text:
How could burying a copper and zinc plate underground connected by a surface wire (earth-based Galvanic cell) promote crop growth?
Background:
To quote a source:

In the 1840s, W. Ross of New York reportedly obtained a severalfold increase in the yield of a field of potatoes when he buried a copper plate (5 ft x 14 ft) in the earth, and a zinc plate of the same dimensions 200 ft away. The two plates were connected by a wire above ground, thus forming a galvanic cell. In similar experiments by Holdenfleiss (1844) with battery-charged zinc and copper plates, yields increased up to 25%.(7)

The same source also claims a temporary benefit form electrified seeds and more interestingly from the application of a low-power direct current, to quote:

P.V. Kravtsov, et al., reported that the population of ammonifying bacteria (especially the sporogenous type) increases about 150% when soil or compost is exposed to continuous low-power DC.

And, as ammonifying bacteria convert nitrogen into ammonia, which in the presence of O2 and free electrons, some of the ammonia may be converted into beneficial nitrite/nitrate (see, for example, On the aqueous reactions of the aminyl radical with molecular oxygen and the superoxide anion).
The source also claims benefits due to magnetic fields as does this paper: The effects of magnetic fields on plant growth and health.
[EDIT] To quote from this controlled study:

A study was conducted to test the hypothesis that a magnetic field can affect plant growth and health. The study divided plants into three groups. The first group of plant seeds grew in a low magnetic field. The second group grew in a high magnetic field. The third group grew in the absence of a magnetic field, serving as a control group. Several growth parameters were measured, including the germination rate, plant height, and leaf size. In addition, the health status was measured by leaf color, spots, the stem curvature, and the death rate. Plant growth was observed continuously for four weeks. The results showed that magnetism had a significant positive effect on plant growth. Plant seeds under the influence of the magnetic field had a higher germination rate, and these plants grew taller, larger, and healthier than those in the control group. No adverse effects of magnetism on plant growth were noticed.

What I am trying to understanding are possible paths, that is, causation arguments. So far, my only thoughts relate to increased oxygen kinetics (see my comments and references here), galvanic cell surface wire induced magnetic field, which could promote a charge alignment leading to increased radical activity, and including, for example, associated bacteria leading to the formation of beneficial nitrite/nitrate.

First, these seem to be anecdotal claims, perhaps pseudoscience, not repeated, carefully controlled, scientific experiments.
That said,

There are bacteria that use electricity as an energy source.
Copper can be a needed nutrient, having nothing to o with electricity.
Zinc can be a needed nutrient, having nothing to o with electricity.
An electric current might help soil minerals to become more accessible... or not.
It is extremely unlikely that a weak magnetic field has any effect on plant growth (though it might cause some insects to deviate from their path).

However, all that is supposition. The best answer is to perform a controlled experiment yourself, measuring changes in plant growth, soil chemistry, etc. That seems to be safer than experimenting with Orgone Energy, though.

There was a definitely controlled experiment on plant growth by Edward Fu cited, paper: The effects of magnetic fields on plant growth and health, https://www.ysjournal.com/wp-content/uploads/Issue11/The-effects-of-magnetic-fields-on-plant-growth-and-health.pdf . However, I obviously downplayed this study, so I will edit and place his conclusions for review.

@AJKOER - hmmmm... your definition of controlled and mine don't match up very well.

I won't list world patents citing the employment of  magnetic related paths to promote various reactions.  I have experimented with a Magnetizer -positive results. Apparently, amazing results with magnetic electrodes in battery cells. To quote, Zou, P.; Leddy, J., 'Magnetized Nickel Electrodes for Improved Charge and Discharge Rates in Nickel Metal Hydride and Nickel Cadmium Batteries', Electrochemical and Solid-State Letters 2006, 9, A43-A45. To quote: "magnetized electrodes yielded higher currents than the corresponding nonmagnetic electrodes by 70 and 230 %, respectively."

Let be honest, if you do not have background or expertise in this area, please delete your comment/answer. Having personally experimented with a Magnetizer and have reviewed the literature, I am barely able to ask the right questions.

DrMoishe Pippik: Your answer is, honestly, remiss. Your best counterclaim for any effects is that Zn/Cu are nutrients? Really?? You also cite an issue with weak magnetic fields, what about strong or alternating?

There is a minuscule chance it is not alchemy .Over 40 years ago Amoco did real chemical tests of many midwest US corn belt farm fields and found there were  a few locations where zinc level in the soils was below the level for optimum corn growth.  Amoco then offered anhydrous ammonia containing a trace of zinc ,I don't  know if there was a charge for this and I don't remember any reports of improved yields that could be traced to zinc. Zinc was the only element that corn doctors thought was limiting.

Well, "minuscule" is not the right term to explain possible chemistry. For example, in the presence of e- (from DC) + O2 -> .O2- (superoxide radical anion). Then, Cu(2+) + .O2- = Cu+ + O2,  where Fe, Cu, Mn,.. in there lower valence states foster, for example, fenton-like reactions, Cu+ + 2 H+ +  .O2- -> Cu(2+) + .OH + OH-,  introducing more reactive oxygen species,...The chemistry just accelerates...NH3 + .OH = .NH2 + H2O and .NH2 + O2 leads to nitrite (as I noted in my question)...nothing small or 'minuscle' here.


text:
Apparent paradox regarding spontaneity
Consider the following analysis ( rev and irrev respectively refer to a reversible and an irreversible path, between the same initial and final states):
$dU_{rev}=dq_{rev}+dw_{rev}$,and,$dU_{irrev}=dq_{irrev}+dw_{irrev}$.
$dU_{rev}=dU_{irrev}$. Therefore, $dq_{rev}+dw_{rev}=dq_{irrev}+dw_{irrev}$. Rearranging,
$$dq_{irrev}-dq_{rev}=dw_{rev}-dw_{irrev}\tag{E01}$$
Now: If the volume increases, then $dw_{rev}$ and $dw_{irrev}$ are negative: with $|dw_{rev}|>|dw_{irrev}|$. Thus $dw_{rev}-dw_{irrev}<0$.
On the other hand, if the volume decreases, then $dw_{rev}$ and $dw_{irrev}$ are positive: with $|dw_{rev}|<|dw_{irrev}|$. Thus, again, $dw_{rev}-dw_{irrev}<0$.
We can thus conclude, in general, $dw_{rev}-dw_{irrev}<0$. E01 thus becomes:
$$dq_{irrev}-dq_{rev}<0\tag{E02}$$
Now the Gibbs free energy (G) is defined as $G=H-TS$. Thus, $dG=dH-TdS-SdT$. At constant pressure and Temperature, $dH=dq$ and $dT=0$. Thus,
$$dG=dq-TdS= dq-T(dq_{rev}/T)=dq-dq_{rev}\tag{E03}$$
If the process is reversible , then E03 becomes $dG=0$. If the process is irreversible, E02 implies $dG<0$. Combining these together, we get $dG \leq0$.
This analysis ensures that $dG$ is never positive...... Which seems to suggest that (if we take $dG$ as the criteria for spontaneity) every process is spontaneous.
The logical conclusion is that there's something in my analysis that isnt general: it holds for spontaneous processes only. What is it? Is it that the first-law requires some modification to be general?

Will the down-voter please explain?

I am not the downvoter, but, dH cannot be dq, as dH is the total differential of the state function, while dq is not. Also, if we consider non reversible change, it would not be at constant T and p, so dH_irr= dU_irr + p.dV + V.dp.

Your analysis isn't general because you yourself restricted it to contstant T and p (and implicitly assumed there is no non-pV work).  And, under those conditions, dG_sys = -TdS_universe.  And since, for any real process, -TdS_universe < 0, we have that dG_sys > 0.  As to why dG_sys = -TdS_universe at constant T and p, with no non-pV work, see my answer here:  https://chemistry.stackexchange.com/questions/124412/what-is-wrong-in-this-argument-that-dg-must-always-be-zero/124470#124470

@Poutin  dq = dH under the conditions he's defined (constant p), if we further assume no non-pV work.

The problem is that you are starting the derivation from conditions that ensure reversibility or spontaneity, but you are not considering conditions that lead to the third possibility.
The following expression from your post is generally true:
$dq-dq_{rev}=dw_{rev}-dw $
I've dropped the "irreversible" labels and assume that $dq$ and $dw$ refer to some process that you have dreamed up.
Now the condition that leads to non-spontaneous processes must be
$dw_{rev}-dw>0$
since you have shown that the other conditions are for spontaneous processes, which leads to
$dq−dq_{rev}>0$
and therefore
$dG>0$
which is what you are looking for.
If you forget about $dG$ for a second and focus on  how you can get to the equation $dq−dq_{rev}>0$ (since you already showed the next step to an inequality in terms of $dG$), you see that to arrive at this equation you have to say that the system can do more work in a process you dreamed up than in a reversible process  ($dw_{rev}-dw>0$), which violates the 2nd law. To carry out a process in which $dG>0$, you have to correct this violation and make up the work difference by having the surroundings do additional (more) work on the system than was assumed in the original analysis.
By the way, you should be careful using PV-work in your examples, as for an ideal gas in the absence of reactions, Boyle's law says that the pressure will change during an isothermal expansion and Charles' law says that the temperature changes in an isobaric process. In neither case would the Gibbs free energy provide the criterion for spontaneity.

I see: So the first law should be  $dU=dq-P_{ext}dv + dw_{additional}$. By ignoring $dw_{additional}$ i am implicitly assuming spontaneity.

@satan29  No, it's not that you've left out dw_addional that causes, in your analysis, dG<0 for all real processes.  That's not the issue with your analysis.  I explained the issue in my comment on your question: It is true that, at constant T and p, and with no non-pV work, and if the system is closed (another implicit assumption you've made), dG<0 for any real change.  If you remove those constraints, dG can be equal to or greater than zero.  Simple as that.  It's the constraints you've added that force dG<0.

@satan29 That's not quite how I would explain it. The last paragraph in my answer is a warning because you mention pV work in your question. I edited my answer to try to explain.

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text:
Preparation of Grignard reagent. Free radical rearrangement

Mechanism of formation of Grignard Reagent
$$\ce{R-X + ^.Mg^. ->[\text{r.d.s.}] R^. + ^.Mg+X-} $$
$$\ce{R^. + ^.Mg+X- -> R-Mg+X-} $$

As the mechanism of preparation of Grignard reagent suggests, it involves the formation of free radical. And so, it is observed that racemic mixture is formed, in case the carbon on which the free radical is formed is chiral. But, why there occurs no rearrangement of this free radical ($\ce{R^.}$) ?

You need to be more specific. What kind of radical rearrangement?

Grignard reagent is an organomagnesium halide in which magnesium and halide bond is ionic. It is not like magnesium is always inserted in between carbon and halogen bond. Once formed, magnesium-carbon bond should be configurationally stable under the conditions of the reaction (Ref.1), and is polar covalent (at least partially) so that $\ce{Mg}$ has to be connected to its bonding partner $\ce{C}$ in the first place. In this polar covalent bond, carbon ends up being the negative end of the dipole, giving its nucleophilicity. The reaction undergoes in free radical mechanism (yet it does not involve a radical chain mechanism) to form Grignard reagent:
$$\ce{R-X + ^.Mg^. ->[\text{r.d.s.}] R^. + ^.Mg+X-} \tag1$$
$$\ce{R^. + ^.Mg+X- -> R-Mg+X-} \tag2$$
Magnesium metal has two valance electrons. The first step is the rate-determining step in this two-step mechanism, which happens on the surface of the metal. In this first step, $\ce{Mg}$ metal transfers one of its two valance electrons to the halogen radical, cleaving carbon-halogen bond homolytically. The process in step by step mode to explain the outcome:
$$\ce{R-X -> [R^. + X^.] ->[^.Mg^.] R^. + ^.Mg+X-} \tag3$$
This first step forms two radicals: $\ce{^.Mg+}$ and $\ce{R^.}$. These two radicals then get terminated by coupling each other to give the Grignard reagent in second step. The rearrangement of $\ce{R^.}$ (if there any) during the second step depends on the rate of rearrangement compared to the rate of the radical termination. However, according to the Ref.1, both chiral cyclopropyl-magnesium bond and vinyl-magnesium bond do in fact retain their configurations from parent alkyl halides. The authors carefully chose non-ionizable alkyl halide in solvent for this study so that any rearrangement should happen during the reaction (e.g., cyclopropyl tosylate solvolyzes at a rate of $\pu{5.7 \times 10^{-13} s-1}$ in actic acid at $\pu{25 ^\circ C}$, Ref.2). The excistence of $\ce{R^.}$ intermediate is proven by trapping it using $\ce{TEMPO^.}$ radical (Ref.3).
References:

H. M. Walborsky, "Mechanism of Grignard reagent formation. The surface nature of the reaction," Acc. Chem. Res. 1990, 23(9), 286–293 (https://doi.org/10.1021/ar00177a004).
P. von R. Schleyer, W. F. Sliwinski, G. W. Van Dine, U. Schöellkopf, J. Paust, K. Fellenberger, "Cyclopropyl solvolyses. III. Parent cyclopropyl derivatives and methyl-substituted cyclopropyl tosylates," J. Am. Chem. Soc. 1972, 94(1), 125–133 (https://doi.org/10.1021/ja00756a024).
Karen S. Root, Craig L. Hill, Lynette M. Lawrence, George M. Whitesides, "The mechanism of formation of Grignard reagents: trapping of free alkyl radical intermediates by reaction with tetramethylpiperidine-N-oxyl," J. Am. Chem. Soc. 1989, 111(14), 5405–5412 (https://doi.org/10.1021/ja00196a053).

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text:
Why is H in OH group is more shielded than in H in CH2 group (ethanol NMR spectroscopy)?
I am confused with this $\mathrm{^1H}$-$\mathrm{NMR}$ spectroscopy result:

Shouldn't $\ce{H}$ in $\ce{OH}$ group the singlet be more towards the left of the $\mathrm{NMR}$ spectrum?
It is near the highly electronegative atom $\ce{O}$ rather than the two $\ce{H}$ in $\ce{CH2}$ group they have a weak eletronegative atom $\ce{C}$ so it shouldn't be that desheilded.

What about that same $\ce{CH2}$ attached to very electronegative  $\ce{O}$ atom?

But the two H's has to throw the C atom and they see the O atom but in H in OH group its directly next to it.

Exactly, it's directly next to the electron-rich O atom.

Electronegativity is not the only thing matter here.

It is a myth that electronegativity is the only things that effect chemical shifts. However, electronegativity is only one of many things that effect chemical shift of a proton. Other things effecting chemical shift are magnetic anisotropy (e.g., large deshielding effects on aromatics and alkenes), steric compression (e.g., measurable $\Delta \delta$ values of alkene protons on cis- and trans-isomers), anisotropy of double bonds (e.g., measurable positive $\Delta \delta$ values of methyl group of methylcyclohexene and  methylcyclohexane, $\delta_{\mathrm{sp^2}(\ce{CH3})}-\delta_{\mathrm{sp^3}(\ce{CH3})} \approx 1.60-0.92= 0.68$), solvent effects, etc.
All of above mentioned factors effect chemical shifts of proton attached to electronegative atom such as $\ce{OH}$ and $\ce{NH}$. In addition, solvent used and solute concentration are also fators changing chemical shifts of these protons. For example, let's consider alcohol $\ce{OH}$ protons. In dilute solution of alcohols in non-hydrogen-bonding solvents such as $\ce{CCl4, CDCl3,}$ and $\ce{C6D6,}$ the $\ce{OH}$ signal generally appears at $\delta \ 1\!- \!2$. For instance, residual peak in $\ce{CDCl3}$ is always appeared at around $\delta \ 1.6$. At higher concentrations in non-polar solvents,  the $\ce{OH}$ signal moves downfield ($+\delta$) because when concentration increase, the $\ce{H}$-bonding increases as a result. For instance, the $\ce{OH}$ signal of ethanol comes at $\delta \ 1.0$ in a 0.5% solution in $\ce{CCl4}$, and it appears at $\delta \ 5.13$  in the pure ethanol (5-HMR-2 Chemical Shift):

The solvent used to take the given spectrum is unknown. Based on the chemical shift of $\ce{OH}$ signal, one can assume solvent might be $\ce{CDCl3}$, which is the most common among NMR solvents, which does not promote $\ce{H}$-bonding. As you can judge by the image above, it might have taken in the concentration of 1-5% range.

@Jan: What I meant to say is non H-bonding. I couldn't find the word at midnight before I finished. It's my mistake, but corrected it.

Ah, aprotic might be the word you’re looking for? ;)

Won't work since DMSO  and acetone would promote H-bonding.

What exactly is the effect of H-bonding that leads to deshielding? After all, the H-atom involved in the H-bond becomes a bridging atom and among other things you might expect local charge to either increase or be shifted so as to increase symmetry about the H. But that does not seem to be the case.


text:
How to find bond lengths from .cif files of crystal structure
I am confused about how to calculate the bond lengths within a given structure given its .cif file. I have highlighted the bond of interest in the image below. The atom coordinates are shown in the dialog box at the bottom of the screen. I used the formula for calculation of distance between two points in 3-D space (square root of the sum of squares of the differences between corresponding coordinates) and got $\pu{0.2372 \mathring A}$. VESTA gives the bond length as $\pu{1.93 \mathring A}$ angstroms which seems correct given the lattice parameters (cubic, $a=b=c= \pu{8.134 \mathring A}$). Am I missing something here?

If you multiply your 0.2372 by the cell parameter of 8.134, you'll get the right answer. Those are fractional coordinates, you see.


text:
Where to find a comprehensive list of chemical/biochemical etymologies?
I've found lists like this, which explain what different root words mean. But I'm looking for explanations of why those roots mean those things.
For example, why does carbonyl refer to a carbon double bonded to an oxygen? What is the historical rational behind why the -yl implies the oxygen double bond?
I'm hoping this gives me a little more intuition to remember what different chemical terminology means without brute-force memorization.

Huh, and can you learn a foreign language using such method? No, no you can't. And even if it's possible, it would be waste of time when one just needs to communicate. Etymology of like every single old name of compound, or group, element etc. is very different and such knowledge is not useful at all. What you should learn is nomenclature.

That's quite a sweeping declaration to say that such knowledge is not useful at all. In chemistry at least, some compound or element names are based on their macroscopic observable properties. And you can learn a great deal about languages based on etymologies. Memorizing nomenclature is useful but I'm looking for supplemental information.

@Mithoron "Etymology of like every single old name of compound, or group, element etc. is very different and such knowledge is not useful at all. " This is indeed a sweeping statement. English is apparently not your first language nor mine but it always helps to learn tidbit of etymologies.

Why the downvotes? As far as I can tell this question is on-topic for this site and hasn't been asked before.

It's too broad and existing answer only showcases that. @M.Farooq Well, I do know a bit on this topic and can't say it's not interesting, but my point is that's rather "additional" then practical.

@Mithoron, What is the harm in asking for reference(s) on chemical etymologies? To the best of my knowledge there are no other books which need to be mentioned. All he needs to know are these list four books for his entire life. Should the OP generate one thousand posts for each word used in chemistry. In my humble opinion, this question has no problem. This topic is close to my heart as well that is why I wrote a detailed response including my own contributions.

It is a good question and many comprehensive works are available specifically on chemical etymologies. Many scholars have spent on portion of their life on chemical etymologies.  Ignore these useless down votes or close votes. The answer essentially boils down to how money are you willing to spend? I would say, if you are really keen, go for OED.
After going through so many etymological dictionaries of science (mainly chemistry & mathematics), I still feel that the Unabridged Oxford Dictionary  (> 20 volumes) is the best (https://www.oed.com/). It is not available for public use but one can buy subscription for a year at an affordable price. Most universities have it. There you can find the etymology for sure, the best part is that you can see the earliest use with original reference.
I will give you a taste of it. Say you want to find out where does the word alcohol come from?  You can extend this that the suffix -ol indicates alcohol group. Perhaps it is too much for you but you can see the extent of depth of analysis. It is sometimes surprising that there might be no connection of older meanings with current meanings.

post-classical Latin alcohol, alcool, alcol, alcofol kohl (galena
(lead sulphide) or stibnite (antimony sulphide)) (from 13th cent. in
British sources), spirit, essence obtained by distillation (a1527 in
Paracelsus in alcool vini , also alcohol vini ) (perhaps via Spanish
alcohol: see below) < Arabic al-kuḥl , Spanish Arabic al-kuḥul < al
the + kuḥl , Spanish Arabic kuḥul eye cosmetic, also denoting various
specific substances used as eye cosmetics or eye unguents (compare
kohl n.1) < the same base as Arabic kaḥala, Hebrew kāḥal (in the Bible
only in an isolated attestation in Ezekiel 23:40), both in sense ‘to
stain, to paint’, Akkadian guḫlu antimony (used as eye paint).
Compare Middle French alcohol, (rare) alcofol, French alcool,
†alcohol, †alcol, †alkol, †alkool, etc., the chief senses of which
are: ‘kohl, very fine powder’ (c1370 in a translation of Chauliac; the
precise sense is often difficult to determine in early quots.),
‘essence obtained by distillation’ (1620, originally and chiefly with
reference to spirit of wine), ‘ethanol’ (1792), ‘any of the members of
a similar class of chemical compounds’ (1834: see note below). Compare
also Spanish alcohol (c1200 as †alcofor; also †alcofol, †alcool,
†alcol, etc.), the chief senses of which are: ‘powder used as eye
cosmetic’ (c1200 as †alcofor), ‘galena, sulphide of lead’ (1541),
‘antimony, also any of various minerals containing antimony,
especially stibnite’ (a1555 or earlier; in a number of early instances
denoting minerals (from the second half of the 12th cent.) it is
impossible to tell whether the word denotes galena (sulphide of lead)
or antimony sulphide), ‘spirit of wine, ethanol’ (1730), ‘any of the
members of a similar class of chemical compounds’ (1865).

On your specific query
Carbonyl

< carbon n. + -yl suffix. Compare French carbonyle (1856 or earlier),
German Carbonyl (1854 or earlier).

Further probing of -yl

Formerly occasionally -ule, a terminal element of chemical terms, <
German -yl, < Greek ὕλη wood, matter, substance (see hyle n.), used
for ‘chemical principle, radical’. It was introduced by Wöhler and
Liebig ( Ann. der Pharm. (1832) III. 262), and first used by them in
the term benzoyl; other early names were éthyle (éthule), élayle
(Berzelius), dadyle, peucyle, citronyle, citryle (Blanchet and Sell).
Some fifteen in anglicized form, including acetyl, amyl, cinnamyl,
glyceryl, salicyl, appear in the Elements of Chemistry by T. Graham,
1842,

If you find OED too much to digest, the next best etymological dictionary of chemistry is

Elsevier's Dictionary of Chemoetymology (1st Edition) The Whys and Whences of Chemical Nomenclature and Terminology by Alexander Senning.
It has 400 pages. For example, the entry aniline shows

aniline C6H7N, derived from al-nil (Arabic: indigo plant), ultimately
from nila (Sanskrit: dark blue) – referring to the fact that aniline
was first obtained by degradation of indigo

The next one, which focussed on organic chemistry is

Organic Chemistry: The Name Game Modern Coined Terms and Their Origins (1st Edition)
by Alex Nickon and Ernest F. Silversmith,
This has plenty of fancy organic molecules from modern organic chemistry but at the end, they do list traditional names for example for benzene

benzene from Styrax benzoin, a tree native to Sumatra and Java. The
bark yields a resin, gum benzoin, from which "benzoic acid" was
obtained. Peligot (1833) and E. Mitscherlich (1834) heated benzoic
acid with lime to form C6H6; Mitscherlich named it "benzine." Liebig
preferred "benzol," but Laurent (1835) proposed "benzene."

Coming to free rides. Dr. John Andraos made a series of webpages by the name Modern Coined Glossary of Coined Names & Terms Used in Science to which I had heavily contributed more than a decade ago.
Here is a link: http://www.careerchem.com/NAMED/Glossary-Coined-Names.pdf
Another free source is Dr. William B. Jensen's lovely book "Ask the Historian". Simply Google "Ask the historian Jensen" and the first result is his book. He explain a lot of names, and even history of concepts. It was kind of him to make this book free of all of us. He has retired and he is no longer active.

(+1) “You know the opinion of Cervantes? He said that reading a translation is like examining the back of a piece of tapestry.” Carl Sagan, Contact (the movie)

Very true, Ed. As a translation promoter, I can say this is really true for art and literature indeed but fortunately not for science.

This is awesome, thanks so much for putting together this comprehensive list and adding examples. I'll give it 24h before marking as answered.


text:
Why can’t I use solubility product for normally soluble compounds?

Solubility product only applies to sparing soluble ionic compounds (chemguide)

I read this in a website. But I don’t see what this should be the case. Taking $\ce{NaCl}$ for example, there’s an equilibrium established as it dissolves. Why then does solubility product not apply to this scenario?

You could calculate it, but it's wouldn't be all that useful.

At high ion concentration, solubility products are activity driven and there are not good models for activity coefficients of such solutions. Using ion concentrations, the effective solubility product would be widely solution composition dependent.

In addition to Mithoron's and Poutnik's comments, the solubility product may not be relevant to your solution chemistry anyway.  Solubility products enter into solution equilibria when the precipitates are present, as in this answer where relative solubility products predict whether mostly magnesium or mostly iron will be dissolved in water when their hydroxides are present.  With readily soluble materials, you likely do not get the precipitate so solubility products, even if you can compute then on an activity basis, have no bearing on the equilibrium.


text:
Is the given statement regarding Radial Probability Distribution Function correct or not?

Radial Distribution Function $(4πr^2R^2(r))$ gives the probability of the electron being present at a distance $r$ from the nucleus.

Answer: The given statement is correct.
My Query:
According to me, isn't the Radial Probability Distribution Function ($RPDF$) used for finding the probability of an electron being present at a distance $r$ from the nucleus within a region of thickness  $dr$?
Therefore, shouldn't the above statement be wrong?

The emphasized part is implicitly assumed. Anyway, the probability to be exactly at a distance $r$ is $0$.

You are correct it should be $r \to r+dr$ but it is not often written that way.

@IvanNeretin are you saying that both answers are correct?

@porphyrin so then mine is correct too, right?

Conceptually you are right as the commenters have mentioned, but since we are on a thread about nitpicking, we might as well go the extra distance.
Technically, $r^2R^2$ itself is not a probability but a probability density. In order to get the actual probability, you need to integrate it over a region. The probability of finding an electron between $r = r_1$ and $r = r_2$ would therefore be
$$P = \int_{r_1}^{r_2}r^2R^2 \, \mathrm dr \tag{1}$$
For an infinitesimal region, it suffices to multiply by $\mathrm dr$, such that the infinitesimal probability of finding an electron in the region $(r, r + \mathrm dr)$ is
$$\mathrm dP = r^2R^2 \,\mathrm dr \tag{2}$$
Notice how if you integrate $(2)$ over a finite region you get $(1)$.
Also, some books give this extra factor of $4\pi$, which I never really understood. You have two choices: either the probability density function is $r^2R^2$ (which works for all orbitals), or $4\pi r^2\psi^2$ (which works only for spherically symmetric orbitals, i.e. s-orbitals). I've written about this in more detail at  What is the exact definition of the radial distribution function?, so won't repeat myself here.

I've read your answer before posting this question, and thanks a lot1


text:
Products of electrolysis of Ni(NO3)2
During the electrolysis of $\ce{Ni(NO3)2}$ using nickel electrodes, it's said that the molarity of the solution remains constant as nickel's active, and dissolves and deposits in equal measure.
But water has an oxidation potential of 0.828 V, as compared to nickel's 0.24. So while the cathodic deposition seems fine, shouldn't it be oxygen that's liberated at the anode?

That's not how it works at all. Look again at that 0.828: what is the corresponding reaction?

Isn't it $2H_2O → O_2 + 4H^+ + 4e^−$, as opposed to $Ni →  Ni^2+ + 2e^-$

At the cathode, the ion $\ce{Ni^{2+}}$ is reduced into metallic nickel. At the anode, the metallic nickel is more easily oxidized into $\ce{Ni^{2+}}$ ion, than water into $\ce{O_2}$.

Some electrolysis experimets are best first performed and then an attempt  at explanation.
For example, the products of the electrolysis of aqueous nickel nitrate with graphite electrodes are reportedly per a source (with video) as:

Electrolysis of a nickel nitrate solution produces oxygen at the anode, and hydrogen and nickel at the cathode.

Looking at related metal electrolysis, for example, Cu(NO3)2, per this source with again graphite electrodes:

Electrolysis of a copper(II) nitrate solution produces oxygen at the anode and copper at the cathode.

Lastly, with the transition metal manganese, a source on the electrolysis of Mn(NO3)2, reports, to quote:

$\ce{Mn(NO3)2 + 2H2O ± 2e- -> H2 + MnO2 + 2HNO3}$

Based on all of the above, and that hydrogen gas is forming per the reactions:
$\ce{H+ + e- -> .H }$
$\ce{.H + .H  -> H2 (g)}$
I would expect similarly some transient nitric acid formation with Ni(NO3)2 per its potential interaction between the nitrate ion and the hydrogen atom radical as follows:
$\ce{ NO3-  + .H  -> .NO2  + OH-}$
$\ce{.NO2 + .NO2 <=> N2O4 }$
$\ce{N2O4 + H2O -> HNO2 + HNO3 }$
Note, my claimed introduction of HNO3 (and its likely subsequent removal by nickel metal) can explain the comment "the molarity of the solution remains constant as nickel's active, and dissolves and deposits in equal measure", and is also consistent with the electrolysis of Mn(NO3)2 resulting in the creation of nitric acid.
And lastly, you are correct, oxygen is produced at the anode.


text:
Does periodic acid cleave acetals?
Recently, I started reading about the cleavage reactions done by $\ce{HIO4}$. I came across a statement which read

$\ce{HIO4}$ can't cleave acetals but it can cleave hemiacetals.

But when I tried to reason out, I thought acetals should have more tendency to get cleaved in acidic medium, because the reverse reaction of acetal formation will have more favourable entropy change than reverse reaction of hemiacetal formation. But this doesn't explain the above statement.
Which part in my reasoning is wrong?


text:
Why are acetic acid and sodium bicarbonate mixed to unblock a drain?
I have always wondered about the rationale behind combining white vinegar and sodium bicarbonate to perform household cleaning jobs, and to unblock drains. I would have expected white vinegar to act well on its own as an acidic cleaner, and sodium bicarbonate to act well on its own as a base cleaner.
Does combining them not simply neutralise the potential of each to perform its job?
The same combination is often recommended for unblocking slow drains. Again, depending on what is blocking the drain, I would have expected that an acid or base would act well in isolation. And given that hair and soap are the likely culprits, I would have anticipated that sodium bicarbonate, a base, would be the more effective.
What is the justification for this? Is it, perhaps, the mechanical action of the carbon dioxide produced that is intended to lift whatever is blocking the drain? Or is there something else at play?
Thank you in advance.

drain cleaner related: Understanding sodium hydroxide drain cleaner's exothermic dissolution of aluminum; does the aluminum need to be oxidized before it dissolves? and Is it known for sure that bases feel slippery because of the production of soap/surfactant?

Quoting text from this source under the section "Why could baking soda and vinegar clean clogged drains?" :

Bicarb soda and vinegar react because of the acid-base reaction. Bicarb soda is bicarbonate $(\ce{NaHCO3})$ and vinegar is acetic acid $(\ce{CH3COOH})$.

When bicarb soda and vinegar react, they fizzle and sizzle and they expand. This is why they are able to remove clogs from drains. The pressure from the expanding product shoves the clog down as it moves along. The vinegar solution (vinegar plus water) can remove soap residue that clings to the walls of the pipes.

So it seems like your conjecture is justified, that it is not actually the reactivity of these two substances with the waste that is responsible for majority of the cleansing, but simply the mechanical action of the expansion of the carbon dioxide gas and the forming product that pushes the waste down the drain, and leads to unclogging.
The relevant chemical equation is:
$$\ce{CH3COOH (aq) + NaHCO3 (s) -> CH3COONa (aq) + CO2 (g) + H2O (l)}$$
This is similar to the mechanism by which baking powder makes a cake fluffy on addition of a little vinegar as explained here.

Please don't trust random websites. There is no such thing as carbonic acid at least in the solution. Your second equation is better.

@M.Farooq Yeah even I was skeptical about the carbonic acid part, I have removed that now. Also, while citing it, I felt that the website above is somehow related to an official Australian plumbing body, but I guess it's just a random company. Will keep that in mind

$$\ce{NaHCO3(s) + CH3COOH(aq)  ->  CO2(g) + H2O(l) + CH3COONa (aq)}$$
From what I can see in the reaction, an amount of water and carbon dioxide is produced which has the potential to move around blockages in the drain similar to using club soda to remove stains from a shirt or table cloth. The resulting sodium acetate is a weak conjugate base and water is too weak to react with anything blocking a drain to begin with. By deduction, I think your hypothesis of the drain cleaning being an action of the gases produced is correct.

Please try to provide references whenever possible while answering,especially when you are trying to prove or disprove the OP's hypothesis for a situation

Yusuf, the general rule for references is that certain factual stuff does not need references in scientific writing. For example, if we write the Sun rises from the East. We won't provide reference but if I say, the composition of Sun's atmosphere is a certain gas, yes this fact needs a "support."

@M.Farooq Ok, I see...The only reason I commented was that this answer had not provided any references whatsoever,even for where the chemical equation had come from. But I see your point, will keep that in mind

The exothermic reaction of the acetic acid on $\ce{NaHCO3}$ provides heat and I do agree that the vigorous nature of the reaction forming $\ce{CO2}$ is likely instrumental in clearing the drain.
Also, I suspect employing an excess of NaHCO3 is probably beneficial as on warming it releases CO2 and creating more alkaline Na2CO3, which will also attack grease.
Reaction:
$\ce{2 NaHCO3 -> Na2CO3 + H2O + CO2 (g)}$
Note: Baking Soda is a common ingredient employed in dishwashers owing to its ability, especially on warming, to dissolve fats/grease (see discussion here).


text:
Rate of solvolysis of allyl and alkyl halides

Suppose I have 2 compounds, 3-chloropropene and 1-bromo-1-methylcyclohexane. Without knowing the values, how can we compare the rate of solvolysis?

My attempt: Resonance is quite dominant character thus alkyl chloride resonance I think should dominate over hyperconjugation but then we have an exception that 3 degree carbocation is more stable than benzyl carbocation hence there might be some disparity in orders.

Resonance effect is more stronger than hyperconjugation irrespective of the number of $\alpha$H. In first case allylic carbocation is formed which is stabilised by resonance. So solvolysis rate is more for the first compound.

3 degree carbocation is more stable than benzyl carbocation - only for trimethyl carbocation; other 3 degree carbocations are less stable than benzyl carbocation. See https://chemistry.stackexchange.com/questions/74943/is-the-t-butyl-carbocation-more-stable-than-the-benzyl-carbocation

Now problem is that in answer of this question they have given the reverse order but I think concepts matter more so what is reality that should be right answer. Thanks

Rate of solvolysis depends on two factors:

$\ce{C-X}$ bond energy: $\ce{C-Br}$ bond is weaker than $\ce{C-Cl}$ bond because, $\ce{C-Br}$ bond is longer than $\ce{C-Cl}$ bond.
Stability of the carbocation:  The stability of a carbocation can be measured (or compared with others) with the help of their hydride ion affinity(HIA) values. More is the HIA value lesser is that stability of the carbocation. Here are some HIA values for some carbocations:
$$\begin{array}{|c|c|}
\hline
\textrm{Carbocations} & \textrm{HIA values} \\
 & (\textrm{in Kcal/mol}) \\
\hline
\ce{CH2=CH-\overset{+}{C}H2}& 256 \\
\ce{(CH3)2\overset{+}{C}H} & 246 \\
\hline
\end{array}$$

The previous answer states that,

Resonance effect is more stronger than hyperconjugation irrespective of the number of α-hydrogens.

I feel this is wrong. Because hyperconjugation is a type of resonance. Effect of hyperconjugation (with one alpha hydrogen) might not be effective as resonance effect but, it is comparable with resonance effect with increase in the number of alpha hydrogens. This can be understood from the HIA values.
Now let's compare the stability of 1-methylcyclohexyl cation and isopropyl cation. Stabilization of cation by inductive effect is more in case of 1-methylcyclohexyl cation because, in 1-methylcyclohexyl cation it is surrounded by more number of carbons than that of in isopropyl cation. As there are more number of alpha hydrogens in 1-methylcyclohexyl cation than in isopropyl cation, the effect of hyperconjugation will be more in case of 1-methylcyclohexyl cation. By combining these two effects we can say that 1-methylcyclohexyl cation is more stable than isopropyl cation. So the HIA value for 1-methylcyclohexyl cation will be less than 246.
So by having a rough idea on the HIA values we can say that 1-methylcyclohexyl cation is more stable than allyl cation. So by combining these two factors we can say that the rate of solvolysis is more for 1-bromo-1-methylcyclohexane than 3-chloropropene.

Resonance effect is more stronger than hyperconjugation irrespective of the number of αH. In first case allylic carbocation is formed which is stabilised by resonance. So solvolysis rate is more for the first compound


text:
Colour due to transmission and reflection
It makes sense to me that when looking through a sample (observer | sample | light), it should appear as the opposite of the light absorbed, but it does not make sense to me to expect the same when not looking through it, just standing by the same side as the light source (observer | light | sample, or light | observer | sample).
In the first cenario, the observer sees light emitted - light absorbed (the transmitted, that barely interacts with the sample). In the second, the observer sees light reflected (to my understanding, light emitted from the de-excitations of excitations caused by the light source).
However, when I see a solution of $Cu^{2+}$, it looks the same when observed in both settings. Thereby implying that transmitted colour = reflected colour. Why is that?
Somewhat related question but without a satisfactory answer: Transmission, absorption, and reflection of light

Since your experimental observation with copper salts negates your original hypothesis, it implies that the way you are trying to explain it is wrong. The main culprit and the source of all these problems is the color wheel which is taught in schools. Misconceptions persist for long. When you are looking at the copper solutions in two different settings, it is misleading to think that copper solution is reflecting blue light back to you. It is not.
In each case, copper(II) solution is absorbing a small portion of the red light from the visible spectrum and it appears to you like a pure blue solution.
Indeed it is our brain which has been created in such a way that when the visible spectrum has a certain red portion missing, it perceives the remaining spectrum as "blue".
Hint: Water in the ocean also appears blue? Water also very very weakly absorbs the red portion of the visible spectrum. You just need tons of water to perceive this effect.
There is a beautiful book by the name of The Physics and Chemistry of Color. The same author wrote an article "The fifteen causes of color: The physics and chemistry of color." It is certainly worth consulting. Article-behind paywall

The fifteen causes of color derived from a variety of physical and chemical mechanisms are summarized in five Croups in this article. Vibrations and simple excitations explain the colors of incandescence (e.g., flames), gas excitations (neon tube, aurora), and vibrations and rotations (blue ice and water). Ligand‐field‐effect colors are seen in transition‐metal compounds (turquoise, chrome‐oxide green) and impurities (ruby, emerald). Molecular orbitals explain the colors of organic compounds (indigo, chlorophyll) and charge‐transfer compounds (blue sapphire, lapis lazuli). Energy bands are involved in the colors of metals and alloys (gold, brass), of semiconductors (cadmium yellow, vermillion), doped semiconductors (blue and yellow diamond), and color centers (amethyst, topaz). Geometrical and physical optics are involved in the colors derived from dispersive refraction (rainbow, green flash), scattering(blue sky, blue eyes, red sunset), interference (soap bubbles, iridescent beetles), and diffraction (the corona aureole, opal).

Now you can imagine that all is not that simple.

(+1). Then there is human perception and such curious things as Impossible color and also https://en.m.wikipedia.org/wiki/Color_constancy.

That's pretty nice, I'll be reading that article soon. However, this answer does not answer my question. The solution is absorbing red, nice, but why is the transmitted colour the same as the reflected? I thought the two process followed different mechanisms, with transmission being "barely interacting" and reflection being excitation -> de-excitation by photon emission.

Dear LocoAsker, as I said there is no reflection. You are still seeing transmitted light. Reflection is not de-excitation by photon emission. This topic which I tried to cover is at least one hour lecture. StackExchange is not a substitute to reading books. Your confusion is arising from mixing of several phenomena. Why don't you read the article first, and then we discuss further?

If you really want to see true reflected light from copper solution in a flat glass tumbler, shine white LED from an angle, do not look at the solution, but let the reflected light beam fall on a white wall or white paper. What color do you see? Don't forget to do a blank, i.e., try with water first.

@LocoAsker also see it this way : the light you see in both cases is that from the source minus what it is absorbed (plus scattering in general). Solid layered films, for instance a transparent colored material cast on a glass slide, can behave differently and similar to your expectation.

So even if I stand between the light source and the solution, I'm still not seeing reflection but transmission? That's unexpected.

@LocoAsker, You are not doing these experiments in a pitch dark room. Do you? Light is coming from every side.


text:
What does the alcohol functional group (specifically oxygen) have to do that causes an alcohol to have a lower energy content than an alkane?
My textbook said that the carbon is "partially oxidised" (don't really know what that means) and I just wanted to confirm.

Search for what oxidized means and what partially means.

Energy released from organic compounds by burning is based on oxidation,  breaking $\ce{O2}$, $\ce{C-C}$ and $\ce{C-H}$ chemical bounds and creation of $\ce{O-H}$ and $\ce{C=O}$ bounds. The latter are stronger than the former, need more energy to be broken, but OTOH more energy is released when they are created. This energy difference of chemical bounds leads to releasing energy.
If a substance already has some oxygen in its molecule, part of this bond rearranging is already done, so we can say the substance is already partially oxidized. The energy released by burning rearranges just the rest of the bonds and is lower.
Another factor affecting the specific burning enthalpy $\Delta H$ in $\pu{MJ/kg}$ is the relatively higher mass of oxygen atoms, compared to hydrogen and carbon atoms. Oxygen acts here like a "dead weight".
An extreme example are explosives, that are very rich in oxygen and need not an external oxygen to be burned. E.g. TNT releases about 10 times less energy than burning the same mass of coal. ( But much faster, of course ).


text:
NGP in Finkelstein reaction
The question is to find the product in the following reaction-

In the answer given, they have replaced Cl with I and inverted the configuration at that carbon(Finkelstein reaction).
But $\text{I}^-$ being a good base, wouldn't it first take away O's acidic H and then $\text{O}^-$ will show Neighboring Group Participation attacking the carbon with chlorine and finally resulting in retention of configuration?

I- is not a good enough base to deprotonate OH

How do you say that?

I- is really not a good base. The pKa of the OH proton is around 16. Iodide in EtOH does not give EtO- and HI

Ok, now it makes sense

Iodide is the very opposite of a good base. HI is the strongest hydrohalogenic acid.

You are thinking too far ahead for a simple reaction. First of all, $\ce{I-}$ is not a very strong base to abstract a protone from alcohol. $\ce{I-}$ is the conjugate base of very strong acid, $\ce{HI}$. Yet, $\ce{I-}$ is a very good nucleophile. Therefore, the condition given ($\ce{NaI}$ in anhydrous acetone; $\ce{NaI}$ dissolves in anhydrous acetone and therefore exists as $\ce{I-}$ in solution) is suitable only for nucleophilic substitution reaction and also favored $\mathrm{S_N2}$ mechanism. Since the byproduct form ($\ce{NaCl}$) is not soluble in acetone, the reaction favors only the forward reaction:
$$\ce{R-Cl + Na+ + I- -> R-I + NaCl_{(s)}}$$
Since it is a $\mathrm{S_N2}$ reaction, inversion is the result ($(1R,2S)$ to $(1R,2R)$ with closed to $100\% e.e.$).
However, note that there is a neighboring group participation (NGP) possible in 1,2-halohydrins in certain reactions (Ref.1). As OP suggested, if NGP is contributed here (without strong base I doubt it though), the product from NGP is going to be the retention product with $(1R,2S)$ configuration. Thus, total $\%e.e.$ of final product is $\lt 100\% e.e.$.
References:

T. A. Geissman, Richard I. Akawie, "Rearrangements of Halomagnesium Derivatives of Halohydrins," J. Am. Chem. Soc. 1951, 73(5), 1993–1998 (https://doi.org/10.1021/ja01149a029).

While OP's mistaking iodide for a strong base is mind-boggling, he has a point - why NGP wouldn't change the picture here?

@Mithoron: I agree with you. There is a possibility but it is very faint. But worth mentioning it. :-) Accordingly I change my answer

@Mithoron What do you want it to do? Form an epoxide?

@Zhe -OH can well enough participate in SN1, but it's effect is much more transient then that, more like first step on the road to epoxide and then step backwards.


text:
Does scotch tape only use van der Waals forces?
Does it also make use of covalent bonds to stick to surfaces?

Essentially yes. You can probably upscale the analysis (surface tension etc) but at the smaller scale it will be VdW forces and no covalent bonds.

Curious, It is way more complex. Read here https://www.scientificamerican.com/article/what-exactly-is-the-physi/


text:
Acid catalyst choice on reaction of carbonyl group and O/N nucleophiles
For oxygen nucleophiles, e.g. $\ce{ROH}$, when reacting with a carbonyl group to form acetal, the acid catalyst used is protonated alcohol.
for nitrogen nucleophiles, e.g. $\ce{RNH2/RR'NH}$, when reacting with a carbonyl group to form imine/enamine, the acid catalyst used is $\ce{NH4+}$ instead of protonated alcohol.
According to both reaction mechanisms, both catalysts are just acting as $\ce{H+}$ donators & acceptors, so why 2 different acid catalysts need to be used? Can't I use protonated alcohol all the time?

Where did you get the idea that NH4+ is the preferred catalyst? I find Ti(IV) species such at Ti(OiPr)4 the most effective.

@Waylander from 'Organic chemistry as a 2nd language 2nd semester', it told me ammonium ion is generally used as the catalyst. And sorry that I am just a beginner, when I look up the structure of Ti(OiPr)4 in Google, I don't know where is the ionizable proton.

Titanium(IV) is a Lewis acids.

@234ff You need a better textbook

Seems all my talking under earlier question was in vain...

@Mithoron Yea sorry since the book always tell me that 'we generally use ... as the catalyst' without any explanation why it is 'generally' when C=O bond is attacked by nucleophiles. When I search it online I got online seems no one uses them, so I am so confused.


text:
Metal deposition on passing electricity
A question asks about the results on electrolysis involving metals, and the answer says that Mg and Al would be deposited on the cathode.
Aren't their reduction potentials lower than that of water? Aluminium has a reduction potential of a bout -1.66, and water's is about -0.8.
How does that work out?

It should not work, unless it is molten salt, or there is a mercury cathode.

Got it; can you specify why water seems to have two different electrode potentials for oxidation and reduction? http://www1.lsbu.ac.uk/water/electrolysis.html says it's -0.828 V at the cathode and 1.229 at the anode.

Isn't one usually the negative of the other? (Sorry if this's really dumb.)

Opposite potentials for opposite half reactions. These are not the opposites. Oxidation of H2O to O2 is not the opposite to reduction of H2O to H2. The opposite of reduction of H2O to H2 is oxidation of H2 to H2O.

@Poutnik: can you post that as an answer so I can mark it as one?

Mentioned deposition of metallic magnesium or aluminium does not work, unless it is molten salt, or there is a mercury cathode.
Opposite potentials are for the mutually  opposite half reactions.  Water oxidation and water reduction are not the opposite half reactions. Oxidation of $\ce{H2O}$ to $\ce{O2}$ is not the opposite to reduction of $\ce{H2O}$ to $\ce{H2}$. The opposite of reduction of $\ce{H2O}$ to $\ce{H2}$ is oxidation of $\ce{H2}$ to $\ce{H2O}$.


text:
How will the X ray diffraction pattern change if the crystal changes its lattice structure
(I asked this question on another community without getting much help there,I hence post it here as it may be more of a chemistry question. I apologize if my question has an obvious answer, I'm only a beginner)
A text says:
An X - ray diffraction experiment is carried out on a crystalline solid having  FCC structure at room temperature. The solid undergoes a phase transformation on  cooling and shows orthorhombic structure with small decrease in its unit cell  lengths as compared to the FCC unit cell lengths. As a result the 311 line of the XRD pattern corresponding to the FCC system will split into a triplet
Now this is what i could think about it:
We know that the interplanar distance for an orthorhombic lattice is given as $$
d=\frac{1}{\sqrt{\frac{h^{2}}{a^{2}}+\frac{k^{2}}{b^{2}}+\frac{l^{2}}{c^{2}}}}
$$ and using Bragg's law we can find the angle at which constructive interference occurs as $$
\sin \theta=\frac{\lambda}{2 d}=\frac{\lambda}{2} \sqrt{\frac{k^{2}}{a^{2}}+\frac{k^{2}}{b^{2}}+\frac{l^{2}}{c^{2}}}$$
using the values of$${h k l}$$$$={3 1 1}$$ we obtain the diffraction angle which is unique as given by the above equation. Beyond this I cannot understand what to do,how does any splitting occur, what am I missing.

So $a,b,c$ are different in orthorhombic vs an fcc cell, right?

I  believe  that's correct.

So what does that mean for the 311?

please check your second equation and rewrite it as it is not consistent with the first.

See, there are 311, 131, and 113. In FCC, they are all the same. In orthorhombic, no longer so. That's the splitting.

@Ivan Neretin, thank you. Isn't it that the text should have put the indices inside angular brackets like <3 1 1> to indicate that it meant a family of directions?

Maybe, but we got used to understand it anyway.

@Ivan, for beginners like me?

You'll get used to it, too. Then again, how would a beginner know the meaning of angular brackets (which is not universal, mind you)?

Oh! I thought it was universal. Thanks again.


text:
What does spontaneous mean in term of forward and reverse reactions?
I just learnt about Gibb's Free Energy and was thinking about an exothermic reaction with a negative change in entropy. This means that the reaction would occur at low temperatures. But that means that if the system the reaction is taking place in has heat removed from it the reaction would occur. I am not sure if this next assumption is true or not, but if the reaction is not happening that means the rate of reaction is negative (rate being defined as Δ[Product]/Δtime). Since the rate is also defined as R = k[reactants], and concentration cannot be negative, this means that k must be negative, and also that k increases when temperature decreases. However this violates the Arrhenius equation, where an increase in temperature results in an increase in the rate constant. The only solution I see to this conundrum is my assumption was wrong and maybe spontaneous just means the rate of reverse reaction>rate of forward reaction, so they both have positive values for k. I wanted clarification if this was correct or not, or if I completely messed up the definition of Gibb's Free Energy in the first place.

The net reaction rate is zero when the reaction is not happening, since the concentration of all species is constant. Both forward and reverse rates occur, and it is possible to write rate equations such that rate constants are only positive. What changes when you are not at equilibrium is that either forward or reverse rate is larger.

If it's true in the forward direction, it's also true in the backwards direction. The magnitude of those changes may be different which is why there is ultimately a dependence on temperature, at least in the case you're describing.

I think you should re-frame your third sentence which is not clear. Other than that (i) If the reaction is not happening, than Δ[Product] is zero, thus, rate of reaction is zero. (ii) Value of rate constant can never be negative. (iii) Value of $k$ increases with temperature for both exothermic and endothermic reactions but for exothermic reactions, the rate of backward reaction will increase with increase in $T$.


text:
Separating Silica from silicates
Does anyone know how one could extract Silica $\ce{SiO2}$ from a one or more of the following compounds in either ground or powder forms?

Plagioclase: $\ce{Ca2Al2Si2O8}$
Pyroxene: $\ce{(Ca,Mg,Fe)2Si2O6}$
Olivine: $\ce{(Mg,Fe)2SiO4}$

Ideally the process would only involve energy and mechanical processes rather than adding many other chemicals.

While we may write silica as $\ce{SiO2}$, it neither is a small molecule (like water), nor a salt (like sodium chloride).  Instead, it forms networks of silicates. Think of little tetrahedra (https://en.wikipedia.org/wiki/Silicate) which may be connected with each other over their tips, or sharing a rim, or a face with the next one.  So there is no simple solvent capable to extract these separately from the other metal ions mentioned for the three minerals.  Silica isn't something you could attract with a magnet, either.

Mechanically speaking, it is not possible to separate $\ce{SiO_2}$ from these minerals. The only way to do it is carrying out a chemical transformation. Here is an example of such a treatment.
Mix a sample of one of these minerals with $\ce{Na_2CO_3}$ in excess, and heat it in a platinum crucible with a Bunsen burner. When the substance is liquid, wait $10$ minutes. Let it cool down. The mineral is now transformed into sodium silicate $\ce{Na2SiO3}$ plus some oxides and carbonates. If I choose the specially simple olivine $\ce{MgFeSiO_4}$ as an example, the following reaction may happen : $$\ce{MgFeSiO_4  +  Na_2CO_3 -> Na_2SiO_3 + MgCO_3 + FeO}$$ The reaction is somewhat more complicated, because usually $\ce{FeO}$ is oxidized in air into $\ce{Fe_2O_3}$, and $\ce{MgCO_3}$ is partially decomposed into $\ce{MgO}$ and $\ce{CO_2}$. Furthermore, the sodium silicate $\ce{Na2SiO3}$is partly transformed into $\ce{Na4SiO4}$. Fortunately this side reactions do not really modify the rest of the process.
Anyway, put the crucible in a small beaker. Add $20$ mL water. Let it boil for a couple of minutes. When the substance is pulverized, add an excess of diluted $\ce{HCl}$. Heat on a water bath until the substance is dry. Add $2$ mL concentrated $\ce{HCl}$. The following reactions happen : $$\ce{Na_2SiO_3  +  2 HCl ->  H_2SiO_3 + 2 NaCl}$$ $$\ce{MgCO_3 + 2 HCl-> MgCl_2 + CO_2 + H_2O}$$ $$\ce{Fe_2O_3 + 6 HCl-> 2 FeCl_3 + 3 H_2O}$$ After $10$ minutes, add some hot water and filtrate. $\ce{H_2SiO_3}$ remains on the filter. Calcium, magnesium and iron chlorides are soluble into water and are eliminated.  Then $\ce{H_2SiO_3}$ can later be heated to obtain $\ce{SiO2}$ $$\ce{H2SiO3 -> SiO2 + H2O}$$  This is one way for obtaining $\ce{SiO2}$ from the proposed silicates. Other processes are known, using for example fluorhydric acid $\ce{HF}$. But, of course, using this acid is not recommended, because of its toxicity.

I've just asked How to understand the solubility of glass in water; 2 ppm? 400 ppm?


text:
When, if ever, can standard enthalphy be subsituted for standard chemical potiential?
I'm reading through a combustion modeling paper by Gordon & McBride which is a bit scant on some of the algebra & calculus used to derive their equations. My advisor and I have derived nearly all the equations missing clear derivations except for a few where the chemical potential seems to be have been completely substituted for by standard enthalpy.
We are mathematicians, and we figure this is likely some clever algebraic trick using a chemical equivalence we are unaware of, but which is so well-known to a chemistry audience that the authors did not bother to include the pertinent equation. The only thing I can think of is that somehow the Gibbs definition was used. However, there is no entropy in the final equation, as I would expect if the Gibbs definition were used to substitute for chemical potential.
Do you chemistry folk have any pointers for a lost math fellow?

No wonder you are lost. I have yet to find a chemist who can explain the concept of chemical potential in words. Every textbook regurgitates some circular definition of chemical potential without ever mentioning how it can be practically measured.

@M.Farooq Come now, that's hardly true that it can't be explained in a non-circular fashion.  It's an intensive property that determines the gradient for the flow of matter -- a substance will naturally flow from a region of higher to lower chemical potential, just as thermal energy will naturally flow from a hot region to a cold region.  And both occur because that process increases the entropy of the universe.   Nor is it true that it can't be measured. Saying you can't measure the chemical potential is tantamount to saying you can't determine the Gibbs free energy, since $\mu$ = G/n.

You need to provide more detail regarding what sort of process you are modeling. You might be justified to drop the entropic contribution, obviously where the enthalpy change is large in comparison.

@neph It would help if you could give us a reference to the paper, and reproduce the specific equations that are troubling you.
Are you referring to the following?: Gordon, Sanford, and Bonnie J. McBride. "Computer Program for Calculation of Complex Chemical Equilibrium Compositions, Rocket Performance, Incident and Reflected Shocks, and Chapman-Jouguet Detonations. Interim Revision, March 1976." (1976).


text:
Are my calculations correct?
The highlighted portions are my calculations. I am unsure whether i am going in the right direction or my sig figs are off. Am I supposed to use only 3 sig figs or what the calculator shows?
volume delivered=(final volume from trial 1) – (initial volume from trial 1)
7.50 mL – 0.00 mL = 7.50 mL
mass water=(mass beaker + water) – (mass of empty beaker)
29.691 mL – 22.109 mL = 7.582 g
density water  (mass water from trial 1) / (volume delivered from trial 1)
7.582 g / 7.50 mL = 1.01093 g/mL
average density= (density 1) + (density 2) + (density 3) + (density 4)/ 4
1.011 g/mL + 1.003 g/mL + 0.999 g/mL + 1.002 g/mL / 4 = 1.004 g/mL
% error= |theoretical value - experimental value| / theoretical value x 100
percent error = 0.997324 - 1.004/ 0.997324 = -0.00468 * 100 = -0.47%

This is not formatted in a way where people are going to be able to help.

additional details have been added

Question titles should be like book titles. They distinguish the book from other books, give a hint about the content, but do not tell the full story. The content then should elaborate the topic to full depth the author is able to do.

Oscar has already explained that your calculations are fine. I see a serious practical problem though. I guess this is a lab experiment for teaching the concept of calibrating a pipette.
I am not aware of any typical 10 mL graduated pipette which can read volumes like 0.03 mL. Are you sure it is a pipet not a (micro)buret? How did someone read 0.03 mL as the initial reading on a pipet? How can one deliver 7.48 mL or 7.51 mL?
Proper calibration work of glassware in classical analytical chemistry requires buoyancy corrections.
Simple rule, don't worry about significant figures until the end. Your balance can read up to 3 decimal places. The calibrated volume by this hypothetical pipet can also be reported to 3 decimal places.

Those numbers were given to me. I did not perform any measurements.

Okay, then it is just "made-up" numbers from someone's mind.

Our professor provided the data that is not highlighted.

No problem. He just made up numbers during Corona times. In reality no pipet can deliver 7.48 or 7.51 mL. Consider it is an exercise.

@M. Farooq: I think what they meant is a burette, not a pippette.

@MathewMahindaratne, This is what I suspected. Even if it is a buret, the procedure is poorly executed. Most likely the instructor never performed any real calibrations.

They look right to me except for the percent error calculation.  You should render
Precent error = (Measured - theoreticel)/(theoretical) ×100%
Here that would be positive, but you have a negative number so it looks like you turned the formula around.


text:
Difference in mechanisms between formation of acetal and enamine
When forming acetal from a C=O group with alcohol, the alcohol will act as a nucleophile and attack twice as the carbon is made very electrophilic twice.
I wonder in the last intermediate before forming the enamine, is it able for the 2' amine to act as a nucleophile and attack the carbon bonded to a highly electronegative N with a positive formal charge? (like the attack from the alcohol when forming acetal) Then with the help with 1 proton transfer, forming a product that has 2 RR'N?
My thought is that it may be too sterically hindered compared to the alcohol when attacking. But is it valid?

It’s really unsettling to find a relevant paper in Scifinder but to be unable to access it because apparently Bull. Soc. Chim. Fr. is not digitalised post-1917 D=

Well, aminals certainly do exist.

This does indeed work. One such reference includes US patent 3696080A in which propionaldehyde is reacted with dimethylamine to give the N,N-acetal as shown in the figure below.

The described reaction procedure involves cooling both reagents to $\pu{-78^\circ C}$ before addition, letting the temperature rise to around $\pu{-35^\circ C}$ at which point anhydrous potassium carbonate was added, letting the temperature rise again to about $\pu{0^\circ C}$ at which point the cooling apparatus was switched from dry ice/acetone to ice/water, and then stirring it overnight allowing the ice to melt in the process. The workup the inventors describe is again rather complex but they claim to isolate up to $\pu{73\%}$ of the bis-amino product.
SciFinder also gave a number of other references but one should serve sufficiently as a proof of concept.
It’s worth noting that $\pu{2.4 eq}$ of amine were added which probably assisted in driving the reaction to the product side.

@234ff I’m not sure if the entire procedure is bizarre from the point of view of an organic chemist. Maybe slightly unusual here and there but overall not too far from the ordinary. That said, in cases like this where two products (enamine and N,N-acetal) are possible it is usually the conditions that shift the reaction towards one of the possible products. The proof of concept ($\pu{73%}$ yield) shows that it is possible but any of many factors may prevent this particular pathway from occurring under ‘standard’ conditions (whatever they may be).


text:
What is the significance of heating value?
Say I burned for example methane and propane to boil some water.
Methane has the heat of combustion of $\pu{890kJ/mol}$ and propane has the heat of combustion of $\pu{2200kJ/mol}$
What does the difference in heat of combustion tell us? In other words, propane's heat of combustion is bigger. Does that mean that methane emits less energy and therefore is less "environmentally harmful"? Is a higher heat of combustion better or a lower heat of combustion?

Why would releasing less energy make it less environmentally harmful? That just means you need to burn more of it.

Note that the mentioned values are molar heats of combustion. Propane is a larger molecule than methane and it is therefore obvious that more energy per mole is released.
If we replace the molar amount by molar mass, the resulting specific heat of combustion has the opposite relation:
$$\begin{align}\text{Methane:}&&\frac{\pu{890kJ/mol}}{\pu{16g/mol}} &=  \pu{55.6kJ/g}\\[0.5em]
\text{Propane:}&&\frac{\pu{2200kJ/mol}}{\pu{44g/mol}}&=\pu{50kJ/g}\end{align}$$
For the enviromental aspect of $\ce{CO2}$ emissions, the relevant value is the heat of combustion per mole of carbon atoms. It is $\pu{890 kJ/mol}$ for methane, but $\pu{2200 kJ/mol}$ for propane, which is roughly $\pu{730 kJ/mol}$ for carbon. For methane, both numbers are the same, as a propane molecule contains 3 carbon atoms, while a methane molecule just 1.
So methane has a an advantage as energy source, as it has higher energy production per the same $\ce{CO2}$ emission.
PLA is partially already oxidized, therefore it has significantly lower energy content per 1 kg of produced $\ce{CO2}$, compared to  PP.
PLA has an energy content of $\pu{16.3MJ/kg}$
PLA has structure [-O-CH(CH3)-CO-]n, with basic unit containing 3 C + 4 H + 2 O
$\pu{1 kg}$ of PLA contains 36/72 kg of carbon
PLA has an energy content of $\pu{16.3 * 72 / 36 MJ/kg} \simeq \pu{32.6 MJ/kg of C}$ .
PP has an energy content of $\pu{46.4MJ/kg}$
PP has structure [-CH2-CH(CH3)- ]n, with basic unit containing 3 C + 6 H
$\pu{1 kg}$ of PP contains 36/42 kg of carbon
PP has an energy content of $\pu{46.4 * 42 / 36 MJ/kg} \simeq \pu{54.1 MJ/kg of C}$ .
To get the same energy by the burning of the stuff,
PP releases $54.1/32.6 \simeq 1.66$ times less $\ce{CO2}$ than PLA.
Or, with the same emission of $\ce{CO2}$, PP releases 1.66 times more energy than PLA.

Thanks for the answer, when you say "methane has slight advantage here" what do you mean? Why is higher energy content the "environmentally better"?

Lets say I am burning  2 types of plastics, PP (Polypropylene) and PLA (Polylactic acid). After some research I found out that PLA has an energy content of  16.3 MJ/kg and PP has an energy content of 46.4 MJ/kg. Which one is more favourable in terms of "environment"?

Also where did you get 730kj/mol from? https://www.engineeringtoolbox.com/standard-heat-of-combustion-energy-content-d_1987.html - from this website it says 2200 $kj/mol$

See the answer update.

The values that I gave is 16.3 Mj of heat produced for combusting per 1 kg of polymer. So greater energy content = more superior energy source, but now I am talking about energy content in plastic. So they would not use as a energy source right? This pdf file says something about energy recovery potential https://www.plasticseurope.org/application/files/7415/1747/5136/eco-profile_methodology_version2-0_April2011.pdf. type ctrl+F and search "energy recovery potential"

Also thanks for trying to help, do you have a patreon account? Maybe I can donate.

See the update. I do not have a Patreon account yet, but it is an interesting idea.

Thank you so much, I just got one last question, you said "PLA has an energy content of $16.3 ⋅72/36 MJ/kg≃32.6 MJ/kgofC$" What do you call this units? It is still called energy content? Or Energy content in terms of Carbon? Thanks

It is energy content recalculated to 1 kg of contained carbon. PLA contains 50% if carbon, so if there is 16.3 MJ/kg PLA, It is 32.6 MJ/kg C.

@Negrawh You have removed a part of the question I have answered, and later even the answer acceptance. I do not care about 15 points, but what was the reason ?

Oh Poutnik I sincerely apologise, I thought since I have commented the question under your answer, there is not need to display the question on the main question box any further. I must have mis-clicked the answer acceptance, because the answer you gave was spot on.  I re-accepted, Sorry!

The point of SE sites it to keep the Qs and As to help others with the same or similar questions.


text:
Can aged acetylcholinesterase be reactivated?
it is a well-known fact that certain phosphorus compounds can inhibit the enzyme acetylcholinesterase, and that the treatment of poisoning with such compounds is administration of atropine and some kind of oxime-containing compound. This needs to be done quickly, since another reaction, called "Aging" usually takes place, whereby a part of the phosphorus compound that does not act as the initial leaving group is hydrolyzed off, leaving a negatively-charged phosphonate or phosphate, which forms a salt bridge with the protonated, positively-charged histidine residue at the active site of the enzyme. Aging renders AChE resistant to reactivation by oximes. Is it possible to somehow reactivate AChE that has aged?

Until the late 2010s, it was considered impossible to reactivate acetylcholinesterase that has undergone aging. However, in recent years, new advancements have been made that show promise in this area.
A group of researchers investigated several sulfonium-containing alkylating agents that were supposed to alkylate the phosphonate moiety bound to the serine of aged AChE, in hopes that this would allow the enzyme to be reactivated by traditional methods.

Computational simulations suggested that all three of these compounds would be able to alkylate the aged enzyme.
Another group of researchers studied two aminoalcohols that would act by transferring a proton to the aged adduct. One of the compounds showed promise in reactivating aged AChE, as it was able to do so according to computer simulations. This is the molecule:

What is notable about this aminoalcohol is that it is predicted to be capable of reactivating the enzyme without alkylating the adduct.
Yet another series of compounds was shown to be successful in reversing the aging of acetylcholinesterase. These compounds reverse the aging reaction by alkylating the phosphonate/phosphate-AChE adduct, and also exhibit the potential to reactivate the enzyme entirely.

Unfortunately, they were quite slow to reverse the aging of acetylcholinesterase, with the most active compound, C8, resurrecting 12% of methylphosphonate-aged AChE in one day at a concentration of 4 mM.
In conclusion, while several compounds have shown promise in reactivating acetylcholinesterase that has undergone aging by traditional nerve agents or pesticides, therapeutic methods of administering those compounds have yet to be developed. Furthermore, these compounds have not yet been tested in vivo, under physiological conditions, or against other anticholinesterase agents such as Tabun or the Novichoks.
Sources:
Franjesevic, A. J., Sillart, S. B., Beck, J. M., Vyas, S., Callam, C. S., & Hadad, C. M. (2018). Resurrection and Reactivation of Acetylcholinesterase and Butyrylcholinesterase. Chemistry - A European Journal. doi:10.1002/chem.201805075
Khavrutskii, I. V., & Wallqvist, A. (2017). β-Aminoalcohols as Potential Reactivators of Aged Sarin-/Soman-Inhibited Acetylcholinesterase. ChemistrySelect, 2(5), 1885–1890. doi:10.1002/slct.201601828
Zhuang, Q., Franjesevic, A. J., Corrigan, T. S., Coldren, W. H., Dicken, R., Sillart, S., … Hadad, C. M. (2018). Demonstration of In Vitro Resurrection of Aged Acetylcholinesterase after Exposure to Organophosphorus Chemical Nerve Agents. Journal of Medicinal Chemistry, 61(16), 7034–7042. doi:10.1021/acs.jmedchem.7b01620


text:
what happens to the intramolecular bonds within water molecule during phase changes?
Does intramolecular bonds within water molecule change during phase changes?

You should provide more information about what exactly you mean. Yes, the bonds change.

The only intramolecular bonds in water are the O-H bonds. They are not modified during phase changes. The hydrogen bond is not an intramolecular bond. It is an intermolecular bond. These hydrogen bonds are modified during fusion or ebullition.

not entirely true. Since the protons are shared between oxygen atoms in the solid and liquid phase, whereas in the gas phase there are two H atoms strictly associated with each O, there are definitely changes in the O-H bonds during phase changes.

@ Andrew. If I follow you, hydrogen bonds are similar to covalence bonds. They are not.

I did not claim the bonds are similar, rather that a covalent bond between an O and an H which are each participating in H-bonds (as in water) is different from a covalent bond between O and H that are not involved in H-bonds (as in gas phase). This is readily observed in a comparison of the vibrational and rotational spectra of H2O in different phases. For example, the O-H stretching frequency changes substantially between liquid and gas.


text:
Correct representation of Lewis structure for sulphate ion
sulphate ion has tetrahedral geometry as predicted by VSEPR theory; the four S-O bonds are indistinguishable and have the same bond order, so the Lewis structure is better represented avoiding S=O double bonds; if instead of double bonds there are covalent dative bonds, what about the hybridization of S and O ? My idea is the following: Sulphur has hybridization sp3, each oxigen atom involved in dative bond has to provide an unfilled p orbital

Both structures cn be accepted. Lewis preferred the structure with 4 single bonds. L. Pauling preferred the mesomeric structure with two double bonds S=O

Both structures might be accepted but the best way to draw it in the Lewis formalism is to put a double positive charge on sulphur and have all four oxygens be negatively charged. That also ensures complete equivalence of the oxygens without the need for resonance structures.

See https://chemistry.stackexchange.com/a/1094/103633 and the Wikipedia page linked therein.


text:
Battery acid solution preparation from 98% ACS Sulfuric Acid
I'm trying to prepare some battery acid for activating a flooded lead acid battery I had purchased. The battery concentration should be around 36-28% sulfuric acid solution. I have decided to go with 37% acid solution.  I would like to confirm if the volume of acid to be added is correct. So, using a 98% ACS reagent sulfuric acid the volume of acid to make 100mL solution should be 37.57% right? The remaining volume is distilled/deionized water to make up for the 100mL.

The answer begins with the general calculation of mixing solutions with various densities. If you are interested only in the particular numeric results, or if you are scared by algebra, , skip the theoretical part between the horizontal lines.

There is the mixing cross rule, deriving the ratio of mass of 2 mixed liquids from their mass percentage.
$$\frac { m_2 }{m_1} = \frac{|w_1 - w_0|}{|w_2 - w_0|}=k\tag{1}$$
respectively
$$m_2 = k\cdot m_1 \tag{2}$$
and
$$m_0 = m_1 + m_2 = m_1 + k \cdot m_1 =  m_1 \cdot \left(1 + k \right) \tag{3}$$
where
$m_1,m_2, m_0$ are respective masses of mixed solution 1 and 2 and  ofthe final solution 0
$w_1,w_2,w_0$ are respective mass percentages of mixed solutions 1 and 2 and the final solution 0
As we are interested rather in volumes, we involve solution densities:
$$V_1 = \frac{m_1 }{\rho_1}\tag{4}$$
$$V_2 = \frac{m_2 }{\rho_2} = \frac{m_1 }{\rho_2} \cdot k\tag{5}$$
$$V_0 = \frac{m_0 }{\rho_0} = \frac{m_1 \cdot \left(1 + k \right) }{\rho_0}\tag{6}$$
We have the given final volume $V_0$ of the final solution of the density $\rho_0$.
Then, for the solution 1:
$$m_1  = V_0 \cdot \rho_0 \cdot \frac{1}{\left(1 + k \right)} \tag{7}$$
respectively the volume:
$$V_1  = V_0 \cdot \frac{ \rho_0}{\rho_1 } \cdot \frac{1}{1 + k}\tag{8}$$
For the solution 2:
$$m_2  = m_1  \cdot k = V_0 \cdot \rho_0 \cdot \frac {k}{1+k} \tag{9} $$
respectively the volume:
$$V_2  = V_0 \cdot \frac{\rho_0}{\rho_2 } \cdot \frac {k}{1+k} \tag{10} $$

Applying the particular data:
Solution 0: 37% $\ce{H2SO4}$,The density  $\pu{1.267 g/mL}$,  target volume $V_0  = \pu{100 mL}$
Solution 1: 98% $\ce{H2SO4}$.The density  $\pu{1.8361 g/mL}$
Solution 2: water = 0% $\ce{H2SO4}$, the density $\pu{1.000 g/mL}$
From the equation (1):
$k = \frac{|98-37|}{|0-37|}=61/37 \simeq 1.649$
We plug in the data into equations (8) and (10)
For  98% $\ce{H2SO4}$:
$$V_1  = \pu{100 mL} \cdot \frac{ \pu{1.267 g/mL}}{\pu{1.8361 g/mL} } \cdot \frac{1}{1 + 1.649} \simeq \pu{26 mL}\tag{8a}$$
For water:
$$V_2  = \pu{100 mL} \cdot \frac{ \pu{1.267 g/mL}}{ \pu{1.000 g/mL}} \cdot \frac {1.649}{1+1.649} \simeq \pu{79 mL} \tag{10a} $$
Be aware volumes are not additive. After cooling down, the total initial volume shrinks from $\pu{105 mL}$ to $\pu{100 mL}$.
For subsequent dilutions, it is an advantage to keep some volume of the residual 37% solution and to do the dissolution in 2 steps.
First, dilute the calculated portion of 98% H2SO4 in this rest of 37% H2SO4. After cooling, pour this mixture to the calculated volume of water.
The amount of heat per dilution would be divided into these 2 steps.
DO NOT forget to pour acid into water, as not vice versa !! Pouring water into acid would very propably end
by a hot acid splash into your face, as mixing is very exothermic. Mix them carefully, slowly, continually mixing and cool it by a water bath.
Pour the acid along the glass rod and not directly. Wear a lab coat or at least an apron, gloves and shield if possible.
Be aware even tiny drops can cause after some days/weeks holes in your clothes.

Comments are not for extended discussion; this conversation has been moved to chat.

I upvoted the answer by @Poutnik, but here is a slightly different take. I assume the 98% and 37% concentrations are by mass, i.e., 100 g of 98% concentrated sulfuric acid contains 98 g of pure acid plus 2 g of water. I also assume the 98% and 37% are exact, to avoid initial fussing with significant figures. Round off to two digits is near the end.
Take 100 g of the 98% concentrated sulfuric acid. It contains 98 g of pure acid, which is to be 37% of the diluted acid. Since 98 is 37% of 264.865, the mass of water required is 164.865 g, i.e., 264.865 g total minus 100 g of 98% sulfuric acid.
From Poutnik's answer, the density of the 37% acid is 1.267 g/mL. Hence the total volume of the 37% acid is 209.05 mL. Since the OP wanted 100 mL of 37% acid, just scale down by 100/209.05, which is 0.4784. Therefore 47.84 g of 98% sulfuric acid is slowly added to 78.864 g of water. For the water, 79 mL is an adequate approximation. For the 98% sulfuric acid, the density is 1.8361 g/mL, as per Poutnik's answer. Hence the volume required is 47.84 g divided by 1.8361 g/mL, which is 26.055 mL. So 26 mL is an adequate approximation.
Summary: As per the safety precautions given in Poutnik's answer, 26 mL of 98% concentrated sulfuric acid is slowly added (with stirring) to 79 mL of water.


text:
Mulliken Spin Density
I would like to know if a Mulliken population analysis to calculate spin densities is in general a valid choice. I see that it is made use of, for example here1.  So up to-date researches apparently use it. On the other hand there seem to be more advanced methods (natural population analysis). Moreover Wikipedia states:

In principle, a complete basis set for a molecule can be spanned by
placing a large set of functions on a single atom. In the Mulliken
scheme, all the electrons would then be assigned to this atom.

So there is some ill definition for this method. This would not only account for charge analysis but also spin, if I am not mistaken.
Knowing that my basis set is reasonable; is there anything to consider that would make Mulliken spin density analysis doubtful?
References

H-Bonding on spin centres enhances spin–spin coupling for organic diradicals by Francis Kirby B. Burnea,  Yeonsig Nama  and  Jin Yong Lee, Journal of Materials Chemistry C, Issue 10, 2020 (link)

Great first question! Someone here may be able to provide an answer specific to spin densities, but I expect the spin densities from a population method have similar issues to  the charge density. For some details on the pros/cons of different charge analysis methods, see https://mattermodeling.stackexchange.com/questions/1439/what-are-the-different-types-of-charge-analysis

Mulliken analysis seems to be considered a method of random number generation because there is no basis set limit: https://mattermodeling.stackexchange.com/q/1439/5, so if you keep increasing the basis set (getting more and more accurate wavefunctions) you will keep getting different numbers, forever... even if your basis set is infinitely large.


text:
Hypervalency of orthonitrate
I have always believed that it is impossible for the atoms of period 2 elements (when they are the central atoms in polyatomic ions/molecules) to accommodate more than 8 electrons in their valence shell. This is because of their small atomic size, which results in a significantly higher electron density being borne by the central atom and thus, instability can easily arise if too many electrons are held by it. Having seen many posts on this forum on the concept of hypervalency, I understand that many seemingly "hypervalent" molecules (e.g., $\ce {PCl5}$, $\ce {SF6}$, $\ce {ClO4^{-}}$) do not really have their central atoms possessing more than 8 electrons in their valence shells. This is due to the electron-withdrawing effect of the electronegative atoms that are bonded to the central atom, often resulting in there being less than 8 electrons populating the valence shell of these central atoms (Gillespie and Silvi, 2002).
However, my belief has recently been challenged by the existence of the orthonitrate ion, in which the $\ce {N}$ atom is claimed to possess $\ce {8.65}$ electrons in the valence shell (as written in the "Alternative definition" section of this article). This seems rather nonsensical to me as the $\ce {O}$ atoms surrounding it are more electronegative than the central $\ce {N}$. I would like to request for a verification of the claim regarding $\ce {N}$ possessing more than 8 electrons in the valence shell. Additionally, I would also like to ask if the ionically-bonded structure with one $\ce {N^+}$ and four $\ce {O^-}$ is an accurate representation of the bonding in the ion.
Reference
Ronald J. Gillespie, Bernard Silvi, "The octet rule and hypervalence: two misunderstood concepts," Coordination Chemistry Reviews 2002, 233-234, 53-62 (https://doi.org/10.1016/S0010-8545(02)00102-9).

Yes, thinking it's nonsensical is quite reasonable. In fact all this "counting electrons" is a stupid idea.

I personally prefer to think of it in terms of which orbitals are non-negligibly occupied. Introductory-level hypervalency means using d orbitals. When that's not feasible, you're limited to 8 by the simplicity of your model of lone pairs and two-electron bonds. More advanced approaches to electron counting that allow for more realistic delocalization and can therefore get >8 still without using the d orbitals are okay.

And the specific definition of "valence electrons" matters too.

When you learn molecular orbital theory, you learn to throw the octet rule out the window in favor of identifying the molecular orbitals that form the actual bonds.  Take a look at the seemingly innocuous case of carbon dioxide discussed in Chemistry LibreTexts (illustration from this reference):

There are really two related ways in which the molecular orbital reality blows the octet rule away even though you thought the familiar Lewis structure $\ce{O=C=O}$ obeyed this rule.
First:  The octet rule assumes that electrons in covalent bonds are shared between pairs of atoms at the ends of the respective bonds.  In the above diagram they're not.  Most of the occupied orbitals share electron density between all three atoms and those that don't are concentrated at the two oxygen atoms, not at the ends of either carbon-oxygen bond.  This three-way sharing is especially notable with the pi bonds; the structure I drew above is really not an accurate way to describe the pi bonds.  You really need a combination of pairwise valence bond structures even to begin to approach the molecular orbitals:
$\ce{\overset{-}{O}-C#\overset{+}{O}}$
$\ce{O=C=O}$
$\ce{\overset{+}{O}#C-\overset{-}{O}}$
In effect, the carbon-oxygen bonds with this multiple-atom sharing are not just double bonds.  They are a mixture of single, double and triple bonds providing more overall bonding than just plain pairwise double bonds could have.
Second:  Given this multi-way electron sharing the number of occupied orbitals that are at least partly shared by any atom can go well beyond four.  In the case of carbon dioxide this is especially true of the oxygen atoms, each of which has a finger in the pie of all sixteen valence electrons.  In the case of carbon dioxide it is true that each oxygen atom cannot have more than half of that electron density close to it, but that is a property of the specific compound carbon dioxide (which has only sixteen valence electrons to work with among all three atoms) and not a universal law of nature.
This should give you some idea of what goes on with a more complex structure such as your orthonitrate ion.  Electrons are not shared pairwise but among multiple atoms in molecular orbitals, and with this multi-way electron sharing the bonds are not really confined to specific pairs of atoms.  What looks like single, sigma-only bonds between the nitrogen and each oxygen atom are really a combination of sigma and pi bonding interactions, delocalized through a complex molecular structure; and each $\ce{N-O}$ interaction is thereby a combination of single bonds, multiple bonds and even purely ionic interactions.  Assigning a meaningful atom-by-atom electron count to such a structure is, as Mithoron suggests, fuzzy math.

I have carried out a calculation of NO43- (B3LYP/ccpVTZ) followed by the ELF population analysis. This method provides a partition of the electron density which is consistent with the Lewis's picture. The valence shell of N is the union of 4 V(N,O) basins each with a population of 1.375 e-, the total population of the nitrogen valence shell is, accordingly 5.5 e-. The valence shell of each oxygen centre consists of three V(O) basins (2.16 e-) and one V(N,O) basin, therefore the population of the O valence shell is 7.86 e-. The octet rule is almost satisfied for the most electronegative atoms, the valence shell population of the central atom which is the less electronegative contains fewer than eight electrons in agrement with the conclusion of our 2002 article. The attempt to calculate weights of mesomeric structures satisfying the octet rule from AIM is a non-sense: the AIM partition yields non overlapping atomic basins and therefore forbids the sharing of electron density corresponding to the Lewis's bonding pairs.


text:
Is there a geometrically intuitive explanation of the equivalence of the numbers of irreducible representations and symmetry classes?
In group theory, as used in chemistry, we’re familiar with the idea that the number of irreducible representations of a symmetry group equals the number of classes of symmetry operations, e.g. three in the case of the $C_\mathrm{3v}$ group describing the molecular structure of the ammonia molecule.
There are proofs of this fact, ranging from the properties of the Great  Orthogonality Theorem and discussions of the dimensionality of the vector space of characters e.g. as given in R. Mirman’s Group Theory: An Intuitive Approach, through to more mathematical proofs relying on the application of Brauer's permutation lemma to Galois automorphism on conjugacy classes, as shown in
https://groupprops.subwiki.org/wiki/Number_of_irreducible_representations_equals_number_of_conjugacy_classes.
But given the physical nature of molecular structure and the geometric properties of the molecular orbitals corresponding to the symmetry group’s irreps, is there no geometrically intuitive explanation of exactly why the number of irreps and classes are the same?

Group theory exists without any connection to molecules, and is used to describe the relationship between points in space which we then associate with the atoms in molecules, so your question is akin to asking 'why mathematical models can be used to describe real phenomena'; a very much bigger issue and I'm not sure that anyone has an answer to this.

@Porphyrin Thanks for your stimulating comment – please see reply in my own answer (having appreciated your https://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613 )

SUMMARY
The geometrically intuitive justification for why the number of irreducible representations and classes are the same is because:

The irreducible representations of a molecule’s symmetry group correspond to a resolution of the molecule’s spatial structure (visualised even as a simple ball and stick model) into a variety of quite different, component geometrical objects – the electronic molecular-orbital components, each obeying the whole set of the symmetry group’s operations in a distinct (orthogonal) manner
As vectors in the Hilbert space of the molecule’s electronic structure, the molecular orbitals occupy (and are symmetry-bound by) the invariant subspaces of the Hilbert space that are reached by the operations of the corresponding classes of the symmetry operations of the group on any of the vectors.
Hence the number of irreducible representations of the molecule equals the number of symmetry classes of its point group.

NB A detailed reply to @Porphyrin's comment under the question, which is too long to insert as a comment in reply, is given at the end of this answer.
DETAIL
For those who, like me, think in more of a visual/intuitive than abstract algebraic manner:
In chemistry, we can easily visualise the groups of the physical symmetries of molecules as combinations of symmetry operations transforming an image of the molecule (for atoms, it is simply the group SO(3) of rotations in R3).
The irreducible representations of the group illustrate the transformation properties of the component molecular or atomic orbitals under the operations of the molecular group, and can also be visualised as corresponding directly to both geometrical objects in physical space (i.e. the spatial distributions that are modelled as the MOs or AOs themselves) and, a bit more schematically, as vectors in invariant subspaces of a Hilbert space of functions representing AOs and MOs.
I think of this in terms of the visualisable spatial structure of the molecule overall (even as a ball and stick model for the purposes of understanding its symmetries) being analysed by group theory (or, if you prefer, resolved into) a set of different component geometric/physical structures – the molecular or atomic orbitals (all having the correct symmetry properties, regardless of the approximation method used in their computation).
[Please note that I’m choosing my words very carefully as there’s much confusion in discussions of representation theory due to frequent abuses of language, especially by mathematicians and by others who are so familiar with the material that they cease to be aware of how easily students can be confused by such abuses.]
So the transformation properties of the component MOs under the overall molecule’s group operations are encoded in the irreducible representations of the group and, for most purposes, conveniently summarised in the associated character tables, listing the traces of matrix representations of the group operations on suitable basis functions).
Character tables can be considered as a shorthand version of the irreps since they contain most of the information of use to chemists.
Indeed, one can say that “the molecular orbitals are the irreducible representations” of the group, in much the same way that Wigner’s Classification leads to the rather more mysterious statement that “fundamental particles are the irreducible representations of the Poincare’ Group”; in that case, it’s the symmetries of transformations in spacetime, under the Lorentz Group plus translations, yielding the Poincare’ group).
Just as we can visualise the orbital geometries described by the irreducible representations, likewise the classes of a molecule’s symmetry operations are also easily visualised.
Therefore the answer to the original question comes by considering the other way of imagining the MOs, as vectors in invariant subspaces of a Hilbert space of functions.
The characters of the irreps will be the same for each symmetry operation of a given class; so for $C_\mathrm{3v}$, for example, all three mirror planes σ will have characters of 1 for the A1 rep, -1 for the A2 rep and 0 for the 2-dimensional E rep.

This says to me that the molecular-orbital wavefunctions (i.e. irreducible representations) each span a subspace of the Hilbert space of molecular orbitals defined or bounded by the classes. No symmetry operation will take an MO with characters = 1 for the mirror planes (i.e. those MOs unaffected by reflections in any of the three mirror planes) into any subspace in which reflections invert the sign of the MO function.
There are three such classes for $C_\mathrm{3v}$, so the MOs appear to be limited to, or bounded by, three subspaces of the Hilbert space.
I conclude therefore that the geometrically intuitive justification for why the number of irreps and classes are the same is that:

the irreps correspond to a resolution of the molecular structure into a variety of molecular orbital components, each obeying the whole set of the symmetry group’s operations in a distinct way
the same molecular orbitals are distributed across as many invariant subspaces of the Hilbert spaces as there are classes of the group.

APPENDED REPLY to @Porphyrin’s comment:
i)  I’m aware of group theory’s abstract nature, for example permutation groups; but even they can have a physical realisation. When group theory is applied to material systems, there should be some sort of physical interpretation.
ii) Some people are content to work with a purely abstract sense of groups, while others find it helpful to have a more intuitive, often geometric sense. You might be familiar with the famous mathematician Jacques Hadamard’s On the Psychology of Invention in the Mathematical Field in which his survey of leading mathematicians and theoretical physicists’ in the mid C20th found a propensity to think in visual imagery.
iii)    It’s reasonable to claim that mathematical models often usefully represent physical realities. Recall the title/quote from the 1960 lecture by one of the supreme masters of symmetry in physics, Eugene Wigner: The Unreasonable Effectiveness of Mathematics in the Natural Sciences