text:
What is the relation between strength of a base and the thermal stability of its salts?
In Concise Inorganic Chemistry, $5^{th}$ edition, by JD Lee, while discussing oxides and hydroxides of Group $2$ metals, it has been stated that:

The increase in basic strength is illustrated by the temperatures at which the carbonates decompose.

What is the basis for this statement? How does the fact that carbonate thermal stability of Group $2$ carbonates increases down the group related to increased basic strength? If I am not mistaken, the explanation for increased stability of carbonates is related to increase in lattice energy due to high coordination number facilitated by larger cation. On the other hand, basic strength should be determined by the $\Delta H_r$ of the reaction:
$$\ce{M(OH)_2}(aq)\to \ce{M^{2+}}(aq)+\ce{2OH^{-}}(aq)$$
Clearly, a high lattice energy for base should decrease the magnitude of $\Delta H_r$, but it seems that this factor is compensated for by other factors. What is the explanation for this?

In ma opinion there is no correlation between the basic strength of hydroxides and the decomposition temperature of the carbonates. Or : if you find such a correlation, it would probably be the result of a coincidence.

Even I was surprised by this claim, but the author mentions this multiple times while discussing group $1$ and group $2$ salts. He even goes so far as to explain the higher stability of Group $1$ carbonates using higher basic strength of alkali bases.

Why compare carbonates and hydroxides ? Why not choose sulfates ? or nitrates ? I would be pleased to discuss this point with the author. Why not ? ?

Even so, does the statement: "stronger bases form stabler salts" have any basis?

Also, I'd like to add this: In the following section, on sulphates, the author says- " In the same way as with the stability and thermal decomposition of the carbonates, the more basic the metal is, the more stable the sulphate is. This is shown by the temperatures at which decomposition occurs".

There can be correlation between Arrhenius basicity of $\ce{M(OH)2}$ and Lewis acidity of $\ce{M^2+}$  which affects salt stability. More acidic cation will have bigger tendency to bond with the anion as the Lewis base and eventually steal $\ce{O^2-}$ to form oxide.

@Poutnik OK, but how does this answer my question?

Comments do not answer questions. If they did, they would be posted as answers.// Level of polarization of anions by cations has impact on anion thermal stability.

And how is defined (or measured) the basic strength of hydroxides ? Has it anything to do with solubility ?


text:
Why is an allyloxycarbonyl (Alloc) protection group more acid-stable than tert-butyloxycarbonyl (Boc) protecting group?
Why will Boc get cleaved under acidic conditions while Alloc does not? Can Alloc somehow be resonance stabilized? If so, how? Any help would be appreciated.

All else being equal, t-Bu carbocation is easier to form than allyl carbocation (see https://chemistry.stackexchange.com/a/87701/16683); so the easiest explanation may be "carbocation stability", although "stability" needs to be carefully defined here: let's say $\Delta E$ for the reaction $\ce{R2NCO2X + H+ -> R2NCO2H + X+}$, where X = allyl or t-Bu. (Of course, $\ce{R2NCO2H}$ will decompose to $\ce{R2NH + CO2}$, but the energetic contribution there is independent of X.)

Your original assertion that Alloc will not get cleaved by acidic conditions is incorrect. If you look at the reactivity charts in Greene' s "Protective Groups in Organic Synthesis" the reactivity of Alloc is very similar to tBoc. It is merely that acid condition are not routinely used in the literature for Alloc removal

@Waylander The "more acid-stable" was given as an explanation in an organic chemistry workbook which discussed the total synthesis of Minfiensine by Overman and colleagues (Overman 2005).  Quote directly from the Overman paper: "Because of the propensity of the aminal fragment to open under acidic conditions, the tert-butoxycarbonyl (Boc) protecting group had to be exchanged for an allyloxycarbonyl (Alloc) group prior to transforming the epoxide to an allylic alcohol." Indeed, they did not deprotect Alloc with acid later in the synthesis, but with Pd-catalyzed Tsuji–Trost reaction

It would be worthwhile to include this context, including a link to the paper, in your original question by [edit]ing it.

@orthocresol Will do when I have the time soon.

Did Overman state what conditions the Alloc survived that cleaved the tBoc? As @orthocresol states tBoc is more readily cleaved by acid because of the greater stability of the tBu cation. The allyl cation is also reasonably stable so will Alloc also cleave in acid conditions. With this difference in cation stabilities it may be possible with careful selections of conditions to differentiate between them, but I would not expect Alloc to be stable to std tBoc cleavage conditions such as TFA/DCM or HCl/EtOAc

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text:
Why does HCl act as a weak base in HF solvent, even though it is a stronger acid compared to HF?
I've read that $\ce{HCl}$ acts as a weak base in $\ce{HF}$ solvent, which means $\ce{HCl}$ is supposed to lose $\ce{H+}$ ion to $\ce{HF}$. $\ce{HF}$ is a weak acid compared to $\ce{HCl}$, then shouldn't $\ce{HF}$ be accepting the proton and HCl be losing it. Why doesn't this happen?

See https://en.wikipedia.org/wiki/Hydrofluoric_acid

related https://chemistry.stackexchange.com/questions/34818/is-hf-the-least-acidic-hydrogen-halide

As Poutnik hints in his reference, it is not accurate to call $\ce{HF}$ a weak acid in water. What appears in dilute aqueous solution as a weak acid is the hydrogen-bonded ion pair $\ce{H3O^+F^-}$ that forms when $\ce{HF}$ reacts with water. If this ion pair is placed in concentrated $\ce{HF}$, it will act as a strong base, forming $\ce{H3O^+}$ and $\ce{FHF^-}$ where the latter is the characteristics basic species in $\ce{HF}$. Thus $\ce{HCl}$ is a stronger acid than the aqueous $\ce{H3O^+F^-}$ solute in water, and $\ce{HCl}$ is also less basic than $\ce{H3O^+F^-}$ in $\ce{HF}$.
The Hammett acidity function of neat $\ce{HF}$, unsullied by water, is close to that of 100% $\ce{H2SO4}$. So, anything with appreciable acidity in $\ce{HF}$ would cross over into superacidity. $\ce{HCl}$ does not make it.

Note that the Hammer function of liquid HF is comparable to the value of H2SO4.

So is the Hammett function.  Autocorrect got you maybe?

:-) Definitely.  Google knows better than you what  want to say. :-D

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text:
What makes an electrolyte strong - solubility or dissociation?
I'm having some confusion related to the terms mentioned in the title. Suppose we have a strong electrolyte but with a very low solubility.
Therefore a given amount of it will not be able to produce lots of ions because less of it will go into solution and dissociate, so what do we call this type of electrolyte overall: a weak electrolyte as it didn't produce many ions or a strong electrolyte as whatever that did dissolve produced as many ions as it can dissociate into?

Not dissolved part does not count. If you mix 100 mL of water and 100 kg of NaCl, does it make it weak electrolyte,because majority is not dissolved ?

@Poutnik if you dissolve 100kg of NaCl in 100L of water then also it will be a strong electrolyte and solution will be fairly conducting but if you dissolve equimolar amount of electrolyte of type OP considers in question in same amount of water, it will not be a fairly conducting solution , In that case  how can you say that the considered electrolyte is strong?

Does this answer your question? How can an insoluble compound be a strong electrolyte

@Lllt  By evaluating molar conductivity, related to total solution concentration. Strong electrolytes have molar conductivity corresponding to ion conductivities at given concentration. Weal electrolytes have molar conductivity much lower. // The mistake is evaluation of conductivity instead of molar conductivity. By this way 0.000001 M NaCl would be weak electrolyte as well.

@Poutnik therefore we can conclude that the electrolyte of the type that OP considered in solution shall be a strong electrolyte as whatever amount of it is dissolved it will completely break into ions

@Lllt Yes, similarly as listed in the referred duplicate question.

See also: https://chemistry.stackexchange.com/questions/131070/is-there-any-difference-between-getting-dissolved-and-getting-dissociated?rq=1 and https://chemistry.stackexchange.com/questions/27863/can-a-strong-electrolyte-be-sparingly-soluble?rq=1&noredirect=1

@Poutnik do you have an example of substance which is strong electrolyte in one solvent and weak electrolyte in other solvent though it dissolves in both

@Lllt You should post it as the question.


text:
Why don't we consider the attached hydrogens in inductive effect?
Consider the molecule 1-chloroethane

Let carbon attached to $\ce{Cl}$ be $\ce{C1}$ and other one be $\ce{C2}$. While learning inductive effect I was told that $\ce{Cl}$ due to high electronegativity pulls the $\ce{C1–Cl}$ bond electrons towards itself leading to development of partial positive charge on $\ce{C1}$ and partial negative charge on $\ce{Cl}$ .Now this $\ce{C1}$ which has became a little electron-deficient tries to pull electron of the adjacent adjacent carbon $\ce{C2}$ leading to development of a further smaller partial  charge on $\ce{C2}$. I have following doubts:

Will $\ce{C1}$ only pull electrons from $\ce{C2}$ or from the hydrogen atom attached to it also. I think it does pull, but then why don't we take that pull into consideration?

When $\ce{C1}$ pulls electrons from $\ce{C2}$, the partial positive charge developed on it due to the pull of electrons by the chlorine atom will be reduced a little but why do we then show $\ce{C1}$ to have same partial positive charge?

Kindly see this and use mathjax again, Thanks! https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange

The effect exists but is very small and insignificant.

Is not the inductive effect taken as the relative effect of atom/group, compared to hydrogen ? By other words, a bound hydrogen is taken as the reference zero by definition.

@Poutnik do you mean the same as  first line in the given link https://en.wikipedia.org/wiki/Inductive_effect#Relative_inductive_effects

@Poutnik therefore you want to say partial charge developed on $C_2$ by $C_1$ will be relative to hydrogen?

@Lllt No, I want to say that inductive effect of the methyl group or the chlorine atom is compared to inductive effect of hydrogen, if it was there instead of them.

@Poutnik therefore a more electronegative atom than hydrogen will pull electron more towards itself will said to have -I effect and a less electronegative atom than hydrogen will not pull electrons more towards itself than Hydrogen and will said to be electron donating or having +I effect

@Lllt It is more general, as it applies to atomic groups as well. E.g. Even if carbon is more electronegative than hydrogen, a methyl in acetic acid pushes electron density toward the carboxyl carbon, compared to the hydrogen atom of formic acid.

@Poutnik Ok I understand that but is my above statement correct and complete if I incorporate atomic groups in them as well

I will show you how I understand this effect. I like to think that partial positive charge on $\ce{C1}$ attracts the electrons from $\ce{C2}$ because of the electrostatic force. With this model your questions can be answered this way:

Yes, $\ce{C1}$ will pull electrons from every adjacent atom, including hydrogen. It may pull them, however, to less extent from hydrogen because it has less electrons than carbon (or rather alkyl group), and smaller charge will result in weaker attraction according to our model.
What you are saying is right. The reason why we still write just $\delta +$ instead of some kind of $0.7\delta +$ is mostly convenience. This writing can be used in organic chemistry to show mechanisms, for example, and in this case we don't need super-accuracy in determining the magnitude of the charge itself (like it could be in some physical model), we just need to show that it is positive. It will be different in every case (molecule) and calculating the exact charge seems like waste of time, but we still are aware of the inductive effect, although don't explicitly show it.

thankyou for replying. I completely understood the answer of my second question. I'm having few doubts related to first one but ,why does small size of hydrogen cause Carbon to pull electrons to a less extent from it

Oh, sorry, I am formulating my sentences not in the best way. Actually, what I misinterpreted as two things is just one — carbon has more electrons. It is the same as the first one because larger atoms have more electrons in them (this is also how dispersion forces are explained). And what you have is that higher charge means greater attraction and electrons are pulled with greater force from carbon than from hydrogen.

@Azamat Hydrogen has less electrons than carbon, why will this fact cause carbon to pull electrons from it to a less extent

@Lllt, it's the electrostatic force. The law is $F = k \ \frac{q_1 q_2}{r^2}$. Let's say that $q_1$ is the charge of $\ce{C1}$ carbon, the force now depends on the charge $q_2$. If there are more electrons around, this charge is greater, therefore the force is also greater. There are more electrons at the carbon (because it has another hydrogens from which to take them), so the force is greater. However, I don't think that this gives a huge difference between the carbon and hydrogen, and teachers should actually note the effect on hydrogen too.


text:
How exactly does UV light break bonds?
For example, in the photochemical decomposition of hydrogen peroxide, the weak $\ce{O-O}$ bond is broken by UV light. Ultraviolet light can also dissociate relatively strong bonds such as the double oxygen bond ($\ce{O=O}$) in molecular oxygen ($\ce{O2}$) and the double $\ce{C=O}$ bond in carbon dioxide ($\ce{CO2}$). My question is, how does this process occur? How exactly does UV light break bonds?

UV breaks bonds in a fashion similar to heat.

@NisargBhavsar please elaborate. I know what dissociation energies are and that stability is inversely proportional to a species's energy, but that doesn't answer my question.

UV is just another way of providing energy to the system. It is used because sometimes it is inconvenient to heat stuff  or it is very slow to do so. Amount of energy provided per package can also be controlled in case of UV and thus prevents side reactions.

Briefly an electronically excited state is produced (distinct from the ground state) and if the uv energy is sufficient it is produced above its dissociation energy , which is less than in the ground state.  Alternatively in the excited state there is another excited state that the  energy from the first one crosses over into and this dissociates.

All starts with quantum interaction of electrons and photons. An UV photon passes enough energy to a bond electron to break a bond, frequently together with ionisation of fragments.

@NisargBhavsar not really. Photochemical and Thermal paths can be totally different leading to different reactions, and products. The only similarity is the energy input.

Perhaps is not what are you looking for but worth reading https://www.britannica.com/science/photochemical-reaction/Consequences-of-photoexcitation

Briefly an electronically excited state is produced (distinct from the ground state) and if the uv energy is sufficient it is produced above its dissociation energy, which is less than in the ground state. Alternatively in the excited state there is another excited state that the energy from the first one crosses over into and this dissociates. The figure tries to illustrate this.

Franck - Condon

@porphyrin What do you mean "If the UV energy is sufficient it is produced above its dissociation energy, which is less than in the ground state"? Shouldn't the photon match the energy difference of the states?

The molecule has a spectrum so experimentally you can choose the uv energy to match one of many possible energy levels so you can choose to dissociate or not.

In my opinion, a good way to understand it is the molecular orbital (MO) theory. In this theory we assume that atomic orbitals (AOs) of two atoms overlap and create MOs. The number of AOs interacting is the same as the number of MOs formed. This is the representation of MOs formation:

In this case, two 1s-orbitals of hydrogen atoms overlap to form two $\sigma$ MOs. One MO has lower energy than the starting AOs, therefore it's called bonding, the other MO has higher energy and is called antibonding. We represent antibonding orbitals with the asterisk (*).
Now, why is this molecule formed? In this diagram we can see that the two electrons will have a state that is lower in energy if AOs overlap and MOs form. That means, if the $\ce{H2}$ molecule is formed, the energy of the system decreases. That's why $\ce{H2}$ will form.
One key thing is that the energy difference $E(1 \sigma ^*) - E(1s) > E(1s) - E(1 \sigma)$. That means, the antibonding orbital is slightly more antibonding than the bonding orbital is bonding. That is why if there's equal number of electrons both in $1 \sigma ^*$ and $1 \sigma$, the molecule will not form, as the overall energy of the electrons will be higher than that of the starting 1s-orbitals.
Now we can return to your question. What happens when the UV light is radiated (or heat is applied)? Of course, we add the energy to the system. And when we add that energy, if $E(\text{UV light}) = E(1 \sigma ^*) - E(1 \sigma)$, then one of the electrons in $1 \sigma$-orbital can go from its current state to $1 \sigma ^*$-orbital.

After this happens, we will have one electron in $1 \sigma$ and one electron in $1 \sigma ^*$. What we've said before? We've said that the overall energy in this case increases and the formation of the molecule is now not favorable. In other words, $\ce{H-H}$ bond will break.
This is hydrogen molecule. Same thing applies to oxygen but it will have slightly more complicated scheme.

You need to deal with states not just orbitals and and how these states change in energy vs bond extension.

@porphyrin, sorry, what exactly do you mean?

What you say is that a single photon can bring the bond into a state where it has more energy than if there was no bond. Correct, but that does not explain why the molecule dissociates.

@Karl, I am confused now. Can you recommend any book or article that explains this?


text:
Do lyophilic sols have an electrical double layer?

"There are two factors which are responsible for the stability of lyophilic sols. These factors are the charge and solvation of the colloidal particles".

Quoted from NCERT book of OC
But we studied the electrical double layer only in the case of lyophobic colloids.
Question:
Do lyophilic colloids have an electrical double layer? If so, why?
Also since the ions should not even have a place to stay on the particles as they are already solvated by the dispersed medium molecules.
So where could the charge of lyophilic colloids come from?

Begin by answering the question "what is a lyophobic colloid?" and include some examples (I encourage you also to include this information in your post)


text:
How to implement SDF format?
I have a program that returns a file with SDF format molecule. Now I want to transfer it to SMILES with Open Babel.
But I don't get the result I expect.
I have visited different websites that show how the SDF must look like, which I think my file does.
Can one find a mistake?

CheMGen 1.0

  9  3  0  0  0               999 V2000
    0.0000    0.0000    0.0000 C  0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 C  0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 C  0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H  0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H  0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H  0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H  0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H  0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H  0  0  0  0  0  0  0  0  0  0  0  0
  1  3  1  0  0  0  0
  2  3  2  0  0  0  0
  1  4  1  0  0  0  0
  1  5  1  0  0  0  0
  1  6  1  0  0  0  0
  2  7  1  0  0  0  0
  2  8  1  0  0  0  0
  3  9  1  0  0  0  0
M  END
$$$$

This is what I get from Open Babel:
CC=C.[HH].[HH].[HH].[HH].[HH]
And this I expect: CC=C.
In Open Babel I have selected "Delete Hydrogens", but they are shown.

Sorry, for rolling back your edit, but what you did invalidated your own question (Q&A with corrected code is not possible) and so I decided to rollback. Thanks for understanding. Consider accepting Buttonwood's answer and upvoting since he did get the code corrected. You can accept by hitting the green check mark near the votes..

(Chemisty.se is a bit sloppy representing the first line of code if it is an empty one.  This equally applies to copy-paste of a.sdf file.)
There is a problem, because the full stop / period is used to concatenate multiple independent molecules into one SMILES string.  It may be useful e.g., to describe multiple molecules in a .sdf about a co-crystal; here, it does not seem suitable.  Possibly there is a problem generating the connectivty table (below the atom coodinates, regardless if these (still) are all zero) with the program (which one?) you use, how you use it, or a combination of the two.
I assume propene is the structure you target.  The freely accessible  sketcher of PubChem reports C(=C)C,  ChemDraw's JS sample page C=CC as a SMILES string about this molecule.  The later is what Openbabel converts the former as canonical SMILES when running
obabel -:'C(=C)C' -ocan

Provision of the canonical SMILES string generates a .sdf with all-zero coordinates by OpenBabel (obabel -:'C=CC' -h -osdf -O test.sdf) similar to the following (there is a leading empty line prior to OpenBabel, but no trailing empty line past $$$$):

 OpenBabel

  9  8  0  0  0  0  0  0  0  0999 V2000
    0.0000    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.0000    0.0000 H   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  2  0  0  0  0
  1  3  1  0  0  0  0
  1  4  1  0  0  0  0
  2  5  1  0  0  0  0
  2  6  1  0  0  0  0
  3  7  1  0  0  0  0
  3  8  1  0  0  0  0
  3  9  1  0  0  0  0
M  END
$$$$

If you diff both files, you recognize that the line specifying the format in further detail (V2000) as well as the connectivity table differ.

I corrected the number of atoms in my example. Thanks for the answer.
It was 3 in my file, I forgot the hydrogens.

"I forgot the hydrogens" ... hopefully, you don't create .sdf files with pen and paper, but with an algorithm within a program (which one?). MDL and now Biovia/Dassault document CT, e.g. here.


text:
Electrolytic cells optimum conditions
I am wondering why acidified copper sulfate is used instead of regular copper sulfate solutions in electrolytic cells? What benefit does the acidified solution give?

It prevents hydrolysis of copper(II) into copper hydroxide.

Your own reasoning, based on searching, reading and analysis, is supposed to be present to avoid the question closure for lack of own effort.

The solutions of $\ce{CuSO4}$ are slightly acidic. This shows that a small proportion of the ions $\ce{Cu^{2+}}$ do react with water in an equilibrium : $$\ce{Cu^{2+} + H2O <=> Cu(OH)_2  +  2 H^+}$$ If some acid is added to such a solution, the equilibrium is displaced to the left-hand-side by action of mass, and no precipitate of $\ce{Cu(OH)2}$ appears in the solution.


text:
Why does iron carbonyl have five carbonyl groups instead of six or four?
When finely divided iron combines with carbon monoxide, we get the complex $\ce{Fe(CO)_5}$.
Why the number 5? It doesn't even relate to the orbital theory wherein we can see the number of empty and filled orbitals.

Although not accurate, the concept of EAN yielding a noble gas configuration can be applied.

In organometallic chemistry there is something known as the 18-electron rule, which essentially suggests that stable complexes have an 'electron count' of 18. A general discussion of electron counting is beyond the scope of the current answer, but in the specific case of $\ce{[Fe(CO)5]}$, we have $\ce{Fe(0)}$ which accounts for 8 electrons, and five $\ce{CO}$ ligands which provide 2 electrons each, for a total of 18.
What is the basis behind this rule? I don't like the argument based on 'noble gas configuration': that makes little sense because the orbital structure is not the same as a plain atom.
A better argument is to draw out an MO diagram for the complex. The metal 3d, 4s, and 4p orbitals ($5 + 1 + 3 = 9$ orbitals in total) are involved in bonding with the ligands, and can form bonding + antibonding MOs. Generally there will be 9 bonding MOs which are low in energy, and these can be filled with 18 electrons.
For example, here's the standard MO diagram for an octahedral complex (from Wikimedia Commons). I'm aware that $\ce{[Fe(CO)5]}$ isn't octahedral, but the same principles apply; or you can imagine for a while that we're talking about $\ce{[Cr(CO)6]}$ or $\ce{[V(CO)6]-}$, which are octahedral and obey the 18-electron rule.
The 'low-energy' MOs are the bottom 6, plus the $\mathrm{t_{2g}}$ orbitals which are unfilled in this diagram. Generally the 18-electron rule holds true for strong-field ligands, especially π-acceptors like $\ce{CO}$. The reason for this is because these ligands push the $\mathrm{t_{2g}}$ orbitals down in energy (which is not illustrated in the diagram here, but can be found in any standard inorganic chemistry textbook).

"3d, 4s, and 4p orbitals (5+1+3=9 orbitals in total) are involved in bonding with the ligands", umm, I'd expect you to say this rule is dumb, because while that was an idea, it's thorough debunked.

@Mithoron I'm not sure what you're trying to say; the 4s and 4p orbitals are the a1g and t1u orbitals in the diagram above. Of course, the resulting MOs are mostly ligand-based; but it doesn't affect the conclusion that there are 9 bonding MOs (with π-acceptor ligands).

I mean this 18 electron rule smells very much of this dsp hybridisation nonsense. Wasn't even a 12 electron rule coined because of that?

I don't know about any 12e rule, although the Wikipedia page on the 18e rule seems to mention something about it. I never mentioned hybridisation in my post, so I think it's unfair to criticise me on this basis; if you want a $D_\mathrm{3h}$ trigonal bipyramidal analysis, you can look in pages 471–473 of Albright et al. "Orbital Interactions in Chemistry" 2nd ed. Exactly the same text that I wrote applies: the metal 3d, 4s and 4p orbitals overlap with ligands, giving rise to 9 relatively low-lying MOs, just that there isn't as nice a 1-to-1 correspondence as in the octahedral case.

To be clear, this is the sort of thing that we agree is outdated, etc. etc.; but I don't think I ever said anything remotely like that.


text:
Textbook on Food Additives?
My Question. Is there a comprehensive textbook on (synthetic) food additives?
Many textbooks on nutrition omits any discussion of food additives, so I was hoping there is a book devoted to the subject. In particular, I would like to learn about the following:

How the additives interacts with the human physiology.
(Lower priority:) how the food additives are synthesized.

I would say near any additive or small group of them would need an extra chapter, as they have near nothing in common but being added to food. And, it does not mean synthetic ones are the bad ones and natural ones the good ones. This criterium goes across both. Switching to poison topic, many of the most devious poisons are of natural origin.

@poutnik Is there a reasonable classification of additives then, other than natural vs synthetic?

Surely there is, their purpose, reflected by their numeric grouping.  See e.g. https://en.wikipedia.org/wiki/E_number

This may help to certain extend: Michael Ash and Irene Ash, In Handbook of Food Additives, Second Edition; Synapse Information Resources, Inc., 2002, 1079 pages (ISBN-13: 978-1890595364).


text:
How would I go about developing a linker molecule that binds SiO2?
I'm trying to develop a molecule that acts as a bridge between a protein and surface of SiO2 lattice. The end that binds the protein is simple and easy but I can't figure out a way to draft up how I can get the other end to bind SiO2.
So my question is, anyone have an idea on what particular molecule or reaction I can look at that targets Silicon molecules or has a high affinity for binding specifically SiO2? It doesn't even have to be a chemical bond that links the bridge to the SiO2, it could be an electrostatic/polar interaction too.
Thank you.

You should search silane chemistry. Molecules can be easily bound to silica using functionalized triethoxysilane.

It is not easy to bind an organic molecule to silicium oxide. It is necessary to synthesize the Si-O bond. It cannot be done starting from $\ce{SiO2}$. A possibility is to react $\ce{CH3Cl}$ over elementary silicium at high temperature to carry out the reaction :  $\ce{2 CH3Cl + Si > Si(CH3)2Cl2}$. Another approach is to make methyl sodium $\ce{NaCH3}$ react with gaseous $\ce{SiCl4}$ to produce : $\ce{SiCl4 + 3 NaCH3 -> Si(CH3)3Cl}$. These compounds can react with an alcool, like for example : $\ce{Si(CH3)3Cl + ROH -> Si(CH3)3OR + HCl}$. And here you obtain a molecule belonging to the class of the silicone. It contains the bond Si-O-C that you were looking for.

thank you friend, you knew exactly  what I was looking for!


text:
Why do polar solvents seem to be of lesser viscosity than non-polar ones?
For instance, in Russian there is an adjective "oily", which means a liquid of high viscosity. Is this assertion even true in general or statistically?

Water and benzene, as an example of polar and non-polar solvent, do not differ much in their visosities reported here.

Non-polar pentane is much less viscous than water. So the assertion is clearly not true in general. It might still be true as a very vague and imprecise rule of thumb, much like "all liquids have density 1".

Having long chain or ring molecules that get tangled up easily sounds like a good way to increase viscosity, compared with a small molecules like water. If we compare liquids whose molecules are more or less similar in size and shape we probably find that polar ones are more viscous.

@OscarLanzi but is there a single highly-viscous polar solvent?

When I first read this I thought you had gotten things the wrong way around. I usually associate higher viscosity with H-bonding liquids like water. Consider just pentane and hexane in eg https://en.wikipedia.org/wiki/List_of_viscosities. Non-volatile oils are not volatile because of their high b.p. which is due to their high MW, as a result of which they are also viscous.

The key is to compare comparable molecules. Consider n-pentane as non-polar solvent,and replace the terminal CH2 by O to get polar n-butanol. A huge increase of viscosity.

I think I got what you mean. But has to do more with "wetting power" and adhesion rather than viscosity. And you can find always a solvent that is more oily looking but polar and the opposite, and all aspects mixed.

Viscosity depends on the intermolecular forces between molecules. Typically, the more viscous a substance or liquid is, the stronger its intermolecular forces are. In polar solvents, there are dipole-dipole interactions while in non-polar solvents, there are London Dispersion Forces. Sometimes, when individual molecules are large, LDF can be stronger than dipole-dipole interactions in some polar solvents. The point is that having stronger intermolecular forces will lead to higher viscosity, and being polar doesn't mean it will have stronger IMF than non-polar substances.


text:
A physical meaning of the Danckwerts boundary condition in terms of concentration profile (adsorption)
Can someone please explain to me in a simplified way, what exactly does a well-recognized Danckwerts boundary condition do to the concentration profile of adsorption phenomena? I just cannot find a relevant source - everyone says they used this condition in their simulation, but never mention any explanation.
As far as I know, the definition of the Danckwerts boundary condition says that after the gas is introduced into the column, the convection transport mechanism is associated with the diffusion, which causes a concentration step change at the inlet. However, when I went through multiple peer-reviewed articles of adsorption modelling, I haven't noticed any effects of the concentration change at inlet on the plotted graphs like I would expect if the Danckwerts boundary condition was used, e.g.,

This pictures (taken from this article) shows a concentration profile in the gas phase along the adsorption column length (Danckwerts B.C. was assumed).
Here, on the next picture is a concentration profile at column inlet/outlet, which describes what one should expect (taken from this presentation). The Danckwerts condition is at the left side for the column inlet...

Why I cannot see something like this on the first picture (I wouldn't expect a different measuring scale of both curves to neglect the Danckwerts B.C. in the first picture)?
My point is also my current work - in my model, I have specified that the concentration at the column inlet is constant (given by the Dirichlet B.C.) such that the concentration at inlet equals the concentration of component in the feed gas (concentration before the gas enters column).
How much of a error do I get by assuming this?
Thank You!


text:
Which is better leaving group out of iodide and water?
In hydrolysis reaction of an tertiary alkyl halide , $\ce{H_2O}$ will replace the $\ce{I-}$ through $\ce{S_N1}$ mechanism and then $\ce{H+}$ will be removed from $\ce{H_2O}$ in order to form an alcohol.
But in the reaction of an alcohol with $\ce{HI}$ , the water leaves as leaving group and $\ce{I-}$ act as an nucleophile replacing $\ce{H2O}$.
So both reactions seems contradicting to me , even the solvent in both case is polar protic , So what actually favours the reaction in a particular direction?

It is probably the concentrations of the various reagents that is the major force in pushing the equilibrium in one direction or the other. For example, when you are hydrolysing something in the lab, you add a whole lot of water and a small amount of reagent. Whereas when you are reacting alcohol with HI, you would probably use a lot of HI with a small amount of the reactant (though I have never done this particular reaction in lab)

@ShoubhikRMaiti I also thought upon it  , but can't seem to get how the concentration of reagents will affect  whether Water will replace Iodide or vice-versa ?

The reaction is an equilibrium—when you add one reagent in excess, the reaction is pushed towards the opposite side (Le Chatelier). When water is in excess, the reaction goes towards the alcohol and free iodide and vice versa.

@ShoubhikRMaiti Thanks ! It seems right , though can you tell which side is favoured thermodynamically , I mean which direction has ΔG < 0 ?

That is hard to say without looking at experimental data.


text:
Doubt while using Le Chatelier's principle
Consider the following reaction
$\ce{H2}+\ce{I2}\rightleftharpoons\ce{2HI}$, for this reaction $\Delta H_r<0$
Now suppose I introduce more amount of $\ce{H2}$ in it, therefore temperature of system increases as total pressure increases. Now by Le Chatelier's Principle, since concentration of $\ce{H2}$ increases , therefore reaction moves in direction where it is consumed i.e. in forward direction as said my book but one thing that troubles me is that when we introduce more $\ce{H2}$ the temperature of system increases, therefore reaction should move backward contradicting the fact given in my book.
Where I'm getting confused?
I think that when my book says that equilibrium shifts forward as we introduce more $\ce{H2}$, it must also say that temperature of system remains constant.

For complex changes with more, mutually opposing factors, quantitative evaluation may be needed. But my guess is, the change of reaction quotient has much bigger influence than change of temperature and therefore of equilibrium constant ( see van't Hoff equation

I introduce more amount of $\ce{H2}$ in it, therefore temperature of system increases as total pressure increases. Why do you think that the temperature is increasing? I believe you are missing that $n$ is also changing.

@nisargbhavsar The OP may mean adiabatic compression by injection of hydrogen.

@nisargbhavsar do you mean to say that pressure increases due as number of  moles of $\ce{H2}$ increases keeping temperature constant

I think in problems with Le Chatelier's Principle we change only one thing and other factors are held constant. In this case, if you want to include more factors at a time, I think more information should be included in the problem. For example, it could be that the reaction is preceding inside a container with a piston, so it's possible to leave temperature constant by increasing the volume (moving the piston).

The principle says temperature increase would push reaction toward left side, while adding a reactant would push the reaction toward right side. It does not say which effect is stronger. You need quantitative computation for that.

You could introduce more gas isothermally.


text:
Rotation of partially double bonds in cinnamaldehyde

Is rotation about that bond free? I learned that sigma bonds rotate and I want to know if there are any exceptions for a bond like that.

Yes it rotates just fine. Sigma bonds that can't rotate (e.g. substituted biphenyls) are pretty unique.

@orthocresol There would probably be a higher barrier to rotation due to the conugation between the double bond and the benzene ring.

@ShoubhikRMaiti Yes, it's slower, exactly like in a conjugated diene; but it's still quite far away from the 'restricted rotation' category.

If the bond didn't rotate and there was an ortho substituent present, then there would be two possible ortho substituted cis isomers (phenyl in plane) or a single racemate (phenyl out of plane).

The molecule shown by the OP is cis (Z) cinnamic aldehyde - not the ordinary trans (E) cinnamaldehyde. Trans cinnamaldehyde melts at -7.5C, so is liquid at room temperature, and while the melting point of the cis wasn't available, it has been described as a yellowish liquid. The trans isomer is available from fine chemical suppliers at about 50 USD/kg, but the cis isomer is considerably more expensive: $750/100 mg from a Canadian company.
If rotation were hindered in cinnamaldehyde, there might be enough intramolecular attraction to raise the melting point. Let me shift the focus a little bit - to cinnamic acid.
Ordinary (trans) cinnamic acid has a melting point of $133$–$\pu{136 ^\circ C}$, so rotation of the sigma bond in the solid trans isomer is probably very hindered, but could perhaps rotate in the liquid or in solution. NMR spectra[1] don't suggest a hindered rotation.
The barrier to rotation in a simple molecule like ethane is about $\pu{2.8 kcal/mol}$[2], and between $2$–$\pu{3 kcal/mol}$ for most $\ce{C-C}$ bonds[3], vs about $15$–$\pu{20 kcal/mol}$ for amide ($\ce{N-C}$) rotation, and $\pu{37 kcal/mol}$ for allyl cation. Allyl cation has quite strong pi bonding over the three carbon system, pretty much the maximum possible for a "sigma" bond - although calling the bond a sigma bond is really a stretch.

The rotation, or more generally, the molecular mobility, in cis-cinnamic acid, seems to be significantly greater than in the trans form. The CRC Handbook gives melting points for three forms of cis cinnamic acid: $\pu{42^\circ C}$, $\pu{58^\circ C}$ and $\pu{68^\circ C}$. Apparently the purification method is the distinction[4], but this just suggests that $\pu{68^\circ C}$ is the best melting point. Still, in the solid, free rotation of the phenyl ring seems unlikely, but steric crowding of the phenyl and carboxyl make the system more unstable and more easily rotatable, whether by melting (at the lower temperature) or in solution.
Knowing exactly how long the bond is between the phenyl and vinyl groups would suggest how high the barrier to rotation is. Structural data for the cinnamic acids did not come up in a search, but a larger molecule (N-(3-nitrophenyl)cinnamamide) with a trans cinnamic group at one end had a $\ce{C^3-C^4}$(or $\ce{A^1}$) bond length (using their numbering system) of $\pu{1.466 Å}$, somewhat shorter than I would expect[5].

But it would seem that you can rotate the bond fairly easily in the liquid or in solution, but you can freeze it in the solid.
References:
[1]: PubChem. Cinnamic acid https://pubchem.ncbi.nlm.nih.gov/compound/Cinnamic-acid (accessed Jun 28, 2021).
[2]: Quijano-Quiñones, R. F.; Quesadas-Rojas, M.; Cuevas, G.; Mena-Rejón, G. J. The Rotational Barrier in Ethane: A Molecular Orbital Study. Molecules 2012, 17 (4), 4661–4671.
[3]: Conjugation and resonance in organic chemistry https://www.masterorganicchemistry.com/2017/01/24/conjugation-and-resonance/ (accessed Jun 28, 2021).
[4]: allocinnamic acid https://www.chemicalbook.com/ProductChemicalPropertiesCB71268435_EN.htm (accessed Jun 28, 2021).
[5]: Lee, J.-S.; Zeller, M.; Warkad, S. D.; Nimse, S. B. Synthesis, Characterization, and Crystal Structure of N-(3-Nitrophenyl)Cinnamamide. Crystals (Basel) 2019, 9 (11), 599.

The structure shown by the OP is not an acid at all but rather cinnamaldehyde.

@Curt: Thanks! What an oversight on my part! I will edit to correct.


text:
I stored CaCO3 in a PET container and it went through the plastic and made the container brittle. What is the reaction?
I stored CaCO3 in a PET container and it went through the plastic and made the container brittle. What is the reaction ?
It also accumulate on the outside of other PET containers and made them brittle.
Does someone know what happened ?
Thank you
/Paul

That's not a thing that CaCO3 would do. Think of other suspects.

What do you mean by "CaCO3 went through the plastic" ? Have you observed a white powder on the outer surface of the bottle ? Are you sure that the powder so collected is calcium carbonate ? How long have you maintained this substance in the container, before observing this transformation ? Which test have you done to state that the container is brittle ? What do you mean with "other containers" ? Were these "other containers" also filled with calcium carbonate ? Where were these containers stored ? In a cupboard ? In a  basement ? With other chemicals ?


text:
Why do poisoned catalysts (Lindlar, nickel borate) result in partial hydrogenation of alkynes?
I read that alkene is more reactive than alkyne, so in hydrogenation of alkynes, it's difficult to isolate the alkenes.
But with poisoned catalysts like Lindlar's catalyst or Nickel-Boron (Ni2B), they only catalyze the alkyne-to-alkene reaction, and won't catalyze the alkene-to-alkane reaction.
I'm curious about the reason. Why won't they catalyze the halogenation of alkenes? What is the mechanism?

They do; it's just much slower. If you leave a Lindlar on for too long it goes all the way to the alkane. Now as to why it's slower... I found a reference on this before... I'll try to dig it up.

I don't think the exact mechanism is very well understood; but here are a few references to check: Chem. Rev. 2013, 113 (3), 1313–1350 — J. Org. Chem. 1983, 48 (23), 4436–4438 — Theor. Chem. Acc. 2011, 128 (4-6), 663–673 — J. Mol. Catal. A 2001, 173 (1-2), 185–221. Sorry, I'm really not sure how helpful this is, but it's what I have.

I suspect the answer can be found in this book chapter from Comprehensive Organic Named Reactions but it is behind a paywall  https://onlinelibrary.wiley.com/doi/10.1002/9780470638859.conrr395  A key sentence from the abstract gives a clue "The addition of lead acetate has been suggested to block the most active catalyst sites and prevent overreduction of alkenes to alkanes."


text:
What is the average distance between electron 'shells' in an atom? Or between the innermost shell and the nucleus?
Somehow, I've never come across any mention of the actual, physical distances between, say the 1s and 2s shells in an atom, whether large or small....
I know that p, d and f (and g?) shells are oddly shaped, and that quantum mechanics tells us that there is only a probability of a particle being found in a particular location, etc, etc....
I am also curious as to the (average) distance between a uranium nucleus and its 1s electrons....

Since the 1s orbital is radially symmetric, the 'average' is r=0.

@JonCuster That would be the average position but not the average distance from the center.

The shells are overlapping; in essence they occupy the same space. Your "etc, etc..." can't be trivially blown off as something unimportant. It begs the question, why would you care to attempt to compute something like this, and, what is it exactly you would be computing? Take a swarm of flies and a swarm of bees centered on the same pot of honey, then compute the actual physical distance between them. Where do you start?

@NisargBhavsar - indeed, but perhaps they can find the homework answers on the internet now...

Actually, this is one of those times where the good old Bohr model isn't that bad for providing insight. The 1s orbital roughly scales with 1/Z, so that would be 1 for hydrogen and 1/92 for Uranium in units of the Bohr radius (53pm and 0.58pm) if we ignore relativistic effects.

Also, a related question here: https://chemistry.stackexchange.com/questions/136453/how-to-compare-size-of-subshells

I am also curious as to the (average) distance between a uranium nucleus and its 1s electrons

I am sorry to say but that is computationally too hard to accomplish. Just the sheer number of interactions will overcomplicate the equation. For such big atomic numbers, relavistic effects will also have to be taken into consideration.
But this calculation can be done for a hydrogen atom as the number of interactions are limited and the ignoring the relavistic effects doesn't affect the end result much.
Wavefunction of the hydrogen $\mathrm{1s}$ electron is[1]:
$$\psi=\frac{1}{\sqrt{π}}\left(\frac{1}{a_0}\right)^{\frac{3}{2}}e^{-\frac{r}{a_0}}$$
Now the probability of finding an electron at an distance $r$ from the nucleus is $\psi^2\left(4πr^2\mathrm dr\right)$. This is because $\psi^2$ is the probability density and $4πr^2\mathrm dr$ is the volume in consideration.
$$
\begin{align}
\psi^2&=\frac{1}{πa_0^3}e^{-\frac{2r}{a_0}}\\
\psi^2\left(4πr^2\mathrm dr\right)&=4\frac{r^2}{a_0^3}e^{-\frac{2r}{a_0}}\mathrm dr\\
\mathrm dP(r)&=4\frac{r^2}{a_0^3}e^{-\frac{2r}{a_0}}\mathrm dr\\
\end{align}
$$
Now mathematically,
$$
\begin{align}
\bar r&= \int_0^\infty r\mathrm dP(r)\\
\bar r&= \int_0^\infty 4\frac{r^3}{{a_0}^3}e^{-\frac{2r}{a_0}}\mathrm dr\\
\end{align}
$$
Now using integral by parts,
$$
\begin{align}
\bar r&= \left[-2\frac{r^3}{a_0^2}e^{-\frac{2r}{a_0}} -3\frac{r^2}{a_0}e^{-\frac{2r}{a_0}} - 3re^{-\frac{2r}{a_0}} - \frac{3}{2}a_0e^{-\frac{2r}{a_0}}\right]_0^\infty\\
\bar r&= \left(0 + 0 + 0 + 0\right) - \left(0 + 0 + 0 - \frac{3}{2}a_0\right) \\
\bar r&= \frac{3}{2}a_0≈\pu{79.35 pm}\\
\end{align}
$$
Therefore an electron in $\mathrm{1s}$ orbital of an hydrogen atom is at an average distance of $\pu{79.35 pm}$ from the nucleus.
As you would have noticed, things got complicated even when we started with the wavefunction already ready. Finding the wavefunction for atoms like uranium is very tedious as well.
Still there has been a research[2] that you might be interested in, which has approximated the wavefunctions for the $\mathrm{6d^6}$, $\mathrm{5f^26d^27s^2}$, $\mathrm{5f^46d^2}$ and $\mathrm{5f^6}$ configurations of metallic uranium.

References:
(1) Libretexts. 11.10: The Schrödinger wave equation for the hydrogen atom
https://chem.libretexts.org/@go/page/41377 (accessed Jun 29, 2021).
(2) Ridley, E. C. A Qualitative Investigation of the Wave Functions of Metallic Uranium. Proc. R. Soc. Lond. 1958, 247 (1249), 199–213.


text:
Can a Grignard reagent form magnesium halide with an alkyl halide?
$$\ce{RMgX + R'X' -> RR' + MgXX'}$$
Is this reaction possible?
I am not able find any reference to it in the books that I have. Anyone has a reference?

Kindly go through this! Thanks https://chemistry.meta.stackexchange.com/q/86/106377 and this as well, https://chemistry.stackexchange.com/help/how-to-ask

The reaction can also be done if catalyst is applied:

[...] we developed a unique cross-coupling reaction of alkyl halides with organomagnesium or organozinc reagents catalyzed by using a 1,3-butadiene as the additive. This reaction follows a new catalytic pathway: the Ni or Pd catalyst reacts first with R−MgX to form an anionic complex, which then reacts with alkyl halides. Bis-dienes were also effective additives for the Ni-catalyzed cross-coupling reaction of organozinc reagents with alkyl halides. This catalytic system tolerates a wide variety of functional groups, including nitriles, ketones, amides, and esters.

Source: Cross-Coupling Reaction of Alkyl Halides with Grignard Reagents Catalyzed by Ni, Pd, or Cu Complexes with π-Carbon Ligand(s), Jun Terao and Nobuaki Kambe
Accounts of Chemical Research 2008 41 (11), 1545-1554
DOI: 10.1021/ar800138a

This reaction does not work for Grignards due to competing deprotonation and reduction reactions, however it is possible if you first modify the Grignard with copper to make the cuprate (Gilman reagent). There is a detailed explanation here

Extension: What are the pros and cons of using Gilman reagent over Grignard reagent to synthesize a hydrocarbon?


text:
Alkyl halide nomenclature, flouthane
The IUPAC name for flouthane is 2-bromo-2-chloro-1,1,1-triflouroethane. i am confused why did not we assign the numbering to the carbon chain the other way possible that would make the name as 1-bromo-1-chloro-2,2,2-triflouroethane? Why would this answer be wrong?

From Section P 14-3.5 of the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, IUPAC Blue book, we have,

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is
lower than ‘3,4,6,8’ and ‘2,4,5,7’.

Because of this rule, comparing the two options of locants we have,

1-bromo-1-chloro-2,2,2-trifluoroethane [1,1,2,2,2]
2-bromo-2-chloro-1,1,1-trifluoroethane [1,1,1,2,2]

The first point of difference between these two sets of locants is the third locant value. Now, we take the lower value of the two. Therefore, option 2 is the preferred IUPAC name.

Therefore the PIN of the compound becomes 2-bromo-2-chloro-1,1,1-trifluoroethane.

Hi, I have a few questions about BITS P CSE. I gave jee this yr and am trying to decide between IIITH CSE (4yr) and BITS P CSE. Would it be possible for you to spare some time to answer some questions about bits? Stackexchange chat/discord/etc. anything is fine. (I actually sent you a friend request on discord yesterday (the Infern.. account), Aniruddha Deb gave me your discord id. I'm double_fault#3535.) If you are too busy then no worries I will find some other student to trouble :D I will delete this comment after a day or two anyway.


text:
In solvent extraction from plant cells, what is the static organic solvent film formed?
I am researching the mechanisms behind solvent extraction of oil from microalgal cells, and it is stated in one paper that in the process of transferring the lipid out of the cell, the solvent-lipid complex must diffuse through a 'static organic solvent film', which is supposedly formed from 'the interaction between organic solvent and cell wall' and 'surrounds the microalgal cell and remains undisturbed by any solvent flow or agitation'.
Attached is the diagram from the paper which shows how the solvent-lipid complex moves out of the cell, the dashed perimeter is the static film.

Here's the link to the paper if it helps: Halim, R.; Danquah, M. K.; Webley, P. A. Extraction of oil from microalgae for biodiesel production: A review. Biotechnology Advances 2012, 30 (3), 709–732. DOI: 10.1016/j.biotechadv.2012.01.001.
I am looking for an explanation with more details on why this occurs or any sources of information. I have had a search online, but there seems to be surprisingly little on the underlying processes of solvent extraction (for biodiesel production) despite its popularity.


text:
Why is 4-hydroxyacetophenone more acidic than 3-hydroxyacetophenone?
At 25 °C, the $\mathrm{p}K_\mathrm{a}$ of 3-hydroxyacetophenone is 9.19, while the $\mathrm{p}K_\mathrm{a}$ of 4-hydroxyacetophenone at the same temperature is 8.05. Why is that so?

3-hydroxyacetophenone

4-hydroxyacetophenone

Have you heard of mesomeric effect?

Please note that names of chemicals should not be capitalised, unless at the beginning of a sentence. Additionally, there is no space between substituents ('hydroxy') and the parent compound ('acetophenone').


text:
Is there any specific conditions for Neighbouring group participation?
I got to know about what neighboring group participation is, but does it always occur if there are lone pair donating groups in beta or delta positions. Or does it only occur in some kind of medium or is there any specific condition that is required?
For example, I have attached a picture below. I was solving some problems on this topic and came across this one.

The product I expected an epoxide(diazotization followed by attack of lone pair of oxygen) ,

But the answer given was a Ring contraction followed by formation of an aldehydic group,

So, to be clear, Why didn't Neighboring group participation take place, or in this case why an epoxide wasn't formed?
The mechanism I predicted:

Can you draw out the mechanism for the epoxide formation?

@orthocresol I have added the mechanism I predicted that would take place. Can you please help me out?

The Tiffeneau rearrangement has been addressed here: https://chemistry.stackexchange.com/questions/116689/tiffeneau-demjanov-rearrangement-products/116706#116706 Remember that the t-butyl group anchors the chair conformation.

As to the epoxide ring contraction, this may help: https://chemistry.stackexchange.com/questions/108388/ring-contraction-when-cyclohexene-oxide-is-treated-with-methyl-grignard/108402#108402

Good work with the mechanism. Now, you've drawn what's basically an SN2 reaction. You remember that the nucleophile must approach from a certain angle in an SN2 reaction...?

@orthocresol Yea I have that the nucleophile must attack from behind i.e is from the rear side, I thought here the OH group is below the plane and the NH2 group above the plane (if I am not wrong in reading the conformations) so I thought SN2 would take place.

@user55119 Thank you and I did refer to both the posts, so does it mean that there is no way of formation of an epoxide rather ring contraction followed by elimination would take place as this one is a trans form??

Certainy not in your case because of the stereochemistry. Rearrangement products are the usual outcome.

You are right in saying that it must approach from the rear. Only that in your case, the $\ce{OH}$ group is not rear to $\ce{NH2}$ - both of them are faced to the right side. Since $\ce{NH2}$ is to pointing left, the attacking/migrating group from adjacent carbon must point to the right. The right pointing group is the alkyl part, which migrates causing ring contraction. It's always the opposite-side-pointing that matters - not up/down or cis/trans for the cyclohexane substituents.

@TRC Oh.. that was the mistake I made, Thanks for pointing it out.

In the given compound the $\ce{-OH}$ group and the $\ce{-NH2}$ and later $\ce{N2^+}$ group are not in anti position with respect to each other. Equatorial positions in cyclohexane are not anti with respect to each other.
Below I have drawn newmann projection of the reactant which should help you visualise that a carbon (encircled) is instead in anti-relation to the leaving group, instead of $\ce{-OH}$. The cyclohexane cannot flip here either since $\ce{t-Bu}$ has locked the conformation would cause heavy steric repulsion in axial location making the ring unstable if it were to flip.

To show neighbouring group participation it is important that the lone pair donor be anti with respect to the leaving group. Hence here this is not availiable. Instead the anti-carbon attacks (undergoing ring contraction) because the resulting carbocation is stabilized by resonance with $\ce{-OH}$ as shown by this mechanism
Footnote:- The reaction does not involve hydride transfer to give ($\ce{t-Bu}$) substituted cyclohexanone as a major product because here $S_n2$ mechanism is faster than $S_n1$ (for most cases of secondary carbon in polar solvents and we can't have direct attack of hydrogen on $\ce{N2+}$ because it is not anti to leaving group) and there is not a great deal of difference in stability of 5 and 6 membered rings. However it still does occur to some extent.


text:
Hydrogen as a reference in inductive effect
What does it mean when we say that I-effect of hydrogen is taken as zero ?
I think that it means that partial displacement of sigma bond electrons between two atoms of different electronegativity is measured with respect to hydrogen i.e.
Relative partial displacement of $\sigma$ bond electrons in polar bond =
actual partial displacement of those electrons- displacement of electrons that would happen if hydrogen was a part of bond
Is my reasoning wrong? If so, why?

See: https://en.wikipedia.org/wiki/Inductive_effect It is relative inductive effect where we consider hydrogen as reference.

@0.00 in that case only what does it means when we say that hydrogen is reference?


text:
Is a change in delta-delta-G typically notated as delta-delta-delta-G?
I'm currently writing a research paper, and need to refer to a difference between $\Delta{\Delta}G$s. The naive way to notate this would be $\Delta{\Delta}{\Delta}G$, but that looks cumbersome and silly to me for some reason, and I can't recall ever seeing that written down anywhere before. Is that a reasonable/conventional way to notate this? Or is a difference between $\Delta{\Delta}G$s still considered a $\Delta{\Delta}G$ (albeit with some different subscript or something)?

In crystallography having many stacking operators (e.g. ΔΔΔΔΔΔ∇∇Δ) in a row is a norm.

It probably isn't too necessary unless you go past three, but you could introduce a notation like $\Delta^{(3)}G=\Delta \Delta \Delta G$

What physical meaning do you associate with the change in delta-delta-G?

I shouldn't have used the word change - it's a difference. I'm trying to compare two delta-delta-Gs associated with the same process in two similar systems. I'm trying to assess how similar the behavior is, and this is one of my metrics.

Maybe you could write like $\partial (\Delta G)$


text:
What is the isomeric relation between bicyclo and spiro hydrocarbons?
Take for example octahydropentalene (bicyclo[3,3,0]octane) and spiro(3.4)octane:

I would think that they are chain isomers of sort but according to nomenclature counting rules, they are both 8-carbon main-chains.
However they could also be considered ring-chain isomers but I thought that is only classified satisfactorily by a main chain being closed and open.
I wouldn't think they are functional isomers although it may be true. Since alkadienes and alkynes (this is also a bit grey to me since I couldn't find a good source for it) are considered functional isomers.
Also I would appreciate any primary source that determines the type of isomers two compounds are called and the rules which determine their relation.

It might be useful to identify the type of reaction that turns one into the other: ring expansion and contraction, e..g. https://en.wikipedia.org/wiki/Spiro_compound#Specific_methods

I don't think there is a "primary source" for types of isomerism.  The types of isomerism are human-defined categories that different authors find conceptually or pedagogically useful in different ways. It isn't like isomerism classification emerges from some unsupervised analysis of all possible structures.
If it were my opinion, I'd follow the lead of the Master Organic Chemistry web site and call these "constitutional isomers".  They have different connectivities: the left structure has two tertiary and no quaternary carbons, while the right structure has no tertiary and a single quaternary carbon.

So I take that you mean there is no official further classification of structural isomers( another way to call constituent isomers)?

Also according to wikipedia functional isomers is a classification. Would you consider these two functional groups? (Ex. Alkynes and Alkadienes are classified as functional isomers but I couldn't find a good source for it)

If you go by the Gold Book, then there is plain old isomer: One of several species (or molecular entities ) that have the same atomic composition (molecular formula) but different line formulae or different stereochemical formulae and hence different physical and/or chemical properties. || And then there are only a hand full of special definitions, i.e. which are widely adapted, e.g. constitutional i., stereoi., tautomers. These are well-defined, but something like functional i. is very dependent on the author. @napstablook


text:
Finding the molar ratio at equilibrium
Consider the reaction $\ce{SO_2_{(g)} + \frac{1}{2} O_2_{(g)}\longrightarrow SO_3_{(g)}}$.  What effect is there on the molar ratio $\frac{n_{SO_3}}{n_{SO_2}}$ at equilibrium if the pressure is increased by reducing the initial volume by half?
Here's what I did: I wrote $\ce{SO_2_{(g)} + \frac{1}{2} O_2_{(g)} \longrightarrow SO_3_{(g)}}$
Before the reaction starts, we have $1$ mole of $\ce{SO_2}$, $\frac{1}{2}$ moles of $\ce{O_2}$ and $0$ moles of $\ce{SO_3}$
During the reaction I think we have $x$ moles of $\ce{SO_2}$, $\frac{x}{2}$ moles of $\ce{O_2}$ and $x$ moles of $\ce{SO_3}$.
At equilibrium I think we have $1-x$ moles of $\ce{SO_2}$, $\frac{1-x}{2}$ moles of $\ce{O_2}$ and $x$ moles of $\ce{SO_3}$.
Now, I am not sure if I put those moles correctly. And I also don't really know how to proceed from here, so I guess that I am stuck. I found this similar question Chemical Equilibrium - Le Chatelier's Principle, Change in Volume, but I don't really know how to obtain the molar ratio from there.

Here's a hint:  How is the equilibrium constant expressed in terms of the partial pressures?

Some information is missing. $1)$ You state that at the beginning there was no $\ce{SO3}$. This is an arbitrary decision.  $2)$ You arbitrarily decide that the mixture of $\ce{SO2}$ and $\ce{O2}$ is stoichiometric. Why ?

@Maurice 1) I stated that because I wrote those moles before the reaction took place (so this is why I would have no $SO_3$. 2) I don't know, I thought that this was the only way that I may be able to solve this

Let:
A = $\ce{SO2}$
B = $\ce{O2}$
C = $\ce{SO3}$
The only information given is the change in moles for the reaction and the volume decrease factor:
$$\Delta n=c-a-b=1-1-0.5=-0.5$$
$$\frac{V_2}{V_1}=0.5$$
For simplicity, the molar ratios of interest before and after compression are:
$$\Theta_1=\left(\frac{N_{\ce{SO3}}}{N_{\ce{SO2}}}\right)_1=\frac{N_{C_1}}{N_{A_1}}$$
$$\Theta_2=\left(\frac{N_{\ce{SO3}}}{N_{\ce{SO2}}}\right)_2=\frac{N_{C_2}}{N_{A_2}}$$
The initial (before compression) equilibrium system in terms of moles is given by:
$$K_{N_1}=\frac{N_{C_1}}{N_{A_1}N^{0.5}_{B_1}}=K_C\;V_1^{\Delta n}$$
The final (after compression) equilibrium system in terms of moles is given by:
$$K_{N_2}=\frac{N_{C_2}}{N_{A_2}N^{0.5}_{B_2}}=K_C\;V_2^{\Delta n}$$
Dividing the final state by the initial state:
$$\frac{K_{N_2}}{K_{N_1}}=\frac{\frac{N_{C_2}}{N_{A_2}\;N_{B_2}^{0.5}}}{\frac{N_{C_1}}{N_{A1}\;N_{B_1}^{0.5}}}=\frac{\Theta_2}{\Theta_1}\left(\frac{N_{B_1}}{N_{B_2}}\right)^{0.5}=\frac{\require {cancel} \cancel{K_C}}{\cancel{K_C}}\left(\frac{V_2}{V_1}\right)^{\Delta n}=0.5^{-0.5}=2^{0.5}$$
In other words:
$$\frac{\Theta_2}{\Theta_1}=\left(2\frac{N_{B_2}}{N_{B_1}}\right)^{0.5}$$
It's not possible to proceed without knowing the ratio of moles of B at both equilibrium states:
$$\frac{N_{B_2}}{N_{B_1}}=\;?$$
But we at least know that due to Le Châtelier's Principle, an increase in pressure with $\Delta n <1$ leads to the consumption of some A and some B in order to produce more C, which means necessarily:
$$0<N_{B_2}<N_{B_1}\implies 0<\frac{N_{B_2}}{N_{B_1}}<1\implies0<\frac{\Theta_2}{\Theta_1}<\sqrt{2}$$
If we evaluate the special case in which the ratio of B is 0.5, then:
$$\frac{N_{B_2}}{N_{B_1}}=0.5\implies \frac{\Theta_2}{\Theta_1}=1\implies no\;change$$
Which means we now know the conditions under which the ratio of interest either increases or decreases as a function of the ratio of B:
$$0.5<\frac{N_{B_2}}{N_{B_1}}<1\implies 1<\frac{\Theta_2}{\Theta_1}<\sqrt{2}\implies increase$$
$$0<\frac{N_{B_2}}{N_{B_1}}<0.5\implies 0<\frac{\Theta_2}{\Theta_1}<1\implies decrease$$

In the beginning, the equilibrium partial pressures of $\ce{SO2, O2, SO3}$ are respectively called $\ce{p_1, p2, p3}$, and the total pressure $\pu{P}$ is given by $\ce{P_o = p_1 + p_2 + p_3}$. If suddenly she total pressure is multiplied by $2$ it means that some amount $2x$ of $\ce{SO3}$ has been produced. Then the final pressures of $\ce{SO2, O2, SO3}$ are respectively $\ce{p_1 - 2x, p2 - x , p3 + 2x}$. The final pressure is : $\ce{P_f = p_1 - 2x + p_2 - x + p_3 + 2x =  P_o - x = P_o/2}$. So $\ce{x = P_o/2 = (p_1 + p_2 + p_3)/2}$. The final pressures are :
$\ce{p_f(SO2) = p_1 - 2 x = p_1 - 2(p_1 + p_2 + p_3)/2 = p2 + p3}$
$\ce{p_f(O2) = p_2 - x = p_2 - (p_1 + p_2 + p_3)/2 = -p1/2 + p2/2 + p3/2}$
$\ce{p_f(SO3) = p_3 + 2 x = p_3 + 2(p_1 + p_2 + p_3)/2 = p1 + p2 + 1.5 p3}$

if pressure is doubled by reducing volume to half wouldn't the partial pressure of all components be doubled?

Initially yes but then they will change until they reach equilibrium.

So it should be 2p(1)-2x


text:
What is the pI of lysine tripeptide?

What the the $\mathrm{pI}$ of a tripeptide of lysine?
$$
\begin{array}{l|ccc}
\hline
\text{Group} & \ce{Lys{-}α{-}NH3+} & \ce{Lys{-}ε{-}NH3+} & \ce{Lys{-}α{-}COOH} \\
\hline
\mathrm{p}K_\mathrm{a} & 8.0 & 10.5 & 2.2 \\
\hline
\end{array}
$$

I have written out the tripeptide, and I know when the hydrogens would come off of it depending on the $\mathrm{pH}.$ I know it has to do with the $\ce{α{-}NH3+}$ and the $\ce{ε{-}NH3+}$:
$$
\begin{align}
\mathrm{pI} &= \frac{\mathrm{p}K_\mathrm{a}(\ce{Lys{-}α{-}NH3+}) + \mathrm{p}K_\mathrm{a}(\ce{Lys{-}ε{-}NH3+})}{2} \\
  &= \frac{8.0 + 10.5}{2} \\
  &= 9.25,
\end{align}
$$
but it was incorrect. Can anyone please help and explain the process to me?
Swiss-Prot — Compute pI/Mw calculator gives $\mathrm{pI} = 10.30$ for KKK sequence in single letter code.

The tripeptide has 1+3 NH2 and 1 COOH. Calculate the average partial charge for each of them as function of [H+] and of respective acidity constant. And then calculate [H+] resp. pH when their charges are balanced with zero net charge.

Another thing is acidity constants for alpha NH2 and COOH will be different, compared to a single lysine, as the peptidic bond has different induction effect than amino or carboxy group. But the task may ignore this.

I think there is not enough data to provide an exact answer theoretically.
You can figure out an estimate value by virtually titrating the tripeptide and figuring out after what step an isolectric point likely situates:
$$
\def\VEQ{{\scriptsize-\ce{H+}}{\Large\downharpoonleft\!\!\upharpoonright}\scriptsize{+\ce{H+}}}
\begin{array}{rcc}
\mathrm{pH} < 2.2 &\quad &\ce{H3\overset{+}{N}-\overset{+}{Lys}-\overset{+}{Lys}-\overset{+}{Lys}-COOH} \\
  & & \VEQ \\
2.2 < \mathrm{pH} < 8.0 &\quad &\ce{H3\overset{+}{N}-\overset{+}{Lys}-\overset{+}{Lys}-\overset{+}{Lys}-COO-} \\
  & & \VEQ \\
8.0 < \mathrm{pH} < 10.5 &\quad &\ce{H2N-\overset{+}{Lys}-\overset{+}{Lys}-\overset{+}{Lys}-COO-}\\
  & & \VEQ \\
10.5 < \mathrm{pH} = \mathrm{pI} &\quad &\ce{H2N-\overset{+}{[Lys-Lys-Lys]}-COO-}\\
\end{array}
$$
As you can see, the formal net zero charge doesn't occur between any given $\mathrm{p}K_\mathrm{a}$s.
Rather, at the highest $\mathrm{p}K_\mathrm{a}(\ce{Lys{-}ε{-}NH3+}) = 10.5$ the tripeptide is still going to have $ 0.5 + 0.5 + 0.5 -1 = +0.5$ formal charge considering $50\,\%$ deprotonation of residue amine groups at $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$ and $-1$ charge at the $\ce{-COO^-}$ group.
This implies that $\mathrm{pI}$ of trilysine must be slightly higher, about $10.8:$

For polylysines $\mathrm{pI}$ should increase with the chain length.
ThermoFisher's Peptide Analyzing Tool suggests theoretical $\mathrm{pI}$ values of $11.0,$ $11.2$ and $11.3$ for tetra-, penta- and heptalysine, respectively.
This tool, however, refuses to calculate isoelectric points for tripeptides (Input sequence is too short).
This can be an indication of insufficient quality of applied algorithm resulting in apparently wrong data suggested by Swiss-Prot in your question.
Note that we didn't corrected $\mathrm{p}K_\mathrm{a}$ values for the tripeptide.
I wouldn't assume they change drastically, but it's not the identical values either.


text:
How are cyclodextrins used to form inclusion complexes with drug molecules?
I'm talking about practical application. A cyclodextrin molecule hosts a drug molecule inside of it in order to increase solubility and stability of the drug.
But how do we make sure this inclusion complex forms correctly? Is it as simple as mixing the drug along with a cyclodextrin in a solvent? Is it common for drug makers to include excess cyclodextrin (more than the 1:1 ratio needed) in order to ensure that all molecules of the drug are encapsulated in inclusion complexes?

There are books and plenty of reviews on cyclodextrins and their drug applications. Go to " Google Scholar" https://scholar.google.com/, and search this keywords.

Sounds like you read the title but not the questions.  Of course I've read papers on these things, but you'd be surprised at how many brush over the practical application.


text:
Why is [M(AB)2] a geometrical isomer?
Our teacher told us that for coordination number 4 and tetrahedral shape $\ce{[M(AB)2]}$ doesn't have a plane of symmetry, so it shows optical isomerism.
What if we take a plane passing through the two $\ce{A}$'s so the $\ce{B's}$ will be the mirror image of each other? Am I wrong?

AB is implied to mean a bidentate ligand with a bridge between A and B. See where the bridge will go in the mirror.

Do we see the bonds as well while considering the mirror image?

In all likeness, the bridges consist of quite a few atoms, not just bonds.

Yes, you are right, I understood.

Just to record the question as answered:

If you had separate A and B ligands with no ligand-ligand bonds, then the complex would have a plane (actually two planes) of symmetry and would not be chiral.

But when you have a bridge between each A ligand and a separate B ligand, the bridges do not conform with the above plane(s) of symmetry and so those symmetry planes are lost. Thereby the complex becomes chiral.


text:
How do I expand a selection in Jmol to include entire residues?
I know how to select by distance in Jmol:

select within(5, ATP)

This selects atoms that are within 5 Angstroms of ATP residues. How do I expand the selection to include the entire residues of the selected atoms?
I checked the Jmol documentation and the Jmol wiki, searching for "expanding selection" but did not find a solution.

To expand the selection, use

select within(group, selected)

This will select all groups (i.e. residues, nucleotides, etc) that have at least one atom selected already.
Examples can be found at https://chemapps.stolaf.edu/jmol/docs/examples-12/new2.htm after searching for "within(group".


text:
Phase equilibrium & phase diagram
Q) Why does phase equilibrium/inter-conversion  ONLY  occur at boundary lines between phases in phase diagrams?
Example: According to water phase diagram, at 200 C and 200 atm  ONLY  liquid water exists. So equilibrium: H2O (l) ⇌ H2O (g) doesn't exist at this point? Why?

Any clarification will be appreciated...

How would it exist in the first place? Imagine a bubble of vapor. The outer pressure is 200 atm. The inner pressure is whatever the diagram gives for that temperature, which is much lower. What would happen to the bubble?

This chart is about boiling not simple vaporising.

Water is so ubiquitous that we can hardly imagine it except as a liquid with at least a little bit of vapor, and the solid is not far from our thought. We are familiar with all three states of matter for $\ce{H2O}$: solid, liquid and gaseous. But the phase diagram contains a lot more information.
From Wikipedia[1]:

In physics, a state of matter is one of the distinct forms in which matter can exist. Four states of matter are observable in everyday life: solid, liquid, gas, and plasma.

We don't often confront plasma, but you could imagine it on the surface of the sun.
A phase diagram is a very compressed data presentation, compared to our usual experience.
From Wikipedia[2]:

In the physical sciences, a phase is a region of space (a thermodynamic system), throughout which all physical properties of a material are essentially uniform.

The area of the graph with which we ordinarily interact is a short ($\text ~\pu{-20°C}$ to $\text ~\pu{+200°C}$), thin ($\text ~\pu{0}$ to $\text ~\pu{3 atm}$) line that you can hardly see, and it is in this area that two phases can exist because equilibrium has not been completely established throughout the entire system.
The purpose of a phase diagram is to show what conditions make it possible for a given state of matter to exist or not exist. There are conditions where only one state of matter is stable. And there are exceptions. For instance, consider supercooled water at $\pu{-10°C}$. It's liquid, not in the right state, according to the phase diagram - but also not at equilibrium. When equilibrium occurs, it will conform to the phase diagram.
Consider a different material: zinc. Zinc is a solid; we essentially ignore its liquid state unless we heat it to temperatures greater than $\pu{420°C}$. We especially ignore its vapor pressure; however, at temperatures greater than $\pu{980°C}$, it can be and sometimes is, distilled as a manufacturing process.
A phase diagram for zinc on the internet doesn't even show the vapor. What would you get out of this simple phase diagram? A melting point and its variation with imposed pressure. The line is the condition of temperature and pressure where two states of matter (solid and liquid) can exist together for a long time i.e in equilibrium.
If you are some slight distance farther on one side of the line, you have only a solid; at some slight distance away on the other side, only the liquid is stable. With only one state of matter, only one phase to deal with, there is nothing stable to equilibrate to.
If you wonder about zinc atoms vaporizing, they will not affect your investigation - they are the same as not even being there. (Unless a few atoms do affect your investigation, in which case you get to submit a research paper!)
Zinc phase diagram from Wikimedia Commons[3]:

References:
(1)   Wikipedia contributors. State of matter https://en.wikipedia.org/w/index.php?title=State_of_matter&oldid=1030288068 (accessed Jul 1, 2021).
(2)   Wikipedia contributors. Phase (matter) https://en.wikipedia.org/w/index.php?title=Phase_(matter)&oldid=1013386008 (accessed Jul 1, 2021).
(3)   File:Phase diagram of zinc (1975).png https://commons.wikimedia.org/wiki/File:Phase_diagram_of_zinc_(1975).png (accessed Jul 1, 2021).

A point in the phase diagram off a line means, for a ONE component substance, that is the only phase that exists. In the case of water at 200Cand 200 atmosphere there is no head space and the pressure is maintained by a solid piston or such device. If the pressure is reduced by lifting the piston and maintaining temperature,  when the pressure reaches the line gas will form and water will continue to evaporate as head space is increased until the water is all gone; then the gas will be in the space below the line.  If the pressure is caused by an inert-gas-head-space the added degree of freedom means that water will evaporate until its appropriate partial pressure is reached. This means that if the external pressure is caused by an inert-gas-head-space a new "line" is generated for every inert gas pressure. [Imagine a 3D phase diagram.,] Since the change in vapor pressure with changes in inert gas pressure is relatively small this change is usually ignored in simple treatments of phase changes.


text:
Rock salt as a green, non-electrical dehumidifier?
Has anyone tried using rock salt as a dehumidifier? Ive read a few DIY bits on it. I live in Georgia and its swampy and humid. I have a couple small dehumidifiers but I'm looking for some green and non-electrical dehumidifier options for my closets and storage spaces. Mold grows everywhere. Thanks!
https://byemould.com/2017/09/04/homemade-dehumidifier/
https://dengarden.com/cleaning/Ways-to-Dehumidify-Your-Greenhouse-andor-Home-Environment

Once the salt is wet / went into aqueous solution, the effect passes away.  And your options then?  At best getting new salt, but on this way you open the door to the porch and humidity enters again. Assuming you already took measures for thermal insulation, you now look for a more efficient method based on physics like an heat engine (which heat pumps are) to freeze the humidity out of the air (like a fridge does) before humidity condenses at your walls.  Thus, the question might be better suited for physics.se.

Same can be said about other hygroscopic salts. Problem like effectiveness and disposal apart, it would be better to look for CaCl2. It is a cheap side product (though I guess that in that form is already hydrated, so it would absorb less). Commercial cartridges do exist, but I don't think that they are greener if a total energy and CO2 balance is considered.

Your suggested use is impractical to dehumidify a living space, though it might help keep humidity down to ~75% in an enclosure.
The issue is that air holds a fair amount of water. At 30°C, 100% humidity (i.e., dewpoint of 30°C), not untypical for a "mild" Georgia (US, not Republic thereof) day, that's 30 g/m3. If you were to buy a 22 kg block of salt for ~US$8, it might last a few days, never dropping the humidity much below 80%, and leaving you with perhaps 28 kg of saturated brine.
Disposing of concentrated brine is hardly ecologically sound -- misquoting Cato, "Ceterum autem censeo Georgia esse delendam." Regenerating it to solid salt by heating will take copious energy, releasing a lot of heat, and likely putting more $\ce{CO2}$ into the atmosphere from your heat source..
[If someone wishes to edit the spelling of Georgia (accusative?), please do so, since I declined to do so, in this case.]

This way may not be much useful for a house of a flat with extensive air exchange, but could be applicable for a small, mostly closed space with limited air exchange.

It might work, if the goal is ~75% humidity. Libraries aim for ~30-50% humidity, but 70% is listed as good for cigar storage. But cigars are not good for us.


text:
Why wouldn't a standard addition fluorometric method return the same starting concentration of a solution used to make the standards?
I am trying out a new standard addition method and having some very basic problems. I am hoping someone could point me in the right direction.
I am using a fluorometric method to quantify ammonium in solution. The calibration curve is produced using the standard addition method, in which known amounts of ammonium chloride is spiked into collection tubes containing the unknown test solution and a fluorometric reagent. I am generating an evenly spaced 6-point calibration curve using this method, with an R2 of 0.99.
When using purified water, values are close to zero, as they should be. I am also getting appropriate values in the natural/environmental samples I am testing. However I would like to assess accuracy. This is where the difficulty has been.
To test accuracy, I prepared a concentrated ammonium chloride stock solution in pure water, and used this to 1) create a diluted "dummy" solution to test the method with, and 2) a solution to be used for standard spikes. For some reason, I am having difficulty reproducing the original dummy concentration, with values that are 10-30% higher than expected (i.e., 3.3-3.7 uM instead of 3 uM).
I am using volumetric flasks, pipettes, and serological pipettes for precision, and acid rinsing all materials before use. I don't believe contamination is a problem since the blanks are very low. Because the measured concentrations are consistently too high, I think I could be losing ammonium to bottle adsorption, and fluorescence values are therefore inaccurately reflecting the standard concentrations (not accounting for the wall loss). However, this happens even after testing freshly made solutions, using Teflon bottles (instead of glass or polycarbonate which supposedly minimize adsorption), and switching to a salt water base in case ammonium behaves unpredictably in deionized water. I don't believe the fluorometer is the problem, since the curve is linearly increasing and only 5 years old.
I am concerned I am missing something basic. Please let me know if you have an ideas, and I am grateful for your thoughts.

It seem you are doing everything carefully. The key question is are the standards and this dummy solution are prepared from the same ammonium chloride standard? Where do you get ammonium ion standards?

If you were losing ammonium ion due to ion absorptio you would not be getting higher values but rather lower values.

Yes, both the standards and dummy solution are made from the same concentrated stock, which was prepared by weighing out solid ammonium chloride powder.

Even if there was slight error in this molarity calculation, it should have affected the standards and dummy solution evenly, since they were prepped from the same solution. 2. Loss of ammonium in the standard solution (adsorption) would decrease fluorescence, but since original standard concentrations are used in the calibration curve, it would make the measured concentrations appear artificially high

Could you provide a reference to your method. I mean the original protocol. Where was it published.

Sure, thank you for taking the time to help. The protocol was first published by Holmes 1999, and later modified by Taylor et al. 2007:
http://docs.turnerdesigns.com/t2/doc/appnotes/S-0103.pdf
https://bioone.org/journals/journal-of-the-north-american-benthological-society/volume-26/issue-2/0887-3593(2007)26[167:ITFAMM]2.0.CO;2/Improving-the-fluorometric-ammonium-method--matrix-effects-background-fluorescence/10.1899/0887-3593(2007)26[167:ITFAMM]2.0.CO;2.full

Just for a sanity check, you said you are using different volumetric and serological pipettes. Have you calibrated your own pipets and your own pipetting process? You may be surprised by the errors. Just use deionized water and transfer required 5.0 mL and 10.0 mL and weigh it carefully. I would say, try serological one first. It hardly takes 10 min but it adds and extra layer of confidence. I would still think that adsortion by glass is negligible. The ion exchange capacity of glass is very very low. Also do you give enough warm up time to the fluorometer's lamp (30 min minimum)?

The pipettes are new, but yes good idea I will check them. No, I have not been waiting 30 mins. I will try this too. Thank you very much.

Lamp warm up is necessary for xenon arc lamps.

Hello again, thank you very much for your suggestion to wait a full 30 mins for the lamp to warm up. In this case however, I am still seeing values >10% higher than they should be. It seems I may not figure this out. I am very grateful already for your feedback nonetheless.

Thanks for sharing. I am surprised that you have a large systematic error. If you can share exact details of the experiment step by step, we can try again.


text:
Why is silicon dioxide acidic in nature and not amphoteric in nature?
Silicon is a metalloid which means it has
both metallic as well as non-metallic
characteristics. But if silicon is a metalloid
then why is silicon dioxide acidic in nature?
If silicon has both metallic as well as
non-metallic characteristics that means its
oxide should have both acidic as well as
basic characteristics which means that
silicon dioxide should have both acidic and
basic characteristics which means that
silicon dioxide should be amphoteric in
nature, right? Then why is silicon dioxide
acidic in nature?

Being a metalloid doesn't mean having both properties but having some properties of metals and some of nonmetals.

See this question and links therein: https://chemistry.stackexchange.com/questions/91852/is-silicon-dioxide-more-acidic-than-the-oxides-of-nitrogen

The question assumes that if silicon were a metal, it's $\ce{MO2}$ oxide would be basic. But in general $\ce{MO2}$ oxides of most metals are not purely basic but amphoteric. Titanium, for instance, is known to form titanates and its dioxide reacts with sodium hydroxide solutions even at ambient temperature [1]. Purely basic metal oxides almost always either have less oxygen than $\ce{MO2}$ or, where that stoichiometry is reached, are peroxides or superoxides (of $s$-block metals). Add the fact that silicon dioxide does show limited basic character through its reaction with hydrofluoric acid (WP does list the oxide as amphoteric), and we can fairly say that silicon dioxide does show intermediate characteristics between metal and nonmetallic oxides having the same stoichiometry.
Reference
1.
Kostrikin, A.V., Spiridonov, F.M., Lin’ko, I.V. et al. "Interaction of components in the NaOH-TiO2 · H2O-H2O system at 25°C". Russ. J. Inorg. Chem. 56, 928–934 (2011). https://doi.org/10.1134/S0036023611060131

The acidic nature of $\ce{SiO2}$ comes from the $19$th century. In these old times, ions and electrons were unknown. Redox equations were not well established. The best known equations were the neutralizations corresponding to the general equation $$\ce{Acid + Base -> Salt + Water}$$ This equation is valid for all mineral and organic acids. And the "base", at this time, was what we call today our present hydroxyde or our metallic oxide. A base was a substance that turns the indicators basic, and reacts with an acid to produce a salt and water. An acid was a substance that turns the indicators acidic, and reacts with a base to produce a salt plus water. As a consequence, there was a tendency to generalize, and to include all inorganic substances in three categories :  "acids", "bases", or "salts". Unfortunately, some substances like silica $\ce{SiO2}$ were difficult to put in any of these categories, as they do not react with acids and bases. But some day, somebody was able to make silica react with molten $\ce{NaOH}$ at high temperature, in an equation like $$\ce{SiO2 + 2 NaOH-> Na2SiO3 + H2O}$$ So here $\ce{SiO2}$ is not an acid, but it reacts with a base as if it was an acid : it produces a salt plus water after reaction with a base. This is why silica $\ce{SiO2}$ was classified as having an " acidic character".

SiO2 is acidic in nature due to a number of reasons:

Si has vacant d-orbital hence can act as Lewis acid(according to Lewis acid-base theory)
Si-O double bond is formed due to 2pπ-3pπ orbital overlap of O and Si respectively but due to the large size of Si, this Si-O double bond is not possible hence it completes it tetravalency by forming 4 Si-O single bonds(you will see that SiO2 I. E. Cassiterite is a 3D network solid in which SiO2 is in solid state and the Si here is sp3 hybridized). This Si-O single bond formed is much more stable because hybridized orbitals are much more stable than unhybridized orbitals due to lesser energy possesed by them hence Si hybrid orbital can overlap with p orbital of O
This Si-O bond formed is more polar and covalent and any compound with polar and covalent bond(s) tends to be acidic
This would be the explanation of acidity of SiO2 at the electronic level

Hope my arguments are reasonable enough to justify the question


text:
How does the value of the equilibrium constant give information about the extent of reaction at equilibrium?
Pretty much everywhere I've read, the numerical value of the equilibrium constant gives you information about the relative concentrations of products and reactants in the equilibrium mixture; the larger the value.
However, for reactions with an unequal number of moles on the reactant and product side, the equilibrium constant has units in terms of concentration or pressure, allowing me to make the numerical value arbitrarily large or small by choosing the appropriate units. Does the numerical value give no qualitative information about the position of the equilibrium in this case? Or are there a specific set of units that I'm supposed to use?
I am aware of a thermodynamic equilibrium constant that uses activities and is thus dimensionless, but equilibrium constants are only defined in terms of molar concentrations or partial pressures at my level

For a gaseous mixture, the equilibrium composition fits at the same time the value expressed in partial pressures and the value expressed in molar  concentrations. I see no big deal.

There is also an equation that relates the two which is $\ce{K_p = k_c(RT)^{\Delta n}}$

I am not asking about the relation between them. Kc and Kp for many reactions have units (instead of being unitless) so their numerical values can be altered arbitrarily by choosing appropriate units. In this case, how does their magnitude give any information about the position of equilibrium?

Does the following post answer your question: https://chemistry.stackexchange.com/questions/117636/why-is-the-equilibrium-constant-unitless?rq=1 ?

@BuckThorn I know that the thermodynamic equilibrium constant is dimensionless but Kp and Kc, which do have units, are used much more often and used to predict the positon of the equilibrium

I agree with @BuckThorn this is effectively a duplicate. This topic has been raised numerous times before: https://chemistry.stackexchange.com/questions/1137; https://chemistry.stackexchange.com/questions/85529; https://chemistry.stackexchange.com/questions/107374.

Yes you can't compare extent of reactions directly based on the numerical value of the equilibrium constant.
A very good example of this is the comparison of solubility of different salts based on their solubility products.
Let's take the example of calcium and silver carbonates:
$$
\begin{array}{|c|c|c|}
\hline
\text{Salt} & K_\text{sp} & \text{Solubility} \\
\hline
\ce{CaCO3} & \pu{8.7 \times 10^{-9} M^2} & \pu{9.3 \times 10^{-5} M}\\
\hline
\ce{Ag(CO3)2} & \pu{6.2 \times 10^{-12} M^3} & \pu{1.15 \times 10^{-4} M}\\
\hline
\end{array}
$$
As can be observed the solubility of silver carbonate is more than that of calcium carbonate even though it has a smaller solubility product.
This contradiction occurs due to the difference in the units of the solubility product. This all sums back to the elementary teaching, "Don't compare things with different units."
We can mathematically compare only the concentration of products and not their equilibrium constants if they have different units.
Directly comparing equilibrium constants is not a problem, in case both the constants have the same dimensions and the comparison is done in the same units. For example you can't compare two equilibrium constants if one of them is in pascals and other in bars.
So if you want to compare extent of two reactions either calculate the final concentration/pressure and compare them or if their equilibrium constants have the same dimensions than compare them in the same units.

The equilibrium constant has to be dimensionless otherwise expressions such as $\Delta G^0=-RT\ln(K)$ are not possible.  In a reaction such as
$$\ce{A_2 <=> 2A}$$
if $\alpha$ is the degree of dissociation and $1-\alpha$ of $A_2$ reacts and $2\alpha$ of A is produced using the partial pressures $(P_A=2\alpha P/(1+\alpha); P_{A_2}=(1-\alpha)P/(1+\alpha))$ and as $K_p=P_A^2/P{A_2}$ it follows that
$$K_p=\frac{4\alpha^2P}{1-\alpha^2}$$
where $P$ is total pressure divided by 1 atm (or 1 bar) and is dimensionless, and so the degree of dissociation is dependent on the equilibrium constant as
$$\alpha =\sqrt{\frac{K_p}{4P+K_p}}$$
which is what you are asking about. This ($\alpha$) has to be the same irrespective of any units used.

$\Delta G^0 = 1.$


text:
Why does Br2 brominate CH4 over CH3Br?
I was reading about the bromination of CH4, and saw that since Br is "selective" in its bromination, when you brominate CH4 with Br2 you predominately get CH3Br, and not CH2Br2. What about CH3Br makes it less "preferable" for Br2 to brominate and create CH2Br2 as a product?

This paper is relevant: https://pubs.acs.org/doi/pdf/10.1021/jp061558h

Bromine will react with methane to give a kinetic or thermodynamic mix of products basically because the HBr bond is moderately weaker than HCl and the reaction with the bromine radical is more selective.  This article seems to discuss this in detail: https://www.mdpi.com/2227-9717/8/4/443/htm.  I did not find the relative reaction rates between CH4 and CH3Br[but didn't look too hard] but I suspect  they are close to 4:3 at some temperature. To answer your question one would need to know the context in which the statement was made since halogenation reactions are generally controlled by reactant mixes and strong temperature control and can give very variable product mixes depending on temperature, inert gas concentration and relative reactant concentrations. If I had to optimize methylene bromide production there is probably a methane-bromine mix, photon flux and temperature that will maximize mono and di bromination and minimize bromoform production.  This is why chemical engineers study so hard and charge the big bucks! to solve problems like this.