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Mysterious precipitate in reaction of copper sulfate and calcium hypochlorite
So I can’t seem to figure this reaction out. I’m pretty new to chemistry and am still learning it in high school so please don’t think I’m stupid. I have reacted 2 solitons (copper sulfate) and ( calcium hypochloride / dry chlorine) and got a light blue  precipitate that resembles copper carbonate. This precipitate is what I cant seem to identify. I don’t think it’s copper chlorate because the precipate is too light blue. Any answers will be helpful thank you in advance.

Mixing two solutions at random is generally a bad way to learn chemistry. Copper chlorate is pretty well soluble, so it would never precipitate. I'd say it is copper hydroxide or some basic salt.

It would be a mixture of calcium sulphate and basic copper salt or even copper hydroxide, as "calcium hypochlorite" is quite basic, being complex mixture of calcium chloride, hypochlorite and hydroxide.I guess similar mixture with copper sulphate is used as a classical antifungi protection of Vitis vinifera, the common grape vine.

Your precipitate has nothing mysterious. It is simply a mixture of calcium sulfate CaSO4 and copper hydroxide Cu(OH)2, made by the following reaction : $$\ce{Ca(OH)2  +  CuSO4-> CaSO4(s) + Cu(OH)2(s)}$$The Ca(OH)2 is coming from the Calcium hypochlorite, because Calcium hypochlorite is a technical product containing much Ca(OH)2. CaSO4 is a white precipitate, and Cu(OH)2 is a blue precipitate, that gets dark at 100°C, because it is decomposed by the following reaction : $$\ce{Cu(OH)2 -> CuO  +  H2O}$$ and the copper oxide CuO is black. CaSO4 is known to be found in the mineral gypsum.

The creation reaction for copper hypochlorite per Mellor:
$$
CuSO4+Ca(OCl)2 → CaSO4(s)+Cu(OCl)2
$$
However, copper hypochlorite is not stable and on standing breaks down. I believe Mellor is likely accurate on describing the reaction and products:
Reference: "A comprehensive treatise on inorganic and theoretical chemistry", Volume 2, by Joseph William Mellor, page 271. To quote:
"R. Chenevix notes the ready solubility of cupric oxide in chlorine water, and P. Grouvelle found that the soln. obtained by passing chlorine into water with cupric oxide in suspension possessed bleaching properties, and these were retained even after the soln. had been boiled for a quarter of an hour. A. J. Balard found that the distillation of P. Grouvelle's liquor furnished some hypochlorous acid and a green oxychloride, 3CuO.CuCl2.4H20, was formed in the retort. A. J. Balard prepared a soln. of cupric hypochlorite by dissolving cupric hydroxide in hypochlorous acid. It is also made by the action of cupric sulphate on calcium hypochlorite. A. J. Balard found that copper filings are partially dissolved by hypochlorous acid, the soln. after standing some time contains cupric chloride, and deposits a green pulverulent cupric oxychloride."
Link: https://books.google.com/books?pg=PA271&lpg=PA271&dq=zinc+hypochlorite&id=7XoGAQAAIAAJ&ots=R0byHnamKL#v=onepage&q=zinc%20hypochlorite&f=false
So, the cupric oxychloride product maybe mixed with CaSO4.
Note, Copper oxychloride, per Wikipedia, is now referred to as dicopper chloride trihydroxide with the formula Cu2(OH)3Cl (or, more informatively, I would write as Cu(OH)2Cu(OH)Cl).
[I have personally prepared dibasic magnesium hypochlorite on another forum by mixing MgSO4 with NaOCl (aq) and adding NaOH. A white precipitate of dibasic magnesium hypochlorite results (which I have collected and experimented with). It is also NOT Mg(OH)2. Link: https://www.sciencemadness.org/whisper/viewthread.php?tid=26741#pid303723 There is also the reputed compound 'copper carbonate' which actually is also always basic copper carbonate per Wikipedia https://en.wikipedia.org/wiki/Basic_copper_carbonate ].

Mellor's treatise is amazing but it is slightly outdated. Certainly it is not simple Cu(OH2) but far more complex solid.


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How would I separate a solution of sodium iodide and water in order to obtain the sodium
I am an amateur in this field of science and I am trying to separate a solution of Sodium Iodide and water. How would I separate the solution so I can obtain sodium?

You can't and shouldn't obtain sodium.

You could use electrolysis with a mercury cathode, taking advantage of its huge hydrogen over potential that makes sodium reduction kinetically favored.  The sodium is amalgamated into the mercury.
Now all you need is to separate the sodium.  Unfortunately, just evaporating the mercury would (1) probably get you into the news headlines when the toxic emissions are identified and (2) leave you with an intermetallic compound (see the phase diagram below from this reference).

Older chlor-alkali cell technologies used this method to make sodium amalgam, but not to refine the sodium.  Instead the amalgam is catalytically reacted with water in a separate chamber to retrieve sodium hydroxide without contacting the anodic products from the electrolytic cell.

You cannot separate sodium from a solution of sodium iodide NaI. If you try to do it by electrolysis, the sodium produced at the cathode will immediately react with water with the following reaction : $$\ce{2 Na + 2 H2O ->  2 NaOH  +  H2O}$$ and you will just obtain some sodium hydroxide NaOH in your sodium iodide solution

don't criticize me i might be wrong in what I say but, what if you were to evaporate the solution to leave behind pure sodium iodide. after doing that is there a way to separate it?

Nobody criticizes. To separate the sodium Na  in sodium iodide NaI, or in sodium chloride NaCl, there is only one way : electrolysis. Electrolysis can be done in solution with a cathode of mercury. Or it can be done with liquid NaI, i.e. at a temperature higher than the melting point of NaI, which is 651°C. It may be done, but it is not really easy to do.

Some more details. At 651°C, sodium will burn in air. So it must be done in an argon atmosphere. And don't forget that simultaneously, an intense vapor of iodine will be produced on the other electrode. For 1 g Na, you will also produce 5.5 g of iodine vapor, which is corrosive to sodium and other metals.

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Are the reaction rates of these equations equal?
$$\ce{CH3CH2CH2Br + OH- -> CH3CH2CH2OH + Br-}$$
$$\mathrm{rate} = k[\ce{CH3CH2CH2Br}][\ce{OH-}]$$
If I change the $\ce{Br}$ with any element from halogens (without changing concentration, volume, mass etc.), does the rate of reaction change? If so, how does it change? What factors change it?

Why should the rates be the same ? Even rates for molecules with different isotopes of the same holegen are slightly different.

@Poutnik Could you please explain why they must be different with examples? What makes them change? It seems like they will be the same.

I will let you think about it for a while.... What major factors influence the kinetic rate constant ? One is kinematic, one is geometric, one is energetic.

@Poutnik All I know is that it has an equation $k = Ae^{\frac{-E_a}{RT}}$.

Start with that. And replace rather the rate by the k. Why should be A and E_a the same for all 4 reactions ?

@Poutnik k makes the difference, thanks. I would like to find a relation between halogens according to their reaction rates with the same substances, I do not know if there is or not. Can this be investigated?

Well, that is a different, and rather a broad question.

For the reaction kinetic rate constants, there is the well known Arrhenius equation:
$$k=A \cdot \exp{\left(-\frac{E_\mathrm{a}}{RT}\right)}$$
$A$ is sometimes called frequency factor, interpreted as the rate of collisions with the proper orientation of molecules.
It has 2 terms:

The rate of general collisions, that is function of temperature and molecular masses, which determine speed of molecular motion (close relation to the kinetic theory of gases). Note that the temperature dependence is much smaller than for the exponential Boltzmann term.
The probability of the proper orientation of molecules, what depends on the molecular geometry. For molecules of  otherwise the same geometry, it depends on covalent atom radii.

The exponential term follows the Boltzmann statistical distribution, determining the probability molecules would have enough kinetic energy to overcome the reaction activation energy barrier.
All 3 terms (2 for $A$ and the exponential term) Are different for different halogen atoms.

The mass molecules increases fluorine < iodine, so collision frequency as the rate constant term is the lowest for iodine.
As the reaction mechanism I suppose SN2. The geometrical aspects of $\ce{-CH2X}$ is tricky to determine from basic principles. Bigger halogen is more sterically blocking, but is also farther from the central carbon. More polar bond of the smaller halogen should cause stronger  repelling of the other 3 bonds, so the other side is more open for sn2 reaction.
Activation energy would decrease ( and the Boltzmann term for the rate constant increase) in order F ... I

That about the principles. To compare the particular rates, it is matter of experimental data.

Poutnik's explanation is perfectly correct. But it does not answer Mathrix's question.,

Well, if you Taylor concentrations in such a way the rates are the same, the will be the same. Otherwise, they will be for the same  concentrations different for reasons above. The same for different concentrations. All will add comments for oarciularveffects for the halogen serie, But I reserve the right to be partially wrong, as I am not an organic chemist. :-)

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Does clove oil/eugenol interact with nylon type plastics?
So I am aware that the eugenol in clove oil readily melts #6 type plastic (polystyrene). Polypropylene has a higher resistance, while HDPE type plastics seem to have the highest resistance.
My main question is whether the clove oil will interact with nylon. Say, the bristles of a toothbrush.

I would search for plastics/solvent resistence table, looking for aromatic hydrocarbons and their derivates. If eugenol is not used in pure form, it would be strongly affected by the composition of mixture.

Nylon 612, primarily used for filaments and toothbrushes, is resistant to clove oil at 20 °C [1, p. 2034]:

Nylon 12, used for coatings, is also resistant to clove oil at 20 °C [1, p. 1829]:

The tech data on nylon at newmantools.com also suggests there is no effect of clove oil on nylon in general.
References

Chemical Resistance of Thermoplastics; Woishnis, W. A., Ebnesajjad, S., Eds.; PDL handbook series; William Andrew: Norwich, N.Y, 2012. ISBN 978-1-4557-7896-6.


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Mechanism involved in bromine water test

$$\ce{CH3CH=CH2 ->[Br2] CH3-CHBr-CH2Br}$$

In the reaction mechanism, why does the nucleophile attack at the 2nd carbon atom? Shouldn't it attack at the 1st one, so that in the transition state, the carbanion is relatively more stable?

Which site is the more stable carbocation in the transition state?

There is no carbanion in this reaction mechanism.

There are no carbanion in this reaction.
In bromine water, $\ce{Br2}$ is partially made of $\ce{Br+}$ and $\ce{Br-}$ ions. Then $\ce{Br+}$ attacks the double bond, producing a carbocation, with $\ce{Br}$ at the outer $\ce{C}$ atom, and the + charge on the central $\ce{C}$ atom. This is the most stable carbocation. Then it will react with the $\ce{Br-}$ ion which will be attached on the central $\ce{C}$ atom.
It also should be mentioned that a secondary reaction may happen where $\ce{H2O}$ reacts with the transition state carbocation. As a result, the molecule $\ce{CH3-CHOH-CH2Br}$ is produced.

Your explanaion is okay but mechanism is wrong. Did you hear about cyclic bromonium ion mechanism?

Of course I know the bromonium ion mechanism. I have not mentioned it, because I think it might have been avoided here.


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5-methylhexa-1,3-diene reaction with HBr
What will be the major product of the reaction?
I guess that rearrangement will take place and give 5-bromo-5-methylhex-2-ene as a product, but I'm completely not sure

Why do you think this will be the product?

In this reaction if peroxide is not present then reaction follows markovnikov addition in which carbonation is formed as intermediate. By stability due to resonance and 2° carbonation
5-methylhexa-1-ene cation with C+ at 3 is formed. Hence 3-bromo-5-methylhexa-1-ene is formed as product.


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Electronegativity of heavier elements of Group 15
While reading about p-block I got to know that in Group 15 elements

electronegativity value decrease down the group but amongst the heavier elements difference is not that much pronounced.

I understood that electronegativity decrease down the group as the atomic size increase but I don't understand why in heavier elements electronegativity difference is not significant.
What is the reason for this?

electronegativity value decrease down the group but amongst the heavier elements difference is not that much pronounced.

The reason for this is that in heavier elements much more filling of inner d and f orbitals starts. For example as we move down from Sb to Bi, just 6 (3+3) more electrons fill p sub-shell, and 2 electrons enter to s sub-shell. Whereas on the other hand, 24 (10+14) electrons make their way to d and f sub-shell. This results in much poor shielding of electrons in outermost shell and effective nuclear charge thus remains approximately same in heavier elements. So, they have nearly same value of electronegativity on electronegativity scales.

Thanks for the answer.  I was also thinking something like this.

For period 6 there is an additional effect: because the 1s electrons are moving so quickly, they experience relativistic effects, causing the atomic orbitals to become contracted and lower in energy. This increases period 6 elements' electronegativity.


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Source of highly nucleophilic iodide
What is be the best commercially available source of highly nucleophilic iodide in non-polar solvent? Solvents used are cyclohexane, benzene, chlorobenzene.
I can get ahold of bromide quaternary ammonium salts at best. So I'm looking for equal substituent of Bu4NI. Any ideas?

How about cesium iodide? It's soluble in acetone. But no data for less polar solvents.

Benzyltributylammonium iodide is pretty good

@waylander "I can get ahold of bromide quaternary ammonium salts at best".  Is your suggestion commercially available?

@OscarLanzi  https://www.sigmaaldrich.com/catalog/product/aldrich/293016?lang=en&region=GB

A musing, since we had for long a bottle of a solution of 18-crown-6 KCN in the hood only the boss used from time to time, what about the analogue with KI (a bit more than just adding 18-crown-6 + KI + THF, as seen in Acta Cryst 2003, E59,  m1184-m1186 (http://scripts.iucr.org/cgi-bin/paper?S160053680302676X) Aldrich offers some 18C6 with apolar groups attached, eg. dicyclohexane, or dibenzo, etc.


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What is the difference between stoichiometric coefficients and number of moles?
Please clear my doubt,
Thanks!

The stoichiometric coefficients are defined once and for all for a given reaction. They are integers. It is 2 for H2 and 1 for O2 in the reaction 2H2+O2->2H2O
The number of moles depends only on the particular experiment. It may be any number, integer or not.


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Why is the composition of a liquified vapor the same as that of the vapor?
In Organic Chemistry Lab Techniques (Nichols), I found this diagram.

I understand that when going from Point $a$ to Point $b$, the composition becomes richer in the more volatile component (component A).
However, why does condensation of the vapor at Point $b$ not change the composition? Shouldn't the condensate of a vapor be richer in the less volatile component (the reverse of vaporization), by the principle of microscopic reversibility?

There is a misconception in using this diagram. You cannot go from Point a to Point b This diagram should not be read as you do. It should be done this way.
Suppose you start from a liquid made of 25% A + 75% B. It is a mixture. Its representative point on the diagram is on the edge of the rectangle between "Composition" and "100% B". When heating this mixture, the corresponding point moves upwards, but stays liquid. When reaching the curve "Liquid", the first bubble of gas is formed. We are at the point a (red) But this bubble has another composition. The composition of the gas corresponds to b (in red). So it is practically pure A. This is the beginning of the vaporization phenomena.
If you continue heating the liquid, more gas will be produced. But, as A is more eliminated than B by vaporization, the remaining liquid looses some A and practically no B. So in percentage, the liquid will contain more B. Its composition will soon be 76% B and 24% A, then 77% B and 23 % A, etc. But the temperature has to be elevated to allow this richer mixture to vaporize. The representative point of the liquid will move but stay on the sloping curve. After some time, the point of the liquid will be just above the capital L from the word "Liquid" At this moment the gas is mainly A but it starts containing 1 or 2% B. The composition of the gas could be determined by drawing a horizontal line from the point above the letter "L". This horizontal line will cut the curve "Gas" at an abscissa equal to about 1% or 2%. This defines the composition of the gas.
If you continue heating, the liquid will contain more and more B. Its point will stay on the curve "Liquid". The temperature must be higher and higher. The amount of liquid decreases. Simultaneously the composition of the gas which is eliminated contains more and more B. Its composition is always determined by drawing a horizontal line from the liquid to the curve on the left-side of the diagram.
The last drop of liquid disappears when the liquid reaches the Point "bp B".
Hopefully you have understood that passing from point a (in red) to point b (in red) has no meaning. a and b are always present : one describes the composition of the liquid, and one the composition of the gas.

You seemed to have danced all around the fact that $a$ and $b$ are in equilibrium at a particular temperature. That point should have been made first.


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Hybrid orbitals, photoionization and time dependence
Wikipedia, in the article on Orbital Hybridisation, states the following:

One misconception concerning orbital hybridisation is that it incorrectly predicts the ultraviolet photoelectron spectra of many molecules. While this is true if Koopmans' theorem is applied to localised hybrids, quantum mechanics requires that the (in this case ionised) wavefunction obey the symmetry of the molecule which implies resonance in valence bond theory.

(Source; emphasis mine)
I am wondering about the bolded part. I interpret it as meaning that the wavefunction is an eigenfunction to both the Hamiltonian and to the symmetry operators of the molecule's point group. To me, the bolded part then seems equivalent to stating that the molecule emerges from the ionization process in one of the eigenstates of the ionized system, and I can understand how the symmetry requirement would hold true in that case.
However, my intuition wants to think about the ionization process in such a way that the departing electron leaves a "hole" somewhere in the wavefunction, which then evolves over time as a superposition state; i.e., the system does not emerge in an eigenstate, and the "hole" left in the wavefunction will "travel" through the molecule in some way. I am under the impression that the time-dependent wavefunction at any point in time could in principle break the symmetry imposed by the molecular point group.
So this leads me to two questions:

Am I wrong in thinking that superposition states do not necessarily have to obey the stated requirement about molecular symmetry? And
is my picture of a time-dependent molecular wavefunction immediately after ionization viable?

Update: I am of course aware of the argument that we can make unitary transformations of the overall wavefunction as we please, as it leaves the total electron density distribution and energy unchanged; this is also how hybrid orbitals are formed when starting from an MO picture. However, because the orbitals involved in hybridization belong to different energy eigenvalues, this creates a time-dependent superposition state with no defined energy. In extension, hybridization silently implies that the MO energies themselves are not observables that would describe the initial states in the ionization process. And, in addition, the hybrid orbitals do not necessarily form irreducible representations of the underlying molecular point group anymore, and do thus not obey the bolded symmetry requirement in the quote above.
So here are two questions which are quite closely related to the two I had originally formulated above:

What is the justification for the assumption that the MO energies are not observables? And
how is it justified from the perspective of the Wikipedia article that we have to obey the symmetry of the molecule in the ionized product state, but for the hybrid orbitals in the ground state, we can for some reason disobey the same principle? Isn't this highly inconsistent?

This paper might be helpful: https://www.researchgate.net/publication/279990609_Structural_evolution_of_the_methane_cation_in_subfemtosecond_photodynamics

There are really two parts to your question - one is to interpret the wikipedia entry, the other is the more general question about the transition of methane from the parent molecule to the ionized $\ce{CH4+}$.
For the first question, I believe the author has muddled the explanation a bit (though it's of course just as likely that I'm the one who is muddled). It sounds to me like they are explaining the reconciliation of molecular orbital (MO) theory and valence bond (VB) theory for $\ce{CH4}$ rather than for the cation or the ionization process itself. Here is what I think they are trying to say:
Electronic structure of $\ce{CH4}$

MO theory
If we carry out an analysis of the delocalized molecular orbitals of methane, we find that the four highest energy occupied orbitals are spread between two energy levels, with one orbital at the lower energy level and the other three at the higher level. The orbitals meet the quantum mechanical requirements of being orthonormal and matching the symmetry operations of the molecule. The energy levels are eigenvalues of the wave function energy operator, so we'll call them eigenstates.
This representation is supported by the experimental data from PES, which shows two ionization energies, with a relative intensity of 3:1.
VB theory
In contrast, the valence bond view of the molecule has four orbitals of equal energy, each one corresponding to a specific C-H bond. These orbitals can be constructed as linear combinations of the MO theory orbitals and are orthonormal to each other. As such, they form an alternate basis set for the same space as described by the MO theory orbitals (that is, any linear combination of the MO theory orbitals can be described by a linear combination of the VB theory orbitals). However, they do not match the symmetry of the molecule, and their energy levels are not eigenvalues for the energy operator.
This representation seems to be in disagreement with the experimental evidence, since it seems to suggest that only one ionization energy should be observed. However, it is not incorrect, as I will explain.
Reconciliation
The reconciliation between the VB description and the experimental result comes from realizing that "a linear combination of MO theory orbitals" is just a different way of describing a superposition of states. All quantum systems exist as a superposition of states. In the mathematical edifice that is quantum mechanics, we require that when we make a measurement of an observable of a superposition of states, the result must still be an eigenvalue for one of the states in the superposition. The probability of a specific eigenvalue appearing is a function of the contribution of the corresponding state to the superposition.
The simplest example of this is measuring the spin of an electron. Unless a constraint is imposed, we expect to measure ($+\frac12$) 50 per cent of the time and ($-\frac12$) 50 per cent of the time. In neither case do we conclude that the electron was fixed in that spin state prior to the measurement. It was in a superposition of both spins until we made the measurement.
Getting back to the methane molecule, we say that all 8 electrons in the valence orbitals can be viewed as existing in superpositions of states, with the only requirement being that within each individual molecule, the electrons are in states that are orthonormal to each other. The VB and MO theory representations are just two of an infinite number of possibilities. When we do the PES experiment, the results must always be eigenvalues, and the intensities will always have a 3:1 ratio, because the average contribution of the eigenstates to the population of orbitals across the population of methane molecules must remain 3:1.

Ionization of $\ce{CH4}$
With the above in mind, we can move on to the harder general question about time evolution of states during and after ionization. Since the measurement in the PES experiment returns an eigenvalue, it would seem that it must force the molecule into an eigenstate, just as measuring the spin of a free electron forces it into a single spin state.
We know that initially the molecule retains the tetrahedral geometry of $\ce{CH4}$, but within a short time it relaxes to a $C_{2v}$ symmetry. This change in symmetry must be accompanied by a change in the wave function and the energies (eigenvalues) of the states. I would assume that this time-evolution of the wave function also allows for scrambling of the eigenstates back to a superposition of states, but I don't know that process well enough to give a clear description. Perhaps someone else can jump in with that.
Conclusion
Referring to your specific questions -

1.Am I wrong in thinking that superposition states do not necessarily have to obey the stated requirement about molecular symmetry?

As I described, I would say you are correct. Superpositions do not need to meet the symmetry requirement. The only requirement is that they are linear combinations of the symmetry-consistent basis set and that as a group they form an equivalent orthonormal basis set.

And is my picture of a time-dependent molecular wavefunction immediately after ionization viable?

I think so, but I'm not certain.

What is the justification for the assumption that the MO energies are not observables?

I don't understand this question. Energy is definitely an observable, and the PES experiment measures it. An observable is anything that can be measured.
UPDATE: Rereading your updated question, I think the confusion is between "not having a fixed value" and "not being an observable". Just because the electrons do not have defined energies does not mean that their energies are not observables. In chemistry, we use language that refers to fixed values of observables like energy, but a fundamental concept of QM is that observables do not have fixed values until they are measured. The measurement process itself collapses the wave function to a specific eigenstate. I tried to clarify that with the example of the electron spin. Position is another common example. Ideally, we would preface lots of statements with "if you were to measure x. . " but that would be really cumbersome.

And how is it justified from the perspective of the Wikipedia article that we have to obey the symmetry of the molecule in the ionized product state, but for the hybrid orbitals in the ground state, we can for some reason disobey the same principle? Isn't this highly inconsistent?

As I tried to explain, I believe the Wikipedia article was mixing up the parent molecule and the ionized state. The immediate effect of the ionization (ie measurement) is to force the molecule into an energy eigenstate. However, that is an eigenstate of the immediate product, which still resembles the methane parent in symmetry and orbitals. There is a time evolution to the ground state of the cation, which has a different symmetry ($C_{2v}$) and different eigenstates. Since we have performed no measurements on this molecule, it is not fixed in any of those eigenstates. Since it has evolved away from the initial $T_d$ symmetry, it is also no longer an eigenstate of the parent.
To summarize, the only point in the process in which the electrons are in a fixed eigenstate is immediately after ionization (before relaxation), and this eigenstate is described by the set of symmetry-allowed MO theory orbitals that correspond to the $T_d$ symmetry of $\ce{CH4}$, not the $C_{2v}$ symmetry of $\ce{CH4+}$.

Thank you for your answer! I have updated my original question with two more questions which basically tackle the same problem from slightly different angles. If you could perhaps comment on those perspectives, it would be much appreciated.

There is also something in your answer that I struggle with: You claim that the PES experiment must end in an eigenstate of the ionized system. Why? Then at the end, you judge my perspective on a time-dependent product state as possibly correct, but that directly implies that the ionization does not immediately end up in an eigenstate, so I'm wondering if your answer contradicts itself. In either case, my updated question 4 becomes relevant: If the process must end in an eigenstate, isn't it inconsistent to claim that it can start from hybrid orbitals, i.e. non-eigenstates?

I'll update to try to address new questions. To answer your comment, I did not claim that the PES ends in an eigenstate of the ionized system. It is instead an eigenstate of the parent CH4 molecule. This is a requirement of any quantum mechanical measurement. Not only must the result be an eigenvalue, but the measurement itself forces the system into the eigenstate corresponding to the measurement output. Simple example is electron spin - once measures as +1/2, the electron remains in +1/2 spin state until another event occurs which scrambles it back to a superposition.

Thanks for your updates. I'll have to go other things in my head a bit, I feel that the issue is still not settled for me somehow. Would you be open to perhaps moving the discussion to a chat sometime later?

Sure. Feel free to open a chat whenever you want.

I've made a chat room here: https://chat.stackexchange.com/rooms/info/102449


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Triphenylphosphine (PPh3) vs. 1,2-bis(diphenylphosphino)ethane (dppe) ligand?
What about $\ce{NiCl2(dppe)}$ makes it a more efficient catalyst for Suzuki cross-coupling as opposed to $\ce{NiCl2(PPh3)2}$ or $\ce{NiCl2(PCy3)2}$? How does the dppe ligand increase the reactivity of the catalyst?
Literature claims:
The experiment I am following is from J. Chem. Educ. 2017, 94(6), 786-789 (https://doi.org/10.1021/acs.jchemed.6b00273). The supporting information contains class data which says that the turnover number using a dppe ligand is roughly 4 times that of $\ce{PCy3}$ or $\ce{PPh3}$. The paper suggests that increasing electron ability increases catalytic activity, so what about electron donating ability changes catalytic activity and what makes dppe more electron donating?

You ask this question: What about $\ce{NiCl2(dppe)}$ makes it a more efficient catalyst for Suzuki cross-coupling as opposed to $\ce{NiCl2(PPh3)2}$ or $\ce{NiCl2(PCy3)2}$? However, you didn't gives any literature support to show this is, in deed, a true claim. Can you provide any site mentioning this so-called fact?

Apologies for not including a reference. The experiment I am following is from J. Chem. Educ., 94 (2017), 786-789. The supporting information contains class data which says that the turnover number using a dppe ligand is roughly 4 times that of PCy3 or PPh3. The paper suggests that increasing electron ability increases catalytic activity, so what about electron donating ability changes catalytic activity and what makes dppe more electron donating?

Would you include these literature finding in your question?

I'm going to include these literature finding in your question. I hope you won't mind.

The simple explanation for why dppe is more electron-rich is because alkyl groups are more electron-donating than aryl groups.

As far as I am aware, the selection of the best ligands for transition-metal catalysed reactions is still trial and error. In fact, not only is the selection of ligand trial and error but so is the choice of starting materials to result in that metal–ligand combination.
What the original discoverer of a reaction method typically do is run a number of screens keeping most parameters constant but varying one; this one varying parameter will be ligand, solvent, time, temperature, additives and metal salt. (Usually, they will have decided before running the experiments which metal they want to use for their reaction so the only reason to include another one is to directly compare the new method to an existing one and e.g. show that nickel is better than palladium in this reaction.) Whichever reference you’re looking at, the authors most certainly performed such a screening and received a slightly higher yield for one set of reactants compared to another; they then published that as their final choice.
After having found the optimal conditions, many researchers are tempted to ‘explain’ why a certain ligand is better than another ligand. Our knowledge does include how electron-rich or electron-poor the donor atom of a given ligand is, how sterically demanding a set of substituents is and so forth. If ligands of one general type form a general trend, one could well be tempted to draw from that general trend and exclaim that e.g. an electron-rich phosphane works best for this reaction. In ideal cases, that prediction is then tested experimentally with a previously unused but more electron-rich phosphane; a reaction that may or may not work. The point, however, of this paragraph is that more often than not the ‘explanation’ is little more than a post-factum rationalising ballpark and should be treated as such. Years to decades of research will likely be required until we actually fully understand the intrinsic properties at play and are able to make good predictions.


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Fires with chlorine
could a fire with it's fuel consisting of mostly carbon like coal or wood use chlorine instead of oxygen. My thinking goes that when fire burns it sucks in oxygen and that bonds with carbon to make carbon dioxide. So... if you had chlorine instead of oxygen would the chlorine bond with the carbon and make Carbon tetrachloride instead of carbon dioxide? Would this work or am I missing something?

Consider that wood is mainly composed of cellulose, which is a polymer of glucose ($\ce{C6H12O6}$), so when it burns in the presence of oxygen, the reaction that occurs is basically the same as in aerobic respiration: $$\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}$$ Check out this related question: https://chemistry.stackexchange.com/questions/96139/can-glucose-burn-in-chlorine

It's «only transfer of electrons» ... said once a teacher of mine, and demonstated how sodium burnt in chlorine gas to form NaCl.

The combustion of a stream of hydrogen gas in contact with chlorine gas is a commercial path to high purity $\ce{HCl}$.
Fires will continue to burn in chlorine. Also, plastics containing $\ce{H}$ and $\ce{Cl}$, will burn in air with the formation of acidic hydrogen chloride.
Note: The combustion of a jet of methane in air, which is moved into a container filled with chlorine, will continue to burn:
$$\ce{CH4 + Cl2 -> CH3Cl + HCl}$$
However, the products now are corrosive, toxic and likely carcinogenic.
Further, similar to a gas mix of $\ce{H2}$ and $\ce{O2}$, a mix of $\ce{H2}$ (or $\ce{CH4}$) and $\ce{Cl2}$ will explode if ignited (strong/red light may also be an initiator absence the presence of any oxygen, which apparently terminates a radical chain reaction, which otherwise may produce a violent explosion).


text:
Origin of the term "millamolecule"
I came across a term "millamolecule" to represent macrocycles with a molecular weight between 500 and 1,000 daltons. Google Scholar show several resources which use this term e.g., Google Scholar Results
What could be origin of this term or is it just a buzzword invented by pharma companies?

In their 2013 Annual Report titled Evolving to a Specialty Care Biopharma Company, a well-known pharmaceutical company, Bristol-Myers Squibb has introduced the following statement under the title, Developing Innovative Drug Platforms:

As R&D evolves its focus, it is also investing in technology platforms that concentrate on new ways to affect disease targets, including antibody drug conjugates, which combine the targeted benefits of biologics with the cancer-killing ability of traditional small-molecule chemotherapies. R&D also will further expand the potential use of millamolecules, which are larger than small molecules but smaller than biologics. These millamolecules may be able to better exploit novel targets and mechanisms by retaining the desirable properties of small molecules with the high degree of selectivity, especially against antigens, that biologics and small molecules have had difficulty targeting.

And then, Bristol-Myers Squibb Company has filled a patent under World Intellectual Property Organization (WIPO) claiming Pet-imaging immunomodulators, abstract of which states that:

The invention relates to the synthesis and use of $\ce{^{18}F}$-labeled millamolecules for imaging various processes within the body, for detecting the location of molecules associated with disease pathology, and for monitoring disease progression are disclosed.

Thus, to my understanding, millamolecules is the word invented by Bristol-Myers Squibb Company's scientists to describe special set of molecules with different activity other than those exhibited by small molecules and higher polypeptides.
Finally, Medicinal & Pharmaceutical Chemistry glossary & taxonomy describes it as:

Millamolecules: Mid-range compounds that fall in size between small molecules and biologics. Size is not the only consideration for this class, as millamolecules should also be orally available and able to interrupt protein-protein interactions.

Unsurprisingly, their reference has also directed to Bristol Myers Squibb: Areas of Focus.

Thanks for the references. You are right that BMS must have introduced this term, but I am wondering what does milla- mean? Is it from mill- for thousand?

Your welcome. About milla meaning, I also checked for the definition, but came up with: a small genus of tropical American cormose herbs. On the other hand meaning of milla as: Milla is a girl's name of Polish origin meaning "young ceremonial attendant"! :-)

However, I thought about medium (size), may be!

Okay, found a footnote in paper by BMS _it is just a buzz word. Editors should discourage buzzwords until and unless authors provide an etymology. This will be good for the long term history. The footnote   "Bristol-Myers Squibb refers to cyclic and acyclic compounds with a molecular weight between 600 and 6000 Da as millamolecules. Millamolecules, especially macrocyclic millamolecules, are ideal for pursuing the so-called “undruggable targets” such as complex protein−protein interactions." https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4499820/

You can see the problem right away with buzzwords. The BMS paper suggests a range of 600 to 6000 Da. In the original post, another reference said 500-1000 Da. Big difference!

Good finding. I glad we solved it. I think original suggestion was 500-1000 Da. Small organic molecules cut off point is 500 Da.

It looks like the term was exposed at least three years earlier in 2010 BMS report. As for the name, I like to think it comes from the German Latin-based name Milla meaning "industrious", but it's a pure speculation on my side.

Someone in Germany also wondered "Explanation:
Das scheint ein von BMS erfundener Begriff zu sein, er taucht praktisch nur in Zusammenhang mit dieser Firma auf. Ich würde ihn daher wörtlich übernehmen." https://www.proz.com/kudoz/english-to-german/medical-general/4861004-millamolecular.html

This is the third definition..."BMS is developing its own macrocycles and collaborating with a third, unnamed, biotech. Its interest in macrocycles
is part of its millamolecular strategy, which is focused on compounds with a molecular weight between 500 daltons, considered a small molecule, and 10,000 daltons, or a macromolecule."


text:
Is it possible to have a negative overall activation energy for a whole reaction mechanism?
I am trying to find out correlation for ignition delay for alkane fuels. The activation energy part of the correlation gives me a negative value.
For the ignition delay formulation, is it possible to have an overall activation energy part being negative?
$$\tau = P^aT^b\phi^c\exp\left(\frac{E_\mathrm{a}}{RT}\right)$$

Well, it sure is possible and does happen sometimes for complex reactions, dunno if it applies for your case, though.

This might be helpful:  https://chemistry.stackexchange.com/questions/95343/is-negative-activation-energy-possible


text:
Software to assist drawing of complex, three-dimensional skeletal formula from scratch or from existing crystal structure
Is there a tool that would help in creating rather intricate line drawings (not pre-rendered balls-and-sticks/wireframe/sticks/ORTEP diagrams) for a skeletal formula in 3D space?
To illustrate the question a bit further, here are two examples:

Organic molecules — An all-benzene catenane and a trefoil knot [1]:

Inorganic molecules — proton transfer in polyoxoniobate $\ce{[H_xNb10O28]^{(6-x)-}}$ [2, p. 111]:

I also saw a much larger line-art structure of an entire peptide drawn the similar way, but unfortunately I cannot find the reference at the moment.
Line-drawn structures like these are usually presented alongside with the determined crystal structures, so I naturally assume that there is a tool that would allow to convert a CIF file to an illustration like the ones shown above.
However, I wasn't able to find anything relevant by screening the IUCr software list, and this already concerns me for several years.
Do people really use molecular editors like ChemDraw for this, or there is a more optimal way?
Also, if ChemDraw is the tool for the job, are there any general recommendations as to how to efficiently draw structures like this in 3D space without loosing one's mind on estimating z-buffer for proper bonds' greyscaling/thickness/type based on the distance from the viewer as well as overlapping atoms and bonds?
I would really appreciate if someone who published similar work would share some tips regarding their workflow.
References

Segawa, Y.; Kuwayama, M.; Hijikata, Y.; Fushimi, M.; Nishihara, T.; Pirillo, J.; Shirasaki, J.; Kubota, N.; Itami, K. Topological Molecular Nanocarbons: All-Benzene Catenane and Trefoil Knot. Science 2019, 365 (6450), 272–276. DOI: 10.1126/science.aav5021.
Polyoxometalate Chemistry, 1st ed.; van Eldik, R., Cronin, L., Eds.; Advances in Inorganic Chemistry; AP, Academic Press, an imprint of Elsevier: Cambridge, MA San Diego, CA Oxford, 2017; Vol. 69.

If you ask this question at Mathematica stackexchange, the wizards there will come up with something. However, their solutions may be ad hoc, whereas you seem to searching for more generic software.

@JosephO'Rourke I never thought this question would attract a math scientist:) Unfortunately, you are right, I'm looking for a more generic software, preferably an addon for a common chemical app or a free program, as sorting out licensing for Mathematica is a pain.

There are editors that have 3D mode and they can display a 3D molecule in 2D manner. MarvinSketch is first that comes to my mind, but there are probably more.

@vapid Good point, I use MarvinSketch quite a lot, but I keep forgetting it has 3D mode out of the box (although all the operations in 3D are slow on my machine, probably due to Java). I have to think about it, thanks for the tip.

This sounds like asking for a new (I think, then sixth) installment of ChemDraw Magick by ChemDraw Wizard Pierre Morieux (a chemist by training, working at PerkinElmer around ChemDraw).  The address of the channel in question: https://www.youtube.com/user/PierreMx3/videos.

@Buttonwood Ironically, one the videos on drawing denfrimers related to My ChemDraw hell was a turning point for me to abandon ChemDraw and look for a programmatic way (chemfig, OpenBabel) of creating my structures:)

Peptide handdrawn http://pdb101.rcsb.org/sci-art/geis-archive/gallery/geis-0512-lysozyme

@KarstenTheis Thank you, but the structures from 10.1038/scientificamerican1166-78 appear to be sticks diagrams: even though they were hand-drawn at that time, these days they can be replicated with crystallographic software. The only skeletal formula there I found was for polysaccharide chains, but that's not really what I mean.

Thanks to vapid and their comment

There are editors that have 3D mode and they can display a 3D molecule in 2D manner. MarvinSketch is first that comes to my mind, but there are probably more.

I figured a crude solution, which I hope might as well serve as a starting point for other answers.
First of all, I completely forgot that MarvinSketch (free (registration required), available for Windows, Linux, macOS) has support for 3D.
Drawing in 3D from scratch with MarvinSketch is probably not the best idea as even though it has a support for optimizing (cleaning) 3D structures, it sometimes gives weird results for organics and practically not suitable for inorganic complexes and especially clusters.
Arguably, the best and the most affordable third-party tool for creating such structures is Avogadro (open source, available for Windows, Linux, macOS).
If there is a crystal structure at hand (e.g. CIF data), then one must do two things first:

crop/grow the fragment of interest;
assign proper bond orders and export the structure in one of the formats that preserve bond orders (such as MOL, MOL2, PDB or CML).

I believe Avogadro can do both, but I personally like to do part 1 in Olex2 (free (registration required), available for Windows, Linux, macOS) by force of habit.
As a proof of concept, I took the crystal structure for the trefoil knot, allowed it to grow from assymetric unit to a complete molecule as well as cut out solvent and guest molecules, so I was left with the initial shape:

which I immediately exported to the MOL file by running file trefoil-knot.mol.
Next, I opened this MOL file with Avogadro, which automatically assigned bond orders:

and the file was exported to MOL once again, and subsequently opened in MarvinSketch (I reduced fog in settings, made implicit hydrogens hidden and used F7 button to align the molecule on screen):

And the final result came out to look like this:

The image is, however, not quite as perfect as the one in the paper.
For instance, there are some artifacts about inflection points as if the program was confused how to place double bonds.
Also, if you look closer, the grayed out part which was supposed to be further from the viewer, actually appears in front, which I attribute to the fact that Marvin was designed as a 2D molecular editor first and foremost, so it doesn't always handles depth (or the aforementioned z-buffer) adequately.


text:
How do I convert an Excel file with SMILES and IC50 to .smi file?
Not much else to say, I have an Excel file with two columns and I need to create a .smi from it.

Could you at least present a chunk of the data as an example as well the expected text output?

In its present form, your question is a bit vague.
This being said, if your spreadsheet contains one column about SMILES strings, a second one about the IC50 values, than 1) simply copy-paste the content of the column about SMILES strings into a text file, and 2) substitute the file extension .txt by .smi.  So far,  .smi I encountered which were processed by babel (vide infra) did not contain IC50 data, however.
Note 1:  Some programs will recognize .smi files because of the file extension or their content.  For others (like babel), you may instruct them that a file  to work with is about SMILES strings even if the file extension is, for example, .txt text.  Have a look at their documentation, e.g., here, on-line or here, off-line.
Note 2: It is legit to have entries about multiple molecules expressed simultaneously in one SMILES string.  An example is exporting structure data from the CSD crystal data base with their conquest interface (available to subscribers) for entries about co-crystals, or hydrates.  In these instances, individual SMILES strings are concatenated by the . (a period on the floor).  Hence, the string CCCCNC.CC expresses the simultaenous presence of methyl-butylamine altogether and ethane in one entry.
Note 3, probably beyond the scope of your question: Consider to 1) add headers to your spreadsheet (e.g., structure, above the column of SMILES; IC50 above the column of the IC50 values), 2) export the table as .txt file.  Then load this file into DataWarrior (initial publication doi.org/10.1021/ci500588j).  DW is a platform-independent program allowing you to continue to work side-by-side with structure data expressed as SMILES, along with (med)chem relevant data -- some you may compute in the program, some the program may fetch for you  from e.g., Chembl, COD -- and to perform analysis with them.


text:
Acidity comparison on R-PO3H2
Question

Compare the acidity $(K_\mathrm{a1})$ of the following three acids:
  $\ce{H3PO4},$ $\ce{Ph-PO(OH)2}$ and $\ce{H3C-PO(OH)2}.$

Thoughts
These three acids are of type $\ce{R-PO3H2},$ where $\ce{R} = \ce{OH}, \ce{Ph}, \ce{CH3}.$ The ionization equations of them are
$$\ce{R-PO3H2 <=> R-P(=O)(OH)O- + H+}$$
Comparing with hydrogen, benzene ring can reduce the charge of the anion because of the delocalised $\pi$ bond. Also, methyl group is an electron donating group, which will destabilize the anion. Hence, the $\mathrm{p}K_\mathrm{a1}$ I suggest is $\ce{CH3} > \ce{OH} > \ce{Ph}.$ This answer is analogue to $\ce{R-SO3H}.$
One thing I'm not sure is that this logic does not apply to $\ce{R-CO2H}.$ Clearly, the acidity of carbonic acid is weaker than methanoate acid. I don't know which explanation should be applied in the situation of phosphorus.

"Clearly, the acidity of carbonic acid is weaker than methanoate acid" - not really, it's slightly stronger, actually, it's just that it's equilibrium with CO2 messes things up. https://en.wikipedia.org/wiki/Carbonic_acid


text:
A "function of proportionality" in a rate law
I am currently studying rate laws and the determination of the order of a reaction. So first-order reactions remind me of linear functions $f(x)=kx$ while second-order reactions remind me of quadratic functions like $g(x)=cx^2$ in high-school algebra. Since we are dealing with reactants here, the input of the functions would be the concentrations of our substrates $[\ce{S_1}]$ and $[\ce{S_2}]$. I am wondering if there are any fancier functions that arise in "real-life" chemistry?

What about https://en.wikipedia.org/wiki/Virial_expansion?

I like the idea behind the question, but at this point it seems a little broad without a clear criteria of what the "right" answer would be. Is there anyway you could narrow down/specify what you are looking for

If by "fancier" you mean "uglier", then absolutely yes. Real chemistry is messy, much more than we can fully handle. The simple rate laws we learn in undergraduate physical chemistry are just approximations.
I've written elsewhere that, in general, a full description of chemical kinetics for any overall reaction requires studying an entire tree of simultaneous chemical reactions. There can be dozens, even hundreds of intermediate species involved, all mutually linked in complex ways, and with more than one final product ("by-products"). The difficulty in performing this full description has lead to the whole field of chemical reaction network theory. I hope you like matrix algebra.
To make chemical kinetics tractable in most cases, two major approximations made are:

Forget about the whole reaction network, and look at only the overall kinetically fastest path from reagents to products
Forget about all the steps and intermediates in the fastest path, and describe only the bottleneck - the slowest step

Even with these two crushing approximations, it's possible to describe many useful reactions with a fair level of accuracy. This is partly luck, and partly unavoidable (due to simplicity naturally arising in randomness). This level is what most undergraduate and even graduate work sticks to. Even when it's a bad fit, we'll sometimes stubbornly fit a system into a particular rate law, do some variational analysis, and hope the error isn't too great.
A nice example of a reaction whose kinetics is slightly more complex than the standard "nice" rate laws is the radical reaction between hydrogen and bromine.
$$\ce{H2 + Br2 -> 2HBr}$$
According to experimental data, it's possible to obtain the following rate law (which may still be approximate!):
$$\mathrm{r_{HBr}=\frac{k_1[H_2][Br_2]^{3/2}}{[HBr]+k_2[Br_2]}}$$
The deduction is nicely described in this site. As you can see, depending on the reaction parameters, the denominator can potentially be simplified down to a single term, with some degree of approximation, and a simple-ish rate law could be used. But in the general case, it's messier. Because of these reaction networks, even the concept of a "reaction order" does not exist in general. An example of a natural but extremely difficult reaction to describe kinetically in full would be the combustion of a moderately sized alkane, such as octane ($\ce{C8H18}$).
When taking full chemical networks into account, there appear cases with pretty unique kinetics. For example, if you haven't heard of oscillating reactions, they're a topic of considerable experimental and theoretical study. With a particular set of coupled reactions, it's possible for chemical systems to display the hallmarks of mathematical chaos. Even if individual steps can be described by simple kinetics, the global process cannot. There's a nice discussion of oscillating kinetics in this Chem.SE question (and possibly others).
I'm sure there are other ways to get strange rate equations. Some reactions just don't follow kinetic theory in its simplest form. This can lead to weird conclusions, such as negative activation energies. Furthermore, rate laws almost always assume macroscopic amounts of matter. When a tiny amount of reagents are involved, the discrete nature of matter calls for statistical corrections to chemical kinetics, because an integer number of molecules can't follow continuous analytical curves.

This is very informative, more than what I can comprehend. Thank you! With regard to the rate laws taught to undergraduate, I tried constructing a more mathematically rigorous argument, but I just can't show that the rate laws should be as stated. I'll try to show my argument in another post and ask for help. Anyway, it's great to know that it's an approximation and great to know chemical reaction network theory is helpful because I have to learn it for my degree's final project.

Chemical reaction network theory is a very specialised field, almost no-one knows or uses it. I don't really think it would be beneficial to you at your current level, unless you have an amazing mathematical background.


text:
Petroleum Benzine or Normal-Hexane for removing a silicone based adhesive?
I’m trying to remove the glass panel wich is glued to the polarizer of an LCD display (an Apple iMac display with broken glass but working LCD wich I would like to repurpose).
It seems to be fixed with a soft, non-curing, silicone-based adhesive . But the process of removing it is terribly tedious and I was wondering if there was a solvent that might help me.
LG, the manufacturer states in the display's data sheet that touches from bare hands, grease, some cosmetics, acetone, alcohol and toluene will harm the polarizer‘s surface.
Some people claim that they removed the glass successfully using baby oil but given that info from the data sheet I'm not sure if the oil may be harmful to the polarizer.
LG recommends to use petroleum benzine or normal-hexane for cleaning the polarizer surface.
Now, my question is if one of these two might be a helpful substance for the task of dissolving/weakening the silicon adhesive and removing the glass?
Thank you!

You may want to ask Louis Rossmann:)

Yeah, though isn't he more an electronics guy?


text:
Reactions which confirms the presence of cyclic structure of glucose
Are there any reactions that can be explained only on the basis of the cyclic structure of glucose rather than the open chain thus confirming its presence in the solution? Like a reaction that can only be given by hemiacetals thus confirming its presence and hence confirming the existence of the cyclic structure.
Any help would be appreciated.

An NMR measurement can confirm that, or do you specifically need a chemical reaction?

This is possibly a duplicate. You can find a answer here: https://chemistry.stackexchange.com/a/110456/61578

@Mathew Mahindaratne No! It does not. It only shows some NMR scans. I am looking specifically for chemical reactions that confirm their presence.

Tony: Try this. https://chemistry.stackexchange.com/questions/119863/haworths-research-of-glucose/119919#119919

Yes, such a reaction exists. In fact, it can not only confirm the presence of ring forms in solution but also how prevalent they are.
In 1962, Walker Jr, Gee and McCready published Complete Methylation of Reducing Carbohydrates, in which they subject solutions of sugars in dry DMF to silver(I) oxide  and methyl iodide. The reaction proceeds according to scheme 1 below.

Scheme 1: Methylation of glucose with $\ce{Ag2O}$ and $\ce{MeI}$ as published by Walker Jr, Gee and McCready.
Before reporting on the products, they note:

It was somewhat unexpected that methylation of
  aldoses would proceed so smoothly in the presence
  of silver oxide. Glucose, shaken with silver oxide in
  DMF at room temperature, is partially decomposed
  in several hours.

Usually, silver(I) salts in solution are used in the Tollens test in which reducing sugars such as glucose reduce the silver ions to give elemental silver which forms a silver film on the inside of the glass. The presence of methyl iodide completely suppresses this reaction as they state in the next sentence:

When methyl iodide is present
  in the reaction mixture, however, no silver mirror
  (indicating oxidation-reduction) is formed

As for the two step process that I have indicated, the authors state:

In this reaction methylation
  of hydroxyl of the hemi-acetal group of glucose is
  essentially complete in one hour (as shown
  by loss of reducing function), but complete
  etherification of the remaining hydroxyls requires
  considerably longer.

With the experimental section detailing that the reactions were run ‘at least $\pu{16h}$’ which permits drawing the mechanism as a two-stage process as 1 can be isolated if the reaction is cut short.
Mechanistically, this reaction process is different from an acid-catalysed acetalisation with methanol. In the latter, an entire $\ce{OMe}$ group is attached to the sugar and the reactive species is an aldehyde form. This could be the open-chain aldehyde form (requires an additional reaction step) but most likely occurs via a cyclic aldehyde as depicted below. When using $\ce{Ag2O/MeI}$, the acetal oxygen is supplied by glucose as methyl iodide does not have any available oxygen. The two competing mechanisms are shown in scheme 2.

Scheme 2: Mechanisms of the acetalisation of glucose. (a) $\ce{H+/MeOH}$ and cyclic aldehyde intermediate; (b) $\ce{H+/MeOH}$ and chain-form aldehyde intermediate; (c) $\ce{Ag2O/MeI}$ single-step mechanism.
As these mechanisms show, acid-catalysed acetalisation need not start from the cyclic form to get acetal 1; it is possible to draw the reaction mechanism as originating from the chain form of glucose. But methylation with $\ce{MeI}$ requires the oxygen to be in place as an $\ce{OH}$ function that is deprotonated and attacks $\ce{MeI}$ which is not possible if the chain form were present in solution.
As to why the $\ce{C{1}}$ oxygen is methylated first, and the explanation is the higher acidity of an acetal hydroxy function when compared to an alcoholic one as the authors state:

This difference in rate is expected
  from the difference in acidity (and ease of
  anion formation) of the hydrogens on the two different
  types of hydroxyl groups.

Furthermore, it bears stating that this reaction not only proves cyclic forms are present but also that they are prevalent. Methylation of the $\ce{C{1}}$ oxygen is rapid compared to all the other hydroxy groups meaning that they should be present in at least appreaciable ratios. Furthermore as mentioned above, under these conditions the reduction of silver(I) to a silver mirror – which requires the open-chain aldehyde form mechanistically – is practically entirely suppressed. Given that the Tollens test is a standardised method to detect glucose, the complete absence of all traces of silver film serves as a good indicator of the low concentration of open-chain aldehyde in solution.
The original paper is: H. G. Walker Jr, M. Gee, R. M. McCready, J. Org. Chem. 1962, 27, 1000–1002. DOI: 10.1021/jo01053a047.


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Copper (II) chloride reduction with Hydrogen gas
I've been looking into making pure copper (I) chloride, but using ascorbic acid is relatively expensive for me. I'd also like to avoid having water in any part of the synthesis to minimize the contamination with copper (II) chloride. Therefore, I came up with the idea of doing the synthesis in ethanol and bubbling through H2 according to the reaction
$$\ce{2 CuCl2 (aq) + H2 (g) -> 2 CuCl (s) + 2 HCl (g)}$$
So my main question is: could this work?

I suppose it would be kinetically blocked. BTW CuCl2 in ethanol would not be (aq).

Sounds possible as the solubility of CuCl2 in EtOH is 53g/100ml  http://www.sciencemadness.org/smwiki/index.php/Copper(II)_chloride. There is a reduction procedure using aq sulfite here:  http://wwwchem.uwimona.edu.jm/lab_manuals/c1901exp35.html

You will end up in metallic copper if H2 is heated. Hydrogen is a powerful reducing agent at high temperature. Bubbling H2 won't do anything. These experiments have a safety issue and should never be performed without proper qualification.

There are better ways to do it.  Copper (II) can be reduced to copper (I) with a variety of reagents, especially in the presence of a soft-base halide ion such as chloride which precipitates or complexes with the copper (I) (which is a good way to prevent "overshooting" to metallic copper).  Some specific approaches can be seen here.


text:
How to calculate the amount of sodium sulfate needed to lower the silver ion concentration of a saturated silver sulfate solution?

How many moles of $\ce{Na2SO4}$ we have to add into a saturated $\pu{0.5 L}$ solution of $\ce{Ag2SO4}$ so that the concentration of $\ce{Ag}$ is $\pu{4.0 \times 10^-3 M}$? ($K_\mathrm{sp} = 1.4 \times 10^{-5}$)

I found out first $\ce{[SO4^2-]}$ in a normal saturated solution:
\begin{align}
K_\mathrm{sp} &= [\ce{Ag+}]^2[\ce{SO4^2-}],\\
1.4 \times 10^{-5} &= 4s^3,\\
s &= 0.015
\end{align}
So we have $\pu{0.015 M}$ of $\ce{SO4^2-}$ in this solution.
Then, I find out $\ce{[SO4^2-]}$ needed for a saturated solution of $\ce{Ag2SO4}$ but with $[\ce{Ag}] = \pu{4.0 \times 10^-3 M}$:
\begin{align}
1.4 \times 10^{-5} &= (4.0 \times 10^{-3})^2 \times [\ce{SO4^2-}],\\
[\ce{SO4^2-}] &= 0.875
\end{align}
So we need $\pu{0.875 M}$ of $\ce{SO4^2-}$ in such saturated solution.
So the difference of $\ce{SO4^2-}$ is $\pu{0.875 M} - \pu{0.015 M} = \pu{0.86 M}$. So we need to add $\pu{0.86 M}$ of $\ce{SO4^2-}$, that is $\pu{0.43 mol}$ of $\ce{SO4^2-}$ ($\pu{0.86 M} \times \pu{0.5 L} = \pu{0.43 mol}$).
Eventually, we obtained we need to add $\pu{0.43 mol}$ of $\ce{Na2SO4}$.

The above are my calculations, can anyone tell me if my steps can be justified?

Why divide 0.86 moles by 2? It is $\ce{SO4^{2-}}$ that is needed in the solution, not $\ce{Na+}$.

Oh, I just added a note to that, and I changed all mol/L to M so it might be more visible. It is 0.86 mol/L, so I multiply it by 0.5 L in order to obtain the mole needed which is 0.43 mol

Dah... I didn't read the problem correctly. Dividing by 2 was correct.

Haha. No worries, it happens

In the given data, $\ce{K_s_p}$ = $\ce{1.4 x 10^{-5}}$. But two lines later, you write another value for the same constant. You write that $\ce{K_s_p}$ = $\ce{4.0 x 10^{-3}}$. Why ? All the following calculations are wrong.

@Maurice sorry, it might be a typo. I calculated all the data using the right number. Thank you for noting that! I have just changed it

Your calculations are true to a certain extend. However, you neglected the amount of $\ce{SO4^{2-}}$ ions precipitated as $\ce{Ag2SO4}$. And also, since we are calculating in the means of concentration terms, we have to assume that addition of $\ce{Na2SO4}$ solid did not change the original $\pu{0.5 L}$-volume.
You calculated correctly the concentration of $\ce{[SO4^2-]}$ ions ($s$) in the original saturated solution, which is equal to the solubility of $\ce{Ag2SO4}$ ($s$):
$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{SO4^2-}]= (2s)^2 \times s= 4s^3= 1.4 \times 10^{-5} $$
$$\therefore \; s = \left(\frac{1.4 \times 10^{-5}}{4}\right)^\frac{1}{3} = 1.52 \times 10^{-2}$$
You also calculated correctly the concentration of $\ce{[SO4^2-]}$ ions in the solution after certain amount of $\ce{Na2SO4}$ solid was added (suppose it is $m$ mols) to make  the concentration of $\ce{Ag+}$ ions is equal to $\pu{4.0 \times 10^{-3} M}$:
$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{SO4^2-}]= (4.0 \times 10^{-3})^2 \times [\ce{SO4^2-}] = 1.4 \times 10^{-5}$$
$$\therefore \; [\ce{SO4^2-}] = \left(\frac{1.4 \times 10^{-5}}{16.0 \times 10^{-6}}\right) = \left(\frac{1.4}{1.6}\right) = \pu{0.875 M}$$
Thus, you have $\pu{0.875 M}$ of $\ce{SO4^2-}$ ions in this solution. However, note that some of $\ce{SO4^{2-}}$ ions would be precipitated as solid $\ce{Ag2SO4}$ when you keep adding $\ce{Na2SO4}$ solid (vide supra) in order to maintain the equilibrium (Le Chatelier's principle). This amount is equal to the half of $[\ce{Ag+}]$ ions precipitated with it:
$$\therefore \; [\ce{SO4^2-}]_{ppt} = \frac{1}{2}\left(2 \times 1.52 \times 10^{-2} - 4.0 \times 10^{-3} \right)= \pu{1.32 \times 10^{-2} M}$$
Suppose you add $m$ amount (in moles) of $\ce{Na2SO4}$ to the original solution. which already had $\pu{1.52 \times 10^{-2} M}$ of $\ce{SO4^{2-}}$ ions. Thus, $[\ce{SO4^{2-}}]$ remain in the solution:
$$[\ce{SO4^{2-}}]_{remain} = \pu{0.875 M} = m + 1.52 \times 10^{-2} - 1.32 \times 10^{-2} $$
$$\therefore \; m = 0.875 - 0.20 \times 10^{-2} \approx \pu{0.873 M}$$
Thus, you need to add $\pu{0.873 mol/L}\times \pu{0.5 L} \approx \pu{0.437 mol}$ of $\ce{Na2SO4}$ to the original solution.


text:
What makes this molecule achiral?
I don't understand why this molecule is achiral:

I've circled what I think is the sole chiral center in red. I thought that any molecule with exactly one chiral center is chiral without exception. I built the molecule and its mirror image with my kit and can't find any superimposability.

I think it is achiral because one and the same group is attached to the circled carbon center. Yes the substitution is different but it is too far away.  The problem with big molecules like these is mirror image criteria becomes difficult because these bulky substituents can have plenty of conformations by rotation. You can only "guess" the chirality of one conformation.

This could be a case of cryptochirality, but it still is chiral. See this relevant question.

It is chiral. $~$

Thank you. I probably should have guessed since my professor makes mistakes like this a lot. My apologies for posting.

@orthocresol: I agree with you, this is a chiral molecule. Unless p-bromobromophenyl group is meant to be a bromo-group. I think OP's teacher made a drawing mistake, if he/she still thinks it's achiral molecule.


text:
How do molecules with muons and electrons bond?
Is bonding specifically for and between electrons? Why cant two atoms share muons, which are different particles of same charge, spin and different mass?
Why there aren't muon-electron bonds?
Why is the octet (or eighteen valence electron) rule only for electrons, and not for all particles with similar charge and spin compared to the electron?

I wrote 32 valence electron rule for f-block elements but then I remembered f orbitals dont participate in bonding.:)

for lanthanides at least .

Remember that the Pauli exclusion principle applies to identical fermions. A muon is not identical to an electron, so if you introduce a muon into an atom it will decay into the lowest possible orbital. This orbital is 1s like of course, so in principle you could make a molecule out of protons and muons; however muons decay very quickly and are captured by the nucleus on a similar time scale.

Also, wouldn't the fact that muons "orbit" much closer to the nucleus mean that it would be very hard to get two nuclei close enough together to have enough interaction of both nuclei with the muon that there is a net benefit? I think the cost of bringing the nuclei together would likely be greater than the benefit of bonding.

Di-positronium

https://en.wikipedia.org/wiki/Muon#Muonic_atoms

I've tried adjusting the question title and content, let me know if it's not what you meant.

You are correct; electrons and muons are fermions with different quantum numbers (specifically, they differ in the electron number and the muon number), so the Pauli exclusion principle does not apply between them (though it of course applies among electrons and muons separately). A somewhat similar case happens with protons and neutrons (also fermions) in the nuclear shell model, which attempts to describe nuclei as containing proton and neutron shells, analogous to electron shells. The proton and neutron shells are filled independently.
Because (negative) muons are the second generation Standard Model equivalent of the electron, whatever electrons do, muons can copy - since there are "electronic" orbitals, there are also "muonic" orbitals. However, there are two main differences.
First, for electron orbitals, it's a good approximation to assume the nucleus is stationary with respect to the electrons (the Born-Oppenheimer approximation), due to the great difference in their masses (the lightest nucleus, a proton, has approximately 1836 times the mass of an electron). Because the muon is approximately 207 times heavier than an electron, now a muon has an appreciable mass relative to a proton (approximately one-ninth), and therefore the BO approximation is considerably worse. This is a scenario in-between a regular atom and positronium, where an electron "orbits" a positron "nucleus", which has the same mass. See Phys. Rep. 1982, 86 (4), 169-216 for a quantitative analysis of BO approximation errors in a simple muonic molecule.
The second and more striking difference is that, again due to the muons being 207 times heavier, muonic orbitals are accordingly around 207 times smaller meaning a muonic orbital has a typical radius of around 0.5-1 pm compared to 100-200 pm for an electronic orbital. The energies are also 207 times larger in magnitude - the 1s electronic orbital in hydrogen has an energy of -13.6 eV, whereas for "muonic hydrogen", the 1s muonic orbital has an energy of -2815 eV. These facts can be determined simply by solving the Schrödinger equation, except inputting a mass 207 times greater for the negatively-charged particle.
As you can see, this leads to a severe mismatch between the realms of electronics and muonics. There is no kind of shared electron-muon bond - at best, there would be a separate electron bond and a muon bond. However, because the energetics involved in the muon bond are so much higher, the system in the ground state is essentially equivalent to just having the muon bond, plus a small correction due to a grossly warped electron bond. The muon bond is formed normally, and the electron bond has to deal with whatever geometry is forced by the muons, however crazy.
As an example, imagine a neutral atom of regular hydrogen and a neutral atom of muonic hydrogen interacting. The muonic hydrogen atom basically pierces into the depth of the electron cloud of the regular hydrogen atom (the electron can hardly repel the muon efficiently, since it is so tightly bound to its nucleus), until both nuclei get quite close. Then the muon latches onto the other proton. The system is stabilised when the two hydrogen nuclei are approximately 0.5 pm apart, and the lone muon forms half of a muonic sigma bond.
From the "point of view" of the muon, the system looks like a singly-ionised muonic dihydrogen molecule ($\ce{\mu-H_2^+}$), with slight corrections due to the electron buzzing around, most of the time far away.
However, from the "point of view" of the electron, it sees a bizarrely elongated nucleus (from the typical ~1 fm sphere to a ~500 fm spindle) with a total charge of +1e (since the muon almost perfectly screens out a full positive charge). This spindly nucleus is still quite small relative to the electron cloud, and so the electron likely behaves similarly to a normal isolated hydrogen atom, with some corrections due to the non-spherical distribution of charge at its "nucleus" (the two separate protons bound by a muon). The electron still provides a slight amount of bonding between the protons, but it is much less than normal due to the odd geometry forced by the muon.
Muonic chemistry in its full glory would be a fascinating (and extremely dangerous!) copy of the electronic chemistry we know, but the two would operate almost completely independently. The muon is actually one of the most stable subatomic particles, with a lifetime of 2.2 µs. That sounds like almost nothing, but it's many orders of magnitude more than what is necessary to observe "chemistry". Unfortunately, it's just too difficult to produce for how ephemeral it is...


text:
What would happen if we supercool and then superheat an atom very abruptly?
Now scientists have found ways to cool atoms to temperatures far lower than we ever could before and maintain them for longer, but what if we had one of those particles and then heated it as fast we could and then to the highest temperature we could as instantly as possible. Like how substances react to rapid freezing and unfreezing what would happen to an atom? something cray cray? a little boom boom possibly?

You bring up the idea of "rate of heating a single atom". Why do you think the rate is important? This is not clear. For instance, you can cause electronic transitions using light, and afaik they occur practically instantaneously. Probably what is important in the context of speed is the uncertainty (pseudo) relation between time and energy.

See https://www.nobelprize.org/prizes/chemistry/1999/zewail/lecture/

Sadly nothing fancy happens that can make headlines. Atom is not a solid in a classical sense that it will break or crack if it is rapidly heated or cooled like glass. Ordinary glass will crack when liquid nitrogen is poured into it. The Sun is an example of superheated atoms or ions. Atoms start to emit light at high temperatures.

I couldn't help it! Has a concept of temperature ever been defined in the context of a single atom?


text:
Has a concept of temperature ever been defined in the context of a single atom?
I wonder if this answer to What would happen if we supercool and then superheat an atom very abruptly?  goes far enough.
I almost wrote the comment:

I don't think we can even talk about the temperature of a single atom, at least based on its kinetic energy. Now if there were several electrons in one atom simultaneously in higher orbits at the same time, then it might be possible.

but then decided to cast this in the form of a question.
One way to think of the temperature of a group or ensemble of atoms can be the average of a particular energy. There can be more than one temperature; e.g. kinetic energy, the distribution of excited states and nuclear spins (What is the ortho/para issue with LH2 as a fuel?) If they are different we can say that they are out of equilibrium, but we can still talk about each as a different temperature in cases where coupling between different partitions is slow compared to the timescale of interest.
In a strongly interacting quantum system like an atom, there can be some processes that can excited several electrons to higher energy states at once.
Question: Has a concept of temperature ever been (usefully) defined in the context of a single atom?
This would likely be an extreme situation and may not happen in conventional reactions, maybe an intense and ultra-short laser pulse or nuclear decay or reaction, or possibly in the case of some explosive reaction.
update: comments mention a related question in Physics SE Can a single molecule have a temperature? with several answers, though they don't directly answer this question because I'm trying to find out if this "has ever been defined in the context of a single atom", and not asking if you think it should/shouldn't or can/can't be. It is possible that at some point in some, (and possibly extreme) experimental contexts this may have been workable.
The answer there that I do find most helpful is @JohnRennie's. Still though, I'm asking primarily about excitations within a single atom rather than the rotational/vibrational states of a large molecule.

This is already discussed in detail https://physics.stackexchange.com/questions/65690/can-a-single-molecule-have-a-temperature

@M.Farooq I've link back there and addressed those answers. Thanks!

It wouldn't make sense. So it is hardly that someone used that idea.

To extend your comment on @JohnRennie's answer, in spectroscopy the term vibrationally ‘hot’ molecule is used sometimes. And, of course, there are hot atoms too.

@Alchimista perhaps that's what they said about Onsager in 1949 as well?  ;-)

No. That is about system of N particles. A bound N particle system might be a single atom but then a concept of T would refer to its electrons. This "reduce" to the electronic energy of the atom and its quantistic treatment. Hot molecules and atoms pointed out by @andselisk still refer to members of a Boltzmann distribution.

@Alchimista It was meant as humor but perhaps it was too oblique, that the concept of "negative temperature" might have also elicited a similar "doesn't make sense" first reaction, but in hindsight can have meaning in a certain context.

The important aspect of temperature is equilibrium. In system not in equilibrium, different kind of energies may be distributed according to different temperatures, e.g. by laser excitation. But if the system is in equilibrium, these formal "temperatures" are equal, or rather they fluctuate around the same value. For a single atom, one can speak about an equivalent temperature only, as the same atomic state may happen in huge temperature range.

@Poutnik thank you for your comment, it's quite helpful! If each electron in a high-Z atom had its own set of excited states unrelated to the other electrons, then an excitation temperature could be defined in a similar way to how it would be done for an ensemble of protons each with one electrons. But I'm thinking that once several electrons are excited, it may not be so easy because it may be impossible to say "electron #1 is 0.8 eV above its ground state, electron #2 is 1.2 eV above its ground state..." etc.

Thus I've asked if this has ever been done because I wonder if it's possible to try at all, no matter how useful it may or may not be.

P.S.:Additionally,, such 1 atom ( or molecule ) equivalent temperature would have different values for different kinds of atom k molecule ) energies.

Aside of equilibrium and the mean energy value aspects, another important aspect is the energy distribution. 1 entity has just 1 point value like the Delta function.

@Poutnik so in this case is would each entity be an electron, or an occupied orbit within the atom?

The kinetic energy of the atom and the electron excitation energy, that would be compared to the mean energy of the particular statistical distribution with the equivalent temperature.

Let us continue this discussion in chat.


text:
Create Sodium Hydroxide from Sodium Silicate
During a reaction I use sodium hydroxide and create sodium silicate as a by-product.
I am curious if there is a way to recycle that sodium silicate into sodium hydroxide?
I am not a chemist, just have to use chemistry and I love it.
My best guess is to use Hydrogen Peroxide and Heat.
or Hydrogen gas.
or maybe even Aqua Ammonia...
For any reaction I have abundant electricity available, to create heat.
I have thought about this for months and stuck in the do-nothing phase until I can get some help from experts and see the reaction on paper.
Any help would be greatly appreciated.

The reduction to Si needs extreme reduction conditions at which H2O2 is not available. Neither ozone neither oxygen can be produced at these conditions. By CH2 you probably mean H2. I suppose very high temperature would be needed for the 2nd reaction and it would end up as Na2O. Si would probably and up as sodium silicide.

The reactions you proposed are pure fantasy. They will never happen. Na2SiO3 does not react with either H2O2 or Hydrogen gas or CH2 (which does not exist, by the way).
The only possibility would be to have silica precipitate by bubbling CO2 in the solution. This will produce a precipitate of Si(OH)4. But this precipitate is so gelatinous that it cannot be easily filtrated. If you are able to filtrate (under pressure), the remaining clear solution contains Na2CO3.
$$\ce{Na2SiO3 +CO2 + 2 H2O  ->  Na2CO3 + Si(OH)4}$$
which could be transformed into NaOH by adding calcium hydroxide (or chalk) Ca(OH)2. This addition creates a new precipitate, CaCO3, which can be filtrated, and the residual solution contains NaOH.
$$\ce{Na2CO3 + Ca(OH)2 -> CaCO3 + 2NaOH}$$

Methylene is a thing, just not very stable or isolable.  See here.

Methylene exists as an intermediate of rection, I know. But you cannot store it in a flask, or make it react with a solution of sodium silicate.

Noted "not isolable".

The one simple chemical reaction that could work is with calcium hydroxide (limewater):
$\ce{Na2SiO3 +Ca(OH)2 -> 2 NaOH + CaSiO3(s)}$
This reaction is used to seal concrete and so is chemically a thing.  Now the bad news:  the reason the sealing process works is the calcium silicate comes out as a hydrated gel.  It would be hard to separate for your purposes, like the silicic acid mentioned in Maurice's answer.  Silicates in general are messy in aqueous systems.
Sodium hydroxide is cheap, the silicate not very hazardous with respect to disposal.  Buying the sodium hydroxide new seems more feasible than either of our reaction schemes.


text:
Which one is the most stable canonical structure?And why?
According to my book the answer is given that structure III is the most stable because all the atoms have complete octets.
But I think it should be II because of,
(a) Structure III has positive charge on a highly electronegative atom and
(b) In structure II, the positive charge can be stabilized by the Indicutive effect of Oxygen as well as resonance.
Can anyone provide a more convincing explanation as to why III is more stable?

The principle of resonance structures is that they are all equally stable, or rather that you cannot separate them from each other as only together do they show the character of the overall compound. There is no energy change between these, all three describe the same compound. That said, there is a most contributing structure which is III.

@Jan your comment is at least obscure.

Whoever wrote this oversimplified problem/answer deserves not to be killed but an even worse fate, showing up in Green Bay with Chicago Bears gear after the recent NFL football game.

Why this is off topic?

You're supposed to recognize that oxygen which has more valence electrons than carbon can take on the positive charge without giving up its octet, thus favoring Structure III.
Now, is it really true that the oxygen atom has more positive charge than the formally positive carbons in the other two structures, when oxygen is also more electronegative versus the octet rule?  I suspect that is beyond the scope of your course material.  You have to accept a higher priority for the octet rule in this situation whether that is real or not.
I hate oversimplified textbook problems :-( .

"Now, is it really true that the oxygen atom has more positive charge than the formally positive carbons in the other two structures, when oxygen is also more electronegative versus the octet rule?"What does this statement mean?


text:
Finding the concentration of each ion in saturated solution of lead(II) hydroxide
Question

If the solubility of $\ce{Pb(OH)2}$ is $\pu{0.155 g/L},$ then the concentration of each ion in a saturated solution of $\ce{Pb(OH)2}$ is
A. $[\ce{Pb^2+}] = \pu{0.155 g/L}$ and $[\ce{OH-}] = \pu{0.155 g/L}$
B. $[\ce{Pb^2+}] = \pu{0.052 g/L}$ and $[\ce{OH-}] = \pu{0.103 g/L}$
C. $[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$ and $[\ce{OH-}] = \pu{1.29E-3 mol/L}$
D. $[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$ and $[\ce{OH-}] = \pu{6.43E-4 mol/L}$

Answer

 C. $[\ce{Pb^2+}] = \pu{6.43E-4 mol/L}$ and $[\ce{OH-}] = \pu{1.29E-3 mol/L}$

My Approach
What I did is that I wrote the dissociation equation:
$$\ce{Pb(OH)2 <=> Pb^2+ + 2 OH-}$$
Then I used to mole ratio to discover that:
$$
\begin{align}
γ(\ce{Pb^2+}) &= \pu{0.155 g/L}\\
γ(\ce{OH-}) &= \pu{0.31 g/L}
\end{align}
$$
I then converted the mass concentration $γ$ into molar concentration $c$ since the $\pu{mol/L}$ is commonly used to describe solubility:
$$\pu{0.155 g/L} × \frac{\pu{1 mol}}{\pu{207.2 g}} = \pu{7.48E-4 mol/L}$$
However, that's not the answer. What did I do wrong?

The mistake you made is that you assumed mass concentrations of the ions from the stoichiometric relation, which is only applicable to the amount of substance, and, as a consequence, amount concentration.
You have to convert mass concentration $γ$ to the molar concentration c first:
$$c(\ce{Pb(OH)2}) = \frac{γ(\ce{Pb(OH)2})}{M(\ce{Pb(OH)2})} = \frac{\pu{0.155 g L-1}}{\pu{241.21 g mol-1}} = \pu{6.43E-4 mol L-1}$$
And, since it's a saturated solution, then a nearly complete dissociation can be assumed:
$$[\ce{Pb^2+}] = c(\ce{Pb(OH)2}) = \pu{6.43E-4 mol L-1}$$
$$[\ce{OH-}] = 2 × c(\ce{Pb(OH)2}) = 2 × \pu{6.43E-4 mol L-1} = \pu{1.29E-3 mol L-1}$$

RE: "And, since it's a saturated solution, then a nearly complete dissociation can be assumed:..." Another way of saying this is that the concentration of $\ce{[OH-]}$ is high enough that the autodissociation of water can be ignored. In yet other words, essentially all the $\ce{OH-}$ is from the dissolution of the $\ce{Pb(OH)_2}$.


text:
Is there an edible version of nylon string?
I'm working on a project where I have to check the feasibility for product ideas. It has to connect other parts and float on the water's surface.
My friend thought that maybe glass noodles would be thin and transparent enough to work in this product. Maybe there is another option that maybe is even water-soluble.
I also looked at PVA, but I don't think it is edible.
Thanks for any suggestion.

A search of "bioplastics" may find what you are looking for

Ok, I've looked a bit. I can't seem to find any information about the edibility of the products. I also didn't find any real strings or threads.

"Edible" is more or less another word for "biodegradable". And water soluble and *high tensile strength" is a contradiction in terms. Spider silk springs to mind, or generally protein fibers, aka. wool.

Surgical thread made of chitin?


text:
Reaction energies of solids from quantum chemistry?
I was reading a paper by Catlow et al. [1], and they mentioned the dissolution of copper into copper ions, which is given by this equation:

$$\ce{Cu2O(s) + 2 h+ <=> Cu+(aq) + 1/2 O2(g)} \tag{2.5}$$

where $\ce{h+}$  is a hole.
I am wondering whether it is possible to calculate the free energy of this reaction using quantum chemistry software? And which software will be able to perform this? I believe that this is not possible using periodic DFT due to the presence of charged ions.
References

Catlow, C. R. A.; Guo, Z. X.; Miskufova, M.; Shevlin, S. A.; Smith, A. G. H.; Sokol, A. A.; Walsh, A.; Wilson, D. J.; Woodley, S. M. Advances in Computational Studies of Energy Materials. Philosophical Transactions of the Royal Society A: Mathematical, Physical and Engineering Sciences 2010, 368 (1923), 3379–3456. DOI: 10.1098/rsta.2010.0111.

Ad 1, you equation is not clear and not in line with your text. Ad 2 the answer to the question is strongly depend on what accuracy you need, what effects you want consider etc.

@koroma Thank you, I corrected the reference (we are using ACS style whenever possible) and the equation number. In the paper the reaction you provided is labeled under (2.5), not (2.4). I also tweaked formatting a little so that the symbol for a hole is upright and easily recognizable. If you are interested in further details, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
Examples of Lewis base nature of sulfur dioxide
I have read that sulfur dioxide acts as both Lewis base (due to presence of lone pair over $\ce{S}$ and/or $\ce{O})$ and Lewis acid (due to presence of vacant d orbital). I found many examples of $\ce{SO2}$ acting as a Lewis acid (towards $\ce{NMe3},$ $\ce{H2O},$ etc.), but could not find any for its Lewis base behavior. Could someone provide some reactions/equations, please?

Sulfur dioxide doesn’t use ‘vacant d orbitals’ in bonding.

See here for Lewis acid behavior: https://chemistry.stackexchange.com/a/99928/61578

You generally want to find a much stronger Lewis acid (typically forming a solvate/adduct) or make a complex with a transition metal: this way either unshared electron pair of oxygen or sulfur can be donated.
Canonical examples are given in Shriver & Atkins’ Inorganic Chemistry [1, p. 135]:

To act as a Lewis base, the $\ce{SO2}$ molecule can donate either its $\ce{S}$ or its $\ce{O}$ lone pair to a Lewis acid. When $\ce{SbF5}$ is the acid, the $\ce{O}$ atom of $\ce{SO2}$ acts as the electron pair donor, but when $\ce{Ru(II)}$ is the acid, the $\ce{S}$ atom acts as the donor … $(\ce{[RuCl(NH3)4(SO2)]+}).$

$\qquad$
Figure 1. Left: Crystal structure of $\ce{SbF5 · SO2}$ (ICSD-67273 [2]). Right: Crystal structure of $\ce{[RuCl(NH3)4(SO2)]+}$ (ICSD-26061 [3]). Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#3050F8}{\Large\bullet}~\ce{N}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#90E050}{\Large\bullet}~\ce{F}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#248F8F}{\Large\bullet}~\ce{Ru}$; $\color{#9E63B5}{\Large\bullet}~\ce{Sb}$.
Reactions for the synthetic procedures adapted from the corresponding articles [2, 3]:
$$
\begin{align}
\ce{SbF5(s) + SO2(l) &->[\pu{-40 °C}] SbF5 · SO2(sln)} \tag{R1} \\
\ce{[Ru^{II}(NH3)4(HSO3)2](aq) + 2 HCl(aq) &->[reflux][15 min] [Ru^{II}(NH3)4(SO2)Cl]Cl(aq) + 2 H2O(l) + SO2(aq)} \tag{R2}
\end{align}
$$
References

Atkins, P. W. Shriver & Atkins’ Inorganic Chemistry; W.H. Freeman and Co.: New York, 2010. ISBN 978-1-4292-1820-7.
Minkwitz, R.; Molsbeck, W.; Preut, H. Kristallstruktur von $\ce{SbF5 · SO2}$ / Crystal Structure of $\ce{SbF5 · SO2}.$ Zeitschrift für Naturforschung B 1989, 44 (12), 1581–1583. DOI: 10.1515/znb-1989-1218. (Free Access)
Vogt, L. H.; Katz, J. L.; Wiberley, S. E. The Crystal and Molecular Structure of Ruthenium-Sulfur Dioxide Coordination Compounds. I. Chlorotetraammine(Sulfur Dioxide)Ruthenium(II) Chloride. Inorg. Chem. 1965, 4 (8), 1157–1163. DOI: 10.1021/ic50030a015.


text:
What is the maximum number of electrons that can be found in an atom when its last electron has the following quantum numbers?
The question is as follows:

What is the maximum number of electrons that can be found in an atom when its last electron has the following quantum numbers: $n=4, m_{l}=+3$

Since $m_{l}=+3$ I know that $l=3$ and then I wrote them all:
$$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}$$
which means that the maximum number of electrons is $70$, but the actual configuration in the answer was:
$$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 5d^1, 4f^{14}$$
which means that the maximum number of electrons is 71. What didn't I count for?

Here are the electron configurations of the relevant elements:

Source: https://chemistrytheperiodictable.files.wordpress.com/2017/11/electronic-config-lanthanides.png?w=474&h=399
The first $f$ electrons get filled in the 6th period of the table of elements. First, the $6s$ get filled, then (for lanthanum) the $5d$ gets filled for the first time, and then the $4f$ subshell. As you go across the lanthanides, the $5d$ is sometimes filled with one electron and sometimes empty.

What is the maximum number of electrons that can be found in an atom when its last electron has the following quantum numbers: n=4,ml=+3

The answer depend on how you interpret "last electron". Going from Thulium to Ytterbium, you could argue the last electron is going into $4f$, whereas from Ytterbium to Lutetium, the last electron is going to $5d$. On the other hand, the $5d$ is filled before $5f$ going from Barium to Lanthanum. Comparing Lanthanum to Lutetium, all 14 electrons seem to be going into the $4f$ subshell, supporting the notion that the answer to the question is 71.
In the end, it is not the best question - it does not have a clear answer, and you don't really learn anything from the answer other than that the energies of $4f$ and $5d$ are close to each other.