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Silver recovery from sodium hydroxide solution
I have washed graphic film with sodium hydroxide solution removing the silver emulsion. Is there any way to now recover the silver from solution?

You could add salt water and precipitate silver chloride, collect that by filtration and then follow the procedure outlined here https://www.finishing.com/195/29.shtml

In relatively high alkaline environments, silver precipitates as dark-brown silver oxide with traces of metallic silver. If this sparingly soluble form is still too insoluble for you (due to the low concentration), maybe adding concentrated $\ce{HCl}$ dropwise might help (you don't want to add too much, as it can form the water-soluble complex $\ce{HAgCl2}$).

A Pourbaix diagram might also be helpful to highlight the species that exist in solution: https://www.researchgate.net/figure/Pourbaix-diagram-for-silver-The-graph-shows-the-speciation-of-silver-as-a-function-of_fig1_284173017

Insoluble silver compounds are not soluble in $\ce{NaOH}$ solutions. If they react, they are transformed into silver oxide $\ce{Ag2O}$ which is a dark brown precipitate.

As @Maurice mentioned in a comment, it is unlikely that you have silver in the solution after washing with $\ce{NaOH}$. In alkali the silver will rapidly form $\ce{Ag2O}$ and this cannot be solved in an excess of $\ce{OH-}$. So I would look for a precipitate.
Instead of washing with NaOH, it is better to use $\ce{NH3}$ instead. This will form a soluble silver complex that you can later make react with acetaldehyde or another aldehyde to recover solid $\ce{Ag}$. This is a very common reaction known as Tollen's test. More info here Tollen's test. If indeed you have a precipitate, you can also use this method by adding an excess of $\ce{NH3}$ to solve it. Note that $\ce{NH3}$ needs to be handled with care and under a hood.
If what you need is to recover a large amount in the order of kgs, then your best alternative could be to use an electrochemical method. And instead of washing with $\ce{NaOH}$ use an acid solution.
And yes, you can also heat to decompose the oxide or the $\ce{AgCl}$ salt, but this requires high temperatures and it's dangerous if you do not have safety measures in place.

Silver nitride can form if ammonia is too concentrated.

Yes, but the concentration at which you will form $\ce{ [Ag(NH3)2+]}$ is much lower than what you need for the nitride.

One must consider chemical costs, but your approach is sound.

Yes.  Silver can be recovered by heating the silver oxide in a crucible, yielding metallic silver and oxygen by the reaction:
2 Ag2O + heat ----->  4 Ag (metallic) + O2
If you want to go for higher purity, acidify with HCl, which will drop silver as AgCl while other metals, such as copper or iron, remain in solution.
The AgCl precipitate can be washed, dried, and flamed exactly the same way as the oxide with excess baking soda (hydrated NaHCO3).
2 NaHCO3 + heat ----> Na2CO3 + H2O + CO2
4AgCl + 2Na2CO3 ----> 4 NaCl + 2 CO2 + 4 Ag (metallic) + O2
Or, as Maurice$^1$ might have it:
4 AgCl + 2 Na2O ----> 4 NaCl + 4 Ag (metallic) + O2
However, the potential of extremely hazardous HCl$^2$ or Cl2 fumes as a side reaction exists.  This step should be done outdoors or in a fume hood.
$^1$ thanks for the input.  2 Ag2O to elemental 4 Ag and O2 balances, and the major reaction yields metallic silver and NaCl.
$^2$ several MSDS sheets list an HCl fume hazard under fire fighting procedures.  This may be attributed to AgCl reacting with water to form Silver oxide (AgCl has low water solubility) and HCl.

Silver chloride cannot be flamed. It melts at $455°$C and boils without decomposition at $1550°$C

@Maurice um, I've done it.  Try a little baking soda along with it. You'll get HCl off and silver carbonate to silver oxide to silver.  BTW AgCl decomposes to metallic silver and Cl2 in the presence of light.

It is impossible to produce $\ce{HCl}$ when heating anything mixed with soda. $\ce{HCl}$ destroys soda.

@ Robert DiGiovanni. I have been specialized in and teaching practical inorganic chemistry during all my life. And I perfectly know about Le Chatelier's principle. I have carried out (a lot of times) calcinations of products with soda ($\ce{Na2CO3}$). I know that it can produce some gas, but this gas is never $\ce{HCl}$. Never ! Another information : When soda is heated to T >> 1000°C, it is decomposed into $\ce{Na2O + CO2}$, but it does not produce $\ce{NaOH}$ as you state.

@Maurice appreciate your comments.  Some of the chemistry may seem a bit unorthodox, but silver + is a voracious seeker of electrons.  Water (oxygen) becomes a reducing agent (or in its forms NaOH, Na2O). With outgassing, I remain convinced there are hazards from side reactions.

Plenty of mistakes in this message ! $1$) $\ce{NaOH}$ will never coexist with $\ce{NaHCO3}$. Both will react according to : $$\ce{NaOH + NaHCO3 -> Na2CO3 + H2O}$$ $2$) Furthermore, at T > $290$°C, $\ce{NaHCO3}$  becomes $\ce{Na2CO3}$ $$\ce{2 NaHCO3 -> Na2CO3 + H2O + CO2}$$ So $\ce{NaHCO3}$ will never occur under a propane torch. $3$) $\ce{Cl2}$ will never be emitted out of a hot mixture containing $\ce{NaOH}$. $\ce{Cl2}$ will react with $\ce{NaOH}$ and produce $\ce{NaCl + NaClO3}$ $$\ce{6 NaOH + 3 Cl2 -> NaClO3 + 5 NaCl + 3 H2O}$$ and $\ce{NaClO3}$ may lose its $3$ O atoms at high T values


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I am confused That what is actually equilibrium constant. WHY THERE was a need to introduce equilibrium constant?
Please answer I am literally stucked into it

Why there was a need to introduce Mount Everest?

Life would be so easier without arithmetics too. No mammoths, few mammoths, many mammoths.

https://chemistry.stackexchange.com/questions/44894/why-is-the-equilibrium-constant-defined-that-way-intuitively?rq=1

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In the nomenclature of Ethyl 2-methyl-5-oxocyclopentane-1-carboxylate, why is the ketone given a smaller Locant than the methyl group?
Ethyl 2-methyl-5-oxocyclopentane-1-carboxylate

The rules say that functional groups are given more priority than side chains but if there is more than one functional groups then we consult the priority table for nomenclature...
Hence the ester group is given more priority over the ketone group. But then while naming the cyclopentyl group, shouldn't we give the ketone group the smaller number than the methyl group?
Why didn't we call it ...5-methyl-2-oxocyclopentane...?

I think your argument is correct. Pubchem names may not always be correct. For example, look here for correct related name.

@Loong please take a look at this name: https://www.achemblock.com/ethyl-4-4-dimethyl-2-oxocyclopentane-1-carboxylate.html

@AgrimTripathi The name ethyl 4,4-dimethyl-2-oxocyclopentane-1-carboxylate is correct; however, it's not related to your question.

@Loong how is it not related? It literally contains the exact same functional groups and prioritizes them in the way one would expect. Ester > Ketone > Methyl Instead of the Ester > Methyl > Ketone prioritization in the original post. I think Mathew is correct in saying it's a Pubchem naming error.

@GrundleButt There is no order of functional group priority for substituents that are cited as prefixes in the name. The name ethyl 4,4-dimethyl-2-oxocyclopentane-1-carboxylate is not related to the question since it can be easily explained in the usual way using Rule P-14.4 (f). This rule, however, doesn't explain why ethyl 2-methyl-5-oxocyclopentane-1-carboxylate is correct; you need to consider Rule P-14.4 (g) in this case.

The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

For the compound that is given in the question, you have correctly identified the principal characteristic group as ester. It is expressed as suffix ‘carboxylate’. Thus, a low locant is assigned first to the ester according to Rule (c), which yields ‘…1-carboxylate’.
The structure doesn’t have multiple bonds that wound be expressed using ‘ene’ or ‘yne’ endings. Thus, Rule (e) is not relevant in this case.
Next, low locants are given to any substituents that are expressed as prefixes (e.g. ‘chloro’, ‘methyl’, or ‘oxo’) according to Rule (f). Note that all prefixes are considered together; i.e., there isn’t any order of functional group priorities at this point. In this case, however, two names are possible, ‘ethyl 2-methyl-5-oxocyclopentane-1-carboxylate’ as well as ‘ethyl 5-methyl-2-oxocyclopentane-1-carboxylate’, since the locant set ‘2,5’ for all prefixes considered together in a series of increasing numerical order is the same for both names. Therefore, Rule (f) is not sufficient to decide which name is correct in this case.
Finally, the lowest locant is given to the substituent cited first as a prefix in the name according to Rule (g); i.e., the locant ‘2’ is given to the methyl group since it is cited first.
Therefore, the complete name is ethyl 2-methyl-5-oxocyclopentane-1-carboxylate.

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Molecular Orbital Diagram of NO
When we draw the molecular orbital diagram for molecules
 For those that have less than or equal to 14 electrons we use the order "σ1s σ∗1s σ2s σ∗2s π2px,π2py σ2pz π∗2px,π∗2py σ∗2pz"
and for molecules that have electrons from 15 to 20 have the order "σ1s σ∗1s σ2s σ∗2s σ2pz π2px,π2py π∗2px,π∗2py σ∗2pz".
Now when we consider the case of NO or nitrogen monoxide the number of total electrons is 7+8=15 so by the above stated logic we should use the second order but from all the sources I have gone through all of them have used the first order as shown below

 Why is this?

Can you provide a source for "For those that have less than or equal to 14 electrons we use the order "σ1s σ∗1s σ2s σ∗2s π2px,π2py σ2pz π∗2px,π∗2py σ∗2pz"
and for molecules that have electrons from 15 to 20 have the order "σ1s σ∗1s σ2s σ∗2s σ2pz π2px,π2py π∗2px,π∗2py σ∗2pz". ?

Problem is, the amount of molecular orbitals will depend on the amount of electrons (basic maths here), yet their order is a matter of energy only (I think that's the point of @Andrew's question)

there isn't a very solid source really this is a portion that is covered in the coaching institution I go for and the same had uploaded this video (https://www.youtube.com/watch?v=JtQPh49fWeI&list=PL7AAT-ai0VD6O25hUm8BpberRzhdBK9ej&index=5) about the same and around the time stamp 24:10 they get into the molecular orbital diagram and just a disclaimer the video is partially in hindi though


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Reactivity of central carbon of Ketone vs Alchohol
I am trying to measure the reactivity of the carbon (of C=O and C-OH bonds) as described below. Lets assume the groups $R,R'$ are identical in both molecules.

Now I am trying here to to figure out the reactivity of the central carbon towards nucleophilic attack. Now from steric point of view, the ketonic carbon is less hindered, hence more reactive.
Now on charge separation (polarity) point of view, and this is where I am not clear, in case of alcohol, the O-H bond is polar, but to make the central carbon electron deficient the oxygen of O-H group has to pull electrons from $\sigma$ bond which is harder than $\pi$ bond of ketone.
Hence I think central carbon of ketone (all else identical) is more susceptible to nucleophilic attack.
Can someone kindly comment on my reasoning?

You're reasoning, while not incorrect, is simply missing the point. There no reactivity of alcohol to nucleophilic attack. Unless you transform -OH into a better leaving group, it simply won't react.

You are throwing snippet facts about and refusing to look at a totality. Sure ketone is less hindered [draw more appropriate structures to compare] but do not compare it only to an alcohol but also to an alkene. Next you are concerned about the polarity of the OH making the C-O bond more polar. This is confounding separate things and makes conclusions more difficult. In both cases the Carbon-Oxygen bond is polar with the carbon more positive. Your reasoning makes it at best a draw although thinking about the pi bonds has merit. You must think more about the possible mechanisms of carbonyls and monosubstituted alkanes, the intermediates that might be involved, and the possible transition states. Instead of alcohols consider sulfonate esters


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What is the difference between "smell/odor" and "vapor" of a substance?
A solid chemical substance with a very high vaporization temperature can still produce potent smells at room temperature. What, then, produces smell, and could smelling some potent solid substance pose any danger, even if it hasn't vaporized at all?
Thanks!

Too smell something, molecules have to reach the nose. Sometimes it is not the molecules you expect, but there is always a molecule that triggers the response.

There is no threshold temperature below which a substance does not evaporate. The rate decreases about exponentially with temperature, but does not reach zero. A nose is very sensitive to some molecules.

Additionally, there can be flying microparticles of fine dust.

What is the difference between "smell/odor" and "vapor" of a substance?

It is assumed that the vapor of a given compound/element is the gas phase of the same pure compound/element. By condensing the vapor, you can obtain the same stuff in liquid or solid form.
Smell on the other hand is a human/animal perception. Although it is thought that plants can also smell. It is a biochemical interaction of that component with our smell receptors inside our nose. A very interesting example is that of a enantiomeric molecules. These are two different compounds with identical vapor pressure at a given temperature yet they smell very differently. One enantiomer of limonene smells like lemons and the other smells like oranges! Same elemental composition, same boiling point but a different orientation in space.
Your next point is how something which has very high pressure can still have a smell. For example, a very high boiling point metal like osmium (the name means smell) has a very bad odor, but this is not due to osmium metal itself but rather due to its very thin volatile oxide coating which forms on the metal.
Alternatively, there is another possibility which has been linked by Karsten in the comments. Metals can react with sweat components and body oils and make volatile smelly compounds. This is why we have a metallic odor, again this has nothing to with the metal vapor itself, but some organic compounds which have originated by the reaction of the metal with our fluids present on our skin in very very small amounts. Human nose is an excellent sensitive detector and it is used in analytical chemistry in something called Olfactory Chromatography! Perfume makers and food flavor chemists love this technique.

A vapor is a gas made by evaporation of a liquid or of a solid (by sublimation). It may have an odor or not. Most substances in a gaseous state do have an odor. Water exists in a gaseous state, but it does not have any odor.

In the contexts I'm familiar with, a gas is not necessarily a vapor. "Vapor" is reserved for compounds below their boiling point, and above the boiling point the terminology is "gas". I'm sure it varies by field, but the first sentence of this answer I think is overly absolute.

@electronpusher. You are right. I have rewritten my first sentence.

In some sense, a gas is either vapour, either supercritical fluid. As vapour is defined as gaseous state at T < T_critical.

@Poutnik. You state that a gas is either vapor either ssupercritic fluid. Well ! A gas like $\ce{H2}$ or $\ce{O2}$ is neither a vapor nor a supercritical fluid.

@Maurice  OK, correction, He at p > 2.3 bar, H2 at p > 13 bar, O2 at p > 50.5 bar are supercritical fluids.

@Poutnik. Yes, but to get an odor, a gas or a vapor has to be at a pressure equal or smaller than 1 bar.

@Maurice But have very different critical parameters.


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Are there any specific requirements for ipso substitution in aromatic compounds?
In one of the questions in my textbook, (bromine + water) is able to replace the -SO3H group attached to phenol via ipso substitution, but in another question, (bromine + Fe) cannot replace the same group when attached to toluene.
Why does this happen? Is it because the water causes greater activation of the ring in the first case by phenoxide ion formation? Or is it because -OH is a better activating group in general?

I suspect this is because you have a different bromine species present. In the reaction with bromine and iron, the bromine first reacts with iron to give Fe(III) bromide which acts as a Lewis acid, giving a Br+ species as the active intermediate.

I still don't understand why it happens. In the first case, one of the double bonds in the ring (namely, the one closest to SO3H) breaks and attacks onto one of the bromine atoms in the Br2 molecule. In the second case, the same thing happens except this time, the bromine is extracted from FeBr4 anion. That doesn't explain the disparity though.

In each case you have an electron donating (Me/OH) and withdrawing (SO3H) group playing a tug-of-war in terms of activating or deactivating the ring towards EAS. A methyl is significantly less EAS activating than hydroxyl, which can formally donate electron density (anion) into the ring through resonance. Methyl's are just slightly activating by induction so toluene rarely undergoes EAS efficiently (personal experience) without forcing conditions, and this is without a deactivating group making it worse.
The introduction of iron would generate Fe(iii) bromide, as the other commenters have said. This species is usually great for helping out EAS brominations on the electrophile's side (increases polarization of remaining Br2). HOWEVER, even this great electrophile setup cannot overcome how poor the "nucleophile" is, being a weak tolule to start off with, further deactivated by the sulfonic acid.


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Do NaHCO3 and cyanoacrylate react?
I just saw this video about using baking soda + cyanoacrylate(CA) glue as a substrate+resin.

Neat, but do CA and $\ce{NaHCO3}$ react in any way?

If so, what does the equation look like and what is the resulting structure?

If not at room temperature, do they react at higher temperatures because CA curing is exothermic.

Is the CA + $\ce{NaHCO3}$ stable once it dries, is there any residue that would be toxic or irritating?

(not sure what to tag this with, feel free to add tags)

"Baking soda"(*) speeds up the cyanoacrylate curing, as it's basic anion initiates anionic polymeration. Similarly as benzoyl peroxide initiates radical polymeration of styrene to polystyrene.
$$\ce{HO-CO-O- ->[CH2=CR1R2] \\
HO-CO-O-CH2-CR1R2- ->[CH2=CR1R2] \\
HO-CO-O-CH2-CR1R2-CH2-CR1R2- ... etc}$$
While curing, there is expected stronger evaporation of monomeric cyanoacrylate, compared to standard curing, due faster reaction, which may be irritating. When cured, the safety concerns are the same as for ordinary cyanoacrylate curing and usage, which has medicinal applications too.
Note that advices about medical safety, acute and especially long term one, is explicitly off-topic on this site, which would comment just chemical aspects of safe manipulation and precautions.
The higher temperature is issue for the structural stability, as the remaining encapsulated baking soda may decompose, forming gaseous carbon dioxide.

(*) -  "Baking soda" = sodium bicarbonate, $\ce{NaHCO3}$. Not to be confused with sodium carbonate, $\ce{Na2CO3}$ resp. $\ce{Na2CO3 . 10 H2O}$ aka 'washing soda", which is caustic, while baking soda is not.


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Lewis structure of XeF2 that comes closer to reality?
Meanwhile, the hypervalency of Xe in XeF2 is explained by the 3c4e bond, where the p orbitals of the 3 atoms overlap to form a delocalized system of 3 MO's in which 4 electrons (two from the Xe p orbital and one each from the p orbitals of the F atoms). This leads to a bond order of 1/2 per Xe-F group.
Obviously, the lewis representation that can be found everywhere on the Internet cannot be correct:

That would indicate a bond order of 1 per F-Xe Group, not 1/2.
As far as I know, resonance drawings are used to indicate delocalized bonds, so they also represent bond orders of 1/2. Shouldn't XeF2 then be represented as a lewis diagram as follows:

And I also wonder what orbitals are those 3 lone pairs at Xe? s,p,p? Or 3 sp2 hybride orbitals (which would make more sense for the geometry but less sense because hybridisation just makes sense for bonds)?

Lewis structures are not intended to be accurate representations of the electronic structure of molecules. They follow conventional rules that make the drawings simple, but informative. According to those conventions, expanded octets are preferred in the drawing for elements larger than fluorine, whether or not they are truly hypervalent.

@Andrew Thank you! Btw. what do you mean by "whether or not they are truly hypervalent"? Are there compounds that are/ aren't truly hypervalent?

@Andrew Or more precisely, are there really any that are truly hypervalent? https://www.ch.imperial.ac.uk/rzepa/blog/?p=2599

@Andrew Maybe I should consider hypervalency as a synonym for hypercoordination?

sure, if that helps you understand it.

Hypervalency is essentially a concept that can only work within the framework of expanded octets. In that framework multicentre bonds do not exist. So hypercoordination is indeed a better term for what is at play here. Even though it is used as a synonym often, I would actually advise against this. However, you can easily substitute the latter for any intends and purposes.

Just after XeF2 was discovered, D. F. Smith published a short communication Xenon Difluoride J. Chem. Phys. 38, 270 (1963) mentioning the electronic structure of your second drawing.
Smith conducted infrared spectroscopy of XeF2 and wrote:

Then C. A. Coulson published The nature of the bonding in xenon fluorides and related molecules J. Chem. Soc., 1964, 1442-1454, which considered the validity of four models (plus some sub-models) of XeF2 bonding.
Coulson adds more resonance structures in addition to the two mentioned by Smith, including a purely ionic F- Xe2+ F- resonance structure.
For a more recent article on the topic, see The essential role of charge-shift bonding in hypervalent prototype XeF2 Nature Chemistry 5, 417–422 (2013). (official link)
Overall, ionic contributions do need to be considered.


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Why are the vapor pressure of toluene and benzene different?
We learnt to apply Raoult's law for the ideal solutions. When the questions are asked, they tell that the intermolecular forces of toluene and benzene are equal. But in the same question, they provide two vapor (saturated) pressures for toluene and benzene.
Question: If the intermolecular forces of toluene and benzene are equal, how do they have different vapor pressure?

Most likely your "they" are exam-paper setters in some public exams. The fact is that the moment you deal with a different molecule, the intermolecular interactions cannot be the same. Benzene and toluene are different molecules just like two different persons on Earth, they have different molecular weights and not surprisingly different molecular interactions. Benzene boils at 81 °C and toluene boils at 110 °C. In general, a bigger/heavier  organic molecule has a lower vapor pressure and a higher boiling point. Intermolecular forces is an umbrella term, so there is no single number, hence this can never be equal for two different molecules.
In the comments you asked about molar mass. In general heavier molecules have a higher boiling point but one cannot make a prediction. Compare benzene with a molar mass of 78 g/mol, and another metal such as platinum with a molar mass of 195 g/mol. The former boils below 100 °C and platinum boils at several thousand degrees. Molar mass alone cannot let you predict a number related to melting or boiling. Many software do predict the boiling and melting point of compounds, more often that is quite wrong.

thank you so much for your answer. I have one more question related to this and hope you can help me. How does the molar weight affects the boiling point. I guess that it increases the london forces. Am I correct?

The benzene/platinum example doesn't seem very helpful. I would expect the boiling point of a "lighter" liquid to be less than the boiling point of the "heavier" liquid. With only two data points, it's not which of the two is surprising: are you saying benzene should have a higher boiling point than it does or that platinum should have a lower boiling point, given the two molar masses? (If you wanted to use lithium, an element with a molar mass of just ~7 g/mol, as the liquid with a high boiling point, that might make more sense for the point you are trying to make.)

@KavinduChamodhNethsara Yes, the mass itself does not affect the vapor pressure directly. A larger mass typically correlates with larger molecules that have stronger interactions per molecule. Krypton pairs have stronger London dispersion forces than neon pairs because Kr has more electrons. Octane pairs have stronger dispersion interactions than butane pairs, because they have twice as many carbon atoms. However, for other types of molecules, other interactions might matter. Larger polysaccharides or polypeptides can form more intermolecular hydrogen bonds than smaller ones.

@chepner I think the point is that the masses are not dramatically different but the bp are. Then again the intermolecular interactions in the two cases are very different. In the OPs example of toluene and benzene, intermolecular interactions would be dominated in both cases by London dispersion. This is what confused the OP.

Perhaps, by saying benzene and toluene have "equal intermolecular forces", maybe they meant they both have London dispersion forces, NOT that they have the same degree of LDF.  Then it would be up to you to explain why they have different vapor pressures by saying the toluene would have more sites for LDF than benzene.

Benzene and toluene do not have exactly the same intermolecular forces. Both have Van der Waals - London forces from temporary induced dipoles and toluene has also a slight dipole-dipole interaction from its very small dipole moment[toluene is small while benzene is vanishingly small effectively zero].  The difference in intermolecular attractions is not the force but the energy of attraction or the work required to separate the molecules. This energy is expressed in the heat of fusion and evaporation. If the forces are about equal, then the area in contact is determined by the molecular size and shape and these affect the energy of attraction. The differences in molecular sizes between toluene and benzene are apparent. Therefore very similar intermolecular forces result in different energies of attraction.
Finally, the boiling point or temperature of equilibrium depends on the heats and entropies of the transitions: at equilibrium delta H = T Delta S;   T = Delta H/Delta S


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Correct IUPAC name of 3 - chloro-1-phenylprop-1 -ene

In this molecule, what should I take the principal group as - the benzene or the propene chain?
I feel that we should take Benzene as the prinicipal ring and the name should be 3-chloropropen-1-yl benzene. However, I am also confused about 3 - chloro-1-phenylprop-1 -ene being a correct IUPAC name .
What takes priority in this case , the unsaturation of the chain or the size of the ring?
What is the relevant IUPAC rule here?
Can someone please help me out on this?

Be aware that a IUPAC name and a systematic (IUPAC) name are not synonyms. Molecules have often several IUPAC names, not all are systematic ones and neither their  preferred IUPAC name must be the systematic one. The typical example is the systematic IUPAC name ethanoic acid versus the preferred IUPAC name acetic acid.

Related: What is the priority of a phenyl group in a compound such as this?, Nomenclature of an organic compound with benzene and double bonds, and Is the ring or the chain chosen as the parent structure in octan-2-ylbenzene?

The compound that is given in the question doesn’t have a principal characteristic group, which would be expressed as a suffix and that would determine the senior parent structure. Therefore, the ring has seniority over the chain when selecting the preferred IUPAC name. In general nomenclature and depending on the context, however, the chain may be favoured to recognize the unsaturated acyclic structure.
The corresponding subsection in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains
(exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
(…) For selection of a preferred IUPAC name, see P-52.2.8.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Method (1): [(1E)-3-chloroprop-1-en-1-yl]benzene (preferred IUPAC name; ring preferred to chain)
Method (2): (1E)-3-chloro-1-phenylprop-1-ene (unsaturated acyclic structure may be recognized)
Therefore, the preferred IUPAC name for the compound that is given in the question is [(1E)-3-chloroprop-1-en-1-yl]benzene.

Note that parentheses are used around substituent prefixes to separate locants referring to different structural elements.


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Silicon crystal orientation
I'm confused about how [110] direction is determined for (100), (110) or (111) wafers. I found a book chapter which just confused me even more. From the image below, I understand how [110] is determined on the (110) wafer but not the other two. I'm also having a hard time understanding what different planes would look like on the (111) and (110) wafer. I would appreciate a resource for silicon wafers specifically (not necessarily crystallography). Here are things I'm not understanding.
How is the [110] direction determined and why is it different for each of the three wafers below? For the (110) wafer [110] is between x and y axis which makes sense. But for (100) it seems to go from z to y and for (111) it's between x and z. From my understanding [110] intercepts x and y axis at 1 and doesn't intercept z axis but that doesn't seem to be the case for (100) and (111) wafers.
Does the definition of x,y and z axis change when we're talking about (100), (111) or (110) wafers? I know that whichever wafer it is, that direction will be pointing out from the wafer surface but when I draw planes with respect to [110] and [111] (for (111) wafer), I don't know where the (100) plane should be. (110) plane would be perpendicular to the [110] direction and would be normal to the surface and (111) would be 35° from it. I can't find a drawing for it which makes me think we don't need to show the (100) plane? But if not then what if I need to align my mask to (100) plane on (111) wafer?
How would I define higher miller index planes on the 3 wafers? For example I need to align my mask to (122) or (411) plane, how would I start with it? I understand that this might become clear once I learn about the primary planes on the wafers.

I would appreciate a resource for silicon wafers specifically (not necessarily crystallography).

This video is fun to watch (the difference between a [111] and a [100] wafer is striking) and it points at further resources.
This interactive Jmol site lets you select a plane while also showing the unit cell orientation. For the image below (which is an interactive 3D model on the live site), first select to show 5x5x5 unit cells. Then, enter (1,1,1) for the plane, increase the thickness with the slider and click "draw atoms in the plane". Once you have a model of your wafer, change to another plane and click "show HKL plane". If you look closely, you can see the unit cell in the center in thin black lines (you have to turn it on manually, by right clicking and selecting Style->Unicell or Axes), allowing you to verify the orientation of the two planes.

How is the [110] direction determined and why is it different for each of the three wafers below?

Silicon has cubic symmetry, so the three directions [100], [010], and [001] are equivalent. What is confusing is that the images in the question show three different planes intersecting with the wafers: (0,1,1), (1,1,0), (1,0,1) from left to right, but they are labeling all of them <1,1,0> because they are indistinguishable once you take away the axis notation.

Source: https://journals.aps.org/authors/crystallographic-notation-h1

Does the definition of x,y and z axis change when we're talking about (100), (111) or (110) wafers?

No, in all three pictures, the x axis goes along [100], the y-axis along [010], and the z-axis along [001].

For example I need to align my mask to (122) or (411) plane, how would I start with it?

Let $\vec{a}$ be a vector perpendicular to the surface of the waver, and $\vec{b}$ be a vector perpendicular to the surface used in the cutoff, and $\vec{c}$ be the vector perpendicular to the desired alignment plane. The cross product of $\vec{a}$ and $\vec{b}$ is parallel to the cutoff edge. The cross product of $\vec{a}$ and $\vec{c}$ is parallel to intersection of the wafer surface and the desired alignment plane. To get the angle between the two, you use the usual dot product method.
For a (111) wafer (rightmost diagram), for example, the cutoff line is parallel to (1,1,1) x (1,0,1) = (1,0,-1). The intersection between the wafer surface and the (122) plane is a line parallel to (1,1,1) x (1,2,2) = (0,-1,1). The angle between these two is 120 degrees. Because of the symmetry of the wafer, other solutions are 0 or -120 degrees. (You would get those by dialing in the (212) or the (221) plane.)
Why does it matter?
From uhoh's comments, slightly edited:

In one of my previous lives I had to align contact photomasks to silicon wafers in order to open holes in photoresist, used to etch through a thin silicon nitride layer (10 to 100 nm) which was then used as a mask for directional wet etching of the silicon. KOH can have a 100:1 or larger wet etch anisotropy for certain crystal directions, and you can get pyramidal or even very vertical (wrt wafer surface) sidewalls from anisotropic wet etches if you rotate your mask to get specific orientations of the pattern edges with respect to crystal directions. [...] Anisotropic wet etching (Orientation dependent etching)

Lithographic mask I presume.

+1 re "mask": In one of my previous lives I had to align contact photomasks to silicon wafers in order to open holes in photoresist, used to etch through a thin silicon nitride layer (10 to 100 nm) which was then used as a mask for directional wet etching of the silicon. KOH can have a 100:1 or larger wet etch anisotropy for certain crystal directions, and you can get pyramidal or even very vertical (wrt wafer surface) sidewalls from anisotripic wet etches if you rotate your mask to get specific orientations of the pattern edges with respect to crystal directions.

Anisotropic wet etching (Orientation dependent etching)


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How can one obtain the same-spin and opposite-spin terms of the MP2 energy by integrating out the spin?
Just as specified on the psi4 website here: https://psicode.org/psi4manual/master/dfmp2.html, the Moller-Plesset Perturbation Theory states that the second order correction to the ground state energy, $E_0^{(2)}$, is given by
$$E_0^{(2)} = \sum_{i<j}\sum_{a<b} \frac{\left|\,[ia\|jb]\,\right|^2}{\epsilon_i - \epsilon_a + \epsilon_j - \epsilon_b} = \sum_{i<j}\sum_{a<b} \frac{\left( [ia|jb] - [ib|ja]\right)\left( [ai|bj] - [bi|aj]\right)}{\epsilon_i - \epsilon_a + \epsilon_j - \epsilon_b}\, ,$$
where $[ia\|jb]$ is the antisymmetrized two-electron integral over spin-orbitals and $\epsilon$ are the different orbital energies. As far as I know, one can integrate out the spin variable, $\sigma$, of said integrals in the following way:
$$\int\int d\omega_1 d\omega_2 [ia|jb] = \int\int d\omega_1 d\omega_2 \sigma_i^*(\omega_1)\sigma_a(\omega_1)\sigma_j^*(\omega_2)\sigma_b(\omega_2) (ia|jb) = \delta_{\sigma_i,\sigma_a}\delta_{\sigma_j,\sigma_b}(ia|jb)\,,$$
thus obtaining chemist’s integrals over spatial orbitals, $(ia|jb)$. Following this path, $E_0^{(2)}$ can be divided into a same-spin, $E_{0,SS}^{(2)}$, and an opposite-spin, $E_{0,OS}^{(2)}$, terms by integrating the spin out such that
$$E_{\rm 0,\,SS}^{(2)} = \sum_{ij}\sum_{ab}\frac{(ia\mid jb)[(ia\mid jb) - (ib\mid ja)]}{\epsilon_i - \epsilon_a + \epsilon_j - \epsilon_b}\, ,\qquad
E_{\rm 0,\,OS}^{(2)} = \sum_{ij}\sum_{ab}\frac{(ia\mid jb)(ia\mid jb)}{\epsilon_i - \epsilon_a + \epsilon_j - \epsilon_b}\,.$$
By doing this, I was able to get the expresion for $E_{0,OS}^{(2)}$ without much trouble. However, when trying to perform the integration for same spins, I haven't been able to get rid of one of the terms and get the expresion I want (which has three terms only). This happens because all four orbitals have the same spin function, so every term should mantain after the integration. I haven´t been able to find ways to get these expressions without just using algebra like many demonstrations have.


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How do common histological stains work?
In histology there are commonly used acidic (= negatively charged) dyes, like eosin, erythrosine, orange G, and basic (= positively charged) dyes, like haematoxylin, toluidine and methylene blue, azure II. Acidic dyes stain positively charged structures, like some cytoplasmic proteins, and basic dyes stain negatively charged molecules, like RNA and DNA. Structures stained with acidic dyes usually appear pink, with basic - blue.
That's what commonly offered in histology course. I could not find any information on what type of bonding these dyes use (is there a covalent bonding or non-covalent interaction?). Nor do I know why cell structures stain that way. Can you explain these properties of the dyes?


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How are polyatomic ions formed?
I know that there should be a covalent bond between polyatomic ions, such as $\ce{SO4^2-}$. But what causes it to suddenly gain electrons and become an ion?

$\ce{H2SO4}$ is electrically neutral. When two protons are extracted from it, the electron pair that belonged to the two $\ce{H-O}$ bonds are donated to the oxygen atoms, resulting in a negative charge for both of them.

This is a wrong question to ask, in that the literal answer to it is useless. What causes things to suddenly gain electrons? Electrons, that's what. You see electrons, you grab them. There ain't no such thing as a free electron.

@IvanNeretin I answered the question in the title, which is not a wrong question. It is  very broad, however, so I kept my answer specific to the example mentioned.

@Karsten Your answer is fine. The body of the question, though, still raises my objection.

But what causes it to suddenly gain electrons and become an ion?

In principle, you can get a polyatomic ion by adding atoms to a monoatomic ion, or by changing the number of electrons in a (neutral) molecule, or by splitting a (neutral) molecule into positive and negative ions, or by reactions of other ions.

[...] such as $\ce{SO4^2-}$

I am not aware of the existence of $\ce{SO4}$, so just adding two electrons to a molecule probably doesn't happen in this case. Here are some formal scenarios to get sulfate, some of them more conceptual than realistic:
$$\ce{HSO4-(aq) <=> H+(aq) +  SO4^2-(aq)}$$
$$\ce{S^2-(aq) + 2O2(g) -> SO4^2-(aq)}$$
$$\ce{Na2SO4(s) -> 2Na+(aq) + SO4^2-(aq)}$$
$$\ce{S2O8^2−(aq) +2I-(aq) -> 2SO4^2−(aq) + I2(aq)}$$
$$\ce{SO3(g) + 2OH-(aq) -> SO4^2- + H2O(l)}$$
Some of these reactions are redox (electron transfer), some are acid/base (proton transfer), some are dissociation or hydration, and some are combinations of these.
Here is a reference for the second scenario (which looked unlikely to me so I looked it up), with a lot of different sulfur-containing ions as intermediates: https://doi.org/10.5935/0103-5053.20160197

Groan, you shouldn't answer something like that without clarification. OP probably just doesn't know what dissociation is. There are good reasons why we close questions like this one.


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In the synthesis of azo compounds, given the strongly basic reaction conditions, why is the azo product favoured?
I was reading about the chemical synthesis of aromatic azo compounds, wherein, first a diazo cation is formed from an aromatic primary amine:
$$\ce{Ar–NH2 + NaNO2 + 2HCl -> Ar–N=N+Cl– + 2H2O + NaCl}$$
This cation is subsequently reacted with a phenol under strongly alkaline conditions. EAS takes place with the diazo cation as the electrophile and we get the azo compound at ortho and para positions.
What I was wondering is, since $\ce{OH-}$ is strongly nucleophilic as well (in fact, I would guess more so than ortho and para positions of the phenol anion since the electron density is being distributed via resonance), how is a favourable yield of the product we want obtained in reactions in like this? Also, why would the the ortho and para product be preferred over the cation reaction with the $\ce{O-}$ of the phenol anion?

It's not done "strongly alkaline conditions". That would lead to deactivation of diazo.

@Mithoron A pH of 10 is fairly alkaline I'd say

Maybe "fairly" but not strongly. You could drink a solution with such pH and main problem would be bad taste ;)

@mithoron Hmm, okay. So the pH isn't high enough to where the  product with OH- exceeds the o/p product. Also isn't high enough to where enough phenols are converted to anions, and therefore the side product from that reaction is not too much either?

Coupling of diazo group with oxygen from OH or O-Ar anions is reversible at such pH. With the ring much less so.

@Mithoron Okay, so not is there a small proportion of those species (OH-, Ar-O-), but the reaction is more reversible as well. But why is that the case?

@Mithoron The reason i ask, is in the reaction O-Ar anion + Ar–N=N+ -----> Ar–N=N-O-Ar, it appears to me that the products are a lot more stable than the reactants. So I don't understand why you are saying it would be reversible at this pH, or the role of pH in the case of something like this (apart from the actual generation of the O-Ar anion)

@Mithoron My bad, I forgot to include the Cl- on the Ar–N=N+. Okay, so since it's an ion exchange reaction, it is reversible. And decerasing/increasing the anion concentrations on the reactant side through varying the pH would modulate the extent of the forward reaction. If I understand correctly, this is the reason you said that? and also that EAS isn't as easily reversible due to its mechanism.


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Carbon Monoxide
In the compound $\ce{CO}$ I know that the carbon has an oxidation number of $+2$ and the oxygen has an oxidation number of $-2$. However up until now I have thought of oxidation number and charge being the same, and it is apparent that the $\ce{CO}$ molecule is most definitely covalent and has no ionic character hence the molecule is neutral and each atom should not have an actual "charge" or we would call the molecule ionic. So are oxidation number and charge different or do $\ce{C}$ and $\ce{O}$ have charges in carbon monoxide.

Let us continue this discussion in chat.

@Poutnik These are formal charges you cite. Partial charges (in whatever partitioning scheme you may chose) are still far away from those. See for more: How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?

@Martin OK, interesting, I do not object.

Related: Does formal charge affect bond polarity?

"one carbon atom and one oxygen atom connected by a triple bond," https://en.wikipedia.org/wiki/Carbon_monoxide

https://chemistry.stackexchange.com/questions/69798/an-introduction-to-oxidation-state-oxidation-number

This is a great question!  Welcome to the site.  I'm not sure why you are getting so many downvotes.  Try not to let the grumpy people around here bother you.

Some possibly helpful related questions are HERE and HERE

Oxidation State and formal charge are most definitely not the same thing; in fact they make opposite assumptions to exist.
The molecule CO is, as you correctly identified, covalent, and you’re likely to only see oxidation states used when it comes to ionic molecules, so focusing on covalent molecules:
Formal charge totally neglects the existence of electro negativity in covalent bonds. It makes the assumption that it any covalent bond, shared electrons belong equally split between the two atoms that share them.
Hence the formal charge of an atom can be defined as :
Number of valence electrons in uncharged atom - number of electrons in lone pairs in the atom in the molecule - (1/2 * the number of electrons shared in covalent bonds by the atom in the molecule).
Oxidation states, however, make the assumption that electronegativity is an all-powerful effect, such that in a covalent bond with an electro negativity difference, shared electrons belong entirely to the more electronegative element.
Hence the oxidation state of an atom in a covalent bond can be defined as:
Number of valence electrons in uncharged atom of element - number of electrons in atom in the molecule (assigning electrons to more electronegative atoms)
The ONLY real similarity between the two concepts is that in any molecule/ion, the sum of formal charges is equal to the charge of the molecule, and so is the sum of the formal charges. But the individual charges on atoms will often differ from the individual oxidation states on atoms.
Considering the molecule CO, where there is a triple bond between oxygen and carbon , and each atom has one lone pair:
Oxygen is more electronegative than carbon and oxygen normally has 6 VE and carbon has 4VE. So for oxidation states, all 6 electrons in the triple bond as considered to belong to the oxygen; for formal charge they are split 3 each. Hence:
Formal charges:
Carbon: 4 - 2 - (0.5*6) = -1
Oxygen: 6 - 2 - (0.5*6) = 1+
Oxidation states:
Carbon: 4 - 2 = 2+
Oxygen: 6 - 2 - 6 = -2
Note that although these are different, both pairs of values sum to 0 as your molecule is neutral.
Finally, you’re right that the molecule is neutral and does not have an overall charge, although it does obviously have a dipole towards the oxygen nonetheless.
But the truth is that these formal charges and oxidation states are both just numerical but artificial concepts. They’re mainly taught as they can be extremely useful in other areas of chemistry, such as organic synthesis in the case of formal charge or knowing when acidic conditions are needed in the case of oxidation states.

The dipole in CO is actually with the negative end towards carbon, basically formal charge beats electronegativity difference.


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What is the mechanism of Salkowski reaction of сholesterols?
I recently came across the Salkowski reaction of cholesterols. Cholesterol in chloroform is treated with concentrated sulfuric acid. A positive test exhibits two distinct layers, the upper chloroform layer gets a red to violet colour, while the lower sulfuric acid layer exhibits a greenish glow.

The chemistry behind the reaction is dehydration using $\ce{H2SO4},$ forming cholestadiene.

In the second step, the cholestadiene undergoes dimerization.

In the last step, the cholestadiene undergoes sulfonation at the 7,7' position. (I believe I drew the correct structure)

However, I couldn't find a valid mechanism for the second and the third steps (dimerization and sulfonation). Can someone explain the mechanism?

I'm not sure, but this mechanism looks very "hand-wavey" to me, possibly just to make it a little easier to follow. This paper makes me think it might go through a carbonation intermediate.

@SendersReagent: However, this article before yours has claimed the isolation of final dimer without sulfonic acid groups under Zac-Henry conditions.

I think this article may give you the insight for given chemical formations.

The mechanism of dimerization is now clear, Thanks to Mathew Mahindratne.
Ref.Yoshihisa, et.al
There are two different isomers of Cholestadiene present, namely 3,5 Cholestadiene [ I ], and 2,4 cholestdiene [ IV ]. The plausible mechanism of their formation is shown below. They can undergo dimerization to yield [ II ] & [III].

Dimer [ II ] was formed from two pathways, which were the conversion of the dimer [ III ] and the dimerization of  [I]  (3,5-cholestadiene).
The dimer [ III ] was formed from [ IV ] (2,4-cholestadiene).
The mechanism of their formation is shown below.

The formation of the dimer [ II ] and/or the dimer [ III ] varies with the kind of acids employed.
Using a Bronsted acid medium results in the formation of [ II ].
For eg: In presence of Trichloroacetic acid and hydrochloric acid (10:1 ratio), dimer [ II ] was obtained.
Using a Lewis acid results in the formation of [ III ]. For eg: Zinc chloride and Acetyl chloride (Tschugaeff reaction),
Antimony trichloride and Acetyl chloride,
Ferric chloride and Sulfuric acid (Zak-Henly reaction),
Ferric chloride, Perchloric acid, and Phosphoric acid results in the formation of [ III ].
If the solution contains both Bronsted and Lewis acids, then a mixture of products is formed.
Eg: Both [ II ] and [ III ] are formed from the reactions with

Sulfuric acid and Acetic anhydride (Liebermann-Burchard reaction)

sulfuric acid (Salkowski reaction)

Trichloroacetic acid-SbCl3

Since sulfuric acid, having both Bronsted and Lewis acid character is used in our reaction, a mixture of products will be formed.
Ps: All Bronsted acids are Lewis acids, but not vice versa.
However, the third step (Sulfonation) needs further explanation.


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Why is polyester hydrophobic as well as insoluble in various organic solvents?
Polyester is known to be hydrophobic as shown by it's low moisture regain.
However, esters are at least somewhat hydrophilic and soluble in water. This decreases with the hydrocarbon chain length increase, but in the case of polyesters, this is accompained with a proportionate increase in the number of ester groups.
My attempts to explain this were as follows

The reason nonpolar molecules are not soluble in water is the H-bonded structure of water is disrupted - there's a loss in entropy; water molecules have to orient themselves in the form of a cage around the non polar regions.

In the case of a polymer chain of PET, the loss in entropy is greater than if we tried dissolving the individual monomer esters that constitute it in water. Therefore one is somewhat soluble, the other is not.

Since like dissolves like, and a polar aprotic solvent would not have the H-bonded structure of water that would be disrupted (and in fact since they are both polar aprotic, one would presumably become well incorporated into whatever bonding structure the other has), I would think a polar aprotic solvent would dissolve polyester.

Turns out that this isn't the case. In fact, "polyester fibres are generally resistant to organic solvents" (A. Bendak and El-Marfasi; pdf link)
Perhaps, I thought, this might be because of the relatively crystalline structure of polyester that doesn't allow solvent molecules to penetrate it below its glass transition temperature.
However, polyester fibres are soluble in phenol and m-cresol. They here also appear to be soluble in THF and DMF, but not the remaining organic solvents.
Why wouldn't the same issues that were occurring with water occur with these polar protic solvents (phenol and m-cresol), and why are polyesters soluble in THF&DMF but not other polar aprotic solvents.
Also, what is a good general framework to be thinking about the solubilities of substances in various solvents?

The solubility is one of the phenomena that are not entirely understood in chemistry. Why is silver perchlorate $\ce{AgClO4}$ soluble in toluene ? Why is silver nitrate soluble in water, ether and glycerin, but not in ethanol ? Look in the Handbook of Chemistry and Physics. You will find plenty of other surprising facts about solubility.

@Maurice Interesting. What is then the best way to think about solubilities? Do we just leave it to experiments & measurement?

You have written the usual tendencies about solubilities. But I am afraid the only sure method is experiment and measurement. Maybe some day in the future somebody would discover THE law. He will become the next Nobel Prize. Try ! Why not ? But you must know that you will not be the first scientist to have tried, and failed...

@Maurice Hmm. Alright.


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What will be the major product in the nitration of (2-methoxyethyl)benzene?
In this question, I am not quite sure what will be the major product in the reaction and on reacting (2-methoxyethyl)benzene with $\ce{NO2-O-NO2}$ ($\ce{N2O5}$). I know $\ce{NO2}$ will act as electrophile and gets attached to ortho or para. In this will it be ortho position instead of para due to hydrogen bonding with $\ce{CH2CH2OCH3}$?

And what "the reaction" really is? You don't provide the context.

I edited the ques can you pls review it again.

@Poutnik I know it would be a nitration reaction but what should be the major product.

@VanshBhardwaj Compare to the nitration of toluene. 2-Nitrotoluene It is made by nitrating toluene at above -10 °C. This reaction affords a 2:1 mixture of 2-nitro and 4-nitro isomers

This is a case of the nitration of alkylbenzene by dinitrogen pentoxide (DNP), which can be considered as an anhydride of nitric acid. The substrate in the question is (2-methoxyethyl)benzene, which is closely related to toluene except for the extra bulkiness of the substitution. Thus, one can except nitration mechanism would be closely related to that of toluene.
What's lacking in this question is the conditions for the nitration. The alkyl group is basically a o,p-director. However, DNP-based nitration reactions are dramatically influenced by the medium (solvent) used (Ref.1 & 2). That means, the ortho to para products ratio is different from condition to condition (e.g., solvent). For example, it was 1:2 if the reaction performed at $\pu{-10 ^\circ C}$ as Poutnik pointed out in his comment above (in this case, I assumed the reaction is in toluene). When the nitration of toluene by $\ce{N2O5}$ is done in an organic solvent other than toluene such as liquefied 1,1,1,2-tetrafluoroethane (TFE), the product ratio varies (Ref.2):

If the reaction is conducted in the presence of a solid Zeolite with small pores and with DNP as the nitrating agent and dichloromethane as the solvent, the reaction products have a very high proportion of para-nitro substituted isomer (Ref.3).
References:

R. W. Millar, M. E. Colclough, P. Golding, P. J. Honey, N. C. Paul, A. J. Sanderson, and M. J. Stewart, "New synthesis routes for energetic materials using dinitrogen pentoxide," Philos. Trans. R. Soc., A 1992, 339(1654), 305-319 (DOI: https://doi.org/10.1098/rsta.1992.0037).
Alexandr K. Kharchenko, Ruslan V. Fauziev, Mikhail N. Zharkov, Ilya V. Kuchurov, and Sergei G. Zlotin, "Nitration of aromatics with dinitrogen pentoxide in a liquefied 1,1,1,2-tetrafluoroethane medium," RSC Adv. 2021, 11(42), 25841-25847 (DOI: https://doi.org/10.1039/D1RA04536A).
Reddy Damavarapu, Keerthi Jayasuriya, and Thomas J. Kwok, "Regioselective nitration of aromatic compounds by dinitrogen pentoxide and the reaction products thereof," US Patent 1999, 5,977,418 (PDF).

Interesting, for o- and p- isomers, I would expect the preference difference would be small and 2 : 1  was the result of statistical availlability ( NO2+ meeting o-C 2x more frequently than p- carbon. ). So for 1 : 1, preference of p-C must be 2x stronger, what seems suspicious. Perhaps the steric effect of methyl are more important than I have thought. /// The solvent influence must be really tricky. // But I have just residual knowledge of organic synthesis.

I think the trick is DNP stays as ion pair in aqueous solvents and as molecular form in organic solvents.

And as a neutral molecule, it is then weaker electrofil and therefore more selective and possibly more sterically picky (my guess)

I see it must vary according to solvent in which reaction is carried out.Thanks a lot.

Nitrogen pentoxide is a good source of $\ce{NO2^+}$ (in fact the solid form directly contains $\ce{NO2^+}$). So, as you imply, we would expect a nitration reaction especially with an activating, ortho-para directing substituent such as the alkoxy group shown.
The ortho product would be favored by hydrogen bonding, but here there is no good mechanisms for that. Hydrogen bonding with hydrogen attached to carbon may actually be possible, but it is weak, and the nitrogen group will not provide more strongly protection hydrogen. Therefore the more sterically favored para product would be major.

Or, you might want to ask the professor why it would be ortho, as it would not be evident to me.


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Salt analysis (cation)
I'm reading about salt analysis and I've got some questions:

Image source: pdf
Why are some sulphides (group II) precipitated in presence of HCl while others (group IV) in presence of ammonia solution? Can NaOH be used in place of ammonia solution? What's the role of NH4Cl in testing for group III? Will Na2CO3 work for group V?
(The groups here are not related to groups of periodic table.)

Too many questions in one post.

Am I the only one who noticed "not to be republished" on a "republished" photo here?

@ Mithoron. Yes. You are right. I had not seen this advice...

Let me answer your three questions.

Some sulphides are soluble in acidic solutions (group IV). Some are insoluble in the same sort of acidic solution (group II). It is a question of solubility product. It can be taken as an experimental result, a fact.

If no $\ce{NH4Cl}$ were present in the solution tested for Group III, the precipitates $\ce{Al(OH)3}$ and $\ce{Fe(OH)3}$ may remain in a colloidal state, and will cross the filter paper. This is a serious drawback. In the presence of enough ions, like $\ce{NH4^+}$ and $\ce{Cl-}$, the colloid is transformed into a precipitate.

$\ce{NaOH}$ can replace ammonia solution, but it will prevent the precipitation of $\ce{Al(OH)3}$ in group III and $\ce{Zn(OH)2}$ in group IV. These precipitates will redissolve in $\ce{NaOH}$ to form a solution of aluminate $\ce{Al(OH)4^-}$ and zincate $\ce{[Zn(OH)4]^{2-}}$. So the method would not detect $\ce{Al}$ or $\ce{Zn}$.

NH4Cl + NH3 forms a buffered solution to give the proper pH and [S=].  Using NaOH will make adjustment difficult.

Will any other soluble carbonate like Na2CO3 work for group V? @maurice

@ Shub Yes. You may replace $\ce{Na2CO3}$ by any soluble carbonate. But the choice is limited. The only soluble carbonates are ammonium, sodium, potassium, rubidium and caesium carbonates.

@Maurice you wrote NaOH will prevent the precipitation of Al(OH)3 in group III and Zn(OH)2 in group IV. But in group IV isn't it the sulphide that gets precipitated?

@Shub. You are right. The elements belonging to Group IV are precipitated by $\ce{H2S}$ in the systematic analysis of cations. But the same elements may also have other reactions which have nothing to do with the cation analysis. For example, they may react with $\ce{NaOH}$, and with plenty of other reagents. Here I was interested in reactions of cations with $\ce{NaOH}$. This is all, and it has no relation with the reactions of $\ce{H2S}$ or $\ce{(NH4)2S}$ in the systematic analysis of cations.


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Dalton's law clarification
I'm looking at Elements of Physical Chemistry by Atkins and de Paula. In section 1A.3, they state Dalton's law as

The pressure exerted by a mixture of perfect gases is the sum of the pressures that each gas would exert if it were alone in the container at the same temperature:
$$ p = p_A + p_B + ... \tag{1}$$
In this expression, $p_J$ is the pressure that the gas $J$ would exert if it were alone in the container at the same temperature. Dalton's law is strictly valid only for mixtures of perfect gases [...].

They then go on to define the partial pressure as
$$ p_J = x_J p  \tag{2}$$
where $x_J$ is the mole fraction of $J$ and $p$ is the total pressure of the mixture.
So my question is this: Eq. 1 holds for all gases given the definition in Eq. 2. So when they say Dalton's law only holds for perfect gases, do they mean that because $p_J$ in Dalton's law is not the same as Eq. 2, Dalton's law doesn't hold for all gases? Using the same notation for the two different meanings for $p_J$ seems to be widespread and potentially quite confusing.

It's the same meaning, just insert Eq 2 into Eq 1: $p = \sum p_i = \sum x_i p = p \sum x_i = p$.

They are definitely not the same meaning. $p$ for a real gas may be considerably lower than $nRT/V$. If $p_J$ is the pressure the gas would exert if it were alone in the container, then it is not the same as $px_J$

Hmmm, I think that Eq 1 can be satisfied simultaneously with the condition that "$p_J$ is the pressure that the gas J would exert if it were alone in the container" only if Eq 2 is true. Otherwise the equality in Eq 1 would not hold true. But you can write $\sum p_i = \sum p x_i$ without requiring that Eq 2 hold for each i.

The updated related question, quoting the same book, gives a hint that Atkins and de Paula mean by "perfect gas" what is now called an ideal gas, with the current term "perfect gas" having narrower meaning as an ideal gas with constant heat capacity.

Perfect gas usually means a gas above its critical T that cannot be liquified; it is usually not a synonym for an ideal gas that anyway is a mathematical figment. the question seems to be [and a good one] "are the deviations from the ideal gas law the same in a pure gas as they are in a mixture of gases?" A test might be to measure the vapor pressure of a precise amount of completely evaporated liquid in a vacuum and in an atmosphere of helium and then in air.

@jimchmst, i used perfect gas because that is the term used in Atkins and de Paula. I haven't seen it used much elsewhere, but in the few instances I have seen it, it wasn't the definition you gave. Do you have any examples of that usage? Wondering if it is field-specific.

Also, that's not quite my question. Deviations from ideality are definitely different in mixtures than in pure substances. Look at deviations from Raoult's law and azeotropes as examples.

@BuckThorn, whether or not $\sum p_i =\sum x_i p$ depends on how you define $p_i$. $x_i$ and $p$ have well-defined thermodynamic meanings. $p_i$ in the question has two different, and non-equivalent, meanings. Only one of the meanings satisfies $\sum p_i = \sum x_i p$ universally.

I think I agree (almost, and my first comment was not entirely correct)! $\sum p_i =\sum x_i p$ is always true (it becomes a mathematical identity because $\sum x_i = 1$ and if $\sum p_i = p$ by fact or definition), but $p_J = x_J p$ is either strictly true (a property of the gas in the case of an ideal gas) or a definition (in the case of a real gas). And $\sum p_i = p$ is meaningless without some definition of $p_i$, of which $p_J = x_J p$ is presumably only one possibility.

If you want to consider real gases then you should consider fugacity see e.g. https://en.wikipedia.org/wiki/Fugacity

@BuckThorn I think we are mostly in agreement, but I take issue with the statement that $\sum p_i = \sum x_i p$ is always true. If $p_i$ is defined as $n_i RT/V$ and $p$ is the actual pressure of  the gas, then for any real gas, $\sum p_i = \sum x_i p$ is not true. See my answer below for further clarification.

@jimchmst See Perfect gas

I read that and don't remember where I read my simple def. The flurry of definitions have convinced me more that physicists do not understand thermodynamics any more than chemists do.

After discussion with other commenters and answers, here is a summary of the issues with the text quoted in the question.
First, there are two non-equivalent definitions of the partial pressure $p_i$ being used.
Definition 1
The first is given only in the text immediately preceding and following Eq. 1 in the quoted text in the question. In mathematical form, this definition is
$$ p_i = \frac{n_i RT}{V} \: .$$
Using this definition for the partial pressure, we can show that
\begin{align}
p_\mathrm{tot}^{(id)} &= \frac{n_\mathrm{tot}RT}{V} \\
&= \frac{\sum_i{n_i}RT}{V} \\
&= \sum_i p_i \: . \tag{A1}
\end{align}
This is Dalton's law, which is only true for ideal (perfect in Atkins and de Paula's parlance) gases.
Definition 2
The second definition of $p_i$ is based on the actual pressure of the real gas and the mole fraction of species $i$, and it is given by Eq. 2 in the question
$$p_i = x_i p \: .$$
From this definition, and using the fact that $\sum_i x_i = 1$, we can express the total pressure $p$ as
\begin{align}
p &= p\sum_i x_i \\
&= \sum_i x_i p \\
&= \sum_i p_i \: . \tag{A2}
\end{align}
This expression holds for all gases, ideal or not.
Summary
Note that the final lines of Eqs. A1 and A2 look similar, but A1 is the ideal gas pressure for $n_\mathrm{tot}$ molecules, whereas A2 is based on the true pressure of the system and does not depend on any assumptions about the form of the equation of state. The confusion that led to my question is the fact that $p_i$ was used with these two different meanings in very close proximity and no special attention paid to differentiating them.

There is no inconsistency as far the pasted text goes. Before studying any derivation or law, we should know their assumptions. Once we accept those conditions and the assumptions, the derived equations become valid. With the following assumptions, Equation (1) and (2) are perfectly consistent as long as the following hold.
So what are assumptions for Dalton's law?
(0) The gases in the mixture will not react, even if they are ideal gases.
(1) There is no interaction among gas molecules
(2) Molecular volume is negligible. This does not hold true at higher pressures.
(3) Temperature is not too high or too low (say, we are not at the Sun's surface or Pluto)
If we start talking about real gases for equation (1) and ideal gases for equation (2), it is comparing apples and oranges.
BTW there are better textbooks on Physical Chemistry and it is a good habit to read the same topic from multiple books! It sometimes gives interesting insights.

I agree, this isn't my favorite P Chem text. I suppose the problem I have is that the presentation in the text as quoted above seems to imply that Eq. 1 is Dalton's law, but possibly more accurately, Dalton's law should be given mathematically as $p = n_A RT/V + n_B RT/V + ... $ to make clear that the partial pressures given are assuming ideal gas behavior, whereas the definition they give immediately below is a general definition for all gases.

@scmartin, I agree. I am not exactly sure (or aware of exceptions) when equation 2 is not valid either for real gases.

I don't believe it is possible to have exceptions to equation 2. $\sum_i x_i = 1$ so $p = p \sum_i x_i = \sum_i p x_i = \sum p_i$. This is why I think the implication in the Atkins & de Paula text is a bit unclear and even possibly misleading.

But you are implicitly using Dalton's law in summations. I mean nothing is absolute in gases laws. There is no universal equation of state. All laws have a limited validity range. This is what I am trying to convey, say you have a mixture of gases, you keep increasing pressure, then suddenly one of the gases starts to liquefy...then equation 2 may break down. I do not know about the universal truth of (2).

No, the summation $ \sum_i x_i =1$ doesn't have anything to do with Dalton's law, it is part of the definition of mole fraction. So if we define $p_i$ to be $ x_i p$, then $p = \sum_i p_i$ is always true. $p$ and $x_i$ are well-defined thermodynamic properties for any system. That is why I thought the Atkins explanation was so confusing. They seem to use $p_i$ to mean both $n_i RT/V$ and $x_i p$, but these are not equivalent. In the first instance, the use it in the Dalton's law sense of ideal gas behavior, but in the second instance they use it in the more general sense of $x_i p$.

To your point about condensed phase formation, as long as the $x_i$ are the mole fractions in the gas phase, $p = \sum_i x_i p$ still holds. Relating it to the condensed phase is more tricky (Raoult's law or Henry's law or something between).

Hi @AChem, so does that mean that Dalton's law does hold for all gases, real or ideal?


text:
Why is Fe-P in NaOH part of Hieltjes & Lijklema (1980) P fractionation?
Background:
In order to separate the different phosphorus (P) species in lake or marine sediments, one can use sequential P fractionation method. Its main idea is to use extractants with weak or strong affinity to P to extract the sediment sample step by step. By doing this, different operationally defined P fractions are extracted sequentially.
There are many methods available, and my question is about the method proposed by Hieltjes & Lijklema (1980). Its procedure is shown in the first screenshot below.
Question:
I'm wondering why Fe-P is in the "NaOH extraction" part, which is "17 hours with 50 ml 0.1N NaOH".
In the Table 2 (see second screenshot below) of the original paper (Hieltjes & Lijklema, 1980), we can see that 98.9% of Fe-P is extracted from the 0.1 N NaOH solution. The authors mentioned that this synthetic Fe&P compounds is "... a suspension of freshly precipitated ironhydroxy-phosphate, aged by standing or moderate heating." However, they didn't explain why this synthetic Fe hydroxides can be dissolved in the NaOH solution.
I understand that Al oxides can be dissolved by NaOH, so Al-P is in the "NaOH extraction" part. But Fe oxides can not be dissolved by NaOH, for example in this posted question. Then why Fe-P is in this extration?

Why don't you give us more information about the problem ?  You say that you will extract phosphate from "sediments". Well ! Which sort of sediment ? Organic sediments ? Silicate sediments ? Both ? Anything in Waste Water ?

@Maurice Thanks Maurice! The question has been updated now, adding the Background part and one table from the orignal paper for better explanation.

@ TX. You promise to show the process in Figure 1 and 2 below. And there is no figures below !! Apart from this, you gave us one Table $1$, and two Table $2$. Yes two ! Then, the table $1$ is no use for the present problem. The second table $2$ says that $98.9$% of the phosphate can be recovered form $\ce{FePO4}$ by $\ce{NaOH}$ treatment. The authors do not say that the iron is also dissolved. In my opinion, the iron is transformed into insoluble $\ce{Fe(OH)3}$ which remains in suspension in the liquid.

@Maurice Thanks for your proposal about the Fe-P extraction mechanism! It reminds me that maybe PO4 ion on the solid Fe is replaced by OH ion, thus NaOH procedure can "dissolve" Fe-bound P when solid Fe itself is not dissolved? (though I'm not sure since I'm not major in chemistry) And I have edited my question to remove one repetitve table, and use the word "screenshot" instead of "figure" to avoid misunderstanding.

@ TX. Your text about dissolution of hydrated ferric phosphate can be considered as a repetition  of my last comment 18 hours ago

@Maurice Not exactly. Your comment seems to say (at least in my understanding) that $\ce{FePO4}$ becomes $\ce{Fe(OH)3}$. But my text is that P bound on $\ce{Fe(OH)3}$ (adsorbed or co-precipitated) is released, but there is never $\ce{FePO4}$: the solid Fe is always $\ce{Fe(OH)3}$, with or without P on it.

@ TX. The reaction of $\ce{FePO4}$ with $\ce{NaOH}$ is the following : $$\ce{FePO4 + 3 OH^- -> Fe(OH)3(s) + PO4^{3-}}$$ The ions $\ce{Na^+}$ are spectators in this reaction. They do not take part to the reaction.

@ TX. How can you state that there is never $\ce{FePO4}$ in your substance ? What is the exact composition of Fe-P ? Is it a mixture $\ce{Fe(OH)3 + Na2HPO4}$ ?

@Maurice I noticed that in (Hieltjes & Lijklema, 1980), they said it is "... ironhydroxy-phosphate", which is in the last but one paragraph in my question. Sorry I cannot tell you what is the exact composition since I'm not major in experimental chemistry (or even chemistry ...). Just curious when I read something about eutrophication.

TX. Iron hydroxyphosphate may be $\ce{Fe2(PO4)(OH)3}$ which behaves like a mixture $\ce{1 Fe(OH)3 + 1 FePO4}$, but the ratio $\ce{FePO4/Fe(OH)3}$ may be different from $1$

@Maurice Thanks! This is a possible explanation


text:
Change in chemical properties on polymerization
What I want to know is how should I expect polyesters, polyamides or other polymers derived from a monomer containing a certain functional group to behave and react chemically vs the monomer, as well as the effect of their structure in their fibre form on their chemical behaviour. Differences in reaction kinetics, and general trends in chemical behaviour to expect and be aware of. I.e In what ways would they be similar and differ chemically from the properties of the monomer from which they were derived

Following the logic of the 2nd paragraph, solubility of n-alkanes in non polar solvents should not depend on the chain length. But it does.

@Poutnik In the case of esters and an increasing hydrocarbon chain, it would make sense right, since the nonpolar component is increasing. But in the case of polyesters there is a proportionate increase in the ester groups

In my understanding, acidic hydrolysis is much slower than alkaline one. Kinetics on solid surface is supposed to be much slower that for liquid boundary or homogenic mixture, considering geometry, accessibility and intermolecular bonds.

Imagine a wooden log with equidistantly spaced children. Each child would randomly  try and stop trying to lift the log. Now keep prolonging the log with adding more children. Will be the probability of successful lifting the log independent on its length?

@Poutnik Ah okay. That makes sense. So esters would be in a liquid form & somewhat solubilised and so the reaction would be faster than the reaction on the solid surface of PET fibres. However, sometimes esters do clump together to form tiny solids (for example oligomers of esters, such as cyclic trimers). How would you expect the reaction to go in this case?

@Poutnik In the sense that the surface area and volume of these tiny particles would be very small compared to PET fibres. So would the reaction be slow like on the surface of PET fibres, or faster like if they were in liquid form

There is another factor. The parts and functional groups of the polymer molecule are held in their place quite well. It may affect even the thermodynamics, favouring the ester more.

Let us continue this discussion in chat.

That's pretty much repost of https://chemistry.stackexchange.com/questions/169694/why-is-polyester-hydrophobic-as-well-as-insoluble-in-various-organic-solvents without analysis of the topic, but with another question added.

@Mithoron No it's not. Even if one of the examples is the same, the point in this question is I want to know how should I expect polyesters (or other polymerised forms of functional groups, like polyamides) to behave and react chemically in general compared to the monomers they were derived from.


text:
How are fundamental equations valid for both reversible and irreversible processes?
We know for an irreversible process, $\mathrm dS\gt\mathrm dq/T$.
And if the process is done at constant pressure we can take the equation as $\mathrm dH-T\,\mathrm dS\lt0$.
And we defined Gibbs energy, $G=H-TS$. At constant temperature and pressure $\mathrm dG\le0$.
But the fundamental equation of Gibbs energy $\mathrm dG$, in terms of temperature and pressure is given by $\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$.
And as per our original conditions, i.e. at constant pressure and temperature for an irreversible process the value $\mathrm dG$ should be less than zero.
I cannot understand for the same condition the two equations give different answers.

So, let's look how much is the equation fundamental, using as the starting point the definition of the Gibbs free energy:
$\require{cancel}$
\begin{align}
G  &=  H - TS\\
G  &=  U + pV - TS\\
\mathrm{d}G &= (\delta Q + \delta W) + (p\mathrm{d}V +V\mathrm{d}p) - (T\mathrm{d}S + S\mathrm{d}T)\\
\mathrm{d}G &= (\delta Q -\cancel{p\mathrm{d}V} + (\delta W_\mathrm{nonV})) + (\cancel{p\mathrm{d}V} +V\mathrm{d}p) - (T\mathrm{d}S + S\mathrm{d}T)\\
\mathrm{d}G &= \cancel{\delta Q} + (\delta W_\mathrm{nonV}) +  V\mathrm{d}p - \cancel{T\mathrm{d}S} - S\mathrm{d}T
\end{align}
For reversible processes is $\delta Q=T\mathrm{d}S$, therefore
$$\mathrm{d}G=  V\mathrm{d}p - S\mathrm{d}T \  (+ \delta W_\mathrm{nonV}) $$
For irreversible processes is $\delta Q \lt T\mathrm{d}S$, therefore
$$\mathrm{d}G < V\mathrm{d}p - S\mathrm{d}T \  (+ \delta W_\mathrm{nonV})$$
and for $T$,$p$ constant
$$\mathrm{d}G <  0 \ (+ \delta W_\mathrm{nonV})$$.

The starting equation should read $dS>dq/T_B$, not dS>dq/T, where $T_B$ is the temperature at the boundary between the system and surroundings (usually an a reservoir temperature) through which the heat flow dq occurs, and T is the average temperature of the system (which may not be spatially uniform). The open literature often fails to make this important distinction. With this correction, all the OP issues are resolved.


text:
Does addition of solute affect the rate of condensation?
I know that addition of a non-volatile solute in a pure solvent effects the rate of evaporation as the number of molecules leaving the liquid bulk per unit time decreases. However when the rate of evaporation decreases immediately , rate of condensation should automatically increase , this is because, if we consider a liquid 'X' in equilibrium as:
$$\ce{X(l) <=> X(g)}$$
Now when some amount of solute is added in the pure $\ce{X}$ , this dynamic equilibrium is obviously disturbed. $\mathrm{(Rate)_{evap}} \downarrow$  so to attain equilibrium $\mathrm{(Rate)_{cond}} \uparrow$ as reaction shifts backward.
also when equilibrium is attained the number of molecules entering a liquid has decreased , so at equilibrium rate of condensation has decreased. So, overall it has been affected.
But when it browsed various sites , all agreed that it is unchanged by solute.
So, where's the problem? Isn't my logic correct?

Actually, I am new to this community. So, sorry for poor presentation of the question. Hope you understand what I wrote. Eagerly waiting for a satisfying answer.

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

When a solvent is at isobaric conditions in equilibrium with its gaseous phase, the rate of evaporation and condensation are equal.
If we add a solute to the solvent, the activity, chemical potential, saturated vapor pressure and rate of evaporation decreases. As the rate of condensation remains the same, the net condensation increases from its equilibrium zero value.
If the system is isolated, then while there is ongoing condensation, temperature raises, saturated vapor pressure increases (due both increasing temperature and diluting by the condensate), while pressure of vapor decreases, until there is reached the new equilibrium with higher temperature but lower gas(vapor) pressure. This presure is now equal to saturated vapor pressure for the final solution and temperature.
If the system is also isothermic, then the heat from ongoing condensation is dissipated. The saturated vapor pressure and the gas (vapor) pressure at the new equilibrium are lower, compared to pure solvent, according to the Raoult law and eventual its deviations.

you wrote ".....saturated vapor pressure increases while pressure decreases.." also "......The saturated vapor pressure and the gas pressure...". What is this extra pressure and what here is "gas pressure"?

Vapor pressure = total pressure > saturated ( equilibrium) vapor pressure at given T and solute concentration. therefore net condensation. // small correction - I have not included the dilution of the solution by condensating solvent - addition inserted..

that means you are saying that besides the vapor which are in equilibrium with the liquid at a given T , there exists some more vapor above the liquid surface which are never in the equilibrium with the liquid surface?

There are not 2 kinds of vapor. The vapor either is in equilibrium with liquid, or it is not. It was with pure liquid,but when solute was involved, the same vapor, all of it, suddenly is not.

[Isothermal] addition of a solute lowers the chemical potential [mole fraction etc.] of the solvent. The chemical potential of the vapor[the vapor pressure] is now higher than the solution, so vapor will condense, slowly lowering the condensation rate until equilibrium is reached. The released heat of condensation will balance everything out. Since a solution of a non volatile solute universally has a lower vapor pressure that seems to sort it all out. I am not so sure about the higher T lower VP. If adiabatic T and VP should rise to the original VP. Tough experiment to run!


text:
Calculating acid volumes needed
I collect water quality samples used for irrigation. The samples are sent to a lab to measure $\ce{Ca^{2+}, Mg^{2+}, CO3^{2-}, HCO3^-}$, pH, and total dissolved solids.
I need to use acid to prevent scaling on my water emitters, so I want to calculate, based on the results of each sample, how much HCl I need to add to lower the pH to 6 and to 4.
I understand that the target pH gives the amount of H+ but how do I add to that the buffer capacity of alkalinity?
Thanks

The pH value would be running due progressive escaping of CO2. To what point should the pH value refer ? Immediately after neutralization or after excess of CO2 escapes ( may take long time ). BTW, is not pH 4 too low for irrigation?

Lets start with  immediately after neutralization. When CO2 escapes the pH rises? The soil buffers the low pH as long as we don't spray it directly on the plants

Initial pH or CO2 content would help, as I do not suppose there would be much carbonates, as pKa of $\ce{HCO3^2-}$ is 10.32.

But, nothing beats the operational tests on site and acting on it in the real time. Usual and cheap titration procedures to determine alkality usually use methyl orange ( 4.4 yellow -> 3.1 red ), heating the sample up to get rid of CO2, cooling down and then fine turn the acid volume, that temporarily raised up. For pH6, methyl red could be used ( 4.4 red - 6.2 yellow) or bromthymol blue ( 6.0 yellow - 7.6 blue ) // pH meter is advantage.

For example: Ca - 101 mg/l, Mg - 37 mg/l, HCO3 - 205 mg/l, CO3 - 0, pH - 8.01, TDS - 754 mg/l.
We usually rely on previous years analysis so not possible to do the titration test itself.

Relying on previous years analysis does not imply no possibility to do the titration test. One may use easily aquarium kits to determine the temporary/bicarbonate hardness and recalculate it to the needed acid addition. // The quite high and probably variable TDS may be sources of big error in theoretical estimation.

The measurements mentioned by Sdabach  give $2.52$ mmol/L for Ca, $1.52$ mmol/L for Mg, which makes a total of $4.05$  mmol/L. This corresponds to a concentration of positive charges equal to $8.10$ mmol/L. On the other hand, the corresponding negative charges due to the $\ce{HCO3^-}$  ions are only $2.97$ mmol/L. These two values ($8.1$ and $2.97$) should be equal. They are very different. Where are the missing negative ions ? And how is it possible to get a TDS as high as $754$ mg/L ?

There can be sulfates as the permanent hardness.


text:
Why does copper(II) sulfate react with potassium iodide in aqueous solution?
According to the list of standard reduction potentials [1, p. 5-79] $E^\circ(\ce{Cu^2+/Cu^+}) = \pu{+0.153 V},$ while $E^\circ(\ce{I2/I^-}) = \pu{+0.5355 V}.$ Doesn't it mean that iodine has more tendency to get electron and form $\ce{I-}$? If yes, then why does the reaction between $\ce{CuSO4}$ and $\ce{KI}$ occur?
Reference

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.

The normal potentials are valid if and only if the concentrations of the ions are $1$ M. It is not the case here. For example, the $\ce{Cu^+}$ concentration is never as high as $1$ M.

https://chemistry.stackexchange.com/questions/53768

When potassium iodide, is added to a copper sulphate solution, a redox reaction occurs in which the iodide ion converts cupric ion, to cuprous ion, (blue to white precipitate) before being oxidised to iodine gas.

See this link: https://chemistry.stackexchange.com/questions/138362/why-cupric-ion-oxidises-iodide-to-iodine

This is literally a complex reaction! If I remember correctly when these are mixed there is an immediate color change that I attributed to a fast complex formation. Someone must check this out. I think I remember this.
$\ce{Cu^{2+} + I-  <=>  CuI+}$ ;
$\ce{CuI+ + e- <=> CuI}$
Combine these two reactions to give:
$\ce{Cu^{2+} +I- + e- <=> CuI } ; ~~~~  \pu{ E^\circ = +0.877 V}$ [Langes Handbook]
We have our oxidant!
$\ce{I- <=> 1/2 I2 + e-} ; ~~~~  \pu{ E^\circ = -0.535 V}$
Combine these two reactions to give:
$\ce{Cu^{2+}  +2 I- <=>  CuI + 1/2 I2 } ; ~~~~  \pu{ E^\circ = = +0.342 V}$
This predicts a mixed product. I can't find a way to convert the $\ce{CuI}$ to $\ce{Cu}$ and $\ce{I2}$.

Those are the standard potentials. The potentials are for one mole of electrons regardless  of how the equation is written.  DeltaG[0] = E[0]nF accounts for the energy change with different equations.

Well, whatever the reason for downvoting the post was, its formatting, or rather lack of thereof, would be most obvious one. Another issue is grumbling about downvotes is pointless, or even detrimental for the site.

It would be great if you put some effort into writing chemical equations in a more readable format, especially since the site does make the necessary tool available and it's not difficult to learn.


text:
Can we use potassium hydroxide to decarboxylate potassium benzoate to benzene?
I know the famous reaction:

I was thinking of extension of this reaction to other salts of Benzoic Acids such as:
$$\ce{Potassium Benzoate + KOH ->[CaO,Δ] Benzene + K2CO3}$$
Is this reaction really carried out in Lab or there is not a favorable outcome/different product/unfeasible reaction such as in case of Kolbe's Reaction we obtain different products(basically isomers) for Na containing compound and for K containing compound.

Benzene is cheap and widely available, no-one needs to make benzene this way.

@Waylander Ok , if not for practical purposes , consider this question for theoretical interest.

It certain works for sodium salts so I don't see why it would not work for potassium salts  https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/The_Decarboxylation_of_Carboxylic_Acids_and_Their_Salts

And since expense is no object, try RbOH and CsOH... but not FrOH. Benzene is enough of a health hazard.

CaCO3, not K2CO3.

@Poutnik actually it's wrong in the diagram (at that time I was short of time so I didn't browsed much) it will be K2CO3 only. Refer: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/The_Decarboxylation_of_Carboxylic_Acids_and_Their_Salts

The reaction is famously known as soda-lime decarboxylation because of its use of sodalime which is mixture of $\ce{CaO}$ (75%) and $\ce{NaOH}$ (3%). It also contains trace amount of $\ce{KOH}$ (0.1%). You can however alter the composition where $\ce{KOH}$ is in majority while $\ce{NaOH}$ is in trace and the reaction will work in similar manner since the chemistry is not altered.


text:
Solvay process and its relationship with sodium chloride and calcium carbonate
I’m studying some chemical engineering process and my teacher made a question in class that nobody knew how to answer it and then she asked to made a search after class. However, I don’t even found nothing related.
$$\ce{2 NaCl + CaCO3 -> Na2CO3 + CaCl2 (do not occur spontaneously)}$$
Do you know why $\ce{NaCl + CaCO3}$ reaction do not occur in a spontaneous form in a process? I know that in Solvay Process we need to produce $\ce{CO2}$ from $\ce{CaCO3}$ from heating, but I really don’t know why we couldn’t do this reaction right away.

It is not a direat reaction, but the inputs and outputs of the production process https://en.wikipedia.org/wiki/Solvay_process .

Calcium Carbonate $\ce{CaCO3}$ is almost insoluble

Here a little digging is required as $\ce{NaCl + CaCO3 -> Na2CO3 + CaCl2}$ of the
Solvay Process leaves out a few steps.
The primary driver of the process is that sodium bicarbonate is less soluble in $\ce{NaCl}$ brine than $\ce{NH4Cl}$, and the starting materials are very inexpensive salt brine and limestone. The product is widely used "soda ash" $\ce{Na2CO3}$.
The more detailed steps are as follows:
$$\ce{CaCO3 ->[\Delta] CaO (s) + CO2(g)}$$
$$\ce{NaCl(brine) + NH3 + CO2 -> NaHCO3 \downarrow + NH4Cl (aq)}$$
Ammonia is recovered using highly basic $\ce{CaO}$ as:
$$\ce{2NH4Cl + CaO -> 2 NH3(g) + CaCl2 + H2O}$$
Bicarbonate is calcined to soda ash as:
$$\ce{2 NaHCO3 ->[\Delta] Na2CO3 (s) + CO2 + H2O}$$
Ammonia and $\ce{CO2}$ are thus recovered and recycled.  By product $\ce{CaCl2}$ is road salt.
Production and isolation of desired products by phase change is highlighted in these reactions:  $\ce{limestone -> CaO + CO2 (reactant)}$, $\ce{NaHCO3}$ by precipitation, recovery of ammonia gas, and conversion of bicarbonate to solid soda ash by removing $\ce{CO2}$ and $\ce{H2O}$ as gasses.

The proposed reaction $\ce{2NaCl + CaCO3 -> Na2CO3 + CaCl2}$ does not occur, and cannot occur by simple mixing in solution. On the contrary the reaction will be reversed, because $\ce{CaCO3}$ is the only insoluble product in the reaction. Let me explain the reason why.
If you mix two solutions, one of $\ce{Na2CO3}$ and one of $\ce{CaCl2}$, both solutions contain ions, namely $\ce{Na^+, CO3^{2-}, Ca^{2+}, Cl^-}$, and nothing can prevent the following reaction to occur :$$\ce{Ca^{2+} + CO3^{2-} ->  CaCO3(s)}$$ So the substances formed on the right-hand-side of the proposed equation cannot occur simultaneously in an aqueous solution. If they were present simultaneously in a solution, their ions would interact. $\ce{CaCO3}$ will precipitate and the proposed reaction will occur in the reverse direction $$\ce{Na2CO3 + CaCl2 -> 2 NaCl + CaCO3(s)}$$


text:
Point group of harmonic oscillator
The book Molecular Quantum Mechanics by Atkins and Friedman [1] says the point group of a harmonic oscillator is $C_\mathrm{s},$ composed by the identity operator $E$ and a reflection $\sigma_\mathrm{h}$.
But I don't understand why the rotation of $\pu{{\pi} rad}$ about the $y$ axis is not a symmetry operation in this case. It looks like this operation also leaves the hamiltonian invariant, since $V(x) = V(-x)$ for a harmonic oscillator. So, why can't the point group in this case be $C_2,$ composed by $E$ and a rotation $C_2?$
Reference

Atkins, P. W.; Friedman, R. Molecular Quantum Mechanics, 5th ed.; Oxford University Press: Oxford ; New York, 2011. ISBN 978-0-19-954142-3.

There is not necessarily a plane of symmetry orthogonal to the bond, which you need for $C_{2}$.

...[t]hat might be true for ground state and first couple of overtones, but the symmetry group fails for heteronuclear diatomics (in general) and homonuclear diatomics above some bond energy; and if one atom is an isotope and the other isn't.

I don't have the referenced book at hand but I think it is safe to assume that it is talking about a 1D harmonic potential, without reference to any actual molecule. Please correct me if I am wrong with this assumption and cite a bit more of the book.  Reflection/Rotation/Inversion are all the same transformation when you are limited to one dimension, since you can only have $x\rightarrow \pm x$. You should not try to transfer this directly to molecules. The situation is quite different in a polyatomic molecule and I highly recommend taking a look at a book that is more dedicated to symmetry.

Maybe, go one step further and consider what the difference between $C_\mathrm{s}$ and $C_2$ really is?
Here are the character tables for both point groups:
$$\begin{array}{c|cc|cc}
\hline
C_\mathrm{s} (= C_\mathrm{h}) & E & \sigma_\mathrm{h} & &  \\ \hline
\mathrm{A'} & 1 & 1 & x,y, R_z & x^2, y^2, z^2, xy \\
\mathrm{A''} & 1 & -1 & z, R_x, R_y & xz, yz \\
\hline
\end{array}$$
$$\begin{array}{c|cc|cc}\hline
C_2 & E & C_2 & & \\ \hline
\mathrm{A} & 1 & 1 & z, R_z & x^2, y^2, z^2, xy \\
\mathrm{B} & 1 & -1 & x, y, R_x, R_y & xz, yz \\ \hline
\end{array}$$
The terms to the right describe rotations and translations in 3D molecules, and the harmonic oscillator isn't a 3D molecule, so these are irrelevant. Once we strip that away, we're left with the conclusion that except for the choice of labels, $C_\mathrm{s}$ and $C_2$ are really equivalent.
In group theory terminology they are isomorphic.

There are two symmetry operations, the identity operation $(E)$ and another one $(C_2$ or $\sigma_\mathrm{h})$
There is one irreducible representation which is totally symmetric $(\mathrm{A'}$ or $\mathrm{A})$
There is one irreducible representation which is asymmetric with respect to the non-identity operation $(\mathrm{A''}$ or $\mathrm{B})$

Sure, the labels are not the same, but we can only draw a clear distinction between the labels in 3D anyway. In the case of the harmonic oscillator there is only one non-trivial symmetry element and that doesn't change whether we call it $\sigma_\mathrm{h}$ or $C_2$. (This echoes the point in Hans Wurst's comment.)
So, it's entirely up to you whether you want to call the point group $C_\mathrm{s}$ or $C_2$. The maths works out the same in the end — as it has to.

As @Hans Wurst already mentioned, the authors are referring to 1D simple harmonic oscillator and the symmetry of the potential and wavefunction.
As a chemist the first think that comes to our mind when we think of a 1D harmonic oscillator is a diatomic molecule $(C_{\infty\mathrm v}$ or $D_{\infty\mathrm h}$. But it can also describe an atom adsorbed to a (flat) surface $(C_\mathrm{s}$ symmetry?).
However, I think the authors are referring to the symmetry of the harmonic oscillators potential and wavefunctions irrespective of the system to which they are applied. In this sense, you only can define two elements of symmetry, the identity and the symmetry around the origin, a mirror plane, thus you could say the oscillator belongs to the $C_\mathrm{s}$ symmetry.
This is what I understand from the book [1, p. 158]:

5.17 Symmetry and degeneracy
We have already mentioned (in Section 2.13) that the presence of degeneracy is a consequence of the symmetry of a system. We are now in a position to discuss this relation.
To do so, we note that the hamiltonian of a system must be invariant
under every operation of the relevant point group:
$$(RH) = H \tag{5.40}$$
A qualitative interpretation of eqn 5.40 is that the hamiltonian is the operator for the energy, and energy does not change under a symmetry operation.
An example is the hamiltonian for the harmonic oscillator: the kinetic energy operator is proportional to $\mathrm d^2/\mathrm dx^2$ and the potential energy operator is proportional to $x^2$.
Both terms are invariant under the replacement of $x$ by $−x,$ and so the hamiltonian spans the totally symmetric irreducible representation of the point group $C_\mathrm{s}$.
Because $H$ is invariant under a similarity transformation of the group (that is, any symmetry operation leaves it unchanged), we can write
$$RHR^{−1} = H$$
Multiplication from the right by $R$ gives $RH = HR$, so we can conclude that symmetry operations must commute with the hamiltonian.

Reference

Atkins, P. W.; Friedman, R. Molecular Quantum Mechanics, 5th ed.; Oxford University Press: Oxford ; New York, 2011. ISBN 978-0-19-954142-3.