text:
Is there a known relation between the diffusion of a molecule and the diffusion of the reactants?
I have a plasma consisting of elements A and B (which have a similar mass). These elements can react to form molecules AB.
$$
\ce{A + B -> AB}
$$
Now if my plasma or gas is expanding based on pressure gradients, I will have some sort of diffusion of these two element concentrations. In fact, this diffusion is the dominant process affecting the concentrations A and B. Based on my spectroscopic analysis, the rate of decay of both elements is similar/the same, so I model them as
$$
[\ce{A}](t) = A_0 \exp(-k_0t) \,, \\
[\ce{B}](t) = B_0 \exp(-k_0t) \,.
$$
Then the concentration of AB changes with
$$
\frac{d[\ce{AB}]}{dt} = k_1 A_0 B_0 e^{-2k_0t}\ - k_2 [\ce{AB}] \,,
$$
where $k_2$ is the rate constant for diffusion. The obtained formula for [AB] is then
$$
[\ce{AB}](t) = \frac{k_1A_0B_0}{2k_0-k_2} \left(e^{-k_2t} - e^{-2k_0t} \right) \,.
$$
This actually fits my data quite nicely, but one thing confuses me: When fitting my data for $[\ce{AB}]$ with this model, the parameter $k_2$ is exactly $2k_0$.
Or rather, they are very close, since the formula has a singularity at $k_2 = 2k_0$.  But if the formula is derived with this assumption directly,  one obtains
$$
[\ce{AB}](t) = (k_1A_0B_0)\, t\, e^{-2k_0t} \,,
$$
which also fits my data very nicely.
I don't understand why $k_2 = 2k_0$ and I am worried that I might say something very stupid in my paper when I try to explain this observation. But this is what I am thinking: $k_2$ is the rate of diffusion for AB, which is twice as heavy as A and B. It will have twice the momentum, which could lead to a doubling of the diffusion rate as well. Does this seem like a probable explanation? If not, which other effects will play a role?
Note that the plasma I am looking at is short-lived and changes constantly, which makes it difficult to apply textbook chemistry to it. Basically all signals I am looking at can be modeled as exponential functions or sums of exponential functions, and sometimes multiple processes might be involved. It's a lot of guesswork, and I am worried that I am actually on the wrong track here. But at least intuitively it makes sense to me that the diffusion of a heavier molecule should happen faster.
Since I was asked why I don't use a more complex diffusion/reaction model: Occam's razor, basically. I don't want to assume too much about my plasma. And I fear that a complex model might contain too many free parameters which would lead to overfitting of the data. (If I'm wrong about this approach, feel free to correct me and explain to me how to go about this. What I do not like are downvotes without a comment telling me what I'm doing wrong. None of the downvotes I have received bothered to do this, which is not helpful.)

It might help if you posted a sample of data to play around with. Obviosuly the cancellation of terms makes one wonder how you managed to fit anything with your first model.

@BuckThorn Well, my question is not really about my data, it's about my interpretation of the fit. I am asking if it makes sense that a reaction product has a diffusion rate that is the sum of the diffusion rates of the reactants due to how momentum is conserved. My equations are shown here to demonstrate how I came to this idea. And: there is no cancellation of terms. That is a misunderstanding of my description. If you do the math, you will see that it works out fine.

Sure, but you get some odd results when performing the fits with your standard model, which led you to posit what @porphyrin regards as a flawed or odd model, so you may want to give it a try. You have nothing to loose afai can see. It's called providing a minimal working problem.

What I am suggesting is also that numerical issues may be what's determining whether you can fit the data with a particular model. Cancellation of terms is usually a sign of an overparameterized model.

@BuckThorn This is pretty vague. How can the model be overparameterized? It is rather simple and only has three free parameters (of which two are found to be equal). Again, there is no cancellation of terms; if you could point out what you mean by that phrase, that would be helpful. The only part of my model that porphyrin thought of as flawed was the part that was directly derived from my experimental data. I don't think his argument was very solid, but I also had the feeling we were talking past each other.

More a set of queries and comments than an answer.
I don't understand why you make A and B vary with time irrespective of your reaction scheme.
You should really analyse using Fick's diffusion equations with the reaction scheme added and do so for each species. This will have to be a numerical calculation in space and time. (There are well established methods to do this. e.g. Crank-Nicholson). By the symmetry of your experiment you may be able to reduce this to one dimension and time.
You should be able to 'guestimate' the diffusion coefficient from kinetic theory at least their ratio which should be quite accurate.
Your second equation ('new formula') is that for a reaction $\ce{A + B ->[k_1] AB ->[k_1] }$ where both rate constants are the same. In your case there seems to be no reason why the two rate constants are the same, only you know enough about your experiment to explain this :)

You say there is only one rate constant in the second case, but there are two: $k_1$ and $k_2 = 2k_0$, or am I missing something? Isn't it $\ce{A + B ->[k1] AB ->[k2] }$? As for why I didn't use a more complex model - because that would be very complicated, and it's not really the focus of my work. And I was hoping the simple model would be sufficient, since it fits the data and I don't want to do unnecessary assumptions about the plasma.

When $k_2=2k_0$ then both are the same and your first solution  fails as you point out hence you have the scheme I suggest in my answer.

In your answer you imply that $k_1 = k_2$, and I don't see a good reason for this, since they will also have different units. I guess it was just a small mistake, but I was confused.

They just have to be numerically the same, which was my point. The scheme you propose seems to have no real physical basis unfortunately. You do really need to model this with combined diffusion and reaction.

They do not have to be numerically the same. I don't know how you come to this conclusion, but as you see in the equations, $k_1$ is just a proportionality factor in the end. As for the physical basis: my model is based on diffusion and reaction, but I make the assumption that diffusion is more dominant than reaction, so that the small effect of the reaction on the concentrations of A and B can be neglected. This assumption is justified by the fact that equilibrium calculations estimate the AB concentration as very low.

(But to be clear, I'm very thankful for your comments. Don't take this the wrong way!)


text:
Why will A atoms diffuse from β phase to α phase and B atoms from α phase to β phase?

At the initial compositions 1 and 2 of $α$ and $β$ respectively the chemical potentials of A and B atoms in each phase can be found by extrapolation of the tangents to the free energy curves at 1 and 2 to the corresponding sides of the free energy diagram, as shown above.
All atoms diffuse so as to reduce their chemical potential.
Therefore, A atoms will have a tendency to diffuse from $α$ to $β$ $(μ_\ce{A}^α > μ_\ce{A}^β)$ and B atoms will have a tendency to diffuse from $β$ to $α$ $(μ_\ce{B}^β > μ_\ce{B}^α)$

The point marked 1 and 2 are basically the compositions of two metal alloys that have been welded. Now annealing is performed at a particular temperature $T_1$ to bring about the change now why will A atoms diffuse from $\beta$ phase to $\alpha$ phase and B atoms from $\alpha$ phase to $\beta$ phase?
How will drawing a common tangent be different in interpretation than drawing two individual tangents at compositions 1 and 2?

Maybe i’m missing something, but I don’t see the combined molar fee energy curve on your diagram.

@ChetMiller This is all that I know.

@ChetMiller - the dashed line is the common tangent for equilibrium between the phases. OP - the common tangent line, and the tie to chemical potential, as well as the drive to have the same chemical potential for the components in the phases, is likely discussed in your text. If not, find a copy of Porter and Easterling in the library and read the first chapter or two.

@Jon Custer  I'm not the OP.  But, the part that threw me a little is that each phase has its own total molar free energy curve in the figure.  The two curves have a common tangent at the equilibrium concentration combination.  I'm more used to seeing a single total molar free energy curve for the two species, with 2 minima in the curve, and a common tangent to the two minimum portions of the curves.  The rest of the explanation makes sense to me.

@ChetMiller Ok ignore what you do not know explain the rest to me.

What I feel Is you can treat that as a single curve as well.@ChetMiller

Well, I think you can probably derive the single curve from the two given, but those given have to be weighted properly in terms of the mole fractions to obtain it.

@ChetMiller - interesting. In the Calphad world, one has free energy curves vs concentration for each phase, then go find the common tangents. Showing it as one curve just means you have reduced it to one ‘hull’, but that doesn’t really help understand what is occurring in various complex systems.

Phases are basically open systems within a close system isn't it?

OP - do you understand the connection between the intersection of a tangent to the free energy curve and the free energy of a component in the phase?

@ChetMiller@Jon Custer Please explain me the rest.

In response to "OP - do you understand the connection between the intersection of a tangent to the free energy curve and the free energy of a component in the phase?" No I do not understand its significance.Nothing at all.

https://chemistry.stackexchange.com/questions/19454/boundary-lines-in-phase-diagrams-and-the-lever-rule/19468#19468 Might help.

@JonCuster What if I had a G vs X curve with two minima at the two extremes and one maxima in between?

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text:
Has there been any independent verification of the results from this Chinese nanotube study?
Watts up referred to the South China Morning Post in reporting that a Tsinghua University team led by Wei Fei claimed in 2018 to have manufactured carbon nanotube fibre with tensile strength of 80 gigapascals (corresponding to an engineering tensile strength of 43 GPa).
This would be a result with potentially very significant implications concerning the future of the human race and its relationship with space. Fiber of that strength might be suitable for the cable of a Terran orbital elevator (Estimates of required strength vary from 60-100 GPa.)
I've not yet investigated the difference between "tensile strength" and "engineering tensile strength", so it seems likely they're not quite strong enough, but if it's true, it's still an exciting result.
Has this result been verified or reproduced independently?
The study was published in Nature and is available in ResearchGate, and a pretty nice plain language summary came out of ArsTechnica.

I upvoted your question, but I feel it's unlikely to get any answer and might even be off-topic. Nanotubes are a very specialized topic area inside chemistry, and you are essentially asking us to hunt down references for you. And you haven't told us why you want the result to be verified independently. Are you also researching on nanotubes or is this just an idle curiosity?

This would be a result with potentially very significant implications concerning the future of the human race and its relationship with space. Fiber of that strength might be suitable for the cable of a Terran orbital elevator.

You should write that in the question, so that it's more understandable. For what it's worth, I think an orbital elevator would not be feasible/usable in at least the next 50 years, but who knows what will happen.

The entry on ResearchGate equally lists the publication's unique doi (https://doi.org/10.1038/s41565-018-0141-z) to be get to the landing page at the journal.  At present (July 27, 2021), the publisher knows there are 127 citations of this work, and a service like scite_ 134.  Normally, you would read these (at least briefly) to check if their citation is about a method, a finding of interest mentioned earlier, if these support / extend / criticize the earlier work.

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text:
What is the major product obtained on acidification of substituted epoxide?
The problem

Source : MS Couhan (problems in organic chemistry . Chapter: Alcohols , phenols and ethers).
My Thoughts

If I proceed along $\ce{path 1}$ , $\ce1a$ is formed . It is unstable due to inductive withdrawing effect of $\ce{C=O}$ neighboring to it. Eventually $\ce2$ is obtained.
However, along $\ce{path 2}$ relatively more stable $\ce{3a}$ (compared to $\ce{1a}$) is formed. $\ce3a$ on ring expansion gives $\ce{3b}$ , leading to $\ce{4}$  ($\ce{3b}$ is relatively unstable compared to $\ce{1b}$).
My question
Which of the two paths should I choose that would lead to a major product ?

The reaction of epoxyketone 1 affords diketone 2 under photochemical and thermal conditions ostensibly through the diradical.1 Boron trifluoride provides the diketone 3 through acyl bond migration and not via alkyl group migration as in your intermediate 3a. I have not located a proton-catalyzed reaction. Unfortunately, diketone 3 was not one of your choices.

1) John R. Williams, George M. Sarkisian, James Quigley, Aaron Hasiuk, and Ruth VanderVennen, J. Org. Chem., 1974, 39, 1028.

@ user55119 this solves my problem. The intermediate formed after reaction with Boron triflouride is more stable then 3b or 1a.This could be the major product.

Cation 1a is a bad choice in any event.


text:
Finding atomic number given energies of emitted photons

A hydrogen-like atom (atomic number $Z$) is in a higher excited state of quantum number $n$. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.2 eV and 17 eV. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV. Find the atomic number $Z.$
A) 2
B) 4
C) 6
D) 3

Energy of first transition $= \pu{17eV}+\pu{10.2eV}= \pu{27.2eV}$
$$27.2=Z^2(\frac{1}{2^2}-\frac{1}{n^2})\to \frac{Z^2}{n^2}=\frac{Z^2}{4}-27.2 \tag{1}$$
Energy of second transition $= \pu{5.95eV}+\pu{4.25eV}= \pu{10.2eV}$
$$10.2=Z^2(\frac{1}{3^2}-\frac{1}{n^2})\to \frac{Z^2}{n^2}=\frac{Z^2}{9}-10.2 \tag{2}$$
Combining $(1)$ and $(2)$:
$$\frac{Z^2}{4}-27.2=\frac{Z^2}{9}-10.2 \to Z^2=\frac{17\times36}{5}\to Z=11.06$$
I can’t find where I am going wrong.

Don't add the energies to start with but take two steps, so try, for example  level $m\to n = 10.2$ then $n\to \mathrm{final ;level}= 17$ eV etc where you have to work out m and n  and then get to Z. You have two sets of transitions from the same starting  level.

But why can’t we add the energies? Doesn’t adding it just give the energy of the spectral line?

Also I tried doing that, but I end up with the same answer.

According to the formula you are using, energy should be a dimentionless entity. Rydberg formula called so for a good reason. Also, please try to write a concise and short title reflecting the question you are asking. An abstract call for help isn't constructive (that may explain the downvotes and closing as homework despite you have demonstrated your attempt).

That’s cool, but I tried the way @porphyrin told me, and I still end up with the same equation

Note that you have exactly enough information to determine the difference in energy between energy levels $n = 1$ and $n = 2$.

Karsten had the right idea, but the wrong formula.
$$E = \pu{13.6 eV} \cdot Z^2\cdot \left(\dfrac{1}{3^2} - \dfrac{1}{2^2}\right)$$
$$\pu{17.0 eV} = \pu{13.6 eV} \cdot Z^2\cdot 0.138889$$
$$ 8.99999 = Z^2$$
$$ Z = 3.00$$

Yup I figured that out. That was a real dumb mistake on my part

Oops, thanks for fixing my stupid mistake. Should have written it down on paper...

Edit: I'm leaving this for context, but the correct answer is the one by MaxW.
This problem has four unknowns (n of the excited state, n of the intermediate state going to the first excited state, n of the intermediate state going to the second excited state, and the atomic number Z). You also have four given quantities, the photon energies.
I like the OPs approach of adding the energies - that eliminates two unknowns we don't need for the answer. I would eliminate another unknown (n of the excited state) by asking for the energy to go from the second excited state (n = 3) to the first excited state (n = 2). You get that energy by going from n = 3 to the initial excited state back to n = 2, with an energy of 17.0 eV (see OP's notes).
Now, we just have to plug this into the correct relationship between Z, n(initial) and n(final). The formula the OP used is suspect because there is an energy on the left-hand side, but only unit-less integers on the right hand side. Wikipedia (Bohr model) gives the energy as:
$$E = \pu{13.6 eV} \frac{Z^2}{3^2 - 2^2}\tag{oops, incorrect formula}$$
so
$$Z^2 = E \cdot \frac{3^2 - 2^2}{\pu{13.6 eV}}= \pu{17.0 eV} \frac{5}{\pu{13.6 eV}} = 6.25$$
Finally, we take the square root to get:
$$Z = 2.5$$
This is not an answer that makes sense because we expect Z to be an integer.

So what should be the answer, 2 or 3, assuming we approximate the value to the nearest.

We should either question my answer, or question the question. One thing you could do is just to assume Z = 2, 3, 4, or 6, and calculate n for the excited state.

You are right though, I made a mistake in my formula. I will try it again, but according to you the answer is 2.5, so my question from the previous comment still stands

@Aditya I don't have an answer - I posted my calculation to demonstrate my confusion.


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Why is the hydroxide ion not considered to be amphoteric
Why is the hydroxide ion not considered to be amphoteric.
Why is $$\ce{2OH- <<=> H2O + O^{2-} }$$ not considered ??
Even though the equilibrium would mostly lie towards the left but still the hydroxide ion can accept a hydrogen ion and donate hydrogen ion , but still it fits the definition for an Amphoteric substance - A substance that can both donate and accept an $\ce{H+}$ ion.

Hydroxide ion isn't a substance...

The hydroxide ion can act as an acid, though it is a tremendously weak one, and in almost every case its basic character dominates its behaviour. Ultimately most chemical species are technically amphoteric, if you expose them to strong enough acids and strong enough bases in sufficiently forceful conditions. However, amphotericity is a term most often used in a practical sense.

The $\ce{O^{2-}}$ dianion is very unstable, especially in water. There is too much electronic density for the nuclear charge of oxygen to hold. We generally do not consider $\ce{OH-}$ to be amphoteric because, except in very extreme conditions, it will not give its proton away.

Beside content (+1): For future reference, ChemSE benefits from mhchem's markup which eases input and display of chemistry-related content in questions, answers and comments -- just have a look at https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here.

How about heating a metal hydroxide such as magnesium hydroxide?  At sufficiently high temperature the hydroxide decomposes according to the reaction
$\ce{Mg(OH)2 -> MgO + H2O}$
which is most simply interpreted as a proton transfer from one hydroxide ion to another:
$\ce{2 OH^- -> O^{2-} + H2O}$
This "disproportionation" would not take place at ambient temperature, but upon heating it's driven by the entropy of water vapor and the superior lattice energy of the solid oxide product.
Magnesium is far from unique here.  Most metal hydroxides do the same thing, although the simple proton transfer concept is compromised if the compound has a high degree of covalent character as the oxides would with some transition metals.


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Why doesn't the solute settle out of a solution, such as brine or sugar water?
If I have a solution of salt or sugar dissolved in water, the solute will never settle out or "fall to the bottom", no matter low long I wait. Why is that?
Since salt and sugar (in their pure, solid forms) are denser than water, I would intuitively expect that gravity would eventually pull all the salt and sugar to the bottom. Why doesn't this happen?
Now, I'm guessing that salt and sugar solutions have lower energy than mixtures of solid salt or solid sugar with water. So that may explain why salt and sugar don't spontaneously settle out; reactions tend to proceed from a high-energy state to a lower-energy state, not vice versa. (In "explain like I'm 5" terminology, maybe sugar sticks to water better than sugar sticks to sugar.) Is that accurate?
I've heard that the favorability of a reaction is determined by such things as energy, entropy and enthalpy, but I don't know the details. I'm guessing the answer has something to do with those.
This question is inspired by this question over at the Earth Science site: Why does the salt in the oceans not sink to the bottom?
(I'm surprised that I wasn't able to find a previous question about why solutions don't settle. All I found was a question about why colloids don't settle, which is also an interesting question, but it's not the same thing.)

Same reason why all the air in your room doesn't just fall to the ground. If all the sugar were to sink to the bottom, the top wouldn't have enough sugar, and some sugar would diffuse back up. Having a lower potential energy is great, but isn't the only important thing: entropy is pretty crucial too.

We have the opposite problem called atmosphere escape: "Atmospheric escape of hydrogen on Earth is due to Jeans escape (~10 - 40%), charge exchange escape (~ 60 - 90%), and polar wind escape (~ 10 - 15%), currently losing about 3 kg/s of hydrogen. The Earth additionally loses approximately 50 g/s of helium primarily through polar wind escape. Escape of other atmospheric constituents is much smaller. A Japanese research team in 2017 found evidence of a small number of oxygen ions on the moon that came from the Earth." from wikipedia

I notice I've gotten a couple of close votes as "too broad". I thought that my question was pretty clear and focused (my fundamental question is "why doesn't the solute settle out of a solution?", and the rest is just clarification), but maybe not. Would my question be better if I removed the parts about air and left just the parts about aqueous solutions? Or is there something else I can do to improve it?

I agree with you. The question is just based on faux assumptions. Let aside stirring and dynamics of various nature, the composition of atmosphere isn't constant with altitude, nor is the salinity of the seas. High enough silos suffices, by the way.

Duplicate of https://chemistry.stackexchange.com/questions/58420/why-does-a-distinction-between-homogenous-and-heterogenous-mixtures-exist

There's also https://chemistry.stackexchange.com/questions/61140/why-dont-heavy-and-light-gases-separate-in-the-atmosphere and lots of its dupes.

There are also some good answers here.

Possible duplicate of Why does a distinction between homogenous and heterogenous mixtures exist?

Possible duplicate of Why don't heavy and light gases separate in the atmosphere?

You discuss two different phenomena, solubility and concentration gradients due to gravity.
Solubility can be explained using the concept of entropy. Solubility requires the the free energy of a solid be higher than the corresponding substance dispersed in solution. Surprinsingly perhaps, additional entropy associated with dispersion of the solute in the solvent suffices to drive the process (interactions can certainly be important, but are not at minimum necessary, although much depends on how much you are attempting to solubilize, and at what temperature).
Gravity can (and does) cause concentration gradients, and can therefore under the right circumstances cause precipitation. The problem is that gravity is a very weak force, so it's effects on light molecules are, well, weak. Gravitational effects require large scales or masses to be significant. Density differences are argued to be critical driving force behind water cycling in the oceans. The air atop Mt Everest is so thin people suffocate without access to an auxiliary supply of oxygen. So these effects do happen, and matter. And you can use a centrifuge to separate components in a mixture, which is analogous to applying gravitational forces, just scaled up in strength.

Supplemental answer for additional background/insight:
Biologists generate salt gradients in test tubes all the time using high-speed centrifugation, which allows them to produce much higher potential energy gradients than gravity can, and in an environment free of ocean-scale mixing and temperature gradients.
If the concentration is close to saturation and the gradient is high enough, this can certainly lead to precipitation. Otherwise you just end up with a static gradient in concentration, being the balance between buoyancy forces downward and thermally driven diffusion upward.
See for example Wikipedia's Buoyant density centrifugation
Heavy salts like $\ce{CsCl}$, $\ce{Cs2SO4}$, and $\ce{KBr}$ have been used historically.
However, you can't use this equation which applies to centrifugation because it is based on "gravity" increasing with distance from the center of the rotor or "downward" whereas in the ocean gravity is fairly constant, but can decrease or increase slightly with depth depending on the specific location and how close you are to dense rock at the bottom.


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Solubility product constant of copper(II) bromide in water
One of the questions from chegg.com mentions $K_\mathrm{sp}(\ce{CuBr2}) = 6.3\cdot 10^{-10}$, but I suspect this value might be incorrect.
Could anyone confirm or refute that $K_\mathrm{sp}$ value?
For CuBr I found Ksp values of 6.3x 10-9 and 4.0x 10-8 which is around that range. One explanation would be that the website would have wrongly attributed the Ksp of CuBr to CuBr2.
I wish Springers had the exact same paper for CuBr2, that would be just perfect. It's a solid source for the CuBr dat
As Ivan points out, wikipedia lists the solubility of CuBr2 in water as 55.7 g/100 mL (20 °C), but without listing any source. Though this value is more likely to be correct, I'm still stuck between two undocumented values that aren't close to indicating the same thing.
Buttonwood shared the link to a paper regarding CuBr which is a solid source as to it's solubility. Anything regarding CuBr2 even close to being that good would be perfect!

Sounds like nonsense to me. CuBr2 must be soluble in water fairly well, akin to CuCl2. Maybe it was CuBr?

https://en.wikipedia.org/wiki/Copper(II)_bromide lists a solubility of $\ce{CuBr2}$ in water with 55.7 g/100 mL (20 °C), whereas CuBr is listed as "slightly soluble".

@IvanNeretin I thought so too. For CuBr I found Ksp values of 6.3x 10-9 and
4.0x 10-8 which is around that range. Do you know the Ksp of CuBr2?

@Buttonwood The wikipedia page doesn't list any source. Though more likely to be correct, I'm still stuck between two undocumented values that aren't close to indicating the same thing.

Sigma-Aldrich sells it and they say at their website that it is soluble in water, alcohol and acetone.

I don't think this question is bad or should be closed as homework. OP demonstrated significant effort in finding related experimental data (CuBr, which is also in agreement with CRC handbook) and it's not their fault that the question on the third-party site is likely presenting false data or data for the wrong compound.

@Hans The open-access preview of e-EROS, https://onlinelibrary.wiley.com/doi/abs/10.1002/047084289X.rc207.pub2 «Copper(I) Bromide» states «Solubility: insoluble in H2O and most organic solvents; partially soluble in dimethyl sulfide.»  The entry https://onlinelibrary.wiley.com/doi/10.1002/047084289X.rc206.pub2 «Copper(II) Bromide» states «Solubility: very sol water; sol acetone, ammonia, alcohol; practically insol benzene, Et2O, conc H2SO4.» Springers preview (https://link.springer.com/article/10.1007%2FBF00647017) states CuBr's solubility of 0.00064 M in presence of 0.0989 M KBr at 24.8 C.

@EdV After your comment I looked at the product page, and it does list some data, but no Ksp or numerical solubility values :( This does give an indication as to what values are possible and which ones aren't though

@Buttonwood I wish Springers had the exact same paper for CuBr2, that would be just perfect. It's a solid source for the CuBr data.

@Hans Since it is soluble, there is no solubility product constant. It is just a matter of soluble, slightly soluble, highly soluble, but not a tiny solubility. For me, the hygroscopic is telling. Anyway, the actual solubility must be out there somewhere! Good luck!

@Hans Searching google "solubility CuBr2" yields an excerpt «solubility of inorganic compounds.pdf» from the 91st edition of the CRC Handbook of Chemistry and Physics, section «Aqueous solubility of inorganic compounds at various temperatures» given away by the Colorado State University.  Among the 300 compounds is CuBr2:  55.8% mass percent of solute @ 25C.  Go to the science library and browse through these thick telephone books.  Or search in Gmelin / Reaxys, if you prefer the electronic database (non-free).

@Hans Could you integrate your comment about CuBr into the question? Maybe then someone could write up an answer before the question potentially gets closed.

@KarstenTheis Done! I didn't know exactly which comment about CuBr so I integrated all my comments into the question :)


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error in making a series of standards out of a stock solution
I want to make a series of standards to make a calibration curve. The analyte is a volatile substance so I'll have to use some glass "containers" with a membrane lit so I don't lose too much of the analyte. The glass container has a certain volume so when making my stock solution the amount of solvent I will use is fixed.
So if I use a big amount of analyt to make my stock solution I will have to use a small amount of stock solution to make the desired concentrations of standards
And if I want to use a big amount of analyte to make my standards I will have to make a low concentration stock solution. Which means a low amount of analyte, because the amount of solvent I can use is fixed by having to use the container with the membrane lit.
When I make stock solution:  If I add a greater amount of analyte the error on the concentration I want to make will become smaller
When I make a standard from this stock solution it's the same thing: If I add a bigger amount of analyte, the error on the concentration I want to make will become smaller.
My question:
Which of the two will have a bigger effect? Or, in other words, is it more important that the error on making my stock solution is smaller or is it more important that the error on making my standard is smaller? I ask this because I will have to choose.
Thanks a lot in advance

Now I'm thinking my analyte is ofcourse more volatile then my stock solution so there is a bigger chance of an error there. So better to make  the stock solution more accurate

Sorry is very difficult to understand the logic  under which the restraints you mention emerge. By the way you should use the most concentrated one and operate by dilution just prior  of the measurements.  And work as fast as possible. It can be even difficult but at least you can check if is feasible or not

Thanks anyway :) I'll just try it and see how it works out!


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Can common refrigerants be chemically converted into something that is a stable solid at STP?
I've been thinking how refrigerants - which are highly potent greenhouse gases (hundreds and even thousands of times more potent than CO2) - are likely to be vented to atmosphere eventually either due to system leakage, accident, neglect, theft of chiller units, or deliberate discharge. It makes me wonder if, for climate change mitigation/adaptation purposes, whether we should consider all the refrigerants deployed and in storage today as being something that will ultimately be vented.
Suppose that mass migration, economic collapse or upheaval, or other calamities may lead to the abandonment, destruction, or even careless salvage of thousands or even millions of HVAC/refrigeration systems. Given that supposition, is there a practical chemical process that could be used to convert the refrigerants that are in common use today to another material - like maybe a polymer? - that would be at least somewhat stable at standard temperature and pressure? One that sequesters CO2 in the process would be a bonus but I don't expect it'd matter very much.

Ammonia may once again become a popular choice as a refrigerant in the near future, exactly to avoid the environmental issues with CFCs and HCFCs. Toxicity is an issue, though.

@NicolauSakerNeto - ammonia is also explosive, yet another issue.

Polymerization might be viable (if complex and costly; one important feature of refrigerants is their inertness) for selected refrigerants. It's simpler to convert the refrigerant into gases with a smaller heat trapping capacity, rather than convert it into a polymer. Presuming that, after the fact or prior to the calamity, you'd want to dispose of refrigerant in abandoned units, your best bet might be simply to incinerate it (see Figure 1 in this EPA article, or search for "refrigerant incineration"). The question is how to perform incineration in place rather then remotely at a centralized collection facility. It would probably remain much simpler to collect the gas and incinerate it at centralized locations, rather than travel around with a mobile incinerator.


text:
Number of stereoisomers of trisubstituted cyclopentane
When 4-methylcyclopent-1-ene reacts with bromine, then 1,2-dibromo-4-methylcyclopentane forms. And I am stuck on determining the amount of stereoisomers that can form as a result.

1,2-dibromo-4-methylcyclopentane has two chiral carbon atoms, those of bromine. And an internal mirror plane, we should have 3 optical isomers.
But I am confused upon drawing the following:

1 and 4 appear to be the same, 2 and 3 appear to be the same.
5 and 7 appear to be the same, 6 and 8 appear to be the same.
This gives me 4 isomers. So I draw them different again:

The puckered structure leaves me doubting myself: while I think that the internal mirror plane causes there to be only 2 different cis structures instead of 4 (or instead of 1 if it had been 1,2-dibromo-4,4-dimethylcyclopentane), I am not sure if the trans structures are actually the same.

So propene or pentene?

In either case there's only 4 isomers, because of plane of symmetry.

Thank you Mithoron for your comments: pentene is what I meant, I've edited my question. I will also attempt to upload pictures of what I exactly mean.

Almost everything that needs to be said has been said. But I do have a few comments about configurations. In enantiomeric, trans-dibromides 1 and 2, the methyl groups are chirotopic and non-stereogenic. Inversion of the stereochemistry of the methyl group in either 1 or 2 returns the same structure, respectively. Thus, there is no designation for stereochemistry at the C4 methyl center. On the other hand, inversion of the stereochemistry of the methyl group in 3 gives 4 and the same operation on 4 produces 3. Therefore, the methyl center in achiral isomers 3 and 4 is achirotopic and stereogenic. Accordingly, the C4 centers have lowercase descriptors, r and s, respectively. This site might be of help.

Thanks for this info, really informative and contains the terminology and information I was looking for!

Nice pictures and analysis!
There are four different stereoisomers: one pair of enantiomers (the trans structures) and two achiral isomers (the cis structures).
In a mixture of all four, there will be 3 sets of boiling points, peaks in chromatographic separation (with achiral columns), etc.

1,2-dibromo-4-methylcyclopentane has two chiral carbon atoms, those of bromine. And an internal mirror plane, we should have 3 optical isomers.

This argument works only if both bromine-bound carbons are (R), or both are (S). Otherwise, the methyl-bound carbon has two different substituents, and could be counted as chiral as well.


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Non-zero current flow through an electrolytic cell, when an electric potential lesser than that needed for electrolysis is applied?
For instance, the potential needed for water splitting is $\pu{1.23V}$. However, there is current flow even at lower potentials (say $\pu{0.5V}$). Clearly, there is no redox reaction happening in the cell as evidenced by lack of bubbles. What is the cause of this current?

As you noticed there is a small amount of current which keeps on flowing when there is no electrolytic decomposition. This "problem" is aptly named as non-Faradaic current. Faradaic current refers to current due to oxidation and reduction. Here is a detailed review: The difference between Faradaic and Nonfaradaic processes in Electrochemistry


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Why can solid lubricants operate only at lower temperature in oxidizing environment but higher temperature in reducing environment?

The two main dry lubricants are graphite and molybdenum disulfide. They offer lubrication at temperatures higher than liquid and oil-based lubricants operate. Dry lubricants are often used in applications such as locks or dry lubricated bearings. Such materials can operate up to 350 °C (662 °F) in oxidizing environments and even higher in reducing / non-oxidizing environments (molybdenum disulfide up to 1100 °C, 2012 °F).

Source: Wikipedia contributors, "Dry lubricant," Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/w/index.php?title=Dry_lubricant&oldid=870516892 (accessed July 24, 2019).
Why can solid lubricants operate only at lower temperature in oxidizing environment than in reducing environment?

Likely they can get oxidised and at a certain point even burn. ..

Ever use a charcoal grill? Of what element are charcoal and graphite made?

Graphite is elemental carbon, and carbon is unstable in the presence of oxygen because they react to form carbon dioxide. The rate of this reaction: $\ce{C + O2 = CO2}$ is very slow and has a high activation energy, so a piece of graphite can sit in oxygen at room temperature for millions of years without anything happening to it. Once temperature increases, the reaction becomes much more rapid. From my own experience, a fist sized chunk of graphite can survive for several hours when heated to above 1000 °C in atmospheric oxygen.
Molybdenum sulfide is composed of two things that are likewise not stable in Earth's surface atmosphere: molybdenum and sulfur. It reacts by: $\ce{2MoS2 + 7O2 → 2MoO3 + 4SO2}$. As with graphite, this reaction is strongly temperature dependent.
There isn't a single temperature at which the lubricant is "not stable" any more. It is a balance of the required lifetime of the lubricant, the temperature, and how oxidising or reducing the environment is.


text:
Bond energy in hydrogen
Why is hydrogen-hydrogen bond the strongest of all single bonded compounds?
The bond energy principle here is based on the stability of atom after bonding, so even other atoms, when they combine with some other atoms and form covalent bonds, they would attain stability,
In what way is this different from that?
Is doublet configuration more stable than the octet configuration?
Or is there a totally different explanation lying?

Can you add numbers to your question? Bond energy of H-H, C-H, O-H, ...

Firstly the bond is non polar covalent bond ( as there is no difference in electronegativity of both hydrogen atoms). Secondly , the small size of the atoms also play a major role. Since the size of hydrogen atom is so small so the bond pair of electrons is bound by the electrostatic force of attraction between the electrons and nucleus. It can also be understood as the closer the electrons to the nuclei the more negative potential energy it possess which means more stability resulting in high bond energy.


text:
Calculating Equilibrium Constant from Cell Potential at Nonstandard Temperature
I'm currently working on a problem for a general chemistry class.
This is the voltaic cell given in the problem:
$$\ce{Sn^4+(aq) + Cu(s) -> Sn^2+(aq) + Cu^2+(aq)}$$
$$E^\circ_{cell} = \pu{-0.21 V}$$
What is the value of the equilibrium constant at $\pu{115 ^\circ C}$?
For this I'm using this equation $$-nFE^\circ_{cell}  = RT\log(K)$$ Plugging this into the equation:
$$ (\pu{2 mol e^-})(\pu{96500 C}/\pu{mol e^-})(\pu{-0.21 V}) = (\pu{8.314 J/molK})(\pu{388 K})\log(K)$$ I get a value of $$K = 3.49 \times 10^{-6}$$ However, my instructor's solution is using this equation $$- E^\circ_{cell} = (0.0592/n)\log(K)$$ and gets a value of $8.04276 \times 10^{-8}$.
I thought you could only use that equation if the temperature is $T = \pu{298 K}$ (standard state conditions). Not sure whether I'm right or wrong about this, any insight is appreciated.

You'd want  to use the van't Hoff equation.

What @Buck Thorn said is correct and works fine: I just did the calculation quick and dirty (neglecting entropy, so enthalpy is approximated as the free energy at 298.15K) and got about $1.6 x 10^{-5}$ for the equilibrium constant at 115°C. What your instructor got for the equilibrium constant is fine at 298.15K.


text:
What is the benefit of high surface area electrodes?
Some electrolytic apparatus is equipped with high surface area electrodes. The number of reduced/oxidized atoms is proportional to the total charge flowing through the cell. Does increasing surface area somehow enable use of greater currents?

Yes. The charges all have to go to the electrodes, but the concentration of ions cannot increase much locally, they'd repell each other. Even more: The larger the surface, the lower the local concentration of ions at the electrode, the larger the gradient letting new ions diffuse in there.

A current is simply the rate of flow charge (and hence the rate of flow of electrons); the more electrons flowing, the higher the current.
Anions migrate across the potential gradient between electrodes, carrying the electrons to the anode and the anions are then oxidised.  The oxidised species (often uncharged) must then diffuse across a concentration gradient to move away from the electrode (via Fick's laws).  Until the oxidised species moves away from the electrode surface, they are blocking the surface for more anions to migrate in, and hence the current is eventually limited when the rate of migration of anions is equal to the rate of diffusion of the oxidised species.  This exchange happens at a fixed rate per unit area at the surface of the electrode.
When we use a larger surface area electrode, there is more area for this exchange to happen, therefore more species involved in the exchange, therefore more electrons can be deposited in a given timeframe, hence permitting a higher current to flow.


text:
Nucleophilicity in polar aprotic solvent (DMSO)

The strength of a nucleophile in polar aprotic solvent is proportional to its basicity. Lone pair on O- in the molecules will donate an e- pair to an atom. I'm not sure why the alkene ring in 2 makes it a worse nucleophile compared to 1. Benzene ring is more stable than cyclohexane but the double bond is not the nucleophile in the molecule. Is it due to steric hindrance?

let me start with nucleophilicity of halides. In protic polar solvent, $\ce{F^-}$ is more solvated (hydrogen bonding) than $\ce{I^-}$ and hence less reactive.
In polar aprotic solvent $\ce{F^-}$ solvation is less and therefore more reactive.
Such behavior may not be applicable here, since phenoxide is resonance stabilized while other compound is not. Therefore phenoxide $\bf{2}$  is less nucleophilic than cyclohexanoate $\bf{1}$.

The negative charge is actually not all on the oxygen on for the phenolate on the right. It can be delocalized in the ring, which makes the whole molecule less basic. The negative charge is fully localized on the alcoolate on the left.

A polar aprotic solvent does not hydrogen bond to nucleophiles to a significant extent, meaning that the nucleophiles have greater freedom in solution. Under these conditions, nucleophilicity correlates well with basicity – and fluoride ion, being the most unstable of the halide ions, reacts fastest with electrophiles.


text:
What is the difference between the sugar in a fresh fruit and the sugar in a juice?
In this paper of the WHO, there is a distinction between intrinsic sugars, who are incorporated within the structure of intact fruit and vegetables, and sugars naturally present in fruit juices. (page 15/59)
What are the differences between those two sugars? How an intrinsic sugar can become another one when making a juice?

From the document: "Higher  intakes  of  free  sugars  threaten  the  nutrient  quality  of  diets  by providing significant energy without specific nutrients ". The sugar is the same, but it comes with or without other nutrients (and fiber). Also, it is easier to drink the juice of 4 oranges (roughly a cup) than to eat 4 oranges, so the sugar is released into the blood stream at a different rate.

In addition  juice you buy may not always be pure but contain added sucrose to make it sweeter.

Your question has a broad scope. Yet, I'll try to give a simple answer. Basically, naturally occurring sugars or sugars contained within unprocessed foods including vegetables and fruits are considered intrinsic sugars. Examples for the most common intrinsic sugars are glucose, fructose, lactose, and sucrose. Besides them, other minor amount of natural sugars in foods such as galactose and maltose are also considered intrinsic.
The other type of sugar (e.g., the sugar in a juice) is called extrinsic sugars. Extrinsic sugars include processed sugars or sugars that are added to food. Extrinsic sugars include syrups such as honey, added sugars in soft drinks, and artificial sweeteners. However, there is no clear difference between intrinsic versus extrinsic sugars (Ref.1):

Development of caries in man is associated with dietary sugars, but the relationship between sugar contents of individual foods and their cariogenic potential is unclear. A recent report categorized dietary sugars as intrinsic (mainly fruit and vegetable) and extrinsic (mainly added sugars, milk, and fruit juices) and concluded that consumption of extrinsic sugars (except milk) should be replaced with that of intrinsic sugars. This paper examines briefly the dental consequences of the recommendations.

Yet, Live Strong recommends following:

According to the American Heart Association, intrinsic sugars should be limited to at least half of your daily intake of carbohydrates and extrinsic sugars should be limited to no more then half of your maximum daily amount.

Reference:

W. M. Edgar, "Extrinsic and Intrinsic Sugars: A Review of Recent UK Recommendations on Diet and Caries," Caries Research 1993, 27(suppl l), 64-67 (https://doi.org/10.1159/000261605).

If I correctly understand your answer, there are no extrinsic sugar in 100% pure juice despite it is juice, right?

If the bottle says "no added sugar", then you are right. Otherwise, it contains some extrinsic sugar.

The WHO document does not distinguish extrinsic and intrinsic, but instead recommends limiting "free" sugar. "Free  sugars  include  monosaccharides  and  disaccharides  added  to  foodsand beverages by the manufacturer, cook or consumer, and sugars naturally present in honey, syrups, fruit juices and fruit juice concentrates." So for the WHO document, it does not matter if it is 100% pure juice or lemonade, they recommend limiting intake of both.

@KarstenTheis So, WHO recommends limiting intake of 100% pure juice despite the fact that the sugar in it is the same as the sugar in the fruit, right?

Yes, it’s not just about the kind of sugar, but also what it comes with. To give a current and tragic example, what is in a pulverized oxycontin pill is exactly the same as what is in the original pill. The effect on the person ingesting it is very different, though.


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Why Rutherford choose Alpha particle for his gold foil experiment?
Why did Rutherford choose Alpha particles for his gold foil experiment?

"Choose" is a strong word. There weren't that many options. Remember that this wasn't necessarily an experiment that anyone thought would produce interesting results.

Rutherford studies radioactivity and i just wanted to know why he was so interested in alpha particles only .

Rutherford and his group also did experiments on betas and gammas. Until Cockroft and Walton made their ion accelerator, those were the particles available for experimentation. Irradiating a gold foil with betas or gammas doesn't really do anything exciting, alphas did.

A better question would be "Why did Rutherford choose gold foil for his alpha experiment?"
The sine qua non was the alpha particle. It was a new probe, particles ejected from newly discovered radioactive polonium. What could he probe? Physicist/mathematicians can make conclusions if they have data, and one of the possible results they could gather was the alpha-stopping power of various materials. Bombard the object, look for penetration.
When there is no more penetration, you have a data point. One point. One point after lots of experimental time. But, if you could have an object only one atom thick, the alpha rays would all penetrate, but some that came close to a positive charge (or negative) might be deflected. There were different theories about the position of the positive and negative charge in atoms at the time, so this would be a good probe. Rather than getting one point per object, you can get a data point from each alpha particle!
An issue was getting a sheet of something as thin as one atom in thickness. Gold is one material that can be beaten so thin that you can see thru it - not one atom in thickness, but not very many. Few materials can be beaten so thin. And gold has a large atomic number, so if the protons were concentrated in one nucleus, there could be a large angular deflection as an alpha particle passed thru the thin sheet. And there were some very large angular deflections (~180 degrees) which strongly implied protons were concentrated in a nucleus.
Electrons pass thru matter like a bullet goes thru gelatin: friction with the electron cloud in atoms just slows the electrons down till they stop. Neutrons generally don't interact with matter; difficult to generate, difficult to detect. These particles would not have given such useful information, and this knowledge left alpha particles as the only interesting probe by elimination.

Excellent answer and neutrons were not discovered until about 1932 anyway.


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Why is a TTT diagram called isothermal transformation diagram?
Why are TTT diagrams called isothermal temperature diagram?As far as I know the temperature variation is plotted with the variation of time in a TTT diagram. Then why do we call it "isothermal".What is so constant about the temperature?

They are there to show you what happens while holding a sample at fixed temperature for different times. For non-isothermal conditions they are not particularly useful as normally presented.

“Put up in a place / where it's easy to see / the cryptic admonishment / T.T.T. / When you feel how depressingly / slowly you climb, / it's well to remember that / Things Take Time.” (Piet Hein, T.T.T.)

They are two names for the same procedure.  A steel is heated to a temperature in the austenite range then quenched/quickly cooled to some specific temperature. Held at that constant ( isothermal) temperature ( T) for a specific time (T) then quenched to room temperature for examination of the transformation (T) products.  You may be thinking of CCT ; continuous cooling diagrams ; where a steel sample is austenitized  then continuously cooled at some specific rate.  Then examined for transformation products. The results can have similarities depending on alloy content , etc.


text:
Is it possible to build or embed the SMILES representation of compounds in 3D?
I would like to know if there is a proper way to get the 3D information from the SMILES string of a molecule.

Is there a standard way to embed a SMILES string in 3D space?
Are there other representations of compounds which include their spatial information too?

SMILES is insufficient
SMILES strings do not encode 3D structure information.  They only convey atom type, connectivity and bond types.  InChI is like SMILES in this regard.
Thus, you will need either (a) an algorithm to infer or guess a plausible 3D conformation of a molecule or (b) a file type that has already specified the 3D arrangement of the molecule.
File types for storing, reading, and showing 3D conformations
Probably the most standard way to represent the 3D conformation of a molecules is with a *.mol file.  There are many tools to read such files.  You can read more about the format on Wikipedia.
Estimating a conformation from SMILES
You can also use computational tools to estimate a 3D conformation from a SMILES string. Note I say a conformation rather than the conformation; molecules can in general have many valid conformations.  Also, tools for generating conformations rely on molecular force fields, etc.  These have many implicit assumptions; there is no guarantee that a computationally generated conformation will be the real conformation of a real molecule in the real world.
Here is some code for generating a plausible conformation from a SMILES string using rdkit.
from rdkit import Chem
from rdkit.Chem import AllChem
from rdkit.Chem import Draw
from rdkit.Chem.Draw import IPythonConsole

my_mol = Chem.MolFromSmiles('NC(=N)N1CCC[C@H]1Cc2onc(n2)c3ccc(Nc4nc(cs4)c5ccc(Br)cc5)cc3')

my_mol

my_mol_with_H=Chem.AddHs(my_mol)

AllChem.EmbedMolecule(my_mol_with_H)
AllChem.MMFFOptimizeMolecule(my_mol_with_H)

my_embedded_mol = Chem.RemoveHs(my_mol_with_H)

my_embedded_mol

print(Chem.MolToMolBlock(my_embedded_mol))

The printed result is:
    RDKit          3D

 33 37  0  0  0  0  0  0  0  0999 V2000
   -8.0789   -0.7261   -1.9565 N   0  0  0  0  0  0  0  0  0  0  0  0
   -8.3618   -0.9375   -0.6556 C   0  0  0  0  0  0  0  0  0  0  0  0
   -9.4453   -1.5737   -0.3799 N   0  0  0  0  0  0  0  0  0  0  0  0
   -7.4690   -0.4468    0.2422 N   0  0  0  0  0  0  0  0  0  0  0  0
   -7.8136   -0.1283    1.6244 C   0  0  0  0  0  0  0  0  0  0  0  0
   -6.7632    0.8908    2.0392 C   0  0  0  0  0  0  0  0  0  0  0  0
   -5.5246    0.3855    1.3227 C   0  0  0  0  0  0  0  0  0  0  0  0
   -6.0688   -0.0733   -0.0461 C   0  0  1  0  0  0  0  0  0  0  0  0
   -5.2554   -1.2432   -0.6177 C   0  0  0  0  0  0  0  0  0  0  0  0
   -3.8658   -0.8320   -0.9216 C   0  0  0  0  0  0  0  0  0  0  0  0
   -3.6647   -0.1417   -2.0770 O   0  0  0  0  0  0  0  0  0  0  0  0
   -2.3059    0.1587   -2.1237 N   0  0  0  0  0  0  0  0  0  0  0  0
   -1.8139   -0.3885   -1.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -2.7692   -1.0082   -0.2227 N   0  0  0  0  0  0  0  0  0  0  0  0
   -0.4078   -0.3427   -0.6136 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.0488   -1.0902    0.4772 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.3984   -1.0569    0.8486 C   0  0  0  0  0  0  0  0  0  0  0  0
    2.3307   -0.2688    0.1543 C   0  0  0  0  0  0  0  0  0  0  0  0
    3.6731   -0.3282    0.5615 N   0  0  0  0  0  0  0  0  0  0  0  0
    4.8291   -0.0477   -0.0843 C   0  0  0  0  0  0  0  0  0  0  0  0
    5.9334    0.1757    0.5968 N   0  0  0  0  0  0  0  0  0  0  0  0
    7.0123    0.4129   -0.2413 C   0  0  0  0  0  0  0  0  0  0  0  0
    6.7153    0.3213   -1.5854 C   0  0  0  0  0  0  0  0  0  0  0  0
    5.0623   -0.0682   -1.7942 S   0  0  0  0  0  0  0  0  0  0  0  0
    8.3378    0.7031    0.3040 C   0  0  0  0  0  0  0  0  0  0  0  0
    9.3324    1.3464   -0.4485 C   0  0  0  0  0  0  0  0  0  0  0  0
   10.5913    1.6060    0.1057 C   0  0  0  0  0  0  0  0  0  0  0  0
   10.8633    1.2259    1.4171 C   0  0  0  0  0  0  0  0  0  0  0  0
   12.5638    1.5736    2.1593 Br  0  0  0  0  0  0  0  0  0  0  0  0
    9.8883    0.5951    2.1844 C   0  0  0  0  0  0  0  0  0  0  0  0
    8.6313    0.3380    1.6284 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.8659    0.4742   -0.9343 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.5160    0.4406   -1.3141 C   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  1  0
  2  3  2  0
  2  4  1  0
  4  5  1  0
  5  6  1  0
  6  7  1  0
  7  8  1  0
  8  9  1  1
  9 10  1  0
 10 11  1  0
 11 12  1  0
 12 13  2  0
 13 14  1  0
 13 15  1  0
 15 16  2  0
 16 17  1  0
 17 18  2  0
 18 19  1  0
 19 20  1  0
 20 21  2  0
 21 22  1  0
 22 23  2  0
 23 24  1  0
 22 25  1  0
 25 26  2  0
 26 27  1  0
 27 28  2  0
 28 29  1  0
 28 30  1  0
 30 31  2  0
 18 32  1  0
 32 33  2  0
  8  4  1  0
 14 10  2  0
 33 15  1  0
 24 20  1  0
 31 25  1  0
M  END

A semi-interpretable 2D image of this 3D conformation, also generated by rdkit, is shown below.  For comparsion, the "un-embedded" molecule, optimized to look nice on a 2D display, is also shown.

From the admittedly not-great 2D depiction of the embedded molecule, you can at least tell that the various aromatic rings are not coplanar.  For better visualization of 3D conformations, you would want to use a tool like py3dmol.
RDKit is just one example of a software with these kinds of capabilities.  I used it here because it's the one I know.  OpenBabel is another one, as Martin rightly mentions in the comments.

What about Standard InChI?

InChI has the same limitations as SMILES.  It does not encode the 3D arrangement of atoms, only the atom types and bond types between them.

Good answer. For the sake of completeness, I'd mention Open Babel, especially if you just need a 'conversion' tool.

So what does SMILE give? Is it the molecular configuration or chemical confirmation?

@0x90 SMILES provide constitution, bond order (single, double (=), triple (#) bond and aromaticity with C, N, O, S (C1CCCCC1 is not the same as c1ccccc1)) and may indicate molecular configuration (cis / trans double bond by / or backslash ; S or R atom centered chirality by @ or @@).  It is less frequent to see the notation including configuration, but it is present.

@Martin-マーチン Good idea, done.  Thanks for the suggestion.

Is a SMILES string unique per molecule?

No.  CCO and OCC both represent ethanol for example.


text:
Can anyone please help me how do you determine kinetically the fastest product of sulphonation of $\beta$ napthol?
I was asked a question in a paper where I was asked to identify the kinetically fastest product of sulphonation of $\beta$ naphthol.
Can anyone please help me how do you determine kinetically the fastest product ??

The answer that is provided was that  the ortho position to the hydroxyl group is the position where the fastest attack would take place. Because when the carbo cation is formed, the most stable one initially.

But I did not find this explanation satisfactory because ultimately the whole molecule is in complete resonance and it should not matter whichever Ortho or para position the sulphonation happens because ultimately the positive charge would end up being stabilised by OH group. Please help me.

The solution argued that this is less kinetically fast as the initial carbocation is less stable. Also there are 2 Ortho positions, but it does not specify which is the faster one.
The solution given was that Oxy Tobias acid is fastest but based on above arguments , R acid is equally likely and Schaeffer's acid is more probable ...(as there is no steric factors
Also what would the thermodynamically most stable product be ??

https://chemistry.stackexchange.com/questions/31711/thermodynamic-vs-kinetic-sulphonation-of-naphthalene

Possible duplicate of Mechanism of formation of 2-naphthol red dye (aka Sudan 1)

Electron density is greatest at C-1 and is kinetically-controlled. Only the aromaticity in the phenolic ring is perturbed. C-5, C-6 and C-8 monosulfonation products are formed less readily. The kinetics are not straight forward. See https://pubs.acs.org/doi/abs/10.1021/ie00003a022

Electrophile can substitute 2-naphthol ring on 3 possible positions: 1-position through intermediate I, 6-position through intermediate II, and 3-position through intermediate III (See following diagram). Among these three intermediates, intermediate I is the most stable (You must visit Mechanism of formation of 2-naphthol red dye to get excellent explanation why substitution at 1-position is the preferable one over 3-position, although both are ortho to the phenolic $\ce{OH}$ group). Thus, the kinetically fastest product of sulphonation of 2-naphthol is 2-hydroxynaphthalene-1-sulfonic acid (Oxy Tobias acid). There will be no 3-hydroxynaphthalene-2-sulfonic acid (Ref.1).

The best evidence for this preference has been displayed during specific reaction using Claisen rearrangement (See above diagram in the box). After first rearrangement free phenolic $\ce{OH}$ group was allylated again and subjected to heat. For our surprise, the rearragement is proceeded to give 100% 1,1-diallyl derivative as shown. No 1,3-diallyl-2-naphthol was detected (according to NMR evidence).
Based on Ref.1, some 2-hydroxynaphthalene-6-sulfonic acid (Schaeffer's acid) has also been produced under kinetic control. Yet, longer heating time allowed to give disubstituted product (2-hydroxynaphthalene-1,6-disulfonic acid) preferably, thus thermodynamic product cannot be determined by these evidence.
Reference:

Paolo Beltrame, Giorgio Bottaccio, Paolo Carniti, Giuseppina Felicioli, “Sulfonation of 2-naphthol by sulfuric acid: rate measurements and kinetic model,” Ind. Eng. Chem. Res. 1992, 31(3), 787-791 (https://doi.org/10.1021/ie00003a022).

@ Mathew M: In your penultimate structure,  the -OH should be =O.


text:
Mono chloro substituted products of naphthalene
My chemistry teacher today told that only two mono chloro substituted products are possible for naphthalene. He explained this by saying since the naphthalene molecule is symmetrical  about the axis passing through the C-C bond where the benzene rings are fused, the corresponding positions should be identical & hence only two such products are possible. I personally don't find this convincing & hope to look for a genuine answer. I will be extremely thankful to anyone who helps.
If someone is unable to understand my teacher's explanation, just ignore it. I myself wasn't able to understand it properly so could have mistaken it for something else.

:/ Well, you better try to understand symmetry, because it's one of most important things in chemistry. Hmm, imagine cutting naphtalene like a cake. If you cut it into four parts, their are gonna be identical.

Naphtalene has ten hydrogen atoms. Draw ten naphtalenes, each monochlorinated, and let the Cl wander around the double ring. Now check how many of them are actually different. (And don't draw double bonds, naphtalene is aromatic! Two six-rings, each with a circle in it to represent the delocalised pi-electrons.)

Alternatively, use models (or even cut a sheet of paper into 10 naphthalenes and mark the position of the chlorine by a paper clip).  Superimpose the models, turn them, rotate them, flip them and see how quite some of them are in fact identical.

Your chemistry teacher is partially correct by saying that naphthalene molecule is symmetrical about the axis passing through the $\ce{C-C}$ bond where the benzene rings are fused (going through the common carbons to both rings). Your chemistry teacher has forgot to mention that naphthalene molecule is also symmetrical about the axis perpendicular to the axis passing through the $\ce{C-C}$ bond where the benzene rings are fused (See the top structure of the diagram below). Thus, each quarter of the naphthalene molecule is equal and each quarter contains only two carbon atoms (disregarding common $\ce{C-C}$ bond). I named them as 1 and 2, because they are symmetrical. Therefore, only two different mono-substituted derivatives are possible from naphthalene. If substituent is chloro, then only 1-chloronaphthalene and 2-chloronaphthalene are possible, and so on.

I had this notion that for molecules to be identical they ought to have similar iupac naming. Never knew symmetry had a role to play. Thanks for your answer.


text:
How to convert Gaussian input file .gjf to SMILES?
Is there a way to convert Gaussian input files (.gjf) to SMILES-codes (.smi)?
I'm aware of Open Babel (http://openbabel.org/docs/2.3.0/FileFormats/Overview.html), but this lists .gjf as a "write-only" format. Is there, nevertheless, a way (which ideally can be integrated into a Python application) to convert .gjf to .smi?
Edit:
(as mentioned in the comments there could be two different formats in .gjf?!)
Some example to clarify:
...

Winmostar

0 1
C
N 1 R2
C 2 R3 1 A3
C 3 R4 2 A4 1 D4
C 3 R5 2 A5 1 D5
C 5 R6 3 A6 2 D6
C 4 R7 3 A7 2 D7
C 6 R8 5 A8 3 D8
H 5 R9 3 A9 2 D9
N 1 R10 2 A10 3 D10
H 8 R11 6 A11 5 D11
C 10 R12 1 A12 2 D12
...
R2 1.380897451
R3 1.347679561
R4 1.373428060
R5 1.389879183
R6 1.397784043
R7 1.383546211
R8 1.399306496
R9 1.102440160
R10 1.375509032
...
A3 108.5501889
A4 108.1923111
A5 131.7222115
A6 117.5631443
A7 123.5229334
A8 121.6972504
A9 120.8892161
A10 105.3917963
...
D4 3.7044863
D5 -177.0963408
D6 -178.8876730
D7 178.9809960
D8 0.1218056
D9 0.5014806
D10 -7.3468625
...

Is it expressed as a Z-matrix or in cartesian coordinates? Are you interested in a single file or in batch? Cartesian coordinates can easily be converted by Avogadro (I am not sure about the Z-matrix format). Anyway, the whole gaussian input file can be used as an input on webmo demo server https://www.webmo.net/demo/. Then, you can download the structure in the format you prefer.

You should be aware that a converter from structural information to a linear text format will make some assumptions. These might often be good enough, but I'd recommend using a GUI to check that the generated representation of bonds is actually what you intended.

Martin's warning comment is very true (sadly, only +1 is possible) since there are more than one implementations of SMILES (e.g., unified, inchified, canonical).  Results may even vary if the same .mol2 file is exported/converted as Smiles by different sketchers, and the reverse input of the SMILES to generate a 2D structure might not re-create the original.

@The_Vinz, good point.... apparently Z-Matrix, correct? I will add some excerpt to the question.
If I can do single file (e.g.with Python), batch is not a problem. Webmo might be good for checking but not for batch conversion.

@Martin - マーチン, as I understand there is no bond information in a .xyz file but in the above added example (taken from a .gjf) there is, correct?

@theozh not really. These are z matrix inputs, they only define the coordinates in an internally consistent way. The bond connectivity would be another section entirely, but gaussview could construct that.

@ Martin - マーチン♦, ok, I see. Hmmm, when I want to batch convert a lot of molecules and have to check each one whether this is the intended one, then this is maybe not the way to go. Well, on the otherhand, assuming I have a Gaussian output file, e.g. from a geometry optimization, does it contain information from which I can create a SMILES code without any doubt?

It is not aiming for code golf, but the following offers the conversion to intermediate .xyz which may be processed further (cf. vide infra):
import fnmatch
import os
import sys

gjf_register = []

for file in os.listdir("."):
    # collect all .gjf:
    if fnmatch.fnmatch(file, "*.gjf"):
        gjf_register.append(file)
    gjf_register.sort()

for entry in gjf_register:
    # recover only the atom block:
    per_file_register = []
    with open(entry, mode="r") as source:
        for line in source:
            per_file_register.append(line)
    del per_file_register[:5]  # clip the first lines

    # the other mandatory entry in the .xyz file:
    number_of_atoms = len(per_file_register) - 1

    # write an intermediate .xyz file:
    intermediate_xyz = str(entry)[:-4] + str(".xyz")
    print("File {} written.".format(intermediate_xyz))

    with open(intermediate_xyz, mode="w") as newfile:
        newfile.write(str(number_of_atoms))
        newfile.write(str("\n"))

        newfile.write(str(intermediate_xyz))
        newfile.write(str("\n"))

        for entry in per_file_register[:-1]:
            newfile.write(entry)

sys.exit(0)

The conversion into .smi may then be delegated to obabel, e.g. by
obabel *.xyz -osmi -m

if there are multiple files to consider.

Proof:  The following .cif
data_EFAHAW
_cell_length_a 8.1819
_cell_length_b 10.4285
_cell_length_c 9.2965
_cell_angle_alpha 90
_cell_angle_beta 99.962
_cell_angle_gamma 90
_symmetry_space_group_name_H-M 'P 21/c'

loop_
_symmetry_equiv_pos_site_id
_symmetry_equiv_pos_as_xyz
1 x,y,z
2 -x,1/2+y,1/2-z
3 -x,-y,-z
4 x,1/2-y,1/2+z

loop_
_atom_site_label
_atom_site_type_symbol
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
O3 O 0.19290 0.35524 0.46580
C9 C 0.16780 0.24532 0.42690
O2 O 0.05757 0.21719 0.30246
C7 C 0.05180 0.07991 0.27798
C1 C 0.16866 0.02605 0.40441
C6 C 0.23660 0.12584 0.49162
C3 C 0.35070 0.10279 0.61788
C5 C 0.39278 -0.02130 0.65283
C2 C 0.32245 -0.12294 0.56458
C4 C 0.20840 -0.10008 0.43826

was converted by
babel -icif EFAHAW.cif -ogjf test.gjf

into
#Put Keywords Here, check Charge and Multiplicity.

 EFAHAW

0  15
O           0.82917         3.70462         4.26502
C           0.68636         2.55832         3.90884
O          -0.01540         2.26497         2.76942
C          -0.02324         0.83334         2.54528
C           0.72957         0.27166         3.70291
C           1.14519         1.31232         4.50144
C           1.87569         1.07195         5.65752
C           2.16377        10.20637         5.97753
C           1.73027         9.14642         5.16948
C           1.00028         9.38482         4.01285

The above script converted it into .xyz
10
test.xyz
O           0.82917         3.70462         4.26502
C           0.68636         2.55832         3.90884
O          -0.01540         2.26497         2.76942
C          -0.02324         0.83334         2.54528
C           0.72957         0.27166         3.70291
C           1.14519         1.31232         4.50144
C           1.87569         1.07195         5.65752
C           2.16377        10.20637         5.97753
C           1.73027         9.14642         5.16948
C           1.00028         9.38482         4.01285

and already mentioned babel -ixyz test.xyz -osmi test.smiles offered the SMILES string
O=C1O[C][C]=C1[C].[C][C][C] test.xyz

Due to the definition of asymmetric unit and unit cell, other original input formats than .cif may work even better (i.e., may yield molecules which are not fragmented as here), e.g. «native» .mol2 or .sdf.

In my experience Babel often has trouble placing double and triple bonds correctly. I wrote xyz2mol which often does better. It converts an xyz file to an RDKit molecule, which can easily be converted to a SMILES string using RDKit

Thank you @Buttonwood for the code. So if the format in my .gjf (see edited question) is Z-matrix, then Open Babel should be able to read it. I need to check...

@JanJensen It definitively looks like an interesting alternative.  Like .cif -> .xyz works virtually each time, but .cif -> .mol2 of course is not so smooth and automatic, hence your xyz2mol could help.  Or for OSRA (https://cactus.nci.nih.gov/osra/) related data retrieval (e.g. DataWarrior's supplements, http://openmolecules.org/datawarrior/datafiles.html).  Or for similarity structure packings beyond uniformly single-bond bound atoms (e.g. the helpful calculate_rmsd.py by Jimmy Charnley et al.) which then equally could account for different bond orders, too. I anticipate weekend study ahead.

To summarize my learnings from the comments (thank you all):
It's probably not a good idea to use a Gaussian input file (.gjf) and trying to create a SMILES out of it. It apparently could happen that the bonds will not be the intended ones.
I thought it would have been easier to deal with the small input files (ca. 3-10 kB) instead of the large output files (ca. 10-60 MB).
Although it doesn't strictly answer my original question, my solution is currently using the Gaussian output files and Open Babel.
With the following little Python script, I can convert Gaussian output files to SMILES codes via Open Babel and integrate it into my application.
### Gaussian to SMILES via OpenBabel
import subprocess
from subprocess import PIPE

obabelexe = r'C:\Users\OpenBabel-2.4.1\obabel.exe'
filename = r'C:\Users\Molecule.log'

output = subprocess.run([obabelexe, filename, "-osmi"],check=True, stdout=PIPE, stderr=PIPE)

print(output.stdout.decode("utf-8").split('\t')[0])
### end of code


text:
Is the equilibrium constant calculation only applicable to homogeneous reactions?
In my text book, it states the general formula for the equilibrium constant is:

Then it states that "The equilibrium constant expressions described here apply to homogeneous reactions."
Does this mean that it only applies to homogeneous reactions or that it can also be applied to homogeneous reactions among other types?

That means that heterogeneous reactions don't work like that.

If you introduce the chemical activity $a$ in thermodynamics you can write the general equation.
$$ \frac{a_C a_D}{ a_A a_B} = K$$
which is correct for any case. Under ideal assumptions we have a linear relationship between the activity and concentration in solution.
So with an arbitrary normalizing reference concentration $c_0$ (e.g. 1 mol/l) the equation becomes roughly:
$$ \frac{c_C / c_0 c_D / c_0}{ c_A / c_0 c_B / c_0} = K$$
As you see the $c_0$ cancel each other out and you end up with the textbook equation that consists only of concentrations.
So the equation applies if you can scale your activities linearly with concentrations. This is usually possible for homogeneous reactions and sometimes possible for heterogenous reactions (As can be seen from the other answer).
PS: Activities hence equilibrium constants are dimensionless. You always have a factor 1 (mol/l)$^n$ to cancel out the units of the concentrations, if you derive it correctly.

In the initial equation you provide, do you mean a_C for the first a in the numerator?

Oh yes, sorry ;-)

"This is only possible for homogeneous reactions." I disagree with this statement, see my alternative answer.

Does this mean that it only applies to homogeneous reactions or that it can also be applied to homogeneous reactions among other types?

The latter. If you consider the distribution of a solute S in a mixture of water and octanol, this is a heterogenous system (in this case, two immiscible liquids). The equilibrium reaction (or process) is:
$$\ce{S(oct) <=> S(aq)}$$
There is an equilibrium constant for this process,
$$K = \frac{[\ce{S(aq)}]}{[\ce{S(oct)]}}.$$
If I dissolve some S in the aqueous phase, some of it will go into the octanol phase until equilibrium is established. If I dilute the aqueous phase, some S will go back to the aqueous phase to re-establish equilibrium.

Although the system is heterogeneous, wouldn't the reaction still be homogeneous as the reactant and product are in liquid states but with a solute mixed into them?

@Liam Perhaps a more convincing example, from wikipedia's article on phase-transfer catalyst: "$\ce{C8H17Br(org) + NaCN(aq) -> C8H17CN(org) + NaBr(aq)}$ (catalyzed by a R4P+Br− PTC)". This is classified as heterogeneous catalysis.


text:
Oxidation of Vitamin C
How is ascorbic acid oxidized into dehydroascorbic acid? What is the reduction reaction in this case? Where are the 2 hydrogen atoms going?

We have a policy which states that you should show your thoughts, effort and attempts to understand underlaying principles. It'll make us certain that we aren't doing your homework for you, and that the Q/A is beneficial for broad audience.
Basically any question with the wording your question has is considered homework; it needn't be literally one. It can be self-study questions, puzzles, worked examples etc.
Please [edit] in your full reasoning or thoughts on this. See Homework
Otherwise, the question may get closed.

I think this question is too broad, it would be useful to see it narrowed down. Also, a little searching reveals some good hits explaining possible mechanisms. The underlying question (is there a duplicate?) seems very interesting (if, again, broad): what is the biochemical mechanism of oxidation of ascorbic acid.

Think of oxidation in a classical sense for a minute or so. Oxidation is the addition of oxygen to a molecule or it can also mean the removal of hydrogen. If you carefully look at the structure of ascorbic acid, you will see that the two -$\ce{OH}$ groups are now converted into carbonyl groups. Those two hydrogens can be lost as a proton, $\ce{H+}$.


text:
Is it possible to synthesize permanganate electrochemically?
I have been researching the production of peroxy acids and other strong oxidizers. I was wondering if it's possible to produce permanganic acid if I were to cast an anode containing Manganese Dioxide from a battery?
Update
I was able to produce what I would believe to be the permanganate anion based on the color of the solution and the composition of one of the electrodes.

See What can I do when getting “We are no longer accepting questions/answers from this account”?; also, Why can't I ask questions on StackOverflow?. It looks like the system temporarily blocked you from posting new question, probably due to heuristics based on the former activity (number of downvotes/flagged posts or comments/improper behavior).

From what I can see, neither of Chemistry.SE moderators manually suspended you in any way, and we are probably not going to be able to inform you regarding the exact reason for blocking as we don't have access to the blocking algorithms. Also, please note that edits as "MODERATOR, Why can't I ask new questions?" are not a good fit for the main Q&A site since they have little to do with the question itself and are better suited for Meta.

I'll remove the Edits.

@andselisk, Okay I do tend to keep my questions short, but that's because on the engineering site I was scolded for providing too much detail. If you look at my questions, they are completely legitimate. I asked about some aqueous chlorine, (I know that now), a pink species that I had no Idea was chromium and I has a tough time with some copper oxidation that I asked about. That's all. The chemists can be a bit condescending at times too. I have a very friendly demeanor, but when someone is rude to me, that can illicit an extremely negative response. I was born and raised in New York City.

@andselisk, Once I'm 100% sure of the step by step process, If I wanted to answer my own question on this occasion, because I did indeed produce some permanganate. It appeared like clockwork. beautiful fuchsia solution. Though as it concentrates, it seems to be looking more like Kiwi-strawberry Mystic. How should I format the answer?

It's absolutely fine post a lengthy and detailed question. I think I remember that post of yours which is unfortunately deleted now (the system does automatic cleanup based on the score and number of visitors) and I thought it was totally fine (AFAIR there were purple-tinted pictures and a decent description of the experimental setup). The thing is, sometimes even good questions get closed by users for no apparent reason just because. This is indeed frustrating. Keep in mind every closed question may be reopened after tweaking/improving. Also, it looks like you can still answer questions.


text:
Reactivity of boron towards acids
Boron is a non-metal, i.e. it accepts electrons and acids donate electrons, so boron should react with acid but it doesn't. Why?

To illustrate Oscar's answer further, it should be noted that boron, despite existing in more than ten allotropic modifications, is in general quite chemically inert (especially in crystalline form).
For example, boron doesn't react with hydrogen directly and all boranes are synthesized by other means.
Although boron may be oxidized by fluorine at room temperature, other halogens (even chlorine) require elevated temperatures above $\pu{400 °C}$:
$$\ce{2 B(s) + 3 E2(g) -> 2 BE3}\qquad(\pu{30 °C}: \ce{E} = \ce{F}; \quad\text{above}~\pu{400 °C}: \ce{E} = \ce{Cl}, \ce{Br}, \ce{I})$$
There is no preference in reactivity towards acids or bases either.
Boron indeed doesn't appreciably react either with non-oxidizing acids or with alkali solutions under ambient conditions.
However, concentrated oxidizing acids such as nitric and sulfuric acids as well as aqua regia react all right — especially at elevated temperatures — yielding boric acid:
$$\ce{B(s) + 3 HNO3(aq) ->[Δ] B(OH)3(aq) + 3 NO2(g)}$$
Amorphous boron reacts with concentrated alkali solutions:
$$\ce{2 B (am) + 2 NaOH (conc) + 6 H2O(l) -> 2 Na[B(OH)4](aq) + 3 H2(g)}$$
When fused in air with alkali, $\ce{Na2O2}$ or $\ce{KNO3}$ + $\ce{Na2CO3}$ mix, borates are formed:
$$\ce{4 B(s) + 4 NaOH(s) + 3 O2(g) ->[\pu{350-400 °C}] 4 NaBO2(s) + 2 H2O(g)}$$
All chemical reactions have been adapted from [1, p. 74].
References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)

Who said boron is a base?  The fact that it does not weaken or neutralize most acids ought to tip you off that it isn't.
Boron is slowly oxidized when powdered and exposed to nitric acid (see here), and we may call it "basic" by virtue of the products being much weaker acids than the nitric acid.  But usually and with better precision the boron (or any other element that reacts similarly) is instead interpreted as a reducing agent.

Wellllll....... If you are using the Usanovich definition of acids and bases https://en.wikipedia.org/wiki/Acid%E2%80%93base_reaction#Usanovich_definition the distinction disappears. But that is far from mainstream


text:
Why is Oxygen compulsory for free radical reactions?
Why is presence of oxygen compulsory for free radical mechanism? Can it take place in presence of any other element?

Other elements can form free radicals. Any diatomic gas would do.

AIBN is a common radical initiator, and it contains no oxygen.

There is no such condition that oxygen must be present to generate free radicals. Chlorine forms free radicals, and so do many organic molecules. In fact the presence of oxygen is bad for free radical reactions.


text:
What is the correct equation for calculating the electric potential of an electrochemical cell?
When I'm studying electrochemistry, we were given an equation that all chemistry students are familiar with, which is:
                    Ecell = Eoxidation - Ereduction

However, when I came to study electroanalytical methods, I came upon this equation:
          Ecell = Eoxidation + Ejunction - Ereduction  - Eelectrode

where Ejuntion is the uncertain potential produced from the salt bridge and the Eelectrode is the potential produced from the reference electrode used in the experiment.
My question is, is this the real equation when calculating problems for an electrochemical cells? I'm thinking that they probably cut Ejunction out of the equation because it is impossible to know the exact value for it. In addition, I think that they cut out the Eelectrode in the equation because most of the experiment data came from a SHE electrode where Eelectrode = 0.00 V.

I think your assumptions are right. The first equation refers to an ideal case (Ejunction=0) with SHE reference.

@BuckThorn so, is the equation correct or not?

I am afraid your equations are other way around.

Ecell = Eoxidation - Ereduction

This is not the accepted convention among electrochemists. Either the reference is outdated or there is a typo. All electrochemists in the world now follow the convention of writing
Ecell= Ereduction-Eoxidation or Ecell = Ecathode-Eanode  ---Eq(1)
The reason is historical and I have written details elsewhere, but briefly the main reason is that Americans used one sign convention and the Europeans used another. It took more than 50 years to come to a consistent sign convention, which was originally used by a German chemist, Ostwald, 100 years ago. Given that all electrode potentials are now written as reduction potentials the above Eq. (1) is valid. If you look up the tables, the signs of the electrode potentials correspond to the actual electrostatic sign (+) or (-) of the electrode.
Now this simplistic equation (1) ignores any "connections" among the half cells. In reality we always need an electrically conducting connection to the half cells to keep them separated yet electrically connected. This "connection" could be a salt bridge, or a porous ceramic frit (used in pH probes and so on).
There is a general interesting electrochemical fact whenever there is an interface (the location where two surfaces meet) one can build up difference of electrical potential. This is a fundamental material property. Recall the simple voltaic cell, which has copper and zinc pieces separated by a piece of wet paper containing some common salt.
Liquid junction potential (Ej) is another type of potential which originates at the interface of two electrolytes with different concentration of ions (=charges). In that case we add the liquid junction potential to this Ecell as follows
Ecell= Ecathode-Eanode+ Ej
I have not seen your other equation.  Ecell = Eoxidation + Ejunction - Ereduction  - Eelectrode. Which book is using this expression and what is its year of publication?

The Eelectrode I saw was from an electroanalytical method chapter. The examples shown includes subtraction of the potential of the electrode from the overall potential. When they use a saturated calomel electrode, they always subtract the potential of the calomel electrode from the overall potential to determine the real potential of the cell.

Could you mention the name of the book? It is wrong to use Ecell=Eox-Ered. Your second part "When they use a saturated calomel electrode, they always subtract the potential of the calomel electrode from the overall potential to determine the real potential of the cell." is another story. This is done to change the scale back to the hydrogen half cell. This is just an algebraic operation which is done when your ref. electrode is not the standard hydrogen electrode.

I guess I am wrong then, by your explanation I'm thinking that I will just subtract the potential of the electrode if I am determining the potential of a Half cell.

No don't use intuition or my explanation. You will not learn this way until and unless you see the algebra behind it on paper. Use a piece of paper and draw a number line (0 in the center, positive on the right, negatives on the left). Set hydrogen half cell to zero. Put the value saturated calomel electrode value with respect to the hydrogen half cell on the number line with its sign. As an exercise, determine the half cell of Cu(II) wrt to the saturated calomel electrode. Once you have this value, use the same number line and convert your Cu(II) (vs SCE) back to the hydrogen half cell.

I see now, the subtraction is just done to "calibrate" the result to match it into a potential that will be produced if a SHE was used.

Exactly. This was pure algebra part rather than the electrochemistry and that is how I understood it using the concept of number lines. Most teachers will teach it like a rote memorized parrot without offering an explanation.


text:
Why is phloroglucinol (1,3,5-benzenetriol) not very soluble in water?
Wikipedia says that phloroglucinol (1,3,5-benzenetriol) has a solubility of 1 g/100 ml in water.  This is in comparison to:

phenol (hydroxybenzene) 8.3 g/100 mL
catchecol (1,2-dihydroxybenzene) 43 g/100 ml
resourcinol (1,3-dihydroxybenzene) 11 g/100 ml
hydroquinone (1,4-dihydroxybenzene) 5.9 g/100 mL

Why does it have such a low solubility?  I would have thought the OH group would make it more soluble.

Solubility at what pH?  It is a weak acid and forms anions at basic pH which are more soluble. At acidic pH keto tautomers play a significant role. Further reading here  https://pubs.acs.org/doi/abs/10.1021/ja00056a016

Related https://chemistry.stackexchange.com/questions/87819/why-is-potassium-bitartrate-so-insoluble

Why does it have such a low solubility? I would have thought the OH group would make it more soluble.

Interesting question but there is a logical fallacy. You are connecting the concept of solute's polarity with its water solubility. Note the word polarity is a catch-all term for many interactions. Solubility is the hardest thing to predict or model! It is very hard to predict melting point as well as boiling point. Not sure why general chemistry textbooks only misguide students and they have to unlearn so many things at a later stage.
You made very good observation. Let us look at your list from another angle i.e., chromatography- a special type of chromatography. Lets us say we have a polar stationary phase like silica and use a mobile phase which contains acetonitrile + water at a neutral pH. In this chromatography mode, the most polar compound elutes in the last and the least polar first. If you inject a mixture of phenol, resorcinol and phloroglucinol into a column, what is the elution order?
Phenol, then resorcinol and then phloroglucinol. Lo and behold, this is exactly like your prediction! So there is indeed a trend. However we cannot correlate this single observation with the water solubility.
Let us think of a cyclic sugar, cyclofructan: 21 hydroxyl groups, yet it is insoluble in water! It is insoluble in many organic solvents as well. So it is very hard to correlate the number of individual functional groups and predict water solubility. Exceptions like these will pop up again and again.
However there are softwares, such as Advanced Chemistry Development (ACD) which can predict solubility. No idea what models they use. You can look up their documentation or even contact them.

Well, you point out why such prediction has limited usefulness, but what about actual factors? Most important being intermolecular interactions in pure compounds vs mixtures.  More OH groups can not only make more h-bonds with water but also with each other, sometimes to bigger extent.


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Reductive amination in case of secondary amines
When an aldehyde or a ketone undergoes reductive amination when a primary amine is acting,the mechanism follows a pathway through hemiaminal and imine,what happens when a secondary amine is acting on the given carbonyl ? There will be no hydrogen on the amine nitrogen in this case to facilitate loss of water with the hydroxyl group of the hemiaminal.How does the reaction proceed then?

Secondary amines undergo reductive amination perfectly well via the iminium ion. This explains it clearly https://www.chemistryscore.com/reactions/reductive-amination/

Did you read @Waylander's link? The mechanism is there.

Yes I did , actually I was not able to follow the answer there , what I was able to gather from there was that the reducing agent , LiAlH4 in my case , will substitute the OH and nitrogen will grab a hydrogen from water

Sodiumtriacetoxyborohydride in DCM is my reagent of choice for reductive aminations. LiAlH4 will just reduce your ketone before it reacts with the amine

In modern organic synthesis, reductive amination of aldehydes and ketones is considered one of the important and principal ways to make secondary and tertiary amines. It is a powerful and reliable strategy for the formation of $\ce{C–N}$ bonds. One of the advantages of the reductive amination is it can avoid the problem of over-alkylation that is always the problem with direct alkylation of amines with alkyl halides.
Reductive amination can be achieved following either a one-step or a two-step procedure. In both procedures, an amine and a carbonyl compound condense to form an imine (with primary amine) or iminium ion (with secondary amine). This imine or iminium ion is then reduced in situ (one-step reaction) or separated and subsequently reduced (two-step reaction) to form an amine product. Following is the suggested mechanism for reductive amination:

Most common condition is sodium triacetoxyborohydride ($\ce{NaB(OAc)3H}$) in THF or dichloroethane (DCE) under slighly acidic conditions (common acid used is acetic acid)(Ref.1).
Reference:

Ahmed F. Abdel-Magid, Kenneth G. Carson, Bruce D. Harris, Cynthia A. Maryanoff, Rekha D. Shah, “Reductive Amination of Aldehydes and Ketones with Sodium Triacetoxyborohydride. Studies on Direct and Indirect Reductive Amination Procedures,” J. Org. Chem. 1996, 61(11), 3849-3862 (https://doi.org/10.1021/jo960057x).

A good answer. A useful additive for reductive aminations that proceed in poor yield (or not at all) is Titanium (IV) alkoxide - Ti(OEt)4 or Ti(iOPr)4.


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How to calculate the i-value of known NaCl solution, if it is isotonic with known solution of glucose?

How do I calculate the value of $i$ for 1.17% solution of $\ce{NaCl}$, if it is isotonic with 7.2% solution of glucose?

I actually do not know where to start in this question.
is it possible to calculate the value of $i$ of $\ce{NaCl}$? is value of $i$ of glucose 1? I tried putting the osmotic pressures equal but the value of $i$ remains a suspense.

Have a read through https://en.wikipedia.org/wiki/Van_%27t_Hoff_factor. Generally, in case of questions in books, it is assumed that the dissociation (or association) is complete, i.e. the NaCl exists only as $\ce{Na+}$ and $\ce{Cl-}$ in the solution. See if you get the solution.

Consider $\pu{100 mL}$ solution of each. That means, the number of particles in $\pu{7.2 g}$ of glucose is equivalent to the number of particles in $\pu{1.17 g}$ of $\ce{NaCl}$. So, you can find the $i$ value of  $\ce{NaCl}$ (you may need to assume $i=1$ for glucose).


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Why is the peak-to-peak separation of cyclic voltammetry is 57 mV?
I'm studying CV, and I wonder if I can calculate the peak-to-peak separation of CV for a reversible redox system.
I found that the separation is 57 mV on my textbook, and my teacher taught me that is can be calculated theoretically using Nernst equation,
but I think it is more complicated.
Is it easy enough (for sophomore) to calculate the value?
And if it is, I want to know the way.
(Sorry for my poor English)

The 57 mv is empirical and not related to the similar value from the Nernst equation.  It arises from the numerical solution of the equations in the oxidation/reduction cycle. See Nicholson & Shain Anal. Chem. v36, p 706 1964 for the  details of the calculation: not an easy read. In introduction is given by Mabbot J. Chem. Educ. v60, p697  1983 and Benschoten et al. ibid. v60 ,p772, 1983

The derivation of equation for cyclic voltammetry requires a lot of advanced calculus and differential equations based on the Fick's law of diffusion. This is not to discourage you but to show that the teacher is oversimplifying it. The derivation are provided in Allan J. Bard's Electroanalytical Chemistry. Why are they teaching cyclic voltammetry in sophomore classes?

Actually I'm not a student at a usual university, so I'm not a sophomore, but in an equivalent grade.
I've heard about the Fick's law of diffusion but I don't know the detail, so I want to study more and more, and to understand the derivation.
Anyway, thank you for answering!

Good to know but Bard's book should be consulted. You will find the details there.


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Where can I find the positions of the infrared band for phosphorus compounds in general? Specifically sodium hexametaphosphate
Specifically I am looking for the IR of the sodium hexametaphosphate compound.

Can you elaborate the question a bit, to add more context? And the title should ideally be short, and any additional information is better placed in the description.

Specific for your compound you find at least two public IR spectra, either in KBr:

(source)
or from one of the larger chemical vendors as .pdf

(source, Sigma-Aldrich, product number 305553, CAS-RN 68915-31-1)
More generally speaking for phosphates, you equally find some related entries in NIST webbook chemistry here, too.
For IR-spectra in general, beside the SDBS with focus on organic compounds and sources already mentioned, there are three big multi-volume collections: Aldrich (equally one large about NMR) which is why frequently chemicals in their online catalogue have a link to either IR or/and NMR spectrum, too. The Sadtler collection is equally broad and diverse, organizes the entries by field of application (e.g., pharmaceuticals, forensics).  At least in part Sadtler is electronically available via BioRad / KnowItAll, too.  Finally Hummel, with emphazis on polymers / plastics.  Because many products of chemical synthesis are characterized by IR spectroscopy, too, you may find them indexed in databases like Reaxys (Elsevier) and Scifinder (American Chemical Society), too.