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Uncertainty Equations Confusion
I am confused about the uncertainty equations calculations.
If I have the equation X1 * Y1 = X2 * Y2 and I want to find the uncertainty of X1 , I convert the equation to X1 = (X2 * Y2)/(Y1) .  Now, I would know how to calculate the uncertainty if it was a straight division or multiplication uncertainty but this equation is both.  How do I go about calculating the uncertainty if the equation contains both multiplication and division? Do I just do the multiplication uncertainty first and then the division uncertainty?
Thanks any advice is appreciated.

You have three variables, for simplicity let's call them, x, y, z each with a (standard deviation) uncertainty $\sigma_x, \sigma_y, \sigma_z$. The formula you want to find the error on is then $f=xy/z$ which will have a standard deviation $\sigma_f$ The method that always works is the  the formula that gives the square of the uncertainty
$$ \sigma_f ^2 = \left( \frac{\partial f}{\partial x}\right)^2_{y,z}.\sigma_x^2+\left( \frac{\partial f}{\partial y}\right)^2_{x,z}.\sigma_y^2+\left( \frac{\partial f}{\partial z}\right)^2_{x,y}.\sigma_z^2$$
which looks horrid and intimidating  but is very easy to use as it only involves taking the derivative of f with x, y and z in turn, while keeping the other variables constant.
In your example $\partial f/\partial x = y/z$, $\partial f/\partial y =x/z$ and $\partial f/\partial z =-xy/z^2$. Putting the parts together gives
$$\sigma_f^2= \frac{y^2}{z^2}\sigma_x^2 + \frac{x^2}{z^2}\sigma_y^2 +\frac{x^2y^2}{z^4}\sigma_z^2 $$
You can simplify this a bit but you should have all the values to plug into this. Don't forget to take the square root at the end.

That method doesn't "always" work. Like mine it ignores the covariances, $\sigma_{x2y2}, \sigma_{x2y1}, \sigma_{y1y2}, \sigma_{x2y2y1}$

As long as the variables are un-correlated this method is ok, and I suspect that this is the case in the OP's question. You would need to know all the cross-correlation sigmas of course and then its best to use matrices to do the calculation.

multiply all the terms in the last equation by $$\dfrac{f^2z^2}{x^2y^2}$$ which is equal to 1. then you'll get $$\dfrac{\sigma_f^2}{f^2} = \dfrac{\sigma_x^2}{x^2} + \dfrac{\sigma_y^2}{y^2} +\dfrac{\sigma_z^2}{z^2} $$

nice, I didn't try to simplify this but I'll add thus  to the answer if that's ok with you? I could also add covariance bit so that there is complete answer for others to use.

$\%\text{RSD}_{X1} = \sqrt{\%\text{RSD}^2_{X2} + \%\text{RSD}^2_{Y2} + \%\text{RSD}^2_{Y1}}$
For multiplication and division the error propagates as the % relative standard deviation.
This ignores the covariances which means that there is no interaction between the variables.


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What are aryl chlorides currently used for?
In a recent paper by Gustafson et al. (DOI: 10.1021/acs.orglett.6b02650) the authors open with the argument that

aryl chlorides are both versatile synthetic handles and common functionalities in drug discovery.

However, they fail to give examples of current use in industry and pharmaceutical research.
I know of only one reaction that aryl chlorides are used for, and that would be the Ullmann biaryl (ether) coupling. Are there any other ones for which aryl chlorides are a good starting material?
And then, about the second part of the sentence: Why are aryl chlorides a common functionality in drug discovery? Is it simply that compounds are chlorinated and then compared to their unchlorinated analogues?
All in all, please provide examples for the quoted statement above.

Synthetic handles: for organometallic couplings (Sonogashira, Suzuki...), there is a selectivity ($\ce{aryl-I} > \ce{aryl-OTf} >\ce{aryl-Br} >\ce{aryl-Cl}$) that can be crucial in the total synthesis. You can make the other position ($\ce{aryl-I}$ or whatever) react first, then use modified conditions suitable for aryl chloride couplings (plenty of examples on this website). Or make an halogen exchange before coupling your chloro- position.
Common functionalities: halogen bonds can change binding affinities. Even if they are weaker than hydrogen bonds, they are specific (see C. Bissantz et al., J. Med. Chem., 2010, 53 (14), 5061–5084.). They can make interactions with electrophiles, nucleophiles, or with themselves (ex: $\ce{C-X...O}$ $\ce{sp^2}$ or $\ce{X...X}$). One example of molecule designed with chloroaryl moieties is
represented here (see Furet et al., Bioorg. Med. Chem. Lett.
2012, 22 (10), 3498–3502, which shows the docking model). This is an inhibitor which blocks the binding pocket of the
regulator protein MDM2. The 6-chloroindolyl moiety enables to fill a
subpocket (TRP 23), while the p-chlorophenyl fills another one (Leu 26). The chlorine-chlorine interactions between the 6-chloroindolyl moiety and the triptophan residue in the protein was a key point in the design of this inhibitor, strongly enhancing the binding affinities.

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Carbylamine Reaction
In my lab manual, it is given that alcoholic KOH should be the base used in carbylamine test. However as it was not available, I used 10℅ NaOH. The test results came out positive. (Yes, it was actually a 1° amine) Is there any reason to prefer alcoholic KOH over NaOH?

My memory tells me that KOH is more soluble in ethanol than NaOH. The chemistry is the same, just more efficient at the higher concentration.

What's the use of ethanol? The elimination at the last step?

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What do we have both r.a.m. and r.m.m.?
Read Orthocresol's answer here first: What is the difference between "molecular mass", "average atomic mass" and "molar mass"?
Orthocresol, in his answer, says that atomic mass is the weighted average mass of the atoms of an element and molecular mass is the mass of the molecule of an element/ compound obtained by adding the atomic masses of the constituent element's/ elements' atoms. Both atomic mass and molecular mass have the units of mass which may be kilograms, grams, daltons or a.m.u.
Relative atomic mass/ atomic weight is the ratio between the atomic mass of an atom and 1/12th of the atomic mass of a C-12 atom and similarly, relative molecular mass/molecular weight is the ratio of the molecular mass of a molecule and 1/12th of the atomic mass of a C-12 atom. Both R.A.M. and R.M.M are unitless since they are the ratio of two physical quantities which have the units of mass.
My Question: Atomic masses and molecular masses expressed in a.m.u or u have the same numerical value as R.A.M. and R.M.M respectively. What is the point of having R.A.M and R.M.M?


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formation of σ molecular orbital in H2

In the picture, graphs show the constructive interference of two 1s atomic orbitals. My question is, what does the horizontal axis in wavefunction graph represent, does It represent the distance from the nucleus? We see the pick point of wavefunction on H atom, does It mean that highest probability of electron density is on the nucleus, which is impossible.  I am quite confused about the meaning of a wavefunction, can anyone help me?

We should also keep in mind that nuclei are usually described as point charges in quantum chemistry.

You are mostly right. The horizontal axis represents the distance from the nucleus. The maximum of the electron density of 1s orbitals is in fact at the nucleus. This is not impossible, remember that electron density can be interpreted as the probability that an appropriate experiment will detect the electron there. In fact, electron density "at" the nucleus is one of the major ways NMR spectroscopy works.

Thank you for the answers,  everything is more clear now.  As far as I understand since the size of nucleus is extremely small,  Its considered as a point and electrons can be assumed to have the highest probability on that point.  Even though they are not exactly inside the nucleus It can be assumed that they share same location.


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Reaction of white phosphorus with aqueous sodium hydroxide
What are the products obtained in the reaction of white phosphorus with aqueous sodium hydroxide?
I know one will be phosphine ($\ce{PH3}$).
How can we find the oxidation state of phosphorus in the other product?
Thanks for help in advance.

We can't. We just know it is hypophosphite; that was not obvious at all, nor could it be easily derived from first principles.

Phosphorus reacts with sodium hydroxide to give phosphine and alkali hypophosphite:
$$\ce{P4 + 3NaOH + 3H2O -> 3NaH2PO2 + PH3 ^}$$
The O.S. of phosphorus in disodium hydrogen phosphite is:
$$\ce{ +2*1 + 1 + x + 3*(-2) = 0}$$
$$\ce{ x = +3}$$

Mechanism: The water that is involved is actually moisture that is accounting for the hydrolysis of phosphorus. The reactions can be dissected into three semi-reactions:
$\ce{3P + 3H2O -> 3OPH + 3H}$
$\ce{P +3H -> PH3}$
$\ce{30PH + 3H2O -> 3H3PO2}$
The phosphine so obtained usually inflames spontaneously on coming into contact with the air; each bubble as it escapes forms a vortex ring of smoke. This is due to presence of some amount of $\ce{(PH2)_x}$, the simplest of them being the dimer, $\ce{P2H4}$.
Source: http://phosphorus.atomistry.com/phosphine.html

Can sodium phosphate form?

@Mockingbird if we use some phosphorus compound like phosphorus pentoxide, phosphoric acid or phosphorus trihalides and react with sodium hydroxide, sodium phoshate will form.

You used Chemiday. That is a well-regarded and reliable source, as surely as my running a sub-3 hour marathon (when I tried one it was closer to seven). Can you review and identify primary references?

When Phosphorous reacts with $\ce{NaOH}$, it produces Phosphine and Sodium Hypophosphite.
The reaction is as follows:
$$\ce{P4 + 3NaOH + 3H2O -> PH3 + 3NaH2PO2}$$
Oxidation State of Phosphorous in Sodium Hypophosphite is $+1$, as $\ce{Na}$ is $+1$, $\ce{H}$ is $+1$, and $\ce{O}$ is $-2$.
Solution:
$1+1 \times 2+X-2 \times 2=0$ (As net charge is zero)
Solving for $X$ gives $+1$.

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Steroid nomenclature: β-sitosterol
Can someone tell me why β-sitosterol has the beta prefix? What is it denoting? When comparing its structure to other similar phytosterols I see no need for it.

$\alpha$ and $\beta$ denote the stereochemistry at C-3, the one with the $\ce{OH}$ group, relative to the methyl group at C-10.
$\alpha$ means trans to the reference substituent, $\beta$ means cis (= on the same face of the molecule).

Edit
You asked in a comment

Why don't then campesterol and stigmasterol also have the same designation?

To my knowledge, both compounds are natural products for which only the 3-$\beta$ isomer is known. It is therefore not necessary to use the $\alpha$ and $\beta$ prefixes here.
For sitosterol, on the other hand, both isomers are known and thus need a unique identifier.

Why don't then campesterol and stigmasterol also have the same designation?

@Klaus I wonder if you have a good source for your claims. Note that there's no ‘3’ in ‘β-sitosterol’. (Also note that the Greek letters are not to be italic typeset.)

While one might think that β-sitosterol is an epimer of α-sitosterol, differing by configuration at C-3 (with $\ce{-OH}$ group), it does not seem to be the case.
There are various products and their isomers of phytosterol sitosterol biosynthesis, and their trivial naming seems to be somewhat rudimentary.
All of the sitosterols are 3β alcohols. The Greek letter (beta) designation in ‘β-sitosterol’ seem to be for more general, non-systematic isomers designation. In the field of natural compounds, it has historical reasons – a compound was isolated, described and named, and later was found that it in fact is a mixture of different isomers. (E.g. “α-caryophyllene” and “β-caryophyllene” do even have different number of cycles).
Sitosterols mentioned the Merck Index (14th ed.) are as follows ("natural product" IUPAC name according to last steroid nomenclature shown as well, might slightly differ from those from the M.I.).

(I) α1-sitosterol, (24Z)-4α-methyl-5α-stigmasta-7,24(241)-dien-3β-ol
      (or (24Z)-4α-methyl-5α-poriferasta-7,24(241)-dien-3β-ol)
(II) β-sitosterol, stigmast-5-en-3β-ol
(III) γ-sitosterol, poriferast-5-en-3β-ol

So, β- and γ- sitosterols (II, III) are diastereomers, epimers at C-24 (those differently named parent structure names stigmastane and poriferastane are in fact just (24R)-24-ethylcholestane and (24S)-24-ethylcholestane, respectively; these ethyl carbons are nowadays numbered 241 and 242 instead of older 28, 29). α1-Sitosterol (I) has the double bond not at C-5 but at C-7, additional methyl at C-4, and additional double bond at C-24 (ethylidene instead of ethyl). (And I guess that that ‘1’ subscript might be because of E/Z isomerism if this double bond.)


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Diffusion of carbon from ferrite to cementite in the eutectoid transformation
Why does the carbon of an alloy diffuse from ferrite (0.022% C) to cementite (6.70% C) during eutectoid transformations? I know that the diffusion of a generic species (e.g. of atom A) carries A from regions where the amount of A is higher to regions where the amount of A is lower.
I found this sentence on page 337 of Callister (9th ed.).

The rule you mention does not apply if different phases are involved. Moreover, cementite is iron carbide $\ce{Fe3C}$ that crystallizes during the transformation. The 6,7% C come from its stoichiometry.

Diffusion is perhaps more complicated than initially taught. Yes, all things being equal you will get diffusion from regions of high concentration to regions of low concentration.
But, in the case you mention, all things are not equal - there is a phase change occurring, and that means that there are chemical potential gradients as well as concentration gradients. Now you have an additional driving force for motion of impurities as they prefer to be in the phase with a lower Gibbs free energy.


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What happens with sulfur dioxide reacts with dimethyl trisulfide? What compounds are left over?
I have only found one single reference to this reaction online which states $\ce{SO2}$ will destroy DMTS ($\ce{C2H6S3}$), but I cannot find a reference as to what may be happening or what may be left over.
Can anyone help me solve this? I'm not a scientist or student. I'm actually a home brewer trying to reduce DMTS in a finished beer by potentially adding $\ce{SO2}$, but I don't want worse compounds left over.


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Why is the mass of 1 mole of an atom/molecule equal to its gram atomic/molecular mass?(proof)
Why do we take the mass of 1 mole of an atom to be equal to its gram atomic mass and the mass of 1 molecule to be equal to its gram molecular mass? I want a mathematical rigorous proof. There are other questions on this site dealing with this question but none of them have the "proof".

@Mithoron This is a self answered question. I did not like the answers in the question mentioned and there were many answers there and my answer probably would not get noticed there so i decided to post it separately.

$\def\p{\mathrm{p^+}}\def\n{\mathrm{n}}$
Mass of one proton = $m_\p$ and mass of one neutron = $m_\n$. (Masses are in unified atomic mass unit-$\mathrm{u}$)
C-12 isotope consists of 6 $\p$ and 6 $\n$. Therefore, mass of C-12 isotope is equal to 6 $m_\p$ + 6 $m_\n$. Since $m_\p$≈ $m_\n$, the mass of C-12 can be replaced by 12 $m_\p$. (electrons are not taken into account since they have negligible mass)
1 $\mathrm{u}$ is defined as 1/12 of the atomic mass of C-12 isotope.
⟹1 $\mathrm{u}$ = 1/12 × (6 $m_\p$ + 6 $m_\n$)
⟹ 1 $\mathrm{u}$ ≈  1/12 ×12 $m_\p$
⟹ 1 $\mathrm{u}$ ≈  1 $m_\p$   ...equation(1)
Now,
12 g of C-12 has $\pu{N_A}$ particles and constitutes 1 mole of it. (according to the definition of 1 mole)
⟹ 1 mole of (6 $\p$ + 6 $\n$) have a mass of 12 g.
⟹ 1 mole of (12 $\p$) have a mass of 12 g.
⟹ 1 mole of $\p$ have a mass of 1 g
⟹ 1 mole of $\p$ × 1 $m_\p$ = 1 g
⟹ 1 mole of $\p$ × 1 $\mathrm{u}$ = 1 g
⟹ 1 mole of $n$ $\p$ × 1 $\mathrm{u}$ = $n$ g
Therefore, mass of 1 mole of $n$ $\p$ is equal to $n$ g
Now, the final part,
Let an atom have $x$ $\p$  and $y$ $\n$ . The atom can be considered to be $(x+y)$ $\p$  for our purposes.
1 mole of $(x+y)$ $\p$ have a mass of $(x+y)$ g.
Thus the mass of 1 mole of atoms of an element is equal to its gram atomic mass.
Molecular mass of an element or a compound is equal to the sum of the atomic masses of each type of it's constituent atoms.
Thus the mass of 1 mole of molecules of an element or a compound is equal to the gram molecular mass.

the way you get the mass of carbon-12 is just wrong. $m_p \ne m_n$ and you ignored the binding energy.

@MaxW Firstly, I have clearly said that the mass of a proton is approximately equal to the mass of a neutron and secondly, why should i consider binding energy?

The point is that $N_A$ atoms of $\ce{^12_6C = 12  \text{grams}}$  by definition.

I have stated this in my answer.

Why is my answer downvoted?


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Why does shower gel / shampoo foam a lot more on hairy than non-hairy body parts?
We usually use shower gel by rubbing it on our palms and then applying it to each body part in turn in order to clean oily substances that clung to it since our last shower. This procedure does not cause the gel to foam (if it is applied to a completely hairless place, such as a clean-shaven head or back, it will produce some small amount of foam (also depending on how much you rub it); if it is applied to a place with relatively small clusters of hairs, such as arms or legs, it might foam up a bit more, and so on).
However, when applying shower gel or shampoo to a hairy place, such as the head, the groin, etc. it usually foams up significantly and rather quickly. A clear shower gel or shampoo will become a white foam in a short time.
Why is that? Is it the air trapped among the hairs?
(Please do not confuse this question with this one, they are different enough).

Not foam at all?  I can wash my hands (no hair) with shampoo and I still get at least some foaming.  I think the difference in foaming is a matter of quantity, not of quality.

@CurtF. Yeah, it's too definite to say "not foam at all"... I'll edit.

No need to tell where you have hair and where you don't

It's mainly due to the surface area and the spacial distribution of hair compared to that of hairless skin.
A typical human hair width is around one micron.  Given the various hair length and thickness (space between hairs) on the different hairy regions of the body, there are orders of magnitude more surface area on which soap bubbles (foam) can adhere as compared to smooth (or even very rough), hairless skin.
The second part of that answer, the spacial distribution, just means that the hair gives support to sustain the bubbles so they don't immediately collapse.
Short answer, but I think this pretty much covers it.

A loofah produces more lather than our hands because of its high surface area, which is due to the large number of holes in it.
When the lather comes in contact with the oil particles micelle formation takes place and the lather fades away as the soap particles adsorb on the oil particles. The hair can act as the loofah by increasing the surface area for the soap to act on.
You may have experienced this that when we wash our hands full of oil with hand wash not too much lather is produced but when we wash our hands which are almost oil free or dry a lot of lather is formed.
There could be another reason,
A typical soap is a Sodium salt of higher fatty acids (RCOONa). At low concentrations soap acts as an electrolyte but at a certain temperature (known as Kraft temperature) and concentration (Critical micelle concentration), the soap particles adsorb on the oil particles and form a colloid.
You may have also experienced that if we take a lot of hand wash and a small amount of water it doesnt lather much but if we take the hand wash with a considerable amount of water a lot a lather forms. This may be due to the critical micelle concentration which wasn't achieved in the former case but achieved in the later case.


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What is the yellow precipitate formed from reacting diamminsilver fluoride and potassium iodide?
These two compounds, silver diamine fluoride and potassium iodide (actually a solution called Lugol’s solution), are being used together in dentistry.
When combined they form a yellow precipitate. I cannot find any articles that state specifically what is formed. I would like to know what the yellow precipitate is and the steps involved in the reaction of the two chemicals. The equation is:
$$\ce{Ag(NH3)2F + ssKI -> ?}$$

Note that Lugol’s solution is not $\ce{KI}$ but $\ce{KI/I2}$ or $\ce{KI3}$.

Silver iodide is formed which is yellow in color.

The yellow precipitate that forms is silver(I) iodide, $\ce{AgI}$. It is very, very insoluble and hence will precipitate immediately if silver ions and iodide ions meet in solution.
In fact, all three common halides other than fluoride form insoluble salts with silver: silver(I) chloride is a white precipitate, silver(I) bromide is somewhere between white and yellow and silver(I) iodide is yellow. These precipitates are so characteristic that they were historically (and still in qualitative inorganic analysis lab courses) used to identify the different halide ions in solution. To further improve the distinction, silver(I) chloride can be dissolved in dilute solutions of ammonia while silver(I) bromide needs concentrated ones. Silver(I) iodide does not dissolve in ammonia.
$$\begin{align}\ce{Ag+ (aq) + Cl- (aq) -> AgCl v &->[NH3 (dil)] [Ag(NH3)2]+ (aq) + Cl- (aq)}\tag{1}\\
\ce{Ag+ (aq) + Br- (aq) -> AgBr v &->[NH3 (conc)] [Ag(NH3)2]+ (aq) + Br- (aq)}\tag{2}\\
\ce{Ag+ (aq) + I- (aq) -> AgI v &->[NH3] \text{no reaction}}\tag{3}\end{align}$$
I assume that the original silver solution was the diammine complex to prevent silver chloride from precipitating as soon as chloride ions are added (due to not perfectly deionised water being used).

To add further to your question, for a silver diamine dental product (Riva Star by SDI), the white precipitate disappears after applying more potassium iodide to the mixture. The excess solution in the tooth cavity is then rinsed off with water.
There is some cool chemistry going on here but I am unsure what it is.
Riva Star comes in 2 solutions that are mixed together in the tooth cavity.
Solution (1): Silver diamine fluoride has a pH of 10.
Solution (2): Potassium iodide has a pH of 7.
Tooth: Consists of collagen, proteins, hydroxyapatite, bacteria.
When solution (1) is first applied to the tooth, the high pH has been shown to aid in the formation of covalent bonds of phosphate groups onto proteins and crystallites to grow. It has been postulated that the fluoride usually reacts with hydroxyapatite, by replacing the calcium to form fluoroapatite, while the silver diamine exerts an antibacterial effect. The silver ions also react with the tooth structure to form the dark black silver phosphates.
So, the possible by-products could be Calcium ions, fluoride ions, ammonium ions, silver ions, silver phosphate.
The addition of solution (2) is to remove the excess silver ions that may form the dark silver phosphates. This is to avoid black stains on the tooth. Solution (2) is added into the cavity to form white precipitates and more solution is added until the white precipitate disappears.
What chemical reaction(s) could be going to explain the formation and dissolution of the white precipitate?
https://www.aapd.org/assets/1/7/SDF_White_Paper_Final.pdf


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PAW vs USPP: Pseudopotentials in commercial DFT codes
What is the difference between USPP (ultra-soft pseudopotential) and PAW (projector augmented wave) in commercial DFT codes?

Are you asking the difference in the algorithms or in utility/use?

This question is somewhat ambiguous, do you want a comparison of USPPs and PAWs in general or specifically within the commercial codes ? The more details and the more pointed your question is, the better answer we can give !

Here is the answer for a very similar question to yours: https://mattermodeling.stackexchange.com/q/1986


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Circular dichroism of a hydrazone boron complex
Last year, I came across a quite puzzling article, in which the authors claim to have found a photophysical explanation for aggregation induced emission (doi:10.1038/nchem.2612). The model proposed implied that the lowest-lying excited state ($\ce{S1}$) was a non-emissive ("dark") state, whereas $\ce{S2}$ (or $\ce{S3}$) was emissive, and depending on the solvent, the molecule could be more or less emissive.
They claim to prove their model by observing the "dark state", and among other concerns, one point that particularly retained my attention was that one of their proofs was a CD spectrum (supporting information page 43) of this molecule:

Now, simply put, I don't understand anything about how it is possible. To precise my trouble, this can be broken into different subquestions (please feel free to answer partially if you wish).

I see strictly no reason why this molecule should show such a CD spectrum in solution (77K in Glycerol/Ethylene glycol (8:2 v/v)). Even if it is chiral, it should be a racemate. Why does this molecule show CD?
The maximum of the band is 150 mdeg. Why is it so high?
Absorption and CD spectra are not completely superimposable. The authors claim that this demonstrates that the band they are looking at consists of two different electronic transitions (hence proving the existence of the "dark state"). I don't find this statement straightforward (especially judging by the signal shape). If you have references which elaborate on this aspect, or other references using this same argument, I would be very pleased to read them.

Am I missing something very important concerning chirality, or CD spectroscopy?

I too find their conclusions surprising from their data, but this is probably not the best forum in which to criticise. Molecules with rotors showing viscosity dependent excited states are not new, see work on triphenylmethane derivatives such as malachite green & crystal violet. These (and similar) were studied in the 1980 with proper picosecond pump-probe resolution. (I found this review see 'Reports on Progress in Physics' 1993, v56 p63. see p160)

Thank you very much for your insight, I was precisely lacking knowledge on the time-resolved spectroscopy approach of this topic, this is perfectly useful. Thanks again

On the origin of chirality, you should read something about atropoisomerism. This sort of chirality arises due to a hindered rotation around a chiral axis.
It is most notably encountered in biaryls, because by introducing substituents you can make rotation more difficult and observe a CD. However, it is also present in very simple molecules. Think about H2O2: every conformation where the dihedral angle is between 0° and 90° is chiral. Probably your teacher didn't point this out when explaining butane conformations, but gauche conformers exist in enantiomeric forms. The problem with these small molecules is that to freeze rotation you should basically go to very low temperatures, and then separate the enantiomers. You would observe CD. If you can do that, you got material for a pretty awesome paper in my opinion!
In this case, think about rotating the phenyl ring around the N-Ph bond. Draw the structures, and you will notice that they are chiral. The claim of the authors is that in a high viscosity solvent this rotation is hindered at room temperature, thus you will observe CD (see more on pg. 46 of the Supp. Info).
Concerning the racemic mixture, I'd say you are right. It is not clear to me why they see CD without separating the enantiomers, or maybe they did but they didn't explain it thoroughly. They refer to some previous work, so maybe you could find more details in other papers.
Why is the maximum band so high?
High with respect to what? 150mdeg really doesn't mean anything if you don't have the concentration and cell length and you can't work out the delta epsilon and the g-factor. CD people never use mdeg because it is not an absolute unit. The authors do not report concentration and cell path, so you can't really say whether it is strong or not, also because absorbance is reported in arbitrary units, so you can't really work out the g-factor. Clearly, they're not CD people, they use it as tool in their research, so the information they provide is not really helpful to answer this question.
Concerning the signals, have a look at figure 64 in the Supp. Info. Think about spectra as a series of lines to which you apply some lineshape function (usually Gaussian or Lorentzian). Now, two isolated bands are exactly superimposable if they have the same intensity and the same broadening. If they are not superimposable, one of these two things is not right, or the bands are not isolated. Since you are speaking about two different spectra, you can always normalise them to the same intensity. Additionally, they are recorded in the same conditions (possibly simultaneously), so there is really no reason why the broadening should be different. The only other possiblity is that there is more than one transition under that signal.
The "dark state" means basically that, in UV such state has a low absorption because of some reason. However, some transition that is dark in UV can have a pretty significant CD signal, because maybe it is only magnetically allowed (think about the n-pi* transition in carbonyl groups, or some charge-transfer state).
I tried my best, however the most interesting point (why do they see CD) still remains obscure to me, mostly because it's weekend and I am too lazy to read the paper! If I remember, I'll maybe read it next week.

Edit:
One thing that I didn't mention is that the sp3 N atom is chiral as well, so actually here, when you put in the axial chirality, you're really speaking about diastereoisomers, if you can fix the chirality of the N atom. This happens in very particular cases (see Cinchona alkaloyds), but since we're in a high viscosity medium, who knows, this might be the case, and explain why only one of the two possible configurations of the chiral axis is preferred.

I'm no student, and I think I did point out I saw the chirality. I'm asking because I rarely work with CD in my research. For the rest of the article I can judge by myself and like porphyrin rightly said, I won't discuss this here. I needed to know if I missed some obvious point in CD spectroscopy though, for research purpose. Thanks anyway, it's nice to discuss science.

Also, please consider the geometry of the N atom (in SI), before discussing diastereoisomerism. Partial excitation of enantiomeric structures with circularly polarized light in viscous media had also been reported, and I initially thought the authors did that, but this is not the case. In fact, "go fetch yourself a transient absorption spectrometer to properly prove your point" was very fine for me as an answer, and what I was looking for ;)

Apart from the geometry, there's actually no way to control the N inversion unless it's in some very rigid structure. I don't understand how transient absorption would tell you anything about the CD. I don't know what partial excitation is...do you mean using only one circularly pol. light?
Anyway, if you google the instrument they have used for CD, it states that it can exert magnetic fields, so it might be that they used magnetic circular dichroism instead of regular CD. If that were the case, the molecule need not be chiral to have a signal.


text:
HPLC Lab Report: Why is weight/purity % sometimes over 100%?
I help with the supply chain of a small supplement company, sometimes the HPLC lab reports we get back have a weight % over 100%. What does this mean, how does this kind of thing happen? Obviously the sample can't be over 100% pure.

Rounding error maybe?

So like a rounding error from the weight calculation algorithm they're using? I've seen it occasionally with different contract labs.

How many different parameters are being measured and how much over 100% is the total? Or is it just one compound coming back as >100% ?

I used to get that question all the time from clients when I worked in an analytical lab (also ~"why is the total 99.5% when it should be 100%). Frequently it came down to error propagation when we'd analyzed for half a dozen compounds.  There were also frequently problems regarding sample homogeneity, confusion about reporting on dried vs. as-received basis, etc.  The latter issues only really apply if you are having a number of different analyses run from one or more samples though.

@airhuff I updated with an example of what we see. Sometimes it'll be the purity assay on the COA that is over 100%.

Definitely very helpful in answering your question.  It sure looks well separated and properly integrated (i.e. separated from the little peak next to it).  I could be behind the times, but I just don't see 100.6% as a significantly high result for an HPLC assay.  Usually when you see assay specs like 0.9999% pure for a compound, for example, they have measured for traces of other stuff. It isn't that they've done an assay for just compound A itself and come up with 99.99% compound A, repeatedly.  I don't know if this applies to your expectations and situation but that's my 2 cents ;)

You could ask the lab for specs like uncertainty of the reported values.

Ok so, rounding error is a sufficient explanation?

Not given the additional information you have provided.  I would strongly suggest "within analytical uncertainty of the method", but you would need to verify with the lab what the accuracy of that measurement is.

By far the most likely reason for reported purities of over 100% when using HPLC is the "analytical uncertainty of the method", but let me illustrate exactly how this is likely to arrive.
A standard procedure for estimating purity is:

Run an exhaustive series of known standards with (assumed) exactly known purities, or weight percents, on the HPLC.
Plot this calibration data with peak area on one axis, and the known weight % on the other axis.
Fit a line to this data.  The slope of this line is called the "response factor" or sometimes just the "response".
Run some unknown samples and measure the peak area for the unknowns.
Using the response factor, calculate a weight percent for the unknown sample.

If there are analytical uncertainties in the calibration standards, then the response factor may be randomly off by small amounts from the "true" value of the response factor.  Similarly, if there are uncertainties in the peak area of the unknown sample, even using the "true" response factor may give a value that is slightly off.
Statistically, these errors should be random, so that sometimes when the "true" purity is 100.000 % the measured value will be 100.6%, and sometimes it will be 99.4%.
Chemically, what is the source of the errors?  Well, maybe the autosampler that injected the unknown sample was supposed to inject 2.0000 μL of sample for analysis, but due to instrument imperfections actually injected 2.012 μL instead.  Maybe there was a tiny power surge in the AC lines going to the UV detector that was measuring the peak right when the compound of interest eluted, leading to an ever-so-slightly greater signal at that moment...
The list of possible causes is nearly endless, but none of these things are really cause for concern, unless you really need to know the purity to three or more decimal places.  From the report you provided, you can safely assume that the uncertainty in their analytical method is at least 0.6%, but not much else.


text:
How to make p-chloroanisole from methanol and benzene?
I am tasked to synthesise para-chloroanisole from methanol and benzene. Other reagents are permitted.
Would it be correct to have the benzene attack the methanol $\ce{H}$ and then attack the carbocation that forms with the methoxy ion? From there use $\ce{Cl2}$ and $\ce{AlCl3}$ which would install para and ortho or a combination?
Also if you write one equivalence, does that suffice to say that you are only having it attach once, so that you would only have a max of two groups at the end for each possible product e.g. addition at the 1,2 positions and 1,4 positions?

You should, in general, ask only one question per post.

Benzene is not basic enough to be protonated by methanol, which is precisely the first step you are proposing.
The use of free methanol to generate an aromatic methyl ether immediately inspired me to take a look at the Buchwald-Hartwig coupling. While the website only mentions the applicability to phenol couplings, the bible of organic chemistry, Strategic Applications of Named Reactions in Organic Synthesis also mentions the possibility of using alcoholates as a coupling partner.[1]
This leaves us the question of how to get to a compound suitable to be subjected to a Buchwald-Hartwig coupling. We would need an aryl halide into which the palladium catalyst may insert. Fortunately, we need halogenation anyway, so we might just make it a two-step procedure. Chlorination of benzene is well-known and well-taught, so I decided to check whether bromination of chlorobenzene is also known. Turns out that Ferguson et al. published $89~\%$ yield of the bromination of chlorobenzene using bromine and $\ce{AlBr3}$ in $\ce{CS2}$ — fully para-selectively.[2] This gives us the following sequence:
$$\ce{C6H6 ->[Cl2/AlCl3] Cl-C6H5 ->[Br2/AlBr3][CS2, $89~\%$] Cl-C6H4-Br ->[Pd(OAc)2, NaOMe][base] Cl-C6H4-OMe}$$
Wherein both the latter two are para-substituted.
The required sodium methanolate is easily synthesised from methanol using sodium hydride:
$$\ce{MeOH + NaH -> NaOMe + H2 ^}$$

References:
[1]: L. Kürti and B. Czako: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 70.
[2]: L. N. Ferguson, A. Y. Garner, J. L. Mack, J. Am. Chem. Soc. 1954, 76, 1250–1251. DOI: 10.1021/ja01634a013.


text:
College chem study question, any help?
The molar extinction coefficient of reduced NADH (nicotinamide adenine dinucleotide phosphophate) at 340nm is 6220 L/mole·cm. 11ml of solution containing 0.8 micromoles of NADH were placed in a cuvette of 1.6cm light path. Calculate the percentage light transmission of this sample at 340nm


text:
What are the important implications to the field of chemistry, if any, of the hexavalent carbon species confirmed in 2017?
I’m struggling to understanding the true nature and implications of recent confirmations of the existence of stable species containing carbon atoms bound six (in another case 5) other carbon atoms. Is this really a breakthrough of science or one of semantics and scientific media hype? I've had a hard time understanding what I've read on this subject.
From Moritz Malischewski, Prof. Dr. K. Seppelt. Angew. Chem. Int. Ed. 2017, 56, 368-370, an open, online article:

Herein we report the crystal structure of $\ce{C6(CH3)6^{2+} (SbF6^−)2⋅HSO3F}$, which revealed a non-classical structure for this labile species (Figure 1).

Figure 1.
Molecular structure of $\ce{C6(CH3)6^{2+} in C6(CH3)6^{2+} (SbF6^−)2⋅HSO3F}$, ellipsoids are shown at 50 % probability, C grey, H white; counteranions and co-crystallized $\ce{HSO3F}$ omitted for clarity.
Further discussion of this matter was also given in C&EN: "Six bonds to carbon: Confirmed", C&EN, Volume 94 Issue 49, p. 13  (abstract).
So, I understand how the octet rule is not actually violated. But then, can these molecules really contain carbons with 6 primarily covalently bound carbon neighbors? In what orbitals do all these electrons lie? Should we not be surprised as similar hypervalency occurs elsewhere on the periodic table?  Any clarification at all of these issues would be greatly appreciated.

In the real world how common is anything with a $\ce{SbF6^-}$ anion going to be?

@MaxW, common, well not very. I'm just interested in it's existence, and relative stability, "days at -78C", not just some pico-second transient ;)

This was briefly mentioned in a material science course I'm rolled in. Due to the concurrent exam period, haven't had the proper time to read the article. Looking forward to someone making the summary for me :-)

Come to think of it, carbon in carboranes is also nominally "hexavalent", so what's the big deal?

@IvanNeretin, yea, that's pretty much what I'm trying to assess.  Sounds like a vote for "scientific media hype" from you ;)  The one difference is the carborane carbons aren't bound to 6 other carbon atoms.  Anyway, I'm also not that familiar with carboranes, so thx for the heads up there.

Don't get me wrong, I mean, the cation is very nice and cute, and the guys who made it can be proud all right. But the "hexavalent" carbon per se is hardly a new thing.

Steven Bachrach also commented on this.

@hBy2Py, big thx.  So his take is also (like Martin's answer): "The carbon is therefore not hypervalent, but it appears to truly be hypercoordinate."

No prob! I thought I recalled that Henry Rzepa had also written about it on his blog, but I can't find a post on it now. And, yup: I also concur with this assessment. I also am confident that, as with other supposedly "hypervalent" systems (oxoanions, $\ce{IF7}$, etc.), that some or all of the bonds to the hypercoordinate $\ce C$ have appreciable charge-shift character. I'm pulling some calculations together and plan to post an answer of my own when I get a chance.

https://chemistry.stackexchange.com/questions/10415/hyperlithiated-carbon-species?rq=1

How would you IUPAC-name it?

The carbon is not  hexavalent, it is hexacoordinated. A covalent bond does not equal to a total of two electrons between the bonding partners and the nature of  the chemical bond may lie somewhere between totally covalent and totally ionic.
Examples of this include boranes, with their three-centre-two-electron bonds. But we don't need to stop there; any π coordination has fewer electrons in the bonding orbitals than double the bonding partners.
We struggle understanding the bonding in simple molecules like carbon monoxide, while more complex molecules like many organic molecules have all similar bonding situations which are quite easy to grasp. The concept of aromaticity is still not fully understood and while we believe it is as simple as drawing a MO diagram, it most certainly is not. However, molecular orbitals certainly enhance our understanding.
Chemist in general are interested in unusual bonding situations, since they challenge our understanding of the bonding and the chemistry of such molecules itself.
One of the prominent examples is the 2-norbonyl cation, the structure of which remained a mystery for a long time since it didn't fit within the common constraint of organic chemistry. Such ions are nowadays usually referred to as non-classical ions. Their bonding is different from the more common two-electron-two-centre bonds we expect in organic (and inorganic) molecules. In a first approximation they can be described with resonance (see here: What is resonance, and are resonance structures real?), but the understanding comes with a few misconceptions. A MO description involves multi-centre bonds and typically bond orders less than one.
Another interesting example of unusual bonding are fluxional molecules like bullvalene. (See also here: What is the conformer distribution in monosubstituted fluoro bullvalene?) Similarly the bonding situation in these molecules is quite fluid, which allow them to change shape in such a way and at room temperature we obtain a single signal in the proton NMR (Addison Ault. J. Chem. Educ. 2001, 78 (7), 924-927.).

As such the hexamethylbenzene dication is another representative of unusual bonding situations. While it has been prepared more than forty years ago, it took until now to confirm the actual structure. It is special since it contains a pyramidal coordinated carbon exclusively bonded to other carbons, and it marks the highest observed coordination number for carbon so far. The authors actually state the general motivation behind such approaches in the first two sentences:

The tetravalency of carbon and the hexagonal-planar ring structure of benzene are fundamental axioms of organic chemistry, and were developed 150 years ago by Kekulé. Chemists have long been fascinated by finding exceptions from these rules.

Let's go a little deeper, and look at the questions you are asking step by step:

So, I understand how the octet rule is not actually violated. But then, can these molecules really contain carbons with 6 primarily covalently bound carbon neighbors?

The covalent-to-ionic character of a bonding orbital is completely independent of whether it is occupied or not. I believe there still is no unilaterally accepted criterion for what constitutes a covalent bond and what does an ionic bond.
As previously stated, a bond does quite often deviate from the traditional Lewis-concept. This is to be expected as the Lewis concept is quite crude and cannot account for many chemical phenomena.
We can analyse the electron density with the Quantum Theory of Atoms in Molecules, which defines us a set of parameters on which we can judge whether a bond is predominantly covalent of ionic. First and foremost we can obtain bond paths, i.e. these are paths that connect nuclei in a way that the value of the electron density is maximum. On this path we will find a bond critical point, where the electron density is minimal. The Laplacian at this point gives us a measure whether the bond is predominantly covalent (negative) or ionic (positive). In the picture below areas of positive Laplacian are marked with solid lines, i.e. charge depletion, and negative values with dashed lines, i.e. charge accumulation. The location of the plane is given in the 3D model. The hexa-coordinated carbon is C10.

        rho/a.u.    Lapl./a.u.
BCP44   0.259       -0.811
BCP47   0.294       -0.121
BCP53   0.157        0.226
BCP61   0.270       -0.882
BCP64   0.288       -0.117

From the pure visual inspection we can see that the area between C10 and C12 clearly is a predominantly covalent bond, which is obvious from the dashed lines. This is also supported by the numbers.
Visual inspection of C10 to C4 would suggest the same. However, the bond critical point lies outside of the charge accumulation area. This might well be an issue with the methodology. I have reproduced the calculation from the paper only on their geometry. Therefore the optimum MP2 density might be slightly different and completely describes the bond as covalent. Also note that there are no symmetry restrictions, which could also lead to slight variations. From what we can see, we can derive that there is a significant covalent contribution though.
The values of the electron density (rho) also show that the pyramidal bonds are only about half as strong as the bonds in the 1-ethylium-1-ylidene or the five-membered ring.
(Fun fact: The AIM analysis shows no cage critical point. From a technical point of view, the five-membered ring is no ring. The whole molecule is rather a goblet (of fire, lol).)

In what orbitals do all these electrons lie?

Since it is a cation of benzene, the atomic orbitals involved in bonding are the same as in regular benzene. The total number of molecular orbitals that are formed are one fewer than in the neutral case. Obviously due to the different geometric arrangement they combine also differently.
The canonical orbitals are actually quite pretty and reproduce the (almost) symmetry quite nicely as some of the orbitals are nearly degenerate (for example HOMO and HOMO-1, or LUMO and LUMO+1), but have a look for yourself:

Should we not be surprised as similar hypervalency occurs elsewhere on the periodic table?

Well, no. Hypercoordinate bonding is one of the most common patterns for anything outside of carbon chemistry.
A few words of advice: It has nothing to do with hypervalency. The term is an ancient relic from the times where people believed that d-orbitals are necessary to describe molecules that exceed the coordination number dictated by Lewis theory. The gold book writes on hypervalency:

The ability of an atom in a molecular entity to expand its valence shell beyond the limits of the Lewis octet rule. Hypervalent compounds are common for the second and subsequent row elements in groups 15–18 of the periodic table. A description of the hypervalent bonding implies a transfer of the electrons from the central (hypervalent) atom to the nonbonding molecular orbitals which it forms with (usually more electronegative) ligands.

Note that this orbital expansion is not necessary when considering MO theory. As such, the term valency, and especially hypervalency, should be avoided at all cost.
It is important to know that according to the gold book, valence is an absolute property of an element, it does not change:

The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.

Carbon is tetravalent. Always. (Seriously, from now on never use *valen{ce .. cy .. t} again, please.)

Is this really a breakthrough of science or one of semantics and scientific media hype?

Of course it is. It is always something incredible when we refine our understanding of the universe, challenge our own theories. Is it as ground breaking as the discovery of electricity; probably not. However, with such advances in synthetic chemistry we can obtain more robust data to develop our theoretical models. We can in general expand our knowledge of bonding, think outside the box.
With advances like this we achieve the complete break down of our theories; without such we would probably still be using Lewis structures (for everything, and that would be ridiculous).

Summary: "With such advances in synthetic chemistry we can obtain more robust data to develop our theoretical models."

FYI, I'm loving this answer, I'm just taking my time to work through it. Some of this is stuff I haven't thought about in a long time and it's pushing me a bit, which is exactly what I wanted of course ;)

The covalent-to-ionic character of a bonding orbital is completely independent of whether it is occupied or not. The more I explore bonding with QTAIM and ELF, the more convinced I am that it's erroneous to apply the notion 'covalent' or 'ionic' to specific orbitals, even if they're localized orbitals clearly of, e.g., $\sigma$ or $\pi$ symmetry. Bonding character is a property of the overall electron density, not of any specific orbital(s) making up a subset of a wavefunction.

The ion is "a cation of benzeze" in terms of stoichiometry, but the pyramidal geometry of the central carbon cluster means the molecular orbital are different from those of a planar benzene ring.  They had better be different because the planar orbital basis would give an antiaromatic ring.  Rearranging to a pyramidal cluster is one way for an otherwise antiaromatic cation  to gain more stabilization.

airhuff: But then, can these molecules really contain carbons with 6 primarily covalently bound carbon neighbors?

No, they can't. I actually take issue with what Martin said on this matter:$\require{begingroup}\begingroup$

Martin: A covalent bond does not equal to a total of two electrons between the bonding partners ...

Yes, actually, a covalent bond does entail a total of two electrons, or very nearly, localized between the bonding partners.

Martin: ... and the nature of the chemical bond may lie somewhere between totally covalent and totally ionic.

This half of the statement is true as far as it goes, but it doesn't tell the whole story.  A chemical bond can also be a charge-shift bond, which is a resonance among both covalent and ionic bonding configurations, and which (vide infra) results in localization of far fewer than two electrons between a pair of bonded nuclei. Bearing in mind that I am not an expert in the topic, I am confident that:
In this hexamethylbenzene dication, each bond between the apical carbon and the basal carbons is a charge-shift bond.
For context on charge-shift bonding, please read my answer to a Chem.SE question on the topic. It will hopefully provide a helpful frame of reference for interpreting the data below.

To start, I pulled the hexamethylbenzene (HMB) dication geometry out of the Supporting Information of the article linked in the original question and ran a single-point calculation on it in ORCA v3.0.3. The key parts of my input file, hmb.txt, were:
! RKS PBE def2-SVP def2-SVP/J RI D3BJ GRID5

* xyzfile 2 1 hmb.xyz

The calculation converged readily; I then loaded the wavefunction into MultiWFN v3.3.7, by generating a MOLDEN file from the resulting .gbw wavefunction file and renaming it from hmb.molden.input to hmb.molden (this helps MultiWFN load it correctly, for whatever reason).  Within MultiWFN, I performed a similar critical point analysis to that run by Martin, plus some additional QTAIM and ELF basin analysis, roughly per a subset of the methodology described by Matito and Solà (doi:10.1016/j.ccr.2008.10.003).
I will shamelessly re-use Martin's plane plot, for reference:

$\def\bond#1#2#3#4{\ce #1 #2\!-\!\ce #3 #4}$Below, I'll refer to the various atoms as follows: apical methyl carbon $(\ce C12)$, apical methyl hydrogen $(\ce H29)$, apical carbon $(\ce C10)$, basal carbon $(\ce C4)$, basal methyl carbon $(\ce C7)$, and basal methyl hydrogen $(\ce H17)$.  I'll refer to the bonds between these atoms as: apical $\ce{C-H}$ $(\bond{C}{12}{H}{29})$, apical $\ce{C-C}$ $(\bond{C}{12}{C}{10})$, pyramidal $\ce{C-C}$ $(\bond{C}{10}{C}{4})$, basal methyl $\ce{C-C}$ $(\bond{C}{4}{C}{7})$, and basal $\ce{C-H}$ $(\bond{C}{7}{H}{17})$.  I'll also refer to basal $\ce{C-C}$ bonds $(\bond{C}{4}{C}{\#})$, where $\ce C\#$ is an adjacent carbon atom in the base of the pyramid.  While there are measurable (er, well, calculable) differences in the environments of the carbons and hydrogens in the apical and basal methyl groups, these differences are slight enough that for the most part I won't discuss them here.
My calculated values for the various quantities of interest at the relevant line critical points (LCPs) correspond roughly to those reported by Martin:
$$
\begin{array}{ccccc}
\hline
\mathrm{LCP}  & \rho  & \nabla^2\rho & \mathrm{ELF} & \eta' \\
& \frac{\mathrm e^-}{\pu{B^3}}  & \frac{\mathrm e^-}{\pu{B^5}} & - & - \\
\hline
\text{Apical } \ce{C-C}       & 0.244 & -0.531 & 0.923 &  0.395 \\
\text{Basal Methyl } \ce{C-C} & 0.252 & -0.564 & 0.940 &  0.405 \\
\text{Basal } \ce{C-C}        & 0.279 & -0.627 & 0.936 &  0.437 \\
\text{Pyramidal } \ce{C-C}    & 0.152 &  0.004 & 0.669 & -0.166 \\
\hline
\text{Apical } \ce{C-H}       & 0.273 & -1.047 & 0.988 &  1.088 \\
\text{Basal } \ce{C-H}        & 0.269 & -1.010 & 0.986 &  1.048 \\
\hline
\text{Ethane } \ce{C-C}       & 0.247 & -0.617 & 0.965 &  0.482 \\
\text{Ethane } \ce{C-H}       & 0.276 & -0.939 & 0.984 &  0.462 \\
\hline
\ce{Na-F}                     & 0.055 &  0.453 & 0.047 & -0.857 \\
\hline
\end{array}
$$
The moderate discrepancies likely stem from differences in the DFT functionals and/or basis sets we used, but do not significantly affect the ultimate analysis. For comparison, I also included the data for ethane from my above-linked answer, and for NaF as a representative ionic bond.  The $\eta'$ values are Tian Lu's modified eta index values at the LCPs (see the Multiwfn v3.3.9 manual, p26), which help to provide a further diagnostic of charge-shift bonding character: $\eta' > 0$ implies the presence of a traditional covalent bond (strictly speaking, a 'shared interaction'), whereas $\eta' < 0$ implies either an ionic or a charge-shift bond (strictly, a 'closed-shell interaction'), depending on other properties of the system at/near the LCP.
Other properties helpful for diagnosing charge-shift bonding can be calculated by examination of, in the language of QTAIM, the basins surrounding attractors in the three-dimensional ELF field (click to enlarge):

The volumes of these basins, the integrated (typically non-integer) number of electrons lying within them, and the extent of electron delocalization into other basins all provide insight into the extent of charge-shift character that is present:
$$
\begin{array}{cccc}
\hline
\text{ELF Basin} & \iiint{\rho} & \mathrm{Volume} & \text{% Localized} \\
   &   \mathrm e^- & \pu{Bohr^3}  & - \\
\hline
\text{Apical } \ce{C-C}         & 2.19  &  31.4  & 49.4\% \\
\text{Basal Methyl } \ce{C-C}   & 2.07  &  24.8  & 49.4\% \\
\text{Basal } \ce{C-C}          & 2.25  &  66.6  & 49.7\% \\
\text{Pyramidal } \ce{C-C}      & 0.72  &   9.5  & 21.6\% \\
\hline
\text{Apical } \ce{C-H}         & 1.96  & 525.2  & 66.4\% \\
\text{Basal } \ce{C-H}          & 1.92  & 474.6  & 65.4\% \\
\hline
\text{Ethane } \ce{C-C}         & 1.82  &  17.8  & 46.5\% \\
\text{Ethane } \ce{C-H}         & 2.00  & 693.1  & 67.7\% \\
\hline
\ce{Na-F}               & \mathrm{NO} & \mathrm{BASIN} & \mathrm{PRESENT} \\
\hline
\end{array}
$$
Based on the above data, most of the $\ce{C-C}$ bonds in the system are similar to standard covalent bonds.  The following points, however, argue in favor of significant charge-shift bonding character in the pyramidal $\ce{C-C}$ bonds, using the covalent $\ce{C-C}$ bond of ethane as a reference:

LCP

The electron density is moderately reduced, but not nearly to the extent as in the 'canonical' ionic bond of NaF.
The Laplacian is considerably less negative but, again, not nearly to the extent as in NaF.
As with the density, the ELF is moderately reduced, but is still much greater than the miniscule value in NaF.
The modestly negative value of $\eta'$ implies a balance of covalent and ionic (closed-shell) character is present.

ELF basin

The integrated electron population is severely deficient relative to the expected $\sim\pu{1.8e^-}$.
While direct system-to-system comparisons are not straightforward, the volume is considerably reduced, implying an appreciably lessened extent of localization to the bond.
The fraction of localized character of the electron population in the basin is less than half that of the ethane reference.

The corresponding implication of the charge-shift bonding concept is that electron density/localization is transferred to the "outer regions" of a putative charge-shift bond.  This behavior is indeed observed in the apical $\ce{C-C}$ bond and the basal $\ce{C-C}$ bonds, again in comparison to the ethane $\ce{C-C}$ reference:

LCPs

The electron density at the basal and apical $\ce{C-C}$ bond LCPs is slightly enriched.
The Laplacian is more negative at both types of LCP.
The ELFs are slightly larger, but are effectively equal to the reference.
The $\eta'$ values are appreciably larger, indicating enhanced covalent character.

ELF basins

The integrated electron populations are appreciably but modestly increased.
While direct comparisons are again not straightforward, the basin volumes are substantially increased relative to both the ethane reference and the pyramidal $\ce{C-C}$ bonds.
The fractions of localized character are enhanced in these basins as compared to the ethane reference.

More broadly, note that the total integrated electron population of the five pyramidal $\ce{C-C}$ bonds taken together is only about $\pu{3.6e^-}$, far less than the $\pu{10e^-}$ that would be expected if each pyramidal bond were a proper covalent bond.
The above characteristics are strongly diagnostic of charge-shift character in the pyramidal $\mathbf{C\!-\!C}$ bonds of the hexamethylbenzene dication, with the majority of the displaced electron density/localization residing in the apical and basal $\mathbf{C\!-\!C}$ bonds.

airhuff: Should we not be surprised as similar hypervalency occurs elsewhere on the periodic table?

Nope. The appearance of "hypervalency," as noted repeatedly by Shaik and Hiberty, usually implies the presence of charge-shift bonding. The significant electron depletion in the ELF basins of charge-shift bonds, as in the above data, ultimately compensates for the atypically large number of interaction partners of such hypercoordinate species.  This can be probed by examining the basins of the attractors in the electron density field, for which there is one basin per nucleus in most systems:
$$
\begin{array}{ccccccc}
\hline
\text{Density Basin} & \iiint{\rho} & \text{% Localized} &
   \pu{e^-} / \text{atom} & \pu{e^-} / \text{atom type} &
   \text{% Contrib}  \\
  & \pu{e^-} & - & \pu{e^-} & \pu{e^-} & - \\
\hline
\text{Apical Methyl } \ce C~(1) &  5.93 & 66.9\% & +0.07 & +0.07 &  3.5\% \\
\text{Apical } \ce C~(1)        &  6.04 & 64.5\% & -0.04 & -0.04 & -2.0\% \\
\text{Basal } \ce C~(5)         &  5.99 & 64.9\% & +0.01 & +0.05 &  2.5\% \\
\text{Basal Methyl } \ce C~(5)  &  5.97 & 66.7\% & +0.03 & +0.15 &  7.5\% \\
\hline
\text{Apical Methyl } \ce H~(3) &  0.89 & 40.4\% & +0.11 & +0.33 & 16.4\% \\
\text{Upper Basal Methyl } \ce H~(10)
                            &  0.92 & 41.2\% & +0.08 & +0.80 & 39.8\% \\
\text{Lower Basal Methyl } \ce H~(5)
                            &  0.87 & 39.9\% & +0.13 & +0.65 & 32.3\% \\
\hline
\text{Ethane } \ce C            &  6.04 & 66.6\% & -0.04 & - & - \\
\text{Ethane } \ce H            &  0.99 & 44.1\% & +0.01 & - & - \\
\hline
\ce{NaF~Na}                     & 10.08 & 98.5\% & +0.92 & - & - \\
\ce{NaF~F}                      &  9.91 & 98.6\% & -0.91 & - & - \\
\hline
\end{array}
$$
In the above table, the number in parentheses is the number of that type of atom in the HMB dication, and the "% Contrib" values are the fraction of the overall $2+$ charge residing on all of the atoms of the various types listed.
Interestingly, according to this analysis, only about $\pu{0.23e^-}$ ($12\%$ of the overall $2+$ charge) resides on the carbon atoms, with the remaining $\pu{1.77e^-}$ distributed across the various methyl hydrogens.  Thus, despite its sixfold coordination, according to QTAIM the apical carbon atom is negligibly enriched in electrons relative to the other carbon atoms in the system, and to the carbon atoms of ethane.

airhuff: In what orbitals do all these electrons lie?

First, as noted above, "all these electrons" amounts to only about $\pu{3.6e^-}$ worth of integrated electron density, so the question of whether the electrons have a sufficient number of orbitals to 'fit into' is much less pressing. That said, as I'll show below, there are three highly delocalized orbitals primarily responsible for the electron density in the pyramidal $\mathbf{C\!-\!C}$ bonds.
In order to show relevant orbital images, I recalculated the HMB dication system using a larger basis set and performed Pipek-Mezey localization on the converged wavefunction, using the below ORCA input (again for v3.0.3):
! RKS PBE def2-TZVP def2-TZVP/J RI D3BJ PAL3 GRID5

%loc  LocMet PM  end

* xyzfile 2 1 hmb.xyz

The following are the localized orbitals associated with the various $\ce{C-C}$ bonds (blue and red coloration indicates positive and negative phase, respectively):

As can be seen, the five pyramidal $\ce{C-C}$ bonds observable in the electron density and ELF fields are embodied by only three orbitals, all of which are delocalized over multiple carbon centers. These orbitals look nothing whatsoever like typical bonding orbitals of either $\sigma$ or $\pi$ character, which is unsurprising given their heavily delocalized nature.  This strong delocalization is consistent with the depressed LCP ELF and reduced localization of the pyramidal $\ce{C-C}$ bonds as described above.
The other $\ce{C-C}$ bond orbitals in the system are relatively conventional.  The portions of the apical and basal methyl $\ce{C-C}$ orbitals directed toward the methyl groups are unremarkable, and the entirety of the basal $\ce{C-C}$ bond is essentially a normal $\sigma$-bonding orbital, albeit with some slight curvature.  The positive lobes of the apical and basal methyl $\ce{C-C}$ orbitals are highly unusual, however, providing further indication of the unusual electronic properties of the pyramidal region of the system.

Martin: Fun fact: The AIM analysis shows no cage critical point [CCP]. From a technical point of view, the five-membered ring is no ring. The whole molecule is rather a goblet (of fire, lol).

Indeed.  One good way to visualize this is to plot the magnitude of the electron density gradient ('gradient norm') in the same plane as Martin's figure, above:

The heavy contour line is at $\pu{0.04e^- Bohr^-4}$, and serves to indicate (1) the rough extent of the region of space occupied by the cation (loosely analogous to the van der Waals surface), and (2) the locations of the QTAIM critical points.  The line critical points (LCPs) associated with the various bonds are visible as indicated, along with the ring critical point (RCP) lying in one of the faces of the pyramid (orange dot on Martin's figure).  The indentation in the gradient norm boundary surface toward the base of the central pyramid is evident, and both a basal RCP and a central CCP are conspicuously absent in this region.  To note, in his blog post on the system, Steven Bachrach points out that the pyramidal region of the system exhibits 3-D aromaticity -- it's almost as though the original planar, aromatic benzene ring deformed upward into a bowl shape! $\endgroup$

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


text:
Is flouride an actual chemical or a misnomer?
I often encounter the term "flouride" used as if it is a chemical. From my understanding, it would be incorrect to call rust oxide and not iron oxide or to call salt chloride instead of sodium chloride. Isn't it incorrect to say drinking water or toothpaste has flouride?

Fluoride is the anion $\ce{F-}$ of Fluorine which you can easily confirm with Google.

Yes, I am aware of that. That's not my question. I want to know why is the term fluoride is correct.

You are right in your assertion that fluoride does not exist by itself as a chemical substance. Sodium fluoride ($\ce{NaF}$) and tin(II) fluoride ($\ce{SnF2}$) are the two most common ingredients in fluoride-based toothpaste, and are the source of the fluoride ions. The reason that the cations of the salts are often left off is because fluoride is the active ingredient. Fluoride functions by:

...repair[ing] rather than prevent[ing] damage to the teeth, causing the mineral fluorapatite to be incorporated into damaged tooth enamel.$^{[1]}$

$^{[1]}$Wikipedia, Fluoride Therapy, Mechanism

Yeah, so the municalities can dump chemical waste in the  drinking water and just call it flouride? i.e. Uranium Hexafloiride?  Technically, it is a flouride. Why does the US care so much about children's teeth, yet dont care whether they go to bed hungry. I can now see the reasoning behind the flouride conspiracy theories.

@user148298 Bollocks. Utilities can only use approved sources of fluoride (which mimic naturally fluoridated water). Conspiracy theorists on this are about as rational as the Jack D Ripper character in Dr Strangelove (the one who tries to bomb Russia because they have corrupted our precious bodily fluids).

@matt_black "Bollocks?" Please educate yourself and stop being a sock puppet.  The US is one of the few countries where water is artificially fluoridated and is rare in Europe.  The Cochrane Collaboration found that fluoridation doesn't prevent cavities. Why? Not implying there's a conspiracy, but why add a chemical that is a known toxin. Moreover, why is it not only encouraged, but is a federal mandate in the US?

@user148298 "known toxin" as any decent chemist should know it is the dose that makes the poison. Fluoride is often present in natural water supplies and nothing governments do puts too much fluoride in any water. BTW fluoridation is common in Europe, but often done in milk or salt rather than water. Fluoridation reduces but doesn't stop cavities. And it is present in most toothpaste for exactly that reason.

This Wikipedia article sums it up pretty nicely:

Fluoride [...] is an inorganic, monatomic
  anion of fluorine with the chemical formula $\ce{F−}$ . Fluoride is the
  simplest anion of fluorine. Its salts and minerals are important
  chemical reagents and industrial chemicals, mainly used in the
  production of hydrogen fluoride for fluorocarbons. In terms of charge
  and size, the fluoride ion resembles the hydroxide ion. Fluoride ions
  occur on earth in several minerals, particularly fluorite, but are
  only present in trace quantities in water. Fluoride contributes a
  distinctive bitter taste. It contributes no color to fluoride salts.

Note the term "chemical formula $\ce{F-}$".  In toothpaste and drinking water treatment, fluoride is often added as sodium fluoride.  So yes it is every bit as much a chemical as, say, chloride in sodium chloride.


text:
Why doesn't Sodium (Na) form duplet?
The atomic number of Na is 11 (2, 8, 1). Na loses an electron to achieve octet stability but why can't it just accept an electron to complete duplet stability?

Why, of course it can. The problem is, there are not enough free electrons for everyone.

@Ivan Neretin Would you elaborate "there are not enough free electrons for everyone. "?

I mean, all $\ce{Na}$ can't convert to $\ce{Na-}$. But I guess that's not what the question was about, anyway.

@IvanNeretin Thanks :). Please elaborate on "there are not enough free electrons" and turn it into an answer; I will mark it as accepted.

@IvanNeretin I think the stability and having a stronger reducer than Na is going  to be an issue forming Na-.

Crystalline compounds of sodium anion with polycyclic ether cage like compounds are known. (According to Wikipedia  see Dye et al. J. Am. Chem. Soc. 1974, v96, p608)

related or, more likely, a duplicate.

Possible duplicate of Is it possible to have a diatomic molecule of sodium in gaseous state?

@bobthechemist The questions are completely different even though they might have a similar answer. It is absolutely not the same asking "Why doesn't it accept an electron?" and "Does a dimer exist?"

@Martin-マーチンI agree.  Rereading the question, I see that I misunderstood that the OP is asking about atomic electron configurations.

Explanation: The sodium cannot just accept an electron and continue its duplet stability because that duplet stability only applies to the Helium. However, an atom can tend to make itself a duplet atom, like Lithium, which has 3 electrons and loses its 1 electron and becomes a duplet atom. Hope this helps.

Well, sodium does have non-zero electron affinity, so it surely can acquire another electron, provided it can get one for free. That's what happens when a lone sodium atom meets a lone electron somewhere in deep space. But here on Earth, with all that condensed matter around, we don't have many free electrons (*). They all belong to some other atoms, and if you want to get an electron (that is, to get reduced), you have to oxidize something else.

(*) Then again, some might recall metals in which the electrons are kinda free, but wait: they are not really free. They are bound by the collective field of metal cations, and you need to apply certain energy to take one out.

It turns out that $\ce{Na}$ and other alkali metals can actually be reduced to form alkalides, but those are exotic compounds. They are not particularly stable, and get oxidized with pretty much anything. It is much easier for $\ce{Na}$ to form a cation, thus achieving the stable octet, and call it a day.
There is a whole different dimension to the problem: why wouldn't two sodium atoms share their electrons, establish a bond, and form a diatomic molecule? Again, they would, but only in a vacuum! When you have many such molecules, the sodium atoms would share their other empty orbitals, find out they don't have enough electrons to fill them all, and just resort to being a metal.

Yes, I believe the species $\ce{Na2}$ (and other such diatomic alkali metals) have been observed in the gas phase, as is predicted by qualitative MO diagrams.

Related: Why are lithides not known?

Published a little after all the above: Hyperfine structure of alkali-metal diatomic molecules J. Aldegunde, J. M. Hutson  https://arxiv.org/abs/1708.05734

Sodium not only forms species with two electrons in the $3s$ subshell ($\ce{Na^-,Na2}$). It is also known to form a dimer of the $\ce{Na^-}$ion, even though the individual atoms therein possess closed-shell electronic structures and would not be expected to form a covalent bonds to offset the repulsion between the negative charges.
The dimeric $\ce{Na_2^{2-}}$ anion occurs in barium azacryptand sodide [1]. The crystal structure, pictured as shown below from [1], features the large, blue atoms which are the paired sodium atoms in this dianion (red atoms are barium, the small green atoms are from the cryptand). The large atomic size, corresponding to a highly diffuse electron cloud, reduces the like-charge repulsion while improving polarizability for a VdW attraction.

Reference
1.
Mikhail Y. Redko, Rui H. Huang, James E. Jackson, James F. Harrison, and James L. Dye, "Barium Azacryptand Sodide, the First Alkalide with an Alkaline Earth Cation, Also Contains a Novel Dimer, (Na2)2-", J. Am. Chem. Soc. 2003, 125, 8, 2259–2263. https://doi.org/10.1021/ja027241m


text:
Production process of kevlar
How easy is it to produce kevlar?
What are/is the chemical recation/-s to produce kevlar, and which chemical compounds are required?

https://en.wikipedia.org/wiki/Kevlar

From this Wikipedia page:

Kevlar is synthesized in solution from the monomers
  1,4-phenylene-diamine (para-phenylenediamine) and terephthaloyl
  chloride in a condensation reaction yielding hydrochloric acid as a
  byproduct. The result has liquid-crystalline behavior, and mechanical
  drawing orients the polymer chains in the fiber's direction.
  Hexamethylphosphoramide (HMPA) was the solvent initially used for the
  polymerization, but for safety reasons, DuPont replaced it by a
  solution of N-methyl-pyrrolidone and calcium chloride.

The structure of Kevlar is given from the same page:

However, Kevlar production on a small laboratory scale will likely be an expensive and difficult endeavor compared to purchasing Kevlar products, which are available in about any form you can imaging, many of them patented. Also, as Kevlar is water insoluble, concentrated sulfuric acid is required to keep it in solution during synthesis, molding, spinning, etc.  Good luck ;)


text:
Charge density is the amount of charge per unit volume but what is charge?
I was studying about x ray crystallography and my book says that charge density is measured in electrons per nano-meter cubed and this means that charge is the amount of electrons. But i am not sure if this definition (charge density = the number of electrons in a unit volume) applies when we are studying crystallography only. So to make sure, is charge simply the amount of electrons ?

The electron has a charge measure in coulombs of $\approx 1.602\cdot 10^{-19}$ C. In a given volume  v ($\pu{m^{3}}$) the charge density for n electrons is $ne/\mathrm{v}$ with unit $\pu{C m^{-3}}$

An electron has a charge (lets say $e [C]$). If you have a certain amount of electrons in a certain volume ($V [m^3]$), you have a certain charge per volume $\rho = e/V [\frac{C}{m^3}]$. Which is a density.

The charge of the electron is measure in Coulombs not volts.

@porphyrin The units don't matter to make my point, but I'll change it


text:
Why isn't a polar protic solvent used in the Finkelstein reaction?
Usually, in a nucleophilic substitution reaction, a stronger nucleophile replaces a weaker nucleophile.
In any polar protic solvent, the order of nucleophilicity is $\ce{F-}<\ce{Cl-}<\ce{Br-}<\ce{I-}$ (due to stronger solvation of the smaller anions).
While in any polar aprotic solvent, the order of nucleophilicity is $\ce{F-}>\ce{Cl-}>\ce{Br-}>\ce{I-}$.
Why does the Finkelstein reaction use acetone in spite of the fact that it is a polar aprotic solvent? Wouldn't using a polar protic solvent be better since it would make $\ce{I-}$ the strongest nucleophile of the group.

The key point is driving the equilibrium where you want it. You could do it by relative reaction rates. The Finkelstein reaction, however, uses a better method: precipitation of one of the products. Acetone is specifically chosen because sodium chloride, bromide and fluoride are insoluble in it while sodium iodide is soluble. So the equation becomes:
$$\ce{R-X + Na+ ({acet.}) + I- ({acet.}) -> R-I + NaX v}\tag{1}$$
Nothing beats the equilibrium-driving force of precipitation.


text:
Why is the outermost part of the Bunsen burner flame not the hottest region?
Why is the blue flame the hottest on a Bunsen burner?
I thought the outermost part of the flame must be the hottest region because of complete combustion of gases taking place there. Where am I getting wrong?

I saw a video about this just a little while ago. You may find this answers your question: Bunsen Burner (THERMAL IMAGING) - Periodic Table of Videos on YouTube

When you have the intake valve (at the bottom of the bunsen burner) opened, the gas can mix with incoming oxygen. By the time is gets to the top of the tube, you've got a nice continuously-flowing source of well-mixed gas and oxygen. That allows combustion to occur very quickly because the methane and oxygen molecules don't have to travel very far to react, increasing the reaction rate, and therefore the rate at which heat is released.
In contrast, when the intake valve is closed (or when it's open and you're considering the outer non-luminous flame), the oxygen for the combustion reaction comes from the surrounding air, which is not thoroughly mixed with the gas at the time of ignition. That results in a slower reaction, and therefore a slower rate of heat being released.
Periodic videos just did a nice thermal imaging of a bunsen burner with this explanation: https://www.youtube.com/watch?v=in9BGz958jg

The blue flame is indicative of complete combustion. This means more of the energy of the gas is being liberated as thermal energy in the exothermic combustion compared to the yellow portion of the flame.
A yellow flame indicates incomplete combustion. This means products other than carbon dioxide and water result including possibly carbon monoxide. This is why you should be aiming for a blue flame on your gas stove at home too. http://www.elgas.com.au/blog/1585-why-does-a-gas-flame-burn-blue-lpg-gas-natural-propane-methane

I remember doing an experiment where I exposed wooden dowels to the flame at various heights and noted where the wood burned (outer blue flame).  That was supposed to be the hottest region, but it's actually just the oxidizing region (wood does not react visibly with the inner, reducing part of the flame).  Now I would identify the hottest part of the flame as the boundary between inner and outer flames where the exothermic reaction takes place.


text:
What is the purpose of halogen in oxidation of thiols?
Why do you need molecules like $\ce{Br2, I2}$ as one of the reagents in oxidation of thiols? I am saying this because when you look at the mechanism of reaction, it seems that you can have same product of disulfide by simply joining two protonated thiols (by strong base like $\ce{OH^{-}}$) together. Thanks.

Probably so the electrons have somewhere to go.

Try drawing a mechanism without halogens present, starting with your base DE-protonating one thiol. See how far you get. The problem is that you'd have to somehow kick out H- (hydride), which is a terrible leaving group (enzymes can accomplish this, but standard lab reactions cannot do this easily). The fact is, combing two thiols to give the disulfide is an oxidation reaction, and there is an electron pair being lost from the reactants. It has to go somewhere, i.e. to reduce a diatomic halogen (playing the role of an oxidizing agent).
The thiol-disulfide redox equilibrium you are asking about has great biological significance; see disulfide bridges (for protein structure) and glutathione (an antioxidant).


text:
Explain the voltage drop curve of a set of lead acid batteries
Living on a boat and using batteries to power things over night, I've noticed that the battery discharge pattern seems to be a bit like a pyramid or the point of a sharpened pencil.
The batteries quickly fall back from 13.2 -> 12.7, and 12.7 -> 12.6 is faster than 12.6 -> 12.5 and by the time you get to 12.5 you seem to have hit the meat of the battery and further voltage drop is much slower, much like a pencil without a sharp point doesn't degrade as quickly as a freshly sharpened pencil.
We've never gone below 12.2, but I'm assuming at some point voltage starts falling faster, or maybe it doesn't.
99% of the load on these batteries comes from 2 refrigerators and 1 freezer and can range from 50Watts to 200Watts depending on temperature and how many are running simultaneously.
I'm curious about WHY the battery bank displays this voltage drop that isn't very uniform.  What is going on inside the cells that cause this kind of voltage drop.  What causes the battery to begin shouldering the load with less voltage drop around 12.5v?  I'm sure there is some chemistry behind the reason, but I don't know what that is.
Charging voltage runs as high as 14.4 during the day when the solar panels are refilling them batteries, and we are adding 3.3-5.2KWh/day to the batteries, depending on sun and clouds.

As you suggest, the answer depends on the electro-chemistry of the batteries. This question might get a better answer on Chemistry SE. ... The steep changes in terminal voltage might look smooth if you take readings more frequently, and might be caused by your fridges and freezer switching on and off as the thermostat dictates. If you provide your data, this would help us to understand the problem better.

The behavior of batteries can be (and is) approached from both side of the chemistry/physics divide, but it is a pretty specialized discipline to the point that the specialty battery manufacturer in my town hires from several majors including physics and chemistry for R&D jobs, but expects to train all hirees in the basics in house.

Basically the battery stores charge much like a capacitor. So when you are using the battery you're draining off the charge which reduces the voltage. See https://en.wikipedia.org/wiki/Lead%E2%80%93acid_battery

A comparison to a capacitor in this context is probably not warranted because the voltage on a capacitor drops linearly with charge, while batteries hold high (but not quite constant) voltage for a large fraction of a cycle before starting to drop rapidly only when largely discharged.

Ye, the question is about the non-linearity of the drop.


text:
Separation of a solution of n-butanol and nitroethane
Theoretically speaking, how could one separate a solution of n-butanol and nitroethane? They have a similar boiling point, so fractional distillation would be not possible. Wikipedia also states that nitroethane is more miscible in water than butanol, but that would probably give a water nitroethane azeotrope (?). Can anyone tell how the separation could be achieved? Also, would it be a difference between a 50:50 solution and a 5:95 (vice versa)?

Use counter-current extraction.
I'm not sure about solvents, but let's say that I can separate nitromethane from water by distillation. Also that butanol can be separated from ether by distillation. (as I remember ether and water aren't miscible...)
So I line up 7 sepatory funnels.
Into sepatory funnel #4 I put my mixture and 50 ml of water and 50 ml of ether. Equilibrate then water phase goes to sepatory funnel #5 and ether phase goes to sepatory funnel #3. Sepatory funnel #4 is now empty.

In sepatory funnel #5 I add 50 ml ether, Equilibrate then water phase goes to sepatory funnel #6 and ether phase goes to sepatory funnel #4
In sepatory funnel #3 I add 50 ml water, Equilibrate, then water phase goes to sepatory funnel #4 and ether phase goes to sepatory funnel #2

So basically I have the water extracts going one way and the ether extracts the other way.


text:
Synthesis of para-chloroanisole using diazotisation followed by chlorination with aluminium chloride
Is this a legitimate reaction scheme to synthesise para-chloroanisole from benzene and methanol as the only organic reactants?
$$\ce{C6H6 -> HNO3/H2SO4 -> H2/Pd -> HONO/{Ethanol}\\ -> {Methoxide}/THF -> Cl2/AlCl3}$$
I think this would make ortho-chloroanisole (minor product) and para-chloroanisole (major product).
The methoxide would be generated by:
$$\ce{{Methanol} -> NaH -> {Methoxide}}$$

Yeah, that works. For posteriority: related question of OP’s.


text:
What's the Gibbs free energy of a reaction at zero temperature and pressure?
I heard that Gibbs free energy changes with temperature and pressure as
$$ \Delta G_1 = \Delta G_0 + R T \ln Q $$
But this makes no sense when pressure is zero or temperature is zero.

How so? What's the problem with T=0?

The guy's name was Gibbs, not Gibb.

It makes no sense of either of them odd zero. Pressure can never be zero and neither do temperature.

If the temperature is absolute zero, how is anything going to move to react? If the pressure is absolute zero, then you don't have a molecule in your volume to do any reactions.

$G^\circ = U(0K) = E^{ele}+ZPE$, i.e. the sum of the electronic energy (electrons and nuclei interacting) and the zero point vibrational energy.
You can't use $K=e^{-\Delta G^\circ/RT}$ at 0K because the derivation assumes that a certain amount of thermal energy is available. But you don't need the equation to figure out the relative amount of two molecules. At equilibrium at 0K all molecules are in the lowest possible state.


text:
How to make solution of specific concentration with non-stoichiometric hydrate?
I am currently working with a drug molecule known as Amiloride $\ce{HCl}$ and it seems to be a non-stoichiometric hydrate.
How would I go about making a solution with a specific concentration of Amiloride $\ce{HCl}$? I know its anhydrous molecular weight but if I don't know how much water is in the solid crystals how can I get a specific concentration?

(1) You'll need to make a stock solution and then analyze that for concentration somehow.  (2) It might be possible to dry the drug by heating in a oven (maybe vacuum oven?). (3) Analyze the sample for water content. (4) purify the sample somehow

@MaxW, I swear I'm not just copying your comments into answers today, we're just on the same wavelength or something ;)

Anhydrous Amiloride HCl is commercially available and is stable in air.  This suggests that you should be able to remove the waters of hydration, however many that may be, by repeatedly heating in a drying oven until a constant weight is reached.  If you know the weight before drying you could of course calculate the original degree of hydration after the final drying, if that is of interest.
If you need to further verify dryness, you could have the dried product analyzed by Karl Fischer water anaysis (~$50). Just be sure to seal the sample under a dry atmosphere and label "HYGROSCOPIC" all over the sample and submittal paperwork!
Once you have the constant-weight product, stored under a good desiccator, just use the formula weight of the hydrochloride salt, 266.1, to calculate the concentration of your Amiloride HCl solutions.

The simplest was would be to dissolve a fixed amount in water and then titrate the content of amiloride HCl. A number of different methods would jump to mind:

precipitation of chloride with silver nitrate and gravimetric analysis
potentiometric and/or pH titration with sodium hydroxide to determine the basic nitrogen content
full deprotonation with a strong base and acidic back-titration.

If you choose your titration solution in such a way that the concentration is more than what you need afterwards, you can even dilute it to the desired stock solution concentration afterwards.


text:
Why is % yield in a purification table measured using activity instead of total protein?
I'm trying to understand why % yield is measured with activity instead of total protein in mg. According to my biochemistry notes, activity is "1.0 units of enzyme activity = 1.0 μmol of substrate catalyzed to product under optimum conditions."
Then, yield is defined as "% of total starting protein/enzyme in final purified protein fraction."

With this definition of yield, it would make sense to say that you get 30% yield in the second step since you have 3,000 mg / 10,000 mg of the original protein amount. The definition of yield in my notes mentions nothing of activity.

Because total protein includes other proteins than the one desired.
"% yield" is expressing how much of the enzyme desired was actually recovered.

Oh duh. Thank you. Activity is talking about a specific reaction with your desired protein. Thanks!


text:
Comparing energies of isoelectronic species, but one protonated
Let's say we have the following reaction:
$$\ce{RH -> R- + H+}$$
By modelling it using quantum chemical calculations in the gas phase I would make calculations for both $\ce{RH}$ and $\ce{R-}$, but $\ce{H+}$ has no electrons.
On the other hand, $\ce{RH}$ and $\ce{R-}$ are isoelectronic and zero energy is assigned for infinitely separated particles.
I have a gut feeling that the energetics of $\ce{RH}$ and $\ce{R-}$ can't be compared directly in this case, but I can't think of a good argument for this.
So, can I compare them? Why or why not?

I actually read something somewhere that discussed this, but I can't remember now what the argument/approach is. Glad to see it as a question here, thank you for posting!

only relative energies for balanced chemical reactions make sense

But if a proton has zero electronic energy, isn't the equation balanced?
(I understand that when one considers thermal effects the proton will acquire translational energy.)

Ah OK, in that sense. Yes, the energy difference is the electronic contribution to the proton affinity.

A little bit too theoretical question: while technically it is true what you say, proton as a lone species hardly exist in chemical environment. In other words, charged specii, esp very small ones's energy is strongly influenced by the interaction with environment and it generally doesn't make any sense to compare neutral and charged entities' energy in vacuum.

I think they can and should be compared. Consider the $\ce{H_2^+}$ molecular ion. When dissociating this correctly (that is, including symmetry breaking), one should eventually end up with $\ce{H}+\ce{H^+}$. One can then extract the dissociation energy of the bond from the calculations. This is equivalent to the proton affinity of atomic $\ce{H}$. By applying this logic to larger species, one reaches the conclusion, that yes, $\ce{H+}$ by definition has zero energy and should enter calculations of relative energies as such.
As an aside, the calculation of negatively charged species in the gas-phase can lead to erroneous results, but any argument made for $\ce{A-/AH}$ will certainly hold for $\ce{B/BH^+}$.


text:
Purple rinse water
After forming a white substance while on the stove, I rinsed the Pyrex pot under tap water (from a well), whereupon the water in the pot turned purple. Why did this happen?
Iron and sulfur are present in the well water here.
Revision: I had prepared white tea in a white corning pot on the stove. After pouring out the tea I rinsed the pot with the tap water (well water) which turned a purplish color and then clear.

@airhuff Please remember to check for tags if you are editing a question’s grammar, wording and/or title. See for example here or here.

First of all, welcome to Chemistry.SE! I suggest that you start off by taking the [tour] to learn more about how this site works. Regarding your question, could you give a little more information?  What was in the pot?  Where was the white stuff, etc.?

What were you cooking that formed the white substance? There could be some residue that caused it.

Does it stay purple or change to red, brown, yellow, clear, etc. after a while?


text:
How was Avogadro's law discovered?
What was the experimental setup and/or reasoning that Amedeo Avogadro, in 1811, used to arrive at the conclusion that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules"?

There is at least a little bit on Wikipedia. There was lots of confusion at the time, and Avogadro postulated something that would rationalize the various observations.

See this for a brief overview, though it certainly doesn't answer you question. It's pretty sparse on references too.

See this (https://www.famousscientists.org/amedeo-avogadro/) and this (http://www.chemteam.info/Chem-History/Avogadro.html). It seems he explained experiments by Dalton and Gay-Lussac

This law bears Avogadro’s name because he implicitly provided circumstantial evidence that it is true, by successfully applying it as explained below. But, strictly speaking, Avogadro’s law has not been discovered neither by experiments nor by reasoning since it had already been in the air if one had accepted the permanent chaotic movement of particles in gases as suggested by Daniel Bernoulli in 1728.
Also Dalton reflected about the equal volumes−equal numbers hypothesis before Avogadro, but discarded it as impracticable because he did not factor in the idea that gaseous elements might be polyatomic molecules; in consequence he thought that the equal volumes−equal numbers hypothesis were not useful to explain the known physical and chemical facts about gases. (According to Avogadro, Dalton had additionally some strange caloric objections against this view, see p. 59 of the link in the next but one paragraph.)
Hence, the idea that elements might be polyatomic molecules was not accepted at that time (1811). It was even declined thereafter for a long time (by Berzelius who dominated chemistry then).
Being a physicist and mathematician, Avogadro did not himself accomplish chemical reactions with gases, but he could built on existing gas density measurements and the following two results of Gay-Lussac (1808), which he discussed in his article in the Journale de physique, de chimie …, 1811 (all the following page numbers refer to this text) with the equal volumes−equal numbers hypothesis as starting point (!):

Gay-Lussac's combining law: “Not only do gases combine in very simple proportions” by volume, “but the apparent contraction of volume, which they experience on combination, has also a simple relation to the volume of the gases, or at least to that of one of them” (p. 58), and
his experimental combining volume ratios for different gas reaction, e.g. of the formation of water (1 + 1 = 2), ammonia (1 + 3 = 2) from the elements (p. 60).

When comparing all these data, Avogadro proceeded in two steps:

he discussed that, as a consequence of the equal volumes−equal numbers hypothesis, one could easily bring into line the densities of elementary gases with the relative molecular weights of the gases; for instance the density of oxygen being about 15.074 times larger than that of hydrogen (p. 59).
Then he had (p. 60) the idea that gaseous elements could consist of polyatomic molecules (“molecules composées”). Subsequently he discussed extensively the known combining volume experiments of gases in this light and showed that so all hitherto contradictions thereof could be solved.

In this way, Avogadro had simultaneously also proved by circumstantial evidence that the equal volumes−equal numbers hypothesis, from which he started his eventually successful reflections, must be true because he had demonstrated that it is one of the important factors for the causal explanation of all those gas data by particle theory.
Hence, the so called “discovery” of Avogadro's law consists merely in the success to bring all these gas data into accord, which vice versa substantiates that it is a law, whereas before this application, it was unclear how the equal volumes−equal numbers hypothesis could be generally made use of.


text:
Existence of Dichlorine Pentoxide
I searched the net and a lot of books, but I was unable to find any information about dichlorine pentoxide.

Does dichlorine pentoxide($\ce{Cl2O5}$) exist?
If it doesn't exist, then why so?
If dibromine pentoxide and iodine pentoxide do exit, then why not dichlorine pentoxide?

I've never heard of this site, but here it is with a couple example reactions for what it's worth ;)

Here, it says it is unknown at present.

That is a pretty old book though (Published in 2001).

Ok , for once lets agree that Dichlorine Pentoxide doesnt exist.
Then why is that so?
Dibromine pentoxide(https://en.wikipedia.org/wiki/Dibromine_pentoxide) and Iodine Pentoxide does exit, then why not Dichlorine Pentoxide?

I've seen this reaction given in half a dozen websites I've never heard of before: $\ce{Cl2O5 + H2O -> 2HClO3}$. The only one with any discussion is in Russian and I'm trying to find someone who can read that. It's just a couple sentences though. I've also seen it mentioned in nomenclature lessons, i.e. "$\ce{Cl2O5}$ chlorine dioxide from chloric oxide (V)" . Still nothing solid about it though.

The crux of this (Russian) article is that $\ce{Cl2O5}$ is not stable enough to be isolated in a pure form, and if it were, it would be strongly acidic oxidizer.
This web site is representative of several others that simply give example reactions of the compound (the second reaction being the most common exampe):
$$\ce{NiCl2 + 3O2 -> NiO + Cl2O5}$$
$$\ce{H2O + Cl2O5 -> 2HClO3}$$
The statement in the first article and the simple fact that there is such a dearth of information about this compound suggest that it really only exists as a transient, unstable, oxidizing acid that apparently has not been isolated as a stable compound.  Furthermore, information can be found in Wikipedia and/or PubChem for $\ce{Cl2O_x}$, where $x \in 1, 2, 3, 4, 6, 7$.