text:
Were gold(II) complexes synthesized for the first time in 2017?
Recently in the Google news feed, I discovered this news:

Gold stabilized in very rare oxidation state +II
Missing golden link found: Divalent gold complex isolated for the
  first time in a pure form; new fundamental insight also provides
  possible explanation of the mechanism of action of cytostatic
  gold(+III) porphyrins
Date: August 8, 2017
Source: Johannes Gutenberg Universitaet Mainz
Summary: A team of chemists has been able to isolate and analyze gold
  in the very rare oxidation state +II. This provides the missing links
  in the homologous series of the coinage metal ions copper(+II),
  silver(+II), gold(+II), and in the 'relativistic' triad of
  platinum(+II), gold(+II), and mercury(+II).

And there are loads of similar news- Here, here and here.
Now, I started digging for more information and found out that gold(II) complexes were known previously and has been synthesized but they are unstable and prone to disproportionation.

Coordination chemistry of gold(II) complexes (October 1999)
The number of complexes with gold in a formal oxidation state of
  two have increased considerably and nowadays this oxidation state can
  be considered nearly as a common state in gold chemistry. This
  review covers the evolution of the knowledge of this oxidation state
  from the first dithiocarbamate derivatives $\ce{[Au2(S2CNR2)2X2]}$,
  stable only at very low temperature, to the room temperature stable
  dinuclear gold complexes $\ce{[Au2(L–L)2X2](L–L=CH2PPh2CH2, CH2PPh2S,
 C5H4PPh2-2; X=halogen}$). Although a dinuclear structure with a
  gold(II)–gold(II) bond supported by two equal bridging ligands were
  the first and most common, this review shows the present diversity as
  mononuclear, $\ce{[Au([9]aneS3)2](BF4)2}$, dinuclear with two
  different bridging ligands, $\ce{[Au2(S2CNR2)(CH2PPh2CH2)X2]}$, with a
  non-supported metal–metal bond, $\ce{[Au2(Ph2PC8H6PPh2)2Cl2](PF6)2}$,
  and polynuclear gold(II) complexes either with chains of only gold
  centers or with other metals, such as
  $\ce{[{(C6F3H2)Au(CH2PPh2CH2)2Au}2Au2(CH2PPh2CH2)2]ClO4}$ or
  $\ce{[Au2Pt(CH2PPh2S)4X2]}$
Disproportionation of Gold(II) Complexes. A Density Functional Study of Ligand and Solvent Effects (July 13 2006)
For the disproportionation of $\ce{Au(II)L3}$ complexes with neutral
  ligands, disproportionation is highly endergonic in the gas phase.
  Calculations imply that for synthesis of a monometallic $\ce{Au(II)}$
  complex, a nonpolar solvent is preferred. With the exception of
  $\ce{[Au(CO)3]^2+}$, disproportionation of $\ce{Au(II)L3}$ complexes
  to $\ce{Au(I)L}$ and $\ce{Au(III)L3}$ is exergonic in solution phase
  for the ligands investigated.
Structural Studies of Gold(I, II, and III) Compounds with Pentafluorophenyl and Tetrahydrothiophene Ligands(February 28
  2007)
A golden opportunity was seized to prepare an unbridged, dinucleur
  gold(II) compound, $\ce{[Au2(C6F5)4(tht)2]}$
  (tht=tetrahydrothiophene), without any stabilizing chelating ligands
  (see picture; Au orange, S yellow, F green, C black) and to
  crystallographically characterize the gold(I) and gold(III) complexes
  participating in this conversion. A unique ligand scrambling of the
  latter gold(III) compound occurs in solution.

Gold(II) complexes has been intensly studied previously. They have been trying to synthesize a stable complex but failed miserably. The stable ones are only present in low temperature and only one was found to be stable in room temperature. So, although gold(II) complexes were previously discovered and (synthesized?), why scientists at University of Mainz and those who discovered it in 2017 claims to have synthesized gold(II) complexes for the first time?

The claim is isolating it, which is not the same as just synthesizing it.  You can synthesize $\ce{HNO_2}$ by adding sodium nitrite to hydrochloric acid in water solution, but can you isolate or even concentrate it?

AuXe4 (2+) was known for for quite some time.

The claim might be clarified by inserting the word "stable". So maybe they can legitimately claim the first synthesis of stable gold(II) compounds.

You are correct in noting that there are a number of Au (II) small molecule species synthesized prior to the report of Professor Katja Heinze in 2017.
The heart of the Au (II) claim by Professor Katja Heinze and team as stated in the article you referenced, see reference 1 given below, has to do with the a) relative stability of the mono-nuclear gold-porphyrin complexes compared to other Au (II) systems and b) that the molecule does not have a Au-Au bond.  The latter point being the more important.
Au (II) is a very stable oxidation state for molecules containing Au-Au bonds.  We synthesized and characterized a number of such species years ago, see references 2 and 3 below for two papers describing the work.
For those working in the field, the nature of the claim by Heinze et al. is understandable.  For those outside the niche field of Au (II) chemistry I can see how the distinction is less than clear.
References
https://www.sciencedaily.com/releases/2017/08/170808145949.htm
https://pubs.rsc.org/en/content/articlelanding/1985/c3/c39850001278#!divAbstract
https://pubs.rsc.org/en/content/articlelanding/1986/C3/c39860000321#!divAbstract


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3D similarity search software
What software / webservice can one use to find a compound that has similar 3D structure/dimension to another compound?
eg. say we are tying to find a 'better' version of atropine, so naturally we try to find compound with similar 3D structure to Atropine (antimuscarinic)
and 3d structure of atropine (and all other drugs) is available
so what software can we run to search 3d shape look-alike of atropine among all available 3d structures ?
similar shape and size for starter?

This is pretty difficult to do. Most of the time, people go through crystal structure databases and look for molecule-looking fragments. But there's a reason that industry researchers spend entire careers doing this (and the related modelling).

It's not that there aren't software products that do this. The problem is having the database of accurate 3D molecular structures.

Probably the closest to your need would be pharmit by David Koes, which allows you to search chemical libraries based on structural or chemical similarity to another small molecule, typically for protein binding

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Molar specific heat
What is the difference between molar specific heat and molar specific heat at constant pressure?
It seems that both of them have the same formula, i.e
$$\frac{1}{n}\left(\frac{\mathrm dQ}{\mathrm dT}\right)$$
Then what is the difference between these two and why do they have the same notation?

The difference is in the tiny index $\mathfrak p$ outside of the brackets. Why did you skip it?

Yes but that $p$ does not change the value of the formula right ? So you mean that both of them are essentially the same ? How is that possible ?

Yes it does change the result, and not by negligible amount. Ever heard about the specific heat at constant volume?

Sorry , I think I understood what you meant to say. Is it that the subscript $p$ denotes that pressure is held constant during differentiation and that gives specific heat at constant pressure. But for specifically for molar specific heat we do not include a subscript which means nothing is held constant during differentiation I believe ?

"Molar specific heat" is not a thing at all. Or rather, it is an umbrella term which means both $c_p$ and $c_V$ at once, much like the word "human" means you and me, and quite a few others. It has no value. Say, you know your weight, but what is the weight of a human?

@Ivan then what is the meaning of the notation $C = \frac{dQ}{dT}$ ? Which kind of specific heat does this represent? On Wikipedia https://en.m.wikipedia.org/wiki/Heat_capacity  also they have defined it the same . Is that a general formula for specific heat ?

Unless it is clear from the context or somehow else, this formula means nothing.

First of all it is important to indicate which properties of your system stay constant during the partial differentiation. Therefore you get the molar specific heat at constant pressure as
$$
\frac{1}{n}\left(\frac{\partial Q}{\partial T}\right)_p
$$
and at constant volume
$$
\frac{1}{n}\left(\frac{\partial Q}{\partial T}\right)_V
$$
Now to the question why this even matters:
When putting energy (in form of heat) into the system its temperature (normally) rises. Considering a system with constant volume this is actually all that happens. Therefor all the energy put into the system gets transferred into a temperature rise of the system.
If you hold the pressure constant instead the system can expand while being warmed up. This requires it do do work against the outside pressure. Therefore some of the energy put into the system is used for the expansion and only the rest is transformed into a temperature rise.
Therefore the specific heat at constant pressure is always greater than the one at constant volume (more energy has to be put in it in order to achieve the same temperature rise).

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Synthesis of Benzaldehyde and Aniline Schiff Bases
I have a problem while synthesize the schiff base. I synthesized using absolute ethanol as solvent and use reflux method for 4 hours generally. The problem occur on TLC where there is still two spot on the product which one of them are very clear (top one) (which I guess the ligand formed) and another one were very light, while on NMR spectra, whereby there is a singlet peak at approximately 10.0 ppm which I assumed that it is from the aldehyde itself which probably not completely reacted. I also tried several derivatives based on benzaldehyde and aniline but some of them have to be recrystallize which I havent found the suitable solvent system yet.
So with all due respect, I really need help especially those who have an experience on synthesizing schiff base and also share to me some ideas to recrystallize my ligands.

So basically, you need a method to separate reactant from product, right? Have you tried column chromatography?

No, I havent tried column chromatography, will it work? because based on my understanding, column chromatography usually used for compound which may consist of more than 3 different material.

The std conditions for Schiff base formation is Dean-Stark reflux in Toluene with acid catalysis. You may find this superior to ethanol reflux. I second the suggestion of column chromatography for purification

The reaction needs acid catalysis but ethanol is fine as solvent. There will always be some starting materials left, but since the product is a solid and the reagents are liquid, the best (and cheapest) is to purify it through recrystallization.
For this, dissolve the product in the smallest quantity of hot ethanol and then add water until precipitation occurs.

Schiff's bases are susceptible to hydrolysis. Avoid water in recrystallization and ethanol for that matter. Go with the Dean-Stark trap, acid catalyst and rerecrystallization with non-hydroxylic solvents.

True. It is an equilibrium, but it is favored to the Schiff base. And you have no acid in the recrystallization, so water is not a problem.

Actually I have tried to dissolve my compound in water but it really doesnt dissolve either in cold or hot condition, so I keep on trying to use other solvent combination such as DMF, THF, DCM and etc. Some of my friends did get crystal compound but it need a lot of patience and as it involve try and error step. However regarding recrystallize using ethanol and water, is it okay if we have tried that our compound does not dissolve at all with water because referring to the concept of recrystallization, "Soluble in hot, insoluble in cold", or its actually gonna work out if I keep on going.

The idea is to dissolve it in hot ethanol and then precipitate it with water. The product is soluble in hot ethanol and insoluble in water.


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Reaction of phenol and zinc dust
There is a reduction reaction of phenol with zinc dust which is found in many books.
$\ce{Ph-OH + Zn -> Benzene + ZnO}$
However, a simple google search fails to explain anything about the reaction. So, I put forth my questions:

1) What is the name of the reaction and who invented it?

I have already done a google search, but it yields no results. My textbook also mentions nothing about the inventor of the reaction, or any name of the reaction. I understand that every reaction might not have a name, but at least it must have an inventor. You can't say that the reduction of phenol by distilling it with zinc dust is known from time immemorial. Can you?

2)What is the reaction mechanism of the reaction?

This and this page writes about a mechanism of the reaction. If drawn, I suppose it would look like this:

It would certainly explain the formation of biphenyl and hydrogen ( if they are indeed formed). But the problem is that no reference is given in the answers. There is at least one book or paper which wrote about the mechanism.
So, if you answer, please add a reference if possible. If you think that the above mechanism might be incorrect, then also add an answer. I want your insights on this problem.
[Note: There is another question at Reaction of phenol with Zinc dust which is similar. But I have already drawn a mechanism and I am asking a different question here. So I don't think this question is a duplicate.]

Strongly related though unanswered: https://chemistry.stackexchange.com/questions/8994/reaction-of-phenol-with-zinc-dust

Even you questioned once: https://chemistry.stackexchange.com/questions/81717/effect-of-zn-dust-reduction-of-phenolic-oh-group-on-other-groups

If this gets answer, old unanswered should be closed as dupe. BTW why everyone thinks that mechanisms for heterogenic reactions are known?

@NilayGhosh, Yes I did post a question but that is a different question. And I have already mentioned about the related question in my question in the [note] part.

Ludwig Gattermann's The Practical Methods of Organic Chemistry (translated into English, 1896) has two references to reduction of aromatics with zinc dust:

The first one refers to Liebigs Annalen, issue 140, page 295 (1866):

Ueber die Reduction aromatischer Verbindungen mittelst Zinkstaub, Adolf Baeyer

This roughly translates to "On the reduction of aromatic compounds using zinc dust".
The second one refers to Chemische Berichte, issue 1, page 43 (1868):

Ueber Alizarin und Anthracen, C. Graebe C. Liebermann

This translates to "On alizarin and anthracen". Wiley, who now owns the Berichte, cites the page numbers as 49-51.
Name of the reaction
The reaction is given without name, even in modern textbooks. Maybe it should be called Baeyer reduction of aromatic oxygen-containing compounds.
Mechanism of the reaction
Maybe this would be a good research project because it does not seem like the mechanism is known.


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Is there a single MO that represents the π system of a Möbius aromatic system? If so, what does it look like?
I recently answered the question of whether twisted polyene chains were known by pointing towards Möbius aromatic system. While in conventional Hückel aromatic systems the lowest-lying π-orbital obviously has a nodal plane containing all the nuclei, this is not possible in a Möbius aromat since topologically these only have one side.
A number of Möbius aromatic systems have been both synthesised[1] and calculated;[2] for an X-ray determined structure of one, see the second structure in the figure below.[3]

One can easily see that the π chain would have green and white or plus phase and minus phase overlapping. This got me thinking and wondering: what would this system’s molecular orbitals look like? Specifically: is there a set of MOs that would correspond approximately to the π system of such a Möbius aromat? If not, how are the π bonds seen in an MO analysis? The only calculations I could find searching the literature were both very old and without nice pictures.
References:
[1]: M. Stępień, N. Sprutta, L. Latos-Grażyński, Angew. Chem. Int. Ed. 2011, 50, 4288–4340. DOI: 10.1002/anie.201003353.
[2]: H. S. Rzepa, Chem. Rev. 2005, 105, 3697−3715. DOI: 10.1021/cr030092l.
[3]: Y. Tanaka, S. Saito, S. Mori, N. Aratani, H. Shinokubo, N. Shibata, Y. Higuchi, Z. S. Yoon, K. S. Kim, S. B. Noh, J. K. Park, D. Kim, A. Osuka, Angew. Chem. Int. Ed. 2008, 47, 681–684. DOI: 10.1002/anie.200704407.

Isn't what the colours in the picture represent?

I suppose you know better than many that aromaticity is not about a single MO, but a whole set thereof. We may just solve the system by the simple Huckel method and see what they are like. Sure, it is terribly old-fashioned and won't give you nice pictures (unless you make some yourself), but I don't think we need them all that much. What we need is the qualitative description, and it will be there all right.
From a cursory glance, it seems to me that the eigenvalues of a Möbius system are given by the same Frost circles, only the polygon is pointing down with its side, rather than a vertex. This, BTW, explains the reversed stability rules ($4n$ instead of $4n+2$). The molecular orbitals will look kinda like those in usual aromatic systems, except that the lowest orbital will be doubly degenerate and indeed will have a node plane across the chain.
So it goes.

I actually do want a (set of) pictures though. And in Hückel cases there is a picture of the lowest pi symmetry which is the nice one showing the system.

The best way to understand the Moebius aromatic orbital structure is to render it as a modification of the familiar Hueckel structure.
Look at benzene.  As you know, the single lowest occupied pi orbital has no nodes orthogonal to the ring plane.  One step above is a pair of degenerate orbitals that are commonly regarded as each having one ring-orthogonal nodal plane.  But note that the nodal plane, in each case, passes through the rotational symmetry axis and so can be divided into two half-planes, 180 degrees apart.  Likewise, at the second, unoccupied degenerate level each orbital may be considered to have four nodal half-planes orthogonal to the ring plane, at 90 degrees from each other.  In each case we have, in all, an even number of nodal half-planes orthogonal to the ring plane.  At the highest level we would expect another degenerate pair, but when the number of nodal half-planes ( in thus case six) matches the number of ring atoms we lose one degenerate orbital.  With an even number of nodal half-planes this drop-off of one orbital from the highest "pair" occurs for even size rings.
Now in a Moebius aromatic system, the twist leads to an odd number of nodal half-planes emerging from the center of the ring.  At the lowest bonding level there is one such nodal half-plane per orbital.  To match the molecular symmetry at this energy level there must then be a second orbital also having one nodal half-plane, orthogonal to the first.  To meet the orthogonality reqirement the second orbital must be turned 180 degrees from the first one, and match the molecular geometry th second orbital must have its nodal half-plane twisted by 90 degrees from the first.  The pair of nodal half-planes corresponding to the two degenerate orbitals look like somebody tried to assemble two half-planes into a full plane but did not get them aligned right.
That is the lowest degenerate pair.  The next pair will have three nodal half-planes per orbital, again with the nodes twisted to match the molecular twist.  To meet symmetry and orthogonality requirements we again need a second, degenerate orbital.  As in the first degenerate pair, the second orbital has its nodal half-planes rotated 180 degrees and twisted 90 degrees from those of the first orbital. The same applies for the the third pair with five nodal half-planes per orbital, etc up to the size of the ring.  At the highest level in an odd-size ring such as $\ce{C9H9+}$ (https://en.wikipedia.org/wiki/Möbius_aromaticity), one orbital drops out in the same manner as we saw for even-size rings like benzene in the Hueckel aromatic case.


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Why is trans-1,2-dimethylcyclohexane an enantiomer?
Can't the two be superimposed over each other after ring inversion.

Related: Enantiomers of 1,2-dichlorocyclohexane

Take a look at this answer https://chemistry.stackexchange.com/questions/83877/enantiomers-of-1-2-dichlorocyclohexane

No, they can't.
A model from any chemical modeling kit will convince you right away.  Or maybe a cheaper alternative made out of chewing gum and matchsticks will suffice.
Or better yet, you may do it in your head. Place the molecule in a horizontal plane (or rather, as close to a plane as possible, since the thing is not planar). Now place your eyes in the same plane, so as to look at the molecule from the side. Walk around until you are facing the bond that connects two methylated carbons in the ring. Look at that bond. Look also at the two methyl groups to the right and to the left from it. You either have the "left one up, right one down" isomer of your molecule, or vice versa. You might think "what the heck, I'll turn it around." No way. You turn each enantiomer around, and it still ends up looking like itself, and not like the other one.
Another way to think of it is via symmetry elements. For a molecule, not being chiral implies having certain kinds of symmetry. Earlier revisions of your question suggest that you were looking for some symmetry plane. Just where it might be? Think of it. And if you don't find that plane after a while, consider the possibility that it might be simply not there (which is true).
Ring inversion is irrelevant altogether. It can be done, but it will switch all equatorial substituents to axial and vice versa, so you won't be able to recognize the same molecule until you invert it again to the original form.
So it goes.


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Wade's rules applied to metallic elements
There are many examples of metallic clusters that distort from the structure predicted by Wade's rules to a more 'spherical', closo-like arrangement. For example,

$\ce{Bi8^2+}$ is a square antiprism (WR predict an arancho-derivative of a bicappped square antiprism, with 1 capping and 1 adjacent to an open-face atoms removed, instead of 2 capping atoms)
$\ce{Bi9^5+}$ is a tricapped prism (WR predict a capped antiprism)
$\ce{Ge8R6}$ is a square prism (WR predict an arachno-derivative of a bicapped square antiprism)
$\ce{Ge9R3^-}$ is a tricapped trigonal prism (WR predict a capped square antiprism, i.e. expect nido but get closo)

My lecture notes (not available online) say that

Wade's rules are much less effective at predicting structures of metallic elements because intact polyhedra (spherical arrangements of metal atoms) tend to be a lot more stable for the more metallic elements, as compared to boron.

Could someone please explain why intact polyhedra are more stable for these elements.


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Chemistry toilet paper
This is a stupid question, but I can’t find one of the compounds on this toilet paper:

I have in order:

Salicylic acid
Aceton
Caffeine
?
Urea

The missing compound is 4-[1-hydroxy-2-(methylamino)ethyl]benzene-1,2-diol, also known as adrenaline.


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Change of temperature upon dilution
Consider a concentrated dye solution, say 10µM of Rhodamine 6G in water, which is rapidly diluted down to 10nM in an open container. I'm trying to estimate the (presumably minute) temperature change, assuming both solutions (dye and water) are initially at the same room temperature. I believe dilution increases entropy ever so slightly, but I'm struggling to turn this statement into a quantitative temperature change. Specifically, how do I calculate the change in entropy? Once I know this, the change in temperature should be straight-forward.

You need to measure the heat of mixing of the solute and solvent.

@ChesterMiller sure, I could (try to) measure the temperature change, but I'm trying to estimate it theoretically.

Why are you concerned with entropy but not with enthalpy contribution?

Also 10µM is particularly concentrated. You could get like 10 mM or more depending on counterion.

@Alchimista if enthalpy is more relevant please provide an explanation. I'm after the temperature change, and I assumed it would occur because of a change in entropy but maybe there's another contribution, I'm quite rusty on all this.

@Mithoron yes, I provided those numbers for completeness but hopefully with a formula it'd be easy to estimate various cases, including higher concentrations. I picked 10µM because typically beyond that there's a lot of dimers which we tend to avoid for other reasons (spectral changes), but that's irrelevant here.

I have no idea honestly.  Just T changes can have both nature.  I cannot imagine any experiment in which I would worry for such a T change " if any" . But it doesn't mean that the question shouldn't be asked. I an afraid, however, that without tabulated values then one has to measure that....

At such low concentrations, can’t you just assume an ideal solution? Then the  change in entropy from mixing would be $\Delta S = -R(x_A\ln x_A + x_B \ln x_B)$. But because the solution is so dilute, you can see that both terms are going to be essentially zero. (For a 10 nM solution, the contribution from entropy of mixing is on the order of $10^{-8}$ J K$^{-1}$ mol$^{-1}$.)
The dilution process would thus lower the total entropy, since there’s less uncertainty in the system. (You’d get the value you compute at 10 nM, the final state, minus the value at 10 µM, the initial state.)

sorry what's $x_A$? I do expect the answer to be small, but is it $10^{-10}K$ small or more like $10^{-2}K$

I'd expect something in between, like $10^{-5}$ or $10^{-6}$.

The x’s are mole fractions of solute and solvent.

I’ve included an order estimate in my answer.

A solution typically does not approach ideal solution behavior at low concentrations (particularly for an aqueous ionic solution).  This would only be the case if you were mixing molecules that have very similar molecular characteristics.  The activity coefficient at infinite dilution in a liquid solution usually deviates maximally from unity.

@ChesterMiller Fine. But my basic argument—that the effects is essentially negligible—is valid unless the activity coefficients are enormous. And there’s enough difficulty in the computation of activity coefficients that unless you have the specifics for the material you’re using, you can’t guess or assume them in advance.

The question is about the heat of dilution, and the tiny effect (presumably approaching linear deviation with concentration) of further dilution.   This is governed by the behavior of the activity coefficient variation of the solute as its concentration approaches zero (where the activity coefficient it typically varying very rapidly).  I agree that this would usually be determined only with the use of experimental data.

Thanks, I can see how this works. Experimental data is likely hard to obtain in this regime


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1s orbitals and 2nd period diatomic elements
a general chemistry textbook I am reading says this:

"In second-period diatomic molecules, the 1s orbitals of the two atoms barely overlap. Because the $\sigma _{g1s}$ bonding and $\sigma _{u1s}$*
  antibonding orbitals are both doubly occupied, they have little net effect on bonding properties and need not be considered"

i'm confused as to what it means...

1s AOs don't and won't form MOs due to weak overlap
MOs formed from 1s Aos aren't significant for chemical reactivity

I may be missing something, but if anyone can clarify as to what the text meant that will be great. and also can you mention if it extends beyond the 2nd period or even to heteronuclear molecules.

The energy splitting between a bonding and the corresponding anti-bonding orbital is dependent on the overlap: the greater the overlap, the greater the splitting. But take note that the relationship is not really linear and that the split is not perfectly symmetrical: the anti-bonding orbital usually rises more strongly than the bonding orbital lowers its energy.
This means that the interaction of the 1s orbitals in, say, dinitrogen, is repulsive (in terms of atom-atom distance), but also very small because the 1s orbitals are small (a consequence of the high nuclear charge compared to hydrogen), which leads to poor overlap.

does that mean the poor overlap is insignificant and won't form molecular orbitals or the poor overlap is insignificant ergo the molecular orbitals formed are insignificant as well.

It is the latter.


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Rate Law with Rate Determining Step

The reaction of tert-butyl bromide with azide ion in aqueous solution is proposed to proceed by the following mechanism:
  $$
\begin{align}
\ce{(CH3)3CBr(aq) &<=>[k$_1$][$k_{-1}$] (CH3)3C+(aq) + Br–(aq)}\\
\ce{(CH3)3C+(aq) + N3–(aq) &->[$k_2$] (CH3)3CN3(aq)}
\end{align}
$$
  $k_i$ is the rate constant.
Assuming that $\ce{(CH3)3C+(aq)}$ achieves a steady-state concentration, but making no further assumptions about the relative magnitudes of the three rate constants, what is the rate law for this reaction?

The answer is
$$\mathrm{Rate} = \frac{k_1k_2[\ce{(CH3)3CBr}][\ce{N3–}]}{k_{–1}[\ce{Br–}] + k_2[\ce{N3–}]}$$
But how would I obtain that answer using multistep reaction mechanisms and not differential equations?

Welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center. Also, this is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

Since $(CH_3)_3C^+(aq)$  achieves a steady-state concentration, the equations when directly added give the overall reaction whose rate constant is the product of the rate constants of individual reactions. Hence the rate law comes out to be as it is written.


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Why do orthoesters react with Grignard reagents, but acetals and ketals don't?
My book mentions the synthesis of aldehydes and ketones using alkyl ortho esters. It's called Bodroux-Chichibabin aldehyde synthesis.
Now, later, the book mentions this:
R- displaces EtO-, giving an acetal or ketal, which are unreactive with Grignard reagents, so a secondary alcohol is not formed.
Why do orthoesters react with grignard reagents, but acetals and ketals don't? They have essentially the same functional groups, just two instead of three.

Grignard reactions are performed in ether or THF.
There you have only one -OR group on a carbon and, since it is the solvent, it is not reacting with it.
This shows that not always a -OR group is substituted in a Grignard reaction.
When you have three -OR's on one carbon, the formal positive charge ($\delta+$) on it is higher than when there are only two -OR's.
Therefore, the reaction proceeds on orthoesters and not on acetals or ketals.


text:
Mass of iron from the reduction of iron(III) oxide

Given the equation
  $$\ce{Fe2O3 + 3H2 -> 2Fe + 3H2O},$$
  determine how many grams of $\ce{Fe}$ can be produced from $\pu{16.5 g}$ of $\ce{Fe2O3}$.

I've already determined the masses of each component. I believe the next step is convert the masses into moles, but after that, I'm not sure what formula to use.
Any help is greatly appreciated.

After finding the moles of $\ce{Fe2O3}$ we see that after balancing the equation, $2$ moles of $\ce{Fe}$ are formed for every mole of $\ce{Fe2O3}$.
But that would be considering that the equation moves completely to the RHS, which is assumed if nothing is explicitly mentioned but not true.

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text:
How to find the plane of symmetry in biphenyls and substituted cyclohexanes?
I was going through my text book, (Solomon&Frhyle)
I came across a couple of of examples which I didn't understand.
1.

How is the molecule symmetrical? Upon folding the molecule over its plane of symmetry  the cooh group will be placed over fluorine.
2.

Similarly in this in structures 3,4,5,6 a chlorine atom out of the plane and below the plane coincide.
3.
It's written that for biphenyls to be chiral a can be the same as a'. (Same for b and b')
           But
If a a' and b b' are the same them won't a plane of symmetry lie in the middle perpendicular to the plane of molecule.

https://chemistry.stackexchange.com/questions/35737/axial-or-planar-chirality-in-spiro3-3hepta-1-5-diene

Finding planes of symmetry is simple: try all planes, some (or maybe none) of them will fit.
As for your examples, the first image is just poorly drawn. The plane is supposed to be the one of the ring containing COOH and F, while the other ring is perpendicular to it (which, BTW, I don't believe is the native conformation of the molecule, but whatever).
The second example asserts right things, but for the wrong reasons. (This is not what you would normally expect of a textbook, to put it mildly.) True, each of these structures contains at least one plane of symmetry. Too bad the planes are drawn incorrectly in all structures except for the first two. You may take your time and fiddle around to find out where they really are.
In the third example, you would be right if the molecule were planar, which it is totally not. Steric repulsion between a and a' (also, between b and b') makes that impossible, even if all of the substituents a, a', b, b' are simply hydrogens.

What angle do you suggest between the phenyl rings in the two biphenyl systems?

Anything between $40^\circ$ and $60^\circ$ sounds reasonable to me.

@IvanNeretin : you're right, example 2 is really bizarre. For a minute I thought they meant axes of symmetry by those dotted lines, but that would only apply to molecule 4, as far as I can see. It will forever remain a mystery what the authors meant, but at face value a really poor didactic example indeed...

For the OP. If your point is to check if the molecule is chiral or not, you don't necessarily need to 'find a plane'. Some achiral molecules don't even have (in some conformations) a plane of symmetry, but a rotoreflection axis instead. To  keep it simple, the idea is that a molecule, fixed in a rigid conformation that is as symmetrical as possible yet chemically plausible, is chiral if there is no combination of rotation or translation by which it can be superimposed to its mirror image. So just draw it, make its mirror image and check.


text:
Why do we have the concentrations powered to the coefficients in the equilibrium expression?
$$\ce{A -> 2B} \qquad K_\mathrm{c} = \frac{[\ce{B}]^2}{[\ce{A}]}$$

Why do we have a power that is the same as the coefficient?
Assuming $K_\mathrm{c} = 1$, shouldn't the quantities of the mixture be $2B = A$ (in other words, the amount of $\ce{A}$ in the mixture equals two times the amount of $\ce{B}$ in the mixture)!? So why do we have $[\ce{A}] = [\ce{B}]^2$ rather than $[\ce{A}]=2[\ce{B}]$?

It's a standard result of classical thermodynamics. See any reasonable university level physical chemistry text book, and http://www2.mcdaniel.edu/Chemistry/ch307.notes/Chemical%20Equilibrium.html covers it, but unfortunately that page doesn't display quite properly in my browser - yours may differ.

"If $K_c = 1$ then $2[\ce{B}] = [\ce{A}]$" that is an intuitive guess, but unfortunately not true. Could you explain why you think that must be so?

Not a duplicate - the other question is based in rates for elementary reactions, this is about thermodynamics and is the general case

This is due to the way the equilibrium constant is defined.
A chemical equilibrium is a stable state at some point in the reaction (characterized by the extent of reaction $\xi$). Because it is a stable state the free enthalpy $G$ has to have a local minimum at exactly that place.
Therefore we can say
$$
\left( \frac{\partial G}{\partial \xi} \right)_{p,T} = 0
$$
On the other hand the following equation applies:
$$
\Delta_rG_m = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum_i \mu_i \cdot \nu_i
$$
With the free, molar reaction enthalpy $\Delta_rG_m$, the chemical potential of the component $i$ $\mu_i$ and it's reaction-coefficient $\nu_i$ (To understand this consider this general reaction equation: $\ce{\nu_1A + \nu_2 B <=> \nu_3 C}$ - note that the coefficients for the educts are negative by definition).
Therefore we can say that the following equation applies for the equilibrium state:
$$
\Delta_rG_m = 0
$$
we can rewrite above formula with the aid of the activity-dependency (basically a concentration-dependency) of the chemical potential ($\mu_i = \mu_i^* - RT\cdot\ln(a_i)$):
$$
\begin{split}
\Delta_rG_m &= \sum_i \mu_i \cdot \nu_i\\
&= \sum_i \nu_i\mu_i^* - RT \cdot \sum_i \nu_i\ln(a_i)
\end{split}
$$
Because $a \cdot \ln(b) = \ln(b^a)$ this is the same as
$$
\Delta_rG_m = \sum_i \nu_i\mu_i^* - RT \cdot \sum_i \ln(a_i^{\nu_i})
$$
Now we define the standard, free, reaction enthalpy $\Delta_rG_m^* = \sum_i \nu_i\mu_i^*$ and because we found out (in the beginning) that for the equilibrium state $\Delta_rG_m$ has to be zero we can write:
$$
\Delta_rG_m^* = -RT \cdot \sum_i \ln(a_i^{\nu_i})
$$
Because $ \ln(a) + \ln(b) = \ln(a\cdot b)$ we can rewrite this as
$$
\Delta_rG_m^* = -RT \cdot \ln\left(\prod_ia_i^{\nu_i}\right)
$$
The last step is now to define the equilibrium constant $K$ as
$$
K = \prod_ia_i^{\nu_i}
$$
And that is how the coefficients in the recation equation end up in the exponent when calculating the equilibrium constant.


text:
Reaction of vicinal diol in presence of benzaldehyde
Could someone please help me figure out the products of the following reaction?

Here are few possibilities I've come up with:

The attacking reagent could the given vicinal diol, with its nucleophilic lone pair, attacking the electrophilic carbon of benzaldehyde. However, that doesn't seem to lead me anywhere.
As the reaction is carried out in acidic medium, one of the -OH groups (or both?) could be protonated, to become better leaving groups leading to formation of a carbocation intermediate. What after that?

Chances are it's probably a nucleophilic addition, proceeding through a tetrahedral intermediate, than a carbocationic one.
It'd be great if someone could show a proper mechanism for the reaction, and explain why it should be so. Thanks a lot!

In my opinion, the answer should be (b). The reaction is the formation of acetal.

Yes, this is acetal formation

This reaction is similar to the one where you use ethylene glycol to form a temporary protecting group for an aldehyde that you do not want to react in further steps. The answer therefore should be (b).

I guess you and I know that you are correct but could you expand? e.g. showing OP where there mechanistic proposal was correct and how to move on from there.


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Reconciling Flory-Huggins Theory with Le Chatelier's Principle
(I discuss the Flory-Huggins parameter as chi from here onward). So I had learned in class that when discussing polymers the Flory-Huggins Theory helps us figure out whether mixing is thermodynamically favorable. We had also discussed UCST and LCST behavior. Now from what I understood (and where there might be a misconception), with a chi value less than 0 the solution always mixes, chi value equal to 0 the solution is ideal (with no preferable interactions) and a chi value greater than 0 the solution may or may not mix depending on the entropy related terms in the Flory-Huggins equation.
Looking at the case where the chi values is greater than 0 the reaction (solvation reaction) is endothermic based on the equation in terms of internal energies ((z/2kbT)(2*uAB-(uAA+uBB)) where u represents the energy of interaction between molecules A and B). However, we also know that if I model chi as A+B/T then, as temperature is increased, the chi value can change giving rise to either UCST or LCST behavior. Which means a reaction (of solvation) that is endothermic (chi greater than 0) could either increase or decrease in terms of Gibbs (or Helmoltz's) free energy with rising temperature. However, this cannot be reconciled then with Le Chatelier's Principle which would tell me that for an endothermic reaction of solvation, rising temperature would always push the reaction to be more thermodynamically favorable to the products. Could someone please point out my error in logic?
In advance, thank you!


text:
Is center of symmetry only valid for symmetric substituents?
The given compound is optically active. Initially I had reasoned that the compound has center of symmetry( two xylene groups across each other). Why doesn't it have a center of symmetry?

Source: MS Chouhan, Advanced Problems In Organic Chemistry, Page 116 (Chapter Isomerism Level 2 q6a)

It probably racemises pretty quickly.

@Mithoron I don't think so.

You assume too much, starting with the words "The given compound". There is no given compound. The picture corresponds to quite a few different compounds, and here is why.
As I told you before on a remotely similar question, the biphenyl fragment totally can't be planar. The xylene substituents must be rotated so their ortho methyl group points either up or down relative to the plane of the central ring. Since there are two of them, we have three options: "both up", "both down", and "one up, one down". The first two compounds are indeed chiral, and enantiomers of each other. The third one is centrosymmetric, just as you've reasoned.

It was a question ; the book had provided that picture and asked to state whether it was chiral or achiral. How do I figure out which of the three isomeric states would it be in?  (the question book specifies whether the bonds are in or out of the plane{through wedges and dashes} ). So I was taking the compound to be planar (which is not possible as you did tell me earlier).

If you have wedges and dashes, then you have one specific isomer. If you don't, then you have all three at once. In that case any answer shorter than mine would be incomplete.

@IvanNeretin "one up, one down" would that stereoisomer's m-xylene rings be exactly perpendicular to the central benzene ring? Because if they are not exactly perpendicular, and rather inclined at some angle, then the centrosymmetry would be lost.

The central symmetry would still be there if they are inclined at the same angle in the opposite direction.

Look at the picture, although the two methyl groups are located equidistant from the centre but they aren't in opposite direction. Therefore, no centre of symmetry and so it's optically active.


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does Hydrochloric Acid react with PET?
From someone coming from electronic engineering not skilled in chemistry,
There are transparent electric conductors in form of ITO ( Indium tin oxide ) coated glass or transparent PET films ( Polyethylene terephthalate ). By etching the ITO coating we can create transparent circuit boards and with the case of ITO coated PET also flexible transparent circuit boards.
There seems to be very little info on what can be used for etching ITO PET sheets.
For ITO coated glass, people use HCL with no issue, but will HCL also work to etch ITO but leave the transparent PET film intact, or
do PET and HCL react?

It is prohibited to store hydrochloric acid (or any other acid for that matter) in PET as that PET will eventually break up. The process, however, isn't all that fast, so you might get it good if you are very careful with timing. ITO-coated polypropylene or (better) polyethylene, however, would be fine

I see, but how long are we talking about? Etching shouldn't take more than 30 minutes and ideally no more than 5 minutes. Is that long enough exposure to cause degrading, or are we talking about much longer?

You can etch ITO on PTE by mean of conc. HCl solution without problems. The treatment duration depends on the ITO thickness but is limited to minutes.
Just make few samples (residuals from cutting etc will suffice to make this test) to find out how long it takes to fully remove ITO.
No reference need it as I do etching of ITO coated PTE electrodes since years.
For sake of completeness let me add that I never investigated PTE itself prior and after the etching so I do not know if there are changes which, for instance, may interfere with long (very long) stability etc.
In my case it suffices that PTE maintains its transparency and still behave as a robust but flexible substrate. It seems it could be your case, too.
Just to be pedantic. Obviously after etching the foil must be washed thoroughly.

Thank you. What can be used for the photoresist? This is out of the scope of the question so I'm posting as a comment. Also, what concentration are you using?

Conc HCl is ok. Is what is generally used.  Its dilutions will still work but I do not any practical reason to use them. More handling more hazard more time etc. As masking materials in a pioneristic stage we used nail varnish.  Then I introduced 3M magic tape and similar drawing adhesive tapes. But our structure are at millimetres scale or more,  and easy to draw.  Often I do just define the structure by means of the top electrode. If you need more precision than photo resist tecnology or so will be necessary

adhesive tape is ok, so is 1mm precision, but how can you create a mask out of it without cutting by hand?

That is the point. If  you need a circuit is getting tricky. In my case I just define basic geom. shape so it is easy to cut the tape with a cutter and the help of preprinted drawing that I use as background. For complicated and or finer patterns again you will have to look for technologies used in semiconductors and optics labs.


text:
H-bonding in methanol and methoxyamine
The question was to determine if the substances made H-bonding between themselves and to represent it.
In the following photos i'm showing my doubt about the H-bond in $\ce{CH3OH}$. In the solution (grey highlighted background) it is shown that $\ce{CH3ONH2}$ can make even 3 H-bonds, so can't $\ce{CH3OH}$ make the same? Just as I drew beside its relative solution?

That’s acetamide not methoxylamine


text:
Reactivity of benzyl halides towards nucleophilic substitution
I experienced some difficulty as I was doing question 22 of this quiz on nucleophilic substitution at saturated carbons:

I picked option B but the answer turned out to be C. My rationale for picking B was that I thought that IV was going to be more reactive than I because I has an additional electron-donating methyl substituent which would increase the energy of the transition state in the $S_{N}2$ pathway.
Evidently, II and III are in the 3rd and 4th places respectively because alkyl bromides are less reactive than alkyl iodides and that phenyl bromides are not as reactive towards nucleophilic substitution as the $ \pi$ electron cloud of the neighbouring benzene ring would repel the nucleophile away.
What went wrong with my thinking?

Consider the contribution of the methyl group of 1 towards the stability of the benzylic cation

If you’re just putting it in methanol it’s going to be SN1 not SN2.

@orthocresol How could you tell so surely?

SN2 needs a decent nucleophile, which methanol is not

@orthocresol So I was on the wrong train of thought because I thought it was SN2. On a side note, would the benzylic halides react by SN1 or SN2 if the nucleophile was a good one?

Probably both pathways are significant.

Certainly methanol is a poor nucleophile for SN2 but for an SN1 reaction answer C would be appropriate.

The answer to this question hinges on figuring out whether the reaction occurs by the SN1 or the SN2 pathway.
As noted by a comment, methanol is not a very good nucleophile, and SN2 reactions require a strong nucleophile for the nucleophilic attack to occur.
Also, it is important to realize that the likelihood of the SN1 pathway being taken largely depends on the stability of the carbocation intermediate that forms.
The benzylic carbocation that forms in this case is extremely stable:

due to resonance and
due the the presence of the polar protic substance, methanol.

Thus, SN1 is highly preferred here.
In fact, due to these three reasons, the probability of the SN2 pathway occurring is almost negligible in I, II, and IV.
I believe you have already worked the rest of it out; the answer is C.

As you already noted, phenyl bromide is least reactive towards nucleophilic substitution, so any answer suggestion that does not place III last is wrong a priori. The reasoning you gave does not quite add up with me so here is my version: no $\mathrm{S_N2}$ back attack is possible through the phenyl ring since that pathway is blocked by the other ring atoms; and no $\mathrm{S_N1}$ pathway is possible since that would require a very unstable phenyl cation with the empty orbital being of the $\mathrm{sp^2}$ type.
Thankfully, this also immediately tells us that II is penultimate, since no other options exist and the question boils down to whether I or IV is more reactive.
If you were to assume an $\mathrm{S_N2}$ pathway, then indeed I would lose to IV due to increased steric inhibition: the electrophilic carbon is not easily accessible from behind.
However, both reactive carbons are benzylic and the neighbouring phenyl ring is very good at stabilising ionic charges. Considering that you also have a rather weak nucleophile — methanol — it is very likely that this reaction will proceed under $\mathrm{S_N1}$ conditions, so the rate determining step is the formation of the cation. From this point of view, I gives the better intermediate cation than IV because of the additional inductive stabilisation from the methyl group.
Therefore, the correct ordering is $\mathrm{I>IV>II>III}$ or answer C.


text:
Metal and reactions with different temperatures of water
I knew that in warm water magnesium forms $\ce{Mg(OH)2}$ and $\ce{H}$ gas, and in steam it forms $\ce{MgO}$ and $\ce{H}$ gas (this is also the case for metals lower down the electrochemical table). However I came to realize that for those higher than magnesium all form metal hydroxide when react with water.
What is the mechanism for such a reaction?

Strongly related: https://chemistry.stackexchange.com/questions/6212/why-do-the-reactions-of-magnesium-in-steam-and-cold-water-differ/7302

You have two reactions, the metal reacting with water and the hydroxide decomposing:
$\ce{Mg + 2 H_2O \rightarrow Mg(OH)_2 + 2 H_2}$
$\ce{Mg(OH)_2 \rightarrow MgO + H_2O}$
The second reaction, for magnesium, sets in around 350°C.  Below that you can get the hydroxide, above it any hydroxide decomposes and you have the oxide.
Heavier alkaline earth metals, and all alkali metals, have hydroxides stable to higher temperatures, making it more difficult to "direct" the reaction to oxide formation.  In contrast, most transition metals lack low-temperature reactivity with water and their hydroxides decompose easily, so they can give only the oxide.


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Can i calculate the solution of the problem in more simple way?
and i got a serious question. During my exercises at home, i found that im struggling a lot while solving simple problem`s like this one:
I have an aquatic substance with 0.100 M of $\ce{Mg(NO3)2}$ and 0.100 M of $\ce{NH3}$.
Given that: $K_\mathrm{sp}(\ce{Mg(OH)2})=1.50\times10^{-11}$  and  $K_\mathrm b(\ce{NH3})=1.75\times10^{-5}$
I need to find:

If there will be any precipitation for $\ce{Mg(OH)2}$?
What concentration of ammonia I need to begin the precipitation of $\ce{Mg(OH)2}$ in a solution of 0.100 M $\ce{Mg(NO3)2}$?
If $\ce{Mg(OH)2}$ was a soluble salt, what was the pH in the solution?

And the process of solving this takes me a lot of time. What I did to solve are those steps:
a. To decide if there will me a percipitation I need to know my Q of magnesium hydroxide and compare it to K. Which will give me a "True" for a precipitation to occur.
b. Then for the next question i made an ICE table where X was the initial conc. of ammonia and the other fields were dependent on the hydroxide concentration from the previous percipiation(stochiometry). This step was long and its included about 2–3 tables.
c. If magnesium hydroxide was a soluble salt, probably he would not have an equilibrium constant and will dissolve in one direction reaction. That tells me about the dependence of the pH in solution, which is based on $\ce {OH^-}$ concentration.
It would be lovely for some of you who know much better than i do, to correct me or show me much simpler ways to think outside the box. Maybe it will reduce my amount of time spent on those exercise (in the best case scenario), if not - it sure will help me or somebody else in other ways. Thanks for support.

I don't think that there is any sword to cut cleanly through this Gordian knot, so don't waste a lot of time looking for one. Note however that solving either 2 or 3 gives you 1.

I think you are overthinking this.
Using the solubility product and the nominal $\ce{Mg^2+}$ concentration, calculate the $[\ce{OH-}]$ concentration that must be exceeded to cause precipitation of the magnesium hydroxide:
$$[\ce{OH-}]>\sqrt{\frac{K_{\mathrm{sp}}}{C(\ce{Mg^2+})}}$$
Now use the base constant equilibrium constant (in its simplified form), to calculate the actual $[\ce{OH-}]$ concentration, due to the ammonia present:
$$K_\mathrm b(\ce{NH3})\approx \frac{[\ce{OH-}]^2}{C(\ce{NH3})}$$
$$[\ce{OH-}]\approx \sqrt{K_\mathrm b(\ce{NH3})\times C(\ce{NH3})}$$
If the latter exceeds the former, precipitation will occur.
And as @MaxW noted (in the comments):

Note however that solving either 2 or 3 gives you 1.

For $2$, use the calculated $[\ce{OH-}]$ concentration required for precipitation to occur, to calculate the minimum $C(\ce{NH3})$, using:
$$C(\ce{NH3})\approx \frac{[\ce{OH-}]^2}{K_\mathrm b(\ce{NH3})}$$

Thanks a lot. Your answer definitely organized my thoughts.


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Vapor Pressure, Equillibrium, and Increasing Pressure
I'm trying to make sure my understanding of vapor pressure and transients is correct.
Let's assume I have a thermally-controlled sealed environment which started out as the regular air (i.e. atmospheric gasses and their respective ratios) with no gaseous water or ethanol at 101.325 kPa, 30 C. This container will maintain a temperature of 30 C after the start of this experiment.
Inside this environment there are two containers which can be remotely opened.
When I open the first container, exposing the pure water inside with a saturation vapor pressure (First question: is that the right term?) of 4.2455 kPa (from Wikipedia) and wait until it reaches equilibrium (100% humidity), does this mean this is the new state of the container?

Pressure: 101.325 + 4.2455 = 105.5705 kPa
Partial-Pressure of H2O: 4.2455 kPa

Then I open the second container, exposing pure isopropyl alcohol inside with a saturation vapor pressure of 10.555 kPa (from WolframAlpha) and wait until it reaches equilibrium again, does this mean this is the new state of the container?

Pressure: 101.325 + 4.2455 + 10.555 = 116.1255 kPa
Partial-Pressure of H2O: 4.2455 kPa
Partial-Pressure of Ethanol: 10.555 kPa

So, in the end, the pressure of the vessel increased each time a new liquid was introduced.
In this hypothetically-ideal environment, would the pressures simply rise until it reached that perfectly equilibrium? If so, I expect in a real environment the pressure would rise but there would always be a cold spot somewhere which caused the vapor pressure (in that specific area) to drop, leading to water condensing, leading to the partial pressure being lower as the air mixes, leading to more evaporation; is this correct?
Some notes:

Assume that there is enough of each liquid to reach equilibrium without running out of liquid.
The sealed environment effectively adds energy to counteract the heat of vaporization of each liquid, and also prevents the gasses from releasing/absorbing energy into/from the external environment or from changing due to pressure changes. It is essentially a magic make-everything-this-one-temperature machine.

There are a couple of gotchas' with the problem.
The volumes of the liquids and gases are specified. So 1 drop of water in a million gallon tank of dry air isn't going to be enough water to reach saturation.
You stated that "This container will maintain a temperature of 30 C after the start of this experiment." However the problem statement doesn't explicitly state that the contents of the container will be in thermal equilibrium with the container. That does seem to be a reasonable interpretation however. (If the evaporation had taken place adiabatically, then the liquid and gas would have to cool to account for the evaporation.)
There is another significant factor here. The water would absorb ethanol, and the ethanol would absorb water. You can't really figure this out without knowing all the volumes, or at least the relative volumes.
Now if you assume that there is enough liquid to saturate the gas phase, that the system (gas phase and liquid phase) absorbs heat from outside the system, then the vapor pressure of the liquid at $30\ ^\circ\mathrm{C}$  is added to the internal pressure of the gas phase before the evaporation takes place.

"If the evaporation had taken place adiabatically, then the liquid and gas would have to cool to account for the evaporation" - Yes, my intention with that statement was to say that the container would keep the liquid at that temperature, adding energy to compensate for evaporation. And yes, my assumption was that there was enough water to reach equilibrium without running out of water.

"...is added to the internal pressure of the gas phase before the evaporation takes place" - are you saying that the pressure change leads the actual water vaporization? So, once the first canister is opened, in [comparatively] short order, the chamber would rise to 105.5705 kPa, even though there may not actually be H2O vapor in the air?

I didn't think about the water/ethanol cross-absorption; I'll have to research that part further. Before the ethanol container is opened, is my original question combined with my previous comment a correct understanding of the system?

I didn't look it up, but I'll assume that the vapor pressure of water at 30C is 4.2455 kPa. So once the gas phase is saturated with water vapor, then the pressure will rise to 101.325 + 4.2455 = 105.5705 kPa. How long it takes once the first canister is opened is another matter entirely...

water/ethanol cross-absorption - I think you'd end up with one of the three situations (1) all ethanol migrates to water (2) All water migrates to ethanol or (3) two volumes of ethanol-water at the same concentration. With one of the possibilities would come true would depend on the (relative) volumes.

Ok, so right when the canister is opened, the pressure is still 101.325 kPa, but rises as the water evaporates until it reaches a pressure where the water's partial-pressure is its vapor pressure?

Would I be going down the right path if I say the mixing occurs because the vapor pressure of the final water/ethanol solution is lower, so it finds that state to be an advantageous state to be in?


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What is the preferred IUPAC name of hexamethyldisilazane?
Hexamethyldisilazane (HMDS, $\ce{(SiMe3)2NH}$) and its conjugate base hexamethyldisilazide (as the Li/Na/K salt) find wide use in organic chemistry.
In my quest to only use preferred IUPAC names in my thesis1 I have come across a problem with this compound. According to P-21.2.3.1 of the Blue Book the name disilazane for the parent hydride $\ce{H3Si-NH-SiH3}$ is no longer preferred because

When nitrogen atoms are present, amine names (see P-62) are preferred because of the higher functionality of amines. This concept does not extend to other elements.

and the name given is N-silylsilanamine.2 Just going off this, I would guess that the PIN of the hexamethyl compound is something along the lines of "trimethyl-N-(trimethylsilyl)silanamine".
However, I'm also not sure whether the addition of the methyl groups makes the choice of parent hydride different, as per P-21.2.3.2 (for example). The frequently used name bis(trimethylsilyl)amine is also a possibility, although I've personally not seen a rule suggesting that it could be a PIN.
What is the PIN of this amine?

I'm just kidding; writing "4-methylbenzene-1-sulfonyl" multiple times is quite silly. But I’m still curious about this PIN.
The original publication has this written as N-silylsilan-1-amine but this has since been corrected in the errata, which are published online.

Why not name it just bis(trimethylsilyl)amine?

I would suggest not using IUPAC names for every compound. HDMS is a widely-recognized name in both fully written and abbreviated forms. You want your committee and anyone else who reads your thesis to focus on what work you have done, not get distracted  by why you chose an obscure name for a common reactant.

@buckminst Thanks for the comment, I definitely agree. It was a joke, although I admit didn’t make it very clear. Using PINs everywhere means using ethoxyethane instead of diethyl ether, etc. I’m not planning to go down that route, not least because it’s a huge waste of time for me to search the Blue Book for every compound, and as you said it simply leads to needless confusion.

@buckminst I think it's still a good idea to be aware of it, just in case if someone asks a passionate/spiteful question. (Of course, depending on how much the thesis has to do with this particular compound.) Anyhow, the question on its own still stands, even if it is not required in the original context.

@LinearChristmas LiHMDS and KHMDS are only mentioned essentially in passing; for the real important compounds in my thesis, nobody will be able to tell whether I found the PINs or whether I just used ChemDraw's auto name generator...

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), as already mentioned in the question, the name ‘disilazane’ is no longer recommended for $\ce{H3Si-NH-SiH3}$. Such structures are now named as amines. (Also note that the name ‘hexamethyldisilazane’ is ambiguous and therefore never was in accordance with IUPAC nomenclature.)

P-21.2.3.1 Heterogeneous parent hydrides composed of alternating heteroatoms, i.e., [a(ba)n hydrides], excluding carbon atoms and the halogen atoms.
Compounds containing an unbranched chain of alternating atoms terminated by two identical atoms of the element coming later in the seniority order O > S > Se > Te > N > P > As > Sb > Bi > Si > Ge > Sn > Pb > B > Al > Ga > In > Tl may be named by citing successively a multiplying prefix denoting the number of atoms of the terminal element followed by the ‘a’ term for that element, then the ‘a’ term of the other element in the chain and the ending ‘ane’. The terminal letter ‘a’ of an ‘a’ term is elided when followed by a vowel; the terminal vowel of a numerical prefix is not elided even when the ‘a’ term begins with the same vowel. When nitrogen atoms are present, amine names (see P-62) are preferred because of the higher functionality of amines. This concept does not extend to other elements. These parent hydrides are preselected parent hydrides (see P-12.2) and have priority to receive preferred IUPAC names, as long as they are used to name compounds containing carbon.

Symmetrical secondary amines are named according to Subsection P-62.2.2.1, which describes three different methods.

P-62.2.2.1 Symmetrical and unsymmetrical secondary and tertiary amines are named only by the same methods described in P-62.2.1.2.
(1) substitutively using the retained name ‘aniline’ or the suffix ‘amine’ and the name of a parent hydride with further N-substitution;
(2) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘azane’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.
(3) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘amine’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.
Method (1) generates preferred IUPAC names. Functional parent names like diethylamine and triethylamine are deprecated. The prefixes in names of such secondary and tertiary amines formed by method (2)[sic: read (3)] are set off by parentheses to distinguish them from these deprecated names.

The corresponding names for $\ce{H3Si-NH-SiH3}$ are actually given as examples in the Blue Book:

(1) N-silylsilanamine
(2) disilylazane
(3) disilylamine
(not disilazane; see P-21.2.3.1)

Note that, since $\ce{H3Si-NH-SiH3}$ does not contain any carbon, no preferred IUPAC name (PIN) is indicated in the Blue Book for this compound.
The corresponding names for the complete compound that is given in the question are:
(1) 1,1,1-trimethyl-N-(trimethylsilyl)silanamine
(2) bis(trimethylsilyl)azane
(3) bis(trimethylsilyl)amine
Since P-62.2.2.1 explicitly states that Method (1) generates preferred IUPAC names even for symmetrical amines, the preferred name for the complete compound that is given in the question is 1,1,1-trimethyl-N-(trimethylsilyl)silanamine.

New abbreviation: LTTSA


text:
Comparing Basicity of Aniline Derivatives
Why is 4-hydroxyaniline more basic than 2-hydroxyaniline? Hydroxyl is an electron donating group, and their resonance structures suggest that they both isolate a negative formal charge on the carbon that is bonded to the amine group (which is also an electron donating group) so it would be more unstable. However, there is a clear difference in basicity between the two, the former having a pKb of approximately 8.53 while the latter has a pKb of 9.28. What accounts for this huge difference?
I also have another question. 3-hydroxyaniline avoids putting the negative formal charge on the carbon that is bonded to the amine group, and it has a pKb of 9.83. This makes sense because that means 3-hydroxyaniline is more stable in terms of resonance and thus its conjugate acid would want to liberate more protons than the other two, rendering 3-hydroxyaniline less basic than the other two. However, aniline has a pKb of 9.37, which means that it is more basic than 3-hydroxyaniline. However, hydroxyaniline has a higher electron density than aniline so shouldn't it be more basic than aniline?

This is just my own conjecture: Perhaps, the fact that the two groups are so close together in 2-hydroxaniline allows hydrogen bonds to form between the lone pair on the nitrogen and the hydrogen on the oxygen. Thus, this decreases the basicity of the nitrogen and it is less able to donate the electron pair to acids.

There could be intramolecular h-bonding effects, a simpler (easier to accept) answer would be based on inductive effects. OH is excellent at pi donation but is sigma withdrawing. A 2-hydroxyl group is more electron-withdrawing than a 4-hydroxyl group. (Or less electron-donating, if you want to be precise.) This also explains why the 3-hydroxy compound is less basic: resonance donation to the carbon bearing the amine is lower, so the inductive withdrawal wins. Btw from a nomenclature perspective I think it’s aminophenol, not hydroxyaniline, but that’s not really important.

@orthocresol But if we were to subscribe to the "closer to the OH- to the NH2-, the less basic it is" argument, then how would we explain that m-aminophenol is less basic than p-aminophenol?

@TanYongBoon how do you explain that phenol reacts with electrophiles at o,p but not meta? Same concept, just a slightly different context. It’s already in OP’s post, actually: in the resonance forms for 3-hydroxyaniline, the formal negative charge never resides on the carbon with the amino group. You can quantify these things, e.g. Hammett parameters.

Pkb of aniline is 9.13 and not 9.37


text:
How to unify all the subunits in a PDB file?
A PDB may contain a TER token. For example, hemoglobin has 4 subunits separated by a TER identifier.
If we take hemoglobin's biological assembly pdb file we see:
ATOM   1067  NH1 ARG A 141      26.176   8.362  17.810  1.00 11.11           N
ATOM   1068  NH2 ARG A 141      24.650   9.068  16.200  1.00 13.86           N
ATOM   1069  OXT ARG A 141      26.697  14.784  20.720  1.00 10.99           O
TER    1070      ARG A 141
ATOM   1071  N   HIS B   2       3.670 -13.643  19.447  1.00 38.58           N
ATOM   1072  CA  HIS B   2       2.695 -14.734  19.744  1.00 32.83           C
ATOM   1073  C   HIS B   2       1.379 -14.140  20.199  1.00 30.79           C

Is there an easy way to merge all the subunits into a single VMD frame?

I still think you should be more careful with your wording. The PDB file is a single frame as far as VMD is concerned. Do you want to remove the quaternary structure completely?

@pentavalentcarbon there are 4 subunits I want all of them to be in a single name. I am not sure what do you mean by removing the quaternary structure.

In a PDB file with the TER entries like this, there is both a chemical and textual representation of the quaternary structure (the subunits). If you remove them, of course the protein still has subunits, so there is some chemical structure, but the textual representation is lost.

@pentavalentcarbon what's a textual representation?

I mean the PDB file. It's the text format describing something chemical.

@pentavalentcarbon ah ok. I am ok that some information will be lost.

If you don't care about residue and atom numbering, which you probably do, then sed '/^TER/d' < 1A3N.pdb > 1A3N_combined.pdb will do. Since atom entries must be reordered, it looks like MDAnalysis will strip the TER entries automatically:
#!/usr/bin/env python2

import MDAnalysis

u = MDAnalysis.Universe('1A3N.pdb')

with MDAnalysis.Writer('1A3N_combined.pdb') as writer:
    writer.write(u)

So,
ATOM   1067  NH1 ARG A 141      26.176   8.362  17.810  1.00 11.11           N
ATOM   1068  NH2 ARG A 141      24.650   9.068  16.200  1.00 13.86           N
ATOM   1069  OXT ARG A 141      26.697  14.784  20.720  1.00 10.99           O
TER    1070      ARG A 141
ATOM   1071  N   HIS B   2       3.670 -13.643  19.447  1.00 38.58           N
ATOM   1072  CA  HIS B   2       2.695 -14.734  19.744  1.00 32.83           C
ATOM   1073  C   HIS B   2       1.379 -14.140  20.199  1.00 30.79           C

becomes
ATOM   1067  NH1 ARG A 141      26.176   8.362  17.810  1.00 11.11      A    N
ATOM   1068  NH2 ARG A 141      24.650   9.068  16.200  1.00 13.86      A    N
ATOM   1069  OXT ARG A 141      26.697  14.784  20.720  1.00 10.99      A    O
ATOM   1070  N   HIS B   2       3.670 -13.643  19.447  1.00 38.58      B    N
ATOM   1071  CA  HIS B   2       2.695 -14.734  19.744  1.00 32.83      B    C
ATOM   1072  C   HIS B   2       1.379 -14.140  20.199  1.00 30.79      B    C

where there are now chain IDs in the last column. See how this doesn't renumber residues? Modify the script; resids is a property, so the in-place object mutation works properly:
#!/usr/bin/env python2

import numpy as np
import MDAnalysis

u = MDAnalysis.Universe('1A3N.pdb')
u.residues.resids = np.arange(1, 1 + len(u.residues.resids))

with MDAnalysis.Writer('1A3N_combined.pdb') as writer:
    writer.write(u)

Finally:
ATOM   1067  NH1 ARG A 141      26.176   8.362  17.810  1.00 11.11      A    N
ATOM   1068  NH2 ARG A 141      24.650   9.068  16.200  1.00 13.86      A    N
ATOM   1069  OXT ARG A 141      26.697  14.784  20.720  1.00 10.99      A    O
ATOM   1070  N   HIS B 142       3.670 -13.643  19.447  1.00 38.58      B    N
ATOM   1071  CA  HIS B 142       2.695 -14.734  19.744  1.00 32.83      B    C
ATOM   1072  C   HIS B 142       1.379 -14.140  20.199  1.00 30.79      B    C

Will it still be possible to refer/select only one subunit out of the four?

If you have a tool that selects the chain ID, I believe so.

I mean using vmd.

It works. For "Selected Atoms" under "Graphical Representations", switch from all to chain A or B, C, D, etc.


text:
Which is a good leaving group Br- or acetate ion?
Acetate ion has charge delocalisation and in Br- charge is distributed throughout the atom ,so how should they be compared?

As acetate has charge delocalisation, it should be better leaving group, I guess. Bear in mind, I have not checked my answer.

Br anion is much larger in size than oxygen anion present in acetate ion. Hence Br anion can hold the negative charge better than oxygen atoms of acetate. Again negative charge density in Br anion is less than that on the 2 oxygen atoms of acetate. Therefore , Br anion is a weaker base than acetate and it is very well known fact that weaker base is a better leaving group or nucleofuse.

"it is very well known fact" doesn't make for a very good argument ;)


text:
Nucleophilic acyl substitution leaving group
From Wade's "Organic Chemistry" [1, p. 1001]:

MECHANISM 21-3 Conversion of an Acid Chloride to an Ester

What I don't understand is the elimination of $\ce{Cl}$, even though a much better leaving group with a positive charge on it is present (which results in the backwards reaction).
Why doesn't the alcohol or acid chloride deprotonate the tetrahedral intermediate before step 2 and after step 2 $\ce{Cl-}$ can attack the protonated alcohol/acid chloride to turn into $\ce{HCl}$?
Reference

Wade, L. G. Organic Chemistry; Pearson Education, 2012,. ISBN 978-0-321-81139-4.

I agree that would make a bit more sense.

Again, I added the better picture and full reference; however, page and mechanism numbers would be different.

Alcohol isn't "much better leaving group". Also depictions of mechanisms in books are usually not describing what exactly happens. More precisely they only show constructive pathway and not show surrounding molecules and their involvement.

Indeed, there is no reason why the protonated alcohol should not be eliminated upon the lone pair rebound. It probably happens more often than not. However, it does not lead to an observable product: it just goes back to the reactants. Thus, we can write an equilibrium arrow on the first step.
$$\ce{R-C(=O)-Cl + R'-OH <=> R-C(-O^-)(-Cl)-OH+-R'}\tag{1}$$
Once the chloride has been kicked out, there is no way back as the ester is a much weaker electrophile than the acid chloride (and chloride a weaker nucleophile than the alcohol). Thus, the reason why we only care about the lone pair rebound eliminating chloride is because only that is productive.
$$\ce{R-C(-O^-)(-Cl)-OH+-R' -> R-C(=O)-OH+-R' + Cl-}\tag{2}$$
These arrows are also given in your source like that; if everything up to a single step is given as equilibria while that one step is given as non-reversible, that should be sufficient driving force to let a reaction happen.
At which point in time the proton is eliminated is somewhat up to debate, of little relevance for the overall reaction and dependent on what is in solution. If just a pure acid chloride and a pure alcohol are mixed the only potential bases that could accept the proton of the tetrahedral intermediate are other acid chloride molecules or other alcohols — or the tetrahedral intermediate’s $\ce{O-}$ atom. All of these are very weak bases and I see no preference for the proton to go anywhere. After we have eliminated chloride, that is added to the set of potential bases. It is also not a strong base; maybe even a weaker base than some of the other options. The big difference is that it is volatile and can diffuse out of the reaction mixture. Thus, with no other bases added, I would actually prefer this sequence of steps.
Once you add a base, I would also prefer deprotonation before chloride is eliminated.

Clayden's also writes the mechanism to have the deprotonation happen first. But what is the reason for saying that deprotonation happens before chloride leaves?

I believe the mechanism shown in the question just once to show the deprotonation after chloride leaves because it wants to show that the chloride is used as the base to deprotonate the intermediate. I wouldn't think it is accurate to show that the chloride ion is used as the base to deprotonate the intermediate as it is likely not the most basic species in solution capable of deprotonating the intermediate. Am I right?

@TanYongBoon A proton can only leave if it has somewhere to go. Thus, always ask yourself where (whither) the proton would go. It is not really happy on the alcoholic oxygen if that carries a formal positive charge. It is not really happy on the acyl oxygen, except for the short moment in time when that is in a tetrahedral intermediate. It is also not really happy on an $\ce{R-OH2+}$ group. And it’s not really happy on chloride but at least the molecule is neutral. Frankly, if there is something that can accept the proton, proton transfer will be the fastest reaction in the mix.


text:
Why is this step exothermic?
From Wade's "Organic Chemistry" [1, p. 1003-1004]:

MECHANISM 21-7 Conversion of an Ester to an Amide (Ammonolysis of an Ester)

Step 2, loss of a poor leaving group is justified by claiming the step is exothermic, hence the TS energy isn't very high (as would be the case w/ $\mathrm{S_N2}$ since it's neither exo or endo)
Why is this step exothermic though? No neutralization of charge and no net additional bonds were formed (1 pi bond formed, 1 sigma bond broken)
Reference

Wade, L. G. Organic Chemistry; Pearson Education, 2012,. ISBN 978-0-321-81139-4.

The carbonyl pi bond is reformed, compensating the energy used in breaking the bond between the carbon and the -OR leaving group. Thus, it is overall exothermic

It's just wrong, proton transfer should be written before leaving.


text:
Shelf life of 1,4-Butanediol, colloquially known as BD or BDO
1,4-Butanediol is a chemical cleaning solvent used for industrial purposes and a precursor of GHB. Companies now post the "official" shelf life as "2 years", but a 1997 publication by BASF said GBL had an "almost unlimited" shelf life if stored properly. Is there a reason 1,4-Butanediol would have a shorter shelf life than GBL? Would anyone know the real shelf-life if stored in sealed containers? How much would it chemically degrade after 10 or 20 years? Thanks!


text:
Beryllium carbide formation
Why does beryllium form carbide of formula $\ce{Be2C}$ and not $\ce{BeC2}$ as the other elements of its group?
I am not able to understand the change in formula of beryllium carbide.

Please [edit] your question and add some of your own thoughts.

Related: https://chemistry.stackexchange.com/questions/34099/why-do-different-elements-form-different-types-of-carbides

There could be a thermodynamic aspect to it: see eg https://chemistry.stackexchange.com/q/34109/16683

Briefly, it's a matter of cationic radii values and crystal packing at given $p, T$. Without referring to the detailed review of the corresponding crystal structures, you can imagine $\ce{M^{II}2C}$ as the derivative of methane $\ce{CH4}$, whereas $\ce{M^{II}C2}$ can be seen as the derivative of acetylene $\ce{C2H2}$. Smaller cation is the best fit for the "richer" tetrahedral coordination environment of carbon, whereas larger cation of the same charge won't be able to do so (sterical hindrance). That's why $\ce{Be2C}$ is the primary carbide of beryllium, $\ce{Mg2C}$ is also known to exist at higher pressures, (magnesium, along with lithium, otherwise forms an allene-based $\ce{Mg2C3}$ or $\ce{Li4C3}$ 2), but bulkier cations $\ce{Ca^2+, Sr^2+, Ba^2+}$ only form 1/2-carbides. Aluminium, on the other hand, forms $\ce{Al4C3}$ that hydrolyses to methane. This is a good example of the diagonal relationship between $\ce{Be}$ and $\ce{Al}$, and also between $\ce{Li}$ and $\ce{Mg}$.


text:
Is it possible to neutralize acids with vinegar (e.g. when experiencing heartburn)?
I would like to know if it is possible to neutralize heartburn with a shot of vinegar, or if one should only use basics (like edible soda). What I know from high school, is that you can neutralize acids only with basics, but someone I know claims, that you can neutralize acids with acids, which doesn't make a lot of sense to me, but maybe there is some trick regarding the human body.

Sounds like nonsense to me. Acids don't neutralize acids. It probably may serve as a remedy in the same vein as shooting yourself in the leg (that is, whatever bothered you before becomes insignificant in comparison).

"Neutralize" is a disputable term here, I think. Vinegar is weaker acid and should dilute HCl, however it also acts at physiological level triggering the production gastric juice, doing more harm than good.

People often say to drink a little apple cider vinegar diluted in water and that the reason why people get heartburn is because there is a LACK of acid in their stomach that causes it. But oh, my God, they couldn't be more wrong. When it comes to heartburn, a teaspoon or so of baking soda in 8 oz of water will do the trick. Mix it well and drink it. It is only a temporary fix if you're dealing with a long term acid reflux issue.

Unlikely. I have found about 1/10 of a teaspoon of baking soda is plenty to neutralize acid in the esophagus which causes heartburn.

Do not try to use vinegar on your heart burn/ acid reflux. The only outcome is damage to your mouth and or esophagus. Try baking soda in water. Half a table spoon mixed well in water

This is bordering on a question asking for medical advice but at least the implicit claim should not be left undisputed. Neutralising acids with acids is something that only makes sense in the context of homeopathy — which I prefer to call homeopathetic methods to better underline their scientific valuability.
If you define ‘neutralising’ in the way it commonly is scientifically — namely removing the acidic/basic component from a mixture and moving the pH towards neutral — then additional acids will never neutralise significantly, unless you are using a weakly amphoteric acid to ‘neutralise’ a very strong acid. But no matter how hard you try, you will never arrive at a fully neutral pH using this method.
If you just define neutralising as changing the pH value towards 7, then indeed weaker acids will serve the purpose due to the principle of dilution: a more diluted solution of the same acid is less strong than a more concentrated solution. Or, in other words, $\pu{1l}$ of $\pu{96\%}$ sulphuric acid is stronger than a mixture of $\pu{500ml}$ of $\pu{96\%}$ sulphuric acid with $\pu{500ml}$ of glacial acetic acid. But as mentioned previously, you will arrive nowhere near a neutral pH using this method.
Whether using vinegar to combat heartburn makes sense medically — which I doubt since it is excess protons in a part of the body not designed to handle these — is off-topic here.


text:
Is it possible to make graphite an insulator?
Since graphite is anisotropic electrical conductivity, won't it be possible to make it an insulator instead? Like manufacturing it such that the plates of carbon atoms are mostly parallel to one another.

Tempting, but no.  Graphite is of course less conductive perpendicular to the basal plane but still boasts a conductivity of 330 S/m in that direction, https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity.  Electrons can "jump" fairly easily from one pi-electron "layer" to the next.


text:
Oxidation of aluminum and manufacturing of electrolytic capacitors
My background is in electrical engineering. Recently I saw instructions on how to manufacture electrolytic capacitors at home by mixing sodium bicarbonate (baking soda) to distilled water, submerging two pieces of aluminum foil into the mixture and applying electrical current through the pieces until current dropped to almost zero. The instructions I followed are here:
https://rimstar.org/science_electronics_projects/make_electrolytic_capacitor.htm
The end result was satisfactory.
However, as I have only very basic knowledge in chemistry, I have only a very vague understanding of what happened, and I would like to understand it in more detail. The article did not explain it very well and I haven't been able to obtain good explanation elsewhere either.
I understand that the current somehow oxidizes the aluminium to form an insulating aluminium oxide layer on top of another one of the plates.
So I would like to pose the following questions regarding this experiment:

Why is distilled water required? I understand that distilled water is basically tap water except its impurities such as charged ions have been removed. This, I believe, makes the water much less conductive. But don't we need the water to conduct? I believe the water, as the electrolyte, acts here as the other plate of the capacitor, with the oxide layer insulating it from the other plate in the final product.
What is the role of the sodium bicarbonate? What exactly does it do in the chemical reaction? We are using distilled water, but then we add sodium bicarbonate to it.
What is the reaction formula of the oxidation? Could you show me how the water, sodium bicarbonate and the aluminium react together to form aluminium oxide, and what is the role of the electric current? I understand that this oxidation of metals, such as aluminium, happens in air as well, and this method simply "accelerates" this process to produce a thicker, more insulating layer.

I would greatly appreciate if somebody could help me understand this process!

Excuse me for commenting when I can only guess: 1. distilled water is used to exclude hazardous contaminations 2. sodium bicarbonate makes the water an electrolyte, without Cl that could react with Al in an undesired way. 3. depending on the voltage, Al on the outside will be converted to an insulating layer of Al_2O_3

@GyroGearloose I am aware of the molecular formula of aluminium oxide - what I'm interested in is the actual reaction between all the chemicals involved. On points 1 and 2, I would like more detail. But thank you!

Yes, distilled water has its ions removed. It will still conduct cause you have added sodium bicarbonate to it-those are ions. Regular tap water has many other ions which might interfere in electrolysis, might be electrolytically deposited on your electrodes, etc.

You are performing an electrolysis.
This means that on the (+) electrode de reaction is Al $\rightarrow$ Al$^{3+}$ and on the (-) electrode the reverse occurs.

Distilled water is required to avoid the presence of chloride ions which will form chlorine at the (+) electrode and attack your aluminum foil.
Sodium bicarbonate is used to raise the pH so that the formed aluminum ions precipitate as the oxide instead of dissolving as Al$^{3+}$ ions.
The oxidation is Al $\rightarrow$ Al$^{3+}$. Then, the basic pH due to the dissolution of bicarbonate:
NaHCO$_3$ $\rightarrow$ Na$^+$ + HCO$_3^-$
HCO$_3^-$ + H$_2$O $\rightarrow$ OH$^-$ + H2CO$_3$
the aluminium oxide precipitates:
2 Al$^{3+}$ + 6 OH$^-$ $\rightarrow$ Al$_2$O$_3$ + 3 H$_2$O

The electric current forces the oxidation reaction to continue.
In air a thin layer of oxide is formed and then the rest of the metal is not in contact with the air anymore.
This is called passivation.
Since the aluminium ions are precipitating as oxide, in the (-) electrode the reduction is of sodium ions:
$\hspace{1cm}$Na$^{+}$ $\rightarrow$ Na
which reacts with water to form hydrogen and OH$^-$ so that the cycle is complete:
$\hspace{1cm}$2 Na + 2 H$_2$O $\rightarrow$ 2 Na$^{+}$ + 2 OH$^-$ + H$_2$

As I said I'm not an expert in chemistry so your answer will take me a while to understand completely. A couple of questions meanwhile: How does the increase in pH stop Al ions from dissolving? I understand that pH is a measure of positive ions in the solution, so does their presence (precisely their positive electric field) somehow repel the positive ions making them "stick" to the electrode instead? Second, why does the aluminium ionize in the positive electrode in the first place? Thank you!

@S.Rotos, it is just Chemistry. I understand that you are not in this field so it will be not to easy to explain in a few words. pH is a measure of H$^+$ ions in solution. When it is above 7 it means that there are more OH$^-$ than H$^+$ ions. Then the neutral species Al(OH)$_3$ will be formed, which, in the case of aluminium, will precipitate as Al$_2$O$_3$, the oxide. This oxide is insoluble so it will stick to the electrode.

@S.Rotos, the positive electrode drains electrons. When you take off electrons from metallic aluminium, you get the positive ion: Al$^0$ - 3 e$^-$ $\rightarrow$ Al$^{3+}$.

Alright, I think I have a much better idea now how it works. I will have to familiarise myself better with some of the concepts involved, but overall your explanation makes sense. I will accept your answer. Thanks!

Question (also not a chemist): Does this mean, once all aluminum is consumed, the resulting solution is sodium hydroxide + water + possibly leftover bicarbonate, with an undissolved alumina precipitate, and some released hydrogen gas? Am I understanding this correctly?

Or is there no NaOH because those ions were used to form the alumina? Or... could there still be some if Na was in excess of Al?


text:
Is methyl phenyl sulfoxide optically active?
I am concerned about the stereogenic centres. But in this compound there is no chiral centres present. Sulphur is $\mathrm{sp^2}$-hybridized, so it cannot be chiral anyway. But I can see there is no symmetry present in this compound. Will this be a sufficient reason to tell the compound optically active?

This compound is in fact optically active as the sulfur has 4 different substituents, one of them being a lone pair.

That makes it chiral.  But if the lone pair can umbrella-invert like ammonia you do not see optical activity.

Possible duplicate of Can heteroatoms with lone pairs be chiral centres?


text:
Is it possible to protect molten magnesium with Butane
I am currently experimenting with simple methods of metal casting at home. It is conventional wisdom that magnesium is better left to the professionals due to the unfortunate arrangement of its combustion and melting temperatures. I dont have access to any inert gasses or vacuum furnaces; but like a challenge.
The first thing that occurred to me was to use CO2 cylinders you can buy at the supermarket to protect my casting area. Luckily I didnt try that experimentally, since it is a known bad idea. Same applies to water vapor. Molten magnesium will happily steal the oxygen from either, leaving carbon resp. hydrogen gas behind. And a lot of heat of course.
Which brought me to the following counter-intuitive idea: if magnesium cares so little for carbon or hydrogen, could you protect your magnesium from burning with hydrocarbon gas? If I douse my casting mould in butane, then pour my molten magnesium in, I am likely to set the butane ablaze, sure. But I can deal with that; as long as my magnesium stays put.
Is this the dumbest idea ever? Or might it actually work?
To make my question more concrete: what is the known chemistry of magnesium at 700c and pure hydrocarbons? Note that there will also be a lot of silicon dioxide present, which can form a thermite with magnesium, but normally shouldn't under these circumstances. But perhaps the hydrocarbons could catalyze that reaction?

Molten magnesium requires very special equipment. Do not play around with this.

It's a dumb idea. If you insist, at least use Argon, cheaply available as welding gas.

It is a dumb idea. // Do not try this using argon gas. Yes argon should blanket the magnesium assuming you have the right equipment which you don't. This isn't a "challenge," rather it is playing Russian roulette.

You said: "I am likely to set the butane ablaze, sure. But I can deal with that; as long as my magnesium stays put." But, can the others in your apartment store, deal with that, too? Sounds not only dangerous but also careless.

Thanks for all the folk wisdom so far; still a bit short on arguments based in chemistry though ;)

For clarity; any experimentation will be done in an open field, well prepared for the worst case scenario of an explosion of burning magnesium

You're "prepared" for a worse case scenario of an explosion of burning magnesium? How the !@#$%^ are you going to put it out if some gets on you?

Excuse me, when you say "any experimentation will be done in an open field" I instinctively add "in a dry, woody area". Sorry ;-) . Mixing air with any flammable gas will be critical as you won't have control on the mixing, and explosion will not happen with too low oxygen, or too low inflammables. The pressure wave will be unpredictable, as long a you have not a long experience with bomb laying or other cases of explosions happening.

Max: I'm thinking of donning a basalt fiber blanket. Not quite 2000c rated but should be good enough for transient splashes.

Gyro: thinking about it, I think setting the mould on fire before coming in with the crucible would be the better idea, so at least you take out that element of surprise, and I can judiciously place my crucible under the flame front before tipping the magnesium from under it's flux.

http://www.magnesium.com/w3/data-bank/index.php?mgw=222 Handling molten magnesium.

Hmm, that source also mentions you shouldn't try putting out a fire with sand; still I don't believe typical melt temperatures suffice to start stealing oxygen from silicon. Then again I really don't have a clue what kind of oxides are in the dirt I am currently working with... Replacing my sand with nacl should solve that though; it has been a fine mould material for zinc so far and I'm sure that won't react with magnesium as it is an integral part of my flux...

The autoignition temperature for butane is under 700C, so you WILL have a butane fire once the molten magnesium hits the butane saturated mold; flames mean oxygen is present so it is possible the molten magnesium can still react with O2. You can't do this safely in a normal atmosphere, you would need some sort of purge stream over the working area to keep the environment from supporting combustion.

Ari; thanks, yeah, the way I was envisioning it is to have the casting sit well below the top rim of the casting flask, and have the flamefront rise over that in a way that allows me to stick my crucible under it. I was hoping the natural tendency of butane to evaporate at ambient conditions would be enough to provide a purge stream.

...the nice thing is that I can experiment with that without the involvement of any magnesium; or I can try if I can extinguish small lints of burning magnesium by moving it under such a flame front. By controlling the temperature at the bottom of the flask I suppose I should be able to get an arbitrarily strong purging stream.

answer to: "thinking about it, I think setting the mould on fire before coming in with the crucible ...": I don't know if this is brilliant or crazy, but at least it would circumvent some major hazards and could also contribute some of the heat to keep the magnesium molten. Worth thinking about.

if it might pull the oxygen out of sand , try using magnesium oxide, surely it won't pull the oxygen from its own self. Or using burnt lime, surely it won't pull oxygen away from calcium.
It might react with butane to form magnesium hydride and carbide ....which would be fun to play with . Also magnesium easily reacts with nitrogen to make the nitride...which is an excellent way to make ammonia

Not sure those make good mould materials; not without water but that's not a option. Apparently carbides and hydrites don't form in the reaction with co2 or water; are you aware of any particulars of these compounds that might be relevant to the situation?

This is a largely nonsense answer.

Magnesium most certainly CAN form a carbide when heated with a hydrocarbon! !

gert is the one full of nonsense

You'll likely decompose the hydrocarbon.