text:
Relationship between MO energies in Hartree-Fock theory and the core Hamiltonian
I am looking at Hartree-Fock (HF) theory as presented in Frank Jensen's book "Introduction to Computational Chemistry", Second edition.
The total energy $E$ of the system can be expressed as
$$
E = \sum\limits_{i=1}^{n} h_i + \frac{1}{2} \sum\limits_{ij}^{n} (J_{ij} - K_{ij}) + V_{\mathrm{nuc}}\phantom{..................}(1)
$$
where
$$
    h_i = -\frac{1}{2} \nabla^2_i - \sum\limits_{A=1}^{N} \frac{Z_A}{\lvert \mathrm{R}_A - \mathrm{r}_i \rvert} \\
    g_{ij} = \frac{1}{\lvert \mathrm{r}_i - \mathrm{r}_j \rvert} \\
    J_{ij} = \langle\phi_i(1)\phi_j(2) \vert g_{ij} \vert \phi_i(1)\phi_j(2)\rangle \\
    K_{ij} = \langle\phi_i(1)\phi_j(2) \vert g_{ij} \vert \phi_j(1)\phi_i(2)\rangle \\
    V_{\mathrm{nuc}} = \langle\mathbf{\Phi}_{\mathrm{nuc}} \vert \mathbf{V}^{\mathrm{nuc}} \vert \mathbf{\Phi}_{\mathrm{nuc}}\rangle =  V_{\mathrm{nuc}} \langle\mathbf{\Phi}_{\mathrm{nuc}} \vert \mathbf{\Phi}_{\mathrm{nuc}}\rangle
$$
Then later, after having introduced Lagrangian multipliers to carry out the contrained minimization, he expresses the total energy in terms of the MO energies $\epsilon_i$
$$
E = \sum\limits_{i}^n \epsilon_i - \frac{1}{2}\sum\limits_{ij}^n (J_{ij} - K_{ij}) + V_{\mathrm{nuc}}\phantom{..................}(2)
$$
These two expressions for the total energy share the same form, and I am wondering if there is a physical connection between $h_i$ and $\epsilon_i$.
The naïve thing to do now is to equate the two expressions to obtain
$$
\sum\limits_{ij}(J_{ij} - K_{ij}) = \sum\limits_{i}(\epsilon_i - h_i)
$$
and insert back to obtain
$$
E = \frac{1}{2} \sum\limits_i(h_i + \epsilon_i)\phantom{..................}(3)
$$
I realize that the $h_i$ and $\epsilon_i$ are not completely independent, due to iterative nature of the way we have to solve these equations.
We have that the Fock operator contains the one-electron core Hamiltonian
$$
\mathbf{F}_i = \mathbf{h}_i + \frac{1}{2}\sum\limits_{j} (\mathbf{J}_j - \mathbf{K}_j)
$$
and that the orbital energies are computed as the expectation value of the Fock operator
$$
\epsilon_i = \langle\phi_i \vert \mathbf{F}_i \vert \phi_i\rangle
$$
so clearly a fully separated representation for these two quantities cannot be found.
So I guess my main question is this: Is my rewriting above formally correct? If so, I suppose it is just another, less explicit way of expression the correction due to the overly repulsive (due to double counting) MO energies?

Equation 1) can't be correct, the expression for $h_i$ contains an operator, while the left hand side, E, is just a number. You need to take matrix elements of this operator, and do you know that this operator will be diagonal in the basis you have chosen?

Oh, I have written it as a matrix elements when it should have been  written as the core hamiltonian operator $\mathbf{h}_i$. My bad...

Frank Jensen is on Matter Modeling Stack Exchange so it might help to ask there.


text:
why is energy not required for the first step in the TCA?
I find biochemistry challenging since the basics aren't very well explained, such as why some reactions require energy while others do not. One of the constants I've been able to find is that energy (and large amounts of it) seem to be gained from oxidizing an aldehyde group to a karboxyl group (like the conversion of fructose 1,6-biphosphate to glyceraldehyde-3-phosphate), or a hydroxyl group to a keto group (like when hydroxyethyl-TPP reacts with lipimine and forms acetollipoamide). I rationalize this by thinking carbon must gain more from losing four electrons than from gaining them, hence the reactivity of methane and the non-reactivity of carbon dioxide.
But this does not seem to be constant. In the first step in the citric acid cycle (when oxaloacetate and Acetyl Coenzyme A react to form citrate) the keto group of oxoloacetate actually turn into a hydroxyl group.
If energy is gained by turning a hydroxyl group into a keto group then energy should be consumed while doing the reverse. Why isn't this the case when Oxoloacetate and Acetyl Coenzyme A react to form citrate? Shouldn't this step require the expenditure of ATP?
If the fault lies with my underlying theory then please say so and point me towards a better model!

Not sure about the specifics, but in any reaction, you need to carefully consider all reactants and all products. The textbook depiction of the citric acid cycle focuses on the main intermediates (pyruvate onwards), but there are other things that react and are formed (NADH, CO2, etc.) and all of these come with associated energy changes / conributions to the reaction enthalpy. To really understand the reaction, though, you will probably need to dive into the mechanism.

A lot of the “energy” for that step comes from the hydrolysis of the CoA ester to a free acid

Although your reasoning isn't quite correct, it is true that simply forming citrate from acetate and oxaloacetate is not an energetically favorable reaction under standard biological conditions. What you've overlooked, however, is that after the C-C bond formation reaction occurs, the CoA ester is hydrolyzed to a carboxylic acid, which immediately ionizes to a carboxylate. This hydrolysis and ionization is very energetically favorable, making the overall reaction much more favorable.
You can say, therefore, that the energy of the reaction comes from the earlier investment in converting acetate to acetyl-CoA, which is achieved via net hydrolysis of an ATP molecule to phosphate and ADP, a reaction which is well known to be energetically favorable.

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text:
Can hydrogen peroxide dissolve rust?
WD-40's "help page" on "How To Loosen Rusty Bolts, Screws And Nuts" states:

Hydrogen peroxide can dissolve rust. Applying hydrogen peroxide to the
fasteners and letting it sit is also an effective method to remove
rust and loosen nuts and bolts.

I'm no chemist, but most of what I've found in the last couple of hours tells me that hydrogen peroxide is more likely to cause (create) rust than "dissolve" (remove?) it, for example, see LibreText: "Hydrogen peroxide is considerably stronger as an oxidizing agent than as a reducing agent".
Can someone who understands these things please let me in on what happens when I mix rust and hydrogen peroxide? Is it true that *"applying hydrogen peroxide […] is also an effective method to remove rust […]? I find this claim dubious.

I think its action is mostly mechanical - it decomposes and bubbles disrupt layers of rust.

Also, it's mostly water and that helps a lot in terms of getting stuff into solution.

@Mithoron Are you saying that "there is no reaction between rust and hydrogen peroxide" (as far as you aware)?

@ToddMinehardt But rust is insoluble in water (at least that's what a few posts on stackexhange would have me believe!)

Rather than dissolve read disrupt. That is not a chemistry book. The statement makes sense, context given. Let me add that iron is a catalyst for hydrogen peroxide decomposition, so the latter should be vigorous at the right place.

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text:
Does the amount of a gas increase with pressure?

The graph of $pV/(RT)$ shows that different gases deviate differently from ideal gas behaviour and that the value of $pV/(RT)$ changes as pressure increases.
My question is, if $pV/(RT) = n$ or amount of substance, does this imply that the amount of a gas changes when pressure is applied? This seems a bit confusing to me because I thought that matter cannot be created out of nothing? So where do these extra moles come from/go to?

To nitpick on the question (I've already edited it to fix these things): (1) The correct term for the "number of moles" of X is the "amount of X", or the "amount of substance". The word "amount" is analogous to "mass", whereas "moles" is a unit that is analogous to "kilograms" (you'd ask what is the mass of this apple, not what is the number of kilograms of this apple). (2) Please capitalise beginnings of sentences.

(3) You can use MathJax to typeset mathematical equations; see FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange? for more information. Your question doesn't need much of it, to be fair.

@orthocresol "Number of kilograms" is awkward, but I find "amount of X" to be awkward as well.  E.g., suppose you had a container filled with balls, and wanted to ask how many there were.  Would you ask "What is the amount of balls?"  I wouldn't, and don't know anyone who would.  One would instead ask: "What is the quantity of balls?"  Granted, one would not ask "What is the number of dozens of balls?", which points up the problem with saying "What is the number of moles of balls."

There could be something related to English but I suspect that, like in other languages, the choice of the name "amount of substance" wasn't perfect in this respect. If we stick to definitions all is fine. But in common words, even among scientists, amount can refer to both mass as well as amount of substance. This is not surprising as for there is a linear relationship among the two. To OP: the ideal gases equation is valid for ideal gases only. That is why you call the behaviour "deviation".

The V in the expression on the diagram is obviously molar volume (volume per mole of gas).  Otherwise, the comparison in the diagram could not be made on a consistent basis.

@theorist Balls isn't something that we might traditionally measure in moles, which is probably why "amount of balls" sounds very odd. I find "amount of methane" to be rather less awkward — basically, I read it as asking "how much methane is there" but expecting an answer in mols (as opposed to, say, particles). Perhaps I don't find it awkward because I've embraced it, and it just takes a bit more time to win over the chemistry community. But I think it is better wording than "number of moles" (to me, it's unclear whether the latter should take a unit i.e. is the number of moles 2, or 2 mol?).

The whole point of the graph is that $pV/(RT)$ is no longer equal to $n$ for a real gas.
$n$ isn't changing (as you point out, this is physically impossible), but the ratio $pV/(RT)$ is changing to some value that is not the same as $n$. Once you're freed from the ideal gas equation, there's no need for $n$ to "follow" the changes in pressure / volume / temperature.
If you prefer, you may wish to recast the "ideal gas equation" as a generalised form that applies to real gases:
$$\frac{pV}{RT} = Zn$$
For a real gas, as $pV/(RT)$ changes, it's not the amount of substance $n$ that changes, but rather the dimensionless quantity $Z$. The latter is called the compressibility factor, and in fact is precisely the quantity that is being plotted on the $y$-axis of your graph. Note that this doesn't mean that $Z$ is a constant. Its value depends on $p$, $V$, $T$.

Suggested substitution/extension: «that $pV/(RT)$ is no longer equal to $n$» into «that $pV/(RT)$ is no longer equal to $n$, nor a constant when varying $p$».

Only for ideal gases the equation $pV = nRT$ holds good. Therefore $pV/RT = n$ is valid only for ideal gases. And in the graph, $n$ is constant for ideal gas.
For the real gases $\ce{CH4}$, $\ce{N2}$, $\ce{H2}$, $\ce{CO2}$, while the amount of substance  $n$ is kept constant, the ratio $pV/RT$ is not a constant anymore ($pV/RT \ne n$).
Instead, for real gases
$$\left(p + \frac{an^2}{V^2}\right)(V - nb) = nRT$$
(the Van der Waals equation) is the better description of the observation. So in case of real gases change in $pV/RT$ doesn't indicate change in number of moles.

van der Waals equation is just one of equations ( even if the most famous) trying to describe  real gas behaviour, more or less successfully. It is not given as a law for real gases. A gas, for which we pretend to behave exactly as the equation describes, is called van der Waals gas.

See $CH_4, N_2, H_2, CO_2$ as $CH_4, N_2, H_2, CO_2$ versus $\ce{CH4, N2, H2, CO2}$ as $\ce{CH4, N2, H2, CO2}$ // For more, see  Notation basics and How can I format math expressions or chem. equations on Chemistry SE.

Please note that using "number of moles" in place of proper term amount of substance is as illiterate as to use "number of grams" in place of mass. Make sure your notations are correct, too. Element symbols in chemical formulas are upright, $P$ is typically power, and $p$ is pressure, and your fraction $pV/RT$ is ambiguous.

@Poutnik I realize that due to to recent wave of corrections of the scientist's name people are afraid to capitalize this Dutch prefix, but you actually should so shall it be the first word of a sentence. Also, please don't recommend to block-chain lists under a single macro, this backfires in horribly formatted paragraphs with huge gaps. It should never be $\ce{CH4, N2, H2, CO2}$, rather $\ce{CH4},$ $\ce{N2},$ $\ce{H2},$ $\ce{CO2}$.

@andselisk To further substantiate your argument: The four page excerpt of the Green Book by Stohner/Quack by 2011 has an entry for pressure as $p$ $(P)$.  But this is altogether with the statement that symbols in parentheses (i.e., $P$ for pressure) clearly is not the recommended way, but second class.

@andselisk I am sorry, language-wise I was not sure the initial capitalization applies to the name parts, so it was rather ignorance than omission. // For the formatting, I have not posted it as explicit recommendation. I wanted to merely compare direct application of mhchem via \ce{} on the MJ code as it was. Regarding the splitting, I may want to keep them together in similar ways as the hard(?) spaces keep together words. And, it would be kind of masochism to do so on a phone. Until SE offers MJ/LaTeX editor, one can point out proper ways, but avoid demanding strict-correctness.

@andselisk it is also true that the symbol n for amount of substance doest not help diffuse that sort of literacy.

For an ideal gas, $pV/(RT) = n$ can explain a lot of situations because it contains four variables. If you add gas to an empty Mylar balloon, the volume will increase. If it is still not filled to maximal volume, heating the gas will increase the volume. If you have a full balloon and gently step on it, the pressure will increase; same if you add more gas to it. So it really depends on the specific situation. If you know all four variables and the equation holds, it behaves like an ideal gas, if not, not.


text:
Can use lead(II) oxide instead of lead(IV) oxide in car battery?

Question: If the discharge input equation $\ce{PbO2}$ is replaced by $\ce{PbO}$, will the battery still work? I think the product will not be water anymore but hydrogen gas. I am asking this question because lead(II) oxide, reddish in colour is easier to produce by just burning lead metal in rich oxygen.

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No ! It won't work, because. first,  in the presence of $\ce{H2SO4}$, $\ce{PbO}$ is transformed into $\ce{PbSO4}$. Then the redox potentials of $\ce{PbSO4/Pb}$ is rather weak (-$0.36 $V}. And there will not be any significant difference between cathode and anode. There will not be any difference between between both electrodes.

Thinking about replacement of $\ce{PbO2}$ by $\ce{PbO}$ is like thinking of replacement $\ce{H2O2}$ by $\ce{H2O}$.

Design of a battery isn't only about the chemicals availability (as an ore, or by synthesis), but their properties (e.g., enthalpy of formation, possible reactions and dissolution in the electrolyte, standard electrode potential).  Or, how much energy may be stored reversibly per unit of mass in such a battery, and longevity of such a device.
A parameter where $\ce{PbO2}$ possibly outperforms $\ce{PbO}$ is electrical conductivity.  To quote Wikipedia:

«Although the formula of lead dioxide is nominally given as $\ce{PbO2}$, the actual oxygen to lead ratio varies between 1.90 and 1.98 depending on the preparation method. Deficiency of oxygen (or excess of lead) results in the characteristic metallic conductivity of lead dioxide, with a resistivity as low as $\pu{10^{−4} Ω·cm}$ and which is exploited in various electrochemical applications. Like metals, lead dioxide has a characteristic electrode potential, and in electrolytes it can be polarized both anodically and cathodically. Lead dioxide electrodes have a dual action, that is both the lead and oxygen ions take part in the electrochemical reactions.»
(reference)

While this obviously is not as good as e.g., for $\ce{Al}$ ($\pu{26.5 nΩ⋅m}$, ref), $\ce{Cu}$ ($\pu{16.78 nΩ⋅m}$, ref), or $\ce{Ag}$ ($\pu{15.87 nΩ⋅m}$, ref) a $\ce{PbO2}$ based accumulator may be (still) good enough given the designed temperature of operation and other parameters to take into account for a energy storage.  I interpret the lack of an entry for $\ce{PbO}$ on Wikipedia as indication that the the resistivity is way too high for a potential application as an electric conductor at/around ambient temperature.
Some of the above criteria are subject to change e.g.,  by design, or law.


text:
Hypothetical Alternative for Nitrogen
Recently I was surfing the internet and I found thisWikipedia article.
It states that

The element silicon has been much discussed as a hypothetical alternative to carbon. Silicon is in the same group as carbon on the periodic table and, like carbon, it is tetravalent. Hypothetical alternatives to water include ammonia, which, like water, is a polar molecule, and cosmically abundant; and non-polar hydrocarbon solvents such as methane and ethane, which are known to exist in liquid form on the surface of Titan.

This leads me to think if there is any hypothetical alternative for Nitrogen as well?
I want to know in the context of amino acids and protein, as in is there any element/molecule that is eligible to replace the amino group of amino acid and yet be able to form proteins.
Please consider, I'm expecting hypothetical views. I'm from a biology background and well aware of the basic nature of amino acids.
Thanx for reading :)

There are no serious hypothetical alternatives to the second period elements like $\ce{C},$ $\ce{N},$ $\ce{O}$, unless for very specific cases. They differ too much from elements of other periods as well as mutually between each other. Another issue would be the relative abundance in the Universe. E.g. there is hard to replace the basic aminogroup $\ce{-NH2}$ or RNA/DNA coding.

@Poutnik, going by your words, it is hard to replace the basic amino group because it's highly abundant. I'm assuming that it is not impossible then. If it is just hard to replace and not impossible.., then what element/molecule has minor chances or nano chances to replace aminogroup in that case.

While looking a superficial properties of elements like silicon as possible alternatives to carbon (eg tetravalency) suggests they might be feasible alternatives for life, the detailed chemistry does not. Life needs chemistry that can sustain very highly complex molecules; no combinations of other elements (like second-row congeners of Si, N, O etc.) have ever been shown to achieve that.

@classy_BLINK Those are not my words, I have just added the last sentence at wrong place so they may have looked so. Abundance of N is just additional factor. Hypothetical life founded on an alternative element is even less probable if it is relatively rare.

I've read that among binary compounds of elements with hydrogen, only those with nitrogen are basic in water without reducing it. That makes nitrogen hard to replace where it provides a basic functional group, such as in amino acids, unless you find nonaqueous life forms.

Does this answer your question? Why did life end up with the big 4 elements, and specifically nitrogen?

If you want to make exact analogs of biological molecules, you need an element with a similar valence to nitrogen, so you're pretty much looking at phosphorus as your best option. But phosphorus behaves very differently from nitrogen in ways that will negatively affect the stability and function of those molecules. Specifically,
$\ce{-PH2}$ groups are not basic like $\ce{-NH2}$ groups are, so the cation functionality and potential for ionic interactions and even strong H-bonding is lost
$\pi$ bonds to phosphorus are generally less favored than sigma bonds, so the many iminium-type compounds (especially heterocycles) would not be as stable with P.
Reduced phosphorus compounds oxidize much more easily than corresponding nitrogen compounds
I think it would be much more productive to speculate about a completely different set of macromolecules in a completely different environment rather than trying to reproduce known molecules with elemental substitutions. The macromolecules we have evolved the way they have precisely because of their current elemental compositions and the existing Earth environment. There is no reason to think they are good templates for other elemental compositions or planetary conditions. Maybe start with the constraints of the environmental conditions and abundant elements of another specific planet and work from there.


text:
Why is Denaturated alcohol made poisonous?
I understand the need to denaturate alcohol, as to have a material that can be used for non-food products without the heavy alcohol beverage tax.
So, adding a foul tasting bitter substance like Denatonium seems like a great idea. It will also prevent children from accidentally chugging a bottle of the stuff.
But what I can't comprehend is: sometimes the alcohol is purposefully made poisonous. By adding methanol or other toxic substances. Can anyone tell me the logic behind this decision?
Is preventing tax avoidance so extremely important that they're willing to kill or blind anyone that might drink the illegal drink? Maybe they bought it without even knowing it was illegally made! Do they deserve to die? It just seems insane to me. Perhaps chemists in this site might have an explanation.

The amounts of typical denaturants are not usually large enough to kill you: they are mainly large enough to make the solution unpleasant to drink. Even methanol in small quantities is not that dangerous when consumed alongside ethanol (which is a treatment for methanol poisoning).

I think this question itself is biased and due to that issue rightfully closed as opinion-based; I think -also in accordance with the currently accepted answer- it could be made less biased and the reopened.

Is this some sort of an SJW joke?

@KanghunKim I don't understand your comment. The question is about the purpose of poisoning alcohol rather than simply embittering it. Furthermore, if you are to downvote it would be good to write the reason in the comments

"Is preventing tax avoidance so extremely important that they're willing to kill or blind anyone that might drink the illegal drink?".... sounds like an SJW tho

Perhaps if you could clarify what SJW means? I'm not familiar with the term.

It seems there is not spread enough fact that ethanol is poisonous even before denaturation. "The dose makes the poison." - the old medieval wisdom.
Most of denatured ethanol is mixed rather with relatively harmless chemicals of repulsive bitter taste, like denatonium aka Bitrex.  Often in mixture with other denaturants, as skilled drunkards are inventive in alcohol purification. :

Denatonium ... is the most bitter chemical compound known, with bitterness thresholds of 0.05 ppm for the benzoate and 0.01 ppm for the saccharide.... Dilutions of as little as 10 ppm are unbearably bitter to most humans.... It is used as aversive agents (bitterants) to prevent inappropriate ingestion. Denatonium is used in denatured alcohol, antifreeze, preventive nail biting preparations, respirator mask fit-testing, animal repellents, liquid soaps, shampoos.......   It is not known to pose any long-term health risks.

Methanol is added ( mixture known as methylated spirit ) usually to special lots for lab or technical usage where is implied the educated workers know what they are doing. The reason is methanol is one of the closest chemicals to ethanol, what is needed if usual additives would have disturbing effect due their physical or chemical properties. The methanol content is not very high, so toxicity of respective amount of methanol is comparable with toxicity of ethanol, for the given ratio.
The deadly cases in 2012 in Czech Republic were special, as evil motivated producers/sellers of cheap spirits mixed ethanol with methanol in ratio 1:1, not used in denaturation cases.
Pure ethanol is produced as well, but in limited scope, as danger of abuse is great.
There are various region specific denatured alcohol complex formulations as well, as example:

the formulation for completely denatured alcohol, according to 2005 British regulations was as follows:     Completely denatured alcohol must be made in accordance with the following formulation: with every 90 parts by volume of alcohol mix 9.5 parts by volume of wood naphtha or a substitute and 0.5 parts by volume of crude pyridine, and to the resulting mixture add mineral naphtha (petroleum oil) in the proportion of 3.75 litres to every 1000 litres of the mixture and synthetic organic dyestuff (methyl violet) in the proportion of 1.5 grams to every 1000 litres of the mixture.

The European Union agreed in February 2013 to the mutual procedures for the complete denaturing of alcohol:    Per hectolitre (100 L) of absolute ethanol: 3 litres of isopropyl alcohol, 3 litres of methyl ethyl ketone and 1 gram denatonium benzoate.

So ethanol is not made poisonous to prevent consumption (which was what troubled me), but rather as a side effect of using methanol to keep it acceptable for laboratory use.

Even water is considered poisonous in lab, so it does not matter, if ethanol contains methanol or not.

"Most of denatured ethanol is mixed rather with relatively harmless chemicals of repulsive bitter taste...Methanol is added ( mixture known as methylated spirit ) usually to special lots for lab or technical usage".  Not sure if that's true. According to several sources, including https://www.sciencedirect.com/topics/agricultural-and-biological-sciences/denatured-alcohol, the most common denaturant is methanol.

@theorist It may be region dependent. I have personally never encountered "methylated spirit" anywhere but in the lab. ( Czech Republic/Czechia ). Anytime I bought in my 56-year life denatured ethanol for outdoor ethanol burner, it reeked by pyridin-like smell. ( I did not taste it if it was bitter as hell due denatonium though ). I have no personal experience if publicly available methylated one uses other additional additives or not.

I would say pure ethanol is not produced in big amounts because it is prohibitively expensive, due to taxation. ;)

@Karl: IMHO, Pure ethanol is usually produced with some purity requirement, what makes it more expensive -- together with other pure solvents -- than taxed food-grade ethanol.

@Karl Commercial and industrial companies do purchase food-grade ethanol for various purposes, and when they do so they don't need to pay alcohol taxes (that's only for retail sale), so that's not a factor.


text:
What actually happens to the equilibrium of solid and aqueous ions in a solution when concentrations are changed?
Taking as an example the equilibrium set up when $\ce{BaSO4}$ is added to water:
$$\ce{ BaSO4(s) <=> Ba^{2+}(aq) + SO^{2-}_4(aq) }$$
The solubility product constant is $[\ce{Ba^{2+}}][\ce{SO^{2-}_4}]$.
If I then add additional $\ce{Ba^{2+}}$ ions to the solution, what will happen to the concentrations of the products once equilibrium is restored, since the product of their concentrations is constant. Will the concentration $[\ce{Ba^{2+}}]$ remain larger than its original value before the addition of barium ions and the $[\ce{SO^{2-}_4}]$ become smaller than its original value to compensate and keep the solubility product constant?

Any alternative idea?

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If somebody is going to vote the answer down, please kindly inform me why it is a bad question and where I’ve gone wrong.

(+1)   The answer to your final question is simply yes: the sulfate ion concentration gets reduced and the barium ion concentration is increased. Adding sulfate ions gives the opposite situation: the barium ion concentration is reduced, etc. In barium medical “milk shakes”, given for fluoroscopic imaging of the throat and esophagus, the barium sulfate suspension is in a liquid with excess sulfate ions, as per the answer by Buttonwood. Barium is toxic, but it cannot do harm if it is tied up in barium sulfate and the equilibrium keeps it tied up.

On the microscopic scale, the thermodynamic equilibrium is dynamic, i.e. there are constant back- and forward reaction. Here: dissociation of $\ce{BaSO4}$ to yield $\ce{Ba^{2+}}$ and $\ce{SO^{2-}_4}$, and association to yield again $\ce{BaSO4}$.  Once the thermodynamic equilibrium is installed, these ongoing microscopic reactions do not change anymore the concentration of $\ce{Ba^{2+}}$, nor  the one of $\ce{SO^{2-}_4}$.
If disturbed at macroscopic level (e.g., by addition of further $\ce{SO^{2-}_4}$, an increase of $[\ce{SO^{2-}_4}]$), keeping the product of $[\ce{Ba^{2+}}][\ce{SO^{2-}_4}]$  constant is possible only if the other factor (i.e., $[\ce{Ba^{2+}}]$) decreases.  Thus, in the case of gravimetric determination of barium in the form of $\ce{BaSO4}$, the addition of $\ce{Na2SO4}$, which is water-soluble, to obtain an exhaustive precipitation.


text:
Combustion analysis of hydrocarbon
Combustion analysis of a hydrocarbon produced 33.01 g CO2 and 4.84 g H2O.
I was asked to find the empirical formula of the hydrocarbon. I used these molar mass values for my calculations:
C - 12.0107
H - 1.00794
O - 15.9994
Here is my work:
Molar mass of CO_2 = 12.0107 + 2*15.9994 = 44.0095
Molar mass of H_2O = 2*1.00794 + 15.9994 = 18.01528

We now have 33.01g of CO2 and 4.84g of H2O, which means:
33.01/44.0095 = 0.750 mol CO_2
4.84/18.01528 = 0.26866 mol H_2O
0.26866*2 = 0.53732 mol H

From the calculation we know that the ratio of C to H is 0.750:0.537, which is approximately 3:2. However, the empirical formula C3H2 is not the right answer. Can anyone tell me why I am wrong? Thank you!

@CaptainToad For future reference: When computing with quantities like $\pu{mol}$, $\pu{kg}$, etc. an equation sign indicates both 1) equality of the numbers on the right and left hand side, and 2) equality of the dimensions on the left and on the right hand side. Thus, units (like in $\pu{0.750 mol}$ $\ce{CO2}$ above) do not suddenly appear out of nothing (on occasion, they may yield a dimension less quantity, e.g., Reynold's number, though). Keep track of them (e.g., for dimensional analysis).

Let Hydrocarbon be of form $\ce{C_xH_y}$
Our combustion reaction becomes $\ce{C_xH_y}+(2x+\frac{y}{2})\ce{O_2}\implies x\ce{CO_2}+\frac{y}{2}\ce{H_2O}$
$1$ mole of hydrocarbon produce $x*44g$ of $\ce{CO_2}$
If $\ce{CO_2}$ produced is $33.01g$, then $x=\frac{3}{4}$
Similarly $1$ mole of hydrocarbon produces $\frac{y}{2}$ moles of $\ce{H_2O}$
If $\ce{H_2O}$ produced = $4.84g$, then $\frac{y}{2}=4.84/18\implies y=\frac{1.61}{3}$
Therefore $\frac{x}{y}\approx1.4$ Therefore empirical formula becomes $\ce{C_7H_5}$

0.750//0.53732 = 1.396/1;Hydrocarbon, no heteroatoms, needs even number of H so ratio becomes 2.792/2.  x4 = 11.18/8  X5= 13.96/10  X6= 16.72/12 best fit is 5times so becomes C14H10.  Notice if one rounds off immediately the answer is nice and neat. Don't ever fall into that trap. Next step is MW determination; Rast method or freezing point in benzene or infrared spectrum and mass spec.

Why was this answer downvoted? It is the only answer without guessing [ or a MW that fits] or approximating and does show that empirical measurements and/or made up problems do have errors.


text:
How to identify an error in cut off filter (optical) usage when running spectrophotometry?
In spectrophotometry, often a cut off optical filter is used. With reference to a long-pass filter, which rejects any wavelengths outside a predetermined interval: low transmission in short wavelengths range and high transmission in long wavelength ranges.
Some are orange, orange-yellow, dark orange, may appear similar to a novice if not labelled well.
How can a beginner see/detect if they have used a wrong filter by looking at the spectra before they continue using it? Practical troubleshooting tips and knowledge?

Unless the filters are labelled in some way, e.g., Maker and number, it is not always easy to visually distinguish filters. I have a box of filters, including long pass non-fluorescent filters, and the only good way to know what they will do spectrally is to get a transmittance or absorbance spectrum. Quick and dirty: use a beam of sunlight and a diffraction grating to project a rainbow on a white wall or white paper. Then look at the rainbow using the filters. Obviously no good for UV or IR and it is very crude: a spectrometer is far better in every way.

I don't understand the question. Are you near the spectroscopy set up or what?

@Edv appreciate your insights and recommending your colleague MFarque to comment. Well, with regards to protein assays (spectrophotometer: fluoroscopy), the cut off filters are sensitively crucial. According to your insights, if incorrect filter is used, there doesn't seem to be a way detect it unless actual recheck of the equipment is done.

@Alchimista please see my comment. Any feedback?

Afraid so. Filters are also problematic because of parasitic fluorescence: a long pass filter may block short wavelength light, but fluoresce, thereby ruining the measurements. This is not so easy to check for because the filter would be tested for its spectral properties while in the sample compartment of the spectrophotometer. Then unwanted fluorescence might well go unnoticed. This is one reason to use solution filters, e.g., potassium nitrate solution. But I do not know of a list of such solution filters and adding another cuvette to a sample compartment may not be feasible. Very annoying.

As Prof. Ed advised, take the transmission spectrum of each optical filter on a good quality double beam or single beam scanning spectrophotometer with the shortest possible wavelength step size (Caution: this could be slow, but better in terms of reliability). Keep the filter vertical. Use air as a blank. A cut-off filter is not as good as an interference filter. Note that Beer's law will not apply if you use cut-off filters. It is valid for a single wavelength. Unfortunately, there is no other way to characterize the optical filter. You do need a scanning spectrophotometer.
Edits: The instrument you are using is not called a spectrophotometer but it is a fluorometer. It uses a very bright xenon arc lamp and it needs two filters. One is called excitation filter and the other is called emission filter. Before you do fluorometry you should know

Excitation spectrum of your analyte or absorption spectrum (shapes are same for simple molecules)

Emission spectrum of your analyte.

Transmission spectrum of your excitation filter

Transmission spectrum of your emission filter.

Ideally there should be no overlap in transmission wavelengths of the two filters.

Please see my comment to Ed V. It's a scanning spectrophotometer. So if that's the case, looking at the scan/spectra, can one detect/identify if correct filter is being used for their assays? You see, if a biomolecule based assay is continued, it can go on till the end and wouldn't be so beneficial to realize in the middle to at the end that filter was wrong.

Sorry, I still do not understand the reason for using a filter with a scanning spectrophotometer. Could you describe your instrumentation please?

Spectrofluorometers and spectrophotometry are completely different instruments.

Is there a chance that both Spectrofluorometers and spectrophotometry functions are combined into one?

@bonCodigo, spectrophotometry is not done with cut off filters.

The instruments in the lab is called a spectrofluorophotometer. My typo on my previous comment 9/10/21. There are colourful set of filters for different bandwidths. These are called cut off filters in the manual, however, no further instructions for newbies. So really wanted to set that bit up for clarity sake.


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Total number of Isomers of [Co(NH3)5(NO2)](NO3)2
I encountered this question in my country's Chemistry textbook of class 12. No solutions are published, but the answer given at the back states - "There are 10 possible isomers. (Hint: There are geometrical, ionisation
and linkage isomers possible)"
I think the answer should be 9 by considering geometrical, ionisation, linkage and solvate isomerism.
Recently a problem came in the JEE Advanced Examination (entrance exam for admissions into the IITs) Paper 1 Chemistry section

Complex is [Pt(NH3)4Cl2]Br2
What should be the correct answer? Do we consider solvate isomerism with NH3 outside the coordination sphere?

Out of curiosity: Do you just have to write down the number, or can you write a few lines with your reasoning?

You can write down the number with some reasoning and preferably a good source/paper if possible. In the exam we had to just report the answer as an integer

And if you wrote "9" and they expect "10", you get zero points. Your reasoning will not be taken into account?

Yes we get 0 points if our answer is not as they expect. They don't ask for any reasoning.

So they're testing your status of brainwashery, no actual chemistry knowledge. ;-) OK, that's actually sad.

This is asked recently: https://chemistry.stackexchange.com/questions/158408/does-solvate-isomers-exist-for-coordination-compounds-with-ammonia-solvent-as-li

@Karl things are far too worse than you can imagine. people basically worship this exam. there is immense competition. (this is the second toughest exam of the country.) And this exam is the only method to enter into the topmost colleges.

@Ritil I've read the story from Richard Feynman, when he was teaching on a sabbatical in Brazil in the 50ies, and was very unsettled about the way both the teachers (and books) and most of the students were practising what he thought would never lead to an actual understanding of physics. "Surely you're joking", there are thousands of online copies around on the internet. I think the publishers are happy about the free advertising that keeps a really old and arcane book selling. ;)

@Karl very few people read Feynman's lecture. The thing is that anyone serious about getting into good college (equivalent to heaven for most materialistic students) won't have any time for it. Definitely not a single teacher can advise or refer to that book. Almost no one is going to value that book unless people change their mindset


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What is the maximum amount of carbon atoms that can comprise a fullerene?
As the fullerene sum formula can be defined as $\ce{C_n}$ with $n$ being any even integer number, is there any fundamental theoretical limit as to how big $n$ can get?
If there is no limit to the chemistry, but there are difficulties in synthesizing large fullerenes, what has been the largest fullerene ever made? Would there be any differences in behaviour?

$n$ is not just any integer. It can't be odd, to begin with. Now to the point: just as you expected, there is no fundamental upper bound.

At some point, the molecule would be so big that it cannot exist at a given temperature, because it is also stiff (has a 1D long range order). I will then just snap in two with a certain half-life time. I don't think a controlled synthesis can ever get close to that value.

Btw.: it is extremely hard to produce macroscopic amounts of very large molecules (a substance) with a defined, uniform molecular weight, i.e. number of carbon atoms in this case. For polyethylene, the record is afaik $\ce{C390H782}$, far, far below the average size of a UHMWPE molecule (easily a million carbon atoms).

One can consider a capped nanotube a fullerene.

$\ce{C3996}$. https://en.m.wikipedia.org/wiki/Fullerene

Multiple questions here - you might be more likely to get answers if you split it into (at least) two questions. Anyway to address the part which people haven't discussed so far in principle if you are considering a single molecule you will always get discrete energy levels. Bands are a result of the translational symmetry of a perfectly periodic system such as a crystal. No periodicity, no bands, and a single molecule won't ever display perfect periodicity. That's not to say the energy levels won't get very, very close and look similar to bands. But bands they won't be.

@IanBush the OP recently expanded the question and added the explanation "Added feedback from the answers and questions, which expands the scope of interests that might help narrow down an answer." I agree it may make this harder to answer and this was not the question to which I added the bounty.

Will rollback the answer as suggested.

Thanks for the rollback! I think you wrote a near-perfect question in the beginning, and that the reason there hasn't been an actual answer posted (despite the helpful comments) is that it turns out to be an interesting and not-so-easy question to answer; the best kind in my opinion :-)

I'll give this a day or so, then add another bounty. We'll get this answered!

To be honest I don't think you will get an answer to the first question 'is there any fundamental theoretical limit as to how big "n" can get?' as I suspect large fullerenes will be thermodynamically unstable WRT decomposition to both smaller fullerenes and graphite, and so the question becomes why is this kinetically infeasible, which is very hard. So you are left with a literature search for the biggest fullerene synthesised or detected.

Not a full answer, but possibly the start of one. I came across a paper[1] discussing some mathematical properties of fullerene graphs. In section 4.2, they give a formula for a lower bound on the diameter based on the number of vertices $n$, assuming there are only pentagonal and hexagonal face.
$$\text{diam}(F)\ge\frac{\sqrt{24n-15}}{6}-\frac{1}{2}$$
Given a lower bound on the diameter, it may be possible to prove that a ball of a certain size would be mechanically unstable. I suspect this would give a fairly high upper bound on the size and it wouldn't account for the wide variety of shapes that are possible for fullerenes[2], but this would at least be a starting point.
This could hopefully be refined somewhat by using the largest synthesized/reported fullerene size as a lower bound. Thus far $\ce{C_{500}}$ is the largest fullerene I can find experimental evidence for [3,4]. I suspect this is still pretty far from the largest possible fullerene, even considering just spherical/icosahedral structures.
Based on [2], you will need to look for some kind of mechanical/kinetic stability limit for the upper bound, as the paper suggests that fullerenes should increase in thermodynamic stability as $n$ increases, approaching the structure of graphene or a infinite nanotube.

Vesna Andova, František Kardoš, Riste Škrekovski. Mathematical aspects of fullerenes. Ars Mathematica Contemporanea, DMFA Slovenije, 2016, 11, pp.353 - 379. ffhal-01416354f
WIREs Comput Mol Sci 2015, 5:96–145. doi: 10.1002/wcms.1207
Shinohara, H., Sato, H., Saito, Y., Izuoka, A., Sugawara, T., Ito, H., Sakurai, T. and Matsuo, T. (1992), Extraction and mass spectroscopic characterization of giant fullerences up to C500. Rapid Commun. Mass Spectrom., 6: 413-416. https://doi.org/10.1002/rcm.1290060702
Front. Chem., 03 December 2020 | https://doi.org/10.3389/fchem.2020.607712

really cool paper! I've just asked What does 3-regularity mean in the context of this diagram of a fullerene? in Math SE.

While this is only a start, I found the paper really interesting and salient, and it has got me thinking, and I'm loath to waste a good bounty. +200 :-)

@uhoh Thank you! I'll try to keep expanding the answer as I run into anything that might fit. I can at least now give a (probably very loose) lower bound with experimental evidence of at least $\ce{C_{500}}$.


text:
Questions regarding the Newman-Kwart rearrangement
The Newman-Kwart rearrangement is a reaction that converts a phenol to a thiophenol by forming a phenolic thiocarbamate with two alkyl groups (methyl) on the nitrogen. This is acheived by reacting dimethylthiocarbomyl chloride with sodium phenoxide and then heating it:

(picture credit.  For their seminal publication, see Newman et al. mentioned below.)
In extension to this, Majetich et al. report alkyl isothiocyanates are able to react with saturated alkoxides to form sodium carbonimidothioate salts. The transition state depicted in the Newman rearrangement is strikingly similar to the salt produced from nucleophilic attack of the alkoxide on isothiocyanate where the sulfur atom bears the negative charge in both cases with the exception that in the rearrangement the nitrogen has a positive charge, but the sodium has the positive in the salt in the latter reaction.  The authors illustrate this with the following scheme:

(Majetich et al., loc. cit.)
My questions are:

Can a sodium carbonimidothioate salt undergo a similar rearrangement?
What if we just add crown ethers to the reaction mixture.  Will that remove the sodium and trigger the reaction? I am interested in the 2-mercaptopyridine which is cited as occurring readily at room temperature.

references
Newman, M. S. and Karnes, H. A. J. Org. Chem. 1966, 31, 3980-3984, doi 10.1021/jo01350a023.
Majetich, G.; Irvin, T. C.; Thompson, S. B. Tet. Lett. 2015, 56, 3326-3329, doi 10.1016/j.tetlet.2015.01.086.

Your links don't work

For future reference: 1) If you reference work, cite the original source at the publishers.  (There are nationwide federations which might offer you access to scientific publications [especially more elder ones/keyword rolling paywall] regardless of affiliation to a scientific institute, e.g. at public research/state libraries, too.) 2) Keep in mind a substantial portion of chemistry.se's readers uses the page to improve their understanding of chemistry. The addition of illustrations/reaction schemes may ease to grasp the context of your question.

If you are interested in the 2-mercatopyridine the direct displacement of the halogen from 2-halopridines is straightforward.


text:
Zetasizer - How can the Z-average mean be lower than all intensity measurements?
I recently used a Malvern Zetasizer on samples of Au nanoparticles in order to determine the size of the particles (we are looking for significant shifts in particle size between samples, and as such do not need perfect precision). Here is a screenshot of one of the intensity histograms the machine produced, and the Z-average that it gave is written in the text box below:

As can be clearly seen, the bars of the histogram are at an x value much larger than that of the z-average mean. This is the case for almost all of the samples we took. Why is this? What does it mean about the distribution of sizes in our sample?
Thank you for your time.


text:
Approximation for pH calculation with diluted acids in weak-strong titration curve
I was wondering how one could calculate small pH changes in titrations with dilute acids.
As we know, usually we approximate the pH by saying that the conversion is quantitative. For example, when we titrate i.e 100ml 0.1M sodium acetate solution with hydrchloric acid, the start point would be just pOH=0.5(pKb-log(c)), with pKb=9.25 we get 5.125. So pH=14-5.125=8.875.
If we add 25ml of 0.1M HCl we say that HCl consumes acetate directly:
CH₃COO⁻ + HCl → CH₃COOH + Cl⁻
So n(HCl)=0.0025mol consumes equal amounts of acetate and forms 0.0025mol of acetic acid. 0.0075 moles of acetate remain. We get 125ml of solution, so c(Ac-)=0.06M and c(HAc)=0.02M, so Hendersson-Hasselbalch gives us pH=pKs+log(0.06M/0.02M)=4.75+log(3)=5.227.
That in itself makes sense, but it doesn't if we, for example, only titrate with 10⁻⁶M HCl? If we added 0.5L of that, according to this approximation we would have n(HAc)=0.5*10⁻⁶mol and n(Ac-)=0.0099995mol both in 0.6L of solution, and subsequently a higher pH value than if we just let the acetate dissociate in pure water. So how can we replace this approximation in this, admittedly, extreme case so that the titration curve can predict the real result?

In such cases, you cannot afford to involve implicit simplifications.
In this particular case, the equilibrium ratio $\frac {[A-]}{[\ce{HA}]}$ is not the same as the nominal $\frac {[A-]}{[\ce{HA}]}$ ratio, but reaching acido-basic equilibrium is to be considered. One has to count also with water auto-ionization equilibrium.
If the added  $\ce{HCl}$ converted the full equivalent amount of acetate to acetic acid, $\ce{pH}$ would not change. A small portion of acetate will not convert to acid and the respective portion of remaining $\ce{H+}$ will add itself to acidity of the solution, all together honoring the acidity constant.
Full analysis, involving mass+charge balance/conservation and ongoing equilibria is to be evaluated and only those simplifications that are justified can be accepted.(1)

Charge balance: $\ce{[H+]} + \ce{[Na+]}= \ce{[OH-]} + \ce{[A-]} +\ce{[Cl-]}$
Weak acid mass balance : $c = \ce{[HA]} + \ce{[A-]}$
Weak acid mass balance 2 : $c = \ce{[Na+]}$
Acidity constant: $K_\mathrm{a} = \frac{\ce{[H+][A-]}}{\ce{[HA]}}$
Water auto-ionization: $K_\mathrm{w} = \ce{[H+][OH-]}$

By solving 5 equations with 5 variables, we get $\ce{[H+]}$. Justified simplifications are always welcome.
Substituting  for $\ce{[OH-]}$ and $\ce{[Na+]}$:

Charge balance: $\ce{[H+]} + c = \frac {K_\mathrm{w}}{\ce{[H+]}} + \ce{[A-]} +\ce{[Cl-]}$
Weak acid mass balance : $c = \ce{[HA]} + \ce{[A-]}$
Acidity constant: $K_\mathrm{a} = \frac{\ce{[H+][A-]}}{\ce{[HA]}}$

Substitution with last 2 equations:

Charge balance: $\ce{[H+]} + c = \frac {K_\mathrm{w}}{\ce{[H+]}} + \ce{[A-]} +\ce{[Cl-]}$
Weak acid mass balance : $c = \frac{\ce{[H+][A-]}}{K_\mathrm{a}} + \ce{[A-]} = \ce{[A-]} \cdot (\frac{\ce{[H+]}}{K_\mathrm{a}} + 1) \implies \ce{[A-]} = \frac {c}{\frac{\ce{[H+]}}{K_\mathrm{a}}+1} = \frac {c \cdot K_\mathrm{a}}{\ce{[H+] + K_\mathrm{a}}}$

and finally:

Charge balance: $\ce{[H+]} + c = \frac {K_\mathrm{w}}{\ce{[H+]}} + \frac {c \cdot K_\mathrm{a}}{\ce{[H+] + K_\mathrm{a}}} +\ce{[Cl-]}$

Multiplied by denominators:
$\left(\ce{[H+]} + c \right)\ce{[H+]} \cdot \left(\ce{[H+]} + K_\mathrm{a}\right) = K_\mathrm{w}\left(\ce{[H+]} + K_\mathrm{a}\right) + c \cdot K_\mathrm{a} \cdot \ce{[H+]} +\ce{[Cl-]}\ce{[H+]} \cdot \left(\ce{[H+]} + K_\mathrm{a}\right)$
Separated by the power order:
${\ce{[H+]}}^{3} + {\ce{[H+]}}^{2} \cdot (c + K_\mathrm{a} - \ce{[Cl-]}) - \ce{[H+]} \cdot ( K_\mathrm{w} + \ce{[Cl-]} \cdot K_\mathrm{a}) - K_\mathrm{w} \cdot K_\mathrm{a} = 0$

(1)- The classical related case is
$\mathrm{pH} = \frac 12 ( \mathrm{p}K_\mathrm{a} - \log{c} )$, implying $c \gg [\ce{H+}] \gg [\ce{OH-}]$


text:
How can I understand if my system has degenerate electronic configuration?
How can I understand if my system has degenerate electronic configuration?
I'm running some post-HF methods and in the meanwhile try to understand which is the most accurate method for my system. However, reading about Coupled-Cluster method it is usually reported that this method is not adequate for studying systems with degenerate or nearly degenerate electronic configuration. As far as I understood this means, for instance, that the excited state could be near in energy to another excited state that differs only for some orbital occupation, but how can I check this situation?


text:
How can I prevent and remove oxidation on the surface of liquid Galinstan?
The oxide layer formed on galinstan (or its component of pure gallium) when exposed to air increases wetting and dulls the appearance. Is there a clear liquid material in which a blob of liquid galinstan can be placed that would prevent oxidation of the surface long-term? Ideally the protective liquid would be cheap, relatively safe, not degrade with time, and have a viscosity of about that of water. The goal is to show off galinstan's un-oxidized appearance and surface tension in a sealed or nearly-sealed container.
I know that application of sulfuric acid can break down the oxide layer temporarily, but am unsure if weaker acids accomplish the same, whether this destroys part of the material, and how long the effect would last.
Clear oils could perhaps accomplish the protection goal, but I'm interested if there are less viscous alternatives (e.g., water, alcohol, etc...) and whether additives are necessary to maintain the galinstan's surface appearance.

Those liquids should be in the first place inert. Liquids with active hydrogen like water or alkohols can react with gallium.

If oil is acceptable, but too viscous, try naphtha, kerosene or charcoal lighter fluid (if it contains no alcohols or ketones). Of course, these are flammable and attack some plastics,, so use a strong glass container.

I'd prefer to avoid easily flammable liquids if at all possible, as larger volumes may eventually be used. Safety and clarity would take precedence over viscosity. Would PFCs carry too much oxygen? What about glycerin?

What about an inert gas, like nitrogen or argon?

I use dilute hydrochloric acid to keep my galinstan looking like mercury. The oxide doesn't just increase wetting it its the sole cause of it. You need at least 0.1 M HCl to get maximum surface tension and to keep the surface shiny. Paper on it
0.1 M is about 0.37% concentration. Any concentration of HCl under 5% is "safe" to handle and only over 15% is considered to have the ability to case serious harm. Muriatic acid from a store is usually around 28% - 36% but they don't always give an exact concentration. You'll either have to buy some from a chemical shop that specifies or do a titration on the common stuff.
I just started some trials with 12 sealed tubes of about 1 ml galinstan each ranging from pure water to 1.0 M HCl. Cool stuff. (Sorry for bad quality, phone camera glass is shattered) Only been a few hours and I plan to leave them for months but as of now at 0.1 M and higher concentrations it looks perfect. 0.05 M looks decent, lower and its no longer shiny. The highest (0.5 M - 1.0 M) have visible hydrogen forming from excess acid attacking the metal itself. So far seems like 0.1 M to 0.2 M HCl is best. Water only vs 0.2 M HCl bath
And try to only get clean metal in the acid without tons of oxidation or the dilute acid wont be able to dissolve it all. I always keep my galinstan under water and my syringes full of water when I work with it as soon as its cooled from the alloying heat to slow the oxidation.
In case you don't know the process going on here, the acid can react with the gallium oxide as it forms from the dissolved oxygen in the water. The products of this are gallium chloride and water. Gallium chloride is very soluble in water so the small amounts are invisible. The dilute acid also slowly (very, very slowly) reacts with the gallium, indium, and tin to form the metal chlorides and hydrogen gas. With dilute levels like 0.1 M the metal being lost is negligible. The container is completely full with water so no new oxygen can be dissolved like it would if there was air present above the water.
I'm thinking that if it is in a sealed container the oxygen dissolved in the water will eventually be consumed and converted to H2O and the system will reach an equilibrium. But I'm no expert. If it can last a year or more I'd be satisfied. (Making non toxic liquid metal snow globes with the centerpiece floating on galinstan.)
Liquid metal is fun to work with. I especially like watching it bounce and rebound off the glass when its in a small enough amount to form a bead. I will update this with my results in weeks or months. Best of luck!

Argon or nitrogen should work yes, but getting galinstan with zero oxidation in a sealed container with zero oxygen contamination in the gas is not feasible unless you have the right equipment. I don't, but maybe you have access to some. Even just a small amount of oxygen will dull the metal and the argon can't remove that oxidation. You would still probably need to have the galinstan in a dilute acid bath before putting it in the inert gas anyways to get rid of oxidation beforehand. Putting a few drops of muriatic acid in some water and adding galinstan is easy enough for me.


text:
Two different definitions for chemical shift (NMR)
Looking online I seem to have found 2 different definitions for chemical shift $\delta$ (in the context of NMR). The first definition uses the chemical shift of the reference in both the numerator and denominator:
$$\delta = \frac{\nu - \nu_{ref}}{\nu_{ref}}$$
This can be found on the wikipedia page here: https://en.wikipedia.org/wiki/Chemical_shift
The second uses the frequency of the spectrometer as the denominator, i.e.
$$ \delta = \frac{\nu - \nu_{ref}}{\nu_{spec}}$$
This can be found, for example, here: https://chem.libretexts.org/Courses/Athabasca_University/Chemistry_350%3A_Organic_Chemistry_I/13%3A_Structure_Determination-_Nuclear_Magnetic_Resonance_Spectroscopy/13.03%3A_Chemical_Shifts
Now, I am aware that the "frequency" of the spectrometer is really just an approximate frequency at which protons tend to resonate in that field, and so these numbers would be roughly similar. My question is, is there an actual difference between the two (which is largely ignored because they give similar values) or are they somehow identical?
Anyways, thanks in advance.

First, let's be careful about what the quantities refer to. The reference frequency, of course, refers to the frequency of a reference compound. In the case of $\ce{^1H}$ or $\ce{^{13}C}$, the reference frequency is given by the methyl groups of tetramethylsilane.
The spectrometer frequency, however, could be considered slightly ambiguous. It may either refer to the carrier frequency, which is placed in the middle of the spectral window. If your spectrum runs from $\pu{0 ppm}$ to $\pu{10 ppm}$, then the carrier frequency is $\pu{5 ppm}$, or whatever that corresponds to in terms of $\pu{MHz}$. On the other hand, the spectrometer frequency itself usually refers to the frequency that corresponds to $\pu{0 ppm}$. (In TopSpin, these two parameters are respectively called SFO1 and BF1; this link contains more explanation about these parameters.)
Given that the reference frequency is defined to be at $\pu{0 ppm}$, it follows that the reference frequency is actually the same thing as the spectrometer frequency, which—in a strict sense—makes this question trivial to answer: it can be either one, since they're both the same thing.*
Officially, the definition is with respect to the reference. As per the IUPAC Gold Book:

chemical shift, $\delta$ (in NMR)
The fractional variation of the resonance frequency of a nucleus in nuclear magnetic resonance (NMR) spectroscopy in consequence of its magnetic environment. The chemical shift of a nucleus, $\delta$, is expressed as a ratio involving its frequency, $\nu_\text{cpd}$, relative to that of a standard, $\nu_\text{ref}$, and defined as:
$$\delta = \frac{\nu_\text{cpd} - \nu_\text{ref}}{\nu_\text{ref}}$$

The above equation assumes that the numerator is measured in $\pu{Hz}$, the denominator in $\pu{MHz}$, and the chemical shift in $\pu{ppm}$. (This implicit convention is a bit annoying, but when actually doing NMR spectroscopy it comes quite naturally.)
Why not the carrier frequency, though? The carrier frequency is also proportional to the field strength, so it has no bearing on the argument that the chemical shift should be independent of field strength. I'd argue that it doesn't make sense to use that because that's a parameter that you can change, by varying the centre of the spectral window. If you acquire two different spectra with different spectral windows, it doesn't make sense that the chemical shifts should differ between the two of them.
On top of that, we can motivate this choice by comparing the chemical shift formula against the expression for the relative error of a measurement: the relative error $\eta$ is given by
$$\eta = \left| \frac{x - x^*}{x^*} \right|$$
where $x$ is the measured value and $x^*$ is the true value. Note that the denominator contains the quantity that we are trying to compare it against. In the case of the chemical shift, it is not an error that we are trying to measure, but rather a "relative deviation" from the reference frequency. It is logically consistent, therefore, that the denominator contains the quantity that we are trying to compare against.

Footnote
* Philosophically, however, they're different things. For a given magnetic field strength, the frequency of TMS is a physical constant. That constant is the reference frequency. However, when calibrating a spectrometer, you'll need to determine what that reference frequency is, and then set the spectrometer frequency to that number. If you don't determine it accurately, then the spectrometer frequency will be off (and the resulting spectra will be shifted by a little bit).
    Using an analogy: let's return to 2018, before the redefinition of the kilogram. At that time, the reference kilogram was defined by a block of metal; so you could say that the reference mass was the mass of that block. This is a quantity that is a fixed constant, and can't be changed. On the other hand, if I had to calibrate a new scale, the scale mass might be said to be the mass that makes the scale show $\pu{1 kg}$. It follows that the scale mass should be the same as the reference mass, for a well-calibrated scale. However, for a poorly-calibrated scale, this may differ.

So if you have a signal at 1.000000 ppm, and you choose the wrong denominator, it would come out as 0.999995 ppm ?

@orthocresol The 2001 recommendations in PAC recall that the 1972 rules included a factor $\times 10^6$ for proton NMR (eq. 4), then define $\delta$ as in the first equation by the OP (eq. 5) as more general (and open to ppb «for some isotopic effects») and show one to calculate $\delta$ in ppm in [Hz] (numerator) and [MHz] denominator.  IUPAC 2008 then only shows the first equation of the OP (eq. 1) in lines of «"ppm" is exchangeable with $\pu{\times 10^{-6}}$ [...] as is "%" with $\times 0.01$».

@Buttonwood Thanks for that. Just to be clear, do you just mean to clarify the missing factor of $10^6$ (which I admit, I didn't even think about!), or there something more than that?

@KarstenTheis I actually have some data that illustrates this: https://i.stack.imgur.com/DFa6H.png The red spectrum was divided by a slightly larger value and the blue spectrum was divided by a slightly smaller value (this took place before/after recalibrating a spectrometer frequency). The difference between the peaks is ca. 0.0005 ppm.

@orthocresol Preparing an answer, I encountered this Hz/MHz division in one of the equations.  Maybe this is a reason why we find the computation of $\delta$[ppm] with division over spectrometer frequency (in MHz) $\times{} 10^6$.  But it is not about an error in your answer I'm fine with.

@KarstenTheis Hmm, thinking about it a bit more, I suspect the numerator plays a larger role. I reason about it this way: the denominator is a large, positive number anyway, so hardly matters whether it's 699.90 or 699.95 or 700.00. The numerator, however, refers to a frequency difference (i.e. difference of two large numbers) which can range from anywhere between, let's say, 0 Hz to 20000 Hz. Shifting the reference frequency (or recalibrating) will affect the denominator by a tiny fraction, but the numerator may shift by quite a bit.


text:
How many geometrical isomers will be formed here?

Can someone help me calculate the number of geometrical isomers in this compound? According to me there should be 3 isomers here but according to my teacher, there are 4.I believe that both EZ and ZE cases will be same here.

in EZ or ZE cases the double bond in the middle also becomes stereomeric hence bringing it back to 4

A possible reason you are being led astray is related to the way the substituents on the right side of the tri-substituted double bond are drawn in a linear fashion. However, when the hydrogen and the n-propyl group are drawn with the proper hybridization, only four isomers are possible, as napstablook has stated in a Comment.
In trienes 1 and 2 the tri-substituted double bond is stereogenic with the (Z)-designation in the former and the (E)-designation in the latter. The (Z)-propenyl group has priority over the (E)-propenyl group. The tri-substituted double bond in trienes 3 and 4 is non-stereogenic owing to the presence of identical groups at one end of the tri-substituted double bond.


text:
Standard Hydrogen Electrode Assignment
How did scientists decide that 0V is the Standard Hydrogen Electrode? Was 0V measured through experimentation or was it decided in a meeting/convention? I am a little confused by the cited texts below. Was the principle behind assigning 0V the same as when we would tare a scale?

Unfortunately, we have no way to measure the absolute standard reduction potential of an isolated half-cell. All that can be measured are potential differences when two half-cells are connected. Therefore, to assign values to the various standard reduction potentials, a reference electrode has been arbitrarily chosen and its standard reduction potential has been assigned a value of exactly 0 V. This reference electrode is called the standard hydrogen electrode, SHE (Blackman, 05/2015).
Blackman, A., Bottle, S. E., Schmid, S., Mocerino, M., Wille, U., Brady, J. E., Sen, F.  (2015). Chemistry,  3rd Edition.  Retrieved from vbk://9780730324928

To simplify the collection and sharing of potential data for half-reactions, the scientific community has designated
one particular half-cell to serve as a universal reference for cell potential measurements, assigning it a potential of
exactly 0 V. This half-cell is the standard hydrogen electrode (SHE) and it is based on half-reaction below:
$\ce{2H+(aq) + 2e− ⟶ H2(g)}$ (Flowers, 2019)
Neth, E. J., Flowers, P., Robinson, W. R., Teopold, K., & Langley, R. (2019). Chemistry: Atoms First 2e. Openstax. https://openstax.org/details/books/chemistry-atoms-first-2e

They needed a standard, so chose one. And since voltage is relative (you measure differences in voltage) they chose to define it as 0.

See also Absolute_electrode_potential

The hydrogen electrode potential was set to zero more than hundred years ago by the German Society (Bunsen-Gesellschaft) in 1911 after collecting all the data from hundreds of sources for most of the elements. The original referene is Abhandlungen der Deutschen Bunsen -Gesellschaft für angewandte physikalische Chemie. This work was edited by none other than Nernst, who was a Nobel laureate (its the season!). In 1911, the Society decided

"Wir setzen also willkürlich die Potentialdifferenz zwischen einer von
Wasserstoff von Atmosphärendruck umspülten Platinelektrode und einer
Lösung, die 1 Mol Wasserstoffion in 1 Liter enthält, = 0. Die
Beziehungen zwischen dieser „Normal- Wasserstoff-Elektrode" und den am
meisten gebrauchten Bezugselektroden, nämlich Wasserstoffelektroden in
Lösungen anderer H--Konzentration, Kalomel- Elektroden, Mercurosulfat-
Elektroden usw. sind mit einer für die Berechnung von
Einzelpotentialen hinreichenden Genauigkeit bekannt."

"We set arbitrarily the potential difference between a platinum
electrode surrounded by hydrogen at atmospheric pressure and a
solution containing 1 mol hydrogen ion in 1 liter = 0. The relations
between this "normal hydrogen electrode" and the most used reference
most commonly used reference electrodes, namely hydrogen electrodes in
solutions of other H- concentrations, calomel electrodes,
mercurosulfate electrodes, etc., are available with an accuracy
sufficient for the calculation of sufficient accuracy for the
calculation of single potentials."

Translated with www.DeepL.com/Translator (free version)
Textbooks do a poor job of explaining electrochemistry concepts. For any experimental measurement you a reference. When you quote height of a moutain, the sea level is set as zero height. For pressure, you define vacuum to be reference, zero. When you measure velocity of a car, we set Earth's velocity to be zero or ignore it. Electrical engineers consider the Earth at 0 V potential. In short electrochemists also needed a reference, and the normal hydrogen electrode was a reasonable choice.
Well there is a long story. A very influential physical chemist named G. N. Lewis, had issues with their definitions and the sign conventions of the electrodes. This war between the European conventions vs. American conventions lasted for decades and confused thousands and thousands of students.

You probably mean the sign of half-reaction redox potentials. :-) It is similar like the natural ( on your left/right side) versus geographic ( on your left/right looking downstream ) sense of the left/right bank of a river, or anode/cathode phys.-chem. convention versus engineering convention for rechargable cells.

Yes, Europeans had their logic right, and this was finally adapted. The sign of the electrode is the electrostatic sign of the electrode. It is invariant. G. N. Lewis created problems from his famous book in 1913.

@M.Farooq Similar thing: «European style» vs. «American style» about the wedged and dashed bonds used around stereogenic atoms (should the thick or the thin end of the tapered bond point to the centre, or toward the substituent).


text:
Displacement of salts due to higher solubility
I was wondering why mixtures of salts in the same solution precipitate excactly in order of their solubilities. Imagine a solution with NaCl, NaBr, NaI in equal moles. If AgNO₃ solution is now added in excess, first silver iodide, then bromide, then chloride will precipitate out of solution in that order due to their different solubilities (Ksp=1.8x10⁻¹⁰, 5x10⁻¹³, 8.3x10⁻¹⁷ respectively).
First of all, what is the thermodynamic background of this phenomenon? It's obvious that for example the iodide is less soluble than let's say the silver chloride, but under the given conditions, all of them are basically insoluble, so I don't really get why the silver iodide selectively precipitates first. Secondly, would it matter if the sodium salts wouldn't be there in equal moles?
And on another note; would it be possible to dissolve silver chloride just by adding another, more insoluble salt? For example:
$$\ce{AgCl(s) + NaBr(aq) -> AgBr(s) + NaCl(aq)}$$
A saturated solution of AgCl with already precipitated, solid AgCl is treated with sodium bromide (NaBr) salt . Given the explanation above, would it be possible that the silver(I) ions form a precipitate with Br- ions giving NaBr, and let the AgCl precipitate dissolve and form NaCl? Thanks in advance.

For better site experience, you can find useful How can I format math expressions or chem. equations on Chemistry SE.

You can write $\pu{1.23E-12}$ to get $\pu{1.23E-12}$, or $\ce{H2SO4}$ to get $\ce{H2SO4}$

https://chemistry.stackexchange.com/questions/124923/calculating-equilibrium-concentrations-of-silver-and-halide-ions?rq=1

would it be possible to dissolve silver chloride just by adding another, more insoluble salt? For example:
$$\ce{AgCl(s) + NaBr(aq) -> AgBr(s) + NaCl(aq)}$$

It is possible to get the chloride ion back into solution by adding NaBr. The effect is to lower the silver ion concentration so drastically that chloride goes back into solution. This kind of process is used in analytical chemistry to get ions of insoluble salts into solution. Here is a related example:
$$\ce{Mg(OH)2 (s) + 2HCl(aq) -> Mg^2+ + 2H2O + 2Cl-(aq)}$$
In this case, you lower the hydroxide concentration by neutralizing it with hydrochloric acid.

Imagine a solution with NaCl, NaBr, NaI in equal moles. If AgNO₃ solution is now added in excess, first silver iodide, then bromide, then chloride will precipitate out of solution in that order due to their different solubilities (Ksp=1.8x10⁻¹⁰, 5x10⁻¹³, 8.3x10⁻¹⁷ respectively).

If you add the silver nitrate dropwise and wait a bit inbetween, you will observe solid silver iodide after the first drop. You should not make any statement about what forms first (question of kinetics) but rather what is present at equilibrium. Once the iodide concentration is 360-fold lower than the bromide concentration, you will also form a silver bromide precipitate when adding more silver ions. When most of the bromide is precipitated, silver chloride precipitate will also be observed.
So in that sense, they form stable precipitate in that order, with barely noticable overlap.


text:
Which reaction is happening in the preparation of sulfuric acid through contact process?
I've read that in the preparation of $\ce{H2SO4}$ through the Contact process, on the absortion tower, the gas $\ce{SO3}$ is getting in contact with $\ce{H2SO4} (l)$ , because adding $\ce{SO3}$ to $\ce{H2O}$ is going to cause an exothermic reaction, so:
$\ce{SO3 + H2SO4 -> H2S2O7}$
After this, it is controllable to add water to the oleum in order to produce liquid sulfuric acid.
$\ce{H2S2O7 + H2O → 2 H2SO4}$
But I've seen in other websites that:
$\ce{SO3 + H2O  -> H2SO4}$
So I don't know which is the real reaction of the absortion tower, because there are other pages that state that the liquid entering is 98% sulfuric acid, and 2% water, this water reacts with the sulfur trioxide (this little water avoids the heat problem), producing sulfuric acid.
I would like to know which premise is correct.

All is correct. Reactions of sulphur trioxide and sulphuric acid or residual water are obviously different. The latter can be also considered as the net reaction of the whole absorption/dilution process.

If you carry out the algebraic sum of your two first equations, you will obtain the third equation. So the third equation does not really happens. But it means that if you take $\ce{1 SO3}$ and  $\ce{1 H2O}$, and that these substances are introduced in some unexplained ... machinery, you will always get at the end $\ce{1 H2SO4}$

There are many questions related to "Contact process" in chem.SE. I am sure you can gain some insight from them. Here, have a look: https://chemistry.stackexchange.com/search?q=Contact+process

The issue for the contact process is that, while the final reaction describes the overall result (water plus sulfur trioxide -> sulfuric acid) the constraints of a practical industrial process mean that the earlier reactions are a better overall way to achieve the product. It is all about practical chemical engineering not the chemistry.


text:
What is the English translation of the German term "Aufschluss", a method to get insoluble analytes into solution?
In the analysis of solids, often the first step is to get the analyte into solution in some form. Sometimes, solids are heated in the presence of acids (acid digestion) or molten salts (sample decomposition through fusion according to this source). These methods combined are call "Aufschluss" in German (roughly translated unlocking). There is a joke in the qualitative analysis teaching lab that for particularly difficult samples, you use the "Assistenten Aufschluss", i.e. you have to cheat and beg the T.A. to tell you what is in your sample.
Is there a technical term in English or in other languages?

I don't think there is a single English equivalent word in analytical chemistry that would match Aufschluss. I feel this is more colloquial although Duden describes it for dissolution.  At best it would be dissolution, decomposition methods or digestion methods. For example, there is a book "Decomposition methods in analytical chemistry." which is a German translation of "Aufschlussmethoden der anorganischen und organischen Chemie, published in 1972 by Verlag Chemie GmbH, Weinheim/Bergstr.

There are English translations, but you lose the wordplay. The joke only works because the typical meaning of Aufschluss is to find out about something, to obtain clarifying information. Which is precisely what you would be doing by asking the TA. Then, there is the more obscure technical meaning of Aufschluss, which is to solubilize or dissolve solids into solution. Note the two suitable translations given in the previous sentence. Unfortunately, in English, neither have dual meanings of "enlighten", "investigate", "reveal", etc.

It's not a special case and there is no problem in translating the German term “Aufschluss” into English once the context is given, i.e. disambiguation is performed.
In the absence of one there cannot be a definitive translation since the word has multiple meanings (Duden lists six) allowing for the wordplay as with your example with TA.
Staying in the realm of chemistry and biology, “der Aufschluss” generally refers to a process of transferring to a soluble state that is convenient for processing.
The most universal English term in this context would likely be “dissolution”, followed by more field-specific ones [1, p. 16]:

Aufschluss chem (in Lösung bringen) dissolution, digestion; disintegration, decomposition; dissociation, solubilization; lysis; fractionation; maceration; (paper) cooking, pulping.

Interestingly enough, “Aufschluss” hasn't been borrowed verbatim into other languages as, for example, “Anschluss” has been: another example of how important the choice and use of separable prefixes in German are.
Reference

Cole, T. C. H. Wörterbuch Der Chemie / Dictionary of Chemistry (Deutsch/English), 2., überarbeitete und erweiterte Auflage.; Springer Spektrum: Berlin, 2018. ISBN 978-3-662-56330-4.

«Digest» (the result) and «digestion» (the process) are seen again and again (e.g., already in the abstract here, or revising the state of the art here (first page, right hand column).  Speaking of this university,  «Freiberg digestion» is a common term, too (cf. e.g., Sundermayer in ACIE (p. 228/229). If the intermediate melt is alcaline, «alkali fusion» is used, too.
«Digest» (like in «Reader's digest») applies well on ores and biological samples, though the concept applied on cells seems more often called «lysis» instead.
«Digestion» equally applies well for French, for which Le Petit Noveau Robert (1996, print edition; a one-volume orthography dictionary for the general use, not tied to chemistry) explicitly states the chemical relevance of this noun:

«2. (chim) Dissolution d'une substance dans un liquide à haute température  ou extraction de certains élements de cette substance.»

which translates into

(chemistry) Dissolution of a substance in a liquid of high temperature, or extraction of certain components of this substance.)

backed nicely by the entry in the dictionary of the Academie Française.  Voilà.

I don't understand «dal», nor found yet a reference for this word; is it perhaps a typo?  «Crumbling» sounds for me like a mechanical action (like mortar and pestle, crushers and stamp mills).  But it is true, physicians call digestion all and every action between entry and exits of our body (though with discern between mechanical digestion, e.g. by teeth; and chemical, e.g., by endogenous substances and the many organisms living in the digestive path).

Maybe there is no exact English translation of the word "Aufschluss". The best translation is something between "dissolution" and "crumble". It is used for extracting an element present in a substance that is insoluble in all known solvents and solutions, like acidic, basic and complex solutions.
Let's take the example of Barium $\ce{Ba}$. This element comes from the mineral $\ce{BaSO4}$, which is insoluble in all aqueous and non-aqueous solutions. The only way to transform it into something soluble is to roast it with an excess of dry $\ce{Na2CO3}$. When heated above $\ce{Na2CO3}$ melting point ($851$°C), the following reaction happens : $$\ce{BaSO4 + Na2CO3 -> Na2SO4 + BaCO3}$$ After cooling, the residual solid mixture is washed with water, and $\ce{Na2SO4}$ gets dissolved, leaving the insoluble $\ce{BaCO3}$. This rection has transformed an insoluble Barium compound into another insoluble Barium compound. But $\ce{BaCO3}$ is more useful than $\ce{BaSO4}$ as it does react with hydrochloric acid, producing :
$$\ce{BaCO3 + 2 HCl -> BaCl2 + H2O + CO2}$$ The result of this reaction is a substance that is soluble into water : $\ce{BaCl2}$ The whole process is called "Aufschluss".

If it is just «dry heat», then it is roasting; a term not only used for coffee and meat.  Re $\ce{BaSO4}$, the reduction into sulfide, leaching as a carbonate and reduction seem at least plausible, too, cf. Russian Wikipedia here; but it is a tangential note only.

@Buttonwood. Thank ou for the idea. The word "roasting" seems more appropriate. And the reduction of $\ce{BaSO4}$ into $\ce{BaS}$ by charcoal is also possible. But the odor of $\ce{H2S}$ is repulsive.

Work with tert-BuSH in stoichiometric quantities during a summer project remains in «strong» memory.  But according to a lecture (including samples applied on test stripes) by a representative from Firmenich, scent/odor equally is a question of concentration (e.g. very dilute solutions of skatole convey a rose-like scent, and are used in perfumes [e.g., Peau de Bête]).  Eh bien.

@Buttonwood. Right. But $\ce{H2S}$ smells of rotten eggs, whatever its concentration.


text:
Poisoned Kitchen Sink?
I wonder…I have a 20 years old stainless kitchen sink in good condition.
“Somebody” poured “Green Gobbler” into the sink. Green Gobbler is a Drain Clog Disolver (supposed to be kitchen sink safe and non-caustic but who knows).
Product Description
I cleaned the sink with hot water and soap. Can residual of the “Green Gobbler” be absorbed into the stainless steel and be hazardous, i.e. “poison” the sink?
Thank you!

In my opinion it should be closed otherwise every fear imagi-marginally related to chemistry can find place here.

Unless I have missed it, the vendor keeps the composition secret, giving just hints it is based on natural components. It is not possible to give more particular safety advices than the vendor gives.

I presume the contents are a mixture of hydrolytic enzymes that will digest organic residues in the drain

Maybe the Green Gobbler might gobble you if poured upon you. Other than that, it is harmless. You can't poison steel like that. You can, however, poison your mind with chemophobia. That's a very real danger. We've seen that happen to people, with lamentable results.

The safety data sheet for this product while not disclosing the actual contents lists them as "non-hazardous". The point about stainless steel is that it has a resistant surface that does not react with much so I think it very unlikely that anything can be absorbed that would then cause harm.

The label says that the substance contains no sodium hydroxide, and no bleach. So it must contain trisodium phosphate $\ce{Na3PO4}$. which is corrosive enough to clean blocked pipes. But this substance is rather soluble in water, and it will not remain or stick on the sink or on the surface of pipes. The sink may be in steel, plastic or ceramics. It will not be modified by sodium triphosphate. The trisodium phosphate should not be swallowed, of course. And it will be slowly destroyed by carbon dioxide $\ce{CO2}$ from the air. A good washing will remove all traces of sodium phosphate. The sink will be clean, but not poisoned.

Thank you very much for your responses!


text:
Solvent choice in imine formation from quercetin and poly(allylamine)
I am trying to react quercetin and poly(allylamine) to form imine on the repeating units. Poly(allylamine) is dissolved in water. The problem is the choice of solvent. Poly(allylamine) is soluble in water, lower alcohols, formamide, and ethylene glycol. Quercetin is insoluble in water, and soluble in DMF and DMSO. Quercetin also dissolves in lower alcohols, but very small amounts (approximately 2 mg in 1 mL). What solvent can I choose? Should I try reaction in heterogeneous phase? Or, it is known that DMF is miscible with methanol or ethanol, so should I try cosolvent of these?

These papers look like they may be useful, but I can only read the abstracts https://link.springer.com/article/10.1007/BF00955714   https://pubs.acs.org/doi/abs/10.1021/acsaem.0c02178

DMF is fully miscible with MeOH and EtOH, I would try co-solvents

@Waylander Thank you for advice. I would try it.


text:
Why is meta xylene formed when toluene undergoes alkylation in presence of excess AlCl3?
My book says that in presence of excess aluminium chloride, when toluene reacts with methyl iodide, meta-xylene is the major product. Why aren't ortho-para products favored here?
Edit: The book is "Higher Secondary Chemistry Second Paper", third edition (reprint: 2013) by Haradhan Nag. The reaction is given on page 292.

The o and p xylenes formed rearrange to more stable m-xyle in presence of $\ce{AlCl3}$

See this: https://pubs.acs.org/doi/pdf/10.1021/ja01626a038

meta xylene is more thermodynamically stable, so if you heat enough then you will get meta as your major product.

Huh, that's cool. TIL. The HC Brown paper above pre-dates the Woodward-Hoffmann rules, but presumably the rearrangement is via a suprafacial (thermally allowed) pericyclic 1,5-carbon shift in the intermediate cation?

@TahsinChoudhury Don't forget to add to add reference to the book you refer to; at least author, title, publisher, edition/year of print.  (Though, if is a book, adding the page number would be great.)

@Buttonwood It is a Bangladeshi high-school textbook with many errors. It is not reliable, so I did not add any reference.

@Jayadithya This is interesting because it contrasts with the direction rules taught and a brief run on molcalc.org yields free energies/heats of formation suggesting (at their level of computation engaged), the p-xylene were the most stable one: $\pu{22.83 kJ/mol}$ (o-xylene), $\pu{20.12 kJ/mol}$ (m-xylene), and $\pu{20.01 kJ/mol}$ (p-xylene).

@TahsinChoudhury I don't mind if the source is a small or big author/school/publisher I do not know (yet).  Reading the freely accessible first page of an earlier article by McCauly (1952JAmChemSoc6246) interestingly indicates the product distribution depends (at least) on catalyst, catalyst loading, and temperature and «the right conditions» indeed may favour the eventual formation of m-xylene.

Related: Thermodynamic stability of meta-xylene over ortho- and para-isomers

Alkylation seems to be reversible under duress.  To make a decent analysis product ratios and times would be informative.  The initial formation of ortho and para is because of lower activation energies. If they can equilibrate meta is favored over ortho because of steric effects and over para because there are twice as many meta and the electronic effect that stabilizes the transition state should actually raise the energy of the para ground state. Subtle perhaps but it does not take much energy difference to change close product ratios.


text:
Ideal Gas Equation Extreme Case
I have recently learned about the Ideal Gas Equation and I have this doubt.....suppose I am in a closed container filled with a gas of fixed volume and definite amount of moles. I am assuming myself as a point mass, not having any effect on the gas and just observing the process taking place. What if I increase the temperature to higher and higher values until infinity ? Would there be an infinite amount of pressure on me as well ? I tried doing this by taking initial pressure as 0 and then realized the final pressure would also be zero. I realize that I might be mixing up misunderstood notions and sound crazy but help would be appreciated !

Pressure is exerted on macroscopic objects such as the wall of the container. The particles do not experience pressure but instead experience collisions with other particles, at some frequency and with some average kinetic energy of the particles involved.

@KarstenTheis but doesn't that mean that the collisions a particle will face will increase linearly as pressure exerted on outside increases..................?

The only two things that increase collisions are higher particle speeds (higher temperature) and higher particle concentration (governed by n and V). A change in outside pressure, if it compresses the gas, will decrease V while keeping n constant, so that will indeed increase the number of collisions.

Yes, of course. Rearrange the ideal gas law into the form $$\frac{P}{T} = \frac{nR}{V}.$$ Since by construction the volume and number of moles is fixed, the right-hand side of the equation is constant, so the pressure increases linearly with the temperature in your situation.

$n$ is the amount of substance, not  "number of moles or "amount of moles".

Sure, but you may want to add that the volume would have to be infinite (or the density zero) for the pressure to be zero at a finite T.

@andselisk, in this case it must be the number of moles because the ideal gas constant $R$ is represented on a molar basis.

@a-cyclohexane-molecule "Number of moles" shouldn't be used at all. Not only this term doesn't make sense semantically, but also IUPAC explicitly recommends against applying it.

@andselisk, thank you for the link and information. I disagree that 'number of moles' does not make semantic sense---it is certainly logically accurate in the sense that 'amount of substance measured in moles' is the 'number of moles of substance', but you do make a good point that it is now against current convention.


text:
Which of these protons are equivalent? Which NMR signals correspond to which protons?
Here are is the NMR spectrum and the compound (both images are the same spectrum, just close up). I know the deshielded signals are the aromatic protons and the ones ~6ppm are the alkene protons but the rest is a bit muddy. I'm having trouble correctly assigning the signals to make them agree with the integrations (below the signals). I don't even see 5 equivalent protons anywhere despite the integration suggesting there are. I'm also confused about the multiplicity: the signal at ~1.35ppm looks like a quintet of doublet, but I don't see any protons that would make that. Can anyone help me?

It'd be helpful if you can calculate the coupling constants in each peak.  Expand each region and get peak-picking in Hz.

Next to the aniline-like aromatic H you have the three Me (2 Me with split of the isopropyl group, 1 Me isolated, singlet) for sure, too. If possible, collect complementary 1D data (DEPTQ, if not available an APT) or, with somewhat less resolution but higher signal sensitivity {1H,13C}-HSQC/HMBC to assign the other (more trickier) 6 protons signals in the multiplets (1.2 to 1.6 ppm) d), e), i), j); and somewhat shifted g) of iPr.  Don't forget to integrate HSQC.

The hydrogens labelled d and e  should be designated as four unique hydrogens with different chemical shifts. Certainly the two proximate to the double bond are different from the distal pair. As Mathew has suggested, coupling constants would be useful.

When I approach this type of question, the first thing I ask myself is whether there is a molecule-wide symmetry element present which would transform distant protons into chemically or magnetically equivalent protons. In your case, the molecule is chiral and does not have an axis of rotation so there are no helpful symmetry elements.
Next, I try to identify bonds around which free rotation is possible and which would transform hydrogens onto another.

The most obvious candidate is the $\ce{{H^{(c)}}3C-C}$ bond. Free rotation around it transforms the three methyl hydrogens onto each other. We thus expect that these three hydrogens are magnetically equivalent and give only one signal.
The same is true for the bonds to methyl groups (f) and (h), meaning that at the very least each of these should give a single signal.

Less obvious but I hear you already identified it is the $\ce{N-Ph}$ bond. While rotation is slightly less free than in the previous cases as it is necessary to break the π system for rotation, this is nevertheless a very non-hindered rotation. Thus, the two (k) protons and the two (l) protons are expected to be the same.
In general, this applies to all phenyl groups and almost all substituted benzenes.

Finally, we have the $\ce{C_\mathrm{q}-CMe2H^{(g)}}$ candidate connecting the isopropyl group to the polycycle bridgehead. Free rotation around this bond might allow the (f) and (h) protons to become equivalent or near-equivalent. Often, isopropyl groups show just one methyl signal corresponding to $\mathrm{6H}$.
However, a closer inspection of this group reveals that the methyl groups in this (and many other cases) are actually diastereotopic due to the compound already being chiral. Diastereotopic groups generally do not give the same signal although they may be similar enough to overlap at low resolutions.

As a sanity check at this point, I would go through protons connected to the same carbon atom and check whether I missed any equivalent ones. In your case, on the $\ce{C{H^{(d)}}2-C{H^{(e)}}2}$ remain. However, if you look at a three-dimensional structure (i.e. draw them with proper bond angles), you should see that they are not equivalent: One of the (d) protons points to the vinyl group, the other points to the succinimide. The same is true for the (e) protons. Thus, we expect four individual signals from these four protons (remember that the left and right hand side of the molecule are not chemically equivalent).
Now that we have covered all the protons in the structure, we can try to identify them in the spectrum. Of course, the phenyl and vinyl groups are easy as you have already stated. The two doublets at $\pu{3.1ppm}$ and $\pu{2.72ppm}$ with integral 1 will correspond to (i) and (j). The signal at $\pu{2.66ppm}$ that appears to be a quintet is probably a septet and corresponds to (g). The methyl group (c) must be a single peak (it does not couple with anything) so it must be in the $1.5{-}\pu{1.6ppm}$ multiplet.
We are still lacking the signals for the methyl groups (f) and (h). These must have an integral of 3 each unless they do not resolve in which case they would have a total integral of 6. Each of these signals must be a doublet (coupling to proton (g)). This narrows it down to the signals at $\pu{1.0ppm}$ and $\pu{1.12ppm}$. Given that these two are distinct, proton (g)’s signal is likely a quartet of quartets rather than a septet, but the coupling constants are likely indistinguishable which will lead to the apparent septet pattern.
Two protons (d) and two protons (e) remain to be identified. Thankfully, the area $1.2{-}\pu{1.6ppm}$ still contains an un-accounted-for integral of 4 protons. The four missing protons will be in this region. However, as they are all chemically (and magnetically) non-equivalent, they should each give a rather complicated ddd coupling pattern. With sufficient time and resources, it would probably be possible to identify the individual signals and coupling constants. However, it is probably not worth your time and I would report that area simply as a multiplet of integral 7 (methyl and bridge protons).

Why do you give methyl groups f and h the option of being equivalent? They are diastereotopic groups, as you state,  inherently different. Rotation is not going to make them equivalent. At low field they may fortuitously appear as a 6H singlet but they are not equivalent. Perhaps we are dealing with semantics here.

@user55119 Mostly because I myself was surprised to realise that they are, indeed diastereotopic. I'm too used to isopropyl groups having just one methyl signal.

One upvote. Take a look here; https://chemistry.stackexchange.com/questions/48124/how-many-total-absorptions-would-appear-on-proton-nmr-spectrum-for-this-molecul/87201#87201