text:
Why does molar conductivity depend on the volume of the solution?
I have been searching the answer to this question everywhere and (almost) the only answer I get is to formula based:
‘Since molar conductivity = k / M = k/1/V for one mole = kV
Therefore molar conductivity increases with decrease in concentration.’
A few rare answers explained the physical reasons citing decrease in attractive forces, etc.
My question is:
Conductance of an electrolyte is directly proportional to Area and inversely proportional to Length.
G = kA/l
The Molar Conductivity of an electrolyte on the other hand is dependant on the product of A and l = V
^m= kV
Why is it so despite both being quantities of almost the same type?
What I gathered after researching for quite a few hours is that due to decrease in the attractive forces between the molecules/ions of the electrolyte (and other reasons), Molar Conductivity does not follow the ‘rule’ of Conductance and has its own ‘rule’ in which it is directly proportional to the volume.
That is to say, the formula for Molar Conductivity cannot be derived from that of Conductance. They both have to be derived separately.
Of course I never saw anything suggesting that the second one is derived from the first but it wasn’t stated explicitly that that isn’t the case either. My textbook gave the formulas and definitions of the two in consecutive sections without addressing the contradiction in formulas.
How correct is my understanding?
Edit: On M.Farooq’s request,
Conductance = Degree to which an electrolyte conducts electricity. Inverse of resistance.
Conductivity = Conductance of unit volume of electrolyte.
Molar Conductivity = Conducting power of all ions produced by one gram mole of an electrolyte in a given solution.

Could you show where does it say that molar conductivity increases with the volume of the solution? I mean share the image of the book which says so.

Also write the definitions of conductivity, conductance, and molar conductivity in the edited question. This will help you sort out the issue. *Normal conductivity is not a standard term. Clarify what do you mean by normal?

@M.Farooq Sorry. I should have said dependant on volume. I realised just now that since k also changes when V term changes (or Concentration changes). Does the point about it being independent of the Conductance equation still stand?

@M.Farooq I meant ‘Normal’ as in simply Conductance (=kA/l) since there are many conductances. I’ll edit the question to make it less confusing.

Molar conductivity for strong electrolytes increases due to increase in mobility of ions (as molecular interaction decreases on dilution). For weak electrolytes, it increases as number of ions increases (dissociation of electrolyte increases). For sparingly soluble salts, like AgCl, it remains essentially same with dilution. You can read about Kohlrausch's Law.

You are confusing V (the volume of the solution) and $V_m$ (the molar volume). They are very different properties. One is an extensive property dependent on the dimensions of the sample. The other does not depend on the volume (unless you are not scaling all extensive properties by the same factor).

@BuckThorn And the volume in the present case is not ‘Volume’? It’s ‘molar volume’?

Yes, when you divide the volume by the "amount of substance in moles" then you obtain the molar volume of the substance. Let's say you scale the solution by a factor x so that the volume becomes xV. But you also increased the amount of substance by factor x, so that it becomes xn. Then the molar volume of the scaled solution is xV/(xn) = V/n which is the same as the molar volume of the original solution.

When you computed the volume of solution for one mole you implicitly computed the "molar volume", an intensive property. It depends on the volume to the extent that you compute it as V/n, but it is "scale-invariant" (intensive) for simple thermodynamic systems.

Let us clarify a couple of terminologies. You wrote the conceptual definitions. Now look up the mathematical definitions as well.

Conductance is the reciprocal of resistance of (a solution in your case). It is dependent on the solution concentration, distance between the electrodes, and area of the electrodes. The key point which you should recognize is that by area, we imply the volume enclosed by the electrodes.
In order to include all these variables into one quantity, one defines conductivity. Look up the mathematical definition in your textbook not on Wikipedia. Wiki has a horrible confusing discussion for a beginner.

Conductivity is still concentration dependent, because you will have to quote the concentration of the solution whose conductivity is being quoted. This quantity is independent of the cell dimensions. So no involvement of the volume here.

Here comes the surprise. If you divide conductivity by molar concentration, we get molar conductivity. It should be independent of concentration. Turns out by experiment that it is not. It depends on the concentration. For strong electrolytes, there is linear dependence and for weak electrolytes it is another non-linear dependence on concentrations of the electrolyte.
Kohlrausch found out >100 years ago, that molar conductivity at infinite dilution (a hypothetical calculated quantity) is independent of solution concentration.

In short, molar conductivity does not depend on the volume of the solution.
If this is still not clear, please share the page of the textbook which which specifically tells you that molar conductivity is volume dependent.
So if I assume, 1 take 1L of 0.01 M NaCl solution I will have a different molar conductivity and if I take 500 mL of 0.01 M NaCl solution, I will have a different molar conductivity...this is not true at all.


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MA3B3 isomers - is this same as meridional one?
I'm new to studying stereo chemistry. (Now I'm studying coordination complexes.)
In my class material and other some datas(from google), it says that there is two, fac and mer.
I want to know is whether the image is meridional one.
By the explanation, two A-M-A bond angles are 90 degrees, and the other is 180 degrees - so I guess it is the mer one.
Or not, I want to know why it is not possible, or not stable.
(I'm guessing my picture is the less stable version.)

I'm looking forward to seeing your answer.

Yes, mer, that's him all right.

@IvanNeretin Is there not some energy issue?

Why would that be?

@IvanNeretin like stablity

Please be more specific.

@IvanNeretin ah, in my materials, there was a picture for mer isomer that one a is on top side, and others in horizonal side.

@IvanNeretin So I was wondering if my picture's case is unstable than the material's case.

You realize that there is no top side in the real world, don't you?

@IvanNeretin ah. In my country it's 2 am. I was almost sleeping. Thank you for answering kindly.

Your description of: In my class material and other some data (from google), it says that there is two, fac and mer, is somewhat misleading. These two type of isomers are only possible in $\ce{MA3B3}$ type of complexes (e.g., $\ce{[Co(NH3)3(NO2)3]}$):

In facial isomer (fac-), all three same type of ligands have $90^\circ$ $\ce{A-M-A}$ angles (same is for all three $\ce{B-M-B}$ angles). As shown in following diagram, all three of same types of ligands are clustered in one side of the octahedron so that they looks like the base of a triogonal pyramid (shown as red triangle on $\ce{NO2}$), head of which is the metal itself. The three other kind of ligands are on the opposite side of the  octahedron, arranged in similar manner.
In meridional isomer (mer-), all the three ligands of same type are arranged in a T-shape in same plane including the metal such that there are two $90^\circ$ $\ce{A-M-A}$ angles and one $180^\circ$ $\ce{A-M-A}$ angles, making it 'T' with metal in the middle (same is true for $\ce{B-M-B}$ angles, though on the opposite side). Also keep in mind that $\ce{A-M(A)-A}$ plane is perpendicular to $\ce{B-M(B)-B}$ plane.

Also note that neither is optically active since both have plane of symmetry. For mer-isomer, both $\ce{A-M(A)-A}$ and $\ce{B-M(B)-B}$ planes are planes of symmetry. For fac-isomer, for example, the plane (indicated by dotted line in the figure) going through trans-$(\ce{A-M-B})$ axis, dividing the octahedron to two is a plane of symmetry.
According to above description, your drown structure is mer-isomer (see red dotted lines for T-shape with $\ce{A}$ and blue dotted lines for T-shape with $\ce{B}$), which you have realized after discussion with Ivan Neretin.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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Disadvantages of SIM vs TIC in GC-MS
In GC-MS what are the relative advantages/disadvantages of using Selected-Ion Monitoring (SIM) vs using Total Ion Chromatograms (TIC)?
I know that SIM offers a better S/N ratio, however if the S/N is acceptable using TIC, is it better to use TIC and if so, why?

I think you should search why and when SIM is used? Hints: What do you think if peak overlap in a chromatogram? Will TIC be able to identify what compounds are eluting underneath?

But if there is no overlap using TIC and the peaks are all seen clearly, separately and with little noise using TIC, is their any advantage of using SIM?

If you are 100% sure that there is no overlap, then there is no harm in using TIC. SIM could be used for confirmation.

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Similarities between Co and Ni
Elements $\ce{Co}$ and $\ce{Ni}$ share many similar/same properties and usually if one of them is an example of some group (e.g., ferromagnetism, same possible oxidation states, packing fraction and coordination number in lattice, melting and boiling point,  similar properties of their alloys, etc.), the other is too.
I understand that it may be due to similar sizes and atomic numbers, but many other elements have these properties in common. Can you give any other example of such a pair and why only few of these exist, or why this relationship between  $\ce{Co}$ and $\ce{Ni}$ special?

Cobalt and Iron ions are the only ones to produce colored complexes when mixed with thiocyanate ion $\ce{SCN^-}$. If you mix up a dilute acidic solution of Iron(III) Chloride $\ce{FeCl_3}$ and a similar solution of ammonium thiocyanate $\ce{NH_4CNS}$, a extremely dark red solution is obtained, due to a complex containing the ions $\ce{Fe^{3+}}$ and $\ce{SCN^-}$. If you replace $\ce{FeCl_3}$ by a solution of Cobalt Chloride $\ce{CoCl_2}$, you obtain a blue color due to a complex containing $\ce{Co^{2+}}$ and $\ce{SCN^-}$. No other metallic ions are known to produce such colored complexes with thiocyanate ions, at least among the usual ions studied in chemical analysis courses. The blue Cobalt complex has also the property of being soluble in pentanol, where it can be extracted if the initial solution is colored.
Any way it should be mentioned that :
1) The reaction with Iron is extremely sensitive. The red color may still be seen if the concentration of the complex is 1 - 2 ppm. The blue complex made with Cobalt ions is only visible at much higher concentrations.
2) If this reaction is made with a Cobalt solution containing Iron as an impurity, the solution will turn red, as the blue color of Cobalt will be hidden. In this situation, a couple of milligrams of Potassium fluoride $\ce{KF}$ should be added to the solution. This will produce a colorless complex of $\ce{FeF_6^{3-}}$ and the blue color of the cobalt complex is restored.

This is one relation b/w Co and Fe, are there anymore?Also Sorry, but this is not what the question asks. But thanks, this was new for me.

There are other couples of neighbors in the periodic table. For example, Cr and Mn. They both can be oxidized to their maximum oxidation number by using PbO2. Cr is oxidized in basic medium and Mn only in nitric acide. But these two elements are the only ones that produce colored ions when oxidized by PbO2. Vanadium makes also colored ions, but not with PbO2. Another couple is Ni-Cu. They both produce colored ions in a NH3 solution. Cobalt ions also react with NH3 but it produces a precipitate.


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Why was Nobel Prize given to Baeyer for his "faulty" Baeyer's Strain Theory?
Baeyer Strain Theory said that using the formula
$\frac{(109^ \circ 28' - \alpha)}{2}$, where $\alpha$ is the angle between 2 sides of the cycloalkane, the most stable cycloalkane could be found. Thus he proved cyclopentane to be the most stable.
But later using the value of heat evolved during combustion, cyclohexane was found to be the most stable.
Then why was the Nobel Prize given to Baeyer for a faulty theory, still resides beside his name?
For a theory which no longer could explain the maximum stability of cyclohexane?

Are you asking why Nobel prizes aren't rescinded? Or is this a question about the theory itself?

Yes. Why wasn't it rescinded?

Baeyer's Nobel was principally for his work on dyes. Windhaus received the Nobel Prize in chemistry in 1928 for his work on sterols even though his proposed structure for cholesterol was shown to be incorrect  a few years later. To my knowledge, the prize was not rescinded.

Well, that's more of a question for the Nobel committee. Not for Stack Exchange. You can send them a message at https://www.nobelprize.org/contact

That's not a question for this community.

Instead of being critics to questions, answers are more welcome. And about the sarcasm, I think I found my answer already, thank you very much

Scientists are human beings after all. Nobody gets a Nobel Prize for inventing a "formula". It is the idea that makes an impact and new ideas are respected. Scientists like normal human beings make honest mistakes, but try their best to be correct. The guy who got the Nobel Prize in polarography, Hervosky, in the 1950s made errors in other electrochemistry papers, does this mean the prize be snatched back from him as well? Max Planck when he derived his blackbody radiation law, he did not believe much in atomic/molecular theory (later on he changed his views). He also received a Nobel prize.
If you read the citation/biographer by the Nobel Prize committee, the award was not due to some cycloalkane strain angle but the prize motivation was "in recognition of his services in the advancement of organic chemistry and the chemical industry, through his work on organic dyes and hydroaromatic compounds."
The Nobel Prize clearly says "Von Baeyer’s work was at once pioneering and many-sided. With admirable penetration and extraordinary experimental skill he combined dogged perseverance and, even at 70 years old, a youthful buoyancy in his work. He was careful never to overestimate the value of a theory. While Kekulé sometimes approached Nature with preconceived opinions, von Baeyer would say: “I have never set up an experiment to see whether I was right, but to see how the materials behave”. Even in old age his views did not become fixed, and his mind remained open to new developments in chemical science." (https://www.nobelprize.org/prizes/chemistry/1905/baeyer/biographical/)
You can see here, (good) scientists are open to change and are willing to listen to others. I found very good scientists to be very humble (with a small fraction of outliers), but only the mediocre ones are at times rigid and arrogant.

Nice response! I believe the "hydroaromatic compounds" part of the citation was tied to his strain theory. He assumed that the rings were all planar. He argued that, if van't Hoff and LeBel were correct about the tetrahedral carbon, then there should be two cyclohexane carboxylic acids, "equatorial" and "axial". On Neptune perhaps, not on Earth.

Unfortunately, he did not deliver a Nobel Lecture, otherwise we could have verified the story behind hydroaromatic compounds.


text:
API or database for verifying chemical reactions
Is there a database or API(rest calls) for verifying chemical equations?
Example, Na + Cl -> NaCl + Loss of energy.
So if I want to verify if there is a loss of energy or gain of energy?
Is there any online portal where I can write a chemical equation like above and verify whether its correct or not?
I know there are lots of paid and open source databases. Would be very much helpful if someone can point to me to top databases or sites.

I fear this may attract downvotes. You should add much more details on the why and what of the problem.

Edited the question

For a sense of why this is not easily feasible, should I accept the inputs from your example? Sodium: in what state? Chlorine: as a single atom? Or should it be in standard dichlorine form? Sodium chloride, in what state? Solid, in solution, or gas phase? Who's to say what's right?


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Problem in defining the Inductive effect
Inductive effect is introduced in many textbooks by giving an example of an alkyl halide $R-X$.
The bond C1-X is polar due to electronegativity difference, therefore X gains partial negative charge ∆(-), and C1 gains partial positive ∆(+). Due to some deficiency of electrons in the C1 , C2-C1 bond becomes polar and above process is repeated.
So by this point of view, if we take $R-Li$, as C1-Li bond is polar, Li gains ∆(+) and C gains ∆(-) and above thinking process is repeated again.
Thus we can conclude that Carbon chain can act as both +I and -I depending upon the situation on hand.

So why do we always consider an alkyl chain to be a +I group?

Example, if on a carbon chain, there is a carbanion, we still consider alkyl group to be +I, but it should have been the opposite.

This is going to be a redundant question...I thought this confusion would be finally answered :(

Well, is the $C-Li$ bond an organic bond?

@Firefox1921, what do you mean by 'organic bond'?

Also if you aren't satisfied with C-Li, we can take C-B (boron) as well, or even a carbanion attached to alkyl group.

I think that inductive effect is a relative concept, it depends on the type of substituents attached to the carbon chain. As Li is more electropositive compared to C or C is more electronegative with respect to Li. So, R group shows -I effect and Li shows +I effect.

I agree with @Manu here. A classic example is the Grignard Reagent. The $R$ group in the Grignard Reagent is a $-I$ group, while the Mg gains a partial positive sign.

@B.Anshuman What I meant is, that kind of bond is called an organo-metallic bond, it's not an ordinary $\ce{C-H}$ or a $\ce{C=O}$ bond that is usually found. The organo-metallic bond is also found in coordination complexes, so you have to look at them more carefully, as they maybe different in the way that they behave than, say a regular $\ce{-C-X}$ bond.

@Manu, your statement is what I have written as argument against the defining example of inductive effect.

@Firefox1921, I agree that bond nature might differ but it is still a majorly covalent bond in the case of R-CuLi. I am saying so because R has ∆(-) on it unlike the Grignard reagent or R-Li.

So, what are you trying to say @B.Anshuman. I'm sorry but I'm unable to understand what is the point you're making.

@Firefox1921, Manu has pretty much summarised what I said.

Further I am asking is that in many textbooks, R is always considered a +I group regardless of the substituent. Even though the substituent is a carbanion.

Let us continue this discussion in chat.

@Firefox1921, what do you conclude?


text:
In Parent chain choosing should one consider longest chain or chain with more substituents?
Should one take the longest chain or the one containing the prefix/suffix with most priority.Moreover if a shorter chain has two units of the main prefix will it be taken as parent?
examples:
2-ethyl butanoic acid or
3-carboxy pentanoic acid

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

You have already found the principal characteristic group, i.e. the carboxylic acid group $\ce{{}-(C)OOH}$ that is expressed as suffix ‘oic acid’.
According to Rule P-44.1.1, the principal characteristic group (suffix) has to be considered first in order to decide the parent chain for this compound. Only the second rule (P-44.3.2) is looking at the longest chain.
Therefore, the name of the first example that is given in the question is 2-ethylbutanoic acid rather than ‘3-carboxypentane’ since the principal characteristic group has to be considered before the longest chain.

Note that, according to Rule P-44.1.1, the parent chain has the maximum number of substituents corresponding to the principal characteristic group (suffix). Also note that the second example that is given in the question contains two carboxylic acid groups.
Therefore, the name of the second example is 2-ethylbutanedioic acid rather than ‘3-carboxypentanoic acid’ since the maximum possible number of substituents corresponding to the principal characteristic group has to be considered before the longest chain.

can ene or yne be a principal characteristic group?

meaning 2-propyl but-1,2-diene or 3-methylidene hex-1-ene? thanks.


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Does Fe(CO)6 really exist?
I was given a question to draw the structure of $\ce{Fe(CO)6}$ and tell the hybridization. Hybridization came out to be $\ce{sp^3d^2}$ and structure as Octahedral. But does $\ce{Fe(CO)6}$ really exist? I could not find anything about it on the internet.

Please see: Why is it wrong to use the concept of hybridization for transition metal complexes? Also related: Why is iron pentacarbonyl known, but iron(III) hexacarbonyl is not?

@Martin-マーチン1. Its a textbook question, although in the textbook it is written that using hybridisation is not always correct, but still questions are given.

@Martin-マーチン2.My question is completely different. The compound does not follow EAN rule or 18VE rule and there is no mention of it on the internet. I am just curious.

See this meta post for context. Our mod @Martin-マーチン, clearly explains why questions regarding 'hybridization' aren't accepted on this site.

@DanishJuneja First: Your textbook is wrong. This may make you sad, but that is the truth. You probably still have to use it, but if you are really curious, you keep in min, that it is wrong. Second: I know that your question is different; that doesn't make it unrelated. There was nothing more I pointed out about that. Lastly: I only provided context with the hope that it is helpful. I am sorry if that has offended you in some way.

@Martin-マーチン I understood, you wanted to me understand that the concept is obselete and questions should not be entertained.OK

There is an iron hexacarbonyl, not as a neutral compound but as a dication which satisfies the 18-electron rule. This along with its heavier Group 8 congeners is reported by Finze et al. [1]. The authors who report the tetrafluoroborate salts, also reference earlier work with the corresponding fluoroantimonates.  The reference was found in Wikipedia.
Reference

Finze, M.; Bernhardt, E.; Willner, H.; Lehmann, C. W.; Aubke, F. (2005). "Homoleptic, σ-Bonded Octahedral Superelectrophilic Metal Carbonyl Cations of Iron(II), Ruthenium(II), and Osmium(II). Part 2: Syntheses and Characterizations of [M(CO)6][BF4]2 (M = Fe, Ru, Os)". Inorganic Chemistry. 44 (12): 4206–4214. https://pubs.acs.org/doi/10.1021/ic0482483

So the question was given just for practice, I suppose.

@DanishJuneja Apparently school textbooks still teaches outdated concepts. Anyway, an organoiron complex of formula $\ce{[Fe2(µ-bdt)(CO)6]}$ has been known where bdt = benzene-1,2-dithiolate. See: https://www.unioviedo.es/jaclab/pub/p86.pdf

Thanks @NilayGhosh.  Just to be clear: The compound cited in the comment, with two metal centers, resumably would not have the octahedral coordination of a mono-nuclear hexacarbonyl.

Yes, I know that. I just wanted to show OP that homoleptic, charge-neutral, binary hexacarbonyl of iron does not exist.


text:
Why are van der Waals' forces between gas molecules strong near its critical temperature?
My textbook says

…easily liquefiable gases i.e. with higher critical temperatures are readily adsorbed as van der Waals' forces are stronger near the critical temperatures.

What I do understand is that at a temperature higher than the critical temperature $(T_\mathrm c)$ the gas molecules have higher kinetic energy which is why they're unable to form strong bonds.
But, what if the temperature is lower than the critical temperature? Or is my reasoning incorrect?

You're missing the point. I'm pretty sure this text was about adsorption in specific, rather high temp. and in such temp. when the critical is higher then adsorption stronger. I wonder why an issue stemming from such misunderstanding, got 4 upvotes already.

Critical temperature is  kind of measure of the strength of intermolecular van der Waals force.
Comparison of actual and critical temperature is kind of comparison of average kinetic energy of molecules with energy needed to break the intermolecular bonds.
So temperature itself does not say anything about the bonding strength. It only indirectly determine probability these bonds get broken by thermal molecular motion.


text:
Why molar volume takes range of values along triple line whereas the temperature and pressure are constant along this line?
In the P-V-T surface the triple line is the line parallel to P-V plane on which three states of matter are in equilibrium. Temperature and pressure are same at all points along the triple line. But we know that the value of one of the three state variables(pressure, temperature, molar volume) will be specified if other two state variables are specified. If we specify two state variables then the third state variable can't take range of values. But along the triple line for specific temperature and pressure there is range of values for molar volume. Can anyone explain why? Maybe I am missing something obvious.

The molar volume of solid, liquid, and gas phase at the triple point are different.  So, depending on the proportions that the three phases are present, the average molar volume of the combination will vary.

@ChetMiller But why the proportion of the three phases different, if they have same temperature and pressure. I mean to say if we keep the temperature and pressure constant there would not be any phase change i.e some solid converted into liquid or liquid converted to gas, so the proportion will be always same and the average molar volume is same. Also it will be great if you give an elaborate answer or refer to some material online which addresses this particular question(I couldn't find one). Thanks.

When a vapor and liquid are at equilibrium above the triple point, do we always have equal amounts of vapor and liquid?


text:
Two salts in mutual saturation, which one will precipitate first?
Let's say I have a satuated solution of two salts with a common ion e.g. KCl and KI, and they're at mutual saturation (let's say there are 75% by weight KCl/ 25% KI). If I slowly evaporate some water, which salt will precipitate first? What if there are 25% KCl instead? More generally, how to predict which salt will drop out of solution first?

Mutual saturation is not a thing. If one salt reaches its solubility product, it will precipitate. If another does, it will. If both do, both will.

This question is unsolvable at this stage. For example, you said your saturated solution consists of 75% $\ce{KCl}$ by weight. Is that mean $\pu{75g}$ of $\ce{KCl}$ in $\pu{100g}$ of solution? If so, it is impossible. Maximum solubility of $\ce{KCl}$ in water cannot exceed 25.5% at $\pu{20 ^\circ C}$. That's right, you forgot to include temperature. Solubilities are highly depend on temperature of the solution.

Good question would be this type: https://chemistry.stackexchange.com/questions/72700/can-other-substances-be-dissolved-in-a-saturated-solution. You can learn about a thing or two of solubility in those answers.

@MathewMahindaratne I could have been clearer, I mean weight of KCl : weight of KI is 75% : 25%

@IvanNeretin My bad on mutual satuation. I'm aware of Ksp, and I mean the concentration of both solutes at which one is at its satuation. Mutual satuation is a thing tho, and can be estimated from Ksp. For KI/KCl it should probably be 7.43M/2.14 M

I guess I answered my own question lol. thanks all!

Do not forget there is a common ion effect here as well.

A saturated solution of a mixture of the two salts would have such a large ionic strength that the notion of the Ksp being a constant just doesn't work, even using activities.


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Comparison of basic strengths of hydroxides of alkali metal group
I have read that $\ce{LiOH}$ is less basic than $\ce{KOH}$. But I don't know its reason. I have tried to frame a possible explanation.
My explaination:
I think that as $\ce{Li}$ undergoes more hydration than $\ce{K}$, $\ce{OH-}$ group will stick stronger to $\ce{Li+}$ ion. Due to this $\ce{Li+}$ will be reluctant to loose $\ce{OH-}$ and thus will be less basic. Is my explanation correct?
And, is this explanation correct for any other pair of hydroxides, eg, $\ce{Sr(OH)2}$ & $\ce{Ba(OH)2}$?

@Mithoron I don't think it does -- this question is specifically asking why $\ce{LiOH}$ is less basic; the linked question specifically says it's not asking "why", but just asking for correct $\mathrmp K_\mathrm b$ values. Appropriately, all of the answers at the linked question just then provide data & references. IMO, not a duplicate.

Yes that is it.
To paraphrase the wikipedia article HSAB theory
"Hard and soft (Lewis) acids and bases" (HSAB), also known as the Pearson acid-base concept, HSAB is widely used in chemistry. It assigns the terms 'hard' or 'soft', and 'acid' or 'base' to chemical species.

'Hard' applies to species which are small, have high charge states (the charge criterion applies mainly to acids, to a lesser extent to bases), and are weakly polarizable.
'Soft' applies to species which are big, have low charge states and are strongly polarizable. The concept is a way of applying the notion of orbital overlap to specific chemical cases.

Pearson evidently had revised values in the paper (which is behind a firewall):

Pearson, Ralph G. "Absolute Electronegativity and Hardness: Application to Inorganic Chemistry." Inorganic Chemistry, volume 27, number 4, 1988, pp 734–740. https://doi.org/10.1021/ic00277a030

The data from that paper is on this webpage: http://www.knowledgedoor.com/2/elements_handbook/chemical_hardness.html
\begin{array}{|c|c|} \hline
Ion & Chemical\ Hardness\ (eV) \\ \hline
\ce{Li+} & 2.39\\ \hline
\ce{Na+} &   2.30   \\ \hline
\ce{K+} &  1.92    \\ \hline
\ce{Rb+} & 1.85     \\ \hline
\ce{Cs+} &   1.71   \\ \hline
\end{array}

Is the trend of chemical hardness same for all other groups or are there any exceptions

@Akshat - In chemistry there are always exceptions.

Sir but is there any other explaination of the same

Other explanation is ionic character, which is increase going down on alkali metal group: $\ce{Li < Na < K ...}$.

Yeah its correct.
Basic character directly proportional to degree of dissociation of OH- ions
which is inversely proportional to hydration energy.
Since hydration energy of KOH is low compared to LiOH , therefore KOH will dissociate more into K+ and OH- ions increasing the basic character


text:
In NaCl electrolysis, why does chlorine ion get oxidized but not oxygen ion?
In $\ce{NaCl}$ electrolysis, on the positive electrode (aka. on the electrode where oxidation happens), $\ce{Cl-}$ ions are oxidized to $\ce{Cl2}$. But why don't we see any $\ce{O^2-}$ (naturally present in water's $\ce{OH-}$ ions) oxidized to $\ce{O^2}$ at all?
My pre-existing knowledge, may or may not be correct, that led to this question:
[1] I am under the impression that chemical reaction is a messy business in the real-world with many reactions reversible, many intermediary steps happening and many small reactions happening in the sideways. So the real-world scenario is never as neat as one clean equation.
[2] I am aware of the concept of electronegativity, and I also have feelings that
$\ce{O^2-}$ and $\ce{Cl-}$ should require a similar amount of energy input to get them oxidized. So we should really expect both reactions to happen.
With my pre-existing knowledge laid out above, I find it remarkable that not a trace of $\ce{O^2-}$ is oxidized at all. This led me to guess that the $\ce{OH-}$ ion itself may have imparted some degree of protection to the $\ce{O^2-}$ inside of it?

You would get dioxygen if the concentration of chloride in the vicinity of the electrode were so low that you preferentially oxidized water instead. Depending on the nature of your cell and the concentrations involved, this may or may not happen.

@Zhe But there seems to be a strong preference for Cl- to be oxidized first.

There is. But that can only happen if it's there. If you locally deplete chloride, and it's not replaced quickly, then you'll oxidize something else.

Eliu: Your observation relative to chlorine is correct. For an accurate technical answer, see my comments below, which I would recommend for extra points if you have to explain your experimental results. Else, more simple, and perhaps appropriate for a less advanced course argument, is espoused by my esteemed colleagues below.

@AJKOER I took one course in chemistry, an university first year 101, my whole life. I was initially just trying to understand fermi levels in semi conductors. But, one digression after another... and here I am. Who knew that a simple action of bumping an electron out of place can lead to soooo many equations and disciplines, microscopic and macroscopic.

I am afraid that as stated in another answer there is no simultaneous evolution of oxygen or chlorine during controlled electrolysis. It all depends on the concentration of ions in the solution. Keep in mind that the concentration of water is 55 M, so the ion concentration is nowhere near it no matter what you do!
The key reason is that the electrode potential of chlorine/chloride couple is quite different from that of oxidation of water, so if you have a high concentration of chloride ion, chlorine is oxidized first, once chlorine falls below a certain threshold, water begins to oxidize. Nernst equation can help in predicting that.

I agreed with explanation given by M. Farooq. However, since OP is really a novice to the field, I'd like to clear some point for OP's benefit.
The electrolysis cell has 3 species ($\ce{Na+}$, $\ce{Cl-}$, and $\ce{H2O}$), each of which can be either oxidized or or reduced. $\ce{Na+}$ can only be reduced while $\ce{Cl-}$ can be only oxidized. Water is a unique species, which can be either oxidized or reduced based on the situation:
Relevant reduction half-reactions:
$$\ce{2 Na+ + e- <=> Na^0_{(s)}} \qquad E^\circ = \pu{-2.71 V} \tag{1}$$
$$\ce{2 H2O + 2e- <=> H2_{(g)} + 2OH- } \qquad E^\circ = \pu{-0.828 V} \tag{2}$$
Relevant oxidation half-reactions:
$$\ce{2 Cl- <=> Cl2_{(g)} + 2e-} \qquad E^\circ = \pu{-1.358 V} \tag{3}$$
$$\ce{2 H2O <=> O2_{(g)} + 4H+ + 4e-} \qquad E^\circ = \pu{-1.229 V} \tag{4}$$
Since $E^\circ$ of all four half-reactions are negative, this need external power source. That's why it is electrolytic cell. We can eliminate reduction of $\ce{Na+}$, because it need extra power to proceed ($\ce{Na+}$ can be reduced if you use molten $\ce{NaCl}$). Thus electrolytic cell would be combinations of either equations $(2)$ and $(3)$ or equations $(2)$ and $(4)$:
$$\ce{2 H2O + 2Cl- <=> H2_{(g)} + Cl2_{(g)} + 2OH- } \qquad E^\circ_
\mathrm{EMF} = \pu{-2.186 V} \tag{5}$$
$$\ce{2 H2O  <=> 2H2_{(g)} + O2_{(g)} } \qquad E^\circ_
\mathrm{EMF} = \pu{-2.057 V} \tag{6}$$
The $E^\circ_\mathrm{EMF}$ value of equations $(5)$ and $(6)$ are so close that either redox reaction can happen with an electrolytic cell consists of dilute $\ce{NaCl}$ solution and a $\pu{3 V}$ battery.
Now consider the The Nernst Equation:
$$E = E^\circ - \frac{RT}{nF}\ln Q =E^\circ - \frac{2.303RT}{nF}\log_{10} Q $$
For equations $(5)$, $Q=\frac{[\ce{OH-}]^2}{[\ce{Cl-}]^2[\ce{H2O}]^2}$ and $n=2$, thus:
$$E_5 = E_5^\circ - \frac{2.303RT}{nF}\log_{10} Q = E_5^\circ - \frac{2.303RT}{2F}\log_{10} \frac{[\ce{OH-}]^2}{[\ce{Cl-}]^2[\ce{H2O}]^2}\\ = -2.186 + \frac{2.303RT}{2F}\log_{10} [\ce{OH-}]^2 [\ce{Cl-}]^2 - \frac{2.303RT}{2F}\log_{10} [\ce{H2O}]^2 \tag{7}$$
For equations $(6)$, $Q=\frac{1}{[\ce{H2O}]^2}$ and $n=4$, thus:
$$E_6 = E_6^\circ - \frac{2.303RT}{nF}\log_{10} Q = E_6^\circ - \frac{2.303RT}{4F}\log_{10} \frac{1}{[\ce{H2O}]^2}\\ = -2.057 + \frac{2.303RT}{4F}\log_{10} [\ce{H2O}]^2  \tag{8}$$
Accordingly, it is possible to increase $[\ce{Cl-}]$ such that, $E_5 \lt E_6$. As a result, redox reaction $(5)$ will predominate. As a reaction progress, $[\ce{Cl-}]$ decreases and at one point, redox reaction $(6)$ will predominate again.

Thanks for the expanded answer. Not for the OP though, Bard & Faulkner cringe at changing the sign of the electrode potential because they correctly say that sign of the electrode potential is invariant (it is an electrostatic sign). I also buy this argument after discussing this with a couple of leading electrochemists. I know changing the signs was taught in universities for oxidation potentials in 1970s / 80s books. What is your school of thought?

Now, I felt duty bound to try to understand what this Nernst equation is about. Also just by eye balling the equations, my summary is that, there are two ways to control what's get oxidized: Voltage threshholding and species concentration.

@M. Farooq: I'm one of those guys who learnt from those 70/80s books. I'm an organic  chemist who didn't get chance to learn those new concepts. Yet, I still think this way is easy to explain for novices without making it complicated. I hope you'd use your techniques in next time so we all can learn. I'm eager to learn them. Thanks for your kind comments. :-)

Mathew: Nice argument but it may be technically not appropriate to explain experimental results. So, in the context of test questions for those studying electrolysis, OK,  but I would not dogmatically assert it relating to lab work or advanced graduate studies.

A technical and commercially significant point has not been cited. It directly speaks as to why, in contravention of the comparative electrode potentials (and perhaps even a Nernst concentration equation argument already presented) that chlorine is actually created over oxygen. To quote a source:

Bubble overpotential is a specific form of concentration overpotential and is due to the evolution of gas at either the anode or cathode. This reduces the effective area for current and increases the local current density. An example is the electrolysis of an aqueous sodium chloride solution—although oxygen should be produced at the anode based on its potential, bubble overpotential causes chlorine to be produced instead, which allows the easy industrial production of chlorine and sodium hydroxide by electrolysis.

Both reactions happen at the anode, the proportion of which depends on the concentrations and of the nature of the electrode. The two reactions are :
$$\ce{2 Cl^- -> Cl_2 + 2 e^-}$$
$$\ce{2 H_2O -> 4 H+ + O_2 + 4 e^-}$$
The proportion of the second equation is reduced by using concentrated $\ce{NaCl}$ solutions and electrodes covered by Ruthenium. But there are no ions $\ce{O^{2-}}$ in water and in all its solutions

There are no O2-, but there are O2- in (OH)-. does (OH)- somehow shield O2-?

@eliu. Sorry! I do not understand your question ...

I meant, the O in (OH)-, the hydroxy group, is an ion. Does the hydroxy group somehow prevents O in it to be oxidized?


text:
What is the entalphy change with respect to pressure at Isothermal Process for *Real Gases*?
How much enthalpy of argon changes when the pressure is isothermally increased by $\pu{1 atm}$? (Gas should be considered as real gas under specified conditions) ($\pu{1 mol}$ of Argon gas is available at $\pu{25 ^\circ C}$ and $\pu{11 atm}$ conditions)
-Firstly I took the total derivative of Enthalpy function of pressure and temperature:
$$dH =\left(\frac{∂H}{∂P}\right)_T \cdot dP + \left(\frac{∂H}{∂T}\right)_P \cdot dT$$
Then divided terms with $dP$ because as the question asks I need $\left(\frac{∂H}{∂P}\right)_T$. After this I get:
$$\left(\frac{∂H}{∂P}\right)_V = \left(\frac{∂H}{∂P}\right)_T + \left(\frac{∂H}{∂T}\right)_P \cdot \left(\frac{∂T}{∂P}\right)_V$$
Third term is enthalpy change with temperature at constant pressure means $C_p$, so the equation turns into:
$$\left(\frac{∂H}{∂P}\right)_V = \left(\frac{∂H}{∂P}\right)_T + C_p \left(\frac{∂T}{∂P}\right)_V$$
so I rewrite this like:
$$\left(\frac{∂H}{∂P}\right)_T =\left(\frac{∂H}{∂P}\right)_V-C_p\left(\frac{∂T}{∂P}\right)_V$$
Then I used Chain rule, which is: if $z=f(x,y)$:
$$\left(\frac{∂z}{∂y}\right)_x\left(\frac{∂y}{∂x}\right)_z\left(\frac{∂x}{∂z}\right)_y=-1$$
I used this rule for writing a derivative term, $\left(\frac{∂T}{∂P}\right)_V$, with constants, and my equation turns:
$$\left(\frac{∂H}{∂P}\right)_T = \left(\frac{∂H}{∂P}\right)_V - C_p\left(\frac{\kappa T}{\alpha}\right)$$
where $\alpha$: expansion coefficient; and $\kappa T$: isothermal compressibility coefficient (kappa t).
In this way, I came up to this point but I couldn't find the exact answer. I Lost myself in thermodynamic differentials. Can anyone explain me? Where did I mistake or is my solution method incorrect?

Edit:
$$\left(\frac{∂H}{∂P}\right)_T = V - T\left(\frac{∂V}{∂T}\right)_P$$
If i wasn't wrong rhs of the equation turns into = V - T ( ßV) which equals to partial derivative of enthalpy with respect to pressure at constant temperature. But i'm confused here. There are other terms that can be used for ßV. For example alphaV or Vkappa t. Which one should i use? What is partial derivative of volume with respect to temperature at constant pressure means? Question asked about isothermally increasing pressure so should i use isothermal compressibility coefficient or isobaric compression coefficient or thermal expansion coefficient or what? Really confused here..
$$\left(\frac{∂H}{∂P}\right)_T = V - TßV$$ this is the last point i came to

If my editing is incorrect, please feel free to correct them as you mean.

Thank you for the arrangement.There was a little mistake but i fixed it myself :)

That derivation is challenging, and requires some non-obvious manipulations.  Here's my suggestion:   You can probably find, in thermodynamics textsbooks, a derivation of $\left(\frac{∂U}{∂V}\right)_T = \left(\frac{\alpha T}{\kappa}\right) - p$.  Take the techniques and manipulations used in that derivation, and apply them to your problem. If you are successful, you will get an answer that contains only V, $\alpha$, and T (no partial derivatives).

The easiest place to begin the derivation is from the following definition for the differential form of the enthalpy:
$$dH= dU + d(PV) = VdP + TdS$$
Taking the derivative wrt P at constant T gives
$$\left(\frac{\partial H}{\partial P}\right)_T= V + T\left(\frac{\partial S}{\partial P}\right)_T$$
It remains to evaluate $\left(\frac{\partial S}{\partial P}\right)_T$. To this end you can employ one of the Maxwell relations, starting from the definition of the differential form of the Gibbs free energy
$$dG = VdP -SdT$$
Because $dG$ is an exact differential, it follows that
$$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$
Substituting in the original equation gives
$$\left(\frac{\partial H}{\partial P}\right)_T= V - T\left(\frac{\partial V}{\partial T}\right)_P$$
You can then proceed by making appropriate substitutions and integrating the above equation, as explained in the answer by Chet Miller.
The last equation can also be written
$$\left(\frac{\partial H}{\partial P}\right)_T= V(1 - T\alpha)$$
where $\alpha$ is the thermal expansion coefficient of the substance :
$$\alpha= \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$$
Use of the coefficient is convenient if it does not change much over the pressure interval of interest (you can then treat it as a constant and pull it out of the integral). Sometimes other symbols such as $\beta$ are used; $\kappa$ is the isothermal compressibility and is something else - the coefficients can be related through the heat capacities but this is another topic!

Thanks for your answer, i appreciated!

The equation for the partial derivative of enthalpy with respect to pressure at constant temperature is given by:  $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P$$From the equation of state for your gas, you can evaluate the rhs of this equation.


text:
How can the torsional barrier of a polymer determined by molecular dynamics?
First consider a simple polymer, like polyacetylene:

Its torsional barrier can be determined by DFT calculations by considering "enough" units and performing a simple dihedral scan. "Enough" appears to be 5-7 units (J. Chem. Phys., 2014, 140, 054310)
But now I want to study a more complex polymer:

The same method does not work, because the structure is too flexible. Different values are obtained after multiple turns around the central unit. Is it possible to use molecular dynamics to find the effective/average rotation barrier?
My guess would be that we could simulate this polymer over a certain period of time, then make an histogram of the central dihedral angle. We could then calculate the torsional barrier by looking at the relative frequencies of the angles (e.g. if we get a flat distribution, the barrier is 0, if some angle is very disfavored, the barrier should be high. I'm not sure how to quantitatively calculate the barrier though)

Hm, what does your MD say to polyacetylene? If it can reproduce the results from DFT, I say you have a force-field that might well be up to the task.

Could you do multiple turns using DFT and then use a statistical analysis for each dihedral  angle (mod 2$\pi$) to get an idea about the potential?

@Karl Assuming the forcefield is not a problem, would the general concept be valid? I think that the ensemble used in the simulation would be critical, and I don't have too much experience with MD.

@Paul That's a good idea, although I don't know if sufficient sampling could be obtained with DFT for relatively large polymers. That's why MD came to my mind.


text:
Predicting in which medium will the oxidation occur

The image above contains the redox half-reaction for Mn2+ oxidation and their standard electrode potentials. The first two are in base medium and the second two are in acid medium. How can I predict in which medium the oxidation will actually take place? Thanks in advance!

Umm, it's a simple matter of comparing potentials. Why would you have any trouble here?

Schmidt, half cells don't have a full meaning. We cannot interpret their use without another half-cell. So in a table you see a lot manganese half cell, some of them are in basic medium and some of them are in acidic medium.
First of all your "other" half cell is missing. What do you wish to oxidize with manganese ions? From a practical perspective, the reducing substrate will decide what medium should be chosen.
It is your completely your choice to pick an acidic medium or alkaline medium. Let me give a practical example. If we wish to oxidize iron (II) ions to iron (III). Say, for experimental purposes, no precipitation or solids are required in the products. As a result one would choose an acidic medium to conducting this reaction.
However, if we somehow choose alkaline medium, we will form solid products such as MnO$_2$ or Mn(OH)$_2$.
Have you heard of Winkler's titration? It is used for determining the dissolved oxygen content in water with manganese (II) salt. Here we want a precipitate of Mn(OH)$_2$ to form, which can readily react with dissolved oxygen. So alkaline medium is used.
So the choice of alkaline or acidic medium for a redox reaction is totally up to you. The question of prediction does not arise.

Thanks for your response. To answer your first question, in my post I'm talking about $\ce{Mn^2+}$ oxidation with $\ce{O2}$

Right, so this is a perfect example of Winkler titration. In Winkler's titration, we want a Mn$^{2+}$ to precipitate as Mn(OH)$_2$. So we want a basic medium. This is immediately oxidized by dissolved oxygen. When we wish to have metallic precipitates as hydroxides or hydrated oxides, as products, basic medium is good for that.

I understand, but how can I prove, mathematically and thermodynamically wise, that the reaction is possible only in basic medium?

No you cannot- at least from scratch. There is some chemistry which is involved. You decide what products you want. Math or thermo won't tell you which products will be formed. May be you can do an exercise. Try to calculate the Ecell of oxidation of Mn(II) to Mn (IV) by oxygen gas  in acidic medium and in basic medium. I would choose the medium which shows higher (positive) Ecell. Find the appropriate half cells (not the ones listed in your question).

You would find them here in this book https://archive.org/details/oxidationstateso00lati

Your equation $(2$) and ($3$) are two equivalent ways of writing one and the same equation. Let's start by writing the Nernst equation for your second equation, which I repeat here $$\ce{O_2 + 2H_2O + 4 e^- <=>  4 OH^-}$$
The corresponding Nernst equation is, without the units Volt : $$\ce{E = 0.40 - \frac{0.0592}{4} log[OH^-]^{4} = 0.40  - 0.0592 log[OH^-]}  $$ If $\ce{[OH^-]}$ is replaced by $\ce{\frac{10^{-14}}{[H^+]}}$, the equivalent following equation is obtained $$\ce{E = 0.40 - 0.0592 log \frac{10^{-14}}{[H+]} = 0.40 + 14·0.0592 + 0.0592 · log [H+] = 1.23 + 0.0592 log[H+] }$$ And this last formula is exactly the Nernst equation corresponding to your $3$rd formula, which I repeat here, just for fun $$\ce{O_2 + 4 H+ + 4 e^- <=> 2H_2O}$$

I understand what you explained and thanks, but how exactly does this answer my question?

@Schmitt. Do as you want. You may use your second or your third equation. It does not matter. With any of these you will get the same result. It is not the case for equation (1) and (4). In a basic medium you must use equation (1). In an acidic medium the hydroxydes do not exist as they are dissolved by H+ ions. So in an acidic medium, you must use the equation (4)

It seems to me that we don't understand eachother, and I'm sorry, it might be my bad. My initial question was not which equation corresponds to which medium, but in what medium will the oxidation of the Mn2+ ion is actually possible and how can I demonstrate that.

Agree, "how exactly does this answer my question? "

OK, hand onto your seat, your electrochemical half cell reactions do not necessarily indicate anything about the underlying reaction mechanics.
You have to know the precise species involved to ascertain if they even exist at the specified pH, and, if existing, their solubility and decomposition pathway which can be pH-dependent relating to the medium.
Here is an example of the half cell reactions for the so-called Bleach Battery based on Aluminum metal corroding in acidified bleach containing HOCl and an inert electrode of copper.
At anode:
$\ce{Al -> Al^{3+} + 3 e-}$
$\ce{Al^{3+} + 3 OH- -> Al(OH)3 (s)}$
At the surface of the inert copper cathode:
$\ce{HOCl + e- -> .ClOH-}$
At pH < 5, the decompostion of the radical proceeds as follows:
$\ce{.ClOH- -> .Cl + OH-}$
$\ce{.Cl + .Cl -> Cl2 }$
and there is chlorine gas evolution.
Now, what happens in the alkaline bleach battery variation with Aluminum metal in NaOCl:
$\ce{NaOCl + H2O <=> Na+ + OH- + HOCl}$
So, largely the same except that now the formed radical behaves differently at pH >5 per the reaction:
$\ce{.ClOH- -> Cl- + .OH}$
So, no evolution of chlorine is apparent and just the precipitation of Al(OH)3
[EDIT] Interestingly, I did find this citation on the acidic Bleach Battery half-cell reaction, to quote:

(7) $\ce{HOCl + H+ + e- -> 1/2 Cl2 (g) + H2O Eo = 1.63 V}$

Note, this reaction would relate to the reaction mechanics of:
$\ce{H+ + e- <=> .H }$
$\ce{.H + HOCl <=> .Cl + H2O}$
$\ce{.Cl + .Cl <=> Cl2}$
which is an alternative possible path when H+ concentration is high and HOCl is low.
Further, in alkaline conditions, adding $\ce{OH-}$ to both sides of Equation (7), reduces to my general cited reaction as:
$\ce{HOCl (aq) + e- -> .ClOH- (aq)}$
Source: See Supplement Table S1 and click on Supplement F1 to download.
Followed by:
$\ce{.ClOH- (aq) -> Cl- (aq) + .OH (aq)}$
as I previously cited.

At least Maurice was talking about OP's reaction...


text:
Feasibility question: Is the following presented chemistry leading to very dilute H2O2 with fast created radicals, effective H2O2?
I start by noting that, theoretically starting by the action of oxygen on acidic FeCl2, one can form in situ hydrogen peroxide and other reactive oxygen species, albeit in very small amounts. The latter could then be vented employing microjets. This should result in the formation of powerful (and transient) radicals from the created dilute H2O2.
The underlying chemistry for hydrogen peroxide formation is outlined in this source, see Table S1 here . In the proposed application:
$\ce{ O2 (aq) + Fe^{2+} → Fe^{3+} + .O2− }$ ( moderately fast) )
$\ce{ . O2− + Fe^{2+} + 2 H+ → Fe^{3+} + H2O2}$  (fast )
$\ce{ H2O2 + Fe^{2+} → Fe^{3+}  + .OH + OH− }$ ( slow)
However, the slow nature of this last reaction can apparently be overcome (by a factor of a thousand!) per this 2013 paper: 'Fenton chemistry at aqueous interfaces', through the cited employment of microjets.
So, for bleaching, stain removal,..., is this a viable project to examine?

Slow nature of the last reaction? First reaction is slow, like million times slower, then the last. Seems it's all backwards. Fe is good at catalytically decomposing H2O2 not producing it.

Incorrect, per my reference Table S1, the classic cited Fenton reaction is one of the slowest. See rate for R1 and compare it to Copper chemistry R25, again one of the slower reactions. The metal auto-metal oxidation involving superoxide is reversal. see forward and reverse reaction for Copper R24 & R29. Reverse reaction with superoxide faster than Fenton for iron (see, for example, R7),

Minor reminder: please make sure you don't create tag duplicates. There is no need for h2o2 tag. We already have water tag.

It should be clear that my question has an intended broad range of important applications, not confined to any particular iron salt but the whole spectrum of applications that had been forever changed by Fenton chemistry.

My provided answer upon research is a total acknowledgment of this fact, and to the inventors, patent researchers, ,,,,, good news.

So, for bleaching, stain removal,..., is this a viable project to examine?

You may recall that rust stains on clothes or surfaces are one of the worst stains in the world. It is very hard to remove them. It is impossible to use iron salts for bleaching purposes. Those who bleach clothes run away from iron and manganese ions in water like plague or in modern times corona.
Fenton's chemistry is great for environment purposes because iron is among the less harmful elements for the soil & water, and the final product of hydrolysis of iron salts is a great flocculant-great for removing colloidal junk in water.
Instead of these random papers, I would suggest that you start from the very original, 1894, and see what he observed.
Fenton "Oxidation of tartaric acid in presence of iron". J. Chem. Soc., Trans., 1894, 65 (65), 899–911. It is a free access.

Iron ion can be chelated, not really employing FeCl2, except for possibly applications other than bleaching. Also,  science from 1894 better than 2013 ?

Chelation will not work because the chelate itself is an organic compound. Fenton's reagent or chemistry is used for destroying organic matter by free radicals. As to the comment "Also, science from 1894 better than 2013?" Indeed, reading original paper is often a good starting point for any serious researcher. Like chemistry, many seminal papers in mathematics are to be found in 18th century-19th century. That is why top tier schools often required the reading knowledge of French/Russian/German for math / chem for a PhD. Harvard still requires it for math.

WRONG on chelation here is one of many many references to quote"The enhancement in Fenton-like is due to the desorption of the entrapped pollutant produced by the addition of the chelating agent and its capability to complex natural metals of the soil" at this Science Direct Source: https://www.sciencedirect.com/science/article/abs/pii/S1385894711007492 .

I did find a source confirming the feasibility based apparently on extensive implementation.
It occurred upon my searching for microjets, where I came upon this reference: Tian, Y.; Sun, P.; Wu, H.; Bai, N.; Wang, R.; Zhu, W.; Zhang, J.; Liu, F. ‘Inactivation of Staphylococcus aureus and Enterococcus faecalis by a direct-current, cold atmospheric-pressure air plasma microjet’ in J. Biomed. Res. 2010, 24, 264–269.
It is cited in this 2019 work, published in International Journal of Molecular Science, 2019, 20, 5216; doi: 10.3390/ijms20205216 .
To quote pertinent passages from the work: 'Different types of plasma can be used in various biological fields, including disinfection/sterilization' including a type referred to as non-thermal plasma.  Here is a pertinent extract from Table I:

Classification:
Non-thermal plasma
Discharge Type:
Glow discharge, Corona discharge, atmospheric pressure plasma jet (APPJ), dielectric barrier discharge (DBD), micro-hollow cathode discharge (MHCD), Plasma needle, (MHCD), Plasma needle, Low-pressure plasma etc.
Examples:
Ozonizer, Plasma medicine, Volatile organic compound (VOC) treatment, Plasma modifications (coating, etching, activation, cleaning, nitration, etc.),

The work notes the following:

The potential of plasma technology in medical and dental applications is extremely broad. As well as disinfection/sterilization of medical and dental devices, the technology could be used to treat beds, desks, and floors [90]. Plasma technology may also have therapeutic potential [91–93]. Therapeutic uses include the treatment of skin diseases [94], blood coagulation [95] as well as dental treatment [96] and applications in dermatology such as chronic wound healing [97].

As to an explanation of underlying mechanics to quote:

Plasma treatment is an especially promising method because the mechanism of bactericidal action is unlikely to differ between multidrug-resistant and normal bacteria. The main mechanisms of bactericidal action in plasma are thought to involve exposure to reactive chemical species for which multidrug-resistant bacteria are unlikely to be resistant [55].

And further:

Reactive chemical species seem to be the principal inactivation factor in most cases, although this may vary depending on the method of plasma generation and whether the sample is exposed to direct or indirect plasma treatment.

Where the reference [55] above, in its extract cites:

Reactive chemical products (hydrogen peroxide-like chemicals), ultraviolet light (UV-A) and slight temperature elevations were observed during the operation of the gas plasma device.

which is also a confirmation of my cited reactive oxygen species (ROS) chemistry where, in the proposed case of wound treatment, for example, we have iron-rich blood with in situ apparently created H2O2, likely forming powerful disinfecting ROSs. Another example, is the more classic Fenton remediation chemistry, removing organic VOCs, with the likes, of yes, perhaps FeCl2.


text:
Magnetic moment of trivalent lanthanide cations

The effective magnetic moment $\mu_{\mathrm{eff}}$ of tripositive rare earth elements, is calculated by $$\mu_{\mathrm{eff}}=g_J\sqrt{J(J+1)}\mu_\mathrm{B}$$

Why can't we use normal formula as $\sqrt{n(n+2)}\mu_\mathrm{B}$ for them? I calculated $\mu_{\mathrm{eff}}$ for $\ce{Ce^3+}$ ion and got $2.54\mu_\mathrm{B}$ using first formula and $2.83\mu_\mathrm{B}$ using second one, there's a significant difference in two. How is the first formula derived?

This J(J+1) includes spin orbit coupling and the so-called normal formula considers only electron spin. With these lanthanides, most likely all these formulae are approximations.

The full formula is $\mu=\sqrt{4S(S+1)+L(L+1)}$ where $S$ is the sum of the spin quantum numbers, $S=5(1/2)$ for Mn(II) and $L$ is the orbital angular momentum quantum number., $L=2$ for a D state, zero for an S. The actual values are rarely as high as this eqn predicts due to interaction with other electrons in the molecule but no smaller than the value when $L=0$ which is the spin only value. (I've assumed that $g$ is exactly 2).

Your question, Why can't we use normal formula to find effective magnetic moment of tripositive rare earth elements (REs)? is answered excellently by  porphyrin' comment (vide supra). Thus, I won't attempt to answer it again except for to mention the calculated values from the equation, $\mu_\mathrm{eff} = g_J \sqrt{J(J+1)}\mu_\mathrm{B}$ (where $\mu_\mathrm{B}$ is Bohr magneton and $g_J$ is Landé factor), have good agreement with experimental values.
Table 1: Theoretical magnetic moment of tripositive rare-earth ions
$$
\begin{array}{c|ccc}
\text{Ion} & 4f^n & S^c & L^c & J^c & g^c & \mu_\mathrm{eff} = g_J \sqrt{J(J+1)}\mu_\mathrm{B} & \text{Observed }\mu^a \\
\hline
\ce{La} & 0  & 0 & 0 & 0 & - & 0.00 & -\\
\ce{Ce} & 1 & \frac{1}{2} & 3 & \frac{5}{2} & \frac{6}{7} & 2.54 & 2.4\\
\ce{Pr} & 2 & 1 & 5 & 4 & \frac{4}{5} & 3.58 & 3.5\\
\ce{Nd} & 3 & \frac{3}{2} & 6 & \frac{9}{2} & \frac{8}{11} & 3.62 & 3.5\\
\ce{Pm} & 4 & 2 & 6 & 4 & \frac{3}{5} & 2.68 & -\\
\ce{Sm} & 5 & \frac{5}{2} & 5 & \frac{5}{2} & \frac{2}{7} & 0.85 & 1.5\\
\ce{Eu} & 6 & 3 & 3 & 0 & - & 0.00 & 3.4\\
\ce{Gd} & 7 & \frac{7}{2} & 0 & \frac{7}{2} & 2 & 7.94 & 8.0 \ (7.23^b)\\
\ce{Tb} & 8 & 3 & 3 &6 & \frac{3}{2} & 9.72 & 9.5 \ (9.58^b)\\
\ce{Dy} & 9 & \frac{5}{2} & 5 & \frac{15}{2} & \frac{4}{3} & 10.65 & 10.6\\
\ce{Ho} & 10 & 2 & 6 & 8 & \frac{5}{4} & 10.61 & 10.4\\
\ce{Er} & 11 & \frac{3}{2} & 6 & \frac{15}{2} & \frac{6}{5} & 9.58 & 9.5\\
\ce{Tm} & 12 & 1 & 5 & 6 & \frac{7}{6} & 7.56 & 7.3\\
\ce{Yb} & 13 & \frac{1}{2} & 3 & \frac{7}{2} & \frac{8}{7} & 4.54 & 4.5\\
\ce{Lu} & 14 & 0 & 0 & 0 & - & 0.00 & -\\
\end{array}
$$
$^a$ Rare-earth and metals really show "Spin"?; $^b$ Reference 1; and $^c$ Reference 3.
The two experimental values from Ref.1 are in $\pu{emu mol−1 Oe−1}$ and have been converted appropriately to be able to compare with the values in hand. The reference also point out that crystals of mixed salt have approximately average of the pure monometal salts. The abstract of Ref.1 states that:

Magnetic moment and susceptibility measurements of single crystals of pure and mixed rare earth fumarates of gadolinium and terbium were carried out at room temperature. The experimental values of molar susceptibilities for $\ce{Gd2(C4H2O4)3.7H2O}$, $\ce{Tb2(C4H2O4)3.7H2O}$, and $\ce{GdTb(C4H2O4)3.7H2O}$ are $2.68 \times 10^{−2}$, $3.89 \times 10^{−2}$, and $3.18 \times 10^{−2}$ (in $\pu{emu mol−1 Oe−1}$), respectively. The calculated effective magnetic moments are in good agreement with the theoretical predictions on rare earth ions.

The magnetic properties of rare earth elements (REs) are due primarily to the electrons in the incomplete shell ($\mathrm{4f}$). In fact the magnetic moment per atom is determined by the number of electrons in the $\mathrm{4f}$ shell according to the Russell-Saunders coupling scheme. The derivation of the equation $(1)$ is partially explained in Ref.2:
$$\mu_\mathrm{eff} = g_{J} \sqrt{J(J+1)}\mu_\mathrm{B} \tag{1}$$
According to Ref.2:

The Russell-Saunders coupling scheme is normally adopted to describe the $\mathrm{4f}$ magnetism of the REs. The spins of the individual electrons are coupled to form a total spin $S$ by the exchange interaction and the individual orbital angular momenta are coupled to form a total orbital momentum $L$ by the Coulomb interaction. The state with the lowest energy is found from Hund’s rules by maximizing $S$ and thereafter $L$. The total angular momentum $J$ is given by $J = |L ± S|$, where the minus sign is used for less than half filled shells and the plus sign for more than half filled shells. The total magnetic moment due to the spin angular momentum is $µ_S = 2µ_B\sqrt{S(S + 1)}$, where $µ_B$ is the Bohr magneton. The total moment due to the orbital angular momentum is $µ_L = µ_B\sqrt{L(L + 1)}$. These moments process around the direction of $J$, therefore the magnetic moment due to the total angular momentum looks slightly more complicated: $µ_J = g_J µ_B\sqrt{J(J + 1)}$, where $g_J$ is the Landé factor [12].

The sighted Reference ([12]) is actually Ref.3 here. According to Ref.3:
$$
J =
\begin{cases}
L-S,  & \text{if $4F^n \le 7$} \\
L+S, & \text{if $4F^n \gt 7$}
\end{cases}
$$
Note: The equation $(1)$ does not obey experimental values of $\mu$ found for $\mathrm{3d}$ elements (Rare-earth and metals really show "Spin"?).
References:

B. Want, M. D. Shah, “Magnetic susceptibility measurements of pure and mixed gadolinium–terbium fumarate heptahydrate crystals,” Journal of Magnetism and Magnetic Materials 2016, 401, 391-393 (https://doi.org/10.1016/j.jmmm.2015.10.065).
I. L. M. Locht, Y. O. Kvashnin, D. C. M. Rodrigues, M. Pereiro, A. Bergman, L. Bergqvist, A. I. Lichtenstein, M. I. Katsnelson, A. Delin, A. B. Klautau, B. Johansson, I. Di Marco, O. Eriksson, “Standard model of the rare earths, analyzed from the Hubbard I approximation,” Phys. Rev. B 2016, 94, 085137 (https://doi.org/10.1103/PhysRevB.94.085137)(PDF).
Jens Jensen, Allan R. Mackintosh, In Rare earth magnetism: Structures and Excitations; The International Series of Monographs on Physics, J. Birman S. F. Edwards, C. H. Llewellyn, & Smith M. Rees, General Eds.; Clarendon Press: Oxford, United Kingdom, 1991.

I think, it was bad idea to think about the proof, I don't think there's a rigorous proof that exist.

Just curious, what goes on with europium?  It alone gives a big miss in the table.

I wasn't sure either. Some other reported values of Sm either agree with calculated value (some of them reported around 3). But, I sicked with one source.

@Zenix there is a derivation. It exists on Wikipedia & other sources. Relies on complicated QM


text:
Strong Base and weak acid reaction problem

I got the above problem correct by setting up the net ionic equation and finding the limiting reactant. I found that there were 0.01076 moles of NaOH left, and then calculated it's molarity which led to the PH. I'm confused by when you have a strong base and weak acid, when calculating the pH when all of the acid is used up, why can you disregard the acid's conjugate base in the calculation of the pH. Doesn't the reaction still create a certain amount of the acids conjugate base?

You have 100% butanoate at this point in the titration, e.g. a mixture of weak base and strong base. The strong base lowers the proton concentration by a lot so that there are insufficient protons available to react with the weak base. Or, said differently, the pH is so much higher than the pKa that you can ignore the acid/base behavior of the weak base.

To state @KarstenTheis's comment differently, yes there will be some butanoic acid in the solution. However because of the very high pH the concentration of butanoic acid is vanishingly small compared to the butanoate anion and hence insignificant in the problem.


text:
Which resonance structure would be favorably contributed to the actual structure?
Total number of pairs of resonating structure in which second is more favorable than first:

I know  (d) and (e) have second structure more stable.
(f) has both same.
According me to, (a) and (c) should have second structure unstable as oxygen has positive charge.
I don't have a clue about (b) and (g).
Please help me out.

(a) Positive charge on $\ce{C}$ versus positive charge on $\ce{O}$: Even though $\ce{O}$ is more electronegative than $\ce{C}$, its octet is fulfilled while $\ce{C}$'s is not (only 6 $e^-$s on positively charged $\ce{C}$). Therefore, second structure is more favored over the first one.
(b) Negative charge on $\ce{O}$ versus negative charge on $\ce{C}$: Here, both negatively charged $\ce{O}$ and $\ce{C}$ have their octet fulfilled. Yet, it is always stable when the negative charge is on more electronegative atom ($\ce{O}$) than on comparatively less negative atom ($\ce{C}$). Therefore, first structure is more favored over the second one.
(c) Positive charge on $\ce{C}$ versus positive charge on $\ce{O}$: Here, same argument of (a) applied here. Therefore, second structure is more favored over the first one.
(d) Charge separation versus neutral: it is always stable when the compound has no charge (neutral) than on has charge separation (with net charge is zero). Therefore, second structure is more favored over the first one.
(e) and (f) Kekulé structures of naphthalene and benzene, respectively: Of each pair, first and second structures have the same energy. Therefore, both structures of each pair have same stability equally favored.
(g) Positive charge on $2^\circ$-$\ce{C}$ in both structures: Since positive charge is on a $2^\circ$-$\ce{C}$ of each compound, the relative energy due to carbocation is same on both structures.  Yet, the first structure has its double bond conjugated to the carbonyl carbon while the second structure has isolated double bond to the carbonyl carbon (cross-conjugated). Therefore, first structure is more favored over the second one.

(+1) but in (g) double bond is not isolated, it's in conjugation with vacant orbital (carbocation). Hyperconjugation, inductive effect, cross conjugation would be better, ig.

With (e), the two Kekule structures are not exactly equivalent.  Huckel MO calculations on naphthalene indicate that the shared edge has more pi bonding than the left and right edges in the structures as drawn, so the second structure is favored.

There is no such thing as a more stable resonance structure.

@ Martin - マーチン: I know that but but that's what asking for.

@Oscar Lanzi: Your argument is probably correct. However, OP's level of education must not be able to understand computational chemistry.:-)

@math then that option should not be included until the student's education is appropriate to arrive at the correct answer.  That's my ulterior motive.

@ Martin - マーチン: I edited the question accordingly. But I wasn't sure that's what OP intended. I appreciate your concern.

I'll be giving my entrance exams this year and this was from a test. I wasn't provided with an answer for this. Thank you for the explanation !

Since computational chemistry isn't taught at this level I'll have to go with second structure of napthalene but thanks I'll keep that in mind

In (b) part negative charge is on less electronegative atom carbon so it will hold negative charge loosely in comparison to oxygen. So resonance can happen more betterly in second one.
In (g) part according to me a should me more stable as in a part there is linear conjugation but in b part there is cross conjugation.
Sorry for the English.


text:
What is the maximum number of hybrid orbitals an atom can have?
What is the maximum number of hybrid orbitals and atom can have? It is usually seen that an atom does not have more than 6 hybrid orbitals. Why is it so?

What about hydrated salts of lanthanides? For example, $\ce{[Eu(H2O)9]^{3+},[ReH9]^{2-}}$

The number 6 relates to a central atom in a molecule with octahedral geometry such as sulfur hexafluoride. $\ce{SF6}$

(image from wikipedia)
A central atom with 8 other atoms does occur in ionic crystals (body centered cubic), but not in any molecules to my knowledge (memory...).

(image from website location)


text:
Bond length in cyclic organic compound
Why is bond length $(a)$ greater than  that of $(b)$?

Is this because $\ce{-NH}$ is electron donating group (+M) and this results in a higher electron density in the ring causing repulsion?

Are you sure? The second structure tautomerises to phenol(almost 100%), so the C-O bond becomes single, which should cause its bond length to be more.

Firstly the tautomers of both of them, I think are aromatic. b becomes phenol, which is of course aromatic while a has a lone pair of nitrogen which makes it aromatic. So both of them have bond lengths greater than the regular length expected. But, when the resonance structures are made for a, you would find greater contribution of structures with single C-O bond than those with double bond. So, this may be a possible reason for that.

@user93057 The first structure's tautomer is non-aromatic, since it still has an sp3 carbon.

Tautomerisn should be let apart in such a discussion. The question does compare those two structures and not other molecules. The comments above are just related but dangerous as for they somehow push to treat tautomers as they are limiting forms.

So, although I stated in the comments that the second structure is very unstable and converts to phenol, as @Alchimista pointed out, tautomeric structures are different compounds and we need to solve the problem based on the compounds that are given to us, regardless of how stable they are.
However, resonance is operative in both of the above compounds. A compound such as this exists as a resonance hybrid, so we need to analyse the bond lengths in all the canonical structures and judge their relative contributions to the resonance hybrid.
Read this extract:

As noted above, we can more accurately describe the bonding in a molecule or polyatomic ion using the (weighted) average of its resonance structures. One model for estimating bond orders and charges in a compound is to simply take the average of those values from all (important) contributing resonance structures.

Note that here 'weighted' refers to giving resonance structures which are more stable more weight.
Try drawing out the resonance structures of both the compounds given above. Alternate resonance structures will have more contribution in the first as opposed to the second, as  resonance in the second structure leads to an incomplete octet on the carbon which becomes a $\ce{C+}$, while in the first, only an $\ce{N+}$ results which is still a complete octet. So resonance structures in the first are more stable, and thus, the single bond character of the CO bond in the first structure increases more than in the second, and that causes its bond length to increase. So, the bond length of (a) is greater.
References: Resonance Structures, Chemlibre texts

There is no such thing as a more stable resonance structure. See the comment of  Martin - マーチン in https://chemistry.stackexchange.com/questions/133010/resonance-stability/133017?noredirect=1#comment257877_133017.

Alright, what word do I use to indicate structures that 'have more contribution to the hybrid'?

You can probably used "favored"!


text:
Carbon-Carbon double bond rotation

I was solving this question and I thought the answer was B>C>A as more the resonance, more the pi electrons are delocalized so more free movement.
However the answer mentions it as B>A>C.
What is wrong with my reasoning?

There is no question in the screenshot. How can there be an answer then?

Um.... There is Actually. Maybe you are unable to load it

I only can read "compare c-c bond rotation across A,B,C" and then some answers to choose from. I'd expect a question like "sort the structures in respect of increasing energy barriers for the rotation around the double bond which is connecting the rings" or sth like that.

They have mentioned the rotation around the bond that they have shown with a curved arrow. And compare how freely each can rotate about it. Compare means arrange in either descending or ascending order

bro go for Bayer strain theory! I can see hybridization of all the surrounded atom is mostly same

Here you need to consider two factors.

The one which follows $4n+2$ rule.
The stability of the ring.

When we see the IIIrd one, the structure don't follow the $4n+2$ rule when the double bond breaks. That means, that the double bond will be reluctant towards breaking. Hence its stability will be least.
COMING TO NEXT ONE
The order of stability of the ring is in the order: 6−member > 5−member > 7−member > 4−member > 3−member.
So when we combine these facts we find the correct order is B>A>C.

Why won't it be aromatic? The electron pair can go to either ring right?

if the electron shifts to 7 membered ring, it will follow 4n rule and if the electron shifts to 3 membered ring it will follow 4n rule. I hope you got the answer else draw all possible structure and and check whether it follows 4n+2 rule or 4n rule

Oh..... Yes thanks a lot I get it now

Your stability order is irrelevant here, as it pertains saturated rings.


text:
Find stereo isomers of camphor
The compound given is camphor:

We have to find the total number of stereoisomers for this compound. Now I can see two chiral carbons so I thought the number would be $4$ but it turns out that there are only $2$.
I can't quite see why. I realize that it is a bridgehead compound which would make it non-planar but how does that affect stereochemistry? Any help would be appreciated.

The bridgehead atoms have to be inverted simultaneously. You can't have a hydrogen or methyl group pointing inside. Technically, we should ask how many stereoisomers are reasonable.

To follow-up on @Zhe's answer, the enantiomer arises by inverting both stereoisomers. However, you are correct conceptually. Inverting only one stereocenter at a time gives a pair of "trans" enantiomers that are too unstable to exist.


text:
Is it possible to estimate the abundance of conformers from J values?
I have a system of two disubstituted cyclohexanes and I want to estimate the abundance/ratio of the two based on the J values and if possible, roughly estimate the difference in free energy.

To get the extreme values of the J constants I have been provided with the data from the system on the image below.

This is all the information I have been provided with.
Is it possible to approximate the ratio and free energy difference and if so, how?

I suspect there might be something wrong with the problem, since the averaged values should be bracketed by the extrema. However there is only one extremum larger than the average values. It should be possible to bracket each of the averages by separate pairs of extrema.

Karplus curves should come in handy.


text:
Ozone in ozone layer
If ozone is a gas, and gases have the highest entropy, then how does the ozone gas stay within a few layers of the atmosphere, even though they span for kilometers?

Entropy and ozone at the top of the atmosphere are two unconnected things. Harmful UV rays from the Sun are unable to penetrate deep into the atmosphere because they encounter oxygen molecules. They absorb the UV and form ozone on the top. If ozone reached us, nobody could survive. It is a harmful thing. Read more about the https://en.wikipedia.org/wiki/Ozone%E2%80%93oxygen_cycle

related: https://chemistry.stackexchange.com/q/41261/7951 https://chemistry.stackexchange.com/q/81971/7951 https://chemistry.stackexchange.com/q/43498/7951

@M.Farooq - For clarification - there's plenty of ground level ozone around. It's not good for us, but it's not so bad at low concentrations that we don't survive. I suspect most people are familiar with the smell of ozone, even if they don't know what they're smelling.

Right, Andrew, ozone at part per million levels it is not that great! It immediately hurts the throat if you are sensitive. If you recall the smell of older photocopy machines, the smell is not that pleasant but fresh rain is of wonderful but that must be parts per billion conc.

Plus some rare highly reactive agens like peroxyacetylnitrate can be created, what reportedly acted in past during some Los Angeles smog situations as plant defoliant. Additionally, ozone is the starter of immission damage to conifer trees by SO2.

It does not have much to do with entropy, rather with the way and place of ozone creation.
Stratospheric ozone is produced typically at altitude 20-30 km by UVC radiation with $\lambda \lt \pu{280 nm}$:
$$\ce{O2 + \nu -> 2 O}$$
$$\ce{O + O2 -> O3}$$
See ozone cycle as courtesy of @M. Farooq.
Ozone in lower troposphere troposphere is created by UVB (280-320 nm)  (+UVA >320 nm ??) mostly due catalytic effects of nitrogen oxides that come typically of the oxidative smog of the Los Angeles type:
$$\ce{NO2 + \nu -> NO2^{*}}$$
$$\ce{NO2^{*} + O2  -> NO + O3}$$
$$\ce{2 NO + O2 -> 2 NO2}$$

Technically, the concentration and positioning of ozone in the atmosphere likely relates directly to both ozone formation, stabilization and a destruction cycle.
While formation creation cycles has already been outlined, I would like to add more detail on the chemistry based on this source:
Creation:
$\ce{O2 + ℎν → 2 O• }$
$\ce{O• + O2 → O3 }$
which is followed by the 'ozone–oxygen cycle' where the ozone molecules formed by the reaction above absorbs radiation in wavelengths between UV-C and UV-B.
$\ce{O3 + ℎν(240–310 nm) → O2 + O }$
forming atomic oxygen which can further react with dioxygen:
$\ce{O + O2 → O3 + kinetic energy }$
The cited net effect of the ozone–oxygen cycle is the conversion of  penetrating UV-B light into heat, with no net loss of O3. This cycle reputedly keeps the ozone layer stable and results in the protection of the lower atmosphere from harmful UV radiation.
Removal of Ozone:
$\ce{O3 + O· → 2 O2}$
$\ce{2 O· → O2 }$
Also, per a source, on the net effect of the two reactions, to quote:

$\ce{2 O3 → 3 O2 }$
This reaction proceeds more rapidly with increasing temperature.

So, higher colder altitudes are more conducive to the preservation of created ozone.
On its destruction cycles, to quote a source:

Ozone is a highly reactive molecule that easily reduces to the more stable oxygen form with the assistance of a catalyst. Cl and Br atoms destroy ozone molecules through a variety of catalytic cycles. In the simplest example of such a cycle,[11] a chlorine atom reacts with an ozone molecule (O3), taking an oxygen atom to form chlorine monoxide (ClO) and leaving an oxygen molecule (O2). The ClO can react with a second molecule of ozone, releasing the chlorine atom and yielding two molecules of oxygen. The chemical shorthand for these gas-phase reactions is:
$\ce{Cl· + O3 → ClO + O2}$
A chlorine atom removes an oxygen atom from an ozone molecule to make a ClO molecule
$\ce{ClO + O3 → Cl· + 2 O2}$
This ClO can also remove an oxygen atom from another ozone molecule; the chlorine is free to repeat this two-step cycle.

Now, pollutants in the atmosphere include chloro-(and bromo-) organics, which can, under UV photolysis, liberate a chlorine (or bromine) radical.
So, as one gets closer to the surface of the earth, radiation-induced ozone formation is abated, warmer temperatures are unfavorable and any diffused ozone entering the lower layers, could be further subject to attack by increasing concentration of photosensitized pollutants engaging in catalytic destruction cycles.
And, that is why the ozone layer resides at a high altitude, maintains itself and is not found closer to the surface of the earth.


text:
How to achieve regioselective (base-catalyzed) enolization of a carbonyl group in a diketone?
One can convert a ketone to an enolate, either base- or acid-catalyzed.
Let's say you now have a diketone. You choose to treat this compound with base (e.g. hydroxide anion, alkoxide anion, LDA/THF, NaH).
How can one control the regioselectivity of this enolization reaction such that only one specific carbonyl group is converted to an enolate group?

You should read up on kinetic vs. thermodynamic control. But you're probably going to end up with a mixture.

Selective protection, as shown here might help...

As @Zhe and @Zenix have mentioned, you have to hope that there is a difference in the rate of enolate formation to differentiate the two ketones.
To do this cool a rigourously dried solution of the diketone and a trapping reagent ($\ce{Me3SiCl}$ if you're doing an anhydrous workup otherwise TIPS-Cl or MeI) to $\mathrm{-75 \, ^\circ C}$ and slowly add 1 eq. of LDA keeping the reaction mixture cold. This will trap the first formed enolate (kinetic enolate). Allow to warm, dilute with hexanes, filter (or wash with aq if you used TIPS or MeI), concentrate and analyse.
If this gives a mixture then you try making the more stable (thermodynamic enolate) and trap that. This means forming the enolate without the trapping compound ($\ce{Me3SiCl}$, MeI or TIPS-Cl), giving it time to equilibrate then adding your trapping compound. This might be done with NaH in dry DMF, left to stir at $\mathrm{-30 \, ^\circ C}$ for an hour then the trap is added, worked up and analysed.
Even then with this example I think you'll get a mixture, but you don't know until you try. If you do get a mixture then you have to rearrange your synthetic pathway so that you don't have this problem - you'll probably have a different one but that's the  way it goes!

Better to add ketone to LDA because it minimizes the generated enolate from exchanging with ketone that has not been enolized.

I agree that for trapping the kinetic enolate of a single ketone that is a good way to go, but in this case as it is a diketone I think it risks forming the substituted di-enolate.


text:
Seeing a case where both + and - mesomeric effect is shown
I came up with that -N=O group shows both +M and -M and I understand that. In my textbook it was written that " -N=O group attached to benzene shows both + m and -m depending on the nature of other groups attached to benzene" I really couldn't get it,how are that groups how to see owe to which group N=O will show +m and by which it will show -m?

https://chemistry.stackexchange.com/questions/57044/why-is-the-nitroso-group-a-deactivating-group-for-electrophilic-aromatic-substit

There answer is written that-N=O becomes  o-p directing in both electrophilic and nucleophilic substitution. I didn't get this why it will be o-p directing when ring is electron rich when -NO is showing -M . Shouldn't it be meta directing in nucleophilic substitution when ring is electron rich?


text:
Safe coating to make a 3D printed respirator mask airtight
What is a good example of something safe that could be applied as a coating layer to the walls of a 3D printed respirator mask to make them truly airtight (seal pores and fissures) while (ideally) satisfying the following criteria?:

Not air permeable at the pressures created when a person inhales
No (or low) VOC out-gassing from the coating
Does not impose health risks to the respirator wearer
Can be applied by dipping (or maybe spraying)
Forms a thin layer that is flexible enough not to crack during normal use of the mask

Here are some materials I have been considering for the job:

White glue (school PVA glue): Seems safe. Is is airtight?
Water based liquid rubber: some SDS suggest it's safe. It's marketed as watertight (but is it AIRtight?)
Some sort of common paint?

If someone here could suggest some potential coating materials or knowledge bases where I could check their non-permeability/safety, I would be really grateful.
As I'm not a native speaker, I apologize for any broken grammar.

The problem: It is well known that 3D printed objects are usually not exactly airtight, due to micro-fissures. To make things worse, I ordered a DIY 3D printed respirator mask (composed of ABS plastic) that is visibly permeable to air (one can literally see the tiny gaps all across it's walls' surfaces through which air and viruses could come in when the wearer inhales).
P.S.: Silicone caulk might not be an option, since it's too viscous and would be a mess to apply to the outside of the respirator, rendering an unprofessional finishing. I should note that I could possibly apply it only to the inside of the walls, but I intend to coat all surfaces, just to be extra safe.

How are you going to breathe if its truly air tight? What am I missing?

You need something flexible and airtight. Could neoprene fabric, like it's used for diving suits do the trick, esp. for prototyping? Otherwise, I thought of injection moulded silicone, as it's used in diving goggles.