text:
Colour of diamagnetic compounds using Crystal Field Theory
I know that the colour in coordination compounds can be explained by CFT in terms of d-d transitions.
Are d-d transitions possible in diamagnetic complexes?
Why does $\ce{[Co(NH3)6]^3+}$ show colour?
It is given that $\ce{[Co(NH3)6]^3+Cl3}$ has a yellow colour and that $\ce{[Co(NH3)5Cl]^2+Cl2}$has a purple colour in my textbook. Can the cause of this variation be explained using CFT?

I think this is similar to this one https://chemistry.stackexchange.com/a/60331/115228

Not quite. I was looking for an explanation on coordination complexes using CFT. It would be appreciated if it was explained using [Co(NH3)6]^3 (if that is possible). There are 3 parts to my question.

As Porphyrin already stated in his answer, diamagnetism/paramagnetism has nothing to do with the colour of the compound.

Are d-d transitions possible in diamagnetic complexes?

It depends on the energy of the photon used. If it has sufficient energy then excitation could be possible whether the electrons are paired or not. If the wavelength of the absorbed photon lies in the visible light region then the compound could exhibit colour. So this answers why those two compounds are coloured.

It is given that $\ce{[Co(NH3)6]Cl3}$ has a yellow colour and that $\ce{[Co(NH3)5Cl]Cl2}$ has a purple colour in my textbook. Can the cause of this variation be explained using CFT?

$\ce{NH3}$ is a stronger ligand than $\ce{Cl-}$. So, $\ce{NH3}$ would produce more splitting energy,$\Delta$ (the energy difference between $\ce{t_{2g}}$ and $\ce{e_g}$ set of orbitals) than $\ce{Cl-}$. Therefore, the splitting is maximum in case of $\ce{[Co(NH3)6]^{3+}}$ (since it has a greater number of $\ce{NH3}$ ligands when compared to $\ce{[Co(NH3)5Cl]^{2+}}$).
So more energetic photon is required to excite an electron in $\ce{[Co(NH3)6]Cl3}$ as compared to $\ce{[Co(NH3)5Cl]Cl2}$. But according to the colour wheel, if the observed colour is yellow then absorbed colour could be violet similarly if observed colour is red-violet (purple) then absorbed colour could be yellow-green. So $\ce{[Co(NH3)6]Cl3}$  would absorb violet photon which is more energetic than yellow-green photon. So the splitting energy is more in the case of $\ce{[Co(NH3)6]Cl3}$, which is in agreement with our theory.


text:
Does ninhydrin staining detect triethylamine?
My reaction mixture has amine compounds, so it shows tailing in bare silica plates. Therefore I decided to add small amount of triethylamine in the eluent. Because my amine compound is not detected by UV, I used ninhydrin staining. But something went wrong. The whole plate turned purple. I tried spot only triethylamine in the plate, dipped into ninhydrin solution, and heated it. The purple color showed. What is the problem? Doesn't ninhydrin only detect the amines which have N-H bond? Searching for this problem, I found that staining with ninhydrin shortly after eluting with 1% triethylamine is one of rookie mistakes in TLC. But how can the ninhydrin detect triethylamine which is a tertiary amine?

Judging from your previous question, you did not seem to have this problem before: https://chemistry.stackexchange.com/questions/164299/tlc-monitoring-of-buchwald-hartwig-coupling-reaction-using-ninhydrin-stain.

@Snijderfrey That's right. The impurity issue is convincing, but I am considering another reasons just in case. Also I feel mysterious triethylamine is detected with ninhydrin although it wasn't about two weeks ago.

Please edit to include such info in the question.

You are right that tertiary amines like triethylamine will not react with ninhydrin in the typical thin layer chromatography (TLC) staining conditions. However, when you use triethylamine as received it will not be 100 % pure. Triethylamine is prepared by alkylation of ammonia. As a result, it usually contains small residues of ammonia and partially alkylated derivatives. These can react with ninhydrin, giving a stained TLC plate. Therefore, I suspect that what you see results from impurities in your triethylamine.
Since you did not have the problem before in one of your previous questions (TLC monitoring of Buchwald-Hartwig coupling reaction using ninhydrin stain), impurities may also have been introduced in the meantime by uncareful usage of the bottle.

Seems probable if ammonia is formed during storage after purchase. But in the website I linked, it is written that staining with ninhydrin shortly after eluting with 1% TEA is common mistakes. I think the website is saying that there is another problem other than the impurities. Is the staining 'shortly' after eluting problem?

It takes a while for triethylamine to evaporate, the same is true for its impurities (and the other solvents you use).

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text:
What is the pKa of the O-17 hydronium ion?
At low concentrations of a conjugate acid/base pair in aqueous solutions, the Henderson-Hasselbalch equation gives a good approximation of the ratio of acid and base:
$$ \frac{[\ce{A-}]}{[\ce{AH}]} = 10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})\tag{1}}$$
If I add a low concentration of $\ce{[^17O]}$ water to an aqueous solution with pH of zero, I would think the heavy water behaves similarly to the solvent. In fact, we don't even have to add heavy water because at natural abundance, 0.038% of oxygen atoms are isotope 17.
I am wondering what the ratio of hydronium to water is for the heavier isotope. For the solvent, it is about 1:55. If it is the same for the heavier isotope, the reaction
$$\ce{[^17O]H2O + H+ <=> [^17O]H3O+}\tag{2}$$
or
$$\ce{[^17O]H2O(aq) + H3O+(aq) <=> [^17O]H3O+(aq)  + H2O(l)}\tag{3}$$
would have a $\mathrm{p}K_\mathrm{a}$ of -1.74. However, the $\mathrm{p}K_\mathrm{a}$ of water is sometimes quoted as 0.
So is the $\mathrm{p}K_\mathrm{a}$ of $\ce{[^17O]H2O}$ different than that of $\ce{[^16O]H2O}$, or do I have to use two different values, depending on whether the solvent is something else or part of the acid/base reaction in question. Surely, $\ce{[^17O]H2O}$ and $\ce{[^16O]H2O}$ have similar $\mathrm{p}K_\mathrm{a}$ values in DMSO, where the equilibrium reaction shown above (3) would be written slightly differently and would have an equilibrium constant very close to one:
$$\ce{[^17O]H2O(DMSO) + H3O+(DMSO) <=> [^17O]H3O+(DMSO)  + H2O(DMSO)}\tag{4}$$

I would use -1.74 for O-17 water and 0 for O-16 water if O-16 is the bulk solvent, because it seems like those are the values that would give the right results for any relevant calculation, but I'm not confident that's the correct general answer

The difference between 0 and -1.74 has nothing to do with isotopes. It has to do with what we define as the unit quantities to put into the equilibrium constant. If a unit corresponds to the pure solvent fir water, then the $pK_a$ is zero. If the unit is 1 molar concentration for water as for other species, then pure water is about 55 units and from the logarithm of that we end with -1.74. It's a confusing convention, but at lease we can say that either way using oxygen-17 should have hardly any effect.

@OscarLanzi I agree, but this specific example shows that there are cases where the value of -1.74 does make sense. If you switched the concentrations of heavy and light water (i.e. traces of O-16 water in bulk O-17 water), you would get zero for O-17 and -1.74 for O-16 with the commonly used standard state definitions (or say that in this case, the $\mathrm{p}K_\mathrm{a}$ of O-16 water is not defined, but the Kw is).

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text:
How much Calcium Hydroxide to manufacture calcium formate?
First, I have no experience with stoichiometry.
I currently manufacture calcium formate for a specific use because here in my country, it is very difficult to get calcium formate in small quantities. If you do, its very expensive.
The easiest and cheapest formic acid I can get is CP grade, 85%, which I have a lot of. I use CP grade anhydrous calcium hydroxide.
I would like to know what the correct amount by weight of calcium hydroxide I should dissolve (Too much of calcium hydroxide makes calcium carbonate. I think that from its reaction with $\ce{CO2}$ in the air, which makes the filtration step a nightmare).
My current recipe: To $\pu{4 L}$ of distilled water, I add $\pu{390 mL}$ of 85% formic acid. I turn on the stirring and start adding calcium hydroxide until the $\mathrm{pH}$ is somewhere between 7 and 8. I then filter at room temperature. Is this optimal given the chemicals I mentioned?

One of the interesting observations I have noted in neutralizations with calcium hydroxide on a large scale is that the reaction of a not-very strong acid (like organic acids) with Ca(OH)2 is quite rapid initially, but slows down as the neutralization becomes more complete, say around 98-99%. At this point, the pH is on the acid side, maybe around 5-6, and it just seems to take forever (several minutes, anyway). It's not nearly as rapid as with NaOH or KOH.
Stirring manually, checking every few minutes with pH paper, then stirring the sludge up again will certainly take a lot of time and labor. With a large quantity, like 4 L, mechanical stirring would seem to be necessary so you can let the reaction go to completion while doing something else.
I suspect you have come across this slowing down, after correctly calculating the amount of Ca(OH)2, and concluded that CaCO3 is produced. It is also possible that CaCO3 is present in your supposedly CP Anhydrous Ca(OH)2, and that would dissolve much slower in the nearly exhausted formic acid. Perhaps that is why you need to filter.
Filtration of 4 L is a pain. I propose another plan: Add 90% of the calculated Ca(OH)2 and stir until clear. The pH will be acidic, ~4-5. Any CaCO3 in your Ca(OH)2 should dissolve moderately rapidly; when the liquid goes as clear as it's going to get, add the rest of the Ca(OH)2. This last addition should dissolve completely, except if it actually does contain CaCO3, that CaCO3 may dissolve so slowly, especially if your pH has gone up to 7-8 (because perhaps your formic acid is only 84.6% - rounded up to 85%). But if your pH stays high, the neutralization is done, and your filtration should be 10 times easier.
BTW, I shouldn't do any homework problems for you, but a final concentration of 10.66% calcium formate solution seems quite nice, considering that it is soluble to the extent of 16.2% at 0ºC. And if you need any other numbers for reference, consider 266.4.

Whatever it takes to get the pH to between 8 and 10, maybe?


text:
Lye from human ashes?
I have a friend who requested to be cremated when they died and their ashes planted with a tree.  I am a bit concerned that the ashes will not be good for the health of the tree.
Will ashes from human remains create an alkaline solution when mixed with water?

Google burial pods.

Human bones would likely be chemically similar to any other mammal bones. It is sold as fertilizer (bone meal), primarily a source for phosphorus. It will also contain other elements like calcium. It should not be a problem, especially after mixing with soil.


text:
How does CO2 'dissolve' in water (or blood)?
I'm trying to understand CO2 dissolved in blood, carbonic acid and the bicarbonate ion.  I think of 'dissolving' as in NaCl dissociating into Na+ and Cl- ions sloshing around in water.  CO2 isn't going to break apart.  Do some CO2 molecules just float around in water/blood, or are they immediately latched on to by an H2O to become carbonic acid, H2CO3?
Or does it readily become H+ and HCO3- (bicarbonate) (Is it HCO3 or HCO3- as an ion to make the + - math work out?)  My understanding is that some magical buffering operation is going on (in the body), where the balance between carbonic acid and bicarbonate ions shifts -- perhaps aided by an enzyme.
I'm an EMT trying to understand how 'respiratory acidosis' (presumably carbonic acid) is different from 'metabolic acidosis' (presumably CO2 in the blood in the form or bicarbonate?)  These seem to be two things, not just 'CO2 dissolved in the blood'.  When I ask medical doctors about this, they usually say 'Gee, I used to know that 30 years ago when I went to medical school'.  I'm hoping that real chemists can describe accurately what is going on.

Gases can be dissolved without any changes in their chemical structures. For example, oxygen from air can be dissolved in water : about $10$ mg per liter. It is not much, I know, but when dissolved, it remains as $\ce{O2}$ molecules in solution. This is the same for $\ce{CO2}$, although a small proportion of these $\ce{CO2}$ molecules are transformed into $\ce{H2CO3}$, and then into ions. Unfortunately, it is difficult to calculate these proportions.

It would be better if you picked one - water, as for blood it's more complicated. Red cells bind not only oxygen but also CO2.

Respiratory acidosis is indeed caused by excess carbonic acid in blood, but metabolic acidosis is other acids, eg lactic acid

@Mithoron.  Lets pick blood, as that is where the respiratory vs metabolic acidosis is important.  I understand now that CO2 is transported in blood in 3 ways, as carbonic acid, as bicarbonate ions, and as you said, bound to red blood cells.

@Andrew, 'other acids like lactic acid are part of metabolic acidosis' sounds like part of the answer. Good.    Is another part of the answer the 'buffering' action of the bicarbonate ion 'changing to or from' carbonic acid?

There's hardly any "carbonic acid" - it's just CO2 and bicarbonate. Any H2CO3 is transient intermediate, at best.

There is reportedly less than 1% of free CO2 in form of H2CO3, with measureable kinetic constants of conversion. The rest is just hydrated CO2.

There are three relevant reactions here:

Dissolution of gaseous carbon dioxide into aqueous solution: $\ce{CO2 (g) <=> CO2(aq)}$

Hydration of carbon dioxide to carbonic acid: $\ce{CO2 (aq) + H2O <=> H2CO3 (aq)}$

Deprotonation of carbonic acid to bicarbonate: $\ce{H2CO3 (aq) <=> HCO3- + H+}$

All three are in rapid equilbrium, with rxn 2 aided in the body by the enzyme carbonic anhydrase. In the blood, all three species ($\ce{CO2 (aq)}$, $\ce{H2CO3}$ and $\ce{HCO3-}$) are potentially present, but at quite different levels. The carbon dioxide can also be bound to hemoglobin, but I'm ignoring that.
The apparent pKa of carbonic acid is about 6.4, below the typical pH of blood (roughly 7.5), so most of the dissolved carbon dioxide is in the form of bicarbonate, which necessarily means that a lot of protons have been released into the blood, lowering its pH a bit (ie making it more acidic). That is important for respiration because it promotes release of oxygen from hemoglobin. The lower pH is also one of the primary mechanisms by which the body recognizes the need for oxygen.
Under typical conditions, about 5% or so of the carbon dioxide would be present as $\ce{CO2 (aq)}$ and the rest as bicarbonate or bound to hemoglobin.
When the acidic blood passes the lungs (ie get exposed to air with low $\ce{CO2}$ content, the rapid equilibria noted above result in the bicarbonate being converted back to $\ce{CO2}$ and released into the gas phase to be exhaled, restoring the blood to the original slightly higher pH.
"Respiratory acidosis" occurs when the lungs are not functioning properly, so that the carbon dioxide/bicarbonate is not effectively removed from the bloodstream and the pH remains low.
"Metabolic acidosis" is unrelated to carbon dioxide and instead results from buildup of other acidic compounds, such as lactic acid or acidic ketone bodies, the presence of which also results in a lower than normal pH of the tissue and bloodstream.


text:
Will an electrolyte eventually stop conducting electricity? And if so, gradually over time or suddenly?
I understand that an electrolyte does not conduct electrons, but rather that the positive and negative dissolved ions are attracted by electric charge to the cathode and anode where they receive and release electrons, and thereby turn into their elemental atom form and possibly form new compounds; e.g. NaCl in water would dissolve into Na+ and Cl- and then possibly after gaining/losing an electron turn into sodium hydroxide and chlorine gas. Am I correct in assuming that this means (at least for this electrolyte) that eventually there would be no more ions left to conduct the electricity? And if so would it be more of a gradual (possibly linear) diminishing capacity? Or perhaps more of a sudden stop? And finally, would this pattern apply to all electrolytes?
Thank you

Yes/maybe/kind-of to all your questions. See Peukert's Law and Google "C rate battery" for the rest.

Note that there are two types of electrochemical cells: galvanic, also called Voltaic, and electrolytic. Galvanic/Voltaic cells derive energy from spontaneous redox reactions, while electrolytic cells involve non-spontaneous reactions and thus require an external electron source like a DC battery or an AC power source. Ref 1.
If the cell generates its own voltage (i.e, we are looking at a galvanic/Voltaic cell), when the reactants get used up, current flow through an external circuit will gradually decrease to zero, or an immeasurably small amount. This is the case in a typical 1.5V battery (or a 3 V mercury cell). But you are surely familiar with the situation of batteries getting used up, either partially or completely.
The conduction of electricity may be stopped in an electrolytic cell if you disconnect the power source from the cell completely, or if you reduce the voltage applied to the cell to such a low level that the anions and cations cannot discharge.
Perhaps you wonder if electrolyzing a solution of NaCl uses up the NaCl and could bring about a complete stop of current flow. No, not really, but things do change. If you could completely flush out the chlorine gas as soon as it is produced, the sodium ion will still remain. If Na$^+$ were reduced to metal, it would react immediately to give H2, but that doesn't even happen (although, if you used a mercury cathode, you could get sodium amalgam, which would slowly decompose to Na$^+$ and H2). The Na$^+$ just carries 4 or 5 or 6 H2O molecules close to the cathode where an electron turns a water molecule into a proton and a hydroxide ion. Eventually, all the NaCl could be converted to NaOH - actually, the conductivity would increase! I suppose you could make the current go to zero by electrolyzing all the water away, to O2 and H2, but the electrolyte would still work if you add more water.
Similar situations occur with most of the electrolytes you will encounter, like NaCl, KCl, CuSO4, Zn(NO3)2, etc.
Now it is conceivable that some electrolytes could degrade by leaving the solution, as in the case of chloride ions. Organic acids could be degraded to paraffins and CO2; nitrates could be flushed out as nitrogen oxides, noble metals (Cu, Ag, Au, Pt) could plate out, but carrying electrons to or from electrodes does not require that the ions die or become ineffective.
Ref 1. https://www.khanacademy.org/test-prep/mcat/physical-processes/intro-electrochemistry-mcat/a/electrochemistry

An electrochemical cell requires a cathode reaction, an anode reaction, a conductive path for electrons to flow from the anode to the cathode and an ion path to neutralize charge movement from anode to cathode. IF any one of these is missing or not operative the cell will not function regardless of chemical conditions[the flashlight is turned off, batteries might be good]; If they are all operative and the cell voltage is zero the cell has reached equilibrium [the flashlight is turned on but the battery is dead, no light].
If the electrolyte is involved in the reaction such as the quantitative removal of copper from a solution the voltage must be chosen that the Copper ions remaining must be essentially zero; as copper ions are removed the current flow lowers and if no other reduction reaction is taking place reaches zero at equilibrium. If this were a copper plating bath a copper anode would be used to replace copper ions hopefully on a one to one basis. If copper ion is being removed by reduction there must be a different oxidation reaction at the anode, perhaps oxidation of water to O2 or maybe a sacrificial anode such as zinc isolated to prevent direct reaction.
The conditions are the same whether a galvanic, producing electricity, or an electrolytic, being charged by an external electrical source. The reactions are simply reversed.

Why was this answer given a downvote?


text:
Electrode potential for saturated silver carbonate solution
An electrochemical cell is set up to measure the electrode potential $E(\ce{Ag^+}/\ce{Ag})$ using the saturated $\ce{Ag2CO3(aq)}$ solution $(K_\mathrm{sp}(\ce{Ag2CO3(aq)},\pu{25 °C}) = \pu{6.3E-12})$ with a standard hydrogen electrode.
Equilibrium concentration of $\ce{Ag+}$ in the saturated solution is $\pu{2.33E-4 M}.$
Electrode potential can be calculated using Nernst equation:
$$E(\ce{Ag^+}/\ce{Ag}) = \pu{0.8 V} + \frac{\pu{0.059 V}}{1}\log\frac{[\ce{Ag+}]}{[\ce{Ag(s)}]}.$$
But we can't use the equilibrium concentration of $\ce{Ag+}$ we got above for the concentration of oxidised species to calculate E. Instead, we use the value half of that, or just the concentration of $\ce{Ag2CO3}$. Why is that?
Reference:
Original question, page 14, question 7(a)(iv): https://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/2020/9701_s20_qp_41.pdf
Mark scheme: page 13, question 7(a)(iv)
https://bestexamhelp.com/exam/cambridge-international-a-level/chemistry-9701/2020/9701-s20-ms-41.php
Worked solution here, at 2:08:20, https://youtube.com/watch?v=rDYnDi4k5pU
But I don't understand the reason mentioned: Because concentration of silver ions is sparingly soluble, so we are using the concentration of $\ce{Ag2CO3(aq)}$.

Useful links for text and formula formatting:
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Consider also the Q/A format of SE sites expects Qs to be self-contained, with quoted/reproduced essential content, making sense even without following links. These are provided mainly just as reference to original material, or for reading full content for better context.

But I don't understand the reason mentioned. Quote the reason you do not understand.

Where is the problem?
Let's call $s$ the solubility of $\ce{Ag2CO3}$. As a consequence $[\ce{Ag+}] = 2s,$ and $[\ce{CO3^2-}] = s.$ So the solubility product is
$$K_\mathrm{s} = [\ce{Ag+}]^2\,[\ce{CO3^2-}] = 4 s^3 = \pu{6.3E-12},$$
which gives
$$s^3 \approx \pu{1.6E-12 M^3} \quad\implies\quad s \approx \pu{1.2E-4 M}$$
$$[\ce{Ag+}] = 2s\approx\pu{2.4E-4 M}$$

The problem is with the calculation of E, they are using half the value of your answer, which I am not sure why.

@radastro If A asks B why C wrote anything wrong then B cannot provide the answer. Only C can answer that, but only in cases it was the intention and not an error. Even the most reputable sources can make accidental errors.

@Maurice I'm afraid you are using MathJax, namely \pu{…} and \ce{…} macros incorrectly. Please look up their intended use — we've had this discussion several times before and you seem to get a good grasp on it, but nothing seems to change which is… disturbing. Also, I personally would refrain from using a non-standard symbol for a variable (solubility $s$) that can also be mixed up with another frequently used quantity in this context (mole fraction $x$).

@Andselisk You are right. I have some trouble with using Mathjax. My main problem is how to introduce italics characters in a text written without italics. By the way, how do you call a character which is "not italics" ? And also how to print the symbol percent (%) using MathJax ?

@Maurice The text is upright without any markup. MathJax switches on math mode to enter variables that should be italicized. Math functions, labels, names should be upright and every type should be handled using appropriate macro. Seriously, read up the documentation and check out examples of decent code posted elsewhere, say, on TeX.SE. There is a quick official guide on mhchem for MathJax, and for the most pat MathJax follows the same principles that LaTeX markup that's been around for decades.


text:
How to obtain excited state energies from configuration interaction? (Helium example)
A theory of CI method
For going beyond the Hartree–Fock (one determinant) approximation one introduce configuration interaction (CI). Interaction means the mixing (interaction) of different electronic configurations (states).
Algorithm of usage CI method:

The solution of the Hartree-Fock equations gives a set of $ M $ spin orbitals, and to construct Slater's determinant use only $ N $, which correspond to the minimum orbital energies (Aufbau principle).

Some of the remaining $ M-N $ functions (corresponding to virtual orbitals) are used to construct additional Slater determinants.
These determinants are obtained by substituting a certain number of spin orbitals of the original determinant $ \Phi^{0} $ for the corresponding number of virtual spin orbitals. The obtained determinants are called excited and are denoted as $\Phi_K $.
The wave function has the form:
\begin{equation}
  \Phi = \Phi^{0} + \sum_{K = 1}^{L} C_K\Phi_K.
\end{equation}

The search for $ \Phi $ is reduced to the variational problem of minimizing electronic energy by varying the coefficients $ C_K $.

When forming the set $ \{\Phi_K\}_ {1, \ldots, L} $ often are limited to determinants, single- and double-excited with respect to $ \Phi^{0} $.

Heluim example
SCF for Helium ground state singlet gives
$$
\phi_1 = 0.842 \chi_1 + 0.183 \chi_2, \quad \text{with orbital energy}\quad \epsilon_1= -0.918
$$
and
$$
\phi_2 = -1.620 \chi_1 + 1.816 \chi_2, \quad \text{with orbital energy}\quad \epsilon_2= 2.810.
$$
The total HF ground state energy $E_\text{total} = -2.862~E_\mathrm{h}$.
The ground state Slater determinant is
$$
\Phi^{0} = |\phi_1 \overline{\phi_1}|,
$$
and excited one
$$
\Phi^{1} = |\phi_2 \overline{\phi_2}|.
$$
So I can construct a secular equation:
$$
\begin{vmatrix}
\langle \Phi_0| \hat{H} | \Phi_0 \rangle - E & \langle \Phi_0| \hat{H} | \Phi_1 \rangle \\
\langle \Phi_0| \hat{H} | \Phi_1 \rangle & \langle \Phi_1| \hat{H} | \Phi_1 \rangle - E
\end{vmatrix} =
\begin{vmatrix}
-2.862 - E & 0.2895 \\
0.2895 & 3.226 - E
\end{vmatrix}
=0
$$
and get two $E$'s:
$E_{1,2} =  -2.876, 3.24~E_\mathrm{h}$.
So, using CI method I get a profit in ground state energy: $-2.876$ instead $-2.862$.
Using a CI method in Quantum Chemistry Computational soft (for example, ORCA) I also can get a correct excited state energies of Helium atom.
Now a question. The energy $E_2 =3.24~E_\mathrm{h}$ is not at all like the energy of an excited state, because it is huge and positive. Since ORCA calculation gives:
-----------------------------
CIS-EXCITED STATES (SINGLETS)
-----------------------------

the weight of the individual excitations are printed if larger than 1.0e-02

STATE  1:  E=   1.908913 au     51.944 eV   418958.0 cm**-1 <S**2> =   0.000000
     0a ->   1a  :     1.000000 (c=  1.00000000)
-----------------------
CIS/TD-DFT TOTAL ENERGY
-----------------------

    E(SCF)  =     -2.855160479 Eh
    DE(CIS) =      1.908912949 Eh (Root  1)
    ----------------------------- ---------
    E(tot)  =     -0.946247530 Eh

How can I obtain correct result for energy of a first excited state doing by hand?
The examples are taken from Quantum Chemistry 7th Edition by Ira Levine.


text:
How to dissolve cobalt ferrite spinel for cleaning reactor?
I put some cobalt ferrite powder and did some catalytic test and I can't clean my reactor very well. I have already used aqua regia, hydrochloric and nitric acids, sodium hydroxide at high concentration and also piranha solution, but nothing worked.

What exactly was "some catalytic test", and what your reactor is made of? Please [edit] the details into your question to help others help you.

Be aware that not well described question context often leads to "a crystal ball scenario".

What is the residue? If it's an oxide, try a reducing agent, e.g., sodium metabisulfite. And as mentioned by others, of what is the reactor made? Have you already damaged it with your experimentation?

I would do what steel strip manufacturers do: use hydrochloric acid solution, which pickles off the mixed iron oxides in hot mill scale (including $\ce{Fe3O4}$, which would likely act similarly to your cobalt ferrite). But you may want to allow some time, because Wikipedia reports that $\ce{Fe3O4}$ dissolves "slowly" in hydrochloric acid. (The industrial pickling process is aided by keeping the acid hot, typically 85-90°C; the presence of the steel strip itself and the micron size of the film.)


text:
Why do sp³ orbitals have tetrahedral symmetry when they are linear combinations of orbitals with octahedral symmetry?
I'm trying to understand how orbitals hybridize. What I understand—or at least think I understand—is the following:

The "standard" $\mathrm{s}, \mathrm{p}, \mathrm{d}, …$ orbitals are orthonormal and can therefore be freely linearly combined to form any other solution to the Schrödinger equation.

The squares of the coefficients in the linear combination of the wave functions must sum to $1$. (Not the coefficients themselves.)

Let's look at $\ce{CH4}$: When the $2\mathrm{s}$ and the three $2\mathrm{p}$ orbitals of the carbon atom hybridize, they might form four $\mathrm{sp}^3$ orbitals. Let $\mathrm{s}, \mathrm{p}_x, \mathrm{p}_y, \mathrm{p}_z$ denote the actual wave functions of the orbitals we started with. Then, one of the resulting hybrid orbitals might have the wave function
$$\frac{1}{2}\mathrm{s} + \frac{1}{2}\mathrm{p}_x + \frac{1}{2}\mathrm{p}_y + \frac{1}{2}\mathrm{p}_z$$
and another one might have
$$\frac{1}{2}\mathrm{s} + \frac{1}{2}\mathrm{p}_x - \frac{1}{2}\mathrm{p}_y - \frac{1}{2}\mathrm{p}_z.$$
There are a total of eight combinations of how the signs of the $\mathrm{p}$ orbitals can be distributed. But two of them are always the negative of each other and form the different halves of basically the same orbital.
(Sloppy description, I know, but I hope you know what I mean.)

Now, here's my problem:
The four $\mathrm{sp}^3$ orbitals form a tetrahedron. If you told me, there are four equally shaped orbitals and nothing else, I would immediately accept why they would form a tetrahedron. But since we started with the $p$ orbitals—which form a octahedron—I don't see how we arrive there via these linear combinations:
Let's choose the symmetry axes of the $\mathrm{p}$ orbitals as our actual coordinate system. Then, the sum $\mathrm{p}_x + \mathrm{p}_y + \mathrm{p}_z$ should have its symmetry axis pointing in the $\begin{pmatrix}1\\1\\1\end{pmatrix}$ direction. Scaling with $\frac{1}{2}$ doesn't change that, of course, and adding the $\mathrm{s}$ wave function also shouldn't change this symmetry direction, as $\mathrm{s}$ itself is fully rotational symmetric. But since the linear combinations of the $\mathrm{sp}^3$ orbitals only differ by the signs of the $\mathrm{p}$ summands, the symmetry axis of any $\mathrm{sp}^3$ orbital should always points towards a vector of the form $\begin{pmatrix}\pm 1\\\pm 1\\\pm 1\end{pmatrix}$. None of which form a tetrahedron. Can anyone explain to me, where my mistake is?
TL;DR: If linear combinations of $\mathrm{s}$ and $\mathrm{p}$ orbitals work the way I think they work, they should always have octahedral/cubical symmetry. But $\mathrm{sp}^3$ orbitals form a tetrahedron and I don't see how to get there.

Also note that the p orbitals as a group do not have Oh symmetry because of the phase difference of opposing lobes.  I think the proper point group is C3v.

You can self-answer your question :)

Since you're talking about octahedra, note also that cubes and octahedra are dual, so you can find the 2 sets of 4 faces that each form a tetahedron.

I like the question. Please post an answer instead of editing the original question. (You'll have my up-vote on both.)

Will add a proper answer; didn't know what the preferred method of self-answering is.

Since I understood it myself in the end, here a bit more about my thought process:

Of course, a cube contains a tetrahedron! You can find an interactive visualization of it here: https://cindyjs.org/gallery/main/NestedPolytopes/
Concretely, you can choose the corners $$\begin{pmatrix}1\\1\\1\end{pmatrix}, \begin{pmatrix}-1\\-1\\1\end{pmatrix}, \begin{pmatrix}1\\-1\\-1\end{pmatrix}, \begin{pmatrix}-1\\1\\-1\end{pmatrix}.$$
As set up there, these describe the sign of the $\mathrm{p}$-summands in the wave functions of the $\mathrm{sp}^3$ hybrid orbitals. I.e., the actual wave functions are
\begin{alignat}{3}
\frac{1}{2}\mathrm{s} \enspace &+\enspace  \frac{1}{2}\mathrm{p}_x \enspace &&+\enspace  \frac{1}{2}\mathrm{p}_y \enspace &&+\enspace  \frac{1}{2}\mathrm{p}_z,\\
\frac{1}{2}\mathrm{s} \enspace &-\enspace  \frac{1}{2}\mathrm{p}_x \enspace &&-\enspace  \frac{1}{2}\mathrm{p}_y \enspace &&+\enspace  \frac{1}{2}\mathrm{p}_z,\\
\frac{1}{2}\mathrm{s} \enspace &+\enspace  \frac{1}{2}\mathrm{p}_x \enspace &&-\enspace  \frac{1}{2}\mathrm{p}_y \enspace &&-\enspace  \frac{1}{2}\mathrm{p}_z \qquad\text{and}\\
\frac{1}{2}\mathrm{s} \enspace &-\enspace  \frac{1}{2}\mathrm{p}_x \enspace &&+\enspace  \frac{1}{2}\mathrm{p}_y \enspace &&-\enspace  \frac{1}{2}\mathrm{p}_z,
\end{alignat}
respectively. So, the symmetry axes of the standard orbitals $\mathrm{p}_x,\mathrm{p}_y,\mathrm{p}_z$ have octahedral symmetry; and also an implicit cubical symmetry (since a cube is a dual of an octahedron). This cubical structure becomes visible if you look at all eight possible linear combinations of the $\mathrm{p}$-orbitals, which only differ in the signs of the coefficients.
And if you choose four of them in the right way, a tetrahedron emerges.

My mistake was to not think about it geometrically. I was looking solely at the linear combinations, and my brain wanted to find four of them which look symmetric/cyclic. Somehow I was working under the assumption that if we start with a symmetric configuration and end up with another symmetric configuration, the computations leading from one to the other must also be somehow symmetric. Silly me...

x = 0 is vertical and y = 0 is horizontal in an xy-graph, but x + y = 0 is on a diagonal.


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Does the solvent used affect basic strength?
In this question, it is explained how basicity and nucleophilicity are different and how the solvent used affects nuclephilicity.
What I wondered was whether there would be a difference in the basic strength of say $\ce{Cl-}$ in $\ce{H2O}$ (a protic solvent) and $\ce{DMSO}$ (an aprotic solvent).
My answer would be that due to solvation in $\ce{H2O}$, the $\ce{Cl-}$ ions would get stabilised and hence their basic strength would reduce as compared to those ions which are dissolved in $\ce{DMSO}$.
Am I right in saying this or did I overlook something?

I think you could more explicit by considering the differential stabilization of the species (i.e., $\ce{H2O}$ and $\ce{Cl-}$ versus $\ce{OH-}$ and $\ce{HCl}$ for a better answer.

Hey I'm sorry but I don't really understand what you're trying to say, could you rephrase that somehow?

The basicity you're talking about is thermodynamic, so you should look at the products versus reactants for the acid-base reaction. You can't just say, oh the of the reactants (the base) is stabilized more than the other, because the product (conjugate acid) could be equally stablized.

Ok thanks a lot for explaining @Zhe

Solvents like DMSO or DMF because of the exposed oxygen supposedly solvate cations more efficiently than anions. Bases such as tertiary butoxide are more reactive in DMSO solutions. Solubility is a concern for a simple ionic such as NaCl.  You might research the effect of DMSO on reactions with LiCl or anhydrous HCl or salts that might be more soluble such as tetramethylammonium chloride.


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Why are radicals considered electron deficient?
I have often read that radicals are electron-deficient and are therefore stabilized by things like electron-donating groups (EDGs) and hyperconjugation. However, today I learned that electron-withdrawing group (EWGs) can also stabilize radicals. Does this mean that radicals are not truly electron deficient?
I thought electron-withdrawing groups destabilized electron-deficient species. Why do we consider radicals to be electron-deficient if they are stabilized by EWGs?

Probably because of this stupid electron counting.

Cobaltocene is a neutral isolable radical compound which is very electron-rich. There are many other such 19-electron transition metal complexes.

It is a term to state that they have the ability to accept at least one electron in the half- filled [or is it half empty] orbital or to form a bond either by combining with or generating another radical. Stability is difficult to encompass, a mix of O2 [a diradical and stable molecule] and a hydrocarbon gas is stable, but one should avoid a room full of that mix at all costs. Radicals react.

Actually, they can be considered both electron-deficient and rich, depending upon the environment. You can read these articles about the electrophilic and nucleophilic behavior of radicals which can cause non-covalent interaction. https://doi.org/10.1039/C9CP05374C , https://doi.org/10.1039/D0CP02619K ,  https://doi.org/10.1039/D0CP01933J  and https://doi.org/10.1021/acs.jpca.9b01133 . Hope this helps.

Radicals should not always be perceived as "electron-deficient". They are not defined in this way. You should always keep the classification criteria in mind. What makes a radical electron "deficient," in relation to what? Reflect upon this.
Another dangerous term is "stable" or "stability" in general chemistry, what is the meaning of stability? Stable with respect to which criterion?
A radical is fundamentally an entity with an odd number of electrons in an orbial, commonly one, in organic compounds. In other words, an unpaired electron. For metal centers in complexes they can also have unpaired spins but one does not call them radicals, althought both radicals and metals complexes are electron spin resonance spectroscopy (ESR) active. This is the true definition by IUPAC rather than "electron deficient" species. Odd electron species can exist as radical cations, and radical anions. There are neutral radicals as well. When you will study mass spectrometry, you will encounter all the three types.

I thought it was "at least one unpaired valance electron" rather than "an odd number of electrons". Or are carbene and triplet oxygen not radicals? IUPAC gold book says "possessing an unpaired electron" which I thought meant "at least one".

Right... better wording is typically odd number of electron or an unpaired spin in an orbital. For most of organic compounds, one odd electron is commonly observed but for metal complexes, the number of unpaired electrons is higher.

From high school chemistry, almost 30 years ago, the basic explanation that we were given (not overly scientific) is as follows.
Each atom "wants" to have a full outer electron shell.  For noble gasses, this is the default because they hit the right number and are full, so they don't react.  For everything else, the amount they are short or over dictates how they react.
Covalent bonds:
For each electron short, the atom wants to interact with other atoms that can "share" their electrons so that the outer shell "feels full".  Oxygen is -2 electrons so it can connect with another oxygen and both share 2 electrons so that they "feel full".  Carbon is -4 electrons so finds some way of connecting with things to make up that deficit, which works in $\ce{CO2}$ by sharing 2 electrons each with the oxygen atoms that want to share 2 each.
Ionic bonds:
Everything in column 1 of the periodic table is +1 in its outer electron shell, and thus highly unstable because they really want to get rid of that extra electron.  This makes hydrogen, sodium, and the like to behave (when isolated) as an explosive.  This works well for elements in the next-to-last column of the periodic table that are of the -1 variety in their outer electron shell.  Sodium will freely give up its unwanted electron to form an ionic bond with chlorine, turning an explosive and a poison gas into harmless table salt.


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protecting copper with benzotriazole
I was told years ago that benzotriazole ("slightly soluble in alcohol"), when applied to a copper surface, will protect the surface from corrosion from the air ("forms a monomolecular layer on copper and silver which precludes most other chemistry, especially tarnishing"). I think benzotriazole used to be a common additive to silver polish? I recently tried this out and the treated surface was identical to untreated when exposed to H2S (boiled eggs in a sealed container). Some possibilities:

I was lied to!
The application process requires subtleties that I missed
H2S is not "most other chemistry" and I should have tested with something else
my container of benzotriazole from ebay isn't benzotriazole

Looking for suggestions for further experiments or alternative processes.

Benzotriazole isn't magic. Apparently it is helpful when added to lacquer or wax coatings, and also in combination with K3Fe(CN)6. There is a good 10p data sheet at https://www.copper.org/publications/pub_list/pdf/a1349.pdf

Test it's solubility in warm water. I've heard water but not alcohol. Make sure to dry thoroughly.


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Has tellurium(II) iodide been properly characterized ever?
According to Wikipedia, tellurium tetraiodide decomposes on heating:
$$\ce{TeI4 ->[\Delta] TeI2 + I2}$$
However, the reaction is somewhat wrong as according to this 2008 paper:

Solid tellurium tetraiodide decomposes to form $\ce{I2(g)}$ and $\ce{Te(s)}$. A very small contribution of $\pu{3.3 ± 2.1 mol\%}$ of gaseous $\ce{TeI2}$ was also determined by both GED and MS. The "metallic" $\ce{Te}$ accumulated in the solid phase vaporizes at above ca. $\pu{670 K}$ as the predominately $\ce{Te2}$ molecular species

In the other Wikipedia article, it says that tellurium diiodide has never been isolated in bulk although organic complexes are well known and characterized.
Why has tellurium diiodide never been isolated in bulk? Is it unstable/prone to disproportionation to monoiodide/subiodide? We never got to know its structure nor its physical properties like b.p, m.p, density, color etc. other than the fact that the species is gaseous?
Question: Has tellurium(II) iodide been properly characterized ever?

Just curious, what is interesting about this specific compound? Does your university subscribe to SciFinder and/ or Reaxys? SciFinder shows that in the last 100 years there are only ~35 papers /patents/reports that even mention it. However, my point is that there are perhaps millions of other structures which do not exist...so back to the original question, what is interesting about it?

It isn't just tellurium(II) iodide. Lower chalcogen halides tend to disfavor the chalcogen(II) compound in favor of subhalides where chalcogen atoms are bonded to each other and have oxidation state no greater than +1.
Let us look at the sulfur-chlorine system, where both the "normal" halide $\ce{SCl2}$ and the subhalide $\ce{S2Cl2}$ are known. However, "sulfur dichloride loses chlorine slowly at room temperature and reverts to disulfur dichloride."[1].
The preference for disulfur dichloride in this system may be traced to pi backbonding. The structure of disulfur dichloride is given here from Ref. 2:

With this structure, a pi-symmetry $p$ orbital on sulfur, which is a good pi donor, overlaps with the conjugated sulfur-chlorine antibonding orbital leading the an electron transfer indicated by the blue arrow, weakening the sulfur-chlorine sigma bond but creating a pi bonding interaction between the sulfur atoms:

In effect, what is formally the sulfur-chlorine bond is delocalized into the sulfur-sulfur pi orbitals. A similar interaction involving the second chlorine atom occurs on the orthogonal plane (note the near-right dihedral angle).
The impact of this interaction is seen by comparing bond lengths. The sulfur-chlorine bond length shown above is $206$ pm, longer (meaning weaker) than the $201$ pm given for monosulfur dichloride in 1. The sulfur-sulfur bond length, by contrast, at $195$ pm is shorter (thus stronger) than $205$ pm on molecular octasulfur [3]. While a similar interaction is possible with $\ce{SCl2}$, it would require chlorine instead of sulfur to donate the electron pair and take the positive formal charge, rendering the backbonding less favorable.
In the case of lower tellurium iodides, there are two preferred subhalide stoichiometries, $\ce{TeI}$ (two phases) and $\ce{Te2I}$ [4]. These have polymeric structures in which the tellurium forms chains, allowing for further bond delocalization through the pi backbonding.
References

https://en.wikipedia.org/wiki/Sulfur_dichloride, retrieved 08 May 2022.

https://en.wikipedia.org/wiki/Disulfur_dichloride, Retrieved 08 May 2022.

https://en.wikipedia.org/wiki/Octasulfur, retrieved 08 May 2022.

https://en.wikipedia.org/wiki/Tellurium_iodide, retrieved 08 May 2022.


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Can oxygen atoms be regarded as a Lewis acid?
Although both oxidants and Lewis acids can steal electrons, they work in different ways. Oxidants acquire electrons via ionic bonds or polar covalent bonds, while Lewis acids form dative bonds with lone electron pairs. As a result, redox reactions are generally not considered acid base reactions. However, in some rare cases oxygen atoms can form dative bonds with lone pairs. For example, trimethylamine can donate a lone pair to an oxygen atom and become TMAO. Does that mean the oxygen atom acts as a Lewis acid in TMAO? And what’s the Lewis acidity of oxygen atoms?

Reactions with radicals such as O or O2 can be written as acid-base reactions but the mechanism is more likely radical single electron transfers rather than dative electron pair transfer or sharing. Some reactions of oxidizing agents such as I2, Br2 and even Cl2 such as the Friedel Grafts reaction or the halogenation of carbon double or triple bonds do have a Lewis acid mechanism; the reaction does not involve radicals.


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Why is Formanilide not named N-benzamide in IUPAC Nomenclature
The preferred IUPAC name for formanilide is N-phenylformamide.

However, it is very similar to benzamide, except for the atom to which phenyl is attached

How does the bonding arrangement, in this case, change the parent group?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), amides are considered derivatives of the corresponding acids.

P-66.1.0 Introduction
Amides are derivatives of organic oxoacids in which each hydroxy group has been
replaced by an amino or substituted amino group. (…)

Simple amide names (such as hexanamide) are generally formed by substitutive nomenclature.

P-66-1.1.1.1.1 Alicyclic mono- and diamides are named substitutively by adding the suffix ‘amide’, to the appropriate parent hydride name, with elision of the final letter ‘e’ before ‘a’. The multiplying prefix ‘di’ is used to name diamides.

A few amides, however, still have retained names that are derived from retained names of the corresponding carboxylic acids. This includes benzamide ($\ce{C6H5{-}CO{-}NH2}$, see Rule P-66.1.1.1.2.1) and formamide ($\ce{HCO{-}NH2}$, see Rule P-66.1.1.1.2.2).
Accordingly, formanilide is a substituted formamide and not a substituted benzamide since the acid part of formanilide is derived from formic acid and the amino part is derived from aniline. Thus, the correct name is N-phenylformamide.
Note that the name “N-benzamide” doesn't make sense anyway. N is a locant that does not describe anything here; so it reads like N-what-benzamide. It’s like writing “2-butane”.

Put simply, formanilide is not N-benzamide because it is not a benzamide to begin with.
"Benzamide" refers to a derivative of benzoic acid, $\ce{C6H5 - C(O) - OH}$, in which the hydroxyl oxygen in the carboxyl group is replaced by nitrogen; the carbon atom outside the ring must be bonded directly to the ring as it is in the parent acid (thus, $\ce{C6H5 - C(O) - NR^{(1)}R^{(2)}}$). Formanilide has the ring joined to the nitrogen atom instead of to the seventh carbon atom and thus can never be a benzamide.


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Does yeast, sugar and water really produce any noticable/significant amounts of methanol?
I just read this question on Quora about distilling during pre-industrial age, which reminded me about multiple discussions I have heard or read during the years about whether the fermentation process used to produce alcoholic (ethanol) beverages produces methanol.
As far as I understand it, yeast, sugar and water results in ethanol. A similar process to get methanol would require you to (destructively) distil wood (methanol is also called something like wood alcohol in many languages) and since you usually don't include wood when making beverages, there should be no methanol in the result.
Furthermore, non-hard liquor like beer and wine do never have a "methanol problem" although, especially wine(-like liquids) often is used as a base when producing hard liquor. Besides, when binge drinking, you typically drink similar amounts of pure ethanol independently of if you drink wine or hard liquor.
My guess is that a reasonably controlled fermentation process of grapes/barley/fruits/grain/etc doesn't result in any noticeable amounts of methanol and if you distil it and keep everything and mix it good it is as safe as drink as home made wine and beer. The problem with methanol, my guess again, is that people in the "moonshine business" sometimes dilute the result with methanol because it is cheaper.

By quick search you can easily realize non-distilled alkoholic beverages like wine (red more than white) do contain minor amounts of methanol and there are law limits for it, much lower than level causing problems. For spirits, aside of illegal methanol mixing, there can be problem with amateur distillation, if early low boiling T fraction is consumed, as it is significantly enriched by methanol.

You're right that fermentation of pure sugar should not produce methanol, but crude fruits (like grapes) contain a lot of pectin, and hydrolysis of the methyl esters in pectin yields methanol in the fermentation.

BTW, the large ethanol:methanol ration makes the small amount of methanol less dangerous, since liver enzymes that convert methanol to far more toxic formaldehyde are engaged by the process of metabolizing ethanol. Clinically, this can be used to treat methanol poisoning - https://pubmed.ncbi.nlm.nih.gov/7083094/ . Do not try this at home.

I am really surprised that this question hasn't received any answers at all!

@d-b probably because Andrew and DrMoishe already summed it up perfectly, but see below for my attempt :P

tl;dr: No, but yeast, plants and water do.
Any Pectin you have in your source material will contribute to Methanol production, and distillation (when you're doing it wrong, especially the remixing bit) can easily increase the resulting concentration high enough to be damaging, while a "natural fermentation" would be too diluted.

As far as I understand it, yeast, sugar and water results in ethanol.

First things first, there are plenty yeasts that might do different things, but most "traditionally used ones" (including baking yeast!) would convert monosaccharides (fructose, glucose, galactose) into ethanol and CO2.
Also, they usually contain enzymes or get help by e.g. lactic acid bacteria in SCOBY fermentations (or the brewer has to do preprocessing steps to help that conversion along) that break down di- (lactose, sucrose) or polysaccharides (starch etc) into monosaccharides.
One of those enzymes contained in a lot of yeasts and plants happens to be "pectin methyl esterase", which converts Pectin (a "heteropolysaccharide", so "one of the things that can be broken down to simple sugar", and incidentally also "the stuff plants contain EVERYWHERE as a structural material").
As a side product, this will essentially always produce methanol.
Also there are some other (but rarer) ways to produce it, such as by various strains of various fermenting bacteria producing more or less of it depending on what they ferment.

A similar process to get methanol would require you to (destructively) distill wood

Thing is, this is a completely different process. Fermentation, the thing you (presumably) want to produce ethanol, will produce methanol in the "pectin eating step".
Distilling wood is a dissimilar, alternative process to get high amounts of methanol quickly without the fermentation.

and since you usually don't include wood when making beverages

well you do though. grapes are "wood", grains are "wood", apples are a LOT of "wood", etc, given pectin content is all that matters for the yeasts, not "whether it looks/feels woody" ;)

Furthermore, non-hard liquor like beer and wine do never have a "methanol problem"

they do, almost all of them contain methanol, just usually so little you don't notice. additionally they contain a LOT more ethanol, which literally just dilutes the toxic effect of the methanol in your body.
the main issue is that when distilling, you're increasing the overall concentration so much, it becomes very nontrivial for your body very quickly, especially if you're keeping a lot of the early product.
Note that despite the popular misconception the post you linked to keeps spreading, a distillation will not produce outputs "in steps" ("methanol first, then ethanol, then water"), but it's all "smooth transitions". so the very first drop will contain the most methanol, and the very last drop will contain the most water, but it will also contain some trace of methanol still, as will anything in between (and as will the first drop already contain a trace of water).
So you're essentially dealing with a tradeoff between "the more of the early stuff i get, the more methanol i get" vs "the more of the early stuff i throw out, the more i'm wasting, and the more i might be throwing out flavors i want to keep".

My guess is that a reasonably controlled fermentation process of grapes/barley/fruits/grain/etc doesn't result in any noticeable amounts of methanol

that'll "just" give you multiple percentage points of poison usually, likely won't kill/blind you in that concentration and presuming you have a normal-sized stomach volume.

and if you distill it

Then you "distribute out" the compounds and increase overall concentration.

and keep everything and mix it good it is as safe as drink as home made wine and beer.

I mean it'll taste burnt and nothing as the original, but yeah if you literally distill everything and then remix it without taking anything out (well except for the burnt up particulates i guess), you have more or less what you started with yes.
Distillation doesn't make methanol, obviously, it only makes it easier to concentrate it on accident.

"moonshine business"

the main reason a lot of moonshiners used to poison themselves is actually just because it's a bit suspicious if you have a bunch of decent lab equipment in a forest and therefore they'd literally just taste test to figure out how much methanol there is (and if you do that all the time it adds up), or because they'd be cheap and literally just distill off the water and keep the methanol in because why not, you're doing an illegal thing anyway, might as well be negligent about it (mind you at the same time the government would literally poison random foodstuffs suspected of being used for moonshine!). I doubt they'd literally buy methanol to add it in though, i mean you're already fermenting and distilling here, why buy more alcohol, if you can just use straw or whatever to get more cheap bulk mass that then just happens to be mostly pectin and almost no sugar, so gets turned into a lot of methanol "on accident" that you then just don't bother to throw out if you're cheap like that.

You (humans) can't taste any difference between pure ethanol and pure methanol.

@d-b i rather won't try, so i'mma believe you, i've been told by moonshiners that they do it by taste, so i assume what they actually taste is other volatile compounds in the "heads", and just hope the ethanol follows the same curve, which then might explain why they went blind :P


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Balanced reaction to Cr given H+
The troublesome nature of this post, is that I was given very little information to work with:
"The metal chromium (Cr) reacts with an acid, to create Cr^+3 ions, and hydrogen gas.
Show the balanced reaction."
(paraphrasing)
Addendum:

It is highly implied that there is water involved somehow, in a follow-up question, there is mention of water-vapor that "carries the hydrogen gas".

Any ideas? :)
EDIT: I failed to mention that the core problem of this question, is that I am entirely unsure with what I am entirely working with here.
"Cr + H^+" doesn't seem right, surely there's more to it?

Are you familiar with half-reactions?

@TheRelentlessNucleophile Yes, I am actually! I just fail to see how is it related? :s

Well, then you should be able to tell that chromium undergoes oxidation accompanied by the reduction of hydrogen ions to hydrogen gas. The addendum is (probably) wrong, I don't see how any water vapor could emerge. You can write the global equation without referring to half-reactions, but I figured intermediary steps might make everything clearer. How many electrons does an atom of metallic chromium lose to ionize as the trivalent ion? How many electrons do hydrogen ions accept to become diatomic hydrogen gas? What is the least common multiple of the results obtained above?

@TheRelentlessNucleophile I'll try to be a tad more elaborate, I feel like we're not quite on the same page: I am confused regarding with which materials am I working with here? You noted that the by-product is (probably) wrong, that there's no vapor being produced. However, do I maybe have H2O, or some other type of reagent in the base step? Right now I am working with "Cr + H^+" and I don't feel like that adds up.
Thanks! ><

Can you do the problem of aluminum metal reacting with acid to give trivalent aluminum ions and hydrogen gas? If so, then you are essentially done.

@EdV Hey Ed, no I don't think so...In a follow-up question, it is highly implied that there is water involved, somehow, but it is really not made explicit at all...Thanks for the suggestion though, I'd love to hear if you have anything in mind!

The half-equations are : $$\ce{Cr -> Cr^{3+} + 3 e-}$$ $$\ce{2 H+ + 2 e- -> H2}$$ So doubling the first and multiplying the second by $3$ and adding the results gives the final equation : $$\ce{2 Cr + 6 H+ -> 2 Cr^{3+} + 3 H2}$$ Water is not taking part to this reaction, except if your teacher wants you to replace the ion $\ce{H+}$ by $\ce{H3O+}$. In this case, the second half-equation becomes : $$\ce{2 H3O+ + 2 e- -> H2 + 2 H2O}$$ And the final equation becomes : $$\ce{2 Cr + 6 H3O+ -> 2 Cr^{3+} + 3 H2 + 6 H2O}$$ But there is nowhere any water vapor, except if you admit hat the hydrogen gas is a little wet and contains here and there some $\ce{H2O}$ molecules.

I tried to help the OP by using Al instead of Cr, but they really wanted that water! ;-)

Absolute legend, this should be it! =)
@EdV yeah, my lecturer kept pointing at it...Heh. I didn't know what I could or couldn't use.


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Why does density decrease when pressure is increased on an NaCl type lattice?
I was told that the effect of pressure on an NaCl type structure would make the atoms come closer and therefore change the type to CsCl type making Z (no of atoms per unit cell) go down, hence density decreases. Won't the added pressure increase the density of the structure? Am I wrong in saying so?

You were told wrong. Z is not related to density.

Isn't density = z * M / a^3 where a is edge length and M is molecular mass

True, but that's not important. Every now and then, you would encounter transitions when Z is multiplied twofold. Does that mean that twice as many atoms are now crammed into the unit cell? No, it means that the unit cell itself has changed and has twice the volume it used to have before. Same thing here.

Two things to consider:

Even though the sodium chloride crystal structure is face-centered cubic, it is not 12-coordinate. Face-centered cubic refers to the lattice formed by periodically equivalent atoms or ions, which in sodium chloride is commonly taken to be the chloride ions. But the chloride ions are not directly coordinated to each other; they are instead coordinated to the intervening sodium ions and that makes them only six-coordinate. Switching to what is usually the caesium chloride structure increases the coordination number of each chloride ion (and each sodium ion) from six to eight, which is a normal effect of increasing pressure.

Placing fewer atoms in a unit cell means lower density if the unit cell remains the same size. But when the sodium chloride switches structure, it not only goes from eight atoms/ions in the normal unit cell down to two in the high-pressure cell. The high-pressure cell is also much smaller than the normal one, a condition promoted by the increased coordination number noted in (1). The net result is that the high-pressure unit cell is less than one-fourth the volume of the normal one, so the density will be higher even with only two ions in the high-pressure unit cell.

You've gotten a good answer from Oscar explaining the atomic details of what's actually happening.
More broadly however, it is important to note that, irrespective of the atomic details, it's physically impossible for an isotropic substance to decrease in density when the pressure increases.*  That's because it would enable a violation of the first law of thermodynamics (conservation of energy). If a substance did decrease in density when you increased the pressure, you could use it to build a perpetual motion machine of the first kind.
To illustrate this, assume you did have such a substance.  Imagine you submersed it in an incompressible fluid, and put the substance + fluid in a cylinder with a moveable piston at the top.  Now put a heavy mass on top of the piston, which increases the pressure on the substance, causing it to expand (if its density decreases, its volume must increase).  Thus the piston head would rise, and the heavy mass would be lifted.  Now move the mass laterally, and let it descend to its original height.  Couple it to something (e.g., an electrical generator) so that the mass does work as it descends.  Since you've moved the mass off of the piston, it has also descended back its original height.  So now you can move the mass laterally back to the piston head.  It rises again.
Thus by moving the mass back and forth you'd be creating energy for free.  Specifically, since you keep returning your piston system back to its original state, $\Delta E$ for that is zero. However, each time you move the mass to the other system, it's able to do work as it descends.
*Unless you're Lisa Simpson:
https://www.youtube.com/watch?v=4FN_DGxZ2dM


text:
Reaction of 1,3 diphenyl propane with alkaline potassium permanganate followed by acidification
Will the product be just benzoic acid?
If yes how does this reaction proceed
Normally alkyl group substituted benzene ring seems to give benzoic acid irrespective of chain length
However is it true when 2 benzene rings are connected
Mechanism is appreciated

I think you are going to get two equivalents of benzoic acid from oxidation of both benzylic sites based on the summary here. The mechanism of this oxidative process is AFAIK not completely understood.

If you are happy with the answer, please accept it.


text:
Why isn't lead tetroxide stable?
I have heated lead(II) oxide at about 450 °C and it changed to lead(II,IV) oxide, but as soon as I took it out from hot plate, it changed to $\ce{PbO}$ again. Why does it happen? Is there a better way to synthesize $\ce{Pb3O4}$ from $\ce{PbO}?$
I had the same problem when I wanted to change $\ce{Pb3O4}$ to $\ce{PbO}.$ When I heated $\ce{Pb3O4},$ it got dark. When I took it out, it became red again. I used straight flame to decompose it.

It seems the cooling down is not fast enough to freeze the equilibrium, previously reached at high temperature.

how can I cool it down very fast? is there any tool?

I wonder that when Pb3O4 is made, it won't change to PbO unless it is heated around 600°C so how does it come back to PbO!!!?

Color can be deceiving. Better would by gravimetric analysis.

will it be better if I make PbCO3 and heat it? or it will treat the same?

The oxidation of $\ce{PbO}$ to $\ce{Pb3O4}$ at about $450$°C is slow. You better check your yield by weighing, and re-heat it if the yield is not satisfactory

Heat is very different while conducting such type of reactions because at a specific range of temperature, you will get a different species of lead oxide. The general scheme of lead oxide conversion is as follows:
$$\ce{PbO2 ->[\pu{293 ^\circ C}] Pb12O19 ->[\pu{351 ^\circ C}] Pb12O17 ->[\pu{375 ^\circ C}] Pb3O4 ->[\pu{605 ^\circ C}] PbO}$$
So, around 375 °C-450 °C, you will get lead(II,IV) oxide. As soon as you overshoot the temperature, it will convert to lead(II) oxide.
If you have lead(IV) oxide, then it can be converted to lead(II,IV) oxide at around 450 °C-480 °C but if you overshoot it, lead(II,IV) oxide will decompose back to lead(IV) oxide (reaction completes at 580 °C). Wikipedia took the reaction from Lead Compounds by Dodd S. Carr, 2000
$$
\begin{align}
\ce{6PbO + O2 &->[\pu{450 ^\circ C - 480 ^\circ C}] Pb + CO2}\\
\ce{2 Pb3O4 &->[>500 ^\circ C] 6PbO + O2}
\end{align}
$$
Lead(II,IV) oxide is relatively stable to atmospheric oxidation and insoluble in water and as such, it was used as a pigment for a long time until it was scraped off due to its toxicity. It is temperature resistant upto a certain point.

I control the temperature during the process by using a laser thermometer it never gets more than 470°c  but when the temperature gets 300°c and less, it changes to yellow again

as soon as temperature gets 470°c the color gets red and when it gets 300°c the color gets yellow!!!

@NilayGhosh You may want to review your answer:  1/ but if you overshoot it, lead(II,IV) oxide will decompose to lead(II) oxide.  2/ Rewrite the 2nd reaction, it is really a mess.


text:
Intuitive explanation for EMF of cell being zero at equilibrium?
I am not able to understand why $E_{cell=0}$.
I went through this answer of the site  Electrode Potential at Equilibrium
It says

This eventually comes to a stop when the equilibrium is reached. That is the rate of negative charge build-up is equal to the positive charge. Like you say at this point there is no more drive for the reaction to go one way or the other.

But if rate of negative charge build up is equal to positive charge but we would still have charge separation so why should potential difference be zero?

Like you say at this point there is no more drive for the reaction to go one way or the other.

I know that the Galvanic cells are drove through the Gibbs' energy of redox reaction and I agree that at equilibrium there is no change in Gibbs' energy but we still have charge separation between cathode and anode so why should the potential difference between them be zero ?
I ask my query as a new question here because answerer to the linked question has been inactive for long time.

When a system reaches equilibrium it reaches the minimal free energy and no energy can be spontaneously withdrawn from it. As mechanical analogy, you can draw energy from a ball rolling downhill. When it is at mechanical equilibrium at the bottom of the pit or valley, you cannot draw more energy from it (by mechanical means).

@Poutnik As I mentioned in question , I know why EMF should be zero due to energy considerations , simply because we don't have and Gibb's energy to do a non P-V work but my query is that we do have charge separation and thus a potential difference

No, we do not have, if equilibrium is reached. Note that some systems may need practically infinite time to reach equilibrium.

@Poutnik Can you explain how we do not have charge separation at equilibrium  ?

Rather, can you explain how we can have it? Having potential electrode difference and being at equilibrium are mutually excluding states ( I mean cell equilibrium, not electrode/half cell equilibrium)

@Poutnik Consider the Daniel cell, as we move towards equilibrium the concentration of Zn2+ keeps on increasing while the concentration of Cu2+ keeps on decreasing, clearly this makes Zn rod more positively charged and Cu rod more negatively charged.

Let us continue this discussion in chat.

At equilibrium, the zinc ion concentration does not keep increasing and the copper ion concentration does not keep decreasing: the system is at equilibrium. It is kaput: no driving force. No free energy.

@EdV Yes , the concentration of zinc ion does not keep on increasing but what will you say about charge separation ?

Even open circuit, it is negligible. Continuing with the Daniell cell, is has relatively low internal resistance in normal operation, by design, so there is little stored charge when the cell is dead. Even if copper ions were the limiting reactant, the dead cell would be what is called “stiff”, i.e., attempting to draw current would produce almost nothing.

Can downvoter explain the reason for same?

I will give you an upvote because your question is fine. One thing to think about: when a proper Daniell cell is fully dead, you basically have two metal electrodes in solutions that have the same final concentrations of the relevant cations and anions. It takes time for diffusion to do its thing. But you could speed it up: dump all three solutions into a beaker, mix thoroughly, then stick the zinc and copper electrodes in. You will have a smaller voltage difference that declines to zero as the redox reactions proceed to the end. This is the same final result as in the proper Daniell cell.

Let's discuss the Daniell cell. In the beginning, the cell is made of a zinc plate in a $\ce{ZnSO4}$ solution (maybe $1$ M), and a copper plate in a $\ce{CuSO4}$ solution (may be $1$ M). Zinc is the anode and copper the cathode.
If the cell delivers some current, the amount of zinc metal decreases, the concentration of $\ce{Zn^{2+}}$ increases. On the other electrode, the amount of metallic copper increases, the concentration of $\ce{Cu^{+2}}$ decreases, and the corresponding anions $\ce{SO4^{2-}}$ are crossing the bridge between the two compartments. So that the cathodic region (around Cu electrode) is slowly loosing all its ions. Its potential decreases from $+0.34$ V to less and less. Simultaneously, the potential of the cathode increases from $-0.76$ V to a less negative potential. The importance of these changes is described by Nernst's law.
When the reaction is finished, both electrodes have the same redox potential, which should be around $-0.2$ V. No potential difference. No current any more. The cell is made of a heavily corroded zinc anode in a concentrated $\ce{ZnSO4}$ solution (maybe $2$ M), and a thick cathode made of much copper, in a rather diluted $\ce{CuSO4}$ solution. $\pu{E_{cell}}$ is zero

Just one question, how can we say that as ions in the cathodic region decreases , the potential of cathode decreases?

One thing: eventually the two electrode compartments and the salt bridge will have the same concentrations of all the cations and anions that remain at the end: diffusion, like rust, never sleeps.

There is a difference between equilibrium for the cathode and anode reaction and general diffusion thru out the cell. In constructing a battery the electrode reactions are optimized and the general diffusion reactions minimized[with varying amounts of success]At electrical equilibrium the activities of all the moieties involved in the electrochemical reaction are at the concentrations[activities]  that meet the equilibrium constant. The half electrode potentials for anode and cathode are the same[I don't think zero] It is hard to visualize dynamic equilibrium but theory says it is.

@Lalit. The potential of an electrode (anode or cathode) is described by Nernst's law : $\pu{E = E° + \frac{0.059 V}{n}·ln\frac{[Ox]}{[Red]}}$. In the case of the cathode, made of copper in a $\ce{Cu^{2+}}$ solution, $\pu{[Ox] = [Cu^{2+}]}$. So the potential $\pu{E}$ of the cell increases when the concentration $\ce{Cu^{2+}}$ increases


text:
What happens to polymers under high pressure?
What happens when you have polymers under high pressure? My very sketchy thoughts of polymers are that they can be thought of tangles of carbon chains surrounded by hydrogen, but that some polymers like polyethylene can be ordered or  "crystalline".  So it seems like there could be phase transitions, or if the pressure was high enough the polymer would have to change it structure or if in a tangled structure  perhaps break links and form new links making a new polymer?
Edit: Very high pressures like those used in diamond anvil cell experiments.

You mean polyethylene specifically? Polypropylene (and some others) is similar, but most of them are not.

@Mithoron More generally, polyethylene was just something I knew could have a crystal structure

Well, lets better make it more specific... This could be pretty interesting entry here, if it won't get closed as too broad, unclear or whatever. Some quite unusual and less known things happen under high pressure.

somewhat related https://chemistry.stackexchange.com/questions/131342/c60-fullerene-polymers https://chemistry.stackexchange.com/questions/31313/why-does-addition-polymerisation-of-high-density-polyethene-use-low-temperatur

@Mithoron Edited to be more specific to be about alkanes since they can be of different lengths and at room temperature go from being a gas to a liquid to a solid depending on the length.


text:
How do the spin-matrices change in systems without spherical symmetry?
The (non-relativistic) spin matrices of a free electron, with $z$ as the quantization axis, read:
$$
\boldsymbol{S}_{x}=\frac{\hbar}{2}\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right], \quad \boldsymbol{S}_{y}=\frac{\hbar}{2}\left[\begin{array}{cc}
0 & -i \\
i & 0
\end{array}\right], \quad \boldsymbol{S}_{z}=\frac{\hbar}{2}\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right].
$$
The eigenvalues of these matrices are related to the energies when measuring the particle in a particular spin state. The spherical symmetry is reflected in $\boldsymbol{S}_x$ and $\boldsymbol{S}_y$ which have the same energy eigenvalues. My question now is, what happens if we consider a doublet molecule without symmetry along the $x$- and $y$-directions? Will $\boldsymbol{S}_x$ and $\boldsymbol{S}_y$ then no longer have the same eigenvalues?

Why should the operators be different when the system is changed? You can talk about the (linear) momentum of an electron or a neutron or a molecule or a cow, but in all cases, we still have $p_x = -\mathrm{i}\hbar(\mathrm{d}/\mathrm{d}x)$. Surely, the expectation values $\langle \psi | \hat{O} | \psi \rangle$ may be different, but that's no reason to change the underlying matrices $\hat{O}$ themselves; in fact the Pauli matrices are sacrosanct because they must obey the laws of angular momentum, namely, the commutation relation $[S_i, S_j] = \mathrm{i}\hbar \sum_k \varepsilon_{ijk}S_k$.

Also, this was crossposted from Physics.SE: https://physics.stackexchange.com/questions/706299/how-do-the-spin-matrices-change-in-systems-without-spherical-symmetry

@orthocresol. I realize now that it was a stupid question. The operators stay the same, but the eigenvalues (and expectation values) change because the wave function changes. Thank you.

No, the eigenvalues don't change; the eigenvalues are only dependent on the operator, and they are still $\pm 1/2$! But the expectation values will change depending on the system under consideration. Not a stupid question, btw.

What I don't get is that if the matrices don't change, and the eigenvalues don't change, how can the expectation values change? If we have $\langle\psi|\hat{O}_1|\psi\rangle=\langle\psi|E_1|\psi\rangle=E_1\langle\psi|\psi\rangle=E_1$ for a free particle, and the operators and eigenvalues stay the same, then we should get $\langle\psi|\hat{O}_2|\psi\rangle=E_2=E_1$. I must be misunderstanding something.

I know that the wave function should change, but the wave-vectors must still be a linear combination of the same two eigenvectors.

Well, it's a linear combination, but the coefficients can vary: if you have $|\psi\rangle = c_\alpha |\alpha\rangle + c_\beta|\beta\rangle$ then the expectation value $\langle \psi | S_z | \psi \rangle = \hbar(|c_\alpha|^2 - |c_\beta|^2)/2$, and obviously this value depends on what $c_\alpha$ and $c_\beta$ are and thus what the state $\psi$ is. The same generally holds true of $S_x$ and $S_y$ but the maths is uglier. I'm not sure how you obtained the first equality in your comment; that only holds true if $\psi$ is an eigenstate of the operator with eigenvalue $E_1$, which is not guaranteed.

Yes, it was assumed that $\psi$ was one of the two eigenvectors. And yes, I get it now that the difference in the expectation values comes down to the difference in the weights of the eigenvectors. Thank you for the clarification.


text:
Will osmosis still occur if there is no water on the other side of the semi-permeable membrane?

During the following experiment: Take two peeled potato-halves and
scoop each one out to make potato cups. Put each potato cup in a
trough containing water. Now,

Keep Potato-cup A Empty
Put one teaspoon sugar in Potato-cup B

We will observe that water gathers in the hollowed portion of cup B.
But no water will gather in the hollowed portions of cup A.

Are not both cups examples of hypotonic solutions where the medium surrounding the cell (potato-cup) has a higher water concentration than inside the cell? Why doesn't water enter the cell in the case of cup A?
Please note that I have a basic (9th grade) understanding of osmosis and diffusion.
This question has been bothering me for some time. Please forgive me if this seems like a HW help.

What is your idea about this phenomena ? We will not do all the job of answering your question, if you don't show us that you have tried some beginning of explanation, and failed to finish the task somewhere. We are not a free machine to do your work.

What does "hypotonic" or "hypertonic" mean, in relation to plain water, and water with a lot of sugar dissolved in it?

This experiment is not designed to solve why an observation happened. You have not given or even thought about the purpose of the experiment and your question gives doubt that you actually performed the experiment. Now if you want to learn about potatoes maybe you should study them a bit before using them as a raw material in an ersatz experiment. A question: Is a potato permeable to water? Does a potato contain water, what are the colligative properties of the water in the potato? Devise some tests to answer these or some questions that you might have regarding your observations of potatoes. What happens if you place some sugar or salt on a slice of potato just sitting on a plate?
Back to the original question. Osmosis without liquid on both sides. Go buy a box of Raisins and some Thompson seedless grapes or some prunes and some plums and ponder them. Remember Chemistry and Physics are what makes the world go round so observe your surroundings carefully and find the answers to thought out questions. After you have figgered it out enjoy the snacks; invite a fellow student over and discuss osmosis.


text:
Reactivity of 1-bromo-4-tert-butylcyclohexane isomers with methanol

(1s,4s)-1-bromo-4-tert-butylcyclohexane (1) and (1r,4r)-1-bromo-4-tert-butylcyclohexane (2) are heated (E2) with $\ce{NaOCH3}/\ce{CH3OH}.$ Which isomer will react faster?

What will change when $\ce{CH3OH}$ without the strong base is used?

I drew chair conformation of both with $\ce{Br}$ in axial position, so it can undergo E2. In 1 I got $\ce{^tBu}$ in axial position, in 2 I got $\ce{^tBu}$ in equatorial position.
Would it be correct to say that because 2 is more stable due to bulky group in equatorial position, it would react slower in E2?
For the second part of the question I suggested SN1. But then how would I know which would be faster in this case, since the $\ce{Br}$ does not have to be axial?

Which group has the greater A-value? Is it equatorial or axial? What are the geometrical requirements for an E2 elimination? Now you have the answer.

You are right on the part of E2, 2 will react slower. The reactivity of SN1 follows the same order - the carbocations are same but in 1 due to $\ce{Br}$ being on the axial position, the $\ce{C-Br}$ bond will dissociate easily to form the carbocation and so 1 reacts faster than 2.
Hope this helps. :)

These are geometric isomers, cis-trans compounds so the RS nomenclature is not appropriate. If they were mirror images they would have identical reaction rates in an achiral system
The general idea is that a bulky group such as t-butyl is almost always equatorial. The 1,4-trans has both groups equatorial so is most likely the lower energy reactant [higher transition E] by a SN1-E1 mechanism since the cis and trans carbonium ion are the same. It is a bit different for E2 since the cis is set up for the lower energy anti transition state and is the higher energy molecule. What I am having trouble visualizing is the reaction of the carbonium ion in an SN1 reaction and the cis-trans product ratio in the ether. Models might help here.

The CIP rules assign lower case r/s descriptors to these two achiral, stereoisomers. See examples here.

I didn't know this and after checking your link am thoroughly confused. I learned this years ago when the rules were in their infancy and obviously didn't learn enough. I fail to see why this rates a down vote perhaps a comment. I think my answer should help the questioner evaluate the energetics of the reaction of the two isomers.

It wasn't my down vote. CIP rules have evolved over time. I think the link is quite explicit as to how the assignments are made. This link goes through the CIP rules step-by-step. Your explanation for the E2 elimination, cis faster than trans, is fine.


text:
Why is silver acetate insoluble in water?
As seen in this video, $\ce{AgNO3}$ reacted with sodium acetate $(\ce{CH3CO2Na})$ to form silver acetate $(\ce{CH3CO2Ag})$.
Why is that $\ce{CH3CO2Na}$ is soluble but $\ce{CH3CO2Ag}$ forms precipitate?
Shouldn't both be polar and thus dissolve in water?

Silver acetate is soluble - just not VERY: ~1-2 g/100mL H2O.

There are lots of cases like this, e.g., calcium fluoride is insoluble while calcium chloride is highly soluble. The silver salts are the reverse. Simple explanations do not always work and the reasons why they don’t can be subtle, deep or both. Welcome to Chemistry!

thanks for the clarification, textbooks are really confusion on this subject.

Silver acetate is moderately soluble in water. // BTW, if you got your answer, feel free to self answer it.

Polarity is not guarantee of solubility. Many very polar substances are very insoluble in water.

Probably for the same reason that AgCl is insoluble compared to NaCl and KCl. Trying to research this on the web gives all sorts of explanations about differences in lattice and solvation energies with no reasons why. My explanation is that the relatively exposed valence orbital because of the filled d orbitals inefficient shielding, the ionic bond has some charge transfer feedback increasing the lattice energy. This loosely explains the decreasing solubility and increasing color of AgBr and AgI and the relative solubilities of the PbCl2, PbBr2 and PbI2.


text:
Reduction of unsymmetrical ketones using BINAL-H (Enantioselectivity)
I am struggling with a problem regarding the enantioselectivity of BINAL-H for quite some time now.

The reaction scheme shows the reaction with (S)-BINAL-H as it is observed plus the transition state (green) preferred to avoid repulsion of lone-pair of the oxygen atom at the aluminum and the alkyne function. However I found an alternative transition state (red) that yields the opposite enantiomer. What am I doing wrong?

On paper drawing, you may think you have a different transition state (TS). But in reality, both TSs you have drawn are the same. Try to get both views using models. You'd realize then.

To understand the concept better, read this original article.

Thanks a lot, Mathew Mahindaratne. Due to the oxygen in my modeling kit being maximally two-valent I totally overlooked the effect of its substituent. Case closed!
BUT: The transition states are not the same. Its even shown in the article that they are in fact different. However the red one is not realized due to steric hindrance between O-Et and the binaphtyl.

True to your statement. Of cause, I overlooked the two transition states. Actually, the red one is derived from R-BINAL.-H.  Look at stereochemistry at Al atom (H on Green is going into the plane of paper and H on Red is coming iout of the plane of paper). Thus, Green one is (S) and Red one is (R).

@user55119: I agree with you. But my comment is just to clear OP's misinterpretation.

@Mathew: I took my comment down as soon as I saw your response. I'll revise and repost so your response has an antecedent. The question arises, why isn't the alkynyl group in the TS axial having a smaller A-value than the methyl. Switch the methyl and alkynyl group in the red structure.  Both reagents (green and red) are S-(P)-BINOLs having a two fold axis of symmetry. After all, these TS's are rationalizations. See BINOLs here.

I think our readers need a good explanation for this question based on user55119's comment "The question arises, why isn't the alkynyl group in the TS axial having a smaller A-value than the methyl. Switch the methyl and alkynyl group in the red structure. Both reagents (green and red) are S-(P)-BINOLs having a two fold axis of symmetry. After all, these TSs are rationalizations."
OP's original question was:

I found an alternative transition state (red) that yields the opposite enantiomer. What am I doing wrong?

First, I need to clarify that the reducing reagent we are talking about is $\ce{BINAL-OEt}$ $(\ce{LiAlH(Binol)OEt})$, which gives pretty high enantioselectivity (than when used $\ce{LiAlH2(Binol)}$) for reduction of ketones that have a $\pi$-system directly attached on one side of the carbonyl group.
The chiral $\ce{BINAL-OEt}$ can be prepared by using enanthiomerically pure BINOL $(\ce{Binap(OH)2^{*}})$. The stereoisomer considering here is 2,2'-dihydroxy 1,1'-binaphthyl (BINOL), which shows atropisomerism (Ref.1) defined by $M$ and $P$ notations (alternatively, $R$ and $S$ notations, respectively). Atropisomers are stereoisomers arising because of hindered rotation about a single bond:

For BINOL, $\ce{A = A' = OH}$ (the highest priority group) and $\ce{B = B' = benzo}$ function (the lowest priority group). Thus, stereochemistry of BINOL group of both Red and Green transition states are $P$ (or $S$).
It is known fact that these atropisomers can transforem chirality to substrates it attached to. For example, the aluminum atom in $\ce{BINAL-OEt}$ is a chiral center based on the use of $M$- or $P$-BINOL. Thus, when enanthiomerically pure $P$-BINOL is reacted with equimolar amounts of $\ce{LiAlH4}$ and ethanol would result a diastereomeric mixture:
$$\ce{LiAlH4 + Binap(OH)2^{*} + EtOH -> LiAl^sHBinap(O2)^{*}OEt + LiAl^rHBinap(O2)^{*}OEt}$$
As I'd explain later in the text (vide infra), each of these two diastereomers would give opposite enanthiomer (when used in pure isomer form) in the reduction reaction in hand. Based in these information, as I pointed out in my comment (and as OP agrees on), the two Green and Red Zimmerman-Traxler Chair-Like Transition States (Ref.2) are not the same:

If you look closely, you would realize that the $\ce{Al}$ stereocenter in each of two Green and Red Transition States are not not identical (c.f., the orentation of napthyhyl ring the bottom box, which is identical in both Green and Red TSs). I assigned the on in Green TS as $s$ while that in Red TS as $r$. To my knowledge, these two TSs are arisen from the reaction of same starting material with each of diastereomers (the optical active reducing reagents) I mensioned earlier:
$$\ce{LiAl^sHBinap(O2)^{*}OEt + LiAl^rHBinap(O2)^{*}OEt}$$
But, it is not possible becase only one of these two diastereomers should be using in the reaction. Therefore, one of these two transition states should be negated.
About user55119's comment: Zimmerman and Traxler have suggested these Chair-Like Transition States to explain the resulting stereochemistry of a reaction, which are as suggested by user55119, in deed rationalizations.  Besides, having the alkynyl group in the axial position in the TS is not just because steric reason (even if having a smaller A-value than the methyl). There are other factors such as $n,\pi$-repulsion forces (Ref.1, and Ref.3-4):

As shown in the above diagram, favorism to the accepted TS is govenrned by a repulsive interaction between the $\pi$ system of the ketone (here it is alkynyl group) and the lone pair of an oxygen atom of the $\ce{BINAL-OEt}$.

References:

Jean Michel Brunel, "BINOL:  A Versatile Chiral Reagent," Chem. Rev. 2005, 105(3), 857–898 (DOI: https://doi.org/10.1021/cr040079g).
Howard E. Zimmerman and Marjorie D. Traxler, "The Stereochemistry of the Ivanov and Reformatsky Reactions. I," J. Am. Chem. Soc. 1957, 79(8), 1920–1923 (DOI: https://doi.org/10.1021/ja01565a041).
R. Noyori, I. Tomino, Y. Tanimoto, and M. Nishizawa, "Asymmetric synthesis via axially dissymmetric molecules. 6. Rational designing of efficient chiral reducing agents. Highly enantioselective reduction of aromatic ketones by binaphthol-modified lithium aluminum hydride reagents," J. Am. Chem. Soc. 1984, 106(22),  6709–6716 (DOI: https://doi.org/10.1021/ja00334a041).
R. Noyori, "Centenary Lecture. Chemical multiplication of chirality: science and applications," Chem. Soc. Rev. 1989, 18, 187-208 (DOI: https://doi.org/10.1039/CS9891800187).

Thank you for the effort on this detailed answer. You are right on the electronic repulsion and the favorism that derives regarding axial and equatorial position of the alkyne function. However I want to focus on what you mentioned above. There are no diastereomers of the reducing agent. It is C2-symmetrical. Rotation around the axis yields same position of BINOL-Substituent, but switches the positions of OEt and H in the plane. Therefore both TS, green and red could exist. However the steric hindrance of OEt and Naphthylring disfavors the red one. See TS 9&10 here: doi.org/10.1021/ja00334a041

In short: The transition states are diastereomeric, but the reducing agent itself is not.

@marsem: I stand by my analogy in this case. If you look at the ring systems of BINOL residue, configurations of Al-O of front naphthyl (dark filling) and back naphthyl (no filling) stays same on both Green and Red TSs. Thus, if you switch the OEt and H groups on Al, you get diastereomers. Yet, TS 9 and TS 10 have different configuration on front and back naphthyl groups (O-Al is axial on front naphthyl in TS 9 and it is axial on back naphthyl in TS 10) thus, switching of OEt and H is allowed without changing the stereochemistry of whole system.

When regarding the transition states, they are diasteromeric, as the tetrahedral aluminum is fixed in the drawn structures. But the single molecule BINAL-H itself only shows axial chirality of the BINOL-substituents. When switching OEt and H in the molecule it is the same molecule rotated by 180° (C2 symmetry). However in the transition states drawn it is then fixed and no rotation of the BINAL-H is possible, thus the TS are diastereomeric, but not the single BINAL-H molecule.

If you do not trust me, then build the models of the "two" molecules and you will realize that they are in fact superimposable with each other.


text:
Would laying an atom thick piezoelectric material near a heat source generate current?
From what I’ve read, a piezoelectric material (like quartz) generates current from squeezing the molecule lattice and deforming it to make the positively charged atoms on one side and the negatively charged atoms on the other. If that’s all it takes to create current, couldn’t heat (which is atoms vibrating) also create current by vibrating the piezoelectric material? Especially if we made it only an atom thick so that there is high surface area to volume?

Thermoelectric_effect is quite different from piezoelectric.

tldr: not quite, but there are a lot of interesting applications
You're correct that a piezoelectric material interconverts mechanical energy into electrical energy. Importantly, there's both:

direct effect: converting mechanical deformation into electrical response
converse effect: converting a change in electrical field into a mechanical response

Thus a piezoelectric like quartz can generate current from vibration. This is used in quartz watches, resonators, etc. It's even used in quartz crystal microbalances - adsorbing material onto the quartz crystal will change the resonance frequency.
Importantly, the vibrations or mechanical stresses are on top of the normal vibrations of the atoms.
Moreover, if I apply a force to a piezoelectric .. say I push down on one of my group's piezo films. We do see electrical voltage or current generated through the direct effect.
From a chemistry perspective, this is an application of Le Chatelier's principle .. we have distorted a polar material, so electrical charge forms at the surfaces:

On the other hand, this is dynamic. When the force is removed, the current will flow back. We can measure both positive and negative current in response to a dynamic force.
In other words, if I increase the vibrations of my quartz, I'm not likely to do meaningful work.
Thermoelectrics
Some piezoelectrics are also somewhat thermoelectric. That is, they can interconvert a thermal gradient into electrical voltage.
The trick is that a good thermoelectric figure of merit requires high electrical conductivity $\sigma$ (low resistivity), among other things:
$$
z T=\frac{S^{2} \sigma}{\kappa} T
$$
(S is the Seebeck coefficient and $\kappa$ is the thermal conductivity)
Conclusions and Future Work
High electrical conductivity is not conducive to a piezoelectric - if you see my diagram, the mechanical stress essentially produces a capacitor (i.e., you want to maintain that electrical field across the material.)
On the other hand, a thin piezoelectric can be good - capacitance is inversely proportional to thickness.
Single-atom thick layers can only be piezo-active in the plane, not across it. But we've made some really thin monolayer piezoelectrics:
"Intrinsically Polar Piezoelectric Self-Assembled Oligopeptide Monolayers"](https://doi.org/10.1002/adma.202007486)
As for thermoelectrics, you can find some good ones, but not quartz. You need something with reasonable electrical conductivity.

very recently there have been claims that energy can be harvested from the thermal motion of graphene layers. https://www.sciencealert.com/physicists-build-a-circuit-from-graphene-that-generates-clean-limitless-power

Yes .. it's hard to classify that as piezoelectric, since there's no polar aspect, and I'm not sure it's thermoelectric either. Some nanoscale behavior defies simple categories. But the original question isn't so far off the mark - it's possible to get some energy harvesting through nanoscale films.


text:
Why isn't the Leuckart reaction used to make (meth)amphetamine?
For example, in Breaking Bad, they synthesize meth from phenylacetone.
Why not just use methylammonium formate, if they already have methylamine?

@OscarLanzi There's many methods and both used in the show are fine.

@OscarLanzi iirc they used aluminum with the methylamine (and presumably HgCl2?)

https://en.wikipedia.org/wiki/Reductive_amination#In_popular_culture