text:
What makes a good chlorinating agent?
From what I know, good inorganic chlorinating agents, $\ce{SOCl2, PCl3, PCl5}$ are able to substitute a poorer leaving group (like $\ce{OH-}$) for chlorine. For example, for substitution of the hydroxyl group to a halide to form a haloalkane, we can't just use hydrogen halides (we can if we first dehydrate it into an alkene, but that is besides the point), but instead use strong chlorinating agents like those mentioned above. I know $\ce{Cl2, HCl, ClO-, CCl4}$ etc. are also chlorinating agents, so what makes them "inferior"? Also, why are some other chlorine-containing compounds like $\ce{ClO3-, CH3COCl, C2H5Cl}$ not considered good chlorinating agents?
Is it similar to how we gauge the oxidizing strength by seeing how high the oxidation number is (the higher the oxidation state or electronegativity, the more electron-hungry it is, and the more willing it is to accept electrons)? Maybe I'm overthinking this and it's just because the bond enthalpy is lower?

Consider the properties of the products formed SO2 and POCl3 etc. How strong are the S-O and P-O bonds? How does Le Chatelier's principle drive hte process if one of the products is a gas?


text:
Active mass of solids shouldn't necessarily be unity
I've read that active mass of solids and pure liquids is considered unity. I'm absolutely sure there is some misunderstanding. I know that concentration of solids doesn't change during the course of reaction, but it shouldn't necessarily be unity either. Consider graphite for example. It's density is $2260\pu{g/L}$
$$\text{[C]}=\frac{n_C}{V_C}$$ where $n_C$ is the number of moles of the carbon sample and $V_C$ is the volume of that sample. $$\text{[C]}=\frac{w}{M\times V_C} $$ $$\text{[C]}=\frac{\rho_C}{M}=\frac{2260\pu{g/L}}{12\pu{g/mol}}\\=188.3\pu{mol/L}$$This concentration will be constant throughout a reaction if the phase doesn't change and temperature change is quite small. But this clearly is not unity. Then why is the concerntration of solids considered to be unity?

What is 'active mass' do you mean activity?

Active mass is a an old term used for molarity/concentration.

You might want to review your figures, too. Graphite is surely not lighter than air.

@IvanNeretin The unit of density used is $\pu{g/L}$ not $\pu{g/cm^3}$

Yes, I see, and that makes it much worse.

No it doesn't. The active mass is the concentration of substances taking part in the reaction. But my question is, why would it be unity?

If there are 2 water suspensions of activated carbon, 1 g/L and  4 g/L, the latter activated carbon is not more reactive wrt adsoption. There is just more of it, similarly with 1 L versus 4 L of some solution of concentration 1 M, the latter is not more reactive than former. So why unity ? Why not ?

Obviously there is a mistake in the unity. Graphite has a density 2.26 g/cm3 = 2.26 g/mL = 2.26 kg/L = 2260 g/L = 2260 kg/m3. It can hardly be 0.00226 anything ; maybe 0.00226 kg/mL ?

@Maurice Yes I did a mistake in the conversion.

Rather, active mass of solids need not necessarily be unity, but arbitrarily is unity with advantage. You do not have to know the molar mass nor density.
The common thermodynamic expressions like $K = \frac{[C][D]}{[A][B]}$
or kinetic ones like $\frac{\mathrm{d}[C]}{\mathrm{d}t} = k \cdot [A] \cdot [B]$
should be more exactly written as $K = \frac{a_\mathrm{C}] \cdot a_\mathrm{D}}{a_\mathrm{A} \cdot a_\mathrm{B}}$ or $\frac{\mathrm{d}[C]}{\mathrm{d}t} = k \cdot a_\mathrm{A} \cdot \cdot a_\mathrm{B}$
where $a_\mathrm{X}$ is unitless quantity called activity, defined by the chemical potential of the substance $ \mu_\mathrm{A} = \left( \frac {\partial G}{\partial n_\mathrm{A}} \right)_{T,p}$ by formula:
$$\mu_\mathrm{A} = RT \ln {a_\mathrm{A, unitless}}$$
If we implement the standard chemical potential, we can define unitless activity as ratio of the activity with unit and the standard activity with the unit value.
$$\mu_\mathrm{A} = \mu_\mathrm{A}^{\circ} + RT \ln \left( \frac {a_\mathrm{A}}{a_\mathrm{A}^{\circ}} \right)$$
where $a_\mathrm{A}^{\circ}$ is the activity at the standard chemical potential $\mu_\mathrm{A}^{\circ}$. We can consider the activity $a_\mathrm{A} = x\ \text{"whatever unit"}$ and the standard activity $a_\mathrm{A}^{\circ} = 1 \ \text{"whatever unit"}$.
If we arbitrarily choose $a^{\circ} = \pu{1 mol/L}$, then $a$ then more or less closely follow the respective concentration [A].
Chemical potential of the pure solid substance $\mu_\mathrm{A}^{\circ}$ does not depend on the amount of substance, so we arbitrarily choose $a^{\circ} = x_\mathrm{A}^{\circ}=1.0$ and $a$ then more or less follows the  molar fraction of substance $x_\mathrm{A}$.

Let's start from the equilibrium constant, which may be defined $\ce{K=\frac{[C][D]}{[A][B]}}$. Of course, activities should be introduced here instead of concentration. But it is not the point here. The main use of this constant is to calculate how the composition of the system changes if at least one of the concentrations changes.
But if one of these concentrations does not change, whatever the perturbation of the system, it is no use maintaining it in the fraction defining $\ce{K}$. There would be two constants, one before and one after the sign "equal". It is much better to multiply or to divide the equilibrium constant by this concentration, so as to make it disappear from the fraction defining $\ce{K}$. For exemple if [B] does not change during a perturbation, it is much simpler to multiply $\ce{K}$ by $\ce{[B]}$ and find a new constant $\ce{K' = K·[B] = \frac{[C][D]}{[A]}}$, which is tabulated instead of $\ce{K}$ and can be used for describing any new change in the future composition of the system.
This is equivalent to saying that the concentration (or activity) [B| is chosen equal to $1$.

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text:
Can reversible reactions proceed by different mechanisms in the forward and reverse directions?
It seems odd to me that this would be the case, but it also seems odd that it wouldn't.
Imagine a mechanism for a (forward) reaction that includes a slow reversible step and a fast irreversible step:

$$
\begin{align}
\ce{A &<=> B} &\quad &\text{(slow)} \\
\ce{B &-> C} &\quad &\text{(fast)}
\end{align}
$$

If the second step is irreversible, it's presumably going to be highly unfavorable to run in the reverse direction, and so there ought to be some alternate mechanism with a lower $E_\mathrm{a}.$ Is it possible that the reverse reaction is more favorable by a different mechanism? If not, why not?

In the way you have written the reaction, as soon as B is formed it reacts to make C so  [B] $\approx 0$ and A is soon depleted.  To get C back to A then of course if the activation energy is absolutely vast it may take millenia, however, a completely different route may be possible not involving B. Most species have more than one way of being synthesised, particularly for organic species.

So you can have different mechanisms for each direction?

If the alternative mechanism is available in the reverse direction, it must also be available in the forward direction or you would be able to make a perpetual motion machine. So you can certainly have mulriple parallel reaction pathways, but both the forward and reverse direction will favor the pathway with lowest Ea.

As Andrew writes the reverse reaction is always there but one way to get back to A would involve using a catalyst. This now involves different chemistry to the forward step.

Yes, you can. An ideal Carnot cycle (admittedly not a reaction) illustrates how this is possible. In such a cycle each step is reversible. To return to a state you can either reverse the direction of the process (running say a heat engine as a refrigerator) or follow the full path of the cycle until the state is again encountered.

There are two different answers, depending on whether you mean that the alternate pathway can occur in the same reaction mix (ie under the same conditions) as the described forward reaction is occurring.
If there is an alternate pathway possible under the same reaction conditions, it will be used for both the forward and reverse reactions. Due to the principle of microscopic reversibility, no reaction pathway can preferentially be used for one direction and not the other.
If you allow for alternative reaction conditions, then there are often multiple possible pathways to get from any A to B, especially if the reaction has more than one step. Even for a single-step reaction, a catalyst can be used to lower Ea (as noted in the comments on the question). Such a catalyst will necessarily lower Ea for both the forward and reverse reactions, though, so the equilibrium amounts of A, B and C in your example will be the same if the only difference in conditions is the presence of the catalyst.

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text:
Bond stability of C-C in ethane, ethene, and ethyne

Which one of the three compounds has the highest $\ce{C-C}$ bond stability?

Ethane has $\mathrm{sp^3}$ hybridisation so less $\mathrm{s}$-character and lesser electronegativity compared to ethyne. Ethyne has high $\mathrm{s}$-character and high electronegativity. Therefore the covalent character of $\mathrm{sp^3-sp^3}$ in Ethyne should be stronger and have more stability right? the order would therefore be $\mathrm{sp-sp}$ >  $\mathrm{sp^2-sp^2}$ >  $\mathrm{sp^3-sp^3}$
Some online sites tell so, but others say since the $\ce{C}$ atoms are so close in ethyne that the nuclear repulsion would make it less stable. So ethane's $\ce{C-C}$ would be stronger and that therefore the order should be reversed
Edit: nutshell- sigma bond of which hybridisation is strongest among $\mathrm{sp}$, $\mathrm{sp^2}$ and $\mathrm{sp^3}$?
I took ethane, ethene, and ethyne just as a example

"Stability" is another word for "energy difference between one state and another". Now which are those for each of your three suspects?

In Ethyne has high Electronegativity the name Ethynil should be used instead. Also, the C in Ethyne has high s character, rather than Ethyne.

Better way to look at this problem is compare IR stretching frequencies of $\ce{C-C}$, $\ce{C=C}$, and $\ce{C#C}$.

Another way to look at this problem is to compare the bond dissociation energies. For C-C, it is about $350$ kJ/mol. For C=C in ethene, it is $610$ kJ/mol. And for ethyne, it is $835$ kJ/mol.

Well one can argue about the single sigma bond in the three cases. This reflects on the stability of the whole skeleton. It cannot be accessed experimentally but it is a legit question. Especially writing out C-C. I don't know what OP is asking, but I guess this. Otherwise the answer is really evident. @Mathew Mahindaratne and at Maurice

@Maurice well that's what i found, but isn't that including the pi bonds too in the C−C bond? i wanted to know specifically for the sigma bond, whether there's any difference in stability between a sigma bond of sp, sp2, sp3

@Alchimista oh yes you are right, I forgot to mention that in the question, I am editing. But I am guessing like you said it must not experimentally available

@AdilMohammed it must be workable somehow with a thermochenical cycle plus some assumptions but I have no time to focus on it. The main problem is that it does not come alone, so calculation would be even better. One can say that being the C to C distance shorter in high order bond, also the sigma bond is energetically more stable. I can't conclude if this is really useful but interesting yes,it is.

Note that my comment does not imply any reversal. If the bond is shorter in spite of a big nuclear repulsion, well, it right means that is stronger/more stable.

"Stability" is a dodgy term that means very different things in different contexts. So which version are you looking for here? If you mean bond strength, say so. If you mean ability to undergo different reactions, say so (systems with multiple bonds can be far more reactive in certain reactions even though the overall bond is stronger).

The truth is: in reality, sigma bonds, bond orders and hybridization do not exist (not even bonds or orbitals do). They are just concepts to explain the reactivity, stability and geometry of a molecule (and in the case of orbitals to build approximate many-particle wave functions). There is thus no way to compare a sigma-bond strength without choosing some model which introduces bias.
The least-biased way to get there is probably starting from the experimental dissociation energy already mentioned. These can be divided by the bond order (which, again, is not a thing, but quite well established):

ethane: 350 kJ/mol -> 350 kJ/mol per bond
ethene: 610 kJ/mol -> 305 kJ/mol per bond
ethyne: 835 kJ/mol -> 278 kJ/mol per bond

In general: for any chosen approach, the sigma bond in ethane will be the strongest even though the bond distance of ethyne is the smallest. The additional electrons of the pi bond(s) destabilize the bond.

I do not think it is particularly helpful to point out how everything does not exist. How do you define bond dissociation energy if you have already claimed that bonds do not exist?

A dissociation energy is a thing, it is defined as E(Fragments) - E(Molecule) without any assumption on bonds. The reason I point out, that all of this does not exist, is that chemists (like me) tend to forget that. Students often take these concepts literally.

I think this answer touches on some philosophical and semantic issues. First, I don't see how what you note is sad. It's just facts. What might make it sad is how popular some of these models are and how many of them are abused. Moving on. I don't think it is wise to claim (for all intents and purposes) that bonds do not exist. The electron density is observable, as are nuclei, we can measure the energy it takes to change one molecular structure to another. Using the term bond just makes things simpler. Lastly: Geometry, that's such a fuzzy term for structure (imo).

@Martin-マーチン you are right that bonds are a very useful concept, it is just not clear where to draw the line between bonds and inter-molecular interactions. Even within molecules its not clear, where bonds exist, i.e. see the discussion on an H-H bond biphenyl. As a physicist friend of mine used to say: there is no bond operator.

Libavius, I did down vote because you enter into much philosophy. You can surely saturate carbon atoms in steps and this is an evident bond order. As it seems that you are aware of the sophistication implicit in these treatments, it should also be that you recognise the validity and depth of the question. Equally dividing the energy per bond just tell us that some bonds are less stronger. I can say that the bond in Ethyne is the stronger because is still there in spite of destabilising extra electrons. Still, the idea is good. Like in benzene, for which delocalisation does not come in play bec

.. Because of an aromatic sextet, but rather the latter is forced to delocalize by the strength of the sigma skeleton. This said I am not sure of which sigma bond is stronger, but I am pretty sure applying theory provides the result. The fact that it can be void of practical importance is clear to me, too.


text:
What is the intuition behind 'mol' as a unit 'symbol'
Why are moles written with 'mol' as opposed to something which is easier to write with one character, or even two?

Cuz "m" is for meters and "mo" sounds kinda silly.

There are other alphabets and it doesn't necessarily have to resemble the word 'mole'; for example the unit for distance isn't 'd'

It can be of interest to some of the readers and OP. And it is not up to date. Indeed I am not ashamed to say that I still have to digest the latter changes in the International System. Perhaps one of us can Ask & Answer about the impact of the recent units definition to the Periodic Tables hanging in our room / office / lab, if any. Beside that, here is the link https://www.degruyter.com/configurable/contentpage/journals$002fci$002f41$002f3$002farticle-p50.xml

The letter $m$ is the abbreviation of the unit "meter". The two letters $mo$ are the abbreviation of the unit "month". $mL$ is the abbreviation of "milliliter". So the only way of making up an abbreviation for "mole" is writing $mol$, as ridiculous as it may look.

@Alchimista Thanks for the link. I wasn't aware that they had really promoted the mole to be a formal SI unit now.

@Karl Now ?? The mole was made the seventh SI base unit in 1971 by the 14th CGPM.  Its definition was just updated recently.

@ Yolo. You say that the unit for distance is not $d$, and you are right. The unit for distance is "meter", with an abbreviation $m$. In the same direction, the unit for chemical amount is not "chem something". It is "mole", and it has an abbreviation $mol$.

@Maurice OTOH, I have thought until today "mol" is both the unit and the unit symbol ( like in Czech ), and that "mole" is mispelling, perhaps because of fitting English tongues better.  :-D

@ Poutnik What is the meaning of OTOH ?

@Maurice. In actual fact, mole is short for 'molecule' because it was counting the number of molecules. Despite this I don't think your reasoning works in the context of labelling efficiency. We call it Ampere (A) despite it being nothing to do with current for example.

@Maurice OTOH is the abbreviation of on the other hand.

@Maurice I have observed you repeatedly insert a space between @ and usedID. This leads to notification failure. E.g., I was not notified about your question regarding OTOH.

@Poutnik Sorry ! I did not know this circumstance. Is it better now ?

The symbol mol is due to Ostwald who was a very influential and respected physical chemist more than a century ago. In German, "mole" is "Mol". It is a shortform of Molekül.
I believe the Internet Archive has an English translation of his historically famous book of 19th century. This is where I recall reading this. The original reference is W. Ostwald Hand- und Hilfsbuch zur Ausführung physikochemischer Messungen. This book is available on Google Books (pg 278). A footnote appears there:

Um nicht stets den schleppenden Ausdruck Gramm -Molekulargewicht brauchen zu müssen, werde ich diese Größe in der Folge kurzweg Mol nennen, und erlaube mir, diesen Vorschlag zur allgemeinen Annahme zu empfehlen. Man schreibt ohnedies diese Größe häufig Mol., und braucht daher nur den Punkt fortzulassen.

In order not to always have to use the sluggish expression gram - molecular weight, I will simply call this quantity Mol in the following, and allow me to recommend this proposal for general acceptance. Anyway, this quantity is often written as Mol., so you only have to omit the point.
Translated via (www.deepl.com) with some manual changes.

The mole was promoted  to the status of "unit" in 1971

The 14th Conférence Générale des Poids et Mesures (CGPM),
considering the advice of the International Union of Pure and Applied
Physics, of the International Union of Pure and Applied Chemistry, and
of the International Organization for Standardization, concerning the
need to define a unit of amount of substance,
decides

The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of
carbon 12; its symbol is "mol".
When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or
specified groups of such particles.
The mole is a base unit of the Système International d'Unités.

It had been in use before that for quite some time, and I guess nobody wanted to change the symbol. Think about the ensuing confusion if you made up a new unit symbol today, that means exactly the same thing as the old one.

The mole unit  ( symbol mol , like kilogram has the symbol kg ) is not only ordinary SI unit, but 1 of 7 base SI units since 1971.

@Poutnik In the grand scheme of the cosmos, this is fairly recent though. ;)

@Martin-マーチン Like the whole science, isn't it ? :)

@Poutnik I looked it up now, and you're right of course. I distinctly remember my school teachers calling it "sort of" a unit, reasoning that it is mostly just a unitless number. And I'm sure most of my teachers had gotten their university degrees after 1971. ;-)

You may know as of 2019 some  S.I base units have been redefined [https://en.wikipedia.org/wiki/2019_redefinition_of_the_SI_base_units?wprov=sfla1] if you are interested then go through this.

@S.MUKHERJEE that is what I meant by "I am not ashamed...." etc. It was already discussed here but the idea of a Q self Answer seems good to me. I cannot do it with specifically going through the issue.


text:
Manually changing symmetry
I have two crystal structures, one in R3m and one in Cm symmetry. Both structures also differ in stoichiometric ratios, $ABC_4$ and $A_2B_2C_8$, respectively. Both structures look nigh on identical, with the exception of double the amount of atoms in one than the other.
How would I go about manually changing the symmetry of the Cm structure to R3m symmetry? I normally use VESTA as a visualization tool, and Avogadro as a tool for imposing symmetry.

Never heard of a group R3m

@Maurice space group 160: http://img.chem.ucl.ac.uk/sgp/large/160az1.htm

Your question is hard to answer because it lacks a lot of information.  $Cm$ (IT #8) belongs to the monoclinic which does not say anything about the threefold symmetry axes present in $R3m$ (IT #160).  But it is not unheard to transfer a monoclinic  model into one that belongs to the  trigonal / rhombohedral system.
Sadly, however, your question does not include a hint about the data and programs other than VESTA (for which I lack working knowledge) with you, nor your previous exposure to crystallography.  Of course the best were to start all over with reducing the diffraction data to write a new .hkl file, now with the hypothesis of higher symmetry present, to solve and refine the model again.  However, if you have a) a modern .cif file (those now including the hkl data and even better a good portion of the ShelX .res file), or b) one of the older .cif file plus the structure factors than you could give Platon's addsyminstruction a try. This routine was just written for the purpose of identifying missed symmetries.  It will write you a new .hkl, for lattice constants already adjusted, and an .ins file (with Laue symmetry updated accordingly) to start over with ShelX or Olex2.
By today's standards, Platon neither truly is CLI, nor GUI (mainly a large FORTRAN 77 executable + X-window; executables available Linux, Mac and Windows), but well documented in publications by Spek himself including many worked test data.  For addsym the page in question is here.  Don't forget to check the model if it still is chemically sensible (bond lengths / angles, ADP ellipsoids, void volumes) and significantly better, e.g. by Hamilton's R-factor ratio test.


text:
Combustion of cyanamide in a bomb calorimeter at constant volume and at 25°C

In a bomb calorimeter, cyanamide was subjected to combustion at constant volume, and the heat evolved was found to be $\pu{742.7 KJ}$ at $\pu{25 ^\circ C}$. Find $q_p$.

This is a question in my textbook. Given in the question that the volume is constant so how are we going to find $q_p$? Since it's a bomb calorimeter so pressure isn't constant. Is the question wrong?

In the given question we are asked to find qp

Okay, what is qp?

qp is the heat transfer at constant pressure

Well ! qp is probably the heat $\ce{Q_p}$ produced at constant pressure. The given data is probably $\ce{Q_v = \Delta U}$. And the unit is probably $kJ/mol$ and not $kJ$. So, in my interpretation, the author wants to get $\Delta H$. To pass from $\Delta U$ to $\Delta H$, it is necessary to know the change in gaseous moles in the combustion reaction. And the equation is probably : $\ce{CN2H2 + 1.5 O2 -->  CO2  +  H2O + N2}$ Now please do the calculation of $\Delta n$, then $\Delta n ·RT$, and you get $\Delta H$.


text:
Titration calculations - working out cation
so I was doing this question and I have some questions about what I did. Firstly am I correct to say that the concordant results = 24.1 cm^3 because there are 2 of these values. From this I then calculate that M = 32.9 however this can not be true, so what have I done wrong?

Let us start by writing the balanced equations. Write a balanced equation showing the reaction of NaOH and HCl.

Be aware the intention of Chemistry SE site is not task solving nor proof reading service, but rather help in understanding underlaying principles. Is there any principle or idea you have trouble to apply ? Also,write in detail your calculations. How can we tell you what you did eventually wrong if you do not show us what you did ?

You must have forgotten a factor $2$ somewhere.
You are right is thinking that $24.10$ mL is needed for the final titration. But now let's redo the calculations.
$24.10$ mL $\ce{NaOH} ~0.25$ M contains $24.10·0.25 = 6.025$ millimole $\ce{NaOH}$. Obviously this $\ce{NaOH}$ has neutralized the same amount $6.025$ millimole $\ce{HCl}$. As a consequence, there was $10$ times more $\ce{HCl}$ in the $250$ mL flask. This is $0.06025$ mol $\ce{HCl}$, and was the residual amount of $\ce{HCl}$ remaining after reaction of HCl plus the unknown carbonate. It also means that after adding $0.100$ mole $\ce{HCl}$ in the initial flask, the difference $0.100 - 0.06025 = 0.03975$ mol $\ce{HCl}$ had been used to destroy the carbonate.
But whatever the formula of the initial carbonate, $2~\ce{H+}$ are needed to react with  $1~ \ce{CO3^{2-}}$. So the amount of unknown $\ce{XCO3}$ or $\ce{Y2CO3}$ carbonate in the original powder is half the amount of $\ce{HCl}$ and it is : $0.03975/2$ mol = $0.019875$ mol $\ce{XCO3}$.
As the mass of this carbonate is $2.50$ g, its molar mass is: M = m/n = $2.50/0.019875 = 125.8$ g/mol. As the molar mass of $\ce{CO3}$ is $60$ g/mol, it means that the molar mass of $\ce{X}$ is $125.8 - 60 = 65.8$ g/mol.
This final value may be twice the atomic mass of a monovalent cation, or the atomic mass of a bivalent cation. As there are no monovalent cation with an atomic mass of about $32.9$, the unknown element must be a bivalent cation. And the only element with such an atomic mass is Zinc : A(Zn) = $65.39$
So the unknown carbonate is $\ce{ZnCO3}$

As an analytical chemist I disagree with Maurice's statement that 24.10 is the "right" value for the titrant. Solving for (d) does given a value of 24.10 ml yielding (1) 24.10, (2) 24.45 and (3) 24.10. Getting two values of 24.10 must be viewed as a coincidence. Thus the average of 24.22 should be used. (I'd use 24.2167 in the intermediate calculations to reduce rounding errors.)
So you need to calculate the standard deviation from the three measurements and then use the Student's T test to calculate a 95% confidence interval about the mean. Finally calculating the 95% confidence interval as a % error value will give you the expected error around the final mass.

For the three values the mean is 24.22 and the standard deviation is 0.20. For a two-tailed Student's T test at 95% CI and two degrees of freedom, the T value is 4.3027. Thus the CI is +/- $(\frac{0.2}{\sqrt{3}}\times4.3027 = 0.50)$. Looking at this as a percentage gives +/- 2.06% for the calculated cation mass value.

Finally the problem statement that the sample is an "unknown carbonate" is somewhat ill-defined.

The problem doesn't specifically state that there is only a single cation. It could for example be the mineral dolomite, $\ce{CaMg(CO3)2}$.
Nor does the problem state that the cation has a single oxidation state. There are minerals with Fe(II)/Fe(III).
The problem also doesn't state that the salt isn't a hydrate.

The point is that you must make an assumption that the mineral is not a hydrate and either a monovalent cation, $\ce{X2CO3}$, a divalent cation, $\ce{YCO3}$ or perhaps a trivalent cation $\ce{Z2(CO3)3}$.

This original question is taken from a British chemistry preparation book for pre-university students. I don't think the question writer ever thought to make it very complicated to have mixed valence or mixed carbonates.

@ MaxW. The difference between my value (24.10 mL), and yours (24. 22 mL)  is about 0.5%. This is also the precision of the original data : 2.50 g is supposed to be 2.50 ± 0.01 g, which is also ± 0.5%. On the other hand, I don't know of any carbonate containing a triply charged metallic ion. These ions are too acidic in solution. Al, Fe(III) and Cr(III) do not form carbonates.

I agree with MaxW about poorly worded question. This should at least states that the unknown carbonate is anhydrous. Assuming it is an anhydrous carbonate with $\ce{M2(CO3)_n}$ formula, I did the calculations and found out that:
$$M=32.9n$$ where $M$ is the atomic mass of the metal cation with $n+$ charge ($\ce{M^n+}$). Thus, OP's result of $M=32.9$ probably obtained by assuming the unknown carbonate is an alkali carbonate with molecular formula of $\ce{M2CO3}$ where $n=1$. However there is no metal with $+1$ charge with $32.9$ atomic mass (closest is $\ce{K+}$, atomic mass which is $39.1$). The next best choice is when $n=+2$, thus $M=65.8$. The closest $\ce{M^2+}$ is $\ce{Zn^2+}$, atomic mass of which is $65.4$. This is an acceptable value. Still, if you want to consider the possibility of $n$ being equals to $+3$ (thus $M=98.7$), the closest $\ce{M^3+}$ is $\ce{Tc^3+}$, atomic mass of which is $98.9$. Although this is slightly better value than $\ce{Zn^2+}$, I doubt whether $\ce{Tc2(CO3)3}$ is exist or not (although $\ce{TcCl3}$ is exist). Therefore it is safe to say that the unknown carbonate we are dealing with here is more like $\ce{ZnCO3}$.

I also that Zn is the most "suitable" answer. The posted image is from a test-prep book for pre-university students. They are neither analytical chemists nor they are aware of Tc compounds or mixed valence compounds.


text:
How can we do amide coupling reaction in water?
Does anyone know if it is possible or if there are any protocols for amide coupling in water (and not a normal polar aprotic solvent like DMSO, DCM, ACN, etc.)?

A search on the phrase "Amide Formation in Water" produces a good selection of hits.
Recent papers such as this and this offer conditions.
From the abstract of 1

A general and environmentally responsible method for the formation of amide/peptide bonds in an aqueous micellar medium is described. Use of uronium salt (1-cyano-2-ethoxy-2-oxoethylidenaminooxy)dimethylaminomorpholinocarbenium hexafluorophosphate (COMU) as a coupling reagent, 2,6-lutidine, and TPGS-750-M represents mild conditions associated with these valuable types of couplings


text:
How do I compute the mean dissociation time from the dissociation constant, assuming instantaneous binding reaction rate?
There is a solution with two chemical species, A and B. Given the equilibrium dissociation constant
$
K_D=\frac{[A][B]}{[AB]}
$
how do I relate the mean dissociation time $t_{off}$ to $K_D$, and initial reagent concentrations, assuming that the reaction time it takes for A and B to bind is much lower than the diffusion time from one to the other?
I know that $k_D = t_{off}/t_{on}$, so here I'm looking for $t_{on}$ in the approximation of instantaneous binding reaction time, which I assume should be the time required for a particle of a species to diffuse through the solution from a particle of the other species to the next. How do I do that to solve the question in the previous paragraph?
EDIT: changed notation to fix  typo and to incorporate Poutnik's feedback.

Could you elaborate the question little bit, to avoid confusion, misinterpretation and false assumptions about the question ? As too laconic questions without enough context often lead to too much clarifications effort.

What exactly do you mean by "mean dissociation time" ? Do you mean the reaction half-time, applicable on kinetics of the first order  like $\ce{AB  -> A + B }$ as $t_{1/2}=\frac{\ln{2}}{k}$  ?

Then you have given K, 1 k from the diffusion control based formula,  the other k from these 2,  and from it t1/2.

As AB dissociates as $\exp(-k\cdot t)$ the mean time $\langle t\rangle=1/k$.

Note that convention usually used Capital K for TD equilibrium constants and small k for kinetic rate constants.
Thermodynamic equilibrium constants say nothing about kinetic reaction rate constants, but about their ratio. There can be 2 reactions accidentally with the same equilibrium constant, but one with both forward and backward reaction rate very fast, the other very slow.
If there is reaction $\ce{AB <=> A + B}$ with the equilibrium constant $K_\mathrm{c}=\frac{[A][B]}{[AB]}$
with the forward reaction kinetic $\frac{\mathrm{d}[AB]}{\text{d}t} = - k_\mathrm{f} \cdot [\ce{AB}]$ and the reaction halftime $t_{1/2}=\frac{\ln{2}}{k_\mathrm{f}}$
and with the backward reaction kinetic $\frac{\mathrm{d}[AB]}{\text{d}t} = k_\mathrm{b} \cdot [\ce{A}][\ce{B}]$
then $K_\mathrm{c}=\frac {k_\mathrm{f}}{k_\mathrm{b}}$
All is assuming the reaction has trivial simple reaction mechanism.
Credit to @Porphyrin:

If $\ce{ A + B -> AB}$ is diffusion controlled, the rate constant would be (for equal sized molecules or approximately so ) $k_\mathrm{d} = k_\mathrm{b} = 8000 \frac {RT}{3η}$
where η is the solvent viscosity,
$k_\mathrm{d}$ evaluates to $≈\pu{5⋅1010dm3/mol/s}$ in low viscosity solvents such as ethanol.

Then :
$$kf=k_\mathrm{b} \cdot K_\mathrm{c}=K_\mathrm{c} \cdot \left( 8000 \frac {RT}{3η}\right)$$
and the reaction halftime is:
$$t_{1/2,f} = \frac {\ln 2}{k_\mathrm{f}} =\frac{\ln {2}}{K_\mathrm{c} \cdot \left( 8000 \frac {RT}{3η}\right)} =\frac{\ln {2} \cdot 3η}{  8000 \cdot K_\mathrm{c} \cdot RT} $$

In fact it would be easier if A+B were diffusion controlled as the the rate constant would be (for equal sized molecules or approximately so ) $k_d=8000 RT/(3\eta)$ where $\eta$ is the solvent viscosity. $k_d$ evaluates to $\approx 5 \cdot 10^{10}\mathrm{dm^3/mol/s}$ in low viscosity solvents such as ethanol.

@porphyrin thanks, that's exactly what I'm looking for. Can I ask where have you taken the formula from?

Updated answer with info from @porphyrin , giving him credit.

Smoluchowski has calculated the rate constant for diffusion controlled reactions assuming that the solvent was a continuous fluid, i.e. not made of molecules. This is a good approximation if the solvent molecules are small compared to those reacting and still good (surprisingly) even when this is not true. The rate constant is $k_d=4000\pi NDr$ where $D=D_A+D_B,; r=r_A+r_B$, $r$ being radius. From Stokes-Einstein eqn $D=k_BT/\zeta$ where $\zeta$ is friction $\zeta=6\pi \eta r$ and $\eta$ is viscosity. Substitute for $D$ for both A and B and make $r_A = r_B$.


text:
Are all mixtures linear?
I'm writing a computer program to help a factory produce bottles of mixtures. A mixture is usually alcohol, water and some other stuff, like menthol. But could also be water and NaOH and some oils (think schampoo).
My basic understanding is that if one bottle is 500 ml and contains:
50% water
45% alcohol
5% oils
I need:
250 ml water
225 ml alcohol
25 ml oils
Then if I want to make 1000 bottles I just multiply this with 1000:
250000 ml water
225000 ml alcohol
25000 ml oils
However this is with the assumption that all mixtures have linear dependencies.
So my questions is:

Is all (possible) mixtures linear?
Are all reactions linear?
Are there a difference between mixtures and solutions here?

I suspect this is a very easy question for someone who know chemistry, but unfortunately my understanding is basic and my search skills too bad to find this out.

Not generally. A solution that behaves like that is called ideal. 2. Don't know what you mean, what reaction? 3. "Solution" is used to refer to a particular type of mixture, where one component provides a medium in which others (minor components) are dispersed. Official meaning of a solution in chemistry: http://goldbook.iupac.org/terms/view/S05746

It depends on you somehow... As usual I would be in trouble preparing the above 50 45 5 mixture. Usually it should mean 50 45 5 all in grams, but diffused bad practice is always a source of confusion (indeed you were oriented to volumes...). This said, volumes are generally not additive but others have already answered/commented on that.

The nonlinearity of volumetric additions can be addressed theoretically. However, in practice, calculating how much material is needed to fill 1000 500 mL bottles containing 50% water 45% alcohol 5% oils (% by weight), can be addressed for one formulation by making a small batch and determining specific gravity. In practice, making the theoretical amount of product will not give the exact amount needed to fill 1000 bottles. Weighing errors, evaporation, residue on the mixing vessel, spillage, etc., etc., can easily add up to 2-3% - in either direction!
The specific gravity for each formulation must be specified (I did not say "determined"). Some formulations can be estimated to have the same specific gravity as another very similar product. If you are familiar with the products, you will know what can be estimated and what needs to be measured. A difference of 1% in the actual specific gravity is often not enough to worry about.
The things to worry about are:

how many bottles do you need to fill? Is it preferable to insure you make enough product for 1000 bottles, or to make sure there is no extra product to store or save (perhaps only enough to fill 973 bottles)?

when your computer printout indicates a number of grams of raw material to add, how many digits are needed if you desire 1% accuracy, and how confusing is it to specify 359.739 grams to be added when your balance/scale is only accurate to 1 gram (or less because it is never calibrated) and

using manufacturer's typical values for raw material specifications is frequently +/- 1-2%, so factoring them into some computer program may give you a feeling of control, but you really need to measure the specific gravity of the final product. Measuring the specific gravity could be a quality control test.

Computers can tabulate and keep track and calculate formulations for different size batches, but they do not do product formulation very well.

Mixtures are never linear. Mixing $50$ mL water + $50$ mL ethanol gives a $96$ mL mixture. The masses are additive, but not the volumes. I agree the discrepancy is not huge. If you can tolerate a $10$% error, you may consider that the volumes are additive. Anyway take care of the fact that, usually percentage are indicated in mass, not in volume.

yes but what is 5000 mL water + 5000 mL ethanol? It should be 9600 mL mixture, right?

This is basically volume contraction in solvent mixing. A similar question has already been asked previously: https://chemistry.stackexchange.com/questions/23534/how-does-volume-contraction-in-solvent-mixing-work


text:
Is there any theory more advanced than MOT?
I am currently going through molecular orbital theory(MOT), I should appreciate it's a very nice theory and it's helping me to fill the gaps of normal hybridization concepts and valance bond theory(VBT). But I am unaware about it's limitations seemingly it's the most advanced theory and have the capability of explaining everything. Although we know everything has it's limitations so if exist then what are they?
It will be also interesting to know about some modern viewpoints, accepted or proposed theory, and concepts regarding this stuffs.

maybe not 'advanced' than MOT, but DFT and QTAIM are some other theories for molecular physical chemistry.

MOT is more of a collection of different methods that are using molecular orbital approach in one way or another. Most calculation methods like DFT are MO based, in principle.

MO theory is more of a descriptive guide than a rigorous mathematical construct. The main idea of molecular orbital theory is that the electrons in atoms can be described by linear combinations of atomic orbitals (LCAOs), which themselves are the exact solutions to the Schrodinger equation for single electron atomic systems.
This idea can be applied in a more mathematically quantitative way by use of the Hartree-Fock (HF) method. HF is essentially MO theory put into practice. HF takes as input the geometry of a system (i.e. the spatial coordinates of all atoms in your molecule) and a set of "basis functions" which are usually sets of 3D gaussian functions centered on the atomic nuclei of the system. These basis functions are your "atomic orbitals" in MO theory. The HF method takes linear combinations of the atomic orbitals to generate a set of molecular orbitals which are usually more delocalized. These are the molecular orbitals of MO theory, and we can calculate them using the HF method running on a computer.
There are several approximations involved in HF theory, which the descriptive MO theory inherits.

The Schrodinger equation (SE) itself is an approximation. The SE is not exact as it fails to account for so-called relativistic effects. These are effects originating from Einstein's theories of special and general relativity. Some examples of relativistic effects are spin-orbit coupling and relativistic contraction of orbitals, especially in heavy atoms. The straightforward HF method doesn't take relativistic effects into account. Starting instead from the Klien-Gordon or Dirac equations will recover (some of) the missing relativistic effects.

The HF method assumes a complete separation between the movement of electrons and nuclei, known as the Born-Oppemheimer approximation. Basically, electrons are assumed to equilibrate instantaneously to the positions of atomic nuclei. This usually isn't a problem because electrons are much lighter than atomic nuclei so they do move much faster. Not a bad approximation in most cases.

Electron correlation. Even if we ignore the previous two approximations, the HF answer (and therefore the MO theory prescription) misses out on a very chemically important effect known as electron correlation. Electron correlation arises because a single set of molecular orbitals combined into a Slater determinant (in order to satisfy the Pauli exclusion principle) is not enough to fully describe the true exact solution to the Schrodinger equation for a molecular system. One must take linear combinations of many Slater determinants to improve on the HF answer. Electron correlation is responsible for effects such as the Van der Waals dispersion force. Another way of thinking about Electron correlation is that it is the energy that arises due to the dynamic Coulomb repulsion between moving electrons (although this picture is incomplete).

In summary, an exact molecular theory would need to account for Einstein's theory of relativity, combine the movement of atomic nuclei with electrons, and account for the dynamic correlations between different electrons in a system, in addition to being a proper quantum mechanical theory.
There are some quantum chemistry theories out there that incorporate relativistic effects into them. For example, the Douglas-Kroll-Hess approach is similar to HF theory, but begins from the Dirac equation and so incorporates relativistic effects explicitly. I believe there are also approaches that don't use the Born-Oppemheimer approximation, though I'm unfamiliar with them.
There is a massive amount of research and development into the so-called post Hartree-Fock methods. All of them attempt to account for electron correlation in various different ways. The main issue with the post HF methods is that the results (energies and wavefunctions) are often incredibly computationally demanding to actually compute, except on very small molecules. The most exact of the post-HF methods is the Full Configuration Interaction (FCI). With a sufficiently large input basis set, FCI produces an exact answer to the Schrodinger equation under the Born Oppenheimer approximation. Other post-HF methods include: Møller-Plesset perturbation theory, Coupled Cluster theory, and Multi-Configurational Self Consistent Field theory. All of these post HF methods are more accurate than the MO theory picture, but they are also way more difficult to visualize and understand for new students.

There is, probably more importantly, DFT, which is in principle exact. There are various approaches for orbital-free implementations, too. MO itself is - a quite handy - approximation and only a mathematical approach to a/the wave function.


text:
Understanding the Jahn–Teller elongation and compression for octahedral complexes (stabilisation/destabilisation of d orbitals)
I am currently learning about Jahn–Teller effect. For elongation I wondered why the d orbitals with $z$ component are stabilised when the metal–ligand bonds are getting longer? I thought a longer bond would mean a less efficient overlap of the d orbital on the transition metal with the orbital on the ligand.
And if yes, shouldn't it be raised so the difference in energy between the d orbital and the ligand’s orbital be greater? I thought the ligands’ orbitals would be lower in energy each time.
Where am I going wrong with my thinking?

JT is usually described as elongation along $z$ and a stabilisation of orbitals with a $z$ component. Not sure what your source on destabilisation has to say.

Ahh gosh, I was getting so confused, that is what I meant to put, but my reasoning doesn’t match up with it being stabilised

Before you consider J-T distortion, take a look at the d orbitals of an octahedral complex. Note that the orbitals that are on the axes where the ligands are are higher in energy. This is because these "d orbitals" are actually anti-bonding orbitals comprised primarily of the metal d orbitals. Because they are anti-bonding, they get higher in energy as the bonding interaction between the ligand and metal gets stronger. Thus, a longer distance between the ligand and metal (ie weaker bonding interaction) results in less destabilization.
Another way to think about it is to consider the interaction of a ligand lone pair (ie negatively charged electron density) with a filled d orbital on the metal (also negatively charged electron density). The closer they are together, the more unfavorable the interaction is.

This is a fantastic answer. Thank you for giving me the time to understand this, I really appreciate it.

There are two ways to approach this: simple crystal field theory or more complex molecular orbital theory. Both can describe and predict Jahn-Teller distortion well.
In simple crystal field theory, one always starts with a set of five degenerate d orbitals on the metal. The ligands are considered as a point-sized negative charge. (This consideration, while not being perfect, is okay, because the main thing the ligands contribute to a complex is electron density or an overall negative charge. The main problem is that it assumes point-shaped or spherical charges which is inadequate to explain more complex bonding – but not required to understand Jahn-Teller.) As the ligands approach the central metal – along the coordinate axes, in an octahedral complex – the d orbitals experience a negative electric field. As electrons occupying an orbital are also negatively charged, the more of a field an orbital experiences the less stable it becomes because the electron is repulsed by the electric field of the same charge.
When considering the orientation of the orbitals, two groups emerge. The first group is $\mathrm{d}_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$ whose lobes are oriented between the coordinate axes by convention. The second group is $\mathrm d_{x^2-y^2}$ and $\mathrm d_{z^2}$ which both point in the direction of the coordinate axes ($x$ and $y$ in the first case, all three but primarily $z$ in the ‘canonic’ shape in the second). Thus, these two orbitals experience destabilisation and are raised in energy.
If the electronic configuration of a complex happens to be $\text{hs-d}^4$ or $\mathrm d^9$ (other cases are possible but these two provide the strongest Jahn-Teller effect), the two highest-energy orbitals are populated unevenly despite being degenerate. In general, uneven population of degenerate orbitals is not a problem; but in this case the system can distort to provide an overall stabilisation. Assume that the two ligands in $z$ direction are moved away from the metal. This means, that the point-sized electric charge in $z$ direction is further away, which in turn means the electric field experienced by orbitals pointing in $z$ direction is weaker and thus ultimately they are destabilised less. This destroys the symmetry of the two highest orbitals (making one lower in energy than the other); the higher orbital can now be unoccupied while the lower is occupied. This constitutes stabilisation.

Scheme 1: simplified scheme of Jahn-Teller distortion in the crystal field model.
If you were to bring the ligands closer, the effect would be opposite: rather than reduce the energy of $\mathrm d_{z^2}$ you would be increasing it. This would again break symmetry and provide for a clear orbital to occupy less, but the overall energy of all orbitals would be higher. Thus, moving the ligands away provides a net stabilisation while moving them closer does not.
$z$ axis">
Scheme 2: assuming the $z$ ligands were moved closer, this simplified scheme of a Jahn-Teller distortion would result.
In the two schemes note the dashed lines; one pair of dashed lines is always horizontal. This is not an accurate depiction but it serves to get the point of Jahn-Teller across and it is close enough. Also note that the lower orbitals also lose their degeneracy. This is because strictly speaking they will be raised slightly by the incoming point-sized ligands although I have not included that in the diagram.

The more complex way to understand coordination compounds is molecular orbital theory. It is more extensive than crystal field but it provides an easy way to understand many more aspects which simple crystal field theory cannot explain.
The key to creating a molecular orbital diagram for a coordination compound is to understand that it is essentially the same as for any other covalent molecule. One has to determine which orbitals are available on either side of the bonds, how these orbitals transform according to the compound’s symmetry and point group, which orbitals can thus combine in what ways to give molecular orbitals and how strong those interactions will be.
The simplest molecular orbital scheme will consider the full set of 3d, 4s and 4p orbitals on the central metal (these transform as $\mathrm{a_{1g} + e_g + t_{1u} + t_{2g}}$) and six s-type orbitals on the six approaching ligands (these transform as $\mathrm{a_{1g} + e_g + t_{1u}}$). Immediately, one can notice that $\mathrm{t_{2g}}$ is only present in metal-centred orbitals and thus these will not budge upon bonding. Furthermore, $\mathrm{e_g}$ is present both in the ligands and in the metal, meaning that there will be a bonding-antibonding interaction between these two groups of orbitals.
Simple electronegativity considerations (ligands are more electronegative which usually corresponds to lower-lying ground state orbitals) will tell us that the metal’s orbitals are higher in energy than the ligand. Thus, upon forming a bonding/antibonding pair of linear combinations, the bonding interaction will be more closely associated with the ligands while the antibonding one will be centred on the metal. This means that the metal $\mathrm{e_g}$ orbitals will be raised in energy, as they are now antibonding ($\mathrm{e_g^*}$). We have already derived the same picture as the crystal field model predicts.

Scheme 3: Simplified general molecular orbital scheme of an octahedral complex including only σ-type interactions. Taken from Professor Klüfers’ coordination chemistry course’s web scriptum (LMU Munich) and first presented in this answer.
To explain the effect of increasing or decreasing the distance of ligands along one axis, the simpest way is probably to again derive it from a simple covalent bond. There, increasing the distance between two atoms will reduce the orbital overlap and thus reduce the energy difference between bonding and antibonding orbitals, moving both closer towards the baseline. On the other hand, reducing the bond length will have the effect of increasing bonding/antibonding interactions and raising antibonding orbitals while lowering bonding orbitals.
If you remember that the $\mathrm d_{x^2-y^2}$ and $\mathrm d_{z^2}$ orbitals are antibonding orbitals in coordination complexes, moving the $z$-ligands closer will mean that $z$-containing orbitals are raised (while $x$ and $y$-containing ones remain unchanged). Moving the ligands away will decrease the energy of $z$ orbitals (and not change that of $x$ and $y$ ones). This again is the same result as in the crystal field model albeit using different considerations and arguments.
Strictly speaking, one would have to consider the new complex under the new symmetry which will be $\mathrm{D_{4h}}$ in both cases. This places the orbitals into different irreducible representations ($\mathrm{2a_{1g} + b_{1g} + a_{2u} + e_u}$ for the ligands, $\mathrm{2a_{1g} + b_{1g} + b_{2g} + e_g + a_{2u} + e_u}$ for the metal), meaning a different set of interactions would take place. Ultimately, this consideration leaves $\mathrm d_{xz}$ and $\mathrm d_{yz}$ as well as $\mathrm d_{xy}$ where they are whatever happens (they transform as $\mathrm{e_g}$ and $\mathrm{b_{2g}}$, respectively, and have no ligand counterpart), while everything else is somehow affected. Without doing a full development on paper it is probably impossible to see how a shorter $z$ coordination distance would affect the energies compared to a longer $z$ distance but the final outcome should be equivalent to the above: a longer distance will lead to an overall stabilising effect.

This deserves at least 1 million upvotes. Thank you!

@MarinaCalder This isn’t even my final form yet ;)


text:
Why is the glycosidic bond between sugars always between the anomeric carbon?
Why can't the hydroxyl group of another carbon condense with another OH of the second sugar (like 2,4 glycosidic bond)
Thanks

Well, that would be a "usual" ether, not acetal, wouldn't it?

can it still occur tho?

If hydroxyl groups condensed each other at ambient conditions, then there wouldn't be stable alcohols at room temperature.

This would be a tail-to-tail connectivity, which has not been observed so far according to this source:

Structurally, for ranunculins A-D (Fig. 6), their potential 6-hydroxymethyls of 3-deoxyglucosones (3-DG) connected with another 6-hydroxymethyl group of sugars through a 6,6′-ether-bond (tail to tail ether-bond)10,11,12,13,14,15. In nature, such types of connections are unusual, in contrast to the familiar head to head (e.g., trehalose: 1, 1 connected), head to tail (e.g., isomaltose: 1, 6 connected) and head to body (e.g., sucrose: 1, 2 connected) connections. Coyolosa, a so-called 6,6′-ether-connected sugar, was reported to have been isolated from the root of Acrocomia Mexicana in one study16. However, very soon thereafter, this structure was regarded to be an incorrect and controversial based on the originally reported NMR spectroscopic data and activity studies17,18,19. Indeed, it is impossible that the 13C NMR data of C-6 were still at δC 61.1 when the 6,6′-ether-connection was formed. Nevertheless, inspired by the proposed coyolosa, Takahashi et al.17 and Haines, A. H.18,19 developed novel synthesis methods for 6,6′-ether-connected pyranoses. However, to date, no tail to tail connected sugars have been found from natural products, not to mention the participation of γ-amino acid moieties.

There is no specific reason for tail-to-tail linkages not to exist in nature. If there were enzymes to catalyze the reaction, you would find these compounds. They have been made synthetically: https://pubs.rsc.org/en/content/articlelanding/2004/ob/b407468h#!divAbstract


text:
Solubility of potassium silicate in water
It is somewhat surprising to me that there is almost no data about solubility of potassium silicate in water under normal conditions. The only thing I was able to find is the following:

<0.336 g/L @ 25ΕC, increasing  to 300 g/L @ 80ΕC. Once  dissolved,
material will not  precipitate out
Technical Document for Potassium silicate also referred to as a
BRAD

However, it is not clear what does the last statement mean. Does it mean that the silicate will not precipitate even if the solution cools down to 25°C?
The solubility of sodium silicate can easily be found on Wikipedia. It is mystery in case of potassium silicate.
So let us assume we want to dissovle potassium silicate in water with initial pH of 7. What amount of potassium silicate can be dissolved?
Edit:
Regarding the comment of @M. Farooq about being more specific. The second sentence on Wikipedia says:

The most common potassium silicate has the formula K2SiO3, samples of
which contain varying amounts of water.

We can consider the anhydrous form of K2SiO3 and couple of most common hydrates. Also, if the exact amount is not known then at least knowing the order of magnitude is fine as there is huge difference between 0.336g/L and 300g/L.

The very first line on Wikipedia states that "Potassium silicate is the name for a family of inorganic compounds."  The same is true for sodium silicate. You have to be specific about the composition.

@M.Farooq, thanks for the comment, I edited my post.

About precipitation of cooled down solutions, it is supposed to be very viscous, up to colloidic - and probably forming easily supersaturated solutions.

PQ Corporation sells many compositions of sodium and potassium silicates. Concentrations of potassium silicate range up to 40% in water.
The product literature includes pH, viscosity, SDS and just about every measurable property you could ask for. The reason such data are not in academic literature or book form is probably that they are considered lower-class information, used only by industrial engineering minds. I've got lots of brochures that are priceless for indicating what variety and range of commercial products are available: raw materials, not pure compounds.

https://www.pqcorp.com/docs/default-source/typical-properties/pq-corporation/potassium-silicates/17-2d-potassium-silicate.pdf?sfvrsn=a1fd57fb_3

Thank you James, but I don't see a solubility column in these tables.

@n0p: For the first 5 potassium silicates, if you add the weight % of K2O to the weight % of SiO2, you get a total weight % which is less than 100. These are liquids (indicated by the viscosity); the remainder is water; the maximum is ~40.9% for Kasil 18. These materials are significantly more viscous than water (~1 cps) and don't drop out crystals at some well-defined low temperature. They just stiffen up like a glass when it cools down from a melt.

Potassium silicate is synthesized by melting $\ce{KOH}$ above $400$°C, and adding some solid $\ce{SiO2}$. Its composition depends on the amount of silica used. The dissolution process does not look so. But it is a slow chemical reaction with a rather viscous liquid. The result can be cooled down and gives a solid substance practically insoluble in water at room temperature.
However it is possible to dissolve it with boiling water under pressure. Surprisingly enough, the obtained potassium silicate becomes more easily dissolved at $T>80$°C under pressure. The concentrated solution so obtained can be cooled down. It yields an extremely viscous liquid, more viscous than honey. This liquid is usually not really transparent, but it can be diluted with pure water without producing a precipitate. Such solutions are probably colloidal solutions of $\ce{H2SiO3}$ or $\ce{H4SiO4}$ in a concentrated solution of $\ce{KOH}$. As a matter of fact, if such a solution is heavily diluted, neutralized to destroy KOH, washed and evaporated in a vacuum, hydrated silica is obtained and it is made of $\ce{H2SiO3}$ flakes which are extremely light : the density of the stuff is not much bigger that air, as if it was made of rather long chains $\ce{HO -[- Si(OH)2 - O -]_n-H}$ sometimes directed in all directions around a central Silicium atom.
The chemistry of the silicic acid and its salts is extremely complex. The ternary diagram $\ce{Na2O - SiO2 - H2O}$ contains $11$ different phases.
Ref.: J. G. Vial, Soluble silicates, Reinhold, New York, 1952.
N. Greenwood, A. Earnshaw, Chemistry of the Elements, Pergamon, 1986, § 9.3.4. p. 398

Thanks for the answer, @Maurice. It looks really very complex, what I'm looking for is a rough estimate. However, if there will be no better answer in couple of days, I'll accept your answer. Thank you!

The above posts don't seem to answer the question, James Gaidis comes close by introducing you to the different forms though. The Anhydrous form of Potassium silicate does not dissolve in room temp water easely, and if you don't want to have to wait a day or more under stirring for it to dissolve. You have to heat it up under stirring to get it dissolved quickly-ish (Even under heating it is still a slow process).
The Hydrous forms of Potassium Silicate (Spray dried being the most common) do dissolve in room temp water quickly. Not as fast as normal sodium salt, around double the time of normal sodium salt dissolution. Obviously like any soluble salts, if heated while dissolving it speeds up the process, but under room temp the hydrous forms of Potassium silicate dissolve quite rapidly under agitation/stirring. As said above, the molar ratios of potassium too silica also have an effect on the time of dissolution. The higher the potassium ratio the faster it will dissolve.


text:
Does electrolysis of water alter the electrodes?
Today I used a 12 volt power supply to observe electrolysis of water.
I mixed table salt into the water, nothing else.
The leads coming from the power supply had clamps on their ends.
When in water bubbles started evolving on the clamps.
Does this damage the clamps in any way ?
Is there something I should put in the clamps to improve this experiment ?

It depends on the substance you have used for your electrodes. If they are in platinum, they will not be altered. If they are in carbon (charcoal, graphite), they will not be chemically tansformed, but after a while, they will be slightly corroded, and some powder of charcoal will progressively appear under the electrode on the ground of the flask used for the electrolysis. If they are in copper, there will be oxidized and a chemical reaction will occur at the anode. The copper metal will be oxidized and transformed into $\ce{Cu^{2+}}$ ions, which will produce a precipitate if the solution is basic or contains an anion like phosphate or carbonate. If the electrodes are in iron or zinc, the anode will be quickly corroded and transformed into the corresponding ions. In the beginning, the cathode will not be altered, but in the long run, they will also be corroded, specially if the solution is acidic.
Anyway you have not described the composition of the electrolyte. Common metals like iron and zinc are altered by acids, with or without electrolysis.

This question goes to the heart of production of H2 from water for the "hydrogen economy."  An excellent question.
Although the chemistry at the cathode, H2 production, for water hydrolysis seems to have been worked out, the stumbling block is the long term stability of the anode.  Generally as practiced, both oxidation and reduction occur simultaneously.  Looking at it another way, it is the oxygen production part of water hydrolysis that makes hydrogen production from water difficult.
See  Nat Energy 4, 786–795 (2019). https://doi.org/10.1038/s41560-019-0462-7 for an interesting approach to the question.
Short answer is yes, and the biggest changes seem to occur at the anode.


text:
Bioisosteric replacement using SMARTS (KNIME and RDKit)
I am trying to create a KNIME workflow that would accept a list of compounds and carry out bioisosteric replacements (we will use the following example here: carboxylic acid to tetrazole) automatically.

NOTE: I am using the following workflow as inspiration : RDKit-bioisosteres (myexperiment.org). This uses a text file as SMARTS input. I cannot seem to replicate the SMARTS format used here.

For this, I plan to use the Rdkit One Component Reaction node which uses a set of compounds to carry out the reaction on as input and a SMARTS string that defines the reaction.
My issue is the generation of a working SMARTS string describing the reaction.
I would like to input two SDF files (or another format, not particularly attached to SDF): one with the group to replace (carboxylic acid) and one with the list of possible bioisosteric replacements (tetrazole). I would then combine these two in KNIME and generate a SMARTS string for the reaction to then be used in the Rdkit One Component Reaction node.

NOTE: The input SDF files have the structures written with an
attachment point (*COOH for the carboxylic acid for example) which
defines where the group to replace is attached. I suspect this is the
cause of many of the issues I am experiencing.

So far, I can easily generate the reactions in RXN format using the Reaction Builder node from the Indigo node package. However, converting this reaction into a SMARTS string that is accepted by the Rdkit One Component Reaction node has proven tricky.
What I have tried so far:

Converting RXN to SMARTS (Molecule Type Cast node) : gives the following error code : scanner: BufferScanner::read() error

Converting the Source and Target molecules into SMARTS (Molecule Type Cast node) : gives the following error code : SMILES loader: unrecognised lowercase symbol: y

showing this as a string in KNIME shows that the conversion is not carried out and the string is of SDF format : *filename*.sdf 0 0 0 0  0 0    0 V3000M V30 BEGIN etc.

Converting the Source and Target molecules into RDkit first (RDkit from Molecule node) then from RDkit into SMARTS (RDkit to Molecule node, SMARTS option). This outputs the following SMARTS strings:

Carboxylic acid : [#6](-[#8])=[#8]
Tetrazole : [#6]1:[#7H]:[#7]:[#7]:[#7]:1

This is as close as I've managed to get. I can then join these two smarts strings with >> in between (output: [#6](-[#8])=[#8]>>[#6]1:[#7H]:[#7]:[#7]:[#7]:1) to create a SMARTS reaction string but this is not accepted as an input for the Rdkit One Component Reaction node.

Error message in KNIME console :
ERROR RDKit One Component Reaction 0:40       Creation of Reaction from SMARTS value failed: null
WARN  RDKit One Component Reaction 0:40       Invalid Reaction SMARTS: missing

Note that the SMARTS strings that this last option (3.) generates are very different than the ones used in the myexperiments.org example ([*:1][C:2]([OH])=O>>[*:1][C:2]1=NNN=N1). I also seem to have lost the attachment point information through these conversions which are likely to cause issues in the rest of the workflow.
Therefore I am looking for a way to generate the SMARTS strings used in the myexperiments.org example on my own sets of substituents. Obviously doing this by hand is not an option. I would also like this workflow to use only the open-source nodes available in KNIME and not proprietary nodes (Schrodinger etc.).
Hopefully, someone can help me out with this. If you need my current workflow I am happy to upload that with the source files if required.
Thanks in advance for your help,
Stay safe and healthy!
-Antoine

For ChemSE's community -- it is my hunch only -- this might be a bit too special.  But addressing this question to RDKit's user group mailing list (https://sourceforge.net/p/rdkit/mailman/, free subscription) with what you did and what you aim for may well yield an answer in less time and more detail than here.

Another possibility is Matter Modeling SE, though I admit I don't know how much expertise there is different chemical notations in this community.


text:
Coordination compound nomenclature for [Cu(acac)(bpy)]BPh4
The title says it all really.
I thought it would be acetylacetonatobipyridinecopper(II) tetraphenylborate. Is this correct? I feel as though it looks a bit clumsy and wasn’t sure if I was supposed to include as (2,2-bipyridine)? Is it ok to use brackets like this for naming a coordination compound the IUPAC way or is my first naming correct?


text:
Fermentation of glucose
I have trouble understanding as to what the last part of the increase in liquid level means. I believe this is the foam formed after the fermentation? I am calculating for the pressure but I have trouble as to what volume should I use to find the moles since I don't understand the last part. It would greatly help me to understand this process first.
I've been extensively searching how the fermentation process actually occurs cause I don't understand a thing about it. To find the pressure from what I know, I converted the volume of glucose to $\ce{CO2}$ to find the moles and plugged the value in the ideal gas equation $p=nRT/v$ with the volume being $\pu{0.07L}$ from $(\pu{125mL} - \pu{55mL})$, because it says the $\ce{CO2}$ in solution is negligible.
I don't know where the changes in densities come in and I don't know how to incorporate the pressure brought about by the liquid solution. I think my thought process is wrong really cause I don't understand fermentation.
https://www.homebrewtalk.com/threads/calculating-total-gas-produced-in-fermentation.165020/ This site proposed a $p_1v_1=p_2v_2$ method where $p_1$ is at STP, but I don't think this is needed in the problem.

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

It seems to me like the task description is wrong, like if they maybe do not consider the decrease of mass of the liquid. As the volume increase due lower density is well compensated by the mass decrease.
If I do not count with decrease of the solution mass, I can get the final volume $\pu{55 mL}$.
Note the $x$ is the glucose conversion factor. $x-0$ means glucose only, $x=1$ means ethanol only.
"V wrong" means the volume change, that is accounting with density decrease due glucose -> ethanol conversions, but not accounting with mass decrease of the liquid.
$\ce{CO2}$ calculations are done by Mathew, no need to doubling that.
$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline
\text{x} & \text{H2O[g]} & \text{glu[g]} & \text{eth[g]} &\text{CO2[g]} & \text{liq[g]} & \text{Eth w/w%} & \text{ $\rho$ g/mL} & \text{ V[mL]} & \text{ V[mL] wrong}\\ \hline
0 & 44.0 & 10.00 & 0.00 & 0.00 & 54.00 & 0.00 & 1.080 & 50.00 & 50.00 \\ \hline
0.05 & 44.0 & 9.50 & 0.26 & 0.24 & 53.76 & 0.48 & 1.075 & 50.01 & 50.24 \\ \hline
0.1 & 44.0 & 9.00 & 0.51 & 0.49 & 53.51 & 0.97 & 1.070 & 50.02 & 50.48 \\ \hline
0.15 & 44.0 & 8.50 & 0.77 & 0.73 & 53.27 & 1.46 & 1.065 & 50.04 & 50.72 \\ \hline
0.2 & 44.0 & 8.00 & 1.02 & 0.98 & 53.02 & 1.96 & 1.059 & 50.05 & 50.97 \\ \hline
0.25 & 44.0 & 7.50 & 1.28 & 1.22 & 52.78 & 2.46 & 1.054 & 50.06 & 51.22 \\ \hline
0.3 & 44.0 & 7.00 & 1.53 & 1.47 & 52.53 & 2.96 & 1.049 & 50.07 & 51.47 \\ \hline
0.35 & 44.0 & 6.50 & 1.79 & 1.71 & 52.29 & 3.47 & 1.044 & 50.09 & 51.72 \\ \hline
0.4 & 44.0 & 6.00 & 2.04 & 1.96 & 52.04 & 3.99 & 1.039 & 50.10 & 51.98 \\ \hline
0.45 & 44.0 & 5.50 & 2.30 & 2.20 & 51.80 & 4.51 & 1.034 & 50.11 & 52.24 \\ \hline
0.5 & 44.0 & 5.00 & 2.56 & 2.44 & 51.56 & 5.04 & 1.029 & 50.12 & 52.50 \\ \hline
0.55 & 44.0 & 4.50 & 2.81 & 2.69 & 51.31 & 5.57 & 1.023 & 50.14 & 52.76 \\ \hline
0.6 & 44.0 & 4.00 & 3.07 & 2.93 & 51.07 & 6.10 & 1.018 & 50.15 & 53.03 \\ \hline
0.65 & 44.0 & 3.50 & 3.32 & 3.18 & 50.82 & 6.64 & 1.013 & 50.16 & 53.30 \\ \hline
0.7 & 44.0 & 3.00 & 3.58 & 3.42 & 50.58 & 7.19 & 1.008 & 50.18 & 53.57 \\ \hline
0.75 & 44.0 & 2.50 & 3.83 & 3.67 & 50.33 & 7.74 & 1.003 & 50.19 & 53.85 \\ \hline
0.8 & 44.0 & 2.00 & 4.09 & 3.91 & 50.09 & 8.30 & 0.998 & 50.20 & 54.12 \\ \hline
0.85 & 44.0 & 1.50 & 4.34 & 4.16 & 49.84 & 8.86 & 0.993 & 50.22 & 54.40 \\ \hline
0.9 & 44.0 & 1.00 & 4.60 & 4.40 & 49.60 & 9.43 & 0.987 & 50.23 & 54.69 \\ \hline
0.95 & 44.0 & 0.50 & 4.86 & 4.64 & 49.36 & 10.00 & 0.982 & 50.25 & 54.97 \\ \hline
1 & 44.0 & 0.00 & 5.11 & 4.89 & 49.11 & 10.58 & 0.977 & 50.26 & 55.26 \\ \hline
\end{array}$

Alcohol fermentation is the formation of alcohol from sugar (or glucose). Yeast, when under anaerobic conditions, convert one glucose molecule $(\ce{C6H12O6})$ to two molecules of pyruvic acid $(\ce{CH3C(=O)CO2H})$ via the glycolysis pathways, then go one step farther, converting each pyruvic acid into ethanol $(\ce{CH3CH2OH})$ and carbon dioxide $(\ce{CO2})$ via acetaldehyde:

Thus, each glucose molecule would form two ethanol molecules and two $\ce{CO2}$ molecules:
$$\ce{C6H12O6 ->[yeast] 2C2H5OH + 2 CO2}$$
Let's look at what we have in the flask: $\pu{50 mL}$ of $20\% (w/v)$ glucose solution, density of which is $\pu{1.08 g mL-1}$. Thus, mass of $\pu{50 mL}$ solution is:
$$\pu{50 mL} \times \pu{1.08 g/mL} = \pu{54 g}$$
Since $\pu{10 g}$ of this mass is the mass of glucose, the mass of water in the $\pu{50 mL}$ solution is $\pu{44 g}$. Assuming density of water is $\pu{1.0 g mL-1}$ at $\pu{30 ^\circ C}$, then the volume of water at this temperature is $\pu{44 mL}$ (Note: This assumption is justified by the facts that neither densities of solutions are given in particular temperatures nor the $\pu{50 mL}$ of solution is measured in same temperature).
If the solution have gone to 100% completion, none of $\pu{10 g}$ of glucose $(MW = \pu{108.16 g mol-1})$ will be remaining in the solution and as a result, the volume would decrease to $\pu{44 mL}$ of water. Let's calculate the mass ($m$) of ethanol $(MW = \pu{46.07 g mol-1})$ would be produced by this fermentation:
$$m = \frac{\pu{10 g}}{\pu{180.16 g mol-1}} \times 2 \times \pu{46.07 g mol-1} = \pu{5.11 g}$$
Therefore maximum volume of ethanol would produce by fermentation is: $\frac{\pu{5.11 g}}{\pu{0.789 g mL-1}} =  \pu{6.48 mL}$. Thus, assuming ethanol and water are additive, maximum volume of the solution in the flask is: $\pu{6.48 mL} + \pu{44 mL} = \pu{50.48 mL}$.
Thus, this reaction would increase the volume of the solution only by $\pu{0.48 mL}$, if the reaction went to 100% conversion, so that the data given is incorrect (can't increase by $\pu{5.00 mL}$ as given).
Thus, assuming 100% conversion, we can calculate the amount of $(\ce{CO2})$ produced ($m_1$) in the flask:
$$m_1 = \frac{\pu{10 g}}{\pu{180.16 g mol-1}} \times 2 = \pu{0.111 mol}$$
Assuming no $(\ce{CO2})$ would dissolve in aqueous solution (this is actually incorrect assumption but it was instructed in the question) and $(\ce{CO2})$ behaves as an ideal gas, the partial pressure inserted by the $(\ce{CO2})$ gas ($p$) is:
$$p = \frac{nRT}{v} = \frac{\pu{0.111 mol} \times \pu{0.0821 atm L mol-1 K-1} \times \pu{303 K}}{\pu{(0.125-0.05048) L}} =  \pu{37.1 atm}$$

I see I was misreading w/v % as w/w % :-)

@Poutnik: Do you want to change your answer accordingly? I also didn't understand what $x$-column means.

Somewhen in the morning CEDT :-)  X is conversion factor, few times mentioned in comments.

Thinking by the task author pressurized CO2 would not dissolve is quite funny. Or, combined with the mass issue, it is rather sad, being the task for the task, detached from reality. Interesting would be to know, how [CO2(l)] eventually affect the yeast growth.


text:
Is lauryl glucoside stable in low pH?
I wish to use lauryl glucoside in the solution of citric acid. I tried to add it to 5% w/w solution of citric acid in water but after about 24h in the concoction which was previously clear I observed a kind of sediment. It looked like a "white smoke" disspersed in the clear liquid. I wonder whether it can be the effect of hydrolysis of glucoside due to the low pH (about 2) of the solution...
I used:

distilled water,
clear citric acid,
clear lauryl glucoside.

Can someone provide an information in what pH values lauryl glucoside is stable? Do you know how to explain what happened with my solution?

I have no specific knowledge of lauryl glucoside, but the key feature of the molecule (where the lauryl alcohol joins the glucose molecule) is an acetal and they are certainly hydrolysed in aqueous acid of pH1 or stronger. pH2 for an extended time seems likely to produce a slow hydrolysis.

Thank you for this response @Waylander. I suspect the white sediment to be a dodecanol then... It's interesting because I know some bathroom cleaners that consist of aquenous solution of citric acid, lauryl glucoside and decyl glucoside and the pH is 2,4 - 3,0

Remember pH is a log scale, pH2 is ten times as concentrated at pH3. So the difference between 2.0 and 2.4 is a lot of H+ ions. Acetals are considered stable in the pH2-4 range so I suspect you are just over the edge of stability with your solution

Fair point. I'll try to prepare solutions with higher pH and see how it works. Thanks! :)


text:
Finding the equivalent weights of the oxide and the metal when reduced by hydrogen

A metal oxide ($\pu{1.80 g}$) required $\pu{833 ml}$ of hydrogen at NTP to be reduced to its metal. What are the equivalent weights of the oxide and the metal?

My attempt:
$$\text{equivalent of metal oxide = equivalent of } \ce{H2}$$
$$\frac{\pu{1.80 g}}{E \ \pu{g/eq}} = \frac{\pu{833 ml}}{\pu{22400 ml/eq}}$$
$$E = \pu{48.4 g/eq}$$
Now, since $$E_\mathrm{oxide} = E_\mathrm{oxygen} + E_\mathrm{metal} $$
$$48.4 = 8 + E_\mathrm{metal} \Rightarrow E_\mathrm{metal} = 40.4$$
where $E$ is equivalent weight.
However, my textbook says that the equivalent weight of the metal oxide is $\pu{24.2 g/eq}$ and that of the metal is $\pu{16.2 g/eq}$ (it does not provide a solution). Is the book's answer correct? If yes, what did I do wrong?

The source of this question must be very old. The concept of equivalent weights is obsolete for over a century, and the SI unit symbol for gram is g, not gm. Also, posting the problem as a title is probably a bad idea.

@andselisk The book is ironically called "Modern Approach to Chemical Calculations", by RC Mukherjee. I get that equivalent weight is a dead concept from other answers on Chem SE, still would like to know the solution because it's in my syllabus. As for the title, I was following the norm from Math SE. Why is it such a bad idea?

The title should be brief. Something like "equivalent weights" or "calculation of equivalent weights". A title must not be  a copy of the question.

And I back Andselisk. The equivalents have been abandoned ion the years 1950, because they too often lead to mistakes in calculations. Tell your teacher that it is obsolete.

Anyway, there are no metals having an equivalent of 16.2 g/eq.

Equivalents are part and parcel of Indian chemical education. I don't think college chemistry is going to be updated soon anytime there. I recently heard a complaint by a company to the chemistry chair in an American University. They said that your students do not understand normality! Please teach these concepts to your undergraduate students. Of course, normality concept sounds Greek to modern students.
Anyway let us assume we are in early 19th century and we are doing a classical experiment of chemical analysis. It is not a bad exercise, then. Balanced equations existed then, so we can write:
$$\ce{M_xO_y + yH2 -> M + yH2O}$$
833 mL of hydrogen gas corresponds to 0.833/22.4 = 0.0372 moles of hydrogen. This concept existed in early 19th century.
So what is the definition of an equivalent weight of anything? It is the weight that reacts with 1 equivalent weight of hydrogen (which is half its molecular weight).
Your first step is wrong. Realize that
No. of Equivalents of $\ce{H2}$ in 833 mL are = 2 x 0.0372.
This implies that equivalents of metal oxide which are present in the reaction are also 2 x0.0372.
Equivalent weight of the metal oxide is then = 1.80 g/ 0.0744 = 24.19 or as per your textbook 24.2
Since the equivalent weight of oxygen is 8, the equivalent of metal is 24.19-8= 16.19.
Now, there is no metal whose atomic weight matches 16.19. So what if the metal is divalent = 32 (this is sulfur, not a metal). Is it possible that this is trivalent metal like Ti so  the atomic weight of that M is 16.19 x3= 48.57 which matches with titanium.
One can do back calculation to double check this using moles. I leave that for an interested reader.

Now I see, thanks. That was a careless mistake.


text:
Co-precipitation particle size
If we are creating a precipitate of (say) Barium Sulfate from a solution of Barium Chloride mixed with Sodium Sulfate, what determines the particle size?
I assume one of the factors will be concentration of reactants in the solutions, but what else?

@Poutnik What else? temperature, rate of heating/cooling, local gravitational variations, illumination, purity, electric and magnetic fields, sonics, ionizing radiation which way the wind is blowing...

@Poutnik What I want to know is if there are any general rules, and relative magnitudes of the various environmental influences. Of course, these may not exist in any systematic form and be totally reaction dependent. In which case all you have to do is say so

@Poutnik I come to SE because I want some quick answer, usually to resolve a go/nogo problem. I am a working scientist/engineer and currently have 4 R&D projects on my plate. Now, I could have waited for Dr Daryl Williams, a professor of chem eng at Imperial College to get in, since his specialty is particle science. But, as you say, I'm lazy. And impatient. But thanks for the answer

Kinetic aspects:
Low concentrations of interacting ions decrease rate of precipitation, leading to bigger crystals.
A high ionic strength of spectator salts leads to lower activity coeeficients , and  decreased rate of precipitation, leading to bigger crystals.
Higher viscosity leads to lower collision rates, decreased rate of precipitation, and to bigger crystals.
Low temperaturee leads to bigger crystals, as rate of collisions leading to crystallisation is lower due higher viscosity, lower molecular speed and lower rates of interactions leading to precipitation.
Note that kinetic aspects may be hard to involve in the proces control, aside of concentrations.
Thermodynamic aspect:
Crystals of low sizes have high surface Gibbs energy and therefore bigger crystals are thermodynamically preferred.
This is initially overrun by the kinetics of the precipitation, but in long term there is ongoing recrystallization in favour of bigger ones. This can be extremely slow for very little soluble salts.
This process is speeded up by high temperature, what increases rates of the opposite processes of dissolution and crystallization. AFAIK, the gravimetric procedure to determinate sulphates as barium sulphate involved heating $\ce{BaSO4}$ suspension near below boiling point for I guess 1-2 hours to speed up recrystallization. As bigger crystals are easier to manage, espacially in this particularly case, as $\ce{BaSO4}$ is known to form too fine crystals.


text:
How does a body lose electrons?
If there is a Na and cl in solid form , There will be atoms inside of them.How do they lose electrons ?.We know solid body has a structure and covering.Just like you can touch is the covering of table that is its outer part.Do we touch the electrons or the atom as we touch a body?.When a body loses electrons from a body,does it mean it loses a part of its structure and when it gains , does it gain some part of other structure body.How does this process happen?

What is it that you want to know about that "how"? Do atoms have small robotic arms, much like Mars rovers? No, they don't. The electrons are mostly driven around by electric fields.

When atoms lose electrons, How is it done?

How do the atoms come out of a body ?

What is happening if you visualise it ?

We can't see nuclei and electrons, but we can measure their effect on their surroundings. The theory is that these are independent quantum mechanical particles. You can think of electrons as flying about attracted by nuclei and repelled by other electrons. They are constrained by the balance of attraction and repulsion. They have only so much energy, so they can't go too far from nuclei, maybe in a metallic solid they can hop between neighboring nuclear attractors. If you put enough energy into them though you free them to fly off and interact with more distant atoms.

Well surely has to happen at the interface. I don't see what puzzles you at least in the frame of the question. Say the specific material reacts with something from the surrounding at the interface. If the product still stick to the original material it can even form a kind of skin and eventually protect the bulk. This happens, for instance, in Al. Forget this protective effect, but also a piece of butter gets rancid from the outer surface in...

I can't present you a movie of electron interactions, but maybe a fuzzy picture. Imagine a metal body (start with copper): the atoms are held in a solid structure - in a sea of electrons! Those loose electrons can be pushed around easily, so we can make wires out of copper and transmit electricity thru them. If you put two different metals together, the sea "levels" will even out, and a potential will develop between the two metals. But I digress.
Imagine your sodium clump. It has the same sort of firm atomic structure, held together by this sea of electron density which we could push around if we made wires out of it. Well, we do make wires out of sodium, but not for transmitting electricity. We make the wires for dehydrating solvents like benzene.
Now the non-metals, like chlorine gas, hold onto their electrons more tightly - no sea there. Sort of like dry land. Put a chlorine molecule next to the clump of sodium (or a copper wire!), and all of a sudden, the molecule realizes (well, it would, if it had a brain) that it would be more stable if it just split and each chlorine atom grabbed an electron from this sea in the metal. Who would notice? Well, after this occurs a few million times, the chlorine anions find that they can develop a structure by including some sodium cations (sodium atoms without their outermost electron). This results in a greater stability, so the overall process is that a number of sodium atoms transfer their electrons to chlorine molecules to generate sodium chloride crystals. The actual mechanism is probably way more complex - perhaps the chlorine molecule picks up the electron first, then gets bumped and falls apart into a chlorine anion and a chlorine atom which immediately picks up another electron. The actual process can be investigated, but the result is still formation of $NaCl$.
In benzene, the sodium lies at the bottom until a water molecule comes by and lands on the surface. An electron jumps into the water molecule because hydroxide ion is a nice place to hide, but this frees a hydrogen atom. This hydrogen atom rolls around on the sodium surface until it finds another hydrogen atom, and they combine to form a molecule. When enough molecules find each other, they form a bubble and float away. And the process is probably way more intricate.
But overall, electron transfer occurs when two materials get close enough to sense their different electronegativities and conduct/transfer one or more electrons. We separate the process of electron emission and electron absorption in electrolytic cells (batteries).
You can also move electrons out of metals by heating them. The Edison effect is the name for the process of emitting electrons from a heated electrode in a vaccuum - the basis for electron tubes. You can also rub electrons out of certain materials (static electricity) and make sparks (scuff across a rug and touch a light switch). Here the electrons (the outer covering) of one material scrubs off some of the outer covering of the other material - but this is a non-equilibrium situation, and quickly evens out.


text:
Absorption and emission at same wavelength?
Is it possible for a molecule to absorb and emit at the same wavelength? What is the reason behind it?
I’m working on charged tin porphyrins and got the excitation and emission (fluorescence) wavelength around 400 nm.

Yes, it is possible but it is not a single wavelength of overlap. Molecular spectra are quite wide. You have to see the excitation and emission spectrum and make sure you are not seeing Rayleigh scattering.

Thank you Sir. I have attached the graphs. Can you help me

Since this is a simulation, I have no idea how to verify its correctness. You need to verify it experimentally.

Well, normally one has to explain why is not.

Yes it is very common particularly in the more rigid type of molecule. The best example is chlorophyll and this overlap of absorption and emission leads to energy transfer in photosynthesis.  The reason for the overlap is that the excited state potential energy profile is very similar in shape but slightly shifted (as displacement) to that of the ground state and this results in a 'mirror image' spectrum. The spectra below are from anthracene.

You can see even in this rather crude figure that there can be absorption and emission at the same wavelength.

You should explain that in molecules the electronic states and vibrational states couple to yield microstates and that is why the molecular peaks are broad.

In the vapour phase lines are v narrow and rotational levels can be resolved, limited by lifetime, collisional and doppler broadening, e.g. in benzene, naphthalene etc. (Density of states will ultimately limit this due to overlap/ crowding of levels). In solution the many collisions with solvent (which also contains electrons, of course, and so cause dipole and induced dipole (dispersion) effects) and these cause energy levels to broaden and is the dominant effect. This gives the appearance normally observed.

@porphyrin Say if we measure absorption and emission at one particular λ. I think the spectrum should show net absorption/ emission? So how come we get both spectra showing non-zero reading at that λ? I expect one spectrum to be non-zero, the other to be zero.

Usually absorption is measured through the sample in one experiment, but fluorescence is measured at right angles to this (in a separate experiment) so that the exciting light can be removed. In a laser the balance between absorption and emission determines if lasing is going to occur.


text:
Reaction of 1,3-dibromobutane with zinc dust upon heating
I've studied that when 1,3-dibromopropane (or in general any halogen, not just bromine) is heated with Zn dust, the electrons produced cause the halogens to break away homolytically and two free radicals are formed. These free radicals then combine to form cyclopropane.

My question is, will this also occur for 1,3-dihalides when the carbon count is increased, say in 1,3-dibromobutane?
Everything I found related to this was either talking about the 3-carbon dihaloalkane case, or seemed to be some complicated free radical cyclization which was way out of my reach.

Please don't use MathJax for chemical names; they're only supposed to be used for chemical formulas, and then it's better to use the mhchem package as this provides convenient and correct typography, i.e. $\ce{H2O}$ $\ce{H2O}$ not $H2O$ $H2O$.

@orthocresol ahh, thanks. I didn't see any specific details on the Ask Question page so just ended up using MathJax which was the only thing I had used before.

This mechanism there is more like symbolic representation of what may really happen there, or even outright incorrect.

@Mithoron could you point me towards a reference with the correct mechanism?

Source: James Grimshaw, Electrochemical Reactions and Mechanisms in Organic Chemistry: Chapter 4

The formation of cyclopropane by reduction of 1,3-dibromopropane was discovered in 1887. Dissolving metals, in particular zinc dust in ethanol, were employed as an electron source. Electrochemical reduction in dimethyl-formamide at a mercury cathode has been found to give good yields of cyclopropane. 1,3-dibromo, 1,3-diiodo and 1-chloro-3-iodopropane all give greater than 90% yield of cyclopropane, the other product being propene.

Since chloroalkanes are not reduced at the cathode potential used, it is concluded that these reactions involve generation of a carbanion by dissociative electron transfer to the most easily reduced carbon-halogen bond followed by SN2 attack of this carbanion on the second carbon-halogen bond. Cyclopropane ring formation under electron transfer conditions shows no stereoselectivity...

...Most of the cyclopropane ring forming reactions can be accommodated in the mechanism that involves a carbanion in SN2 displacement of the second halogen atom. An exception is the reactions of 2,6-dibromobornane, which cannot accommodate the transition state stereochemistry necessary for an intramolecular SN2 displacement. Also, the original preparation of cyclopropane in good yield by dissolving metals in a protic solvent is unlikely to involve carbanions. When the reacting orbitals are parallel, addition of the second electron and carbon-carbon bond formation are likely to be synchronous. If these orbitals are not aligned, reduction leads to a diradical after which the carbon-carbon bond is formed.

The mechanism shown in the question is not wrong, but grossly incomplete.
Cyclopropane ring formation has been achieved in many other compounds, including those with high ring-strain. Although I could not find the specific reaction of 1,3-dibromobutane, it is reasonable to assume that it will undergo the same reaction to give methylcyclopropane.

Cyclobutanes and cyclopentanes are formed in low yields by electrochemical reduction of the appropriate $\alpha,\omega$-dibromoalkane in dimethylformamide. 1,4-Dibromobutane affords 29% cyclobutane along with butane and butene at a vitreous carbon cathode... Similarly, reduction of 1,5-dibromopentane affords 28% cyclopentane... 1,6-Dihalohexanes afford less than 5% of cyclohexane at a vitreous carbon cathode. The cyclization step in these reactions is believed to be an SN2 displacement of the remaining carbon-bromine bond by a carbanion formed in the initial reductive bond cleavage step, Cyclization must be very fast so as to avoid protonation of the carbanion intermediate.


text:
Equilibrium constant and stoichiometry
In the expression of equilibrium constant, each concentration term is raised to the power of their stoichiometric coefficient. But this could affect the equilibrium constant.
For example, we can write the chemical equation of ammonia in these two ways
$$\ce{N_2(g) + 3H_2(g)<=>2NH_3(g)}$$
and
$$\ce{\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)<=>NH_3(g)}$$
This affects the equilibrium constant. How do I decide which chemical equation's coefficients to use?
I think maybe the stoichiometric coefficients used in the equation should be in the simplest ratio of natural numbers. Is that so?
Equilibrium constant should not depend upon stoichiometric coefficients because they just tell about how many molecules are striking each other to form a certain product. It shouldn't matter what the coefficients are because they do not give information about the actual number of moles reacting. Then why does the equilibrium constant (if it does) depend upon them?
I think if you could tell me how they ended up in the equilibrium constant expression in the first place, it would make things easier to understand.

You can rephrase "Equilibrium constant should not depend upon stoichiometric coefficients because they just tell about how many molecules are striking each other to form a certain product. It shouldn't matter what the coefficients are because they do not give information about the actual number of moles reacting." putting "does" instead of "should" and that would be an answer. Only coefficients describing how exactly rds looks like have bearing on kinetics.

https://chemistry.stackexchange.com/questions/20140/where-does-this-equation-for-the-activity-in-the-gas-phase-equilibrium-between-n

or https://chemistry.stackexchange.com/questions/42953/when-is-it-permissible-to-use-the-stoichiometry-of-a-reaction-to-determine-rate

So how can one decide which set of stoichiometric coefficients to use. There can be infinite such sets like $(1,3,2), (2,6,4), (3,9,6), (\frac{1}{2},\frac{3}{2},1)....$ When asked about equilibrium constant for, say formation of ammonia, which set of coefficients would one choose?

In general afaik, if we're talking specifically about the formation of one product (here ammonia), we would choose the equation in which one mole of ammonia is formed as the product; this is similar to the case of when we talk about the heat of formation, we take the equation in which one mole is formed. Do note that there is no rigorous rule here, so it's always better if an explicit equation is given. If not, I would pick the one where one mole of product is formed, but this can obviously lead to a wrong answer if the question was designed with another equation in mind.

So basically, they always mention the equation for which the constant is asked.