text:
Which is a better electrolyte for electroplating copper?
I need to know which is a better electrolyte to use when electroplating copper: copper sulfate or copper nitrate?

Better in what way? Copper sulfate is readily available on that well known auction site (e***) and it is very inexpensive. Copper nitrate is not so easily available because it can easily be used to make other nitrates. And nitrates are more troublesome (electroactive) than sulfates when it comes to electrochemistry.

What i was wondering is whether one of them helps influence the mass transfer between electrodes. So basically is the total mass of both electrodes added together closer to the starting mass depending on which electrolyte I use or do they both have the same outcome.@EdV

Just a guess, but I think it is about the same. Nothing more I can say, frankly.

In addition to price advantage, CuSO4 also provides a source of sulfate ions, which are relatively stable. The sulfate anion is also a known scavenger of radicals that may be present in the solution. For example, a reaction between sulfate and the hydroxyl radical (a possible electrolysis product):
$\ce{SO4^{2-} + .OH -> .SO4^- + OH^-}$
And, the creation of Cupric sulfate with the action of the sulfate radical anion acting on cuprous:
$\ce{.SO4^- + Cu^{+} -> CuSO4}$
Related reactions starting with chloride:
$\ce{Cl- + .OH -> .Cl + OH^-}$
$\ce{.Cl + .Cl -> Cl2 }$
$\ce{Cl2 + H2O -> H+ + Cl- + HOCl}$
where the hypochlorous acid is a precursor to the creation of copper oxychloride, an insoluble basic chloride, to be avoided.
With nitrate anion, the essential potential problem is that with associated radical presence (per the ongoing electrolysis), in time, NO3- can be broken down to NO2-,..., even NH3, so probably avoid also.
In the presence of further created active radicals, I suspect, a possible attack of the newly plated Copper, so I will not be surprised if you witness some variation in the quality of results by switching between salts.
More research may be valuable, here is a video that uses CuSO4 also.
My recommended answer is CuSO4.


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Molecular antibonding orbitals energy
When we heat hydrogen the energy of the molecular orbital of H2(1s) is increased .What happens to the energy of the antibonding molecular energy of hydrogen(1s*)?

So is it possible to stabilize an HeH molecule in high temperatures where the 1s* and 2s molecular orbitals will overlap?

Most likely no because you have an odd number of electrons and large nuclear repulsion. HeH(+) can be made and is very unstable. However what is helpful here is the even number of electrons, also remember that the orbitals will couple and mix to form new bonding and antibonding pairs.

The molecular orbitals themselves are not changed on heating, but more of the vibrational and rotational energy levels within the ground state potential-well become populated. Eventually there will be enough vib/rot energy to dissociate the molecule. Excited states can have a potential well also (and so behave like the ground state) but others are also repulsive in which case there are no vibrational/rotational levels. The way to excite an excited state is by absorbing a photon as they are usually too high in energy to be reached thermally, e.g. by collisions.

Energy is not stored in the energy levels themselves but in the states which occupy them.
At normal room temperature the bonding orbitals are occupied according to Maxwell-Boltzmann's statistics $P_i =e^{\epsilon_i /kT}/Z$, and store that chemical energy. At higher temperatures the antibonding orbital will become occupied releasing the chemical energy.

Yes but in hydrogen ground state 1s is filled and 1s* doesn't have any electron.

"Energy is not stored in the energy levels themselves but in the states which occupy them." very well worded!

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How to properly model a mechanochemical reaction of barium titanate and copper(II) complexes?
I would like to understand, in terms of chemical kinetics, how I could equate the reaction rate between $\ce{BaTiO3}$ and copper(II) complexes under ultrasound effect.
Examples in literature would be great!

What do you mean equate the reaction rate between? Force will generally reduce reaction barriers involving bond stretching. However, it can be quite a tricky relationship! For example, we have found that by putting a radical on a polymer causes delocalization of the HOMO, therefore the force isn't efficiently directed along the orbital like it is for the non-radical C-C bonding orbital.

Could I write such a reaction between BaTiO3 and Copper(II) complexes as:
CuIIX2/L + BaTiO3 -> CuIX/L + X-
?
Where X is a halogen and L a ligand.

Reaction rates can be calculated from the Arrhenius equation $k=Ae^{E_a/RT}$ where $E_a$ is the activation energy found at the transition state.
Please check out the Reference 1 for calculating transition states under force. Following is an excerpt from the paper (Ref.1):

As a force is delivered to the mechanophore, the energy provided by the work can be quantified by $\Delta E =−F\Delta x$ where $\Delta E$ is the change in energy, $F$ is the amount of force applied, and $\Delta x$ is the change in distance over which the force is applied ($\Delta x$  in the examples of this article will correspond to the distance between the two atoms where the tensile force is applied…

We have recently developed a python version of the code.
Force can also be added along bonds to simulate ultrasound and we have some work in preparation showing this for polymers. Essentially ultrasound creates bubbles which pop and drag the polymer in opposite directions.
References:

Allison G. Roessler, Paul M. Zimmerman, “Examining the Ways to Bend and Break Reaction Pathways Using Mechanochemistry,” J. Phys. Chem. C 2018, 122(12), 6996-7004 (https://doi.org/10.1021/acs.jpcc.8b00467).

I found an article (https://doi.org/10.1002/ceat.200401905) that Arrhenius equation could be rewrite as k = (g*(1 + d*W)^f)e^(Ea/RT), where W is the ultrasound power, but I don't know if this equation is common in sonochemistry, it is?

Awesome find! I'll check it out and let you know but I've never seen that used before. Also I edited my answer for clarity

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Should I relate the number of moles between substances when converting weight units?
Good day! I graduated from an undergrad degree of chemistry almost 3 years ago but I'm frustrated that I fail to reconcile these things:
In gravimetric analysis, to get the correct amount for the desired way of expressing the analysis result, we convert the mass of the acquired substance to the desired form, and in doing this, we should relate the number of moles of the acquired substance to the desired substance. For example, if you desire to get the grams $Al$ from a sample which you assayed gravimetrically until you have formed $Al_2O_3$, then you have to use the following formula:
$\text{grams } Al = \text{grams } Al_2O_3 \times \frac{2\times MW Al}{MW Al_2O_3}$
wherein the factor $2$ in the numerator was obtained from the term $\left( \frac{4 \text{ mol } Al}{2 \text{ mol } Al_2O_3} \right)$, which arose from the oxidation reaction $4Al + 3O_2 \rightarrow 2Al_2O_3 $.
Now, the question is, should I also use this method of conversion if I just want to express %Al from a %$Al_2O_3$ concentration of a certain substance, even if I don't plan to do a gravimetric analysis to find it? Also, this method of conversion is reaction-specific, but there could be many different ways, not only single-step chemical equations but also multiple-step ones in which the overall mole relation of the substances in a given chemical equation is different from that of a different chemical equation.
Also, what about the expression of ppm $N$ from ppm $NO_3$ and ppm $NO_2$? Does their conversion method work in the same way in which you have to know the reaction involving $N$, $NO_3$ and $NO_2$ to know their mole relations, knowing that monatomic $N$ is not very easy to generate? In this case, I can only see that the conversion method is based on the number of atoms the primary element in each form not on a certain reaction like thee previous case. But if that's the case, then I have another contradicting case to show:
Another one is that of between $NaOCl$ and $Cl_2$ in disinfection solution formulations. If you base it on the last method I mentioned, then the mole ratio should be 2 mol $NaOCl$:1 mol $Cl_2$ because you only have one $Cl$ atom in $NaOCl$ while $Cl_2$ has 2, but according to all of the reputable resources I have found in the internet, the mole ratio here is 1:1. The reason here is supposed to be because of the chemical equation $Cl_2 + 2NaOH \rightarrow NaOCl + NaCl + H_2O$.
If the answer is supposed to be obvious after all these, or I am just completely missing the point, I hope you would be kind to point it out clearly because it's really bogging me down.

All the trick is the equivalence of both considered forms, according a particular reaction process, regardless of if the process is real or just formal.
If there is a real or formal process/reaction
$$\ce{n A + ... ->  m B + ...}$$
or
$$\ce{n A + m B ->  ...}$$
then $n \cdot M_\mathrm{A}$ grams of substance $\ce{A}$ is equivalent to
$m \cdot M_\mathrm{B}$ grams of the substance $\ce{B}$.

So you may calculate the equivalent in atomic $\ce{N}$ without producing atomized nitrogen but just formally.
For particular forms of nitrogen, a particular mass of nitrogen form {in whatever unit} is equivalent to a particular mass of "$\mathrm{N_{N form}$". Then total nitrogen equivalent is then the sum of the partial nitrogen equivalents.
About 62 g of nitrates is equivalent to 14 g of "N-NO3", the nitrate nitrogen.
About 46 g of nitrites is equivalent to 14 g of "N-NO2", the nitrite nitrogen.
About 17 g of ammonium it equivalent to 14 g of "N-NH3", the ammonia nitrogen.
(The same can be expresses in other units as well, like ppm(w). )
as there can be written formal reactions
$$\ce{NO3- -> N + 3 O + e-}$$
$$\ce{NO2- -> N + 2 O + e-}$$
$$\ce{NH3  -> N + 3 H}$$

Similarly, phosphorus in fertilizers is conventionally expressed as $\ce{P2O5}$, even if there is no $\ce{P2O5}$ ( ,because there is $\ce{P4O10}$ )

For the sodium hypochlorite case, it depends on  if you calculate
the total chlorine content :  ( 1:1 as $\ce{Cl}$, formally only )
$$\ce{NaOCl -> Na + O + Cl}$$
or the active chlorine content: ( 1:1 as $\ce{Cl2}$, really)  ( or 1:2 as $\ce{Cl}$, formally ).
$$\ce{NaOCl + 2 HCl -> NaCl + H2O + Cl2 (-> NaCl + H2O + 2 Cl )}$$
The both ways ( 1:1 vs 1:2 ) are equivalent for obvious reason, as the half molar amount of substance with the double molar mass has the same mass.

Also note that by the same formal manner is determined the oxidation number:  $\ce{ H2SO4 -> 2 H+ + S^6+ + 4 O^2-}$

Sorry about this but, I don't get many things from your answer. (A) What do you mean with "equivalence of both considered forms according to a particular reaction process?" (B) Also this sentence, "For particular forms of nitrogen, a particular mass of nitrate {in whatever unit} is equivalent to a particular mass of "NNO3"." (C) And also, what is the difference between total chlorine and active chlorine content? I didn't get what should be obvious here.

See the update. It seems you managed to forget all from the undergraduate studies, as when I was a student, it was all the high school stuff.

Thank you for taking the time to flesh out your explanation. :) We actually didn't learn the chemical equations having a "formal" manner, but I can understand what you're trying to say. Am I correct that you're saying that it's okay to just base it according to the number of atoms between those chemical forms since they are equivalent "formally"?

Yes, that is correct. Also note that by the same formal manner is determined the oxidation number:  $\ce{ H2SO4 -> 2 H+ + S^6+ + 4 O^2-}$

Thanks, I understand. However, for the sodium hypochlorite case, I'm actually meaning to find the wt% available chlorine and its relation with wt% NaOCl, and as we know, the active components in disinfection that we're interested in are hypochlorous acid and sodium hypochlorite, so why do the literature refer to the reaction which produced these active components, instead of the disinfecting action of these active components?

As usually, it is historical convention. As active chlorine may be provided by various ways, e.g. by applying elementary chlorine. Similarly, there is an older unit of water hardness dGH equal to the equivalent concentration  10 mg/L of CaO, even if there is no CaO in water.

Right. You raise an important point, historical convention. What about more complex conversions like converting Fe2O3 to Fe3O4, or MgP4O7 to P2O5? In these cases, what should hinder me from basing it on the oxygen atom instead of the Fe or P?

Let us continue this discussion in chat.


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Writing half equation confusion
The following is the answer for a question that I am doing:

Reduction: $\ce{X^2+(aq) + 2 e- -> X(s)}$

Can I write $\ce{X + 2 e- -> X^2-}?$ Why can't I start with $\ce{X}$ instead of $\ce{X^2+}?$
EDIT:
The question:
Deduce the balanced equation for the voltaic cell constructed using metals B and C as electrodes.
Context:

Those are two different things, you can not substitute one for the other

For starters, the second reaction may be impossible, because $\ce{B^{2-}}$ is not stable. If both reactions exist, they will happen at different electrochemical potentials, or voltages. Consider the real example of $\ce{Cu/Cu^+/Cu^{2+}}$.

What was the original question? Feel free to [edit] it in. Also, please avoid using generic clickbait-alike titles. The title must precisely reflect the essence of your question and don't bee too broad.

@andselisk I think there is a misunderstanding, that element is not Boron, I just mistakenly named it B. I changed the title and renamed that to element x. Sorry about that.

@Negrawh I made no assumption that it is, and it does not matter to my first comment. Now that you have edited in the question: I have never seen a metal-based anion either.

Going by a car from place A to place B can be replaced by arriving at place B from place A, not by going from place B to place C that does not even exist..

@Poutnik Sorry sir, I don't get the analogy.

How so ? Place A = X2+, place B = X,  place C = X2-  :-) We do not know stable metallic anions in water solutions. Unless you mean X = M2+, what would be very unusual and contraintuitive convention and as such strongly not recommended. Chemists are used to symbols carrying their true elementary charge and not the charge transposition.

Can I write $\ce{X + 2 e- -> X^2-}?$ Why can't I start with $\ce{X}$ instead of $\ce{X^2+}?$

The question says that the metals are $\ce{A(s), B(s), C(s) and D(s)}$. That means the letters stand for the uncharged element, not for something more arbitrary. In the table, the letters are used again, in the following way (shown for A):
$$\ce{A(NO3)2(aq)}$$
Because nitrate has a negative charge, the metal A forms an ion in water with a positive 2 charge ($\ce{A^2+}$).
$$\ce{A(NO3)2(s) -> A^2+(aq) + 2NO3-(aq)}$$
With the same argument, you can figure out that B also forms a divalent cation in solution, while C and D are monovalent. There is no flexibility of having $\ce{A}$ stand for an ion, or for the formation of a metal anion; you should stick to the species and symbols in the text and table of the question.
Examples where letters stand for something else
In acid/base chemistry, A and B are used in the following way:
$$\ce{AH <=> A- + H+}$$
$$\ce{B + H+ <=> BH+}$$
If you look closely, $\ce{B}$ stands for a base (with conjugate acid $\ce{BH+}$) and - which might cause confusion - $\ce{AH}$ stands for an acid (with conjugate base $\ce{A-}$). Neither A nor B have to be a single atom, and their charge could be anything. However, $\ce{B}$ and $\ce{A-}$ have one charge less than $\ce{BH+}$ and $\ce{AH}$, respectively.
In redox (bio)chemistry, $\ce{NADH and NAD+}$ are used. Again, the charges are to be interpreted as relative, and the relationship between the two is given by:
$$\ce{NAD+ + H+ + 2e- <=> NADH}$$

Hey sir thank you for the answer! I answered the question: can you check whether it is correct? The balanced equation using B and C as electrodes is: $2C + B^{2+} --> B + 2C^+$ How come the reactant side looks like the product side? Is it still correct and does this occur often? Many thanks!

@Negrawh The reactant side looks like the product side because in a balanced equation, you have the same type and number of atoms on both sides. In this case, the only difference is the transfer of electrons (no hetero-atomic covalent bonds made or broken, just going from metallic element to ions and reverse), so the only indication of the reaction is a change in charges.

A is Zinc or Iron. C is Copper. B is Silver. I don't know what is D.
The reason is that when dipped in Cu2+ or Ag+ solution, a piece of iron or zinc reacts with both and produces a metallic deposit of Cu or Ag.
A metallic strip of silver never reacts with Zn2+ or Cu2+.
A metallic strip of Copper reacts with a Ag+ solution, producing a nice silver deposit. And it does not react with Zn2+.


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What is the general or ideal equation for the radial wave function?
I came across this question on the internet:

What is the value of $(n + l + r)$ in a given wave function, where $n =$ principle quantum number and $l =$ azimuthal quantum number and $r =$ total number of node present in given wave function.
$$= \frac{1}{81(6\pi)^{1/2}}\left(\frac{Z}{a}\right)^{7/2}r^2\mathrm e^{-Zt/3a}(3\cos^2\theta - 1)$$

But I don't know how to get / interpret the values of $n,$ $l$ and $r$ from the given equation.
Is there any general formula for the radial wave function or I need to calculate it through a series of integrals or something.

I've rewritten your question from the screenshot for better searchability and appearance, but there is no equation, the left side appears as non-existing. Also, please add the source of the question, and you might want to consider rewriting the title as IMO it doesn't correlate well with what is being asked.

The question is from my school test papers. But I didn't get what the question is about and what is a radial wave function

But you said in the question body the problem is from the internet; now it's from your school's test papers? Is there any way to have a look at the source?

This looks like a modified form of one the normalized hydrogen wavefunctions.  What the value of $t$ is in your function is a mystery to me.


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Why is o-nitroaniline not more basic than aniline?
We know from SIR effect (Steric Inhibition of Resonance) that in o-nitroaniline, the nitrogen's p-orbitals will change their plane, causing the conjugation/resonance of its lone pairs to stop. So, the only effect acting on it now is −I effect from the $\ce{NO2}$ group.
Comparing with just aniline, the $\ce{NH2}$ lone pairs are in constant resonance with the benzene ring making them unavailable for donation. Thus, aniline should be less basic than o-nitroaniline. But it isn't as per experimental values. So what's going on? What's the logic behind this?

Oh, do we know that? No, it's your misunderstanding - in such cases it's ortho-effect not SIR.

Yes it is true that SIR is observed in o-nitro aniline so it will not show -M effect.
But I think you're confused in the resonance of the lone pair of NH2 group with the benzene ring which in fact occurs in both cases. The thing that makes o-nitro aniline less basic is the -I effect which makes the lone pair on - NH2 group less available for donation .
In fact the complete order of basic strength of nitro substituted anilines is:
[Aniline>m-nitro aniline>p-nitro aniline>o-nitro aniline ]

How can resonance of Lone Pairs of Nitrogen of$NH2$ occur in ortho case?? Resonance completely stops due to SIR effect.

I think it the nitro group goes out of plane


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Chemicals that form coloured complexes with Pb2+
I’m currently conducting a photo-spectrometric analysis of lead concentration in water and was looking into using Eriochrome Black t as an indicator for Pb2+ (due to the complex forming a reddish colour). Because Eriochrome is relatively expensive, would there be any alternate chemicals that can complex lead in solution of the form Pb(OH)NO3, such that colour is present. This is because the source of lead is from a dissociation reaction of lead (II) nitrate when it is dissolved in water. The resulting lead is complexed.
All the best!

Any idea of the ball park lead concentration? The dithizone method is what I used long ago, as an undergrad taking Instrumental Analysis, trying to detect lead in Hudson river water. Even after extensively boiling down the water specimens, to pre-concentrate the lead, the method simply was not sensitive enough (or the river was not polluted enough!). Precipitation methods also have detection limits, so 10 ppm should be relatively easy, but 10 ppb is hard.

Interesting! I’m performing an adsorption experiment where the initial concentration of solution is 20 mg/L (or equivalently 20 ppm) of dissolved Lead Nitrate. However this is only the initial concentration, and the final solution will have substantially less lead due to uptake by the filtration medium. I was thinking about complexing lead so that I can perform a photospectrometric analysis. Is there a detection limit for this process too?

That I do not know. My guess is around a ppm or so. Is there any chance you can do a quick and dirty home lead test using the test kits they sell in hardware stores? That might give you a hint about the detection capability you will need.

In the old book from Hopkin and Williams, Organic Reagents for Mineral Analysis. London (1933), the following information can be found in the Chapter "Diphenylthiocarbazone, or Dithizone".
Diphenylthiocarbazone is a black powder, insoluble in water, but soluble in organic solvants, giving green solutions. Its main application is its reaction with lead. Lead ions produce a red color to its CCl4 solution. To 10 mL Lead aqueous solution, 1 mL KCN 10%, and two drops of ammonia are added. Add 2 mL of a dilute solution of diphenylthiocarbazone in CCl4 (50 mg/L). Stir. A reddish color must appear if the sample contains 0.001 mg Pb. A similar color appears without lead, due to the solubility of the reagent in CCl4. Addition of KCN removes totally any excess of reagent in the aqueous phase.
The only drawback about this method is the necessity to use KCN, which is forbidden today. I have found in the same book a small article about benzidine, where they say that lead solutions give a blue color when mixed with benzidine, if a solution 0,05% benzidine in acetic acid 10% is used.
Lead can also be detected and quantitatively determined by 8-hydroxyquinolein. But the book mentions only the following bibliographic  references :
V. Marsson, L. Haase, Ueber die Bestimmung des Bleis mit Hilfe von o-Oxychinolein, Chem. Ztg, 52, 993, (1928)
V. Hovorka, Sur le dosage du plomb au moyen d'ortho-oxyquinoléine, Coll. Czech. Chem. Comm, 9, 191 (1937)

I’ve looked into it and I believe using chelators to complex lead would be preferable (and safer), so the 8-hydroxyquinolein would be a viable alternative. Is the chelator strong enough to complex lead despite it being complexed to Pb(OH)NO3?

One possible answer, to quote:

you add colourless potassium iodide solution (or any other source of iodide ions in solution) to a solution of lead(II) nitrate, a bright yellow precipitate of lead(II) iodide is produced.

More interesting, to a soluble lead ion, add chlorine bleach (NaOCl). This results in a white transient precipitate of Pb(OCl)2, which on mild heating decomposes, leaving behind red, PbO2. To quote a source:

Initially, lead(II) ion reacts with hypochlorite ion to give a white precipitate of lead(II) hypochlorite:
$\ce{Pb^{2+} (aq) + 2 ClO^{-} (aq) -> Pb(ClO)2 (s) }$
The lead(II) hypochlorite is then oxidized to lead(IV) oxide. Reaction is slow at room temperature but much faster when the tube is placed in boiling water.
$\ce{Pb^{2+} (aq) + 2 H2O(l) -> PbO2 (s) + 4 H+(aq) + 2 e- }$
$\ce{ClO-(aq) + 2 H+(aq) + 2 e- -> Cl-(aq) + H2O(l) }$

What's "$\ce{Pb^2++}$"? Besides, the second URL leads to some sketchy-looking webpage that doesn't seem to cite any reputable sources.

$PbI_2$ and $PbO_2$ are both insoluble. And Thunderbolt114 would like a soluble compound.

As Maurice suggested, I believe this analytical method only works if the solute is present in the aqueous phase. In other words, I can’t test on a precipitate. Are you familiar with any coloured complex formed with lead in solution?

Per Wikipedia on PbI2: "Typically, a solution of PbI2 in an organic solvent, such as dimethylformamide or dimethylsulfoxide, is applied over a titanium dioxide layer by spin coating."

I have performed (and posted online here http://www.sciencemadness.org/talk/viewthread.php?tid=62249#pid429056) the reaction of adding NaOCl (as common Bleach) to  Lead acetate. The reaction proceeds as indicated resulting in the red PbO2.  Quite interesting!


text:
Scale to be considered for comparing electronegativities of nitrogen and chlorine
The Pauling scale gives the $\chi$ values of $\ce{N}$ and $\ce{Cl}$ to be $3.04$ and $3.16,$ respectively, but the Allen scale gives the $\chi$ values of $\ce{N}$ and $\ce{Cl}$ to be $3.066$ and $2.869,$ respectively.
Another point that should be made note of is that the most electronegative element as per Allen scale is $\ce{Ne}$. So, why is it accepted in the first place? And whose electronegativity is actually greater?

It is best to consider them equally electronegative if in different compounds but consider N more electronegative if in the same compound.

Is this just your opinion, or you can provide a reputable source or logical arguments to back up this claim?

Unanswered duplicate: Which is more electronegative: chlorine or nitrogen?

In NCL3 - reactions and uses is listed $\ce{NCl3 + 3 H2O -> NH3 + 3 HOCl}$ what gives a hint of higher N electronegativity, also mentioning moderate polarity.

@Poutnik But doesn't cite source and that would be plenty wacky too get such strong oxidiser and moderate reducer out of such reaction,,,

@Mithoron According to https://en.wikipedia.org/wiki/Monochloramine, various reactions lead to both ways, so it is kind of electronegativity  deuce.

P.S.: NCl3 article cites
Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. ISBN 978-0-08-037941-8.

Please take into account that there are a lot of different ways for calculating electronegativity.
1) Pauling obtains it by comparing the energies of dissociation of the three diatomic molecules AA, BB, and AB, stating that H electronegativity  is 2.1 !  Unit : √energy
2) For Mulliken, it is the average value between the electronic affinity and the first ionization energy of the atom. Unit : energy
3) For Allred - Rochow, it is the force acting between a valence electron and the charge of the atom when the other valence electrons are removed. Unit : force
4) For Sanderson, it is the average electronic density of an atom divided by the average electronic density of the corresponding atom of inert gas.No unit.


text:
Why does covalent bonding satisfy atoms?
Having to retake an introductory chemistry course has gotten me thinking. Covalent bonding is where electrons are shared between two atoms that both want more electrons.
Let us take $\ce{O2}$ as an example.

The red and purple oxygen atoms share 2 electrons. This means that they now both have 8. Fantastic!
However, electrons do not sit around statically but rather buzz around randomly. If the electron moves around, then on average having 4 electrons between 2 atoms means 2 electrons each. This means no one is any better off than from when we started.
My question is, seeing as the electron sharing would actually mean that on average each atom still only has 6 electrons, how can they become stable in this covalent bond?

One thing to keep in mind is that a lone oxygen atom does not want 10 electrons (including the 2 $1s$) around it. As a doubly-negative ion it will rapidly autoionize. But, it does prefer to share a few electrons with another oxygen atom (that shares a few with it) as part of them negotiating to make a molecule.

https://chemistry.stackexchange.com/questions/710/fundamental-forces-behind-covalent-bonding?rq=1

What answer would you want really? It's clear that "having 8 electrons" is a dumb rule of thumb for kids, especially in this case - dioxygen is a biradical and its bonding looks quite differently then you write, but on your level of learning of chemistry... let's just say that "electrons do not sit around statically but rather buzz around" isn't useful thinking there, some would even think it completely incorrect; not me but it's utterly useless when actual bonding is concerned.

You seem to forget that electrons have spins. And that there are two possibilities for the spin : up or down. No other possibilities exist. Now electrons have the tendency to make pairs with one spin up and on down. Sorry for the image, but it looks as if spins up and down represent two different sexes. And that the electron have the tendency to make "couples" of "up" and "down" together, as in the traditional way of being together. On the average, there exist the same amount of electrons "up" and of electrons "down".
Let's take H atoms. Half to the H atoms have spins up, and half down. If you produce somewhere a lot of free H atoms, they will collide. When they have the same spin, nothing happen after the collision. When they have different spins, they will join and stay together, make a covalent bond, which makes a H2 molecule,  after the collision. After a while all H atoms are groups in H2 molecules
In the next atom, Helium, there are two electrons on the first layer. They are of opposite spins and Helium will not join with any other atom, and not produce any molecule.
For heavier atoms, form Li to Ne, there is enough room for 8 electrons. The first four have the same spin. The next four have the opposite spin. Hopefully you may understand your self that Li makes LiH, Be makes BeH2, etc. But Oxygen has 6 electrons, 4 of a given spin and 2 of the opposite spin. In Oxygen, there are two doublets which cannot form covalent bonds. Only 2 can do it.
I will not develop this theory too long. You may do it your self. It is not too difficult to do it.

It doesn't satisfy the atoms. It's intended to satisfy the molecules by placing electrons in the lowest energy states consistent with the electrostatic potential field imparted by all the atoms in the molecule. These criteria do not "satisfy" the individual atoms, they override satisfaction of the individual atoms.
As an example, consider the case of water. To fill the valence shells of the two individual hydrogen and one oxygen atom we would need fourteen electrons including the inner-core ones on the oxygen, but the neutral water molecule has only ten. The atoms in the water molecule are then so arranged that their combined electrostatic potential imparts just enough low-energy orbitals to fit with the ten electrons, overriding the original atomic criteria that added up to fourteen. The "extra" orbitals that would have saturated the original atoms with fourteen electrons are pushed to higher energy in the molecular potential field, and no longer need to be (or should be) occupied in a stable molecule.


text:
How to experimentally determine the elemental composition in fingernails?
I am doing a project in my highschool about analysing C/N and C/S ratio of finger nails of various ages and sexes.
Please suggest me a way to do this. I don't know where to start, I know the various methods of estimating the amount of N or S present theoretically. Please provide info that how should I dissolve the nails etc.
I will be doing my experiment in a nice lab which has got all necessary equipment that I might need, so please answer without considering the limitations of my lab.
I got the idea by this website: https://pubmed.ncbi.nlm.nih.gov/18424895/

Sorry, not a doable project at a high school level. Of course elemental analysis can be done for carbon but sulfur content is very tricky. Sulfur is also done by elemental analysis but when the amount of S is very small, it becomes tricky. Secondly will cost a lot, you will at least to do duplicates to get a reliable measurement. Think of other projects which are doable. Secondly use Google Scholar and see what others have done before.

@M. Farooq OK Sir I might not do the sulphur analysis so can you suggest me a more exhaustive and accurate approach for nitrogen content

Nitrogen is also determined by elemental analysis. There are commercial instruments called CHNS analyzers. They burn the organic compound, and analyse the gaseous products, CO2, NO, SO2, by gas chromatography. Many universities may have this service. You will have to go through your teachers and see if they have access to this service. They need to pay as well. Have a look at the Wikipedia article https://en.wikipedia.org/wiki/Elemental_analysis

There is no nascent hydrogen! It is an incorrect notion: see the short wiki article on “Nascent hydrogen”!

@Ed V where will I find the short wiki article

https://en.wikipedia.org/wiki/Nascent_hydrogen

''I will be doing my experiment in a nice lab which has got all necessary equipment that I might need, so please answer without considering the limitations of my lab."  We actually do need to take into accoount the limitations of your lab, since we didn't we'd just assume you had a machine like this, which is found in labs that do C/N/S analysis:  https://eu.leco.com/product/928-series.  Once you've completed the calibration, the machine is simple to run—just load the sample and press a few buttons.  Analysis takes only about 5 minutes/sample.

These analyses are difficult to carry out at a high school level. Nails are not soluble in any solvent. They must be burned and the gases coming out are analyzed and weighed. To burn them, they have to be mixed with a great excess of copper oxide, put in a long horizontal glass tube, heated from below. A current of oxygen gas should be sent through the glass tube, and the gas should be absorbed first by solid $\ce{CaCl2}$ which absorbs $\ce{H2O}$, then by $\ce{NaOH}$ which absorbs $\ce{CO2}$. By weighing the $\ce{CaCl2}$ and $\ce{NaOH}$ flasks before and after the operation, you can calculate the amount of $\ce{C}$ and of $\ce{H}$ contained in your original sample. To measure the amount of nitrogen of your nails, you have first to destroy them by sulphuric acid, then add them to $\ce{NaOH}$ and some Dewarda alloy (mixture $\ce{Al}$, $\ce{Zn}$) have it boiled to liberate $\ce{NH3}$ gas, and then absorbed this gas containing $\ce{NH3}$ into a solution of an acid like $\ce{HCl}$ or $\ce{H2SO4}$, and then titrate the remaining acid to deduce the percentage of $\ce{N}$ in your original sample. This can not be easily done in a high school lab, slowly.

Maurice, a good rendition, however, could you comment on whether the procedures you outlined are particularly suitable for determining very small differences, in say, sulfur presence.

@ Ajkoer. The traditional chemical analysis I described has been used for all

Thanks for the reply. For others, I would point out that the use of  "NaOH and some Dewarda alloy (mixture Al + Zn) " is also a de facto use of nascent hydrogen.

@ Ajkoer. The traditional chemical analysis I described has been used by all chemists in the 19th and 20th century to obtain the chemical formula of a new compound. But it is not adapted to very small differences as you stated.  There exists several ways of analyzing Sulfur. The simplest is Liebig's approach : A known amount of substance is fixed on a platinum grid and burnt in the middle of a big bottle containing 6 liters $\ce{O_2}$. Later on, a solution containing 10% $\ce{Na_2O_2}$ is added, then the whole is acidified by HCl, and the ion sulfate is precipitated by $\ce{BaCl_{2}}$.

@Ajkoer. Yes. Dewarda alloy produces nascent H in NaOH solution.

Some suggestions:
First, collect the fingernail samples by applying a nail file to a subject's nail. This creates a fine powder.
Second, precisely weigh the sample and place the nail powder in distilled water.[EDIT] Here is a possible dissolution of nail path per a reference that should be verified in its accuracy, which claims:

COCO COLA IS CAPABLE OF DISSOLVING A NAIL
Water is great for your overall health including your nails, whereas fizzy drinks like Coke are not so good.  In fact Coco Cola is kind of scary.  If you place a piece of nail in a bottle of Coca Cola it will dissolve within 4 days, that’s how corrosiveness it is!

where the chemistry is likely conditional on the presence of the Aluminum (some electrochemistry at play). More interesting, do nails, in time, dissolve in an Aluminum can containing just Phosphoric acid, carbonated water, sugar and perhaps some NaCl, in a closed system (no gas escape)?
Third, I like a nascent hydrogen approach here to breakdown the nail powder (research the Marsh Test to detect small amounts of arsenic). So, further add Al/NaOH (or Zn/HCl) and collect the vapors in dilute H2O2 (which has been pH adjusted to a recorded value at over 7). [EDIT] This step may have to be re-evaluated based on revised Step 2, perhaps just neutralize with NaOH/H2O2.
Depending on the test equipment available, a simple comparative test would be to investigate the evidence/concentration of a visible sulfur suspension (a light-based test), as a pH check alone, for example, for any significant decrease in pH (a measure of increased acidity from sulfur-based acid formation) would not likely be sensitive enough for small differences in sulfur concentration.
One could refine the process starting with test samples containing but small differences in the amount of sulfur presence, as a feasibility check.
[EDIT] To be honest, I (and apparently at least one other) have doubts on the potential feasibility of this project. However, with the to-be-confirmed electrosynthesis dissolution of human nails, I would have to upgrade your chances especially based on your possible access to equipment to confirm relative elemental composition.

Thanks. Will I be able to use the gases evolved in the coca-cola exp, also can you edit your answer without considering the limitations of my lab since it has got almost all necessary equipments that I might need,

Also instead of marsh test can you suggest more arsenic confirmation tests


text:
Seeing solubility in various cases

Match the following:
$\begin{array}{|c|c|c|c|} \hline &\textrm{Column-I}&&\textrm{Column-II} \\\hline \textrm{(A)}&\ce{AgBr}&\mathrm{(p)}&\mathrm{(Solubility~in~water~is~more~than~expectation)} \\ \hline \textrm{(B)}&\ce{AgCN}&\mathrm{(q)}&\mathrm{(Solubility~in~acidic~solution~is more~than~in~pure~water~\\(consider~no~common~ion~effect~from~anion~of~acid)}\\ \hline \textrm{(C)}&\ce{Fe(OH)3}&\mathrm{(r)}&\mathrm{(Solubility~in~strongly~basic~solution~is~more~than~in ~pure~water)}\\ \hline \textrm{(D)}&\ce{Zn(OH)2}&\mathrm{(s)}&\mathrm{(Solubility~decreases~in~presence~of~common~anion)}\\ \hline \end{array}$

Link to original image of the question.
The problem is to 'match' options in the two columns (possibly having multiple answers?).
How to determine the solubility?
I tried this through reaction like
$\ce{AgBr + H2O -> AgOH + HBr}$
How do I know its solubility is greater or smaller than expectation?

Feel free to add your attempt (a thought process), and what have you tried/read/analyzed to solve the problem.

A is wrong. $\ce{AgBr}$ is one of the least soluble compounds of $\ce{Ag}$. Why don't you have a look in the table of solubility products to be convinced? Further more, $\ce{AgBr}$ will never react with $\ce{H_2O}$ to produce a strong acid like $\ce{HBr}$ and a base like $\ce{AgOH}$. The reaction goes the other way round. Strong acids react with hydroxydes to produce a salt and water.
C is wrong, because of the Le Chatelier's principle. Try to get the reason your self.
To solve D, you must know that $\ce{Zn(OH)_2}$ is soluble in an excess of $\ce{OH^-}$ ions

The point is to match up entries in the two columns.

@Buck Thorn. OK. I know but I thought we should not give the whole answers without the author showing us some beginning of reasoning.  So let me give him two partial answers. AgBr is poorly soluble. But its solubility decreases in if $\ce{Br-}$ ions are in excess, due to its solubility product, and the absence of any complexation reactions. So $A$ is related to $s$. Now the only product which is soluble in alcaline medium is $\ce{Zn(OH)2}$, giving the ion$\ce{[Zn(OH)4]^{2-}}$. So $D$ is related to $r$. The next substances are easy to match.


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Non-toxic organic bases?
I am trying to find out if there are any non-toxic organic bases that could be given orally on a non-empty stomach to treat acidosis. From my research, histidine and diluted choline hydroxide were the best candidates. Some phosphazenes, including polyphosphazenes, could make the cut, but I don't know which ones exactly.
Wikipedia [1] lists the following compounds as organic bases: pyridine, alkanamines (such as methylamine), imidazole, benzimidazole, histidine, guanidine, phosphazene bases, hydroxides of quaternary ammonium cations or some other organic cations.
The LDLO of guanidine given orally to rabbits is 500 mg/kg [2]. The LD50 of imidazole (rabbits, oral) is 950 mg/kg. Pyridine seems to be in the same range [3]. Benzimidazoles are also toxic.
Histidine, which is considered the most toxic amino acid, can be therapeutic in doses of up to 4.5 g/d, but is toxic in the range 24–64 [4].
The last ones from this list are choline hydroxide and phosphazenes. The UTL of choline is 3.5 g/d; in the 10-16 g/d range it causes fishy smell so it probably overloads us with TMA, which is a carcinogen. Polyphosphazenes are being studied for drug delivery; apparently their metabolism generates low-toxicity products such as urea and phosphates. [5].

Well, that's quite some research, still -1 from me - it's clearly too broad, also not realistic. Everything is "toxic" if dose is too high. This paragraph about histidine in particular shows that it would be all too good already - "toxic" when at least 24 g is eaten? Maybe water is also too toxic for you?

Naturally occuring amino acid with basic sidechain - L-Arginine:

Oral supplementation with L-arginine at doses up to 15 grams daily are generally well tolerated.

source here

Yes,  I am upgrading as this is an excellent answer as it is non-toxic and has associated health benefits. Some may be aware that Al(OH)3 was once viewed as a non-toxic treatment for excess acid, but more research suggests, based on the interaction of the superoxide radical anion with Al3+, it is probably not a good idea to have those Aluminum ions running around in ones body.


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Does choline bitartrate react with hydrochloric acid?
Choline hydroxide is a base and reacts with HCl to form choline chloride salts and water.
I am wondering if choline bitartrate can also neutralize HCl. In case it doesn't, is there any other cheap form of choline that does?

HCL is a stronger acid so it will displace the bitartarate forming choline hydrochloride and tartaric acid


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How would you describe the energy of solution formation involved in the dissolution of sodium chloride in vinegar?
I have found that the lattice energy of dissolving salt in water is 779 kJ/mol. While 774 kJ/mol is its hydration energy. Subtracting the two values gives a positive number therefore dissolving table salt in water is endothermic. I know that vinegar comprises of mostly water but how do you exactly explain the energy absorbed and released in the process if acetic acid is involved? Is there any constant that you know involving the dissolution of salt in acetic acid?
Our teacher wants us to include a discussion comparing water and vinegar in dissolving salt considering the enthalpy of solution but I am not quite sure what that means. I don't know if this question is too basic but I would really appreciate if I would get some answers. Thank you T.T

Vinegar is a solution 6% acetic acid in water. I doubt that the heat of dissolution of salt in vinegar is really different from what it is in water. Dissolving salt in vinegar is not like dissolving salt in pure acetic acid.


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Why is 3,6-dimethylpiperazine-2,5-dione optically active in spite of having a centre of symmetry?
I learnt in my chemistry class that if a molecule has an element of symmetry (either a plane of symmetry or a centre of symmetry), it will not be chiral and hence will not show optical activity.
My professor told me that in alanine anhydride the carbon atoms with methyl groups would be non-planar, hence it will be non-symmetrical:

But if I consider one of the above-mentioned carbon atom above the plane of the ring, and the other below the plane of ring, a centre of symmetry will still exist. Where am I going wrong?

Well You are assuming that the ring is planer which is not the real case

@Pj30 The trans isomer has a centre of symmetry even though the ring is non planar(the answer posted by someone has its image)..How then is the compound optically active?

There is no evidence (e.g. aromaticity) to support your claim that the compound is planar. In fact, crystalographic studies [1] revealed existence of both cis- and trans-3,6-dimethylpiperazine-2,5-dione. Corresponding CCDC numbers are LCDMPP01 and TRDMPP01, respectively — feel free to open the links and play around with the 3D structures in JSmol to understand the difference.
According to Sletten, in cis (ʟʟ) modification the ring is puckered resembling a skewed boat conformation, whereas the ring in trans (ᴅʟ) molecule is nearly planar:

Figure 1a. cis-3,6-dimethylpiperazine-2,5-dione (drawn from LCDMPP01)

Figure 1b. trans-3,6-dimethylpiperazine-2,5-dione (drawn from TRDMPP01)
References

Sletten, E. Conformation of Cyclic Dipeptides. The Crystal and Molecular Structures of Cyclo-D-Alanyl-L-Alanyl and Cyclo-L-Alanyl-L-Alanyl (3,6-Dimethylpiperazine-2,5-Dione). J. Am. Chem. Soc. 1970, 92 (1), 172–177. DOI: 10.1021/ja00704a028. (Open Access)

This is still rather nitpicking then full answer. Thing is if trans isomer is centrosymmetric or not. I wager it is and therefore OP's right. (Perhaps some additional effects would come with interactions in solid state or temperature, though.)

Yes the trans isomer has a centre of symmetry.How is the compound still optically active then?(I got this question wrong in a test where they had asked whether the compound shows optical activity or not)

If you flip the directions of the methyl groups (make the sticky-down one on the left stick up, and vice versa), can you superimpose that on the original? If not, then you have two enantiomers. If you can, then you don't.

The sufficient condition for a compound to be optically inactive is that it should have a centre of symmetry..and i think the trans isomer has one.Can you explain to me why is the trans isomer optically active. I agree that the cis isomer is optically inactive due absence of any element of symmetry

The transfer isomer does not have C2 symmetry or a centre of inversion. If you try to superimpose a mirror image, the carbonyl oxygen would be where the NH is and vice versa. Plus, the methyls will be in opposite configuration. Has the possibility of amide tautomerism been considered too, just to muddy the water ?

@Beerhunter That's a good point, probably yes, in solution this might be the case. Here, I only compared crystallographic data.


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Arsenic bound proteins confirmatory test
I want to confirm the presence of arsenic in some protein samples, apart from the marsh test can I use the following method?--
Arsenic is present as arsenic (III) ion in the samples. Hydrogen sulfide gas when passed in the sample of arsenic (III) ion gives precipitates which are yellow in color.
The yellow precipitates dissolve on adding concentrated nitric acid. This solution on reaction with ammonium molybdate in presence of concentrated nitric acid again gives yellow precipitate.
This confirms the presence of arsenic (III) ions in the sample.
If not, please suggest another method
P.S I have some info about arsenic bound proteins from-
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3797521/

You could just use Hach arsenic test strips: https://www.hach.com/quick.search-product.search.jsa?keywords=arsenic+test&pr.allCategoryIds=35546331612. They are used to detect arsenic in water down to 10 ppb. There is a great deal of literature on this.

@Ed V but I can't use them in my chem project

You may wish to give this suggestion a try, courtesy of the literature on constructing nano-particles.
First, dissolve the protein samples in say ethanol or other flammable solvents.
Next, place it into an alcohol lamp with a wick composed preferably from glass fiber, but one may be able to employ woven cotton or a cotton string.
Lastly, light the lamp and place a bowl over the lamp to capture/condense escaping particles.
My recollection is that the target metal of interest, is slowly draw through the wick and thermally broken down. Still looking for the source, but here is a related article.
This is clearly relates to the final step in the Marsh Test, where the arsenic presence is distinctive.
Note, this proposed procedure, if successful, avoids the conversion of a sample of a compound into deadly AsH3 gas per a nascent hydrogen pathway, which is perhaps not required here, as you already have constructed (via an alcohol lamp), a path to 'burn' the As(III).

Will I be able to see the As(|||) on the wick?

I suspect not, just the flame residue deposited on the bowl. I also updated the source reference link, which is closer in composition to the work I was recalling.


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Finding the pH of a mixture of weak acid and strong base
What are the steps we go about in finding the pH of a solution of strong base and weak acid?
Here is the question that I've been given

What is the pH of a solution made by mixing $\pu{50 ml}$ of $\pu{0.2 M}$ $\ce{NH4Cl}$ and $\pu{75 ml}$ of $\pu{0.1 M}$ $\ce{NaOH}$, when $\mathrm pK_\mathrm b(\ce{NH3}) = 4.74$?
A. 7.02
B. 13.0
C. 9.73
D. 6.31

What I did to solve it was to use the Henderson equation for buffers,
$$\begin{align}
\mathrm{pOH} &= \mathrm pK_\mathrm b + \log{\frac{[\text{salt}]}{[\text{base}]}} \\
&= 4.74 + \log{\left(\frac{0.2 \cdot 50}{0.1 \cdot 75}\right)} \\
&= 4.74 + \log{\frac{4}{3}} \\
&= 4.86
\end{align}$$
and thus $\mathrm{pH} = 14 - \mathrm{pOH} = 9.14.$
The correct answer is C. But on putting values I'm getting the wrong answer. Can you help me figure out why? Also, shouldn't the buffer solution have a common ion?

What is the reaction between $\ce{NH4+}$ a weak acid, and $\ce{NaOH}$ a strong base?

I admit Karsten Theis has given en excellent answer for OP's question. However, I'd like to point out that this could be solve without getting confused by $\mathrm{p}K_\mathrm{b}$, which is common with novices when using the Henderson–Hasselbalch equation for buffers. The equation is derived by dissociation of weak acid ($\ce{HA}$):
$$\ce{HA + H2O <=> H3O+ + A-}$$
Hence, we can derived Henderson–Hasselbalch equation getting log value of $\mathrm{p}K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{AH}]}$ in both side and simplifying it as:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$
This works fine with any buffer solution made with a weak acid and its conjugate base. However, most novices get confused by when the buffer made with a weak base and its conjugate acid. The confusion directed mainly by two fact:

The weak base is usually provided by its $\mathrm{p}K_\mathrm{b}$ value (For example, $\mathrm{p}K_\mathrm{b}$ of ammonia is $\approx 4.3$ while $\mathrm{p}K_\mathrm{a}$ of ammonia is $\gt 34$, not $14 - \mathrm{p}K_\mathrm{b}$).
The value of ($14 - \mathrm{p}K_\mathrm{b}$) is really belongs to conjugate base of ammonia, $\ce{NH4+}$ (made by the reaction with strong acid). It is a good practice that to use this value as $\mathrm{p}K_\mathrm{a}\mathrm{H}$ (See this article).

Accordingly, it is a good rule of thumb that we can use a equation for the dissociation of conjugate acid ($\ce{BH+}$) of a weak base:
$$\ce{BH+ + H2O <=> H3O+ + B}$$
Thus, we can derived Henderson–Hasselbalch equation getting log value of $\mathrm{p}K_\mathrm{a}\mathrm{H} = \frac{[\ce{H3O+}][\ce{B}]}{[\ce{BH+}]}$ in both side and simplifying it as:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a}\mathrm{H} + \log{\frac{[\ce{B}]}{[\ce{BH+}]}}$$
You reacted $\pu{0.010 mol}$ of ammonium salt (conjugate acid of a weak base) with $\pu{0.0075 mol}$ of $\ce{NaOH}$, a strong acid. It resulted $\pu{0.0075 mol}$ of ammonia (a weak base) and $\pu{0.0025 mmol}$ of unreacted ammonium salt in the solution, which is a buffer. Since both species in same volume, the ratio of weak base to weak acid, $\frac{[\ce{B}]}{[\ce{BH+}]}$ is $\frac{0.0075}{0.0025} =3$. Now, since $\mathrm{p}K_\mathrm{b}$ of ammonia is given as $4.74$, $\mathrm{p}K_\mathrm{a}\mathrm{H} = 14 - 4.74 = 9.26$.
If you substitute these values into above equation, you get the answer:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a}\mathrm{H} + \log{\frac{[\ce{B}]}{[\ce{BH+}]}} = 9.26 + \log 3 = 9.74$$
Note: Your error in calculation is miscalculation on $\frac{[\text{Base}]}{[\text{acid}]}$ ratio.

Buffer equation
The Henderson equation for buffers is:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$
$\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{b}$ add up to 14, as do $\mathrm{pH}$ and $\mathrm{pOH}$. So the expression for $\mathrm{pOH}$ is:
$$\mathrm{14 - pOH = 14} - \mathrm{p}K_\mathrm{b}  + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$
or
$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b}  - \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$
Amounts of weak acid and weak base

Also, shouldn't the buffer solution have a common ion?

You start with $\pu{10 mmol}$ of ammonium salt (weak acid), to which you add $\pu{7.5 mmol}$ of $\ce{NaOH}$. The result is $\pu{7.5 mmol}$ of ammonia (weak base) with $\pu{2.5 mmol}$ ammonium salt remaining. So the ratio of weak base to weak acid is 1:3.
If you plug this into one or the other buffer equation, you get the answer.


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True muonium energy levels
In a true muniom(muon-antimuon atom) do the energy levels match with that of the hydrogen's?
The core is lighter for about 100 times but the muon is 100 times heavier than the electron so it would make sense the energy levels of a true muonium and a hydrogen atom to match?
Or is this wrong?

I'm not sure, but it seems the opposite - it's even more different.

This isn’t a chemistry question, really. But since the reduced mass of the system is different, the energy levels will be different.

Core mass is not all that important. Think of hydrogen vs deuterium, they barely differ at all.

The theory for a muon-antimoun "atom" is essentially the same as any hydrogen-like atom, just with masses changed.  See this Wikipedia page for the relevant formulae for relativistic (Dirac) and non-relativistic treatments.
The non-relativistic energy levels are given by :
$$E_n =  \frac { \mu c^2 \alpha^2}{2n^2}$$
where $alpha$ is the fine structure constant and $\mu$ is the reduced mass, in this case because the masses are equal, it's $\mu=\frac m 2$ where $m$ is the muon mass.
Note that the muon mass is about $206.8$ times that of the electron and the reduced mass is then $103.4$ times than of the electron making the energy levels for the non-relativistic state about $103.4$ times that of the Hydrogen energy levels.
As the Wikipedia page explain (in more detail that you asked for), a more detailed relativistic calculation using the Dirac equation includes corrections for the spin-orbit interaction.

The core is lighter for about 100 times but the muon is 100 times heavier than the electron

This is wrong.
Both the muon and antimoun have the same mass, hence the effect of reduced mass being a factor of one-half.
The reduced mass of two particles is given by :
$$\mu = \frac {m_1m_2}{m_1+m_2}$$
Strictly speaking there is no "core" as both the muon and antimoun are of the same mass whereas in Hydrogen the electron is of much smaller mass than the proton.

So a bound muon-antimuon pair would actually be almost exactly described by the mathematics of the electron-positron pair (positronium), up to a scaling factor equal to the electron/muon mass ratio. Is that correct?

@NicolauSakerNeto A scaling factor of half the ratio of the electron and muon masses !  That's the reduced mass factor coming into play.

@StephenG But in positronium you also have this factor, so Nicolau is right.


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Why is L-lysine only sold as L-lysine HCl and not as base (cf. L-arginine)?
Both L-lysine and L-arginine enjoy popularity as supplements, but only the latter is sold either as HCl or as base. Given that L-arginine is even more basic than L-lysine (pH 12.5 vs 10.5), that is surprising! Is there a reason why? And could I isolate L-lysine base from L-lysine HCl with a simple reaction? I am thinking of mixing it with sodium bicarbonate, but then I have to remove the NaCl.

Lysine hydrochloride is an easily handled, stable and well-behaved solid. Lysine freebase is not, it is a sticky oil.

Interesting! I had just thought that the problem might have been the opposite. If lysine freebase were too powdery, then it could be dangerous to the lungs.


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What are some possible classification tests that one can use for o-nitrobenzoic acid?
I have analyzed an IR spectrum and found out that my compound (o-nitrobenzoic acid) has a nitro group, carboxylic group, and benzene ring.
I was curious to know: if regioselectivity effects the solubility of the compound? If/If not regioselectivity effects the solubility of o-nitrobenzoic acid, what would be the best solubility test to run for o-nitrobenzoic acid since it has a carboxylic acid and a nitro group in the ortho-position? And, if I wanted to test o-nitrobenzoic acid functional groups, what would be some good classification tests to run?

Of course the structure of a molecule affects solubility, but solubility isn't a particularly specific test. Old fashion organic qualitative analysis has all sorts of tests for different functional groups. There is also melting point/boiling point for isomers. MPs - 2-Nitrobenzoic acid 147.5 °C; 3-Nitrobenzoic acid 139 to 141 °C; 4-Nitrobenzoic acid 237 °C

All substituted benzoic acids are easily soluble in NaOH solutions. Your acid can be esterified and give perfumes. It also may be reduced by iron in acidic solution to produce o-aminobenzoic acid


text:
How to prepare cyclohexanone without starting with cyclohexanol?
What's an easy way to prepare cyclohexanone if I dont have direct access to cyclohexanol? I'm willing to synthesize cyclohexanol from more common materials, then make cyclohexanone from that. But most of the instructions for one cyclohexano- say to start with the other cyclohexano-.
I'm looking to start with common materials and synthesize cyclohexanone or cyclohexanol. If i have cyclohexanol, i can use bleach and acetic acid to make cyclohexanone.
Since I dont have access to lab deliveries at the current moment, what are the most common chemicals to start with to make this happen? AKA, stuff I can order legally online without a lab cert.

Can you access chloro or bromocyclohexane?

Radicalic bromination of cyclohexane, the one common alkane where this leads to just one product.

I can access potassium bromide and small amounts of cyclohexane. If it matters, i own a centrifuge as well  (cheap, 4000 rpm max)

Found this: https://www.slader.com/discussion/question/because-bromocyclohexane-is-a-secondary-alkyl-halide-both-cyclohexanol-and-cyclohexene-are-formed-wh/

Based on the link above, I can make bromocyclohexane, then mix it with acetate and water to get cyclohexanol (at 60 deg C?). Then i add acetic acid and bleach and mix at the same temp. Then I add hypochlorous acid and mix. And i end up with what % cyclohexanone? What rough extraction / filering steps am i missing at each step?

i just realized a centrifuge probably wouldnt help at these quantities, but for getting a catalyst out from another pathway it might

Do you have access to Dimethyl Sulfoxide?

Your bounty sounds weird. You want to do that in your kitchen, in a quantity of gallons? Why, in the name of whatsit?

Why are you doing this? Cyclohexanone is available commercially. 2L is less than $100. It's kind of a waste of time and money to try to make it...

Ummm why do you care why I'm doing it? Gasoline is 3 dollars per gallon. 2L for 100  dollars is a terrible deal. I need to make plastic for coronavirus masks

Do you have access to phenol? If you have, then use that to make cyclohexanol (hydrogenation of phenol using nickel catalyst):

Then treat cyclohexanol with copper(II) and chromium(III) oxides to convert it to corresponding ketone (Dehydrogenation):

Benzene could work as a starting material but the reaction might be costly and hectic as the catalyst is not a household item$\ce{^{[2]}}$. Secondly, it is used as a starting material in the synthesis of nylon and the reaction intermediate is a mixture of cyclohexanol and cyclohexanone(mixed oil)$\ce{^{[3]}}$ which add to your task of separating them. Better convert it to phenol$\ce{^{[4]}}$.
Notes and References

http://www.leck.co.uk/demo/eci3/polymers/polyamides.html (above reaction synthesis source). It is a part of polyamide synthesis
https://link.springer.com/article/10.1023/B:CATL.0000011080.03035.eb
http://www.greener-industry.org.uk/pages/nylon/4nylonpopup.htm
https://www.quora.com/What-is-the-method-to-convert-benzene-to-phenol
https://russianpatents.com/patent/220/2205819.html

Meta comment about phenol: I think it is not wrong to (still) use the Thiele ring here provided one keeps in mind that it symbolizes the six delocalized $\pi$ electrons.  The Kekulé-pattern as if there were $\ce{C=C}$ double and $\ce{C-C}$ single bonds in benzene (known to represent one mesomeric formula only) however has the advantage to count the $\pi$ electrons in question exactly.  It is too easy to draw two Thiele cycles in the formula of naphthalene and to think «2 Thiele rings with 6 $\pi$ electrons each = 12 in total», while the Kekulé formula shows there are only 10 of them.

What machine do I use to run a reaction at 5 atmospheres, lol? A pressure cooker only gets to 1.02 atm....

@nick Making cyclohexanone requires good quality catalyst and somewhat drastic conditions. If you aren't able to achieve such conditions which I am sure you won't in household environment, you can always purchase cyclohexanone. Also, I believe that the collective price for reagents, catalysts and equipments used for making cyclohexanone will definitely surpass the price of cyclohexanone itself.

There exist different approaches to synthesize cyclohexanol and/or cyclohexanone from cyclohexane. One method is to use O2 and Vanadium oxides at 400°C and 15 atmospheres. Another method is to oxidize with H2O2 at 80°C with a catalyzer made from Silica gel treated by a silicone derivates holding an amino drop at the end of the alkyl chain fixed on Si. The so obtained amino-catalyzer is then treated by eosin, and then with copper  ions, before being used to oxidize cyclohexane. I doubt these processes can be used easily for producing small amounts of cyclohexanol or -one in an ordinary lab.
Another approach is starting from phenol, that gets reduced by H2. This is the way DuPont produces cyclohexanol which is then oxidized by HNO3 and converted into adipic acid, before being then copolymerized  with hexanediamine for synthesizing nylon.

It would be much better if you could cite the sources. It would be really a great answer if you do that..

My source was my organic course I took a long time ago, without any reference !

Not directly relevant to the question, but FYI: although phenol reduction is used commercially, it was never the primary route to cyclohexanol for adipic acid production by Dupont. The majority was produced by cyclohexane oxidation (cf this 1994 EPA report:https://www3.epa.gov/ttn/chief/ap42/ch06/bgdocs/b06s02.pdf). Current production data is available in this report if you have a subscription: https://ihsmarkit.com/products/adipic-acid-chemical-economics-handbook.html

Well as u said you have access to cyclohexane then oxidation of cyclohexane in the presence of oxygen yield a mixture of cyclohexanol and cyclohexanone.You can add a good oxidizing agents such as Pyridinium chlorochromate or potassium dichromate as per availability increase the yield of cyclohexone (as ketones dont get easily oxidised).
Reference:
1)http://www.greener-industry.org.uk/pages/nylon/4nylonpopup.htm
2)https://www.researchgate.net/post/What_is_the_mechanism_of_the_oxidation_of_cyclohexane_like_What_are_the_most_common_products
3)https://en.wikipedia.org/wiki/Oxidation_of_secondary_alcohols_to_ketones
4)https://www.chemguide.co.uk/organicprops/alcohols/oxidation.html


text:
Linear combination of eigenfunctions
In the textbook "Molecular Quantum Mechanics" by Atkins and Friedman [1, p. 11], it says in chapter 1:

… a general function can be expanded in terms of all the eigenfunctions of an operator, a so-called complete set of functions.

Can ANY function be written as a linear combination of eigenfunctions of ANY operator? I think not.
What are the criteria here for which functions can be written as linear combinations of which operators?
References

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

you should probably post this on the math stackexchange. The sentence you quoted reads like the author is defining a function that will be analyzed further along in the text.

In short, any square-integrable function can be written that way.

The complete, formal mathematical answer is a bit complex. It might be better to get that discussion from Math Stack Exchange.
An answer to the world of Chemistry is more straightforward: Yes, for any physically relevant operator and any relevant function, you always can. That’s a fundamental property of quantum mechanics: (functional) states are superpositions of eigenstates.

Just a short note for OP: in introductory texts on quantum physics & chemistry one can often encounter phrases like "physically relevant" or "well-behaved function" with the actual meaning being "square-integrable function". There are a lot of mathematical nuances and subtleties but to start from square-integrability requirement would be enough.

That's enough for me! Thank you to Nick, Bob, and Wildcat for answers/comments and to andselisk for editing!


text:
laser diode problem
I'm a student studying chemistry, i'm working with diode laser to look at emission spectra for certain complexes. i'm using integrating sphere as my setup, when i use the 405nm diode laser output of 4.5mw, i'm getting a small emission peak at 600nm in my background and sample, any idea why that could be, whats causing it? i've used other wavelengths with the same power and had no issues.

Fluorescence? I mean that´s what you´re looking for anyway, isn´t it?

The OP mentioned it is present both in the sample and in the background. So basically he is looking either for some unwanted 600 nm radiation emitted by the 405 nm laser or fluorescence from a contaminant present in the chamber, that would absorb at 405 nm and emit at 600 nm.

thank you PLD, thats exactly what i meant. i don't think its a contaminant because it happens every time i use that laser, i've changed the sample and the cuvette is clean but  it still happens so i believe is the laser, i'm just wonder what could cause that?

If you believe it is due to the illumination by the laser, you could try to get a spectrum of your laser with a calibrated photometer. Or for a quick check, you could in the sample holder something that might disperse more of the excitation light and see if it goes up. Just be careful not to send too much power of the laser to the detector as it might damage it (this is if you use a grating spectrophotometer to measure your spectrum).

i did it with a CCD. i want to find what is happening to cause the laser peak to have another emission peak at 600nm? this is happening with the blank sample, i've ran over 22 samples and it in each one

The simplest way is to use a yellow/orange cut-off filter, the 405 nm will be absorbed and the red transmitted, say, onto a piece of white paper.

porphyrin i'm trying to work out why its there in the first place? are you saying if i use a yellow/orange cut-off filter, that could remove the peak?

If it comes from your laser, you can a) repair the laser, b) buy a new one, c) use a filter, d) ignore it, or e) subtract it with the background. Could be it´s normal with the laser type you have.

karl i've been subtracting it from my background but in my report i need to comment on it and say what might be causing it, any idea?

I just did the experiment suggested by @porphyin. I used a 405 nm laser diode built into a toy "Space Gun" and the orange long pass filter shown in the figure below:

The results are in the next dark figure:

The spots on the paper are not fluorescence of the paper (see the violet fluorescence of the paper around the dark shadow of the filter) and are not due to fluorescense of the orange long pass filter. So it appears that the 405 nm laser diode has some parasitic longer wavelength emission. It can be filtered out using non-fluorescent short pass filters.

thank you! do you know why it might have that?

Actually, no. But unwanted emission is fairly common in LEDs. My 397 nm LEDs have some red emission from the LED and the colorless plastic case emits some green.  So I used non-fluorescent filters to get rid of that. The filter in the photos is not non-fluorescent, but it was in one of my junk drawers and was fit for this purpose.

thank you.  i understand how to remove it from future experiments so thats not an issue for me. i need to write in my report what could be causing that to happen

Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!


text:
Preparation of potassium cyanide from potassium ferrocyanide via HCN
Does anyone have any experience with the preparation of potassium cyanide from potassium ferrocyanide via $\ce{HCN}$ method? I did some math and am going to proceed eventually, any advice would be welcomed. I have access to well equipped lab, respirator with filter rated for $\ce{HCN}$ and I know what I'm doing, I just never previously worked with $\ce{HCN}$, so I'm looking for "holes" in my preparation to do it as safely as possible.
\begin{align}
\ce{K4[Fe(CN)6] + 3 H2SO4 &-> FeSO4 + 2 K2SO4 + 6 HCN} \\
\ce{2 K4[Fe(CN)6] + 3 H2SO4 &-> 3 K2SO4 + 6 HCN + K2Fe[Fe(CN)6]}
\end{align}
Then we pass $\ce{HCN (g)}$ through absolute ethanol medium to precipitate $\ce{KCN}$. An aqueous solvent not feasible due to polymerization.
$$\ce{HCN (aq) + KOH (aq) -> KCN (aq) + H2O (l)}$$
My question is: Why doesn't first reaction work under usual conditions?
There are some major mistakes in the above equations: both $\ce{H2SO4}$ and potassium (III) ferrocyanide couldn't be used, thus should be swapped for concentrated nitric acid and potassium (II) ferricyanide, which make them not feasible.
I did have access to $\ce{KCN}$ for years now, which was given to me by a fellow chemist, now I'm running low on supply and I'm about to prepare some myself. I am aware of how does this thread may sound like, so I can provide any proof of my word upon request.

I don't know if it is wise to answer, because HCN is one of the worst toxic gases. It was used by the Nazis during WWII to kill hundreds of Jews in a couple of minutes. So why are you so interested in producing HCN ? Why ? I will not develop too much this field, as for example the first reaction does not work in usual conditions.

Since you can buy potassium cyanide and it is far easier to handle than HCN, why the heck would you choose this route to it?

Maurice what do you mean by "first reaction does not work in usual conditions" ? Can you specify?

@THF. No. I will not help you synthesize HCN. I will not tell you anything about your reaction for producing it.

If one needs and does not have a licence to buy and manage KCN, it is a very bad idea to want to synthetize it via the more feared HCN, and even worse idea would be providing any advice for that, except not doing so.

Related: How to prepare potassium cyanide from potassium ferrocyanide

I have again cleaned up your post. Please take the [tour] to learn how this site works. This is not a forum, there should not be any visible marks of edits in your question, this should be understood without the context of how it came about; this also includes any of the comments or answers. Please do not use all caps, it is considered rude or offensive by most of us.

I will quote some words from an experienced chemist per this source:

Ferricyanide and ferrocyanide have their cyanide ions very tightly bound to the iron core and 6M HCl will have difficulty releasing the cyanide as HCN. Maybe you get a little HCN if you heat the solution, but even then do not expect a smooth and satisfactory reaction.
  Ferricyanide also has the added disadvantage that the iron(III) is somewhat oxidizing and in practice, the yield will be even further reduced, due to internal oxidation of the ion. You'll get other decomposition products.
Another problem with your suggestion is that you should not bubble the gas through aqueous KOH. The aqueous solution of KCN is quite unstable and this solution quickly turns brown with formation of polymeric species. Better results are obtained with solutions of KOH in absolute ethanol. KCN will precipitate from such solutions if HCN is passed through them.

which should answer your question as to why your first reaction will not work under usual conditions.

Remember you are helping also people who want to obtain KCN for malicious reasons. Not very wise. The is no guarantee the OP does not belong to them either.

I got KCN for years in my lab, which was given to me. Now Im running low on supply and want to actually extract some gold from processor pins which I have too much for what I got now.

@THF You state to be in Germany, so I speculate the processors in question equally are.  Just type «processor recycling germany» into a search engine and you find multiple companies which already have the know-how and equipment in place to do this for you (even now) and offer money in return. By reading their lists, you'll see how little Au still is used today (e.g., https://www.precious-metal-services.com/prices-sorting-criteria-for-electronic-waste/).  Think twice, not only about your safety and your neighbours.  Let professionals do this instead of you.

Dear Buttonwood, didn't you think that everything I do got a reason behind? First off: Im already into reprocessing other metals like silver, platinum or palladium so its gold time now. Im not doing it for money because gold recycling became way less profitable in recent years. My final objectives are always to explore the science myself and learn. I also got every bit of equipment and now I got complete know-how too so?

"Let professionals do this instead of you" sounds like you telling me that doing anything by yourself instead of sending it to someone who does it at large scale is nonsense, especially when you are not a "professional" whatever it means. My neighbours are too far away for them to be affected by anything I do, and my own safety is my own concern. What do you think is the difference betwen self-learning and studying it on university? Someone tells you what to do to do it safely. You know why? Because like 90% of students are complete morons, just a bit educated to help them elaborate.

@THF Have you considered thiosulphate gold/silver leaching instead of cyanide one  ?


text:
Conditions for Complete hydrolysis of salt
My text book says that the cations (or anions) which are stronger than hydronium ion(or OH-) and their conjugate base (or acid) being very much weaker than water show complete hydrolysis.
 While the cations(or anions) which are weaker than hydronium (or OH-)and their conjugate base(or acid) being stronger than water but weaker than hydroxide(or H3O+) exhibit hydrolysis to Limited extent.
I don't quite understand why is this so?

What's to understand? Stronger acids replace weaker acids in their salts. Water is an acid.

The question is, in fact, a good one, but the concept of weaker/stronger is confused - and confusing!
A way to describe the situation, using the terms in the question is this: strong acids and strong bases undergo complete hydrolysis in water. The energy gained from hydrogen bonding with a water molecule is great enough to effect complete ionization in water. Examples would be HCl and NaOH.
Water lies in the middle here, being able to be protonated by strong acids and deprotonated by strong bases. Hydrolysis is a measure of how much the water is protonated to H3O+ or deprotonated to OH-.
There are also materials which ionize in water, but not so extensively, e.g., acetic acid and amines such as ammonia. They dissolve, but do not react so much and thus hydrolyze to a limited extent. Confusing, perhaps, that we call these materials weak acids and bases but they have strong internal bonds which prevent their total ionization. These strong internal bonds mean that if you did form the the conjugate acid or the conjugate base (in a salt), they would react strongly with a water molecule to form a considerable amount of the corresponding acid or base. Thus the salts would ionize and hydrolyze where the original acid or base would not be said to hydrolyze (much).
The "strength" of the cation may perhaps be better described as a polarity, which, if it is a "strong", favors hydrolysis, i.e., complete ionization, meaning it is strongly attracted to the oxygens of water and less so to its own anion. By this terminology, Na+ is a strong acid because it is strongly attracted to water (oxygens) at the same time OH- would be a strong base.
A "weak" cation, like NH4+, would be one which will not protonate water to a great extent. A solution of 1 M NH4Cl will give complete ionization to NH4+ and Cl-, but we do not call that hydrolysis. The ammonium ion is stable (strong internal bonds) and only weakly protonates the H2O molecules around it, so we call it a weak base, with limited hydrolysis.


text:
Descale: Vinegar vs citric acid and temperature
The core of my question: What is the best temperature to descale with vinegar / citric acid?
I often read to use vinegar essence with 25% acidity, dilute it with water and then let it soak. Sometimes you're supposed to use warm, sometimes cold, and sometimes boiling water.
How does the effectiveness depend on the temperature?
The same question I have for citric acid. If the temperature is too high (so I have read), you 're instead making calcium citrate (which is supposed to be harder). But what is too high? Why heat the solution at all?

Clarify your question please. What is the object made or what do you want to descale?

The temperature has only one effect : it decreases the duration of the operation. That's all !

Higher temperature has several effects:

The rate of chemical reactions increases empirically and typically  2-4 times by every increase be 10° C
Heating toward boiling or intensive gasing turbulently mixes solution, what brings fresh solution to limescale.
Faster diffusion in hot solution due lower viscosity and higher ion/molecule mobility.

I never had problems with secondary precipitates with vinegar or citric acid, but I did have precipitated crystals in case of tartaric acid I had once handy.
I do not have data about temperature dependency of calcium citrate solubility, but I guess it may be rather that at high temperature one easier gets to higher concentrations.
What is essential is the $\mathrm{pH}$. With the higher ratio citric acid vs scale, there would be lower final $\mathrm{pH}$. Then the lower concentration of citrate anion would make calcium citrate precipitation less probable.


text:
Heterogeneous ion solution
I am a college student and for my chemistry of solutions class, I had a lab where we had to titrate orange juice to check the calcium content. However, the experimental results are 20% off the labelled calcium content, and in my report I have to provide a "Tentative explanation of the discrepancy."
Is it possible that the orange juice was not completely homogeneous? What I mean is that maybe some of the calcium and other 2+ ions were not well mixed and there were more/less in our sample. Is that possible or are ions always equally dispersed throughout the solvent when in solution?

Well, I do not remember homogenous orange juice. Were you supposed to determine the total, or dissolved calcium ? What was the procedure ?

We had to titrate the solution with an acidic solution (EDTA), a pH buffer and an indicator. The EDTA would react with 2+ ions first, and then with the indicator.

The calcium amount may change from one fruit to the next one, or from one day to another one, or from one field to another one. 20 % is not a significant difference between such a measurement and another one. And this may happen for any constituant, not specially for calcium.

As somewhat of a side note, the FDA must have some tolerance around how much the actual value can vary from the stated value on the label.


text:
can a vapour denser than air evaporate in a closed chamber?
Imagine a closed container filled with air.
There is some liquid substance on a bottle placed on the floor of the container.

Let's heat the liquid until it starts to evaporate. Let's say 20% of the liquid evaporates.
Disregard any effect of pressure buildup in the container which would block the evaporation.
Note: the liquid we are considering generates a vapour that is denser than air, so a "cloud" of vapour would start forming at the bottom on the container, unable to rise. (unlike water vapor for example, which would be less dense than the surrounding air and would start to rise)

If it's not clear enough, imagine what happens when solid carbon dioxide (dry ice) sublimates:

Would this patch of saturated air block the evaporation of the rest 80% of liquid?
I am thinking that the liquid would be soon in contact only with saturated air and the evaporation would stop even though there would be still plenty of unsaturated air in the upper section of the container.

You think right. Of course, this situation won't be stable forever; diffusion will gradually mix everything.

Entropy of molecules won't decrease, rather will increase (cuz, gravity as well as heat both will increase entropy). Thus, there's no question of settling down of vapours, they will diffuse, however might diffuse to within small height first. Continuous heat supply will help vapours to diffuse to higher heights..

SF6 happily mixes with air.


text:
thermodynamics problem
The boiling point of a substance $X$ at 1 atm pressure is $500 k$ . The enthalpy of vapourisation at the boiling point of $X$ (l) is 80 kj/mol .The molar specific heat of $X(l)$ = $5 × 10^{-3}$ kj/mol and $C_p$ of X(g) = $5 ×10^{-4}$ kj/mol
What is the molar latent heat of vapourisation of $X(l)$ at $800 K$ and at $1 atm$ ?
Options :
A: lesser than $80 kJ$ $\\$
B: greater than $80 kJ$ $\\$
C : equal to $80kJ$ $\\$
D: Since 800K is the higher temperature than the boiling point of $X(l)$ ; therefore , enthalpy of vapourisation is not applicable

I'm wondering how this HW question got two upvotes in 3 min after publication.

Should be able to solve by Kirchhoff's Law? Or simply logically, Cp of gas lower, as T increases enthalpy of gas increases less than enthalpy of liquid increases, so the reaction becomes less endothermic.

If its equilibrium vapor pressure at 500 K is 1 atm, how can it also be 1 atm at 800 K?


text:
Can o-phenanthroline be used as a complexing agent for lead(II)?
I’ve been trying to find a method to perform a spectrophotometric analysis to determine lead concentration (on the ppm level). This would require some chelator capable of forming a coloured complex with lead in the aqueous phase.
Fe(II) ion can bind to o-phenanthroline to form an orange colour (with deeper hues based on the present concentration of iron irons).  I was wondering if lead(II) can too complex in the same way.
Note: The source of lead is $\ce{Pb(NO3)2}.$ This means some nitrate ions will be present in solution after dissociation of this chemical in water.

Yes, there are numerous known lead(II) o-phenanthroline complexes with thiocyanate, perchlorate and bulkier organic ligands.
As for nitrate, Bytheway et al. [1] reported synthesis (from lead(II) nitrate in methanol) and crystal structure of bis((1,10-phenanthroline)-(nitrato-O,O'))-lead(II):

Figure 1. crystal structure of $\ce{[(phen)2Pb(NO3)2]}$ (CCDC: 1246661, CSD: RAQMAX).
Unfortunately, there is no UV-Vis/NIR spectra available.
Harrowfield et al. [2, p. 1030] reported that the crystals are colorless, which likely eliminates visible region of detection.
References

Bytheway, I.; Engelhardt, L.; Harrowfield, J.; Kepert, D.; Miyamae, H.; Patrick, J.; Skelton, B.; Soudi, A.; White, A. Lewis-Base Adducts of Lead(II) Compounds. XIII. Synthetic, Structural and Theoretical Studies of Some 2:1 Adducts of 1,10-Phenanthroline With Lead(II) Oxoanion Salts. Aust. J. Chem. 1996, 49 (10), 1099. DOI: 10.1071/CH9961099.
Harrowfield, J.; Miyamae, H.; Skelton, B.; Soudi, A.; White, A. Lewis-Base Adducts of Lead(II) Compounds. VII. Synthetic and Structural Studies of Some 1:1 Adducts of Linear Polyamines With Lead(II) Nitrate. Aust. J. Chem. 1996, 49 (10), 1029. DOI: 10.1071/CH9961029.


text:
What is really the number in the periodic table?
I learning basic chemistry and i have a doubt.
I have this definitions
Molar mass: Is the mass of $6.022\times10^{23}$ atoms of a chemical element and its unit of measurement is $\mathrm{g/mol}$.
In the periodic table, for example, beryllium has two numbers: $4$ and $9.0122$. I know that $4$ is the number of protons, but I have the doubt with $9.0122$.
I know that $9.0122$ is given in $\mathrm{g/mol}$, therefore $9.0122$ is the mass of $6.022\times10^{23}$ atoms of beryllium, and is given by the average of the mass of one mole of each isotope of the beryllium.
But I know that is the mass number and is the amount of protons + neutrons. (And how is possible that is not an integer number?)
So, what is this number really? the mass of one mole of the element or the sum of protons + neutrons?

I'd suggest to give Standard atomic weight a good read. I also would recommend to start using a better periodic table which includes proper definitions, units and up-to-date values, e.g. the one issued by NIST. On top of that, do note that it's not $\pu{6.022E23}$ any longer; since 2019 it's $\pu{6.02214076E23}$ exactly.

Atomic mass is not exactly proportional to the mass number of the nucleus. Let's look at all the complicated factors that appear in one little atomic mass value:

Particles: Protons and neutrons do not have exactly the same mass
Binding Energy:  The mass of a nucleus is less than the sum of the individual protons and neutrons because energy is lost (so mass is lost, $E=mc^2$) when the nucleons bind together.
Averaging over isotopes:  On top of that there are isotopes, and they have different numbers of neutrons therefore different atomic masses; when we see a single number it's usually averaged among all the atoms of a typical sample, usually as the element occurs on Earth.
And in light of all those things, whatever unit we define for atomic mass.  (I believe it is now a gram divided by a defined value for Avogadro's number.)

With all those factors in play, you're basically not going to get a whole number for atomic mass in the Periodic Table.

Isotopic variation is particularly noticeable with sulfur, which has differing ratios due to natural fractionation depending on its source, e.g. https://www.sciencedirect.com/science/article/abs/pii/S0016703720300363 . This caused issues when setting the accepted value.

So, what is this number really? the mass of one mol of the element or the sum of protons + neutrons?

The number is the molar mass expressed in g/mol, i.e. stripped off these units. (The molar mass is the mass of a sample of the element divided by its amount of substance. This explains why the units are g/mol.) Alternatively, you can also say the number is the relative atomic mass (historically known as standard atomic weight). The way the two are defined results in the same numeric value when the molar mass is given in g/mol (the relative atomic mass is dimensionless, so it has no units at all).
Elements with a single stable isotope

In the periodic table, for example the beryllium have two numbers: 4 and 9.0122. I know that 4 is the number of protons, but i have the doubt with 9.0122

Beryllium has a single stable isotope, $^9\ce{Be}$, with 4 protons and 5 neutrons. The mass number of this isotope is 9, and the molar mass is 9.0122 g/mol. The molar mass is not exactly 9 g/mol because the molar mass of protons and neutrons is not exactly 1 g/mol, and because there is a mass defect related to the famous $E = m c^2$ relationship.
Elements with more than one stable isotope
Take boron, for example. The two stable isotopes are $^{10}\ce{B}$ and $^{11}\ce{B}$. You get the molar mass of terrestrial boron by a weighted average of the molar masses of these two isotopes (or more practically by measure the mass and getting the amount of a fairly pure sample of boron).

Source: https://upload.wikimedia.org/wikipedia/commons/9/92/NuclideMap.PNG
See also: https://chemistry.stackexchange.com/a/114484/72973


text:
Methods to quantify lead(II) concentration in water on the ppm/ppb level
While there are several ways to quantify metal concentrations in water, a lot of them are only specific to select metals. For example, there is the orthophenanthroline method for quantification of Fe(II) concentrations via spectroscopy. How can lead(II) concentrations be determined using a similar spectrophotometric method?
If spectroscopy isn’t the most efficient way to quantify these concentrations, is there a better way to do it?
The salt used to release $\ce{Pb^2+}$ ions into solution is $\ce{Pb(NO3)2}.$

Russian interstate standard GOST 18293-72 "Drinking water. Methods for determination of lead, zinc and silver content" (PDF in Russian) suggests to use sulfarsazene (plumbone), which forms orange-yellow complex with lead(II).
Reported method sensitivity (spectrophotometry): $\pu{0.5 μg L-1}.$
A brief review in English [1, p. 315]:

To determine lead in solid particles of industrial facilities, natural waters, biological objects and salts, sulfarsazene is used (4''-nitrobenzene-1'',4-diazo-amino-1,1-azobenzene-2''-arsono-4-sulfoacids).
  While it is less sensitive than dithizone , it has a number of advantages: it is water-soluble, easy to handle and needs no cyanide as masking agent.
  But, it has insufficient reaction selectivity.
  Manganese, nickel, copper, zinc, mercury, cadmium, iron, zirconium, and other elements interfere with the determination. The methods developed on the basis of sulfarsazene require a preliminary ion separation; for instance, lead is separated by extracting with dithizone solution in carbon tetrachloride …
Study of the reaction of sulfarsazene with lead ions points to simultaneous participation of the lead cation with triazene and arson groups in complex formation.
  Polarographic investigation confirmed the existence of lead bonds with the o-arsonotriazene group of the reagent, and cleared up the azo-group contribution to the formation of the complex which may be due to coordinative bonding with lead atoms.
  The role of the nitrogroup as an electrophilic substituent that improves the mobility of the hydrogen atom of the aminogroup was experimentally confirmed.
  The structure of the resultant compound is as follows:

GOST 18293-72 also includes polarographic method for lead detection using the dropping mercury electrode.
Reported method sensitivity: $\pu{10 μg L-1}.$
References

Roeva, N. N.; Savvin, S. B. Spectrophotometric Determination of Minute Lead Contents of Natural and Industrial Objects. Reviews in Analytical Chemistry 1986, 8 (4). DOI: 10.1515/REVAC.1986.8.4.313.

Does the method sensitivity refer to the concentration limit in which the coloured complex follows Beer’s law?

Moreover, is there a purchase link online? The name seems obscure.

@Thunderbolt114 Yes, the limits are for the quantitative determination, and the calibration curve is plotted using lead(II) nitrate at $λ_\mathrm{max} = \pu{500 nm}$.

@Thunderbolt114 As for buying, it seems that you would need to go with Russian or Chinese suppliers. Being listed in CRC Handbook of Organic Analytical Reagents, sulfarsazene still isn't a popular reagent in the West.

I’ve been trying to search for an affordable pricing for this reagent but haven’t stumbled upon any suppliers. Might you know of any site/supplier?


text:
Can Potassium Carbonate be used to reduce Calcium hardness in water
I have a solution of KCl 20g/l (2%) and 500mg/l Ca,
To reduce the hardness, Na2CO3 is normally used, but can I use K2CO3 instead?

I have a drilling fluid, KCl in water, the K+ is to inhibit the swelling of clay, it is better than Na+, but in addition I want to supress Calcium and maintain a pH of 8.5 or more. Normally the following can be added, NaOH, Na2CO3, but I would like to use K2CO3 to precipitate the Ca, I just want to know if it can, and how effective it is in comparison to Na2CO3, ie about the same, better or worse and by how much

I know nothing about drilling mud.
Solutions of both $\ce{Na2CO3}$ and $\ce{K2CO3}$ will precipitate calcium carbonate.
$$\ce{Na2CO3 ->[H2O] 2Na+(aq) + CO3^{2-}(aq)}$$
$$\ce{K2CO3 ->[H2O] 2K+(aq) + CO3^{2-}(aq)}$$
$$\ce{Ca^{2+}(aq) + CO3^{2-}(aq) <=>[H2O] CaCO3 v}$$
Since there is already some $\ce{KCl}$ in solution this would slightly reduce the solubility of the potassium carbonate. However since potassium carbonate is more soluble than sodium carbonate anyway taht wouldn't seem to be a problem.
Solubilities are typically given as grams/100 ml of water as in the Wikipedia table.
\begin{array}{|c|c|c|}\hline
\pu{Temp }^\circ\pu{C} & \ce{Na2CO3} & \ce{K2CO3} \\ \hline
0 & 7 & 105  \\ \hline
10 & 12.5 &109 \\ \hline
20 & 21.5 &111 \\ \hline
30 & 39.7 & 114  \\ \hline
40 & 49  & 117 \\ \hline
50 &  -  & 121.2 \\ \hline
60 & 46  &127 \\ \hline
70 &  -  & - \\ \hline
80 & 43.9 & 140 \\ \hline
90 & 43.9 & 148 \\ \hline
100 &45.5 & 156  \\ \hline
\end{array}
However the important consideration here is moles per liter.
$$\pu{500 mg }\ce{Ca^{2+}}\pu{/L} = \dfrac{\pu{0.5 grams/L}}{\pu{40 g/mole}} = \pu{0.0125 moles/L}$$
The solubility product for calcium carbonate is $3.3\cdot10^{-9}$.
$$K_{\mathrm{sp}} = 3.3\cdot10^{-9} = \ce{[Ca^{2+}][CO3^{2-}]}$$
where the brackets indicate moles per liter. So $\pu{0.0125 moles/L}$ of $\ce{CO3^{2-}}$ would precipitate the calcium. You might want to add say 20% excess to allow for the fact that the calcium concentration might be just more or less $\pu{500 mg/L}$.
Other considerations are the extra water from the potassium carbonate solution which would dilute the mud. Also how to get the potassium carbonate solution mixed evenly with the mud.
It is strictly a guess, but sodium carbonate is probably preferred over potassium carbonate due to price.
\begin{array}{|c|c|c|}\hline
 & \ce{Na2CO3} & \ce{K2CO3} \\ \hline
\pu{grams/mole} & 106.0 & 138.2\\ \hline
\% \ \ce{CO3^{2-}} & 56.6 & 43.4 \\ \hline
\end{array}

just what I wanted to know, thank you very much


text:
If all the points on my graph fit on a logarithmic and polynomial trend line, how do i justify which trend line i should use?
My graph is mass of deposit vs concentration. The mass of deposit increases with concentration.

You need to expand this question and explain the options better. In part it is more of a statistics question than a chemistry one. However I'll grant that models are often based on some chemical/mathematical theory.

There are 2 cases:
If the formula of the theoretical model is known, the approximation follows this formula to determine the formula parameters.
If there is no theoretical model, or the model is not followed, then reasonable empirical formulas are tested and the best fitting and simple enough is chosen.
Aside of polynomials and logarithms, interesting results can be provided by rational functions ( polynomial fractions )
The important thing is to have difference  "number of points" minus "number of parameters", aka degrees of freedom, big enough. This keeps it as an approximation ( usually least squares) rather than an interpolation.