text:
Is constant pressure heat capacity = entropy during adiabatic reversible expansion of perfect gas?
For any process, ∆G = ∆H - ∆(TS).
Apply this on reversible adiabatic expansion of perfect gas, since the entropy of system does not change during this process, S is constant so
∆G = ∆H - S∆T
The process is reversible, so
∆G = 0
and hence
∆H = S∆T
For perfect gas, assume heat capacity independent of temperature,
∆H = Cp∆T
and so
Cp∆T = S∆T.
∆T is not zero during adiabatic reversible expansion- temperature drops. So divide both sides by ∆T gives
Cp = S
Is this legit? If not where is my mistake? I have a hunch that I did something terribly stupid. The final result seems to be so counter-intuitive because
The Cp and S are constant throughout the process, so at the end of the process the two will still be the same. By measuring Cp, even though your system is not doing adiabatic reversible expansion, you will get S. So we are able to measure S?

dH = Cp.dT is valid for isobaric process, but you consider isentopic adiabatic process.

@Poutnik I thought: dH always = CpdT for perfect gas. Just that dH not the same as dq if process is not isobaric. Enthalpy is a state function!

Yes, H is a state function. but dH=Cp.dT describes the state path via the constant pressure path. If the pressure changes with T, the final state will be different and so does H. Additionally, what does the index p in Co mean ?

@Poutnik The index means p is constant. But for a perfect gas, enthalpy is independent of pressure. For perfect gas, enthalpy depends on Tp only. If Tx2 and px2, let enthalpy change be a. If Tx2 and Vx2, the enthalpy change will still be a. Correct me if I am wrong.

For an adiabatic reversible expansion of an ideal gas, dH=TdS+Vdp=Vdp.  So, $C_pdT=Vdp$.  What's the problem?

@ChetMiller The problem has been stated in the title. I am looking for reason that my understanding being wrong (if any), not reason that correct understanding being correct.

CpdT=Vdp. dG=Vdp-SdT=0 for reversibility so Vdp=SdT. Sub into first equation, CpdT=SdT, so Cp=S?

Who says dG is zero for a reversible adiabatic expansion?

Inspired by Chet Miller.
The fallacy in the argument is that, ΔG = 0  because the process is reversible. This is not true and so leading to wrong final result.
ΔG is a signpost for spontaneity of a process only if the process occurs through constant temperature and pressure. Likewise, that ΔG = 0 means reversibility, is true only if the process happens through constant temperature and constant pressure.
Adiabatic expansion clearly involves change in temperature, so ΔG is not a signpost for spontaneity of adiabatic change, and hence ΔG ≠ 0 for reversible adiabatic expansion.


text:
Which compound will have the most stable enol form?

Which compound will have the most stable enol form, Phloroglucinol or 4-Pyridone?

My Thoughts:
Both in their enol form are aromatic. I am not sure how do I compare their stabilities. The answer given is 4-Pyridone though.
Any help would be appreciated.

Related (but, not a dupe): Stability of 4-pyridone vs 4-pyridinol

The most questions about keto-enol tautomerism can be answered by their NMR spectral data (e.g., evidence of existing enol-form of acetylacetone at room temperature). According to its $\ce{^1H}$-NMR spectral data, it was declared that phloroglucinol, in its neutral form, exists nearly exclusively in its aromatic form (Ref.1). However, their dianion form (e.g., disodium phloroglucinolate) favors keto structures. These results were further analysed and confirmed by kinetic studies (Ref.2).
Wikipedia states that:

4-Pyridone is an organic compound with the formula $\ce{C5H4NH(O)}$. It is a colorless solid. The compound exists in equilibrium with a minor tautomer, pyridin-4-ol.

Martin - マーチン has given here a nice answer using computational evidence for this preference  of 4-Pyridone (data are from Ref.3).
Based on these, it is clear that in their neutral forms, phloroglucinol has the most stable enol form, existing in predominantly in phenol form (1,3,5-trihydroxybenzene).
References:

R. J. Highet, T. J. Batterham, “The Structure of the Phloroglucinol Dianion,” J. Org. Chem. 1964, 29(2), 475-476 (https://doi.org/10.1021/jo01025a501).
Martin Lohrie, Wilhelm Knoche, “Dissociation and keto-enol tautomerism of phloroglucinol and its anions in aqueous solution," J. Am. Chem. Soc. 1993, 115(3), 919-924 (https://doi.org/10.1021/ja00056a016).
Peter I. Nagy, Giuliano Alagona, Caterina Ghio, “Theoretical Investigation of Tautomeric Equilibria for Isonicotinic Acid, 4-Pyridone, and Acetylacetone in Vacuo and in Solution,”  J. Chem. Theory Comput. 2007, 3(4), 1249–1266 (https://doi.org/10.1021/ct6002252).

So for this problem, we'll have to rely on the experiments rather than theory?

Theories are mean to change.:-) They changed according to experimental values, which are real.

So,  just to be double clear,  according to you phloroglucinol has more stable enol form than 4-pyridone right?

You are Correct.

In case of pholoroglucinol, its ketoform has 3 $\ce{C=O}$ bond which make its keto form more stable as compared to the keto form of 4-pyridone which has to offer only one $\ce{C=O}$.

On the subject of 4-pyridone, Wikipedia says this:

4-Pyridone is an organic compound with the formula C
  5H
  4NH(O). It is a colorless solid. The compound exists in equilibrium with a minor tautomer, pyridin-4-ol.

So as far as I know, the aromatic enol form of this guy is lesser in equilibrium content than that of phloroglucinol.
The reason for this is somewhat related to intermolecular H-bonding.
An abstract from this paper says:

The N−H···O H-bonding enthalpy between 4-pyridones connected in a chain of H-bonds can achieve 23 kcal/mol for the most central H-bonds, while that between two 4-pyridones is 9.90 kcal/mol based upon DFT calculations on the counterpoise-corrected potential energy surfaces. That the range of enthalpies for N−H···O H-bonds can vary from as little as 2 to as much 23 kcal/mol depends primarily upon the polarizability of whatever internally connects the N−H and CO within the H-bonding molecule, which are two parallel −CC− entities in 4-pyridone. The contribution of covalent or charge-transfer interactions between the π-systems of adjacent 4-pyridones is small.

The following image shows how such a chain may exist

(Red- oxygen, blue- nitrogen, grey- carbon, white- hydrogen)
Hence,the enol content of phloroglucinol would be more than that of 4-pyridone
Also, an interesting point to note about phloroglucinol as well(as per this paper) is that for the neutral compound, the keto tautomer is undetectable spectroscopically. Upon deprotonation, the keto form predominates, possibly due to the conjugation present in the sigma complex formed(but the reasoning is just an educated guess)

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text:
pH implications of redox balancing half reaction method
I have seen similar questions, but I’m asking a new question as far as I know. Why is water instead of hydroxide ions used to balance oxygen atoms in redox half equations? asks about using hydroxide instead of water and that is closest.
I have seen example problems where $\ce{H+}$ is produced or consumed, $\ce{OH-}$ is produced or consumed, and water is used to balance. My thought is that each of these changes affects the $\mathrm{pH}$ or $\mathrm{pOH}$ of each solution.
If the solution is already an acid, there is an abundance of $\ce{H+}$ and it makes sense to me that a reaction may use excess $\ce{H+}$ to make the solution more neutral. One could make the same argument about using up excess $\ce{OH-}$ in a basic solution. What doesn’t make sense to me is adding to the basic or acidic nature. See example below which was specified to be in a basic solution. It appears to be making the solution more basic by producing more $\ce{OH-}$ instead of using it up. Shouldn’t that be less likely to occur? Why not? What am I missing?
$$\ce{6I- (aq) + 4H2O + 2MnO4- (aq) —> 3I2 + 2MnO2 (aq) + 8OH- (aq) }$$

Just because it is catalyzed by a low pOH doesn't mean that the reaction makes the solution more acidic. Often, the pH or pOH of the solution only catalyze the reaction because they favor the reaction mechanism, not the products.

This reaction cannot occur in solution, because it would produce both $I_2$ and $OH^-$. Unfortunately $I_2$ and $OH^-$ cannot exist simultaneously in solution : they would react and produce a mixture of iodide $I^-$ and iodate $IO_3^-$ ions, according to the equation : $$\ce{3 I_2  +  6 OH^- -> 5 I^- + IO_3^- + 3 H_2O}$$

@Maurice that came directly from the textbook. I believe you, but that’s why I’m frustrated. Makes no sense to me half the time.

It may happen that textbooks are wrong. They just propose exercices to see whether the student has understood the theory. It is not the first time that it happens. I have once seen a problem where the student had to calculate the pH of a solution 20 M in H2SO4.

If the reaction happens at neutral or mildly alkalic $\mathrm{pH}$:
$$\ce{6I- + 4 H2O + 2 MnO4- -> 3 I2 + 2 MnO2 (s) + 8 OH-}$$
If the reaction happens at mildly aciding $\mathrm{pH}$, it could be as well:
$$\ce{6I- + 8 H+ + 2 MnO4- -> 3 I2 + 2 MnO2 (s) + 4 H2O }$$
In any case, either spending  $\ce{H+}$ either production $\ce{OH-}$ increases $\mathrm{pH}$. Both iodide and permanganate carry a negative charge while their spent forms are neutral. This negative charge must be passed to other anions like $\ce{OH-}$ or neutralized by $\ce{H+}$.
If one considers further disproportionation of iodine to iodide and iodate in alkalic solutions, then the total reaction is:
$$\ce{ I-  +  H2O + 2 MnO4- ->  IO3- + 2 MnO2(s) + 2 OH-}$$
In strongly alkalic it could be also
$$\ce{ I-  + 6 OH- + 6 MnO4- ->  IO3- + 6 MnO4^2- + 3 H2O }$$
It may be obvious if the equation is written together with spectator ions:
Mildly acidic:
$$\ce{6KI + 8 HA + 2 KMnO4 -> 3 I2 + 2 MnO2 (s) + 4 H2O + 8 KA}$$
resp neutral/mildly alkalic
$$\ce{6KI + 4 H2O + 2 KMnO4 -> 3 I2 + 2 MnO2 (s) + 8 KOH}$$
resp alkalic
$$\ce{ KI  +  H2O + 2 KMnO4 ->  KIO3 + 2 MnO2(s) + 2 KOH}$$
resp. strongly alkalic
$$\ce{ KI  + 6 KOH + 6 KMnO4 ->  KIO3 + 6 K2MnO4 + 3 H2O }$$

Okay. That does help, but KOH is still a base and would still affect the pH. In other words, I add KI and KMnO4 which are both neutral and I suddenly arrive at a base. Why?

Thanks for editing the notation. I’m not familiar with it but I’ll try to figure it out.

Because all anions of acids are removed and nothing but OH- can carry negative charge. Do you see other option ? If suddenly OH- is created, pH suddenly increases.

What you are missing is understanding of general nature of redox reactions. Redox reactions need two half reactions to complete. You can't arbitrarily select these two half reactions. They has to be chosen as instructions given in the problem. For example, for your given problem, it should be noted the medium of the reaction, whether it is acidic or basic or neutral.
Suppose the question asked is: Balance the following redox equation in acidic medium.
$$\ce{I- (aq) + MnO4- (aq) -> MnO2 (s) + I2 (aq)}$$
To solve this, you can directly get two half reactions from a textbook or balance it old-fashioned. The oxidation half reaction is:
$$\ce{2I- <=> I2 }$$
It is already balanced by mass, so add electrons (no mass) to balance the negative charge:
$$\ce{2I- <=> I2 + 2e-}$$
Now oxidation half reaction is completed. The reduction half reaction is:
$$\ce{MnO4- <=> MnO2 }$$
However, it was stated that reaction is in acid medium. Therefore we should add acid to reactants' side:
$$\ce{MnO4- + H+ <=> MnO2 }$$
Now, we have to balance mass first. Since the reaction is in acidic aqueous medium we have plenty of $\ce{H2O}$ molecules around to balance oxigen as well as hydrogen molecules (by $\ce{H+}$ ions as needed):
$$\ce{MnO4- + 4H+ <=> MnO2 + 2H2O }$$
Now equation is balanced by mass. Doing so, we increase net positive charges in left hand side by 3, so we should neutralize it by adding 3 electrons to left hand side because right hand side is neutral:
$$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O }$$
Now reduction half reaction is also completed. To achieve balanced redox reaction, simply add balanced oxidation and reduction half reactions in order to cancel unwanted electrons:
$$\ce{2MnO4- + 8H+ + 6I- -> 2MnO2 + 3I2 + 4H2O }$$
This redox reation is forward reaction because it has a net positive potential (refer reduction potentials of two half reactions). The major point is the reaction starts in acidic medium and stay acidic at the end. Keep in mind that if you have used very strong acidic conditions, $\ce{MnO2}$ would be further reduced to $\ce{Mn^2+}$ according to the half reaction: $\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O }$. But oxidation half reaction would stay unaffected, yet final redox reaction would be:
$$\ce{2MnO4- + 16H+ + 10I- -> 2Mn^2+ + 5I2 + 8H2O }$$
Main point is changing the conditions most certainly change  the reaction. We cannot arbitrarily choose it. For example, to balance the given equation, we have to use what's available for us. We cannot use what is not available or would be available in near future as a product (e.g., as $\ce{OH-}$ in the example in oxidation reaction in neutral medium you are considering).
For the shake of the argument, let's consider that reaction:

Suppose the question asked is: Balance the following redox equation in neutral medium.
  $$\ce{I- (aq) + MnO4- (aq) -> MnO2 (s) + I2 (aq)}$$

Again, to solve this problem, you can directly get two half reactions from a textbook or balance it old-fashioned as described above. The oxidation half reaction does not changed by the conditions and following is the completed equation:
$$\ce{2I- <=> I2 + 2e-}$$
The reduction half reaction is:
$$\ce{MnO4- <=> MnO2 }$$
However, it was stated that reaction is in a neutral medium (the product $\ce{MnO2}$ is possible both in acidic or neutral medium). Therefore we cannot add acid or base to reactants' side but water in either side. Therefore, we can balance loss of oxygen in product side by using water because it is in aqueous medium and we have plenty of $\ce{H2O}$ molecules around:
$$\ce{MnO4- <=> MnO2 + 2H2O}$$
Now we have extra hydrogen in product side, so we don't have a choice but balance that by $\ce{H+}$ in reactant side for now:
$$\ce{MnO4- + 4 H+ <=> MnO2 + 2H2O}$$
It is fact that acid-base neutralization reaction produces $\ce{H2O}$ predominantly. Thus, we can add  $\ce{OH-}$ to both sides of the reaction so that it can keep the mass balance of the reaction. Since the reaction is in aqueous medium this won't affect the reaction:
$$\ce{MnO4- + 4H+ + 4OH- <=> MnO2 + 2H2O + 4OH-}$$
$$\ce{MnO4- + 2H2O <=> MnO2 + 4OH-}$$
Now the equation is balanced by mass. However, we have increased net negative charges in right hand side by 3, so we should neutralize it by adding 3 electrons to left hand side to cancel the charges:
$$\ce{MnO4- + 2H2O + 3e- <=> MnO2 + 4OH-} \tag{1}$$
Now reduction half reaction is also completed. To achieve balanced redox reaction, simply add balanced oxidation and reduction half reactions in order to cancel unwanted electrons:
$$\ce{2MnO4- + 4H2O + 6I- -> 2MnO2 + 3I2 + 8OH- }$$
The major point is this reaction starts in neutral medium, but become basic at the end. Again, we didn't have choice to keep it neutral because that't the way this happens in nature. Keep in mind that we did not add anything unavailable during balancing the equation. If you check literature for reduction half reaction of $\ce{MnO4-}$ in neutral medium, you would get nothing but equation $(1)$ as we derived.

Okay. I’m just making notes. You said “The major point is the reaction starts in acidic medium and stay acidic at the end.” Maybe but if we reduce the pH by 8 moles of H+, aren’t we at least changing it? It seems that by adding these ions I would be changing the pH.

“Now we have extra hydrogen in product side, so we don't have a choice but balance that by H+ in reactant side“ okay, that makes sense. Then you add OH on the reactant side because the solution is neutral. I can see that. The 4H+ had to come from somewhere leaving the OH- behind.

Alright I’ve read it all closely, but I still don’t understand why the pH should change. “The major point is this reaction starts in neutral medium, but become basic at the end.” If I add these ions, I wouldn’t expect any of them to alter the pH. I is not a base. I suppose MnO4- is a conjugate base, but I would expect to see the extra H+ in that case. It does not make sense to me for anything but a base to make the solution more basic. Please explain.

Comment 1: Remember, adding $\ce{H+}$ to equation doesn't mean you are adding to solution. Solution is acidic already by the using conditions, we are just using the exsisting condition to balance the equation.

This is true for the remaining comments. For last one, $\mathrm{pH}$ increases as reaction progress.

For the reaction:
$$\ce{6I- (aq)  + 4 H2O + 2 MnO4- (aq) ->[excess I-] 3 I3- + 2 MnO2 (s) + 8OH-}$$
what you are missing is the Nernst Equation. The two reduction half cell reactions are:
$$\ce{I3− + 2e- <=> 3I− \quad\quad EMF = +0.53}$$
$$\ce{MnO4− + 2H2O + 3e− <=> MnO2(s) + 4 OH−  \quad\quad EMF =+0.595}$$
The gist of the Nernst Equation is that the voltages change as the concentration of the ionic species change. The standard EMF is for 1 molar concentration of all ionic species.
For the $\ce{I3-/I-}$ half cell there is no pH consideration in that HI is a strong acid.
For the permanganate reduction:

As the permanganate is reduced and the hydroxide builds up, the Nernst equation shows that the EMF of the half cell drops towards zero.
If the initial permanganate was 1 molar and the hydroxide was 0.1 molar, then the EMF would be even more positive than 0.595 V.

As a bit of a side note $\ce{I2}$ has a very limited solubility in solution, hence an excess of $\ce{I-}$ is typically used so that the soluble $\ce{I3-}$ complex is formed.

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text:
Conductometric titration curve when KCl is gradually added to AgCl solution
The situation given is that we are performing conductometric titration where KCl is gradually added to AgCl solution and we have to plot the variation of conductivity with the volume of aq. KCl added. I am new to conductometric titration and I have covered the cases of titration of acid v/s bases but I have never come across a case like this.
My thought was that since AgCl is a weak electrolyte it's initial conductivity would be low. And now we add KCl to it which is a strong electrolyte so the conductivity should increase. I also think that common ion effect would come into play and the dissociation of AgCl would be pushed back further but I'm still unsure about how an end point would be obtained here. Can someone guide me? Any help would be appreciated.

Please provide more details about the experiment. We can't guess what experiment you are doing.

It isn't an experiment. It's an MCQ question I was solving in my course material.

Could we have the problem exactly as stated? I think I know what is happening, but I'm somewhat confused by the phrase "AgCl solution."

I think the OP means AgNO3. Nobody can titrate KCl with AgCl. OP check your experimental notebook.

@M.Farooq No it's given as AgCl with KCl, could be a typo. I already know the AgNO3 experiment.

I'm a bit puzzled by the phrase "AgCl solution," since there wouldn't be a solution but rather colloidal size particles of AgCl in the aqueous solution. I don't think you could have a bottle of colloidal size particles of AgCl since once dried the particles would stick together in a lump.

For AgCl, a tiny amount would dissolve hence there would be conductivity slightly better than pure water.
$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$
However AgCl also form a complex with chloride:
$$\ce{AgCl(s) + Cl-(aq) <=> AgCl2-(aq)}$$
So as you add KCl the curve would show one slope as KCl builds up enough concentration to dissolve the AgCl dissolves, then a second slope as the AgCl dissolves, then a third slope after all the AgCl has dissolved.
For ecah change the curve won't look like two straight lines intersecting at a point but more like a parabola. So you'd need to extend linear portion of each and find the end point at the intersection of the linear extensions.

This is beyond the high school level. It is very likely a typo. It must be AgNO3. The whole experiment is described here: https://www.ausetute.com.au/conductpptn.html. It is a simple curve.

@M.Farooq - Super fantastic! Yes, a $\ce{AgNO3}$ solution would make much more sense.

@MaxW Your answer is correct. The graph shown in the answer is just as you described.

@infinite-blank-would you mind sharing the picture of the curve?

While yes, for AgCl, a small amount could dissolve per the reactions
$\ce{AgCl(s) <=> Ag+(aq)+Cl−(aq)}$
$\ce{AgCl(s) + Cl−(aq) <=> AgCl2−(aq)}$
However, I believe the presence of the AgCl as a suspension would have an impact on conductivity per this source, as well, where AgCl particles replace Al2O3–H2O per the abstract below:

In this study, Al2O3–H2O nanofluids were synthesized using sodium dodecylbenzenesulfonate (SDBS) dispersant agent by ultra-sonication method. Different amounts of SDBS i.e. 0.1, 0.2, 0.3, 0.6, 1 and 1.5 wt.% were tested to stabilize the prepared nanofluids. The stability of nanofluids was verified using optical microscope, transmission electron microscope and Zeta potential. After selecting the suitable amount of dispersant, nanofluids with different volume fractions of Al2O3 were prepared. Zeta potential measurement of nanofluids with low alumina and intermediate fractions showed good dispersion of Al2O3 nanoparticles in water, but nanofluids with high mass fraction were easier to aggregate. The stabilized nanofluids were subjected for measuring of rheological behavior and electrical conductivity. The electrical conductivity was correlated to the thermal conductivity according to Wiedemann–Franz law. The results revealed that the nanofluid containing 1% SDBS was the most stable one and settling was observed for the fluid contained 0.75 vol.% of Al2O3 nanoparticles which gave higher viscosity. The rheological measurements indicated that the viscosity of nanofluids decreased firstly with increasing shear rate (shear thinning behavior). Addition of nanoparticles into the base liquid enhanced the electrical conductivity up to 0.2 vol.% of Al2O3 nano-particles after which it decreased.


text:
Why does exchange integral 1s2p not vanish for Helium excited state?
I am studying first-order perturbation theory applied to the first excited state of Helium. I am facing the following problem:
I understand that the Coulomb integral for all 2p functions will not be $0$, but for me, all the exchange integrals involving the 2p functions will be $0$. For example, look at this two integrals for the 1s2pz state:
1s function:
$$1s=\sqrt{\frac{Z^3}{a_0^3 π}}e^{\frac{-Z}{a_0}r}$$
2pz function:
$$2p_z=\frac{1}4\sqrt{\frac{Z^5}{2a_0^5 π}}re^{\frac{-Z}{2a_0}r}cosθ$$
Coulomb integral:
$$J_{p_z}=H_{(1s2pz;1s2pz)} =∫∫\frac{1s(1)2p_z(2)e^21s(1)2p_z (2)}{4πε_0r_{12}}dr_1 dr_2$$
$$J_{p_z}=\frac{Z^8 e^2}{128a_0^8 π^3ε_0} ∫_0^∞e^{{\frac{-2Z}{a_0}}r_1} r_1^2 dr_1 ∫_0^∞e^{{\frac{-2Z}{a_0}}r_2} r_2^4 dr_2 ∫_0^π\frac{senθ_1 dθ_1}{\sqrt{r_1^2+r_2^2-2r_1 r_2 cosθ_1}} ∫_0^{2π}dϕ_1 ∫_0^πcos^2 θ_2 senθ_2 dθ_2 ∫_0^{2π}dϕ_2$$
Exchange integral:
$$K_{p_z}=H_{(1s2pz;1s2pz)} =∫∫\frac{1s(1)2p_z(2)e^22p_z(1)1s(2)}{4πε_0r_{12}}dr_1 dr_2$$
$$K_{p_z}=\frac{Z^8 e^2}{128a_0^8 π^3ε_0} ∫_0^∞e^{{\frac{-3Z}{2a_0}}r_1} r_1^3 dr_1 ∫_0^∞e^{{\frac{-3Z}{2a_0}}r_2} r_2^3 dr_2 ∫_0^π\frac{cosθsenθ_1 dθ_1}{\sqrt{r_1^2+r_2^2-2r_1 r_2 cosθ_1}} ∫_0^{2π}dϕ_1 ∫_0^πcos θ_2 senθ_2 dθ_2 ∫_0^{2π}dϕ_2$$
In all cases, the electron 1 was placed along the z-axis for its $θ$ angle to be equal to the angle between the two electrons.
So, the exchange integral (K) will be zero because the integral:
$$∫_0^πcos θ_2 senθ_2 dθ_2=-\frac{(cos^2π-cos^20)}2=0$$
And the Coulomb Integral will not vanish because:
$$∫_0^πcos^2 θ_2 senθ_2 dθ_2=-\frac{(cos^3π-cos^30)}3=\frac{2}3$$
But in all the materials I have read, both Coulomb and exchange integrals are not zero, but they do not show the calculations. I have searched a lot on the internet, and I still have not found where my mistake is.

Introduction
I start off this derivation more generally and analyize the configuration where $(n_A,\ell_A) \neq (n_B,\ell_B)$. By $E_{ n_A\ell_Am_A,n_B\ell_Bm_B}^{Z_1,Z_2}$. Later, I restrict my attention to the case where $(n_A,\ell_A) = (1,0)$.
Throughout this answer, the first two (of four) integrals in the equations tend to be the direct integrals, while the second two (of four) integrals tend to be the exchange integrals.
Results
Let's calculate the energy of where $(n_A,\ell_A) \neq (n_B,\ell_B)$. By $E_{ n_A\ell_Am_A,n_B\ell_Bm_B}^{Z_1,Z_2}$, I denote the energy of the configuration.
\begin{align*}
&E_{ n_A\ell_Am_A,n_B\ell_Bm_B}^{Z_1,Z_2}
\\
&=
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\int_{\mathbb{R}^{3\times 2}}
\Psi^*(\mathbf{r}_1, \mathbf{r}_2)
\, \frac{1}{ r_{12}}
\,
\Psi (\mathbf{r}_1, \mathbf{r}_2)
\,
d^3\mathbf{r}_1\,d^3\mathbf{r}_2
\\
&=
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\int_{\mathbb{R}^{3\times 2}}
u^*_{n_A\ell_Am_A}(\mathbf{r}_1)
\,
u^*_{n_B\ell_Am_B}(\mathbf{r}_2)
\,
u_{n_A\ell_Am_A}(\mathbf{r}_1)
\,
u_{n_B\ell_Am_B}(\mathbf{r}_2)
\, \frac{1}{ r_{12}}\,d^3\mathbf{r}_1\,d^3\mathbf{r}_2
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\int_{\mathbb{R}^{3\times 2}}
u^*_{n_A\ell_Am_A}(\mathbf{r}_1)
\,
u^*_{n_B\ell_Am_B}(\mathbf{r}_2)
\,
u_{n_B\ell_Am_B}(\mathbf{r}_1)
\,
u_{n_A\ell_Am_A}(\mathbf{r}_2)
\, \frac{1}{ r_{12}}\,d^3\mathbf{r}_1\,d^3\mathbf{r}_2
\\
&=
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\int_{\mathbb{R}^{3\times 2}}
u^*_{n_A\ell_Am_A}(\mathbf{r}_1)
\,
u_{n_A\ell_Am_A}(\mathbf{r}_1)
\,
u^*_{n_B\ell_Bm_B}(\mathbf{r}_2)
\,
u_{n_B\ell_Bm_B}(\mathbf{r}_2)
\, \frac{1}{ r_{12}}\,d^3\mathbf{r}_1\,d^3\mathbf{r}_2
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\int_{\mathbb{R}^{3\times 2}}
u^*_{n_A\ell_Am_A}(\mathbf{r}_1)
\,
u_{n_B\ell_Bm_B}(\mathbf{r}_1)
\,
u^*_{n_B\ell_Bm_B}(\mathbf{r}_2)
\,
u_{n_A\ell_Am_A}(\mathbf{r}_2)
\, \frac{1}{ r_{12}}\,d^3\mathbf{r}_1\,d^3\mathbf{r}_2
\end{align*}
Putting in the separable wave functions.
\begin{align*}
&E_{ n_A\ell_Am_A,n_B\ell_Bm_B}^{Z_1,Z_2}
\\
&=
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\int_{\mathbb{R}^{3\times 2}}
R^*_{n_A\ell_A}(\mathbf{r}_1)
Y^*_{\ell_Am_A}(\mathbf{r}_1)
\,
R_{n_A\ell_A}(\mathbf{r}_1)
Y_{\ell_Am_A}(\mathbf{r}_1)
\,
R^*_{n_B\ell_B}(\mathbf{r}_2)
Y^*_{\ell_Am_B}(\mathbf{r}_2)
\,
R_{n_B\ell_B}(\mathbf{r}_2)
Y_{\ell_Bm_B}(\mathbf{r}_2)
\, \frac{1}{ r_{12}}\,d^3\mathbf{r}_1\,d^3\mathbf{r}_2
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\int_{\mathbb{R}^{3\times 2}}
R^*_{n_A\ell_A }(\mathbf{r}_1)
Y^*_{ \ell_Am_A}(\mathbf{r}_1)
\,
R_{n_B\ell_B }(\mathbf{r}_1)
Y_{ \ell_Bm_B}(\mathbf{r}_1)
\,
R^*_{n_B\ell_B }(\mathbf{r}_2)
Y^*_{ \ell_Bm_B}(\mathbf{r}_2)
\,
R_{n_A\ell_A }(\mathbf{r}_2)
Y_{ \ell_Am_A}(\mathbf{r}_2)
\, \frac{1}{ r_{12}}\,d^3\mathbf{r}_1\,d^3\mathbf{r}_2
\end{align*}
The computation of this integral is well served   by the expansion of $1/r_{12}$ in terms of spherical harmonics.
$$\frac{1}{ r_{12}} =
\begin{cases}
\frac{1}{r_2}\sum\limits_{k=0}^\infty \frac{4\pi}{2k+1}  \left(\frac{r_{1 }  }{r_{ 2} } \right)^{k } \sum\limits_{q=-k}^{k} \,Y^{*}_{k,q}(\theta_1, \phi_1) \,Y _{k,q}(\theta_2, \phi_2),
&
\text{for}~r_2>r_1;~\text{and}
\\
\frac{1}{r_1}\sum\limits_{k=0}^\infty \frac{4\pi}{2k+1} \left(\frac{r_{2 }  }{r_{ 1} } \right)^{k }\sum\limits_{q=-k}^{k}   Y^{*}_{k,q}(\theta_2, \phi_2) \,Y _{k,q}(\theta_1, \phi_1),
&
\text{for}~r_1>r_2.
\end{cases}
$$
\begin{align*}
&E_{ n_A\ell_Am_A,n_B\ell_Bm_B}^{Z_1,Z_2}
\\
&=
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\sum\limits_{k=0}^\infty \frac{4\pi}{2k+1}
\sum\limits_{q=-k}^{k} \,
\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{n_A\ell_A}(\mathbf{r}_1)
\,
R_{n_A\ell_A}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{1 }^{k+2}\,r_{ 2}^{1-k }
\,
dr_1\,dr_2
\\
&\quad\times
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
Y^*_{\ell_Am_A}(\theta_1,\phi_1)
Y_{\ell_Am_A}(\theta_1,\phi_1)
Y^*_{\ell_Am_B}(\theta_2,\phi_2)
Y_{\ell_Bm_B}(\theta_2,\phi_2)
\,
Y^{*}_{k,q}(\theta_1, \phi_1)
\,
Y _{k,q}(\theta_2, \phi_2),
\,
\sin\theta_1\,d\theta_1\,d\phi_1
\,
\sin\theta_2\,d\theta_2\,d\phi_2
\\
&+
\frac{e^2}{ 4\,\pi\,\epsilon_o }
 \sum\limits_{k=0}^\infty \frac{4\pi}{2k+1}
 \sum\limits_{q=-k}^{k}
\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{n_A\ell_A}(r_1)
\,
R_{n_A\ell_A}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{2}^{k+2}\,r_{1}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
Y^*_{\ell_Am_A}(\theta_1,\phi_1)
Y_{\ell_Am_A}(\theta_1,\phi_1)
Y^*_{\ell_Am_B}(\theta_2,\phi_2)
Y_{\ell_Bm_B}(\theta_2,\phi_2)
\,
Y^{*}_{k,q}(\theta_2, \phi_2) \,Y _{k,q}(\theta_1, \phi_1)
\,
\sin\theta_1\,d\theta_1\,d\phi_1
\,
\sin\theta_2\,d\theta_2\,d\phi_2
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
 \sum\limits_{k=0}^\infty \frac{4\pi}{2k+1}   \sum\limits_{q=-k}^{k} \,
 \int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{n_A\ell_A }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{n_A\ell_A }(r_2)
\,
r_{1}^{k+2}\,r_{2}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
Y^*_{ \ell_Am_A}(\theta_1,\phi_1)
Y_{ \ell_Bm_B}(\theta_1,\phi_1)
Y^*_{ \ell_Bm_B}(\theta_2,\phi_2)
Y_{ \ell_Am_A}(\theta_2,\phi_2)
\,
Y^{*}_{k,q}(\theta_1, \phi_1) \,Y _{k,q}(\theta_2, \phi_2),
\,
\sin\theta_1\,d\theta_1\,d\phi_1
\,
\sin\theta_2\,d\theta_2\,d\phi_2
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\sum\limits_{k=0}^\infty \frac{4\pi}{2k+1} \sum\limits_{q=-k}^{k}   \int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{n_A\ell_A }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{n_A\ell_A }(r_2)
\,
r_{2}^{k+2}\,r_{1}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
Y^*_{ \ell_Am_A}(\theta_1,\phi_1)
Y_{ \ell_Bm_B}(\theta_1,\phi_1)
Y^*_{ \ell_Bm_B}(\theta_2,\phi_2)
Y_{ \ell_Am_A}(\theta_2,\phi_2)
\,
Y^{*}_{k,q}(\theta_2, \phi_2)
\,
Y _{k,q}(\theta_1, \phi_1)
\,
\sin\theta_1\,d\theta_1\,d\phi_1
\,
\sin\theta_2\,d\theta_2\,d\phi_2
\end{align*}
Now, since
\begin{align}
& \int Y_{l_1 m_1}(\theta, \phi) Y_{l_2 m_2}(\theta, \phi) Y_{l_3 m_3}(\theta, \phi)\,\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi \\
&\quad =
\sqrt{\frac{(2l_1 + 1)(2l_2 + 1)(2l_3 + 1)}{4\pi}}
\begin{pmatrix}
  l_1 & l_2 & l_3 \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  l_1 & l_2 & l_3\\
  m_1 & m_2 & m_3
\end{pmatrix}
\end{align}
we rewrite as
\begin{align*}
&E_{ n_A\ell_Am_A,n_B\ell_Bm_B}^{Z_1,Z_2}
\\
&=
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\sum\limits_{k=0}^\infty \frac{4\pi}{2k+1}
\sum\limits_{q=-k}^{k} \,
(-1)^{m_A+m_B+q}
\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{n_A\ell_A}(\mathbf{r}_1)
\,
R_{n_A\ell_A}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{1 }^{k+2}\,r_{ 2}^{1-k }
\,
dr_1\,dr_2
\\
&\quad\times
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
Y_{\ell_A,-m_A}(\theta_1,\phi_1)
Y_{k,-q}(\theta_1, \phi_1)
Y_{\ell_A,m_A}(\theta_1,\phi_1)
Y_{\ell_A,-m_B}(\theta_2,\phi_2)
Y_{k,q}(\theta_2, \phi_2),
Y_{\ell_B,m_B}(\theta_2,\phi_2)
\,
\sin\theta_1\,d\theta_1\,d\phi_1
\,
\sin\theta_2\,d\theta_2\,d\phi_2
\\
&+
\frac{e^2}{ 4\,\pi\,\epsilon_o }
 \sum\limits_{k=0}^\infty \frac{4\pi}{2k+1}
 \sum\limits_{q=-k}^{k}
(-1)^{m_A+m_B+q}
\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{n_A\ell_A}(r_1)
\,
R_{n_A\ell_A}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{2}^{k+2}\,r_{1}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
Y_{\ell_A,-m_A}(\theta_1,\phi_1)
Y _{k,q}(\theta_1, \phi_1)
Y_{\ell_Am_A}(\theta_1,\phi_1)
Y_{\ell_A,-m_B}(\theta_2,\phi_2)
Y_{k,-q}(\theta_2, \phi_2)
Y_{\ell_Bm_B}(\theta_2,\phi_2)
\,
\sin\theta_1\,d\theta_1\,d\phi_1
\,
\sin\theta_2\,d\theta_2\,d\phi_2
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
 \sum\limits_{k=0}^\infty \frac{4\pi}{2k+1}   \sum\limits_{q=-k}^{k} \,
 (-1)^{m_A+m_B+q}
 \int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{n_A\ell_A }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{n_A\ell_A }(r_2)
\,
r_{1}^{k+2}\,r_{2}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
Y_{ \ell_A,-m_A}(\theta_1,\phi_1)
Y_{k,-q}(\theta_1, \phi_1)
Y_{ \ell_Bm_B}(\theta_1,\phi_1)
Y_{ \ell_B,-m_B}(\theta_2,\phi_2)
Y_{k,q}(\theta_2, \phi_2),
Y_{ \ell_Am_A}(\theta_2,\phi_2)
\,
\sin\theta_1\,d\theta_1\,d\phi_1
\,
\sin\theta_2\,d\theta_2\,d\phi_2
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\sum\limits_{k=0}^\infty \frac{4\pi}{2k+1} \sum\limits_{q=-k}^{k}
(-1)^{m_A+m_B+q}
 \int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{n_A\ell_A }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{n_A\ell_A }(r_2)
\,
r_{2}^{k+2}\,r_{1}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
\int\limits_{0}^\pi
\int\limits_{0}^{2\pi}
Y_{ \ell_A,-m_A}(\theta_1,\phi_1)
Y_{k,q}(\theta_1, \phi_1)
Y_{ \ell_B,m_B}(\theta_1,\phi_1)
Y_{ \ell_B,-m_B}(\theta_2,\phi_2)
Y_{k,-q}(\theta_2, \phi_2)
Y_{ \ell_A,m_A}(\theta_2,\phi_2)
\,
\sin\theta_1\,d\theta_1\,d\phi_1
\,
\sin\theta_2\,d\theta_2\,d\phi_2.
\end{align*}
Upon substitution,
\begin{align*}
&E_{ n_A\ell_Am_A,n_B\ell_Bm_B}^{Z_1,Z_2}
\\
&=
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\sum\limits_{k=0}^\infty \frac{4\pi}{2k+1}
\sum\limits_{q=-k}^{k} \,
(-1)^{m_A+m_B+q}
\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{n_A\ell_A}(\mathbf{r}_1)
\,
R_{n_A\ell_A}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{1 }^{k+2}\,r_{ 2}^{1-k }
\,
dr_1\,dr_2
\\
&\quad\times
 \frac{(2\ell_A + 1) (2k + 1) (2\ell_B + 1)}{4\pi}
\begin{pmatrix}
  \ell_A & k & \ell_A \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_A & k & \ell_A \\
  -m_A & -q & m_A
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  -m_B & q & m_B
\end{pmatrix}
\\
&+
\frac{e^2}{ 4\,\pi\,\epsilon_o }
 \sum\limits_{k=0}^\infty \frac{4\pi}{2k+1}
 \sum\limits_{q=-k}^{k}
(-1)^{m_A+m_B+q}
\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{n_A\ell_A}(r_1)
\,
R_{n_A\ell_A}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{2}^{k+2}\,r_{1}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\frac{(2\ell_A + 1)(2k + 1)(2\ell_B + 1)}{4\pi}
\begin{pmatrix}
  \ell_A & k & \ell_A \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_A & k & \ell_A \\
  -m_A & q & m_A
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  -m_B & -q & m_B
\end{pmatrix}
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
 \sum\limits_{k=0}^\infty \frac{4\pi}{2k+1}   \sum\limits_{q=-k}^{k} \,
 (-1)^{m_A+m_B+q}
 \int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{n_A\ell_A }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{n_A\ell_A }(r_2)
\,
r_{1}^{k+2}\,r_{2}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\frac{(2\ell_A + 1)(2k + 1)(2\ell_B + 1)}{4\pi}
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}^2
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  -m_A & -q & m_B
\end{pmatrix}
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  -m_B & q & m_A
\end{pmatrix}
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\sum\limits_{k=0}^\infty \frac{4\pi}{2k+1} \sum\limits_{q=-k}^{k}
(-1)^{m_A+m_B+q}
 \int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{n_A\ell_A }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{n_A\ell_A }(r_2)
\,
r_{2}^{k+2}\,r_{1}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\frac{(2\ell_A + 1)(2k + 1)(2\ell_B + 1)}{4\pi}
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}^2
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  -m_A & q & m_B
\end{pmatrix}
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  m_A & -q & -m_B
\end{pmatrix}
.
\end{align*}
From \url{https://en.wikipedia.org/wiki/3-j_symbol}, the Wigner $3-j$ symbol is zero unless (i) $m + q -m = 0$, and (ii) $|k-\ell| \leq \ell \leq k+\ell$. Consequently, the only values of $k$ and $q$ that result in a non-trivial definite intergal are $q=0$ and $k=0,1,\ldots,\ell$.
I now employ the fact\footnote{From \url{https://mathworld.wolfram.com/Wigner3j-Symbol.html}
} that for $J = 2\ell+k$
\begin{align*}
&E_{ n_A\ell_Am_A,n_B\ell_Bm_B}^{Z_1,Z_2}
\\
&=
\frac{e^2}{ 4\,\pi\,\epsilon_o }
 (2\ell_A + 1)  (2\ell_B + 1)
\sum\limits_{k=0}^\infty
(-1)^{m_A+m_B}
\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{n_A\ell_A}(\mathbf{r}_1)
\,
R_{n_A\ell_A}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{1 }^{k+2}\,r_{ 2}^{1-k }
\,
dr_1\,dr_2
\\
&\quad\times
\begin{pmatrix}
  \ell_A & k & \ell_A \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_A & k & \ell_A \\
  -m_A & 0 & m_A
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  -m_B &  & m_B
\end{pmatrix}
\\
&+
\frac{e^2}{ 4\,\pi\,\epsilon_o }
 (2\ell_A + 1) (2\ell_B + 1)
 \sum\limits_{k=0}^\infty
(-1)^{m_A+m_B}
\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{n_A\ell_A}(r_1)
\,
R_{n_A\ell_A}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{2}^{k+2}\,r_{1}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\begin{pmatrix}
  \ell_A & k & \ell_A \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_A & k & \ell_A \\
  -m_A & 0 & m_A
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  -m_B & 0 & m_B
\end{pmatrix}
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
 (-1)^{\ell_A+k+q}
 \,
  (2\ell_A + 1) (2\ell_B + 1)
 \sum\limits_{k=0}^\infty   \int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{n_A\ell_A }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{n_A\ell_A }(r_2)
\,
r_{1}^{k+2}\,r_{2}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}^2
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  -m_A & m_A-m_B & m_B
\end{pmatrix}^2
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
(-1)^{ \ell_A +  k + \ell_B  }
\,
 (2\ell_A + 1) (2\ell_B + 1)
\sum\limits_{k=0}^\infty
 \int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{n_A\ell_A }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{n_A\ell_A }(r_2)
\,
r_{2}^{k+2}\,r_{1}^{1-k}
\,
dr_1\,dr_2
\\
&\quad\times
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}^2
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  -m_A &  m_A-m_B  & m_B
\end{pmatrix}^2
.
\end{align*}
And
\begin{align*}
&E_{ n_A\ell_Am_A,n_B\ell_Bm_B}^{Z_1,Z_2}
\\
&=
\frac{e^2}{ 4\,\pi\,\epsilon_o }
 (2\ell_A + 1)  (2\ell_B + 1)
\sum\limits_{k=0}^\infty
(-1)^{m_A+m_B}
\begin{pmatrix}
  \ell_A & k & \ell_A \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_A & k & \ell_A \\
  -m_A & 0 & m_A
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  -m_B &  & m_B
\end{pmatrix}
\\
&\quad \times
\begin{bmatrix}
+\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{n_A\ell_A}(\mathbf{r}_1)
\,
R_{n_A\ell_A}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{1 }^{k+2}\,r_{ 2}^{1-k }
\,
dr_1\,dr_2
\\
+
\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{n_A\ell_A}(r_1)
\,
R_{n_A\ell_A}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{2}^{k+2}\,r_{1}^{1-k}
\,
dr_1\,dr_2
\end{bmatrix}
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
 (2\ell_A + 1) (2\ell_B + 1)
\sum\limits_{k=0}^\infty
(-1)^{ \ell_A +  k + \ell_B  }
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}^2
\begin{pmatrix}
  \ell_A & k & \ell_B \\
  -m_A &  m_A-m_B  & m_B
\end{pmatrix}^2
\\
&\quad\times
\begin{bmatrix}
+
\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{n_A\ell_A }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{n_A\ell_A }(r_2)
\,
r_{1}^{k+2}\,r_{2}^{1-k}
\,
dr_1\,dr_2
\\
+
\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{n_A\ell_A }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
r_{1}^{1-k}
\,
dr_1
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{n_A\ell_A }(r_2)
\,
r_{2}^{k+2}
\,dr_2
\end{bmatrix}
.
\end{align*}
Now for the case of interest here which includes one electron in the 1s state, I find that
\begin{align*}
&E_{ 1,0,0;n_B,\ell_B,m_B}^{Z_1=2,Z_2=2}
\\
&=
\frac{e^2}{ 4\,\pi\,\epsilon_o }
    (2\ell_B + 1)
\sum\limits_{k=0}^\infty
(-1)^{ m_B}
\begin{pmatrix}
  0 & k & 0 \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  0 & k & 0 \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_B & k & \ell_B \\
  -m_B & 0 & m_B
\end{pmatrix}
\\
&\quad \times
\begin{bmatrix}
+\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{1,0}(\mathbf{r}_1)
\,
R_{1,0}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{1 }^{k+2}\,r_{ 2}^{1-k }
\,
dr_1\,dr_2
\\
+
\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{1,0}(r_1)
\,
R_{1,0}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{2}^{k+2}\,r_{1}^{1-k}
\,
dr_1\,dr_2
\end{bmatrix}
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
   (2\ell_B + 1)
\sum\limits_{k=0}^\infty
(-1)^{  k + \ell_B  }
\begin{pmatrix}
  0 & k & \ell_B \\
  0 & 0 & 0
\end{pmatrix}^2
\begin{pmatrix}
  0 & k & \ell_B \\
  0 &  -m_B  & m_B
\end{pmatrix}^2
\\
&\quad\times
\begin{bmatrix}
+
\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{1,0}(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{1,0 }(r_2)
\,
r_{1}^{k+2}\,r_{2}^{1-k}
\,
dr_1\,dr_2
\\
+
\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{1,0 }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
r_{1}^{1-k}
\,
dr_1
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{1,0 }(r_2)
\,
r_{2}^{k+2}
\,dr_2
\end{bmatrix}
.
\end{align*}
According to the rules for the Wigner-$3j$ numbers, the only values of $k$ that result in non-trivial values here are $k=0$ for the direct integral, and $k=\ell_B$ for the exchange integral. Thus,
\begin{align*}
&E_{ 1,0,0;n_B,\ell_B,m_B}^{Z_1,Z_2}
\\
&=
\frac{e^2}{ 4\,\pi\,\epsilon_o }
    (2\ell_B + 1)
(-1)^{ m_B}
\begin{pmatrix}
  0 & 0 & 0 \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_B & 0 & \ell_B \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  0 & 0 & 0 \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  \ell_B & 0 & \ell_B \\
  -m_B & 0 & m_B
\end{pmatrix}
\\
&\quad \times
\begin{bmatrix}
+\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{1,0}(\mathbf{r}_1)
\,
R_{1,0}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{1 }^{ 2}\,r_{ 2}^{1  }
\,
dr_1\,dr_2
\\
+
\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{1,0}(r_1)
\,
R_{1,0}(r_1)
\,
R^*_{n_B\ell_B}(r_2)
\,
R_{n_B\ell_B}(r_2)
\,
r_{2}^{ 2}\,r_{1}^{1 }
\,
dr_1\,dr_2
\end{bmatrix}
\\
& \pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
   (2\ell_B + 1)
\begin{pmatrix}
  0 & \ell_B & \ell_B \\
  0 & 0 & 0
\end{pmatrix}^2
\begin{pmatrix}
  0 & \ell_B & \ell_B \\
  0 &  -m_B  & m_B
\end{pmatrix}^2
\\
&\quad\times
\begin{bmatrix}
+
\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^*_{1,0}(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{1,0 }(r_2)
\,
r_{1}^{\ell_B+2}\,r_{2}^{1-\ell_B}
\,
dr_1\,dr_2
\\
+
\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^*_{1,0 }(r_1)
\,
R_{n_B\ell_B }(r_1)
\,
r_{1}^{1-\ell_B}
\,
dr_1
\,
R^*_{n_B\ell_B }(r_2)
\,
R_{1,0 }(r_2)
\,
r_{2}^{\ell_B+2}
\,dr_2
\end{bmatrix}
.
\end{align*}
Reducing further
\begin{align}
\label{eq:helium-energy-levels-l2-equals-0}
\boxed{
E_{ 1,0,0;n_B,\ell_B,m_B}^{Z_1,Z_2}
=
\begin{matrix}
+\frac{e^2}{ 4\,\pi\,\epsilon_o }
\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^{Z_1=2}_{1,0}(\mathbf{r}_1)
\,
R^{Z_1=2}_{1,0}(r_1)
\,
R^{Z_2=2}_{n_B\ell_B}(r_2)
\,
R^{Z_2=2}_{n_B\ell_B}(r_2)
\,
r_{1 }^{ 2}\,r_{ 2}^{1  }
\,
dr_1\,dr_2
\\[2em]
+
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^{Z_1}_{1,0}(r_1)
\,
R^{Z_1}_{1,0}(r_1)
\,
R^{Z_2}_{n_B\ell_B}(r_2)
\,
R^{Z_2}_{n_B\ell_B}(r_2)
\,
r_{2}^{ 2}\,r_{1}^{1 }
\,
dr_1\,dr_2
\\[2em]
\pm
\frac{e^2}{ 4\,\pi\,\epsilon_o }
\dfrac{\int\limits_{0}^\infty
\int\limits_{r_1=0}^{r_2}
R^{Z_1}_{1,0}(r_1)
\,
R^{Z_2}_{n_B\ell_B }(r_1)
\,
R^{Z_2}_{n_B\ell_B }(r_2)
\,
R^{Z_1}_{1,0 }(r_2)
\,
r_{1}^{\ell_B+2}\,r_{2}^{1-\ell_B}
\,
dr_1\,dr_2
}{   (2\ell_B + 1)  }
\\[2em]
\pm \frac{e^2}{ 4\,\pi\,\epsilon_o }
\dfrac{\int\limits_{0}^\infty
\int\limits_{r_1=r_2}^{\infty}
R^{Z_1}_{1,0 }(r_1)
\,
R^{Z_2}_{n_B\ell_B }(r_1)
\,
r_{1}^{1-\ell_B}
\,
dr_1
\,
R^{Z_2}_{n_B\ell_B }(r_2)
\,
R^{Z_2}{1,0 }(r_2)
\,
r_{2}^{\ell_B+2}
\,dr_2
}{   (2\ell_B + 1)  }
\end{matrix}
.
}
\end{align}
In the boxed solution above, the first two (of four) integrals are the direct integrals, while the second two (of four) integrals are the exchange integrals.
Discussion
If you look at Appendix B of Atomic Physics by Mark Fox (Cambridge University Press, 2018), it appears that $Z_1 = Z_2 = 2$. However, in Atomic Physics by Christopher Foot, Foot argues that $Z_1 = Z_2 = 2$ is correct when calculating the 1s2s configuration. However, $Z_1 = 2$ and $Z_2 = 1$ is correct when, for example, calculating the 1s2p configuration. Further, Foot states (Oxford University Pres, page 55, 2005) that for the 1s2p configuration, upon plugging in the hydrogenic wave functions $Z_1 = 2$ and $Z_2 = 1$, that the two direct integrals yield
$$-\frac{e^2}{ 8\,\pi\,\epsilon_o\,a_o }\frac{{Z_{2}}^5\,\left(108\,{Z_{1}}^2+16\,Z_{1}\,Z_{2}+{Z_{2}}^2\right)}{2\,{\left(2\,Z_{1}+Z_{2}\right)}^6} = -0.028~\text{eV}.$$ Using the boxed equation above, I obtain that same value.   Foot also states (Oxford University Pres, page 56, 2005) that for the 1s2p configuration, upon plugging in the hydrogen wave functions ($Z_1 = 2$, $Z_2 = 1$), that the two exchange integrals yield $ +0.10$ eV (page 55, 2005 edition). However, using the boxed equation above, I obtain a different value; namely
$$\frac{e^2}{8\pi\epsilon_o\,a_o} \frac{2^{10}\,{Z_{1}}^2\,{Z_{2}}^2\,\left({Z_{1}}^2+5\,Z_{1}\,Z_{2}+{Z_{2}}^2\right)}{3^8\,{\left(Z_{1}+Z_{2}\right)}^5}'
 = 0.52~\text{eV}.$$
In both of my computations, I use the Bohr radius $a_0 = 52.9~\text{pm}$. I do not understand why I get a separation 5 times greater than what I see in the book and on energy-level diagrams for helium.


text:
Ambiguity in oxidation state of phosphorus in H3PO3
Phosphorus(2.19) is less electronegative than hydrogen(2.20), so during calculation of oxidation state of phosphorus the oxidation state of hydrogen should be taken as-1 but in books it's taken as +1 why?

Saying that 2.19 is less than 2.2 is true, mathematically. But only.

Oxidation number has only a loose connection with electronegativity. It's counted based on rules and in this case not only for P, but all 15 group elements H bound to them is on +1.

Oxidation number is rather a convention for gray zones and this convention says that at a formal break of P-H bond in H3PO3 electron stays at P.

There are several electronegativity scales. In the Pauling scale, H and P are at 2.1. In the Mulliken scale, H is at 2.2 and P at 2.3. In the Allred and Rochow scale, it is the contrary : H is at 2.2 and P at 2.1. These scales are contradictory for H and P. So it is necessary to choose for teaching purposes. And book authors have usually chosen +1 for H, because H is rarely at -1, and only in very reactive compounds like NaH.
Ref. : Michael O. Hurst, J. Chem. Ed. 79, June 2002. How we teach Molecular Structure to Freshmen.


text:
Could the Superoxide Radical Anion Be Responsible for the Removal of Chlorine Dioxide Used to Treat Natural Waters?
Some background facts:
$\ce{ClO_2}$ is a gas that is highly soluble in water (some 8 g/L at 20 °C). However, per Wikipedia , to quote:

It does not hydrolyze when it enters water, and is usually handled as a dissolved gas in solution in water.

Although technically, there are many (including important) produced products with water, which
I account for by the dissociation of chlorine dioxide, a stable free radical and subsequent hydrolysis:
$\ce{.ClO2 + .ClO2 <=>  Cl2O4}$
$\ce{Cl2O4 + H2O <=>  HClO2 + HClO3}$
A government source makes the statement:

In air, chlorine dioxide readily dissociates both thermally and photochemically and may form chlorine, oxygen, hydrogen chloride, HClO3, HClO4.ClO, chlorine peroxide, and/or chlorine trioxide, dependent on temperature and humidity. Chlorine dioxide dissociates in water into chlorite and chloride, and to a lesser extent into chlorate (Budavari et al. 1996).

Per a Dutch study: Chlorine Dioxide as a Post-Disinfectant for Dutch Drinking Water, to quote:

In this paper, results are presented of experiments into the consumption and reaction kinetics of chlorine dioxide in a number of (drinking) waters in The Netherlands. It was found that chlorine dioxide consumption is related to the dissolved oxygen content (DOC) of the water and the reaction time.

Lastly, a cited reaction between hydrogen peroxide and chlorine dioxide:
$\ce{H2O2 + 2 .ClO2 ->  O2 + 2 HClO2 }$
Now, based on the Dutch study suggested connection to dissolved oxygen, I start with the last reaction above and use it as a thought tool to arrive at my postulate reaction as to how ClO2 decomposes in natural waters. The first step removes H+, next removes an electron from HO2-, next dissociates the radical $\ce{.HO2}$, then cancel H+ from both sides, merge an electron into ClO2 to produce ClO-, and in the final step cancel part of the chlorite ion present. The manipulation steps are:
$\ce{HO2- + 2 .ClO2 -> O2 + ClO2- + HClO2}$
$\ce{.HO2 + e- + 2 .ClO2 ->  O2 + 2 ClO2- + H+}$
$\ce{H+ + .O2- + e- + 2 .ClO2 ->  O2 + 2 ClO2- + H+}$
$\ce{ .O2- + e- + 2 .ClO2 ->  O2 + 2 ClO2-}$
$\ce{.O2- +  .ClO2 + ClO2- ->  O2 + 2 ClO2-}$
Producing the final postulated reaction of this question:
$\ce{.O2- +  .ClO2 ->  O2 +  ClO2-}$
Namely, I postulate that the presence of generated superoxide from various sources in natural oxygen-rich waters (from sunlight and transition metals,..., interacting with O2), which has been treated with chlorine dioxide, where superoxide concentration is connected to dissolved oxygen presence, is one of the more likely pathways to account for the reported removal of $\ce{ClO_2}$.
Comments welcome or evidence to assert otherwise.
[EDIT] Given interest per comments on the presence of superoxide in natural waters, I refer those interested to Measurement of Antioxidant Activity toward Superoxide in Natural Waters, which mentions a photochemical pathway for superoxide, which is likely the largest source for sunlit waters.
Superoxide can also be created by a so-called metal auto-oxidation reaction with say ferrous (as in $\ce{Fe(HCO3)_2}$):
$\ce{Fe(II) + O2 <=> Fe(III) + .O2-}$  (See also, Figure 1 here)
The presence of ferrous in natural waters is discussed in this paper, where, it should be noted, that absence light, transition metals and dissolved oxygen could be sources of superoxide. If this is but minimal, then chlorine dioxide should not be significantly consumed (as it is reported).
I would like to also propose a modified system, where oxygen and associated superoxide are but catalysts (note, the question as presented remains valid). In particular, the interaction of ferrous with chlorine dioxide:
$\ce{Fe(II) + .ClO2 -> Fe(III) + ClO2-}$
And, the cycling of a soluble ferric via the reversible metal auto-oxidation reaction:
$\ce{Fe(III) + .O2- <=>  Fe(II) + O2}$
assuming such a soluble ferric species exists here (perhaps ferric bicarbonate), else my original path proposal is unchallenged. Note, the net of the last two reactions closely resembles the prior proposal reaction, but cites a transition metal presence:
$\ce{.O2- +  .ClO2 -(Fe(II),Fe(III))->  O2 +  ClO2-}$

I think I´m missing sth here. Where is this supposed to take place? Your municipal tap water should not contain larger amounts of transition metals, and sees no sunlight.

Thanks for the review. And yes, likely limited superoxide presence which is why ClO2 is likely employed (so called post-treatment) and remains largely effective. The science and particulars of superoxide generation are not included (but could be if someone wishes to base a master thesis on the suggested mechanics).

I have edited my question with an article, 'Measurement of Antioxidant Activity toward Superoxide in Natural Waters',  targetting the measurement of superoxide which, per the source, can also be photochemically produced .


text:
Why reverse reaction is not possible in the case of dissociation of strong electrolytes and thus non existence of ionic equilibrium?
In chemistry textbooks there is generally written that ionic equilibrium can’t be achieved in the case of strong electrolytes (because their degree of dissociation is almost equal to 1) but it can be achieved in the case of weak electrolytes(because their degree of dissociation is much much less than 1) Now my question is that what does degree of dissociation has to do with whether the reverse reaction is possible or not. Suppose we have a strong electrolyte which almost dissociates completely now why don’t the reaction occurs in the backward direction(please don’t say that it is so because the dissociation is almost 100%. If so then what, it doesn’t mean that reaction can’t occur in the backward reaction there is possibility that the almost 100% dissociated ions may again recombine to make the reaction go backwards and reach a state where the rate of the forward equals the rate of backward). But this doesn’t happen(acc. To textbooks) why?
According to me(my understanding of Arrhenius’ theory of electrolytic dissociation) the electrolytes dissociate in water to furnish ions and these ions are continuously recombining to form the electrolyte molecules and thus are in equilibrium and therefore the strong electrolytes must also follow the same principle but we can say that in this case particularly the equilibrium is very highly established towards the product side(but it will be still said to be equilibrium however product sided it may be)
One more fact in favour of my argument i.e. the values of equilibrium constants are present on the web for  strong acids so if the equilibrium never exists in the case of strong electrolytes then how come we are able to calculate these constants?
NOTE- I’m a high school student.

Well there can be an equilibrium. If you dump an excess of NaCl into water, then the NaCl in solution is in equilibrium with the solid NaCl that didn't dissolve. $$\ce{NaCl(s) <=>[excess NaCl] Na+(aq) + Cl-(aq)}$$If you don't add enough salt to make a saturated solution, then the system isn't at equilibrium since more NaCl could dissolve into the solution.

Tangible comment

@MaxW, why not make it an answer?

You seem to be fretting over the definition of strong versus weak electrolytes, the notion of chemical equilibria, and what I'll call exactness.
Let's discuss exactness first. Mathematicians have calculated $\pi$ to trillions of decimal places. Chemistry doesn't work like that. Models in chemistry that are good to 3 or 4 significant figures are gospel. So if an electrolyte dissociates to 99.99%, a chemist would say that it is "completely" dissociated.
The whole science of chemistry is based on the fact that reactants combine to form products driving towards some equilibrium. So there is always some equilibrium however lopsided that might be.
Let's now consider hydrochloric acid. Concentrated hydrochloric acid is made by saturating water with hydrogen chloride gas. So there are two different equilibria. First an equilibrium of hydrogen chloride molecules between the gaseous phase and those which are dissolved in water, and secondly an equilibrium between hydronium ions and chloride ions with hydrogen chloride molecules in the liquid phases. Concentrated hydrochloric acid is 12.4 molar, meaning that one liter of solution reacted with $\ce{7.5 x 10^{24}}$  molecules of hydrogen chloride gas. But the concentration of hydrogen chloride molecules in the liquid phase is effectively zero.
If your pour some  concentrated hydrochloric acid into a beaker you can smell the acid in the air. Hydronium and chloride ions don't jump out of solution into the atmosphere to form a hydrogen chloride molecule, so there must be some hydrogen chloride molecules in solution. So the solution is trying to establish an equilibrium between hydrogen chloride molecules in the gas phase and hydrogen chloride molecules in the liquid phase. There are essentially no hydrogen chloride molecules in the bulk atmosphere. So ignoring the water evaporation (or replenishing it daily), this means that eventually all the hydrogen chloride in solution would dissipate into the atmosphere. Eventually in this case being a very very very long time.

Water evaporation cannot be ignored; dilute strong acids will concentrate because the vapor pressure of water is higher, concentrated acids will evaporate acid until an azeotrope is usually formed. This is why acid spills should be cleaned up and neutralized to prevent hidden drops of conc. acid forming.

Certain compounds that are considered ionic as solids can form significant amounts of uncharged (but obviously polar) associated species in water solution, and we can measure equilibria associated with such associations.
One of the best documented examples is magnesium sulfate, $\ce{MgSO4}$. With ions having $+2$ and $-2$ charges, this can readily form uncharged pairs in water solution.
Sebastiani et al.[1] study this solution. Through terahertz spectroscopy ("terahertz" means frequencies between infrared and microwaves), two types of ion pairs are identified: solvent-separated pairs in which the solvated magnesium and sulfate ions are joined by electrostatic attraction but retain their separate solvation spheres, and solvent-shared pairs where some overlap between the solvation spheres occurs. In neither case are the ions directly bound to each other. But the same holds true in the solid salt equilibrated with water ($\ce{MgSO4•7H2O}$), in which the magnesium is coordinated to water molecules and the sulfate ions are hydrogen-bonded to water molecules. Thus we may regard the associated species as hydrated magnesium sulfate molecules in equilibrium with the dissociated ions.

Illustration from Ref. 1. Upper left: solvent-shared ion pair; lower left: solvent-separated ion pair.
Reference

Federico Sebastiani, Ana Vila Verde, Matthias Heyden, Gerhard Schwaab, and Martina Havenith (2020). "Cooperativity and ion pairing in magnesium sulfate aqueous solutions from the dilute regime to the solubility limit". Phys. Chem. Chem. Phys., 22, 12140-12153.

Your thinking is correct. What happens at the extremes of conditions is not understood [or just oversimplified]. Here is a made up example using easy math. Lets give the Ka for a strong acid such as HCl a value of 100 [they probably are closer to 5-20].
HCl = H+ + Cl-;   100  = [H+][Cl-]/[HCl]. Lets make a 1M solution of NaCl
100 = [10^-7 * 1]/[HCl]; [HCl] =10^-9M, effectively zero and difficult to measure so Cl- is a weak base and HCl a strong acid.
Lets take a glance at a 10M solution of HCl; 100 = 10*10/[HCl]; [HCl]= 1M. It looks like an eleven M solution was needed to make a 10M acid level. If you sniff a concentrated HCl solution you would think that was close to reality. That something is going on can be seen with a good pH electrode.
A comment is made that there can not be an equilibrium with a solid until the solution is saturated. This ignores interface behavior and intermediate equilibria.


text:
Can a spectrometer be used to figure out spice ratios in a sauce?
There is a really good sauce that a local restaurant makes. I have had trouble recreating it, but have gotten somewhat close. I had the following idea:
I could take an "image" of the sauce's color spectrum under good light using a USB spectrometer, and compare it to the spectrum of my home made version. I could then slowly tweak the ingredients and resample the color spectrum of the home made sauce until it started converging and eventually matched the restaurant's.
What could prevent this from working well?
Could this realistically outperform taste testing?

No, that wouldn't work. UV-Vis spectroscopy yields broad bands that are non-specific in mixtures.

Unfortunately, this will not even qualitatively work because taste which is a sense of perception and UV-Vis are barely related.. The USB spectrometer would be great for color matching, but you are assuming the taste and aroma comes from UV-Vis absorbing compounds only. This may not be the case. Let us assume, the taste comes from UV-Vis absorbing compounds, as you know from Beer's law, an $n$-component mixture will have  absorbance sum from $n$-components. If some of these molecules are in trace quantities, they might have a little effect on the overall spectrum but they might have a huge influence on the taste and aroma of your sauce. Food chemists prefer more of gas chromatography and mass spectrometry & infrared spectrometer which may cost more than 100K.


text:
If a molecule with 3 chiral centers that are all R, is it the same as the S version?
Consider a molecule with 3 chiral centers all marked R. Is this molecule the same as the S version?

What you think?

No I don't think both will be the same.They will be enantiomers or non superimposable mirror images of each other.I am sharing a simple trick to find the isomers of a compound with 3 chiral centres instead-
RRR and its enantiomer SSS
SRR and its enantiomer RSS
SSR and its enantiomer RRS
SRS and its enantiomer RSR
With makes a total of 8! You can generalize the formula by 2 raised to the power n (where n is number of chiral centres)


text:
What is the net ionic equation for the reaction between Ba(OH)2 and H2SO4?
I know its either:
1.) Ba + SO4 = BaSO4
2.) Ba + 2OH + 2H + SO4 = BaSO4 + 2H2O
But can someone explain why water is or isn't part of the net ionic?

There is no such thing as SO4, OH in chemistry.

The reaction between $\ce{Ba(OH)2, H2SO4}$ is known as acid-base neutralisation, as $\ce{Ba(OH)2}$ is a relatively strong base and $\ce{H2SO4}$ the strong acid.
The molecular reaction is given as,
$$\ce{Ba(OH)2 + H2SO4 -> BaSO4 (s) + 2H2O}$$

Before moving to net-ionic reaction, let's see the reaction of $\ce{BaCl2, Na2SO4}$ [1]:
$$\ce{BaCl2 + Na2SO4 -> BaSO4 (s) + 2NaCl}$$
So, breaking up into ions, we see:
$$\ce{Ba^{2+} + 2Cl- + 2Na^{+} + SO4^{2-} -> BaSO4 (s) + 2Na^{+} + 2Cl^{-}}$$
$\ce{Na^{+}, Cl^{-}}$ remains completely dissociated in the aqueous solution as they're salts of strong acid and strong base (i.e., they're strong electrolytes). Hence, $\ce{Na^{+}, Cl^{-}}$ are spectator ions as they don't participate in ionic reaction. Let's cancel them out, and we get the net ionic reaction:
$$\ce{Ba^{2+} + SO4^{2-} -> BaSO4 (s)}$$

Now, let's see the reaction given in the question. In this, $\ce{BaSO4}$ precipitates out, so $\ce{Ba^{2+}, SO4^{2-}}$ can't be spectator ions. Also, H2O has a very low dissociation contant ($\ce{K_W = 10^{-14}}$, at 25°C), so it can't dissociate into ions. Hence, $\ce{H+, OH-}$ are too not the spectator ion. Therefore, this reaction has no spectator ion and hence ${H_2O}$ will be included in the ionic reaction, which is given as:
$$\ce{Ba^{2+} + 2OH- + 2H+ + SO4^{2-} -> BaSO4 (s) + 2H2O(l)}$$

Another example:
[1]: Complete ionic and net ionic equations, Khan Academy
Note: As mentioned by andselisk in comments, $\ce{BaSO4 (s)}$ is the recommended way to write the precipitate, as per IUPAC. Old school symbol for precipitate was "↓".

Your highlighted equations have some erroneous in ions (e.g., $\ce{Ba^2+ + 2 Cl- -> BaSO4}$). Please revisit and correct.

@MathewMahindaratne: Many thanks for pointing out. I've corrected them now.

"(↓)" is not a standardized notation and it never was. "↓" is an obsolete symbol to denote solid state; consider replacing it with the currently recommended "(s)" notation. This would also make the reactions look coherent since you are already denoting liquid state of aggregation this way. Further on the nitpicking side, it might be worth mentioning $K_\mathrm{w} = \pu{1E-14}$ at 25 °C only, in order not to give a student a false sense that it is some fundamental constant (also, note the typography of "$K_\mathrm{w}$").

@RahulVerma Exactly, like "Cu(s)". Please see Section 2.10.1 (vi), p. 54 of IUPAC "Green Book" (PDF). Alternatively, you can consult ACS Guide to Scholarly Communication if you have access to it.

Since you don't seem to understand the concepts let's break this down more. Rahul's answer is correct, but let's go into more detail. Consider dissolving barium chloride, equation 1, and sodium sulfate,equation 2, in separate solutions. (Note that it is important to show the charge on ions!)
$$\ce{BaCl2(s) ->[aq] Ba^{2+}(aq) + 2Cl-(aq)}\tag{1}$$
$$\ce{Na2SO4(s) ->[aq] 2Na^{+}(aq) + SO4^{2-}(aq)}\tag{2}$$
Now if the two ionic solutions are mixed the reaction is:
$$\ce{Ba^{2+}(aq) + 2Cl-(aq) + 2Na^{+}(aq) + SO4^{2-}(aq) <=> BaSO4(s) + 2Na^{+}(aq) +  2Cl-(aq)}\tag{3}$$

(s) means that the reactant is a solid, or that the product is a precipitate.
(aq) means that the molecule or ion is dissolved in water. Often this it is assumed that the context is an aqueous solution and the designation is just left off.

Since the sodium ions and the chloride ions appear on both sides of the equation they are often canceled out as spectator ions. So for a reaction you might see an equation like equation (4). Typically one would assume that for equation 4 that the $\ce{Ba^{2+}}$ and $\ce{SO4^{2-}}$ ions came from some unspecified soluble salts, and that the reaction was happening in water.
$$\ce{Ba^{2+} + SO4^{2-} <=> BaSO4(s)}\tag{4}$$
Another way to specify reaction conditions would be to put such specifications over the reaction arrows. Such as using $\ce{ ->[CH3OH]}$ to indicate that the reaction is happening in methanol.


text:
Flow tests using safe acids and bases and an indicator
I want to conduct some visualization tests on a new mixing device.  I would like to mix acid with base and use an indicator to see a color change.  These are flow tests and the flow rates of acid and base will be equal.  I'd like to only use a stoichiometric excess of 20% of either the acid or base.
Since I will be conducting these experiments in my basement, I'm thinking I should probably try to use acids and bases that are safe to humans.  I suppose I could use a 0.001M concentration of HCL and NaOH.  But I'd be handling more concentrated solutions prior to dilution to these concentrations.
I could maybe use something like baking soda for the "base" and vinegar for the "acid".  Since these are weak, there will be equilibrium shifts during the reactions in order to use up all of either the acid or base (depending upon which reactant is the 20% excess).  I need the reaction to be very rapid such that the several microsecond mixing that I am expecting will be rate controlling.
Any suggestions for acids/bases and indicator that I might use?  A strongly colored indicator in the weak acid or base would be strongly preferred. It would be best if after reaction, that indicator was colorless or an opposite color prior to reaction (red or blue converting to yellow). Again, please note that the flowrate of acid will be identical to that of base (around 0.2 gpm) except that either the acid or base will have a molarity that is 20% higher than the other reactant.
Thanks.

Baking soda is not a good choice because you will get carbon dioxide gas released, so bubbles could be a problem. Phenolphthalein, like @MaxW suggested in your previous question, is a good indicator choice. Nothing wrong with NaOH as the base.

@Ed V   Yes, I forgot about the carbon dioxide production.  What about this idea.  I will use water as one of the "reactants" with neutral red indicator in the water.  Neutral red is red at pH below 6.8 and yellow at a pH above 8.0.  Hopefully, the color would be red in water and if not, I'd only need to add a few drops of vinegar, probably.  Then I thought I could react this with baking soda water to get it to turn yellow.  But your point about baking soda is a good one.  So I guess I could use a dilute solution of NaOH even down to .0001M to turn the indicator to yellow.

Definitely give it a try! I suggest trying several concentrations of NaOH solution. If 0.1 mM NaOH works well, then try 1 mM. If no change, the go back to the more dilute solution. If there is a change, then try 10 mM. You want a fast reaction and that favors higher concentrations: this chemistry is all about favorable encounters at the ion/molecular level!

Here's what I'm planning to do.  I'm going to use sodium carbonate as my base with a pH close to 12.  Water with red cabbage indicator will be the other reactant.  The blue color in the water/red cabbage solution should turn yellow when mixed with sodium carbonate.

Since I'm using water as the "acid", carbon dioxide shouldn't form at a pH that will probably be 9 or higher.

Is there any reason you can't just use plain water and water colored with food coloring?


text:
When reactants are in their transition state, can they go back into their reactants form instead of progressing to form the products?
When reactants are in their transition state, can they go back into their reactants form instead of progressing to form the products? I don't see any reason why the reaction has to keep progressing forward.

You answered your own question in your last sentence!

It can go back and often does go back.  Nitrous acid/nitrosyl ion can attack a teritary aliphatic amine just like it does with a secondary or primary amine, but with no proton to eliminate it has to turn around and go bsckwards in the tertiary case.  It may keep going forward with a primary or secondary amine to give those nitrogen-nitrogen bonded products.

Oscar Lanza, oh ok so can I draw the conclusion that the reason you don’t have all the reactants react to form products is because you would need to make the formation of the products favourable in order to actually obtain them? Otherwise they would just turn back right?

In the gas phase the reactants have kinetic energy as they move towards one another, so you would think that they would just cross the transition state, but molecules also have vibrational energy which means that it is possible to reach the transition state and then turn back. See the figure below which shows a simple LEPS calculation of the O+H2 reaction.
The solid black line shows the position and vibration of the H2 molecule as it approached the O atom and then 'reflects' at the transition state. It has just enough energy to cross, but the vibrational motion and position on the energy surface prevents it.  The axes show the separation of H and H2 (mid point) and of H and OH products.

In solution (or high pressure gas phase) the situation is a little different because collisions with may solvent inhibit or enhance the motion of the reactants towards one another. The species diffuse together and also diffuse along the transition state pathway so that the transition state can be crossed and recrossed many times before reaction occurs or the reactants are reformed. This approach is named after Kramers and is the most important one to understand when thinking about transition states in solution. (Strangely, it is rarely mentioned in Chemistry textbooks even though it was developed more than 70 years ago.)

That is interesting, I will look up Kramer’s to get a better understanding of this. Thank you for your help

I only have an old reference Chemical Physics, v151, p239, 1991. There will be references in that paper that describe other experiments and theory.

That's even better, thank you once again.


text:
Naming Functional Isomers

For naming the isomers of this compound, (1-chloropropane and 2-chloropropane), is 3-chloropropane a valid isomer? Since 3-chloropropane is the same as 1-chloropropane, is it customary to just mention 1-chloropropane?
Thanks!

Welcome to Chemistry! Do you think 3-chloropropane is a valid IUPAC name?

The answers to your two questions are : "No" 3-chloropropane is not a valid isomer. And "Yes". it is customary to just mention 1-chloropropane.

If you have a basic knowledge of IUPAC nomenclature of hydrocarbons then you should know that we start naming a hydrocarbon by first choosing the longest carbon chain (if the hydrocarbon is a saturated one otherwise we take the chain containing double or single bond) and then by assigning the lowest possible number (which stands for its position ) to the substituent atom (here, chlorine) and followed by other steps(which we don't need for this explanation). So according to this method, in the hydrocarbon which you mentioned as "3-chloropropane", we could assign chlorine the lowest possible number for its position by starting to number the position of the carbon in the principal chain from the carbon atom with whom Cl is attached and hence we name it as 1-Chloropropane and not "3-Chloropropane".

According to the IUPAC nomenclature of hydrocarbon: Number the carbons of the parent chain from the end that gives the substituents the lowest numbers. When comparing a series of numbers, the series that is the "lowest" is the one that contains the lowest number at the occasion of the first difference. If two or more side chains are in equivalent positions, assign the lowest number to the one which will come first in the name. According to this rule, Chlorine gets the first priority and numbering starts from 1-Chloropropane.


text:
Do aprotic solvents "slow down" SN1 reactions?

SOLVENT EFFECTS. Predict if using the polar, non-protic, solvent N,N-dimethylformamide (DMF, $\ce{Me2NC(=O)H})$ will speed up or slow down the following reactions and BRIEFLY justify your answers.
a)

b)

I have no doubt that b) takes $\mathrm{S_N1}$ pathway, but would DMF actually "slow down" this reaction? I want to be meticulous; I wonder if aprotic solvents "slow down" $\mathrm{S_N1}$ reactions or just regarded as not so ideal solvents for $\mathrm{S_N1}$ reactions. The wording of the question made me wonder.

SN1 with a primary cation intermediate is a no-no... You may want to reconsider. Anyway, what is the rate in DMF being compared to? Is it faster or slower than... what? This is a classic organic chemistry question and I can make a good guess at what the author is trying to ask, but it doesn't seem to be phrased in the best way. Also, before I can answer, we need to come to some agreement on what the mechanism is (because the answer depends on the mechanism), and I'm pretty sure it's not SN1.

@orthocresol I know and it is known that primary carbocations are unstable but I thought the one in (b) would be stabilized by BF4- anion. Since DMF, the aprotic solvent, wouldn't be able to dissolve negative ions well, I thought the BF4- anion will be freely hanging and stabilize the primary carbocation.


text:
Benzyl Carbocation Stability: Would electron-withdrawing group on the ring destabilize the carbocation?
In the following question I need to decide whether the reaction would undergo SN1 reaction or SN2 reaction.

I initially thought it would be SN1 since the resulting benzyl carbocation can be stabilized by resonance. Would the strongly electron withdrawing NO2 group destabilize it though? What kind of effect would this NO2 group have on the reactivity?

The reaction shown here is a Finkelstein reaction and SN2 reaction mechanism is followed.is an organic reaction that uses an alkyl halide exchange into another alkyl halide through a reaction wherein the metal halide salt is used. This reaction takes place at an equilibrium process by taking the advantage of poor acetone solubility in metal halide salt that is newly formed. The mechanism of the Finkelstein reaction is single-step SN2 reaction with stereochemistry inversion.
The success of this reaction depends on the below conditions.

Nucleophilicity of your halogen
Nature of leaving group
Carbon-halogen bond
Alkyl Halide reactivity

Holtz and Stock showed that electron-withdrawing groups increase the rate of SN2 reactions. This can be ascribed to stabilization of the transition state by withdrawal of some of the electron density. So NO2 will make reaction faster.
Ref:ADVANCED ORGANIC CHEMISTRY: REACTIONS, MECHANISMS AND STRUCTURE, 4TH ED

Can you give the reference for the Holtz and Stock work?

but isn't C-I bond weaker (i.e. less stable) than C-Cl bond? Would substitution really happen? I mean, if those 4 factors are what I need to consider, I don't see the driving force of the reaction. No reaction was also an option to pick for this question, would it be that?

@chemrese Believe the reaction is reversible however to my understanding it’s because iodide is a much better nucleophile than chloride and when it’s displaced, NaCl is insoluble in acetone whereas NaI is soluble. This essentially removes chloride from the system and so the equilibrium is forced to the right.


text:
Application of the electronegativity effect on bond angles
There's a rule in determining molecular geometry that says that if the electronegativity of the central atom increases, the bond angle of the molecule increases.
Is this is only applicable in the cases where the molecular geometry of the said compounds are similar? For example, considering $\ce{ClF3, BF3,}$ and $\ce{ PF3}$, they're primarily differentiated on the basis of them having different geometries.

Yes, in generality, you can apply this in those cases with similar geometry. You can see why due to the rationale it stems from: imagine hypothetically that the bond pair looks like a ring-shaped cloud, and that ring is on your index finger(for example, let us take one O-H bond of H2O). Now, imagine that the O is present on the aforementioned finger towards the knuckles, while the H is sitting towards the fingernail side. Now, more will the electronegativity of the central atom, more will the ring be pulled towards the knuckles.....

….and it will cause the middle finger to move aside, thereby increasing the angular separation between the index and middle finger(note that the middle finger here rougly represent the second O-H bond, and more precisely, the oxygen of the H2O would be placed on the depression between your index and middle finger). Hence, you can see why the bond angle of H2O would be bigger than, say, OF2 and H2S. This will also help you to understand that you can't do this in cases of radically different molecular geometries, as this visualization will not be very useful there

https://chemistry.stackexchange.com/questions/15620/what-is-bents-rule

(I will simply be converting my comments above into an answer here)
Yes, generally, you can apply this in cases with similar geometry. You can see why from the following rationale:

Imagine that the bond pairs look like a ring-shaped cloud, and one is worn on your index finger and another on your middle finger, both towards the fingertip. For example, to represent the two  $\ce{O-H}$ bonds of $\ce{H2O}$.
More electronegative central atoms will pull the ring towards the knuckles and it will cause the middle finger to move aside, thereby increasing the angular separation between the index and middle finger

Here's an illustrative image for $\ce{H2O}$, with blue rings representing bond pair density :

Hence, you can see why the bond angle of $\ce{H2O}$ would be bigger than, say, $\ce{OF2}$ and $\ce{H2S}$.
In case of the former, the fluorines won't led the ring move towards the depression easily, hence the index and middle fingers can come closer to each other in contrast to $\ce{H2O}$
For the latter, $\ce{S}$ is less electronegative than $\ce{O}$, so again, the rings will show less movement towards the knuckles.
This will also help you to understand that you can't do this in cases of radically different molecular geometries, as this visualization will not be very useful there


text:
Carbon compounds with no covalent bonds
The first line of the Wikipedia article on Organic Chemistry says:
"Organic chemistry is a subdiscipline of chemistry that studies the structure, properties and reactions of organic compounds, which contain carbon in covalent bonding."
This suggests that carbon compounds with no covalent bonds are excluded but what are these compounds?  I am aware that other types of bonds may occur and be important but, even when they do, the compound still seems to be regarded as organic.
What is an example of a carbon compound excluded by the clause "which contain carbon in covalent bonding"?

All organic compounds contain at least one covalent bond coming from a Carbon atom.. Compounds like aluminium carbide $Al_4C_3$ are not organic compounds.

https://chemistry.stackexchange.com/questions/43966/can-carbon-and-titanium-form-an-ionic-bond

Even if they include a covalent bond for example $\ce{CaC2, Mg2C3}$ they aren't organic compounds. These are carbides.

@NilayGhosh Yes, it seems that it does.  I was aware of compounds such as $\ce{CaCO3}$ but it is only partially ionic.  I had forgotten to check the carbides, e.g. $\ce{Al4C3}$, and did not realise that they were ionic.  If I had remembered them then I would have struggled to guess.

See also: https://chemistry.stackexchange.com/questions/22195/is-carbon-dioxide-organic-or-inorganic?


text:
Vibrational modes in BiOCl

These are the vibrational modes in BiOCl. I'm trying to account for the each mode, and where in the spectrum you would find such mode. And I found out that at 1400 cm-1 you have asymmetric stretch of bi-cl (Eu), at 1040 cm-1 there is symmetric stretch of bi-cl (A2u), and at 528 there is a symmetric stretching mode (A2u). Which leaves an Eu mode.
I've looked through many research papers, and they all seem to use microspheres of biocl, and there's often OH bands due to water vapour, but none have modes. What can be a reason for this? and what should the 4th mode be?
This is an example of FTIR spectrum I've seen:

To what point group does this molecules belong? The symmetry species and structure do not seem to fit.

point group is D4h I believe


text:
XRD Peaks in databases
I had a look at the JCPDS card of BiOCl and saw that there were very small peaks (such as 0.1 percent relative intensity) that I think wouldn't be observed in XRD pattern, they would just blend in with the noise, so I'd like to know how do they calculate the XRD peaks and know their relative intensities so well (this card referenced a 1935 paper).
Many thanks

Why do you think they could not measure to 1 part in a 1000?

Because I thought it would just blend in with the noise. Also I looked at modern papers and they don't label many peaks (because they are too close).

See, for example, https://en.m.wikipedia.org/wiki/Geiger–Müller_tube for available equipment. There were a bunch of smart people doing good measurements a 100 years ago.

Thanks for the reference. I'm wondering if you were to plot the values from the database (2 theta) vs intensity, and have very sharp peaks, is that what you would expect to get If only n=1 diffraction would occur and there was no noise?

As a side note, there is no such thing as an "XRD spectrum", it's an oxymoron (although some textbooks use this term improperly) as by definition the radiation is monochromatic. There is an x-ray spectra (like this one of the Cu anode) and x-ray diffraction pattern/diffractogram (what you are asking about, e.g. $2\theta$ vs intensity). I think someone already pointed this out already in the comments to your previous question which is now deleted by you.

@andselisk Your point simultaneously was considered in an edit of the original question.

short answer: The record is based on the old record by Bannister 1934 as first description of this phase . The fine intensity scaling 1:1000 on the record card is based on a more recent re-investigation.
recording the data:
Beside the point detector / Geiger counter technique mentioned in Jon Custer's comment -- still highly valuable in powder diffraction analysis -- a common method was to record the diffraction pattern on a photographic film.
In the case of powder samples, you used for example Guinier cameras:

(source)
with a stripe of a film placed in the inner of the housing.  After the experiment, the film was developed; and the more diffraction intensity, the darker the band on the developed film.  The following image depicts a simultaneous analysis of four samples:

(source).
In case of single crystals, one method was to place the oriented sample in front of a film, for example with a Buerger precession camera

(source)
to yield clichés like

(source)
where the location of the spots provides information about symmetry distance relationships of the lattice.  (Relative intensity of the spots equally has a meaning, but beyond the scope of this question here.)
In both cases, the intensity of the signals were attributed by eye (e.g., strong, medium, weak), or later by densitometry and densitometers in numbers. While these film methods today are superseeded in convenience by positional detectors and CCD arrays (no wet chemistry to develop the films, quick numeric read-out), you still find these noted in volume C of the International Tables of Crystallography, section 7.1 «Detectors for X-rays».

(source)

your phase:
You find the first entry about $\ce{BiOCl}$ as entry #1011175 published 1934 in Nature, and #9011782 published 1935 in Mineralogical Magazine by the same author as indexed by the COD with their .cif files. You may use these to predict how a powder diffraction pattern of this phase should look like:

And indeed, there are some signals expected to be much weaker, than others, e.g., reflection $(103)$ (blue rectangle) would be weak and maybe not spot.
Interestingly, the AMS database equally offers access to the Bismoclite data, including a copy of Bannister's publication in Mineralogical Magazine, too (here).  And table 2 (loc. cit.), based on the photographs and their visual readout lists only 16 entries.  You find all of them in the subsequent re-investigation in 1983 (here, view text file next to download diffraction data), now with a total of observations equal to 37.
Thus, it is plausible the record card you accessed refers either to this set (1983) of data (or an  even more recent study) to list $(h,k,l)$ and their diffraction intensities, but credits Bannister for the first description of the very same phase in the 1930s.


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Why do the splittings in mirrored COSY crosspeaks differ?
I have a COSY spectrum of a mixture of butanal and butyl bromide and I don't understand why the cross-peaks have various/different splitting on the other sides of the diagonal. I think it may have something to do with polarisation transfer but I'm not sure.

Let's focus for now on the two circled crosspeaks. In a COSY sequence, the frequency in the indirect dimension (the vertical axis) corresponds to the spin which the magnetisation is on during the $t_1$ period. The frequency in the direct dimension (horizontal axis) corresponds to the spin which you measure the magnetisation of during $t_2$.
It turns out that these crosspeaks are from butanal, so let's also colour-code the protons in butanal as follows: we have $\ce{C\color{red}{H}_3-C\color{blue}{H}_2-C\color{green}{H}_2-CHO}$.
The red crosspeak occurs at ~1.65 ppm in the indirect dimension, and ~0.95 ppm in the direct dimension. This means that the magnetisation during $t_1$ was on the blue $\ce{C\color{blue}{H}_2}$ proton, and was subsequently transferred to the red
$\ce{C\color{red}{H}_3}$ proton during acquisition (in $t_2$). The coupling between these two sets of protons is referred to as an active coupling; it is the coupling that is responsible for generating the crosspeak.
Conversely, the the blue crosspeak arises due to transfer from $\ce{C\color{red}{H}_3}$ (during $t_1$) to $\ce{C\color{blue}{H}_2}$ (during $t_2$). So, it also has an active coupling between the same nuclei. The difference here is that the final $\ce{C\color{blue}{H}_2}$ protons are also coupled to the green $\ce{C\color{green}{H}_2}$ protons! This is what is called a passive coupling: one that does not generate the crosspeak, but is still reflected in the multiplet structure as a splitting.
The differences in the crosspeak structure arise because of differences in passive couplings to other nuclei.
Exactly how this is manifested is not easy to explain in this case, because there are multiple equivalent nuclei. But if you are keen, read on. (I'm also not going to present any quantum mechanical analysis, because that would really take way too long.)

In fact, the red crosspeak also has one passive coupling – the $\ce{C\color{red}{H}_3}$ group is coupled to not just one, but two protons adjacent to it, and the crosspeak resembles a case where one of the two protons is an active coupling partner and the other is passive.
We first deal with passive couplings. These lead to in-phase splittings, i.e. peaks which are both positive and are separated by the coupling constant, approximately 7 Hz in our case. For simplicity's sake we will stick to just the direct dimension for now. Because we only have one effective passive coupling, this leads to a doublet with a peak separation of ~7 Hz.
(The indirect dimension, broadly speaking, has similar behaviour in that it is also affected by passive and active couplings, but the exact source of these couplings can differ. Also, it is sometimes harder to observe multiplet structure in the indirect dimension because of different resolutions in the two dimensions.)
Active couplings, on the other hand, lead to antiphase splitting, where peaks have opposite signs. So, our doublet is further split into two doublets: one positive and one negative, and these two doublets themselves are 7 Hz apart. We find that the middle peak exactly cancels itself out, and we are left with one positive and one negative peak separated by exactly twice the coupling constant. This is what is observed in the red crosspeak.

On the other hand, the blue $\ce{C\color{red}{H}_2}$ protons have one active coupling (to one of the $\ce{C\color{red}{H}_3}$ protons), but the other two $\ce{C\color{red}{H}_3}$ protons are effectively passive couplings. On top of that, it has two further passive couplings to the $\ce{C\color{green}{H}_2}$ protons.
The four passive couplings lead to the formation of a $1:4:6:4:1$ quintet (we're going to assume that all the couplings have similar size, which is quite reasonable). The single active coupling causes an antiphase splitting of this quintet, which is easier to express mathematically:
1   4   6   4   1
   -1  -4  -6  -4  -1
-----------------------
1   3   2  -2  -3  -1

(imagine that there is 7 Hz separation between every column of numbers). This $1:3:2:-2:-3:-1$ pattern exactly matches how the blue crosspeak looks like.
For more information, see e.g. Claridge, High-Resolution NMR Techniques in Organic Chemistry (3rd ed.), Section 6.1.5.2.

I´ll have to mull this over a bit. ;)  The effect would not depend on the double quantum filter mentioned in the original post?

Yes, this applies to DQF COSY as well. The OP didn't actually specify DQF, I added it in, because I don't think it's very common to run high-resolution COSY without the DQF, but I really shouldn't have done that, so I reverted that edit.

I have done COSY, but only by hand, written the basic pulse sequence myself. ;) I´ll have  to read up on that DQ filter. But wouldn´t it also affect one dimension differently than the other?

Hmmm. The active coupling will always leads to an antiphase doublet in both dimensions, but the passive couplings will show up differently (technically, a different subset of passive couplings will be observed, although there are probably further complications associated with equivalent spins). I should make that a bit clearer in the text. I'm not sure how much to read into the spectrum either; the resolution in both dimensions won't be the same so looks may deceive, especially when it comes to more complicated multiplets.


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How to compare the rate of decarboxylation?

For the following acids compare the rate of decarboxylation on heating
I $\ce{Ph-CO-CH2COOH}$
II $\ce{Ph-CO-COOH}$
III $\ce{CH3-CH2-COOH}$
IV $\ce{HOOC-CH2-COOH}$

I tried this question by forming carbanions and comparing their stability. But I am stuck at comparing I and IV.

Is the answer 4?

@Tony Yeah, (4)

Yes the ans is 4 but please tell me how did you compare between I and IV ?

In the mechanism for intramolecular decarboxylation,if you recall,the EWG bature of the carbonyl is responsible for withdrawing electrons from the -COOH part. It won't be able to do that very efficiently in IV due to cross conjugation presented by the OH ,which will decrease it's withdrawing nature

For comparing the rate of decarboxylation just compare the stability of the carbanion formed. It's an easy jee concept.

First thing to see is to see carbanion stability.
Now the question arises that between  carbonyl and COOH  groups which can stabilize better. For that you can look at the COOH 's already possible resonance,within its molecule, which decreases it electron withdrawal  tendency.
Hence order would be: I>IV>III>II.

Compound 1 undergoes decarboxylation via vinologous intermediate. The intermediate is a vinyl alcohol. Not a carbanion. The comparision is akin to bananas and apples.

So i still haven't got the concept! Call u provide a link from where I can read it?


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Combination of polydisperse polymer mixtures
Say I'm given a mixture of batches of polydisperse polymers. I know the weight $w$, the number average $ \bar M_n$, and the weight average $ \bar M_w$ of each batch that was added to the final mixture. How would I go about calculating the $\bar M_w, \bar M_n$ and PD of the final mixture?
I'm thinking that the $\bar M_w$ value would simply be a weighted average of each of the batches, but I'm not quite sure how to get the number average.
In math, I'm thinking that $\bar M_{w,final} = \frac{\Sigma_{i} w_i \bar M_{w,i}}{\Sigma_i w_i} $
I haven't been able to reason what I should do to account for the differing numbers of molecules/moles of each batch. I think it should also be a weighted average, but using the numbers of each batch for weighting instead of the weight, but how would I calculate that?
Thank you

What do you mean by "numbers of each batch"?

The way I think of number weighted average is molecular weight averaged by weighting by the numbers. It could be number of molecules or moles or any similar unit, I think. In this case Moles probably makes the most sense since that's how the averages are given.

It should not be difficult to derive either of the number-averaged molecular weight or weight-averaged molecular weight by their definitions. To avoid ambiguity, I will use $i$ as the index of each actual length of chains and $b$ as the index of batches.
First, the definitions:
$\bar{M}_\mathrm{n,final}=\frac{\Sigma_in_iw_i}{\Sigma_in_i}$
$\bar{M}_\mathrm{w,final}=\frac{\Sigma_in_iw_i^2}{\Sigma_in_iw_i}$ where $n$ and $w$ are number and molecular weight of each chain length.
The numerator of $\bar{M}_\mathrm{n,final}$ is, in fact, the total mass of the mixture. Hence, $\Sigma_in_iw_i=\Sigma_bw_b$. The denominator is the total number of chains. Hence, $\Sigma_in_i=\Sigma_bn_b=\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}$.
It is slightly more complicated in the case of $\bar{M}_\mathrm{w,final}$ but you got it correct. Feel free to ignore my derivation. The numerator does not have a straightforward definition. It has to be derived from the $\bar{M}_\mathrm{w,b}$ instead. $\Sigma_in_iw_i^2=\Sigma_b\Sigma_{i,b}n_{i,b}w_{i,b}^2=\Sigma_b\bar{M}_\mathrm{w,b}w_b$. Since we have already defined the $\Sigma_in_iw_i$ above, we can have the final equation.
$\bar{M}_\mathrm{n,final}=\frac{\Sigma_bw_b}{\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}}$
$\bar{M}_\mathrm{w,final}=\frac{\Sigma_b\bar{M}_\mathrm{w,b}w_b}{\Sigma_bw_b}$
$\bar{M}_\mathrm{w,final}/\bar{M}_\mathrm{n,final}=\frac{\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}\Sigma_b\bar{M}_\mathrm{w,b}w_b}{(\Sigma_bw_b)^2}$


text:
Is it possible to synthesize a benzene ring with every carbon on the ring substituted?
We recently began covering aromatic electrophillic substitution in my first year organic chemistry course, and we've been discussing the nature of certain functional groups being activating $(\ce{-OH},$ $\ce{-OCH3},$ $\ce{-NH2})$ or deactivating $(\ce{-SO3H},$ $\ce{-NO2}).$
We haven't learned how to do everything obviously, so this probably why I can't answer this question myself, but we've learned how to add nitro groups, sulfonic acid groups, alkyl and acyl groups, and halogens.
In any case, is there way out there in the world of organic chemistry to create a benzene ring with every single carbon substituted?

Hexachlorobenzene and hexafluorobenzene exist

There are simply too many examples. You might want to narrow down your question, e.g. by specifying the exact groups of interest or adding extra restraints such as uniqueness of each group or their order.

https://de.wikipedia.org/wiki/Hexamethylbenzol https://en.wikipedia.org/wiki/Mellitic_acid

The interesting one is also a mellitic anhydride, that is formally a carbon oxide C12O9 https://en.wikipedia.org/wiki/Mellitic_anhydride

I don't really see why this is too broad; it's a simple yes/no question to which the answer is "yes", which is maybe not a super interesting answer, but it's not broad. OP isn't asking anybody to enumerate all the possible compounds or methods.