhenius based viewpoint, do not gives considerable amount of OH-. This will be true even assuming their complete dissociation!
a finer analysis of their equilibria in water does reveal the coexistence of different degree of dissociation, so they can be classified as weak indipendently of their limited or negligible solubility.

Although the OP did not have troubles in cases in which he could easily identify the covalent molecule that goes in solution without (or better, partially) dissociate, let me conclude by pointing to the fact that a weak base it is not necessarily sparingly soluble (as for a weak acid it is not necessary sparingly soluble), as this will not leave room for  ammonia (acetic acid) etc.
Ergo we still need a definition of weak bases independently of their solubility.

I'm still trying to sort out the question "If ionic compounds can dissolve without dissociating, into what sort of particles do they dissolve?"  I get that Me(OH)n can exist in equilibrium with Me(OH)n-1 positively charged ions.  Me(OH)n dissolve into individual particles of Me(OH)n before dissociating into Me(OH)n-1 and hydroxide.  If the metal hydroxides (other than alkaline and alkaline earth) can have "different degrees of dissociation... independently of their limited or negligible solubility," then do they first dissolve into some particle before dissociating?

There are a couple of things you need to carefully think about. You claim that Ca(OH)2 is an ionic compound. Yet I doubt you believe it exists as separate ions of Ca++, O=, and H+. If that is your belief, you are wrong. If it isn't your belief, then you probably recognize that the anion OH- is covalent. Allow me to give you a counter example: NH4OH exists in aqueous solution mostly as NH3 molecules. [perhaps this isn't the best counter-example, since ammonium hydroxide isn't a stable isolable chemical compound; perhaps I should have used tetramethyl ammonium hydroxide, (CH3)4OH instead]. One (counter-)example disproves the assertion. Solubility of "ionic" compounds does not necessarily require dissociation. Since water is such an excellent solvent for screening electric charges, water solutions of ionic compounds are often, perhaps even usually, dissociated. But water is only one of (nearly) countless solvents. (Granted, liquid water comprises over 70% of the surface of this planet, so it's arguably the most important solvent as far as you and I are concerned.) Here's another example: acetic acid. It isn't a strong acid (pKa ~4.76) and neither is it ionic (but, obviously, it does ionize in (aqueous) solution. Are there weak ionic bases which are highly soluble in water? Hmmm. If we agree that anything which contains the hydroxyl radical can not be purely ionic, then I'd have to do a bit of searching to see if I could find a binary (purely ionic) compound which was itself soluble and yet not strongly dissociated in aqueous solution. Hmmm. pKb of LiOH is ~0 meaning LiOH exists in aq. solution as a molecule (but not purely ionic, but certainly purely inorganic). I guess another way to answer your question is to agree that for purely ionic compounds (not that common, as it turns out) dissociating and dissolving in aqueous solution is common, but not necessary. But in general ionic shouldn't be confused with inorganic, dissolved shouldn't be confused with dissociated (nor ionized) and solvent shouldn't be confused with water.

My understanding is that Ca(OH)₂ is an ionic compound composed of one Ca²⁺ ion and two OH⁻ ions.  Each OH⁻ is itself a polyatomic ion held together by covalent bonds.  Are you saying that Ca(OH)₂ can dissolve without dissociating into Ca²⁺ and OH⁻?  If so, what are the particles into which it dissolves?

@Chad after so long time. If you put the smallest grain of Ca(OH)2 in a saturated solution, it will stay as it is. The point is that it will find its companions and separate out. In such a sense is tricky to speak of little solvated particles of ionic compound in solution but not dissociated. Opposite than purely covalent molecule, there is still energetic room for them to build a solid lattice.


text:
How does a molecule search work by drawing in visual editor?
There are a lot of chemical search engines (such as ChemSpider's
Structure search) that allow users to draw and search for a molecule. How does it work? Is the comparison of images used? Could you give me an algorithm or pseudo code?

That title had me confused a bit. How does a molecule search work? It doesn't, it molecules...

This site gives a pretty in depth explanation on how these work and has a series on how to develop such a search engine

Usually, structure search is implemented using a graph model. For example, let's take cyclohexane. On screen this is a hexagon - but this is converted on the server to a simple atom/bond graph model of six carbons in a ring.
This graph can then be compared to all the molecules in the database by various means. For example, subgraph isomorphism or fingerprints or even SMILES string comparison.
As far as I know, ChemSpider allows a variety of searches - with/without hydrogens, substructure search, element fingerprints, and so on. Each of these is a whole question by themselves, so I'm not sure how useful an algorithm that covers them all would be.

Using image based search wouldn't work very well since the same compound can be drawn in a lot of different and correct ways. However, it is possible to convert drawn structures or even 3D structures into text, for example using SMILES or InChI. Those can then be used for standard text based database search.

ChemSpider likely has an algorithm that takes the image and converts it to a SMILES string. It would likely have to do so in order to avoid problems in how the structure is drawn as @DSVA pointed out.
For example, if there's a large molecule (where there could be on the order of billions of slight conformational differences), it would be nearly impossible to find a perfect match in the database. However, by considering only the connectivity of atoms (as a SMILES string would achieve), the software could find a match nearly instantaneously.
ChemSpider has an option on their website to search using a SMILES string, so it's known that their software is capable of recognizing SMILES strings. However, I was unable to find confirmation as to whether SMILES is used as a search method as their software does not appear to be open-source.


text:
Flory–Schulz type distribution with stoichiometric imbalance
In a typical step-growth polymerization, the Anderson–Schulz–Flory distribution is the probability mass function that describes the number fraction or weight fraction of polymers of length $x$ at a given extent of reaction $p$.  The extent of reaction, $p$, is a value between 0 and 1, and $x$ is the degree of polymerization for any given oligomer or polymer present in the system.
$$\begin{align}
\text{mass fraction} &= p^{x-1} \cdot (p-1)^2 \\
\text{weight fraction} &= x \cdot p^{x-1} \cdot (p-1)^2
\end{align}$$
A nice derivation of these equations can be found in this video.  Figure 1 illustrates a plot of the number fraction distribution as a function of the degree of polymerization for several extent of reactions.  Figure 2 illustrates the plot of the weight fraction distribution as a function of the degree of polymerization for several extent of reactions.

Note that high extents of reaction in step-growth polymerizations result in very broad distributions, and are also required for high molecular weight polymers.
Carothers' equation describes how stoichiometric imbalances necessarily limit the possible degree of polymerization:
$$x_\text{average} = \frac{1+r}{1 + r - 2rp}$$
where $x_\text{average}$ is the average degree of polymerization, $p$ is the extent of reaction, and $r$ is the mole fraction of the reacting functional groups. $r$ is always a number between 0 and 1.  The excess functional group is always taken to be in the denominator. When $r$ is equal to 1, stoichiometric balance exists, and the equation reduces to:
$$x_\text{average} = \frac{1}{1-p}$$
For example, if I react a diol and a diacid under step-growth conditions:

My goal is to purposely limit the degree of polymerization of a chemical reaction by introducing a stoichiometric imbalance in order to form short chain oligomers.  I'd like to predict/calculate the statistical distribution of molecular weights for a given extent of reaction, p and stoichiometric imbalance, r.
I'm having trouble convincing myself that the Flory–Schulz distribution is appropriate for modeling step-growth systems that contain stoichiometric imbalances. Is there a straightforward way to incorporate stoichiometric imbalance into these distribution functions? Any help would be appreciated.

I came up with a way to solve my own problem as I was writing my question, but I feel as though it is a bit hand-wavy.  I would appreciate any input regarding the validity of this solution.
The following is a workaround to force the Flory distribution to have a maximum value that is consistent with the average degree of polymerization due to stoichiometric imbalance via Carothers' equation.  Carothers' equation models how stoichiometric functional group imbalances limit the average degree of polymerization for the step-growth mechanism.  This average degree of polymerization $(\text{DP}_\text{average})$ can be calculated very simply:
$$\begin{align}
\text{DP}_\text{average} &= \frac{r+1}{r+1-2rp} \\
\end{align}$$
When there are no stoichiometric imbalances, and the Flory distribution is valid:
$$\text{DP}_\text{average} = \frac{1}{1-p}\\
f(\text{DP}_\text{average}) = x \cdot p^{x-1} \cdot (p-1)^2$$
Let $p_\text{eff}$ be the effective extent of reaction, which is related to $r$ and $p$ via:
$$\frac{1}{1-p_\text{eff}} = \text{DP}_\text{average} = \frac{1+r}{1+r-2rp}\\
p_\text{eff} = 1 - \frac{1+r-2rp}{1+r}$$
This $p_\text{eff}$ can be used to calculate the ideal Flory distribution according to
$$F(\text{DP}_\text{average}) = x \cdot p_\text{eff}^{x-1} \cdot (p_\text{eff}-1)^2$$
The function $F(\text{DP}_\text{average})$ will force the molecular weight distribution to be properly centered about the  $\text{DP}_\text{average}$ that is consistent with Carothers' equation.  I'm not sure if this is chemically sensible, or accurately describes a real-ish polymer system.  Comments/alternative answers would be appreciated.


text:
Aromatic and aliphatic amines
Is benzylamine $\ce{C6H5CH2NH2}$ an aromatic or aliphatic amine? I've tried searching for it however different sources have conflicting information.

The nitrogen is bound to a benzyl group, so it would most aptly be classified as a benzylic amine. The benzyl group is a common protecting group for amines, and can be removed via hydrogenation with a palladium catalyst. As for the basicity of the nitrogen, it would be most similar to an aliphatic amine.


text:
Is it possible to improve electrolysis rate without increasing heat?
I'm separating water into hydrogen and oxygen by connecting a DC current to 99.9% pure nickel electrodes in a solution of NaOH in deionized water.
It works great, except that any attempts to increase the rate of production by increasing current, increasing NaOH, bringing electrodes closer together, etc. also result in an increase in heat.
At the moment the bottleneck for my production rate is trying to keep the water from boiling or otherwise causing heat damage to the apparatus.
I've considered investing in a cooling system, but I'm also interested in knowing whether or not it is possible to somehow improve the efficiency of the electrolysis and thereby improve separation rate without increasing heat.
(In case you're wondering, I'm using hydrogen to fill balloons.  It works quite well, I can fill a normal sized balloon in a few minutes, but then I need to wait an hour for everything to cool down.  (Yes I know hydrogen and fire don't mix well.))

Consider the electrolysis apparatus as a resistor in series with an EMF. Ideally, the cell voltage should be just enough to overcome the EMF (potential needed to split water), but some over-voltage is inevitable to have a decent current flow. Much of the heating is due to this wasted over-voltage. Then, to increase the efficiency and reduce heating, it's necessary to reduce the series resistance.
$\ce{NaOH}$ is already highly ionized, so adding more, beyond a certain point, is not the solution (no pun intended;). Close spacing helps, but you must leave room to physically separate $\ce{H2}$ from $\ce{O2}$. The remaining variable is electrode area: make the electrodes as large as feasible. You can further increase area by using porous sintered, or at least coarsely-sanded electrodes.
Quantum Sphere Inc. claims to produce a more active electrode from $\ce{Ni-Fe}$ particles, though they're probably a bit pricey for filling hydrogen balloons. Perhaps they offer samples?
BTW, I assume you're using a moisture trap for the gas, otherwise bubbling hot caustic electrolyte would inadvertently wind up in the balloon, reducing lift, damaging the balloon and possibly causing severe chemical burns. $\ce{NaOH}$ is called "caustic soda" for good reason!

I know, everyone says that you only need 1.3 volts or whatever to electrolyse water, but I've tried 5V vs 12V and there is much, much, more volume of gas produced at 12V.  I'm using large nickel sheets so surface area is good.  People also say that you only need a spoonful of NaOH, but generally the more I add the more gas (and heat) I produce.

What is a moisture trap?  I ran the output through 2 meters of tubing submerged in cold water to reduce water vapour in the product.  Although it doesn't really catch much so I don't really bother anymore.

A moisture trap can be as simple as a tube filled with glass wool to catch the condensing water vapor and "spitting" from a reaction such as electrolysis, or the tubing you mentioned. For dryer H2, the tube can be filled with CaCl2 or other desiccant, which is not needed for balloons. See https://www.thevespiary.org/rhodium/Rhodium/chemistry/filt/filtration_files/vacfilt7.jpg for an idea of how to make one from a flask. Keep the trap small, and preferably of plastic rather than glass, to reduce mixing the H2 with residual air and to reduce shrapnel in the event of hydrogen igniting.

Oh, right, I gotcha, I remember that water absorbing setup from chemistry class.  I have a load of silica gel packets, will  those work instead of CaCl2?

Yes, even better than CaCl2, in that it won't dissolve in condensed moisture, and is easily dried for reuse in an oven.


text:
Do almonds smell like they do because of cyanide?
I am guessing so otherwise it is quite a coincidence that cyanide smells of bitter almonds. And if so, are large quantities of almonds over a long period safe?

Strongly related: https://chemistry.stackexchange.com/questions/47204/how-do-people-know-hcn-smells-like-almonds

Almonds smell like they do mostly due to the presence of benzaldehyde:

This colorless liquid has a characteristic almond-like odor. Benzaldehyde is the primary component of bitter almond oil and can be extracted from a number of other natural sources.

Hydrogen cyanide also has an almond-like odor, but it is not as pronounced as that of benzaldehyde. From CDC|Facts About Cyanide:

Cyanide sometimes is described as having a “bitter almond” smell, but it does not always give off an odor, and not everyone can detect this odor.

Benzaldehyde and hydrogen cyanide are both byproducts of the enzymatic catalysis of amygdalin, which is present in almonds (and other foods). Have a look at this abstract (Toomey VM, Nickum EA, Flurer CL. Cyanide and amygdalin as indicators of the presence of bitter almonds in imported raw almonds. J Forensic Sci. 2012 Sep 57 (5): 1313-7) for more about the cyanide-amygdalin-almond connection.
In terms of safety:

For a 70-kg human, the lethal dose is estimated at 50 mL. Benzaldehyde has been classified as a hazardous substance by the United States Environmental Protection Agency.

and

Benzaldehyde does not accumulate in any specific tissues. It is metabolized and then excreted in urine.

Safety issues from benzaldehyde consumption $\it{via}$ the almond consumption route appear to be minimal.
From our sister SE site Skeptics in re: "How many almonds does it take to get cyanide poisoning?":

With bitter almonds, 8 - 32 almonds will give you the lethal dosage of cyanide. Bitter almonds yield about 6.2 mg of cyanide per almond and the LD50 for cyanide is 50 mg - 200 mg.
This applies only to bitter wild almonds: considering that you're not dead, you most likely ate domesticated sweet almonds, which apparently do not have this problem. The sale of wild almonds may be illegal, in fact, though I can only find blogs and internet comments alluding to this.
From wikipedia
Wild almonds are bitter, the kernel produces deadly cyanide upon mechanical handling, and eating even a few dozen at one sitting can be fatal.
  ...
While wild almond species are toxic, domesticated almonds are not

Given the evidence and the expectation that you won't ingest several kilograms of almonds in a day or so, I'm comfortable in endorsing the almond as safe for consumption.

But does HCN smells almond too? If it is indeed so, it is curious that both byproducts of amygdalin catalysis smell the same. I just ask you because I was told that HCN smells like almonds all along my chemistry education. Thanks.

@Alchimista - Yes, $\ce{HCN}$ does have an almond-like smell, but it is faint, not detectable by all people, and not the primary source of the characteristic smell of almonds. I've amended my answer to address your (good) point.

As someone who has smelled a lot of HCN during experiments I can tell you that the smell of HCN is nothing like the smell of Benzaldehyde or any almond oils you can buy as flavors. I never really smelled real fresh almonds and if you look a bit further into literature it says it may smell like really fresh almonds (not the ones you buy) so I cannot really compare but its nothing like benzaldehyde at least. It is very characteristic though. I could easily tell if something is HCN or not, it's like with vanilla for example. You don't confuse that smell with any other like you would do for various solvents. Also what literature also does not mention that often is that this characteristic smell of HCN only applies for low conc. you should not forget that it is still a Pseudohalogen and it has the same property as for example HCl. If you get a higher conc. at once it will burn in your nose without any characteristic smell at all, just like when smelling on a bottle of HCl. I once had the unfortunate accident to get a small cloud of higher HCN conc. into my nose when the pressure in my setup became to big.
Also to the original question, I doubt almonds smell like cyanide or vice versa. A friend of mine is working with really big molecules and sometimes he has to attach small reactive groups so he can link them. He was working with two very big aromatic compounds and added an -SH group to both of them. One then started to smell incredibly intense while the other one didn't change at all. Sometimes this is just how we attach a smell to something. So perhaps there is a small relationship that they bind to similar receptors, or the people who extracted HCN from almonds for example didn't fully clean it and the smell of the almonds somehow contaminated the experiment?

Interesting. Maybe a myth that continues due to few people having experience smelling cyanide.

I made the experiment (went to the pharmacy to buy bitter almonds for the purpose): the benzaldehyde smell masks the cyanide smell for me when mashing the bitter almonds with water. When mashing with dilute sulfuric acid, I smell the cyanide before benzaldehyde takes over. And yes, also for me the HCN and benzaldehyde are very different smells.

@Jeff: and instead relying on Agatha Christie...

@cbeleites: If you are joking, that is amazingly funny.

@Jeff: you mean about Agatha Christie? Partially, yes. I'm sure there are lots of (crime) stories around not by Agatha Christie and also making this link "smells like almonds => this is hydrocyanic acid". But OTOH, the scene that comes to my mind with this scentence ;-P is in a Margaret Rutherford Miss Marple movie from the 60s (don't remember which one - may be "Murder she said").

The joke to me is that I would try the chemical experiment you suggest but rereading it, it sounds like you actually did this. No idea you could buy bitter almonds in a pharmacy and why would they stock them?

However, if you are wondering about the bitter almond experiment: before that I already knew a) that I am able to smell hydrocyanic acid (solid NaCN does have a distinct cyanide smell for me). b) actually a fraction of a bitter almond is quite enough for the experiment c) but not enough to pose danger to an adult (I think neither for a kid, but I'd calculate again before the experiment ;-).

Bitter almonds here (Germany) used to on sale in small packages in grocery stores/supermarkets for baking. Package size was restricted ot 50 g maximun in the (1960s after some cases of accidental poisoning)[http://www.spiegel.de/spiegel/print/d-46176140.html] (btw that article says 5-10 pieces can be dangerous for a kid, for an adult the limit is aroung 60 pcs ≈ 60g). Bitter almonds are very bitter, so accidental poisoning is very rare (and also for the case described in the newspaper it is unclear why the lady ate the bitter almonds). Nowadays, supermarkets rarely have it, and in the small...

... town was no dedicated spice store (they'd have it as well), but someone in the supermarket told me to try the pharmacy. And they did have it. Note that pharmacies here  used to have (maybe still have, in rural areas) also e.g. nitrate and nitrite for mixing curing salt. So in a way they still are in some cases the local "chemist". BTW: the pharmacist said that in her opinion selling bitter almonds in the pharmacy instead of the supermarket is overkill - but people do associate them with poison (and don't realize that far more kids are poisoned by lamp oil from the supermarket...)


text:
Why do dithiins have 8 π-electrons?
It is mentioned on Wikipedia that dithiins have 8 π-electrons. Why is this so?
I was thinking it would suffice for just one of the lone pairs on one of the sulfur atoms to participate in resonance delocalization and hence give dithiin a total of 6 π-electrons, but why does the other lone pair on the other sulfur delocalize as well to give a total of 8 π-electrons?

"..suffice for just one of the lone
pairs on one of the sulfur atoms to participate.. " Why would one Sulfur atom donate it's lone pair & the other one won't?

Hi Mockingbird, I was thinking that if only one sulfur were to donate its lone pair, the molecule would be aromatic, and hence the donation of another lone pair by another sulfur would destroy aromaticity, hence being energetically unfavorable.

If only one sulfur atom were to donate a lone pair, then aromaticity is nevertheless broken by the lack of delocalization across the other sulfur atom.

TL;DR Ab initio HF molecular orbital studies of the potentially antiaromatic $8\pi$-electron systems of distorted and planar 6-membered rings revealed that conformational rigidity (calculated or measured folding energies) is most likely the best extension to a $(4n+2)\pi$ criterion. 1,4-Dithiins experience large energy decreases when permitted to fold, and when no external strains are applied, there is no reason for sulfurs to unevenly donate electronic pairs.

Crystal structure of 1,4-dithiin [1] is indeed suggesting a non-planar molecule with the fold angle between "red" and "blue" planes (both plotted on the mirrored $\ce{-S-C=C-S-\,}$ fragments) of $137.14^\circ$:

Same goes for the structure of (1,4)-Dithiino(2,3-b)-1,4-dithiine [2] with the dihedral angle of $131.03^\circ$:

The chair conformation of the molecule prevents conjugation, and the C-C bond distances clearly corresponds to normal unconjugated bonds, (simple double bond = 1.337(6) Å) so the molecule is non-aromatic in the solid state.

Theoretical calculations ($\Delta_fH^\text{MNDO}$) suggest that both molecules remain bent not only in crystal structure, but they also are flexible in gas phase and in solution [3].
At this point one might think that geometrical conformation dictates the aromaticity or the lack thereof, which is in my opinion an example of violation of the law of cause and effect. Under certain "strains" 1,4-dithiins can exhibit planar structure, like in case of 1,4-dithiine-tetracarboxylic N,N'-dimethyldi-imide acridine [4], where $\pi$-stacking causes a coplanar arrangement of both molecules:

The intermolecular distances ... are shorter than the van der Waals distance (3.40 Å). This suggests that $\pi-\pi$ interactions between molecules ... exist in these parts and the molecules form a paired molecular group. The dihedral angle between the least-squares planes of molecules ... is $2.3^\circ$}.
  Distances between paired groups are larger than the van der Waals distances... The stacking of paired molecules is in accordance with the poor electrical
  conductivity of this complex.
Anti-aromatic 1,2-dithiins display properties opposite to those of 1,4-dithiins, whose dications show aromatic stabilization. Unlike other antiaromatic compounds, the 1,2-dithiin derivatives, with eight $\pi$-electrons ..., appear in natural products synthesized by composite plants. The Dewar resonance energies of nonplanar 1,2-dithiins are close to zero. The aromatic 1,4-dithiin dication is generated by two-electron electrooxidation of the neutral 1,4-dithiin. The aromaticity was established by NMR measurements and theoretical calculations.

Conformational criterion for aromaticity and antiaromaticity has been extensively investigated for a number of $8\pi$-electron systems [5], and for 1,4-dithiin the energy required to force it into planar geometry is less than $\pu{2.7 kcal/mol}$ (max. value of 3-21G folding energy), allowing for a high degree of conformational flexibility. This seems to be one of the most recent methods of quantitative determination of both flexibility and aromaticity, as authors propose:

Relative to planar reference geometries of maximum symmetry, folding of aromatic and antiaromatic systems results in large endothermic and exothermic changes, respectively. Aromatic compounds are planar and strongly resistant to
  folding, whereas antiaromatic systems exhibit large energy decreases when permitted to relax (fold) from planar geometries to nonplanar, minimum-energy conformations. According to this criterion, systems ... which have
  shallow folding energy curves, are nonaromatic.

Of course, $6\pi$-aromaticity can be achieved via electrooxidation of non-aromatic 1,4-dithiine, which leads to a planar dication which exhibits aromatic character [6]:

Bibliography

Howell, P. A.; Curtis, R. M.; Lipscomb, W. N. Acta Cryst, 1954, 7 (6–7), 498–503. DOI 10.1107/S0365110X54001582.
Varma, K. S.; Sasaki, N.; Clark, R. A.; Underhill, A. E.; Simonsen, O.; Becher, J.; Bøwadt, S. Journal of Heterocyclic Chemistry 1988, 25 (3), 783–787. DOI 10.1002/jhet.5570250314.
Bock, H.; Roth, B.; Lakshmikantham, M. V.; Cava, M. P. Phosphorus and Sulfur and the Related Elements 1984, 21 (1), 67–77. DOI 10.1080/03086648408073129.
Yamaguchi, Y.; Ueda, I. Acta Cryst C, 1984, 40 (1), 113–115. DOI 10.1107/S0108270184003218.
Podlogar, B. L.; Glauser, W. A.; Rodriguez, W. R.; Raber, D. J. J. Org. Chem. 1988, 53 (9), 2127–2129. DOI 10.1021/jo00244a059.
Balaban, A. T.; Oniciu, D. C.; Katritzky, A. R. Chem. Rev. 2004, 104 (5), 2777–2812. DOI 10.1021/cr0306790.

Quoting directly from https://en.wikipedia.org/wiki/Dithiin:

Being antiaromatic, they [meaning 1,4 dithiins] are nonplanar and can
  be oxidized to their radical cations.

If 1,4 dithiins are nonplanar, then that nonplanarity is there to remove antiaromatic coupling.  We really have a ring rendered nonaromatic by reduced conjugation like cyclooctatetraenes.
In the case of cyclooctatetraenes one could argue that nonplanarity is driven by steric strain rather than avoiding antiaromaticity.  1,4 dithiins with hexagonal rings offer a clearer example of deliberately avoiding antiaromaticity.
The 1,2 isomers tend to lose sulfur forming thiophenes.


text:
What is the chemical equation for wine fermentation?
I am researching about the impact of sugar on wine fermentation, so I am looking for the equation of that reaction. However, there are a few different versions from different sources, so I was wondering which one is the right one to use and what does each equation represent?
\begin{multline}
\ce{Glucose + 2 NAD+ + 2 Pi + 2 ADP ->}\\
 \ce{2 pyruvate + 2 NADH + 2 ATP + 2 H+ + 2 H2O + heat}\tag{1}
\end{multline}
\begin{gather}
\ce{C6H12O6 -> 2CH3CH2OH + 2CO2 + ATP}\tag2\\
\ce{C6H12O6 -> 2C2H5OH + 2CO2}\tag3
\end{gather}

Generally there are two types of fermentation in biochemical processes:

homolactic fermentation
alcoholic fermentation

Both these processes occur under anaerobic conditions to replenish NAD+
by the reduction of pyruvate in an extension of the glycolytic pathway.

The overall reaction in muscles is:
$$\ce{Glucose + 2ADP + 2Pi -> 2 lactate + 2ATP + 2H2O + 2H^+}$$
For alcoholic fermentation:
Glucose is first converted to pyruvate by glycolysis, and the pyruvate is converted to ethanol and $\ce{CO2}$ in a two step process:

Note: For the second one we just started with pyruvate (to show mechanism and enzymes involved)Otherwise, for alcoholic fermentation, the net reaction is:
$$\ce{Glucose + 2ADP +2Pi + 2H+ -> 2CO2 + 2C2H5OH +2ATP +2H2O}$$

Both homolactic and alcoholic fermentation have the same function: the anaerobic regeneration of NAD+ for continued glycolysis. Their main difference is in their metabolic products
Having said that lets go back to experimental wine fermentation, it is essentially alcoholic fermentation (2).
Enzymes in yeast (zymase a mixture of enzymes) and  cozymase  a mixture of cofactors-coenzymes such as NAD+ , ATP, and ADP, as well as metal ions) help carry out the fermentation reactions.

So pyruvate converts to acetaldehyde while giving out CO2, then converts to ethanol, while NADH converts to NAD+? What would be the equation for that reaction? Is this process within the Glucose⟶2CO2+2C2H5OH process?

@Izy instead I have put net equations (starting all the way from glycolysis) but I wanted to emphasize the conversion of pyruvate to ethanol and CO2 in alcoholic fermentation


text:
Does a salt always have a neutral charge?
I know that a salt is the product of a reaction between a base and an acid, but does this mean that there will always be one cation and one anion and therefore a neutral charge?
In relation, are acids and bases charged?

IUPAC defines salt as "a chemical compound consisting of an assembly of cations and anions", so virtually there is no limitation on whether it can be charged or not.

A salt is an ionic solid compound. As such, it needs to have a zero net charge.
However, that doesn't mean it should contain only one cation and one anion. The stoichiometry of the salt depends on the charges of the ions. For instance, iron (III) sulfate is $\ce{Fe2(SO4)3}$; this stoichiometry is required to balance the charges from $\ce{Fe^3+}$ and $\ce{SO4^2-}$.
Also, other types of salt exist, called mixed salts. They contain several different cations and/or anions (e. g. $\ce{KAl(SO4)2}$).
Compounds with a net charge exist, they are just not called salts (look up Complexes).

In relation, are acids and bases charged?

Acids may (e. g. $\ce{NH4+}$) or may not (e. g. $\ce{H2SO4}$) be charged. Similarly, bases may (e. g. $\ce{SO4^2-}$) or may not (e. g. $\ce{NH3}$) be charged.


text:
Determining the structure of N2O using hybridization
I have learnt how to determine the structure of molecules where only the central atom is hybridised like $\ce{ClF3}$, $\ce{C2H2}$, $\ce{PCl5}$, but in $\ce{N2O}$ it seems as though both nitrogen and oxygen have hybrid orbitals. How do I find the hybridization of oxygen and nitrogen in $\ce{N2O}$ and finally determine its structure?
Research Effort: I watched some lectures on hybridization, but all of them included cases in which only the central atom had hybrid orbitals. I also read the topic in my book but this issue wasn't addressed there too.

I don't see how the hybridsation model could yield a structure prediction. In my understanding hybridisation can be used a post Lewis/VSEPR justification of a structure prediction. In that line you 1. draw a proper Lewis representation (possibly even some mesomeric forms) 2. apply VSEPR and 3. then explain it using hybridisation.

Usually you would approach this problem from the other direction. First, find the most likely structure with the correct number of bonds, then deduce the hybridisation. This is also how it works in quantum chemistry: hybridisation is deduced from geometry, not the other way around.
If you try to write the structure of $\ce{N2O}$ you should come up with two equally like structures; these are mesomeric structures as given in the spoiler tag below.

 $$\ce{\overset{-}{N}=\overset{+}{N}=O <-> N#\overset{+}{N}-\overset{-}{O}}$$

From this, we can deduce the most likely hybridisation which will result in a linear molecule. The central atom in a linear molecule typically features two $\mathrm{sp}$ hybrid orbitals and two unhybridised $\mathrm p$ orbitals.
So what about the outer atoms? Well, we cannot say for sure because we do not have enough geometric information. We can tell by the type of bonds that we again need two unhybridised $\mathrm p$ orbitals on the end atoms (remember that the orientations of the double bonds are equivalent!) but we don’t know whether it is a better description for the other two orbitals to be seen as $\mathrm s$ and $\mathrm p$ or two $\mathrm{sp}$ hybrids. In these cases, I would tend to go with ‘unhybridised’ until an experimental or calculative result proves me wrong.

After we have done this, we can indeed look back and realise that we could have determined the structure a priori like this from the general rules; we would just have had to ignore the outer atoms and discuss hybridisation only for the central atom. Indeed, ignoring the hybridisation of terminal atoms gets you very far even in the simplified theory that puts hybridisation first (and is practically wrong).

I am/ was a bit hesitant to up-vote this, but given the two wrong answers, this is much better. Okay, back to nitpicking: The only correct way to use hybridisation is to use it post-geometry determination (so it's not only practically incorrect). In principle there is no reason to assume any other external field than local linear coordination at terminal atoms. Constrained to this sp hybridisation is the max you can get.

To do this, you will need to follow the steps mentioed by R_Berger in the comments. I will use $\ce{CO2}$ as an example.

Draw a Lewis diagram.

Look at bonding & lone pairs on each atom.

Carbon has 2 double bonds which means 2 $\times \ \sigma$ bonds and 2 $\times \ \pi$ bonds.
Therefore, it has sp hybridization (2 $\times$ sp orbitals and 2 $\times$ p orbitals)
Each oxygen has 1 double bond and 2 lone-pairs which means 1 $\times \ \pi$ bond and 3 $\times \ \sigma$ bonds (actually 1 bond and 2 orbitals with lone pairs). Therefore they are sp2 hybridization. (3 $\times$ sp2 orbitals and 1 $\times$ p orbital)

Determine the shape

This can be done directly from the Lewis diagram using VSEPR theory. However, we only need to consider the central atom as that's where the bond angles will be! (You need two bonds to measure an angle...).
So for carbon dioxide, we have got sp hybridization at the central atom (or two areas of electron density with no lone pairs), so the shape will be linear.
Aside: You probably were already fine with the determining hybridization part. You just need to use the same rules on outer atoms!

Oxygen does not have sp² hybridisation. Terminal atoms have at most sp hybrid orbitals. Also note that atoms are never hybridised, only orbitals are. It is perfectly valid, to have a variety of hybrid orbitals at one atom. Just because there is one sp² orbital, it does not mean that there need to be two more. (Have a look here for more on carbon dioxide.)

That link is not one I've found before so I've just spent some time reading it. A few notes (please correct me if needed)... if I understand this, the method I described above (and is used in a majority of places if I search) is outdated? Where do the lone pairs reside? From your link, I see the central argument is that the electron density map is a cylinder which could not happen with my sp2, but does fit an sp hybridisation. Do you have any links for those maps? (Aside: my knowledge is from teaching IB level Chemistry, so much of the advanced theories has gone past me!)

After reading around the topics more, I believe that my answer is correct AND Martin's answer is also correct! I have answered the question using Valence Bond theory (where the hybridisation sp² describes what happens on the oxygen atom itself) whlie Martin used Molecular Orbtial theory (where the sp hydridisation is between the carbon's s and oxygen's p orbitals) (See here)

I am sorry, but you have not answered the question with Valence Bond theory, which is essentially an equivalent theory to molecular orbitals. The sp² orbitals for such atoms used to be part of a VB description, but (at the latest) with the formulation of Bent's rule that basically became obsolete. Mathematically this would still work, physically it simply does not make any sense. For the simple case of water it can be proven, that the lone pairs are not equivalent; this can be extended to more complex molecules. Side remark: MOT does not use or need hybridisation.

Thank you for the above. I'm now reading about Bent's rule... there's a lot to take in / relearn here. I've mainly been involved at secondary/ high school level where the older hybridisations are still considered as the correct answer. Once I've done more reading I'll look at updating my answer.

Based on hybridisation theory, the hybridisation states of the atoms are $\mathrm{sp}$, $\mathrm{sp}$ and $\mathrm{sp^3}$, for the atoms N, N and O, which are connected to each other in this order:

Conventionally, as most textbooks suggest, hybridisation states are derived from electronic geometry of the atoms (i.e. how the electron density regions are arranged around the atoms). For 2 electron density regions, it is sp-hybridised. For 3 electron density regions, it is $\mathrm{sp^2}$-hybridised. For 4 electron density regions, it is $\mathrm{sp^3}$-hybridised etc.
From the Lewis structure I have drawn, we can derive the results as such.


text:
Stability of cyclic isomers of C6H12
I was given the problem to arrange all cyclic isomers of $\ce{C6H12}$ from the lowest to the highest heat of combustion. I found twelve isomers and quickly realised that cyclohexane is the most stable and thus has the lowest heat of combustion, since it has zero strain, and that the only cyclopentane isomer - methylcyclopentane, follows after since it does not have much strain. However, what confuses me is the stability of the other isomers (four cyclobutanes as well as six cyclopropanes). I understand that the cyclobutanes are more stable than the cyclopropanes, but what about the substituents? How do they affect stability?

Not much. That said, multiple and/or branched substitutes would probably make a dense crowd around the ring and thus increase the strain. One long side chain would not do that.

As a sidenote: I think your count is a bit off. I count at least 6 isomers of cyclic butane.

ethyl-, 1,1-dimethyl, cis + trans 1,2-dimethyl-, cis + trans 1,3-dimethyl-


text:
About the protonation of guanidines
I am working with the protonation of guanidines.
The structure of these compounds looks like this:

I know that possibly the protonated-nitrogen will be the double bonded one (that structure conserve resonance), but i am looking for some experimental or theoretical proof, and it is not being easy to find a nice one. Ideas or links about this problem (maybe a paper) will be helpful.

All other lone pairs are delocalized into the $\pi$-system. Try drawing resonance structures.

In solid state, yes: http://scripts.iucr.org/cgi-bin/paper?a04854

Also in solution, there's just one signal by proton NMR: http://www.rsc.org/suppdata/cc/c3/c3cc46518g/c3cc46518g.pdf


text:
Single electrode potential measurement
My textbook says that in a galvanic cell, it's not possible to measure single electrode potential independently. Instead, a Standard Hydrogen Electrode is used as the system is under equilibrium. Can someone explain this concept more clearly?

What is unclear there? Closed circuit is needed, you can't have that with only one electrode in solution.

But what does  the system being in equilibrium have to do with it? Sorry if the question is trivial, just learning these stuff!

Probably that nothing happens so it's sort of "equilibrium".

Equilibrium means that the chemical potential has to be the same for the two side of the cell. One thing cannot be in equilibrium, as it makes no sense, it needs another system to be equilibrium with - which can act as a reference electrode

@Greg Can't the escaping of metal ions from electrode into electrolyte and the deposition of metal on the electrode be considered for equilibrium?

@GauthamShankar Why would it be?

Related: https://chemistry.stackexchange.com/questions/28534/standard-hydrogen-electrode

Through a more complicated technique, it is possible to determine an absolute reduction potential for a single electrode. I don't really understand why this isn't more well-known and why the absolute values aren't given in textbooks.

Electrodes transport electrons produced from one half-cell to another, thus producing an electric charge. You need to have a reference value to compare it to in order to obtain the relative potential. Relative to the Standard Hydrogen Electrode (SHE), that is.
The SHE has a potential that is set arbitrarily to zero giving you a basis with which to compare the potentials from the electrode you're testing.
For further reading, I feel this article on the Standard Hydrogen electrode is pretty descriptive.

Potential is a relative measurement. You will need to set a reference for measuring any potential.
As in this case, Hydrogen electrode is set as the reference.
All other potentials are measured against Standard Hydrogen electrode to get their potentials under standard conditions.

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


text:
The need for excess acid in dehydration of alcohols
Why is there a need to use excess acid in the dehydration of alcohols? My instructor said that excess acid is necessary to minimize the ether formation via the competing SN2 mechanism, but I'm not sure how this works.

Acid is catalyst so I don't see here any meaning in "excess"

@Mithoron Well, excess sulphuric acid would obviously assist in dehydration. Probably, it would even make dehydration faster than etherification.

Alcohols generally are very poor acids and have pka's in the range of 16-17.
Hence, if the medium is only slightly acidic, only some alcohols will protonate, after which the unprotonated oxygen (of ROH) will attack the generated carbocation, following which the H+ ion will leave making H+ act as a catalyst, resulting in the formation of ether.
But, if the solution is highly acidic, we would be generating the carbocation in full and there would be no nucleophile left to attack the carbocation.
We need to maintain the optimum pH as substitution reactions are generally favoured over elimination reactions especially in the case of primary and secondary alcohols.


text:
Why is there a need to rotate the molecule in a sharpless asymmetric epoxidation?
As far as I understand sharpless epoxidation, you first need to rotate the molecule such that the vincyclic alcohol is in the upper right quadrant. Then, the epoxide forms on the top/bottom of the double bond, depending on the chirality of DET. (See image below, from masterorganicchemistry.com: Sharpless Epoxidation)
My question is that why do we need to rotate the molecule when forming the epoxide, because can't we just form the epoxide directly without rotation? And if we can't,  then we still can rotate the molecule back to its starting orientation after a epoxidation, right? Pretty confused about this.

You don't have to. That thing is just to help you, a human, figure out which face the epoxide forms on. The chemicals don't need to work it out that way, and of course they don't need to rotate the alkene into whatever position.

@orthocresol How does rotating the molecule even help a human in determining the structure, if the epoxide is formed above/below the plane either way?

Rotation allows you to pick out a preferred orientation to discuss the molecule. Once the canonical orientation is achieved, then conventional descriptions such as up, down, left, right have unambiguous meaning.


text:
Use of geometry index for the determination of coordination environment
Geometry index $\tau$ is supposed to resolve proper geometry for coordination numbers (C.N.) 4 and 5 based on its extreme values ($0$ or $1$). There is also a web app Geom which handles both cases for a structure in XYZ format.
I'd like to summarize the questionable topics regarding proper and efficient usage of this method:

I'm not sure how to address the intermediate values. Say, for $\tau_5 = 0.33$: is it a square pyramidal geometry with a character of trigonal bipyramid? Or one can just name this coordination environment square pyramid and call it a day?

Are there similar algorithms developed for the higher C.N.s:

capped trigonal prism vs pentagonal bipyramid (C.N. 7);
cube vs square antiprism (C.N. 8)?

Based on the abstract of the original paper that defined the geometry index (http://pubs.rsc.org/en/Content/ArticleLanding/1984/DT/DT9840001349) your first description in question 1 seems to be more accurate. It seems to suggest thinking of the parameter as measuring the percentage of trigonal distortion.

For your first question, the original paper (J. Chem. Soc., Dalton Trans. 1984, 1349–1356) that described the geometry index $\tau_5$ defined it as an "index of trigonality". For example, they write for a compound with $\tau_5=0.48$

By this criterion, the irregular co-ordination geometry of $\ce{[Cu(bmdhp)(OH2)]2+}$ in the solid state is described as being $48\%$ along the pathway of distortion from square pyramidal toward trigonal bipyramidal.

For your second question, while I haven't been able to find a $\tau_7$ or $\tau_8$ used in the literature, it seems possible to define such parameters under the right conditions. To devise a $\tau_8$, we can see that for a regular cube $\ce{MX_8}$, there can only be bond angles of $70.5^\circ$ (between adjacent $\ce{X}$ in the same square) and $109.5^\circ$ (between opposite corner $\ce{X}$ of the same square or between corner $\ce{X}$ of different squares). However, an antiprism instead has an angle of $99.6^\circ$ separating the $\ce{X}$ of different squares.

(Image obtained from Inorganic Chemistry by Miessler and Tarr)
This suggests using a formula reminiscent of $\tau_5$ to define $\tau_8$ as the antiprismatic distortion index. One possibility is $$\tau_8=\frac{\beta-\alpha}{9.9^\circ}$$ where $\beta > \alpha$ are the two largest valence angles and $9.9^\circ$ is a normalization factor to make it between $0$ and $1$. So when $\alpha=\beta=109.5^\circ \to \tau_8=0 \to$ cubic geometry and $\alpha=109.5^\circ$ $\beta=99.6^\circ \to \tau_8=1 \to$ antiprismatic geometry.
This will only work if the structure is a regular antiprism (i.e an anticube). The same is true for defining $\tau_7$ between a pentagonal bipyramid and a monocapped trigonal prismatic. This is because the angles for these will vary if all the attached groups are not the same and so a consistent scheme based on the angles would not suffice.
I also imagine that $\tau_7$ would be harder to define in this way because I don't think there is pair of angles that on its own could describe the distortion between the two geometries.

Yep, these "capped" geometries ($\tau_7$) are tricky, thankfully one can relatively easy spot one by cross-sectioning the crystal space with Voronoi-Dirichlet polyhedra. Thank you for the great answer!


text:
Geometric isomers of substituted methylidenecyclohexane?

My thoughts on the problem:
I am sort compound of sure  C and D will not show geometric isomerism because of the same groups (H and H) attached.
In compound A and B, suppose I swap the positions of H and Cl, I notice that it is not possible to obtain the new compound by rotation of the original one.

My doubt: Does that mean A and B show geometric isomerism.  Also is my method, of interchanging the positions of H and Cl and checking if the same compound can be obtained by rotation, correct?

Related: Why is 1-ethylidene-4-methylcyclohexane chiral?

@orthocresol - Ohh ok. From that post I could conclude that D and C can not be correct answer. But I'm still confused between A and B. The solution given tells me the answer is only B.


text:
Measurement of cell constant
Why is cell constant value measured through indirect methods (like using a solution whose conductivity is already known)? Isn't it easier to  measure the length and area of cross section of a conductivity cell directly?

The mathematical description of the cell constant ($l/A$) does not include factors like fringing fields (the ionic current does not travel precisely in a well-defined path between the electrodes) and the actual surface area for rough surfaces.
The cross section of a round electrode may have different fringing fields depending on its separation from its companion. And if the two electrodes are of different geometry, you have to decide how to average the areas out. I think there was more standardization in the choice of electrode materials, size and spacing than has been carried over into the beautifully simple formula $l/A$, so when the formula is used outside of certain narrow bounds (and I don't know what they are), it's only an approximation, and resistance or conductance can be measured more precisely than the cell constant from geometry. So you might as well calibrate against a known cell or solution.

The conductivity value of any cell changes with the area of cross-section of the electrode. When we use an electrolyte of concentration within the desired area, the conductivity value varies. When a standard value of specific conductance is given, then from the observed conductance value we can calculate the cell constant.