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What is the shape of [Cr(NO)₄]?
I was learning about crystal field theory recently. I am quite confused about predicting the shape of $\ce{[Cr(NO)4]}$ complex because I am not able to conclude whether it will be square planar or tetrahedral.
How to predict the shape of $\ce{[Cr(NO)4]}$ using crystal field theory?
Will $\ce{NO}$ act as a 3-e- donor here?
Will all the four $\ce{NO}$ that are in coordination with $\ce{Cr}$ be of the type nitrosyl or will three of them be nitroso and one nitrosyl (because $\ce{Cr^3+}$ is one of the common oxidation state of $\ce{Cr})?$

The oxidation state is zero, and not +3. Are you sure that this compound exists ? I have not voted down.

@Maurice Thanks for answering one part of my question.Yes,I am sure that the compound exists.You can check that here https://en.wikipedia.org/wiki/Metal_nitrosyl_complex

@Maurice Since the oxidation state is zero and not +3, you mean the shape has to be square planar or tetrahedral?

$\ce{[Cr(NO)4]}$  is very bad for CF theory. CF works well with complexes with strongly charged components. CO, NO and PR3 complexes do not really fit.

I quickly googled and found a paper claiming that the system is best viewed as complex of $\ce{Cr^+}$, but actually is very similar (and isoelectronic) to $\ce{Ni(CO)4}$

$\ce{Ni(CO)}4$ is the obvious analogy, with NO being a 3 electron donor compared to the two electron CO in the Ni case - both leading to 18 electron compounds.

As per my understanding:
In $\ce{[Ni(CO)4]}$ :Ni in the presence of strong field ligand must have d^10 configuration (due to pairing of e-s).So,in the absence of inner d-orbital,I can understand that $\ce{[Ni(CO)4]}$ has to be tetrahedral
But in $\ce{[Cr(NO)4]}$ : Cr in the presence of strong field ligand must have d^6 configuration (due to pairing of e-s).As dx-y and dx2-y2 orbitals are free,square planar option is still there.Also,NO is a strong field ligand,which means it favours formation of square planar complexes which have have higher crystal field splitting energy.

@IanBush I don't understand how $\ce{[Ni(CO)4]}$ is the obvious analogy to $\ce{[Cr(NO)4]}$

Both are 18 electron compounds (https://en.wikipedia.org/wiki/18-electron_rule). The CO ligands donate 4x2=8 electrons to Ni make 8+10=18. The NO ligands donate 4x3=12 electrons to make 12+6=18. They are therefore isoelectronic. The nickel compound is tetrahedral. So therefore is the Chromium compound.

Does this mean all isoelectronic compounds are isostructural?If not,when all can that logic be used?

@permeakra "CO, NO and PR3 complexes do not really fit." This is because CF theory views interaction between metal and ligand as purely ionic right?
I am not getting "the system is best viewed as complex of Cr+"
Does this mean NO- is acting as a ligand here? or is there some electron being pulled by nitrosyl from Cr ?

@Chem-Learner CF theory was designed for rationalization of optical spectra of ions in aqueous solution. It can make semi-quantitive predicitons for mostly ionic complexes like $\ce{Ti^{3+}}$ aquaions. It works well for the first row of d-elemnt aqueous solutions.

@Chem-Learner However, once covalent interactions become dominant, purely exlectrostatic considerations, that are the core of CFT, are not sufficient anymore.

Describing the system in details would be too long, regrettably. I'm not ready for it at the moment.

Based on available theoretical considerations and a available literature, a tetrahedral geometry appears to be a good assumption for $\ce{Cr(NO)4}$.

The NO ligand can be rendered as a three-electron donor, if we take
it as uncharged.  It is actually similar to CO, really, in that it
interacts through its $\pi^*$ orbitals as well as its $\sigma$
orbital; but NO unlike CO has an electron to offer from its $\pi^*$.
That plus the $\sigma$ electrons makes three.
We then may identify the central metal atom as having 18 valence
electrons.  When this happens with multiple ligands we typically see
an arrangement with highest order symmetry, thus tetrahedral for
four-coordination.  This gives the best possible bonding to the
entire set of valence orbitals involved in the 18-electron
structure.  This is seen in the tetrahedral four-coordinationn of
$\ce{Ni(CO)4}$, and in the octahedral six-coordination of the series
$\ce{[V(CO)6]^-}$, $\ce{Cr(CO)6}$, $\ce{[Mn(CO)6]+}$, $\ce{[Fe(CO)6]^{2+}}$.  So, we should expect $\ce{Cr(NO)4}$ to be tetrahedral.
Heberhold and  Razavi found evidence for a tetrahedral structure in $\ce{Cr(NO)4}$ from its IR spectrum.


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Camphor for molecular mass determination
I came across this statement in my book that read, "Camphor is used to determine the molecular mass of substances because it has a high value of cryoscopic constant." What does this have to do anything with the ease of determining molecular masses?
I am not familiar with the process that is being referred to, and hence, do not understand the significance of camphor having a high cryoscopic constant. Please help me out. Thanks in advance.

Doesn't the quoted statement answer the question? Also, please always provide complete references for the sources you quote.

Is the question more on what the cryoscoic constant is, or how it helps if camphor has a high value?

Related: https://nvlpubs.nist.gov/nistpubs/jres/7/jresv7n3p477_A2b.pdf

@JonCuster Its more on how it helps if camphor has a high value

Following wikipedia freezing point depression can be used to determine molecular masses as the freezing point depression depends on the number of particles (a colligative property).
In an experiment, you would dissolve a known mass of the unknown sample in a known mass of a solvent (here camphor) and compare the change in freezing point for sample-camphor mixture to pure camphor. Some worked examples can be found at chemteam
This effect is much larger for camphor than for many other (more common) solvents.

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Isopropyl Alcohol Vs Acetone for Dissolving Organic Contaminants on PCB
Acetone is widely used in the labs as a solvent to clean the vials and tubes as it is good solvent for organic materials. While Isopropyl alcohol is used as a rubbing alcohol for cleaning contaminants on the body before injection. Both are good solvents for organic materials. I would like to know which one is more suitable for cleaning the organic contaminants on the PCB's (Printed Circuit Board) with minimal residue?

Depends if there are plastics attacked by acetone, e.g. polystyrene.

PCB should be OK, but a soldered PCB has components, e.g. polystyrene caps, or a piezo element in a polystyrene case.

@DrMoishePippik I see, make so much sense! I didnt know that there are polystyrene caps. I guess isopropyl alcohol or ethanol is the best alternative then. The reason I asked this question is I tried to use medical isopropyl alcohol swabs to clean my phone screen, I saw residue left over the screen, despite the claim that isopropyl alcohol doesn't leave residue.

The residue may be oils left behind that the iso missed... but I certainly wouldn't use a stronger solvent on a phone. Just gentle rubbing with microfiber cloth will remove most oils.

@DrMoishePippik +1 makes so much sense. I did it on the screen to see if residue is left or not because its hard to check on the PCB. Allright, I guess you answered my question. Acetone isnt recommended for PCB's where as Isopropyl can be used instead.

Acetone aggressively attacks many polymeric components (not just polystyrene) and often completely destroys them. The circuit board itself, for example, may well be damaged as will many electronic components. isopropanol does not and is always a better choice.

@matt_black makes sense. just curious: can isopropanol and ethanol be used interchangeably for cleaning circuit board in your opinion.

@gfdsal yes. But pure dry ethanol in not readily available and comes with legal restrictions. Pure Isopropanol is easier to get and it is easier to dry and doesn’t need to be denatured to stop people drinking it.

@matt_black +1 I believe you made it easier for me to decide what to purchase. I had an option of ethanol or isopropanol 70%, I shall go with isopropanol as pharmacy ethanol is most likely to have ingredients that denatures it and might leave residue on the circuit board which is undesirable.

"Rubbing alcohol" is 70% IPA, you should use 99% for cleaning. FWIW I found that by "upgrading" to acetone for cleaning BARE boards before etching improved the adhesion of the photoresist, which allowed me to do higher resolution etches. But this is bare boards, nothing to damage really at this stage.

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What actually happens in strong acid- weak base reactions?
I'm going back through my notes from a chemistry class, which I did well in despite not having a very good conceptual understanding, in order to hopefully develop that understanding, but there's something I can't find in the book or online. Why is it that, when solving for the pH of a solution that results from the mixture of a strong acid with a weak base or a weak acid with a strong base, we can treat the acid-base reaction as going to completion before solving any resulting equilibrium problem, or completely ignoring any equilibrium in the pre-mixed solutions.
For a concrete example, when mixing 40.0 mL of 0.50 M ammonia with 25.0 mL of 0.30 M hydrochloric acid, the approach the class would have me use is to first solve the limiting reaction problem as if the reaction goes to completion. Doing so, we see that there are 0.0075 mol of chloride ions, 0.0075 mol of ammonium, and 0.013 mol of ammonia. Then, because we don't have excess hydrogen ions, nor excess hydroxide, nor only neutral ions, we solve the equilibrium problem of ammonium with ammonia.
What I don't understand is how we can ignore the fact that the ammonia solution would have already been in equilibrium when mixed with the hydrochloric acid, and as I understand it should be constantly arriving towards equilibrium as the reaction is taking place.
Alternatively, if there had been excess hydrogen ions, I was told to use the concentration of hydrogen ions from the completed reaction, but that would also be ignoring the fact that the ammonium contributes to the pH (because it again ignores it's equilibrium).

The strategy of major species
You correctly describe the strategy. First, let reactions go to completion (in the direction that makes sense, i.e. weak base and strong acid forms weak acid and spectator ion - never the other way around). Then, check what the major species are and estimate the pH. Finally (and this step is often omitted), see if equilibria involving minor species (such as hydroxide for acidic solutions) need adjustment.
Why does the strategy of major species work?
For the example, the OP already calculated the estimated amount of species: 0.0075 mol of ammonium, and 0.013 mol of ammonia. With a total volume of 65 mL, that comes out as:
$$c_\ce{NH4+} = \pu{0.115M}$$
$$c_\ce{NH3} = \pu{0.20M}$$
From this, we can estimate the pH to be a bit more basic than the $\mathrm{p}K_\mathrm{a}$, i.e. 9.49. Now we can think about the minor species, hydroxide and hydronium. At pH = 9.49, there should be an excess of hydroxide over hydronium, by about 0.00003 M. Dissociation of water does not give an excess, so it has to come from ammonia turning into ammonium:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
However, when this reaction makes hydroxide, it also changes the ratio of ammonium to ammonia, changing the pH. The only reason we don't keep running in circles, adjusting one equilibrium and then the other, is that this adjustment is tiny compared to the concentration of the major species. It amounts to decreasing the ammonia concentration by 0.00003 M and increasing the ammonium concentration by the same amount. The shift in pH is so small that it makes no difference when the pH is written with appropriate number of significant figures.
Alternate strategy
For the calculators we use, this is a good strategy to deal with multiple equilibria (in your case, that of the weak acid/base pair and autodissociation of water).
If you had an analog computer that adjusts the concentrations based on a "pH slider" and reports out whether all the constraints are met, there would be no reason to do this in multiple steps. Instead, you would just push your pH slider from one extreme to the other and stop when all equations are satisfied.
In your case, you could calculate the ratio of ammonia to ammonium from the pH and the $\mathrm{p}K_\mathrm{a}$, and the hydroxide concentration from the autodissociation constant of water. Then, you vary the pH until the charge balance (chloride plus hydroxide has to match hydronium plus ammonium) is achieved. It is a fun exercise to program this into a spread sheet.
What actually happens?
All reactions (any of the acids - water, hydronium or ammonium - reacting with any of the bases - water, ammonia or hydroxide) go on at the same time. The further from equilibrium a reaction is, the faster is the net change. Once all the reactions are at equilibrium, there is no more net change. The details depend on the kinetics but are not relevant to the equilibrium state that is finally reached.

Does that mean the strategy of major species gives an approximation of the pH? If so, could you add some explanation as to why the approximation works?

@themathandlanguagetutor I added the math for your example to illustrate why the approximations are allowed in this case.

Sir do you mean its like this : salt is getting formed from the acid base reaction and then the salt goes to dissociation and also the anion part of salt then reacts with water ? If i am right ? Which reactions are the equilbirium ones ? @KarstenTheis

@ProblemDestroyer In reality, there is no back and forth. The reactions proceeds in one direction until they reach equilibrium. The back and forth is an artifact of the calculation. There are two equilibrium reactions here: autodissociation of water and protonation of ammonia.

Given the following problem:

25.0 mL of 0.30 M hydrochloric acid is added to 40.0 mL of 0.50 M ammonia. Calculate the pH, and concentrations of $\ce{NH3 \text{and} NH4+}$.

To solve such problem you must make some assumptions. Some of which are good, some ok, and some bad. In math $\pi$ has been calculated to 50 trillion digits.  Chemistry generally works with 2-4 significant figures.
Note that the concentrations of both hydrochloric acid and ammonia have only been given to 2 significant figures.
Assumptions:

The final concentrations of $\ce{NH3, NH4+ \text{and} H+/OH-}$ at equilibrium are independent of how the solution was prepared.

The equilibrium doesn't depend on how the solution was mixed. So regardless of how 65 ml of solution containing the equivalent of 7.5 millimoles of hydrochloric acid, and 20 millimoles of ammonia was made, it will have the same equilibrium.
Also $z$ millimoles of ammonium chloride could be used ($z \le 7.5$). Then $7.5-z$ millimoles of hydrochloric acid and $20-x$ moles of ammonia would be needed in the mixture.

That the volumes of the hydrochloric acid and ammonia solutions are additive.

Certainly good to 2 significant figures, 3 might be good, but 4 is probably an overreach.

That concentrations can be used instead of activity coefficients.

This is definitely a bad assumption for such concentrated solutions.

That the reaction $\ce{H+ + NH3 -> NH4+}$ is quantitative.

This is ok for 4 significant figures and the given concentrations.

Expanding 4, also assume that all the $\ce{NH4+}$ is due to the reaction with hydrochloric acid.

This assumption is good to 2 significant figures, probably 3, but 4 may be a issue. The issue is that there is also the autoionization of water which shifts the equilibrium $\ce{NH3 + H2O <=> NH4+ + OH-}$
$$\ce{H2O <=> H+ + OH-}$$

There is really nothing that can be done with assumptions 2 and 3 without experimental data and/or some sort of functions for the data. So the only thing that can be done is to note the assumptions in the solution.
From Wikipedia the $K_\mathrm{b}$ for ammonium hydroxide is $1.8\times10^{-5}$ ($\mathrm{p}K_\mathrm{b} = 4.7447$, $\mathrm{p}K_\mathrm{a} = 9.2552$).

Solution 1
Ignoring the autoionization of water yields the following charge balance equation:
$\ce{[NH4+] \approx [Cl-]}$
Thus 7.5 millimoles of $\ce{NH4+}$ and 12.5 millimoles of $\ce{NH3}$.
$$\ce{[NH4+] = \dfrac{\pu{7.5 millimoles}}{\pu{65 mL}}} = \pu{0.1153 M}$$
$$\ce{[NH3] = \dfrac{\pu{12.5 millimoles}}{\pu{65 mL}}} =\pu{0.1923 M}$$
Using the hammer known as the Henderson–Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\left(\dfrac{\mathrm{[base]}}{\mathrm{[acid]}}\right) = 9.2552 + \log\left(\dfrac{\pu{0.1923 M}}{\pu{0.1153 M}}\right) = 9.4773 $$
Now rounding to two significant figures:
$$\ce{[NH4+]} = \pu{0.12 M}$$
$$\ce{[NH3]} = \pu{0.19 M}$$
$$\mathrm{pH} = 9.47$$
Solution 2 Considering the autoionization of water yields the following charge balance equation:
$\ce{[NH4+] + [H+] = [Cl-] + [OH-]}$
To determine if the first solution is adequate, do two checks using the values from the first solution:

Is $\ce{[NH4+] \gg [H+]}\ $?

$$\pu{0.12 M}\ \ce{NH4+} \gg \pu{3.4\times10^{-10} M}\ \ce{H+}$$
So this assumption is good.

Is $\ce{[Cl-] \gg [OH-]}\ $?

$$\pu{0.19 M}\ \ce{Cl-} \gg \pu{3.0\times10^{-5} M}\ \ce{OH-}$$
So this assumption is good to 2 significant figures,
but barely good for 4 significant figures.
Thus the first solution is adequate

You might want to go back and check your MathJax

@themathandlanguagetutor - At the bottom I changed H+ to OH-. Was that the mistake you found?

I'm seeing several lines that just say "undefined control sequence \pu" and a few places where what should be a command is in the output, but if you're not seeing the same, maybe it's an issue with mobile

@themathandlanguagetutor - I looked at the \pu markups and they look fine to me. The answer renders fine for me. I don't know what I can do to fix it for you.

@themathandlanguagetutor The \pu thing does not work on phones, but it is very nice on computers.

Sir in reality what actually going on  its like this : salt is getting formed from the acid base reaction and then the salt goes to dissociation and also the anion part of salt then reacts with water ? If i am right , Which reactions are the equilbirium ones ? @MaxW


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Can Ethanol be used to carry out SN1 reaction?
My book says that the reaction of
$$\ce{Cl-CH2-O-CH2-CH3}$$ with ethanol to form
$$\ce{CH3-CH2-O-CH2-O-CH2-CH3}$$
occurs through $\mathrm{S_N1}$ mechanism, but shouldn't it actually be $\mathrm{S_N2}$?
My understanding of this statement is that the substrate is a $1^\circ$ halide and on top of that ethanol is available to us as $\ce{C2H5-O-}$, which is a considerably strong nucleophile. All possible conditions are pointing towards an $\mathrm{S_N2}$ reaction but then why does the reaction proceed through $\mathrm{S_N1}$ mechanism?

The presence of the oxygen in chloromethylether stabilises +ve charge so Cl- can leave in an SN1.

Ok, so basically what you are trying to say is that the carbocation formed when chlorine leaves overpowers the benefits of SN2 in this case , right?

yes, essentially H2C=OEt+ is the intermediate formed

I still don't by Sn1 even with the alpha hydrogen. The bimolecular pathway also sees a significantly increased rate.

It is well-established that alkyl chlorides where derivatization involves the introduction of a substituent atom such as oxygen, sulfur, or nitrogen on the α-carbon give, on ionization, resonance stabilized carbocations as depicted in equation $(1)$ (Ref.1):
$$\ce{R-X-CH(Cl)-R’  ->[-Cl] R-X -CH^+-R’ <-> R-X+=CH-R’}  \\\quad \text{where } \ce{X = O, S, N} \tag1$$
The formation of the carbocations, with an appreciable amount of delocalized charge, is relatively rapid (Ref.1-Ref.5). This behavior of chloromethyl ethyl ether (where $\ce{R = CH3CH2 \!-}$, $\ce{X = -O \!-}$, and $\ce{R' = -H}$) is well explained in the page 289 of Ref.4:

The rate constant for the solvolysis of chloromethyl ethyl ether, chloromethyl octyl ether, and chloromethyl methyl sulfide have been determined in several pure and binary solvents. Aplication of the extended Grunwald-Winstein equation, $\log (k/k_\circ) = \ell \mathrm{N_T} + m\mathrm{Y} + c$, gave appreciable “$\ell$” values (0.55-071) for the three substrates indicating that there is significant nucleophilic salvation of the developing carbenium ion in the transition state of these reactions. The $k_\ce{Cl}/ k_\ce{F} = 1.2 \times 10^{5}$ found for the hydrolysis of  chloromethyl methyl ether in water is virtually identical to that observed for the unimolecular solvolysis of t-butyl chloride and trityl halides confirming the unimolecular mechanism for these reactions.

The effect of the adjacent lone pairs of electrons can be dramatic (Ref.1). For example, the solvolysis of chloromethyl ethyl ether ($\ce{CH3CH2OCH2Cl}$) in $36\%$ dioxane proceeds at a rate of about $10^9$ times that of butyl chloride (Ref.3, p. 103). Also, Ballinger, et al. (Ref.6) has presented the evidence showing the solvolysis of related chlorodimethyl ether $(\ce{ClCH2OCH3})$ is unimolecular in ehanol and mixtures of ethanol/diethyl ether system, rates of which are at least about $ 10^{14}$ faster than the corresponding reactions of methyl chloride. With the same substrate, the ethanolysis rate is about $10^{13}$ times faster than that of propyl chloride (Ref.3, p.103). Note that it is of interest that when the nucleophile is ethoxide ion, the reaction undergoes in bimolecular mechanism.
On the other hand, the reaction mechanism is completely different when ethoxide ion is involved as a nucleophile in these reactions. For example, note that it is of interest that when the nucleophile is ethoxide ion (when the substrate is $\ce{ClCH2OCH3}$), the reaction undergoes in bimolecular mechanism. Similar to the shown rate enhancement in unimolecular reaction with ethanol, in this bimolecular reaction with ethoxide ion, the second-order rate coefficient is also considerably larger (by a factor of about 105) than that for the methyl chloride (Ref.6).
All of these findings suggest that an unimolecular $\mathrm{S_N1}$ mechanism is the most possible for ethanolysis of $\ce{ClCH2OCH2CH3}$, a reagent of which has commonly used in $\ce{R-OH}$ protection in synthetic procedures.
References:

Dennis N. Kevill, Young Hoon Park, Byoung-Chun Park, Malcolm J. D'Souza, "Nucleophilic Participation in the Solvolyses of (Arylthio)methyl Chlorides and Derivatives: Application of Simple and Extended Forms of the Grunwald-Winstein Equations," Current Organic Chemistry 2012, 16(12), 1502–1511 (DOI: 10.2174/138527212800672592).
C. A. Bunton, In Reaction Mechanisms in Organic Chemistry Series, Volume 1: Nucleophilic Substitution at a Saturated Carbon Atom; Publishing Co.: New York, NY, 1963. pp. 50–51 (ISSN: 0079-9823).
Andrew Streitwieser, Jr., In McGraw Hill  Series in Advanced Chemistry: Solvolytic displacement reactions; McGraw Hill: New York, NY, 1964 (ISBN-13: 978-1258354817).
K. C. Westaway, “Chapter 7: Nucleophilic Aliphatic Substitutions,” In Organic Reaction Mechanisms•2012: An annual survey covering the literature dated January to December 2012; A. C. Knipe, Ed.; John Wiley & Sons, Ltd.: Chichester, West Sussex, United Kingdom, 2015, pp.267-306 (ISBN: 978-1-118-36259-4).
Andrew Streitwieser, Jr., “Solvolytic Displacement Reactions at Saturated Carbon Atoms,” Chem. Rev. 1956, 56(4), 571–752 (https://doi.org/10.1021/cr50010a001).
P. Ballinger, P. B. D. de la Mare, G. Kohnstam, B. M. Prestt, "The reaction of chlorodimethyl ether with ethanol and with ethoxide ions," J. Chem. Soc. 1955, 3641–3647 (https://doi.org/10.1039/JR9550003641).

Thank you so much , cleared all my concepts regarding this topic.

Be careful!  Ethanol is not a strong acid ($\mathrm{p}K_\mathrm{a} \approx 16$). So its conversion to ethoxide anion is not so easy. Rather, it would attack the $\ce{C}$ center by its lone pair and then it would release proton to form the product.
Actually here what I think is, the hetero atom $\ce{O}$ just near to the $\ce{C}$ undergoing substitution will assist by its lone pair to kick out chloride and then being planar ethanol can attack from either side. And finally, you get the product of $\mathrm{S_N1}$ type.
But don't mess up between ethanol and ethoxide ion. They are totally different in nucleophilicity. Treat them as two different nucleophiles.

So, if the reaction would have stated that ethoxide ion is present then would it have undergone SN2?

Yes I think so. Of course ethoxide ion is a strong nucleophile than ethanol and will more bend towards Sn2

Thanks and i had just one more doubt , if it would have been given aq. ethanol then what would  we have considered it as , ethanol or ethoxide ion?

Aq.Ethanol means you have water as solvent which is polar. As polar solvent would bend towards Sn1(stabilize the carbocation by solvation) I think it should be Sn1.

@TheChemist if you choose an aqueous medium then ethanol will be replaced by H2O as nucleophile due to H2O being more available than ethanol added.


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Relative humidity at constant temperatures
It’s known that Relative humidity = (water pressure in a certain volume/ saturated water pressure in the same volume)* 100% “at the same temperature”
So my question is why does the amount of water isn’t max all the time instead it differs so we calculate humidity?
The unusual thing for me is that we didn’t change the temperature so what makes it differs?
Thanks in advance

Why do you think it should be always the max ?

To have 100% relative humidity at constant temperature, 3 conditions have to be met:

There must be water source to saturate space with vapour.
There must not be present materials absorbing humidity below 100%.
There must be allowed enough time to reach saturation.

E.g., if there is solid table salt exposed to air, the equilibrium relative humidity would not be higher than about 75%, even if water is there. As salt absorbs water and starts dissolving in it. Either all water evaporates and  humidity is still <= 75%, either all salt dissolves and continues absorbing vapour and diluting itself.
But even if all salt dissolves, 100% could be reached only if salt solution absorbs so much vapour we can neglect salt content in the salt solution.
For real life air conditions, varying temperature of free air leads to cycling of evaporation and condensation, resulting in air usually unsaturated. If kept closed at constant temperature, without evaporation or condensation, it stays unsaturated.


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To what degree can the Rare Earth Elements be interchanged in its workaday applications?
In their humdrum pedestrian "high-tech applications due to their unique magnetic properties", how interchangeable are the Rare-Earth Elements (REEs) if, e.g., the US lacks one of them?
RRautamaa comments on Pentagon in talks with Australia on rare earths plant.

Rare earths are not rare, but they're chemically very similar. (This is because when you go to higher atomic numbers, they start filling the $\mathrm{f}$-shell, but the $\mathrm{f}$-shell is lower in energy than the $\mathrm{d}$-shell, so it's "hidden" by the $\mathrm{d}$-shell. The $\mathrm{d}$-shell is the valence shell and is the same in all of them, so their chemistry is almost the same.)
Separating two chemically similar substances is expensive. Separating 15 of them is much worse. This means that the mining part is easy, but the processing plant on the site costs about $500 million. This is why rare earth mines are so few in number.

This question was spurred by the Pentagon's halt on May 22 2020 on rare earths funding program pending 'further research'. "Of course, the problem with rare earth elements has never been scarcity. They're actually pretty common. The problem is just that the cost of mining them exceeded their market value everywhere but China, so China ended up with a monopoly and the ability to drive up the price. So it's not clear that this changes much.

It depends on the particular use of the "rare earth". Mischmetal, containing an assortment of mixed lanthanides, can be used as-is for lighter "flints" or fire-starters. On the other hand, ultra-high purity lanthanide elements are needed to get specific spectral lines for lasers, for example.
Other uses, such as in magnets and in high-temperature superconductors, may be somewhat less demanding of purity. Didymium glass, for example, used in glasswork to block yellow sodium light, is a mix containing primarily samarium, praseodymium and neodymium.
Another expense involved in lanthanide production is separation and disposal of the radioactive isotopes from the ore. At one time, for example, gas lantern mantles contained mildly radioactive $\ce{ThO2}$, but those are no longer available in many countries due to concern about the alpha emitting thorium and the alpha and beta emissions from daughters.


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What is the relationship between these isomers of dimethylcyclohexane (what type of isomers are they)?
What is the relationship between these isomers of dimethylcyclohexane (what type of isomers are they)?

Is  (1R,2S)‐1,2‐dimethylcyclohexane  superimposable on its mirror image?  What about  (1R,2R)‐1,2‐dimethylcyclohexane?

In general, any two stereoisomers that are not mirror images are diastereomers, More specifically, these are geometrical isomers or cis–trans isomers. (1R,2S)-1,2-Dimethylcyclohexane is the cis isomer, whereas (1R,2R)-1,2-dimethylcyclohexane is the trans isomer.
For the second part of your question, (1R,2S)-1,2-dimethylcyclohexane is superimposable on its mirror image as it has a plane of symmetry. (1R,2R)-1,2-Dimethylcyclohexane is not superimposable as there are no elements of symmetry and hence is optically active.

The relation between them is Optical isomers.
The relation between them is  Geometrical isomers also.
Any 2  compound showing either of GI or OI are related as diastreomer.
In more appropriate words diastreoismer is used to relate the streoisomers which are non superimpose mirror images of each other.
Hence they are also related as  diasterisomers
Is (1R,2S)‐1,2‐dimethylcyclohexane superimposable on its mirror image?
It is superimposable on its mirror image.The reason being is the presence of events of symmetry POS.
You can verify this even by drawing the mirror image of the compound and try to superimpose on the given structure .They will perfectly overlap (and they would be related as *identical *)

,> What about (1R,2R)‐1,2‐dimethylcyclohexane?
It is clearly non superimposable on its mirror image as there is absence of any of the elements of symmetry like POS and COS .
You can also verify this by drawing the mirror image of the compound and try to superimpose on the given structure.
They would not overlap.. (its mirror image would be related as Enantiomer with it.)


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What is the difference between primitive Hexagonal unit cell and Hexagonal closed packing HCP
There are 14 types of Bravais crystal lattices (Primitive + Centered) in 3D. Every solid in this universe is made by the unit cells of these lattices.

With the help of these Bravais lattice can HCP be formed?.
Is there any relation between Hexagonal unit cell and Hexagonal close packing
As we see that Hexagonal unit cell exists only in primitive AAA form then how can HCP ABA be formed from it .

Primitive Hexagonal unit cell

Hexagonal closed packing HCP

If there is a clear distinction then Should HCP also be placed among 14 Bravais lattice.

With the help of these Bravais lattice can HCP be formed?

Yes. One choice of unit cell for HCP is a primitive hexagonal cell with two atoms per unit cell. If you place the first atom on the origin, the second will be at (1/3 2/3 1/2). Here is a 3D view of the situation and some more information: http://lampx.tugraz.at/~hadley/ss1/crystalstructure/structures/hcp/hcp.php

Should HCP also be placed among 14 Bravais lattice?

No, because it is already in there, just hiding a bit. If you look further into it, the space group is #194, P$6_3$/mmc, see http://img.chem.ucl.ac.uk/sgp/large/194az1.htm


text:
Calculating concentration of OH⁻ ions

Calculate concentration of $\ce{OH-}$ ions in $\pu{0.66 mol L^-1}$ solution of $\ce{NH4+}.$

I think that there might be something missing, as the only law I have for this kind of problems is
$$[\ce{OH-}]^2 = K_\mathrm{b} × C_\mathrm{b},$$
but is there a way to calculate $[\ce{OH-}]$ by using concentration of base only $(C_\mathrm{b} = \pu{0.66 mol L^-1}).$

I don't understand the question either, there is no such thing as solution of $\ce{NH4+}.$ Was it probably translated from another language? If it's ammonium hydroxide solution, then the answer is too obvious: $[\ce{NH4+}] = [\ce{OH-}],$ which would be strange.

Yeah it probably means ammonium hydroxide.

It was a multiple choice question and 0.66 M wasn't one of the choices, so I suggest that this is the concentration of the base itself not the ions.

And since ammonium hydroxide is a weak electrolyte, then concentration of the base must be greater than concentration of hydroxide ions.

Fair enough, but were you supposed to know $K_\mathrm{b}(\ce{NH4OH})?$ It look like you were expected to give a quantitative answer.

Yeah if I had Kb I would be able to calculate [OH-], but I am sure that it was not given.

Unfortunately I don't know any other laws except for this one.

You can't have two laws or ways of calculating giving you different correct answers. So if you need to know the $K_\mathrm{b}$ for one way of solving it, the answer will depend on it for any other way of calculating it. Bottom line, you need to know it.

@Kartsen Theis so it can't be calculated withour Kb , or is there a way to calculate Kb to be used in this law?

@andselisk One simply cannot have that much ammonium cation in simple NH3 solution. That means is an ammonium salt, for example NH4Cl

Your problem seems to have no solution without further information. You may be interested in the the general method for any acid /base calculation is summarised below.
When there is solution of a weak acid and it salt, or just the weak acid or just the salt, i.e. pure HA, NaA + HA or pure NaA, then there is no distinction between these types of solution because of the equilibria involved. ( NaA just represents any salt ).
The following calculations apply for each or any of these solutions.
The full calculation for problems of this type starts by defining the equilibrium constants for the reactions. A molecule dissociates as $\mathrm{HA \leftrightharpoons H^+ + A^-}$, but to be general we let the base be [B] instead of $\mathrm{[A^-]}$ and so write $\mathrm{HA \leftrightharpoons H^+ + B}$
$$K_a=\mathrm{[H^+]_e[B]_e/[HA]_e} \tag{1}$$
where the concentrations are those at equilibrium.
There is also the equilibrium $\mathrm{H_2O \leftrightharpoons H^+ + OH^-}$ to consider and
$$K_w=\mathrm{[H^+]_e[OH^-]_e}$$
(The activity of water is 1 so is ignored in this equilibrium ).
We know the amounts of acid $c_a$ and base $c_b$ added at the start of the reaction. To obtain the pH at equilibrium the amount $\mathrm{[H^+]}$ has to be calculated. To do this work out the concentration of acid and base in terms of the initial amount and the ionized species.
The total concentration of HA is
$$c_a=\mathrm{[HA]_e+[H^+]_e - [OH^-]_e}\tag{1a}$$
and for the base
$$c_b=\mathrm{[B]_e -[H^+]_e + [OH^-]_e\tag{1b}}$$
The e subscripts are now dropped for clarity. (Note that some authors use mass and charge balance instead to work out the concentrations.)
These values can be substituted into the equilibrium constant equation (1) and using $K_w=\mathrm{[H^+][OH^-]} $. Substitution leads to the general equation that can be used to solve all problems of this type;
$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - K_w/\mathrm{[H^+]} }{c_a - \mathrm{[H^+]} + K_w/\mathrm{[H^+]}  }\tag{2}$$
which is a cubic equation in $\mathrm{[H^+]}$ that is best solved numerically in the most general case. This equation can be simplified under different conditions.
Rewriting eqn 2 gives
$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - \mathrm{[OH^-]} }{c_a - \mathrm{[H^+]} + \mathrm{[OH^-]}  }$$
assuming that $\mathrm{[OH^-]} \gg \mathrm{[H^+]}$ then
$$K_A=\mathrm{[H^+]}\frac{c_b  - \mathrm{[OH^-]} }{ \mathrm{[OH^-]}  }$$
As $K_w=K_AK_B$ then
$$K_B = \frac{\mathrm{[OH^-]^2}}{c_B-\mathrm{[OH^-]}}$$
and making a further approximation, if $c_b >\mathrm{[OH]^-}$ then $K_Bc_B\approx \mathrm{[OH^-]^2}$, which is the equation you quote. Therefore there seems that your problem cannot be solved without more information.


text:
When we name an organic compound ,if there is double bond and a functional group which is now a substituent who has priority?
For example if a particular compound contains carboxylic acid, alcohol and double bond and the compound is such that we can either include double bond or alcohol in the main chain then who has priority?  Here alcohol behaves as substitutent as carboxylic acid is main functional group.

Please see our guidelines on how to ask nomenclature questions.

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

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

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

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

Therefore, if a compound contains a carboxylic acid group, an alcohol group, and a double bond, the principal chain has to include the principal characteristic group first, i.e. the carboxylic acid group $\ce{{}-(C)OOH}$ that is expressed as suffix ‘oic acid’.
Next, according to Rule P-44.3.2, the principal chain is the longest chain that contains the carboxylic acid group. (Note that this is a change from earlier IUPAC recommendations given in 1979 and 1993.)
If there is still a choice, the principal chain includes the double bond according to Rule P-44.4.1.


text:
Can old matches combust spontaneously?
I came upon this box of matches, it's probably from a little before 1900 and has significance for my family history, and I would like to add it to my "box o' historical items" but I'm concerned about flammability.  I know that old negatives can combust spontaneously, is there a similar danger with matches?  Can the matches be chemically changed (by dipping the tips in something, for example) that will make them inert?
EDIT
This is a strange question as it straddles the line between history and chemistry.  I've also asked this question on history.stackoverflow.com.

Those are safety matches made with red phosphorus, which becomes more stable with time. If they haven't burned spontaneously by now, there is little reason to fear that they will. Just in case, you might want to check with a phillumeny club.

I checked, there isn't a phillumeny.stackexchange.com ;-)  but I did find the Rathkamp Matchcover Society. Thanks!

Cross posting is frowned up...


text:
How to determine the source of PVA: crude or plant-based?
We're a small startup company looking to make products using ingredients solely derived from plants. One of the main ingredients in one of our products is polyvinyl alcohol (PVA). We have found a company that claims that their PVA is made exclusively from plants, and that no natural gas or crude oil is used in its production.
To provide some extra clarification based on the discussion below this question: PVA is a synthetic polymer prepared by hydrolysis of polyvinyl acetate. Polyvinyl acetate is made from acetic acid and ethylene. Ethylene is often made from natural gas or petroleum. Is it possible to determine whether the original ethylene used to make our PVA is derived from petroleum or plants? If so, can this be done given just the PVA itself, and no other information?
The reason I'm asking is because I want to independently verify the manufacturer's claim that their product is indeed completely derived from plant sources.

I don't think there is any plant or tree which makes polyvinyl alcohol. It is purely a synthetic polymer. Contact the manufacturer and ask them more details.

I agree with @ M. Farooq. PVA is purely a synthetic polymer. Your version of natural or synthetic probably based of its starting material, vinyl acetate?

Nor is there a "natural" source of vinyl acetate as it is a highly unstable with respect to hydrolysis to acetaldehyde. The feedstock for vinyl acetate production is ethylene or acetylene (https://en.wikipedia.org/wiki/Vinyl_acetate). How many industrial processes does a material need to go through before it ceases to be "natural"?

PVA is indeed a synthetic polymer prepared by hydrolysis of polyvinyl acetate. Polyvinyl acetate is made from acetic acid and ethylene. Ethylene is often made from natural gas or petroleum. My question took the shortcut of omitting this process in its description. Perhaps a better way to phrase my question would be: is it possible to determine whether the original ethylene is derived from petroleum or plants?

All oil-based chemicals are, ultimately, derived from plants, just 100m year-old plants.

Did you find the answer to your question from the manufacturer?
I was wondering the same. As far as I can see, PVA is a synthetic polymer and petroleum-based is the only option. Thanks.

It certainly would be possible to get an indication of some of the sources of carbon through mass spectrometry and through radiocarbon dating. For example, fossil carbon sources would have little 14C, sources predating the 20th century, e.g. older trees, would have more. Post-Atomic Age sources might have more yet, beyond that due to natural decay of 14C.
That said, it's a nice question, but are you prepared to pay for the analysis to have it answered? Perhaps your government might perform such testing to verify a claim of being "organically produced".

Thanks, that's an interesting avenue to explore! And yes, I'd be willing to pay for the analysis.

I believe your suggested approach is described in "Determination of the Biobased Content in Plastics by Radiocarbon" by Quarta et al. (2013)

This is actually a well-known technique for testing whether drinkable alcohol is made from industrial synthesis or fermentation.


text:
Which of the following would be an example of something that is quantized?

How do we defind quantized there, why it is climbing a flight of stairs?
Thank you!

I would say it's just a bad question. Besides, there are no options above, so it's also poorly coded.

It's not an amazing question, but do you know what "quantised" is, to begin with? Were you taught it, did you look it up on Wikipedia? https://en.wikipedia.org/wiki/Quantum

These are the four options:A.sliding down a hill. B.climbing a flight of stairs C.walking up a ramp, D. all of the above

Yes, I did, I do not know why only climbing a flight of stairs can be an example of quantized, but why not option D? Thank you!

Very very pooor conceptual question.

Are the hill or the ramp levels quantized ? Is walking or sliding quantized ? Is hill sliding or ramp walking quantized ?

It would appear the instructor is trying to draw an analogy between the quantization of an observable on the sub-microscopic scale and the macroscopic "quantization" of height.
To slightly improve upon your instructor's attempt (because walking up a ramp is ambiguous), let's say we are moving on roller blades. Then we can ask ourselves, for each given path (ramp, hill, staircase), at what points would we be stable? Only on the staircase will you find levels on the way up (or down) where you are stable (where you can stay put and your height is well-defined).
To reiterate, the instructor asked a poor question as it's not a great way to depict quantization. I don't fault you for answering all 3! I'd venture to say the only helpful part of this analogy is the idea of a stable state. Only on the stairs do you have multiple stable states that are discretized. The ramp and the hill have a continuum of states, and are thus do not have quantized states of height.


text:
How many sodium ions is each oxygen atom attracted to in aqueous sodium chloride?
When $\ce{NaCl}$ salt is dissolved in water the $\ce{Cl-}$ and $\ce{Na+}$ ions and the polar $\ce{H2O}$ molecules are attracted to each other such that each $\ce{Na+}$ ion attracts several oxygen atoms and each $\ce{Cl-}$ ion attracts several hydrogen atoms.
Solubility depends on temperature but around room temperature the solubility of $\ce{NaCl}$ in $\ce{H2O}$ is around $100$ g $\ce{H2O}$ to $\pu{35 g}$ $\ce{NaCl}$. This works out at a ($\ce{H2O}:\ce{NaCl}$) ratio of around ($9:1$) for a saturated solution.
So each $\ce{Na+}$ ion is attracted to up to around $9$ oxygen atoms depending on temperature. But is each oxygen atom only attracted to the one $\ce{Na+}$ ion at a time or are they simultaneously attracted to multiple $\ce{Na+}$ ions creating a kind of a web throughout the solution?
Every diagram I can find online suggests only one but I would like confirmation on this.

You're asking about mechanism of NaCl dissolution, or what? Either way you're wrong.

You can make a quick estimate yourself based on observable bulk physical properties:
According to “Aqueous Solubility of Inorganic Compounds at Various
Temperatures”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL., the solubility of $\ce{NaCl}$ at $20\ \mathrm{^\circ C}$ is
$$w_2=26.41\ \%$$
This solubility value is expressed as mass percent of solute, where
$$w_2=\frac{m_2}{m_1+m_2}$$
and $m_2$ is the mass of solute and $m_1$ the mass of water. Therefore
$$m_2=\frac{-w_2\cdot m_1}{w_2-1}$$
For example for a mass of water of $m_1=100.0\ \mathrm g$
$$\begin{align}m_2&=\frac{-w_2\cdot m_1}{w_2-1}\\[6pt]
&=\frac{-0.2641\times100.0\ \mathrm g}{0.2641-1}\\[6pt]
&=35.89\ \mathrm g\end{align}$$
I.e., $100.0\ \mathrm g$ of water can dissolve $35.89\ \mathrm g$ of $\ce{NaCl}$.
Furthermore, you know that $M(\ce{NaCl})=58.43977\ \mathrm{g\ mol^{-1}}$ and $M(\ce{H2O})=18.01500\ \mathrm{g\ mol^{-1}}$. Thus, the molar ratio of water to $\ce{NaCl}$ is
$$\begin{align}r&=\frac{n(\ce{H2O})}{n(\ce{NaCl})}\\[6pt]
&=\frac{m_1/M(\ce{H2O})}{m_2/M(\ce{NaCl})}\\[6pt]
&=\frac{100.0\ \mathrm g/18.01500\ \mathrm{g\ mol^{-1}}}{35.89\ \mathrm g/58.43977\ \mathrm{g\ mol^{-1}}}\\[6pt]
&=\frac{5.551\ \mathrm{mol}}{0.6141\ \mathrm{mol}}\\[6pt]
&=9.039\end{align}$$
This means that nine water molecules are needed to dissolve one $\ce{Na+}$ ion and one $\ce{Cl-}$ ion. Therefore, you may assume that the solvation process is not a simple one to one pairing.

When NaCl is dissolved in water, the sodium ions are surrounded by several water molecules which are all oriented with their oxygen atoms touching the sodium ion. But this number may fluctuate, as these water molecules are not chemically attached by covalent bonds and they are often bumped and even ejected by other water molecules due to the Brownian movement. The same thing happens to chloride ions which are surrounded by several water molecules even though they are not oriented like around the sodium ions. Anyway the ions are independent and can move separately in the solution. It is not a one to one pairing, as you think.

Also search solvation number and hydration number.


text:
Stability of CI4 and PbI4
I have read in my textbook that $\ce{CI4}$ is more stable than $\ce{PbI4}$ due to inert pair effect. But shouldn't we also consider the effect of steric repulsions? In $\ce{CI4}$, four large iodine atoms surround the small carbon atom, which should make it less stable than $\ce{PbI4}$.

You are correct up to a certain point: yes, steric repulsions destabilise CI4. But there is nothing that lets you conclude that that factor will destabilise CI4 to the point where it is less stable than PbI4. It is like saying that I earnt some money at my job, so surely I am richer than Bill Gates now? Well, it's certainly made me richer than I was before; but unfortunately, Bill Gates still has more money. So, yes, sterics sure make CI4 less stable than it should otherwise be, but PbI4 is still less stable. Do you see why the reasoning doesn't make sense?

Steric hindrance does indeed interfere with the stability of carbon tetraiodide, as does poor covalent bonding overlap between the very differently sized carbon and iodine atoms.  Even so, carbon tetraiodide is definitely established and in use as an iodinating agent.  Storage below normal room temperature is recommended.
Carbon tetraiodide is also very red.  Ordinarily saturated carbon compounds are white or colorless because the electronic transitions involving single bonds to carbon require ultraviolet radiation to go, but the relatively weak bonds in carbon tetraiodide enable transitions in the visible light region.  Iodoform is more lightly colored for a similar reason.
Lead tetraiodide, meanwhile, does not seem to be definitely known to exist.

Just curious, why was this unaccepted given there is no alternative answer? (Meaning what's missing?)


text:
Balancing reactions algebraic method limitations
How do I balance this reaction using the algebraic method?
Seems that the equation can't be solved.
How do I balance such reaction and does the algebraic method has limitations and wont work for balancing some reactions?
$$\ce{XeF4 + H2O -> XeO3 + Xe + O2 + HF}$$
$$\ce{a XeF4 + b H2O -> c XeO3 + d Xe + e O2 + f HF}$$
$$4 a = f$$
$$2 b = f$$
$$a =  c +  d$$
$$ b = 3 c + 2 e $$

I get a = 4, b= 8, c = 2, d = 2, e = 1 and f = 16.

Thank you for your comments. I'm dabbling in chemistry as a new found hobby. Seems that I have a lot to read and learn as I'm not familiar with the terms you are using. If you can write the comment as an answer I will accept it.

By the way, it is usually best to wait a day or so to award the green check mark: people are all around the world and maybe you will get an answer that is much better (your call) than my answer. So feel free to take the green check mark back, if you want to do so. But, otherwise, I appreciate it and hope my answer helps a bit.

We'd better check ourselves here.  Look at what I found ... .

Full disclosure: when I saw this question, I just did it the old redox way I have always used. However, as the answers below show, this question had some interesting aspects because it was a combination of a disproportionation reaction and a redox reaction sub-reaction. You never know what nice puzzle will pop up here when you least expect it!

This equation in not determined, because whatever your balanced equation, you can always take a part of XeO3 formed, and have it decomposed in Xe + O2

As per the OP's request, here is a solution. The equation is:
$$\ce{a XeF4 + b H2O  -> c XeO3 + d Xe  + e O2  + f HF  \tag 1}$$
So atom balance yields:
$$4a=1f     \tag 2$$
$$2b=1f     \tag 3$$
$$1a=1c + 1d     \tag 4$$
$$1b = 3c + 2e     \tag 5$$
The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I proceed as follows:

Assume $a = 1$, which is the minimum non-zero value it might have. Then, from equation $(4)$, $1 = a = c + d$, which is impossible because either $c$ or $d$ would have to be zero.
Assume $a = 2$. Then $b = 4$ because equations $(2)$ and $(3)$ show that $b = 2a$. But then equation $(5)$ shows that $4 = b = 3c + 2e$. Then if $c = 1, e = 0.5$, which is impossible. Obviously, $c$ cannot equal $2$, because that would give a negative value for $e$.
Assume $a = 3$. Then $b = 6$, because $b = 2a$, as above. Then equation $(5)$ shows that $6 = b = 3c + 2e$. Then if $c = 1, e = 1.5$, which is impossible. If $c = 2$, then $e = 0$, which is impossible. As before, $c$ cannot be greater than $2$, because that would give a negative value for $e$.
Assume $a = 4$. Then $b = 8$ and $f = 2b = 16$. Then equation $(5)$ shows that $8 = b = 3c + 2e$. If $c = 1$, then $2e = 5$, so $e = 2.5$, which is impossible. If $c = 2$, then $2e = 2$, so $e = 1$. Then equation $(4)$ shows that $4 = 2 + d$, so $d = 2$. This is a solution. Lastly, $c$ cannot equal $3$, because that would give a negative value for $e$.

Almost final result: $a = 4, b = 8, c = 2, d = 2, e = 1,$ and $f = 16$. Hence
$$\ce{4 XeF4 + 8 H2O  -> 2 XeO3 + 2 Xe  + 1 O2  + 16 HF  \tag 6}$$
For a = 5, the solution is a = 5, b = 10, c = 2, d = 3, e = 2 and f = 20. This is the same as Oscar Lanzi's example solution. For a = 6, the solution is a = 6, b = 12, c = 2, d = 4, e = 3 and f = 24.
For a = 7, a double solution occurs: a = 7, b = 14, c = 2, d = 5, e = 4 and f = 28 and also a = 7, b = 14, c = 4, d = 3, e = 1 and f = 28.
There are an unlimited number of valid balanced equations, as expected. So the a = 4 solution is simply the first solution in the infinite series and it is the minimum coefficient solution, since the total number of atoms is 11a.
Bonus: Mithoron provided a link to a problem Balancing disproportionation reactions answered by Ivan Neretin years ago. The equation that was to be balanced was:
$$\ce{a XeF2 + b H2O  -> c Xe + d XeO3  + e O2  + f HF  \tag 7}$$
So atom balance yields:
$$1a=1c + 1d     \tag 8$$
$$2a=1f     \tag 9$$
$$2b=1f     \tag {10}$$
$$1b = 3d + 2e     \tag {11}$$
The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I first note that equations $(9)$ and $(10)$ show that $a = b$. Then equation $(11)$ shows that $b$ must at least equal $5$. Otherwise either $d = 0$ or $e = 0$, which is impossible.
Now I proceed by assuming $a = b = 5$. Then equation $(8)$ gives rise to the following four possibilities: $c = 1$ and $d = 4$; $c = 2$ and $d = 3$; $c = 3$ and $d = 2$; $c = 4$ and $d = 1$. If $c = 1$ and $d = 4$, then equation $(11)$ yields $5 = 12 + 2e$, which is impossible. Likewise for $c = 2$ and $d = 3$ and also for $c = 3$ and $d = 2$. But if $c = 4$ and $d = 1$, then equation $(11)$ yields $5 = 3 + 2e$, so $e = 1$. This is a solution: $a = 5, b = 5, c = 4, d = 1, e = 1,$ and $f = 10$. Hence the balanced equation is:
$$\ce{5 XeF2 + 5 H2O  -> 4 Xe + 1 XeO3  + 1 O2  + 10 HF   }$$
However, as Ivan Neretin beautifully pointed out in his answer, there is also the simpler redox reaction given by
$$\ce{2 XeF2 + 2 H2O  -> 2 Xe   + 1 O2  + 4 HF   }$$
So any linear combination of these, with positive integer coefficients, is also a valid balanced equation. For example, just add the last two equations together together:
$$\ce{7 XeF2 + 7 H2O  -> 6 Xe + 1 XeO3  + 2 O2  + 14 HF   }$$

Actually there is no unique solution.  You have four linearly independent equations and six compounds.  In a typical redox reaction you need one less equation than the number of compounds.  When the difference is greater than one, you have a combination of multiple reactions.
We can get unique numbers only by assuming that just one oxidation product is formed, either $\ce{XeO3}$ or $\ce{O2}$.
If we choose $\ce{XeO3}$, then we can use the redox method to see that two moles of that compound are formed per one mole of elemental xenon.  This leads to the result
$\ce{3 XeF4 + 6 H2O -> Xe + 2 XeO3 + 12 HF}$
If instead $\ce{O2}$ is chosen, one mole of that matches with one mole of xenon being formed:
$\ce{XeF4 + 2 H2O -> Xe + O2 + 4 HF}$
Now any linear combination you want is a balanced reaction, corresponding to some assumed ratio of $\ce{XeO3}$ versus $\ce{O2}$ formation.  If you assume equal molar amounts of both products, for instance, you would have the first component plus twice the second:
$\ce{5 XeF4 + 10 H2O -> 3 Xe + 2 XeO3 + 2 O2 + 20 HF}$
A very different answer from others yet equally well balanced!
In a situation like this you have to measure the relative contributions of the two oxidation products to get the true stoichiometry. And specify the conditions of such an experiment, for no law of nature requires the ratio to remain constant.

I like how Ed V has explained the mathematical solution logically. I'd like to show another way of solving the problem. The equation is as written by OP:
$$\ce{a XeF4 + b H2O  -> c XeO3 + d Xe  + e O2  + f HF  \tag 1}$$
So atom balance of equation $(1)$ yields:
$$\text{Balancing } \ce{F}: \quad 4a = f     \tag 2$$
$$\text{Balancing } \ce{H}: \quad 2b = f     \tag 3$$
$$\text{Balancing } \ce{Xe}: \quad a = c + d     \tag 4$$
$$\text{Balancing } \ce{O}: \quad b = 3c + 2e     \tag 5$$
The four equations relate to the six unknown coefficients $a$ through $f$. As Ed V pointed out, each coefficient is an integer greater than zero, so I proceed as follows:

I assumed $a = 2$, because in right hand side there are two $\ce{Xe}$ atoms. Then, from equation $(2)$, $f = 8$ and from equation $(3)$, $f = 8 = 2b \Rightarrow \; \therefore b = 4$.
Because coefficients are non-zero integrals, I also assumed $c = d= \frac12 a =\frac12 \times 2 = 1$. Thus, from  equation $(5)$, $2e = b - 3c = 4 - 3 = 1 \Rightarrow \; \therefore b = \frac12$. This is not an integer so my original assumption of $a = 2$ is incorrect. Yet, we can complete the task at this point because fraction of coefficients are allowed in chemistry if at least one coefficient is an integral. Therefore, the equation would be:

$$\ce{2 XeF4 + 4 H2O  -> 1 XeO3 + 1 Xe  + 1/2 O2  + 8 HF  \tag 6}$$
If you multiply whole equation $(6)$ by 2, it becomes:
$$\ce{4 XeF4 + 8 H2O  -> 2 XeO3 + 2 Xe  + 1 O2  + 16 HF  \tag 7}$$
The equation $(7)$ is the correct answer, identical to Ed V's. If this is not acceptable, then we can continue as follows:

Since we get $e = \frac12$ by assuming $a=2$, we can start again assuming $a = 4$ as stultification of 2. Then, from equation $(2)$, $f = 16$ and from equation $(3)$, $f = 16 = 2b \Rightarrow \; \therefore b = 8$.
Apply this values to equation $(4)$ and $(5)$:

$$4 = c + d     \tag 8$$
$$8 = 3c + 2e     \tag 9$$
Subtract equation $(9)$ from equation $(8) \times 3$:
$$3d - 2e = 4    \tag 10$$
The smallest values of $d$ and $e$ to fulfill this requirement are $d = 2$ and $e = 1$. Hence, from the equation $(4)$, $c = a - d = 4 - 2 = 2$.
As a result, the values of coefficients would be: $a = 4, b = 8, c = 2, d = 2, e = 1,$ and $f = 16$. Hence the balance equation with integer coefficients:
$$\ce{4 XeF4 + 8 H2O  -> 2 XeO3 + 2 Xe  + 1 O2  + 16 HF}$$


text:
Latent Heat, Vapor pressure and Equilibrium
So after reading about this topic i wanted to share what i learned and i wanted you guys to tell me if what i am writting is correct:
A pressure vessel has propane inside and it is pressurized. The liquid is at atmospheric temperature meaning that it is in equilibrium with the outside temperature. The pressure inside the pressure vessel will change according with ambient temperature.

If I open the valve of the pressure vessel, the gas will come out and the pressure inside the vessel will decrease and the liquid propane inside the vessel will immediately star to boil so it can re-establish the equilibrium. When boiling the liquid goes into the gaseous state and to do this it requires energy in the form of heat (latent heat of vaporization).
Since the liquid has the same temperature that the outside it cannot take heat from the outside and it will need to take this energy from itself, decreasing the temperature of the liquid.
Now there is a ΔT between the liquid and the outside temperature, so the liquid will steal heat from the outside (sensible heat) so it can incease his temeprature in order to have enough energy to go to the gaseous state.
If the flow of gas that is coming out of the vessel is constant the pressure and the temperature of the liquid and of the gas will initially decrease a little but it will stabilize, because the heat that is being taken from the outside is enough to keep the liquid boiling without decreasing more of his temperature and pressure.
If the flow of gas that is coming out of the vessel increases the outside heat may not be enough to maintain the flow of gas required, so the temperature of the liquid will start to decrease (auto refrigeration) (heat absorbed by the liquid < heat used by liquid to go to the gaseous state due to the high gas demand)
If the temperature of the liquid keeps decreasing so will the pressure and if it decreases below the atmospheric boiling point, the liquid will stop boiling and the gas will stop to come out (negative pressure inside the pressure vessel).

Keep in mind the Andselisk's note.

@ Mathew Mahindaratne  what is said in that question is correct?

This is a shorter attempt than your previous post, but it still rambles a little and as @andselisk commented, this is not a proof-reading service. One comment: the following is not logically correct: "Since the liquid has the same temperature that the outside it cannot take heat from the outside.". Separate statements that describe properties or states of the system (adiabatic) from those that describe changes in the state. Heat transfer is a change. A particular temperature or the inability to exchange heat are descriptions of states.

This is your 5th question in last 5 months about the same topic, one would think you have already learnt all the basic knowledge about it, but you still ask the basics.
Instead of asking to verify the story, formulate the questioned principles in pointed out isolated statements, and before asking, search about them.

What do you think will happen if you let propane escape (at whatever rate), but the propane is insulated thermally from the outside temperature?

@ChetMiller the propane will start to boil, his temperature will start to decrease because the heat that is needed to go from liquid to vapor phase is being taked out from the liquid itself, and the liquid is not able withdraw heat from the outside to counter that heat loss. At the same time the pressure will also decrease.

Sounds pretty correct.  Quantitatively the entropy per unit mass of the combination of liquid and vapor remaining in the vessel will remain constant.

All is about  3 basic principles.

Liquid evaporation consumes thermal energy, taken from liquid itself or surrounding. Vapour condensation releases this energy.
Liquid saturated vapour pressure (about) exponentially raises with temperature.
If saturated vapour pressure exceeds external pressure, liquid boils. If vapour pressure exceeds the saturated vapour pressure, vapour condensates.


text:
Why alkyl shift happens during hydroboration oxidation?

In this oxidation step, where alkyl shift happens for 3 times to kick out the OH- to obtain the boron-ester. But why does this happen? Won't the boron atom become less stable as the empty p orbital is regenerated? Won't this cause the OH- (strong nucleophile) to attack the boron (electrophile) again?
What is the mechanism of trialkylborane oxidation with hydroperoxide? (I have read this but sorry I don't get it. before this reaction, isn't the boron atom also bonded to an oxygen atom? Why would this reaction make it more stable?)

But why does this happen?

Well, if it was a plain old carbon in the middle, this migration wouldn't happen. It would sit there forever. Boron isn't carbon, though: it has fewer protons, is less electronegative, and doesn't like holding onto electron density all that much. That electron density gets pushed onto the adjacent carbons, which make them rather more nucleophilic, and happy to jump next door into the $\sigma_\ce{O-O}^*$ orbital.
You can see, after all, that there's something about that boron that's not quite perfect: it has a formal negative charge. While this shouldn't be taken as a sign of great instability, it does suggest that boron won't mind losing that formal negative charge too much.

Won't the boron atom become less stable as the empty p orbital is regenerated?

The energetics of this migration aren't just about the boron atom, it's about the system as a whole. You are also breaking a weak O–O bond, forming a stronger C–O bond in replacement, and so on. And indeed, even if this one step is energetically unfavourable, it might not even matter as long as the overall reaction is spontaneous.
Is the overall reaction energetically favourable? I'll leave it to you to think about it, based on the types of bonds that are broken and formed over the course of the entire reaction. You already saw some of the considerations in the linked question. (I don't get what you mean by the boron atom being bonded to oxygen before the reaction; it isn't, as it has three bonds to carbon.)

Won't this cause the $\ce{OH-}$ (strong nucleophile) to attack the boron (electrophile) again?

Well, yes, but not much happens if you go down that route. It's far more interesting if $\ce{OOH-}$ attacks the boron again, because that means the second alkyl group can migrate. That is precisely the mechanism that you have learnt.

Phenol synthesis using cummene peroxide?

@AnindyaPrithvi that's under acidic conditions. Not the same thing. By protonating the peroxide you get a much better leaving group, and the driving force for migration doesn't need to be so strong. This migration is done under basic conditions, and you are probably aware that $\ce{OH-}$ is not (usually) a good leaving group.

Got it, so if it were given that we have acidic condition in this question as well..then?

@AnindyaPrithvi If you had acidic conditions you would never get a four-coordinate boron to begin with, so that's pretty much it. You'd have water floating around which is quite happy on its own, so isn't inclined to add to the trivalent boron. Maybe eventually you might hydrolyse the borane; I know this happens with carboxylic acids, but I don't know how stable they are to acids in general.


text:
Radioactive decay of ions in solution
What happens when ionic salts in solution decays? For example, if I had a handful of Francium-223 Palmitate(Francium salt of palmitic acid) and I were to put it in solution, how would the francium decay? Since the francium is in solution, the francium would exist as an ion and if it decays, I would think that it would stay as an ion and become either At+ or Ra+. And after that, what would happen to the Palmitate ion? Would it still be ionically bonded with the decay products?
EDIT: I was thinking of the saponification reaction when making this thread, as in the FrOH would react with tripalmitin, but then again I saw some comments saying that the radioactive decay of francium would blow up the Francium Palmitate, and I wasn't thinking too much when creating this thread, so I'll rephrase this. I was thinking about the radioactive decay of salts, like for example if you were to hypothetically dissolve FrCl (yes it will evaporate the water but that wasn't my question) in water(if it exists and/or is soluble in water), since it is in solution, the electron from Francium would end up sticking with the Chlorine atom, creating a Chloride anion and a Francium cation. And I was asking what would happen to the Francium cation, after the radioactive decay. Would it keep it's positive charge? And what would happen to the salt if the Francium cation actually kept its charge after the decay? If it doesn't keep the charge, what would happen to the chlorine radical? (I don't know much about chemistry so pls be nice)

I would doubt francium palmitate is readily soluble in water; what solvent are you proposing? I also doubt there is going to be any ionic bonding, and nuclear processes usually don't care where to occur, in bulk or in solution; the products are either entrapped in the solid matrix or, in case of a soluble compound, released and their fate depends on what solution we are talking about.

If you had a handful of francium palmitate,, it would blow up due enormous heat production rate. If you survived that, what is very improbable, you would die soon from radiation exposure.

Francium itself is unstable. Its most stable isotope has a half life of 22 minutes and then quickly decays to radon, astatine etc. I don't think francium palmitate will exist in any form even if you forcefully react francium and palmitic acid.

A handfull of pure Fr-223 salt would heat up with a power of about 1 MW. That would evaporate the solution within a few seconds and produce significant radiolysis. After 1 s in 1 m distance to the solution, your effective dose would be more than 40 Sv, which is enough to kill you after a few days.

A house is hit directly with a nuclear explosion, and you are worrying about the origami figures on the window sill. Will they get damaged? Er, well...

In saponification reaction, the hydroxide ion is important rather than alkali metal ion. NaOH and KOH are used because they are cheaper. Nobody is bothered using rubidium or caesium hydroxide. Even if we hypothetically were to use francium hydroxide, expect a nuclear explosion to occur. There is a thing called transmutation reaction which will occur at that moment where francium will decay to astatine-216 with crazy release of alpha particles.

Related: https://chemistry.stackexchange.com/questions/20449/does-radioactivity-affect-chemical-reactions

According to Wikipedia, $\ce{^{223}Fr}$ decays with the half-time 22 minute mostly by beta decay to $\ce{^{223}Ra}$ ( 99.994% ) and rarely ( 0.004 % ) by alpha decay to $\ce{^{219}At}$.
If we ignore my prior question comment  and if we pretend francium is reasonably stable and it's compound can be prepared in significant amounts, then:
The effect would be very similar like if you bombarded a soap solution by highly energetic and ionized radium ions and free electrons. As both would be flying fast across the solution like a herd of elephants through a collection of china ( radium just fractions of millimeter, electrons some centimeters ). There would be heavy radiolysis of water and palmitate.
Radicals from the organic matter would tend to recombine, preferably with other radicals or ions, eventually would react with other molecules.
Ions would have tendency to quickly gain such a charge they are used to have by scavenging free protons and electrons, or tearing them from other molecules. Radium would end as ion $\ce{Ra^2+}$ and would (co)precipitate as insoluble radium palmitate.
But better would be to consider more stable isotopes, like caesium palmitate ( $\ce{^{137}Cs}$ with the half-life about $\pu{30 years}$ ) or potassium palmitate ( the natural weakly radioactive isotope $\ce{^{40}K}$ with the half-like $\pu{1.5e9 years}$.) You know all plants and  animals are naturally radioactive, don't you ?


text:
Newman projections: how to tell which groups are bulkiest to determine what is the most stable conformation?
The scenario
Let's say we have a molecule where in the Newman projection the front carbon has a hydrogen, methyl and an ethyl, and in the back carbon there's a hydrogen, something like a S, Cl or Br (bigger than carbon but only one atom), and something like another ethyl or isopropyl:

My understanding
Methyls/ethyls/isopropyls etc. would cause more steric hindrance than something like S, Cl or Br because they consist of multiple atoms and take up more space, so in the scenario above, the front carbon's ethyl should be 180° away from the back carbon's ethyl/isopropyl, and the S/Cl/Br's position would be dependent on the ethyl from rotation. I think the following should be the most stable conformation:

My question
I've seen some similar examples where (in the same type of scenario) the S/Cl/Br was considered "bulkier" and was put 180° away from a big isopropyl, when something like an ethyl could have been there instead, but was put only 60° away from the big isopropyl (see the first illustration). I'm not sure if these examples were just wrong or if I'm completely misunderstanding how to determine what's bulkiest, so any advice on how to determine that would be super. If a table that ranks most-to-least bulkiest exists, please share.

Reading about A-values would be useful: https://www.masterorganicchemistry.com/2014/07/01/substituted-cyclohexanes-a-values/

Well, I think your book is wrong. Read the extract:

A-Values are numerical values used in the determination of the most stable orientation of atoms in a molecule (conformational analysis), as well as a general representation of steric bulk. The utility of A-values can be generalized for use outside of cyclohexane conformations. A-values can help predict the steric effect of a substituent. In general, the larger a substituent's A-value, the larger the steric effect of that substituent. A methyl group has an A-value of 1.74 while tert-butyl group has an A-value of ~5. Because the A-value of tert-butyl is higher, tert-butyl has a larger steric effect than methyl. This difference in steric effects can be used to help predict reactivity in chemical reactions.

I guess what you're looking for are A-values. The higher the A-value, the more "bulky" the group is considered.

So, the $\ce{-SH}$,$\ce{-Cl}$ and $\ce{-Br}$ groups seem to be less hindering than the $\ce{-iPr}$ group.
References: A-value, Wikipedia


text:
The increasing order of reactivity of substituted benz0yl chlorides towards nucleophilic substitution

Arrange the following compounds in the order of their increasing reactivity towards nucleophilic substitution:

The answer to this question is $[3<2<1<4]$. We have to compare the positive charge density on the carbons to which leaving group is connected. In $(1)$, the phenyl ring shows negative inductive effect but my teacher did not mention anything for that. In $(3)$ and $(4)$ the connected groups ($\ce{OCH3}$ and $\ce{NO2}$) show +M and -M respectively. But, they also show -I effect. Why are we ignoring the inductive effect?

You're kind of overcomplicating it. The rate-determining step is attack of a nucleophile on the electrophilic C=O. The more electrophilic it is, the faster the reaction. You don't need to care about the inductive effect of the phenyl group, because all of them have phenyl groups. The only thing that matters is the 4-substituent.

The person who decided it was OK to teach students that +ve and -ve are acceptable abbreviations is one of the worst people of the last century. + and - already mean positive and negative!

rate of reaction is proportional to the concentration of the nucleophile. But in this question, we don't know which nucleophile is being used.

Electronic effects govern in the order
Resonance/Mesomeric > Hyperconjugation > inductive
+M or +R dominate over inductive. That's because +M involves π electrons which are more reactive than the sigma electrons. + I operates through sigma electrons. So therefore +M effect has priority over +I.
Therefore the order of reactivity towards nucleophilic substitution is 4>1>2>3

does that mean we completely ignore the inductive effect?

No, it's just that +M must be considered first. That's because  +M involves π electrons which are more reactive than the sigma electrons. + I operates through sigma electrons.


text:
Why Peroxydisulphuric Acid is also called as Marshall's Acid and Peroxymonosulphuric Acid as Caro's Acid?
I recently come across the oxaacids of Sulphur and got to know to Acids, $\ce{H2S2O8}$ and $\ce{H2SO5}$ as Marshall's Acid and Caro's Acid respectively, but I am just eager to know why they are named respectively so.
I tried searching in Google and couldn't found any relevant information regarding the same.

It is named after respective chemists who discovered it.

@NilayGhosh, It is not necessary that they "discovered" it. You will have to check the original papers from 1890s! In the very first line he says that Bertholet described it initially.

If you are curious about word or name origins like I am, get hold of two books:

Elsevier's Dictionary of Chemoetymology: The Whies and Whences of Chemical Nomenclature and Terminology
Alex Nikon's Organic Chemistry The Name Game.

In older times, it was pretty common to name a compound after a person who described the preparation or because nobody knew the structure. Nothing unusual about it. Here is the entry:
Caro’s acid:$\ce{H2SO5}$, named for the German chemist Heinrich Caro (1834-1910)
Marshall’s acid: $\ce{H2S2O8}$, named for the 19th century Scottish chemist Hugh Marshall


text:
How many grams of sodium phosphate must be added to precipitate as much of one ion as possible?
Problem text:
Solid sodium phosphate is slowly added to $\pu{200 mL}$ of a solution containing $\pu{0.002 mol}$ of aluminum chloride and $\pu{0.001 mol}$ of calcium chloride (assuming no volume change occurs). How many grams of sodium phosphate must be added to precipitate as much of one ion as possible, while the other ion just does not precipitate? What percentage of the ion that will precipitate first is still left in the solution at that time?
My approach:
Two salts will be produced as the reaction goes: $\ce{AlPO4}$ and $\ce{Ca3(PO4)2}$
$K_\mathrm{sp} (\ce{AlPO4})=9.83\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\times 10^{-33}$
Therefore, I can conclude that $\ce{Ca3(PO4)2}$ will precipitate first.
I need to find concentration of $[\ce{PO4^3-}]$ via
$$[\ce{Al^3+}][\ce{PO4^3-}]=K_\mathrm{sp}$$
$$[\ce{Al^3+}]=\frac{\pu{0.002 mol}}{\pu{200 mL}}=0.01M$$
$$[\ce{PO4^3-}]=\frac{9.83\times 10^{-21}}{\pu{0.01M}}=9.83\times 10^{-19}$$
Then I can find $m_\ce{Na3PO4}$ via finding number of moles first and multiplying by molar mass:
$$n_\ce{Na3PO4} = [\ce{PO4^3-}]\times V = 9.83\times 10^{-19}\times \pu{200 mL} = 1.966\times 10^{-19}$$
$$m_\ce{Na3PO4} = n_\ce{Na3PO4} \times \mathrm{Molar Mass} = \pu{3.22\times 10^{-17} g}$$
At this point, my result does not make much sense to me because the mass I obtained is extremely small. Unfortunately, I do not have the correct answers for that task, therefore, I would highly appreciate any help that can be useful in spotting mistakes I made in the solution.

Overall the problem is poorly setup considering the significant figures involved, and very poorly considering the real chemistry. // You calculated the $\ce{PO$^{3-}}$ remaining in solution correctly for the $\ce{Al^{3+}}$. However you need to do the same for $\ce{Ca^{2+}}$ to determine that $\ce{AlPO4}$ will precipitate first. // You calculated the amount of $\ce{PO$^{3-}}$ which will remain in solution, and did not add the amount needed to quantitatively precipitate the $\ce{AlPO4}$ too.

In an answer that is now hidden user Maurice pointed out the complexities of the chemistry. So let's assume:

All Al present in solution as $\ce{Al^{3+}}$ and any hydoxy complexes or precipitates will be ignored.
All Ca present in solution as $\ce{Ca^{2+}}$ and any hydoxy complexes will be ignored.
All $\ce{PO4^{3-}}$ added to solution stays as that ion and doesn't hydrolyze to $\ce{HPO4^{2-}}$
Two salts that will be produced are $\ce{AlPO4}$ and $\ce{Ca3(PO4)2}$
$K_\mathrm{sp} (\ce{AlPO4})=9.83\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\times 10^{-33}$

Concentration of $[\ce{PO4^3-}]$ that will cause $\ce{Al^{3+}}$ to precipitate:
$$[\ce{Al^3+}]=\dfrac{\pu{0.002 mol}}{\pu{200 mL}}=0.01M$$
$$[\ce{PO4^3-}]_\ce{Al}=\dfrac{K_\mathrm{sp}}{\ce{[Al^{3+}]}} = \dfrac{9.83\times 10^{-21}}{\pu{0.01M}}=9.83\times 10^{-19}$$
Concentration of $[\ce{PO4^3-}]$ that will cause $\ce{Ca^{2+}}$ to precipitate:
$$[\ce{Ca^2+}]=\dfrac{\pu{0.001 mol}}{\pu{200 mL}}=0.005M$$
$$[\ce{PO4^3-}]_\ce{Ca}=\sqrt{\dfrac{K_\mathrm{sp}}{\ce{[Ca^{2+}]^3}}} = \sqrt{\dfrac{2.07\times 10^{-33}}{(\pu{0.005 M})^3}}= 1.2890\times 10^{-13}$$
$\therefore \ce{AlPO4}$ will precipitate first.
The molecular mass of $\ce{Na3PO4}$ is 163.941 g/mol.
Note here that you can't just assume a stoichiometric amount of $\ce{Na3PO4}$ to say three significant figures. The solubility of the $\ce{Ca3(PO4)2}$ is so low that the $\ce{Na3PO4}$ would have to be weighed to an absolutely unrealistic precision.
So we'll go father down fantasy lane and suppose that:

There are $0.02\bar0$ moles of  $\ce{Al^3+}$
That we are using God's balance which can weigh an infinitely precise amount of $\ce{Na3PO4}$.
That the molecular mass of $\ce{Na3PO4}$ is exactly 163.941 g/mol (this is so precise that isotopic ratios limit the value...)
$K_\mathrm{sp} (\ce{AlPO4})=9.83\bar0\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\bar0\times 10^{-33}$
That the volume changes in the solution when adding $\ce{Na3PO4}$ or precipitating $\ce{AlPO4}$ are insignificant.

So if the $[\ce{PO4^3-}] = 1.\bar0\times 10^{-13}$ then no $\ce{Ca3(PO4)2}$ will precipitate. At that concentration the amount of $\ce{Al^3+}$ left in solution will be:
$$\ce{[Al^{3+}]}=\dfrac{K_\mathrm{sp}}{\ce{[PO4^{3-}]}} = \dfrac{9.83\times 10^{-21}}{1.\bar0 \times \pu{10^{-13} M}} =9.83\times 10^{-8}$$
So $\pu{0.001999998034000 mol}$ of $\ce{AlPO4}$ will precipitate.
For 200 ml of a $1.\bar0 \times \pu{10^{-13} M}$ solution there are $2.\bar0 \times \pu{10^{-14} moles}$
So the total moles of $\ce{Na3PO4}$ needed is $\pu{0.001999998034020  mol}$ or $\pu{0.327881677695273 g}$.
% Al left in solution $= \dfrac{0.002\bar0 - \pu{0.001999998034000 mol}}{0.002\bar0} =0.0000983\% $
Again the problem requires such a precise amount of $\ce{PO4^{3-}}$ to precipitate "all" the $\ce{Al^{3+}}$ yet leave "all" the $\ce{Ca^{2+}}$ in solution that it is totally unrealistic.

Followings can be calculated from given data:
$[\ce{AlCl3}] = \frac{\pu{0.002 mol}}{\pu{0.200 L}} = \pu{0.01 molL-1}$ and $[\ce{CaCl2}] = \frac{\pu{0.001 mol}}{\pu{0.200 L}} = \pu{0.005 molL-1}$.
Since both in same solution and dissolved, $[\ce{Al^3+}] = \pu{0.01 molL-1}$ and $[\ce{Ca^2+}] = \pu{0.005 molL-1}$.
If solid $\ce{Na3PO4}$ is added to the solution, two salts will be precipitated out: $\ce{AlPO4}$ with $K_\mathrm{sp}^\ce{AlPO4} = 9.83\times 10^{-21}$, and $\ce{Ca3(PO4)2}$ with $K_\mathrm{sp}^\ce{Ca3(PO4)2} = 2.07\times 10^{-33}$ (units ignored, and assume volume stays same during the process).
At first glance, since $K_\mathrm{sp}^\ce{Ca3(PO4)2} \lt \lt K_\mathrm{sp}^\ce{AlPO4} $, one can assume that $\ce{Ca3(PO4)2}$ would precipitate first. However, in reality, it is not the case:
Requirement for $\ce{Ca3(PO4)2}$ to precipitate:
$$Q_\mathrm{sp}^\ce{Ca3(PO4)2} \ge K_\mathrm{sp}^\ce{Ca3(PO4)2} \Rightarrow (0.005)^3[\ce{PO4^3-}]^2 \ge 2.07\times 10^{-33}\\ \Rightarrow [\ce{PO4^3-}] \ge \sqrt{\frac{2.07\times 10^{-33}}{(0.005)^3}} \approx 1.29 \times 10^{-13}$$
Similarly, requirement for $\ce{AlPO4}$ to precipitate:
$$Q_\mathrm{sp}^\ce{AlPO4} \ge K_\mathrm{sp}^\ce{AlPO4} \Rightarrow (0.010)[\ce{PO4^3-}] \ge 9.83\times 10^{-21}\\ \Rightarrow [\ce{PO4^3-}] \ge \frac{9.83\times 10^{-21}}{(0.010)} \approx 9.83\times 10^{-19}$$
Since, $9.83\times 10^{-19} \lt 1.29 \times 10^{-13}$, actually $\ce{AlPO4}$ precipitate first.
The requirement for $\ce{Ca3(PO4)2}$ to precipitate is: $[\ce{PO4^3-}] \ge 1.29 \times 10^{-13}$. Suppose, at this point $n$ amount of $\ce{AlPO4}$ been precipiteted. Thus, $[\ce{Al^3+}]$ is $\frac{0.002-n}{0.200}$. Hence:
$$\frac{0.002-n}{0.200} \times 1.29 \times 10^{-13} = K_\mathrm{sp}^\ce{AlPO4} = 9.83\times 10^{-21}$$
Now, you would be able to solve this equation for $n$. That is the amount of $\ce{Na3PO4}$ in $\pu{mols}$ you have have added to the solution before $\ce{Ca3(PO4)2}$ starts to precipitate. If you multiply this number by molar mass of $\ce{Na3PO4}$, you get your answer. I got $\pu{328 mg }$ for rough calculations. It's up to you.

There is a problem in that you can't just weigh out the $\ce{Na3PO4}$ "roughly." An extra milligram would be more than enough to cause some $\ce{Ca3(PO4)2}$ to precipitate.

The initial acidity due to $\ce{Al^{3+}}$ hydrolysis cannot be ignored. The initial concentration of $\ce{AlCl_3}$ is $0.02$ mol/$0.2$ L = $0.01$ M. In a $0.01$ M solution of aluminium chloride, the ion $\ce{Al^{3+}}$ is hydrolyzed according to :$$\ce{Al^{3+} + H_2O <-> [Al(OH)]^{2+} + H^+}......(1)$$ The $pK_a$ of this reaction is $4.9$. So in this solution, $\ce{[H^+] = \sqrt{K_a·[Al^{3+}]} = \sqrt{10^{-4.9}·0.01} =  3.55·10^{-4}}$. The corresponding $p$H is $-log[H^+] = 3.45$. At this $p$H, any addition of phosphate ion $\ce{PO_4^{3-}}$ will be transformed into monohydrogenophosphate $\ce{HPO_4^{2-}}$, and even dihydrogenophosphate $\ce{H_2PO_4^-}$ as the $pK_a$ of these equations are : $pK_{a2} = 12.3$ and $pK_{a3} = 7.2$
 :$$\ce{PO_4^{3-} + H^+ -> HPO_4^{2-}}.........(2)$$ $$\ce{HPO_4^{2-} +  H^+ -> H_2PO_4^-}........(3)$$ So if some sodium phosphate is added to such a solution ($p$H $3.45$), it will be immediately transformed into $\ce{H_2PO_4^-}$.  The remaining concentration of $\ce{PO_4^{3-}}$ can be calculated .
Suppose we add $a$ mol $\ce{Na_3PO_{4}}$ to the $0.2$ L of this solution. The following concentrations can be calculated :  $\ce{[H_2PO_4^-]}$ = $ a/{0.2} = 5$$a$ ;
$\ce{[HPO_4^{2-}] = K_{a3}·[H_2PO_4^-]/[H^+] = 10^{-7.2}·5$a$ /(3.55·10^{-4}) = 8.76·10^{-4}} a$
$\ce{[PO_4^{3-}] = K_{a2}·[HPO_4^-]/[H^+] = 10^{-12.3}·5$a$ /(3.55·10^{-4}) = 7.055·10^{-9}} a$
So the final solution has a concentration of phosphate ion $\ce{[PO_4^{3-}]}$ = $a·7.00·10^{-9}$ M. The solubility product for $\ce{AlPO_4}$ is obtained for a concentration of$\ce{[PO_4^{3-}]} = K_s/(0.01) = 9.83·10^{-21}/0.01 = 9.83·10^{-19} = a·7,00·10^{-9}$.
So : $a$ = $1,40·10^{-10}$ mol $\ce{Na_3PO_{4}}$

No, this isn't right. As you precipitate the $\ce{Al(OH)^2+}$ then $\ce{OH-}$ is released neutralizing the $\ce{H+}$ initially formed. In fact if the $\ce{Al(OH)^2+}$ is excatly precipitated then the pH should return to neutral. I  have been looking somewhat the same notion, but wondering what pH you'd have to buffer the phosphate solution at. I think it would be between 4 and 5. You want a $\ce{[PO4^{3-}]} \approx 3.6\times10^{-16}$

@ MaxW. $\ce{Al(OH)^{2+}}$ is not a precipitate. It is an ion. Ions are in solution. A precipitate has no charge.

Sorry I didn't express myself better. I was trying to infer as the $\ce{Al(OH)^2+}$ was precipitated as $\ce{AlPO4}$ then the $\ce{OH-}$ would be released.

I'm fiddling around with a different take on the problem. Let's assume that a buffered solution of phosphate species is used. What pH would the buffer have to be so that the $\ce{Al^3+}$ would be quantitatively precipitated but the $\ce{Ca^{2+}}$ would be left in solution?
(1) This gets around the problem of the hydroxy complexes of $\ce{Al^3+}$ and $\ce{Ca}$ since the solution will be acidic.
(2) Seems like a final concentration of $\ce{[PO4^{3-}]} = 3.6\times 10^{-16}$ would be a good target.
$$\log{(3.56\times 10^{-16})} \approx \dfrac{\log{(1.289\times10^{-13})} + \log{(9.83\times10^{-19}})}{2}$$
(3) Assume the total phosphate species is 1.00 molar. Then then the $\ce{AlPO4}$ precipitated can be assumed to cause a negligible change to the phosphate species at whatever pH the solution is buffered at.

Actually there will be some, but we can check to see that the affect is negligible.
We should really be using activities at this concentration, but we'll ignore that complication.

For a buffer with $\ce{[H3PO4] = [H2PO4-]}$ the pH will be 2.14.
For $\ce{H3PO4 <=> 3H+ + PO4^{3-}}$ the equilibrium expression is:
$$\ce{1.8208\times10^{-22} = \dfrac{\ce{[H+]^3[PO4^{3-}]}}{\ce{[H3PO4]}}}$$
or
$$\ce{[PO4^{3-}]} = \dfrac{(1.8208\times10^{-22})\times \ce{[H3PO4]}}{\ce{[H+]^3}} = \dfrac{(1.8208\times10^{-22})\times (0.5)}{(10^{-2.14})^3} = 2.4\times10^{-16}$$
which is just about perfect.


text:
Naming a synthetic molecule from QM9 dataset
I was trying to write a name for a synthetic molecule in QM9 dataset. The molecule is:

is 3-aminooxolane-2-carbonitrile a correct name for this?

I wonder what dataset do you mean. I've found one with $\ce{C7H10O2}$ isomers. I've also searched through huge GDB-17 (no match).

.. I mean, no exact match, bu too many substructure matches :)

(note: The structure representation is suspicious and confusing. Linear nitrile $\ce{-C#N}$ group is "broken". But I see some software products generate such representation automatically.)
If the structure is really meant saturated, no stereochemistry is indicated.

3-aminooxolane-2-carbonitrile
(Watch the typo in your proposal …oxalane.) But it's not present at PubChem. (SMILES: N#CC1OCCC1N)

Maybe, if unsatured, furan derivative is meant:

3-aminofuran-2-carbonitrile


text:
Selective carbonyl protection using ethylene glycol
Can anyone explain to me which position is more favourable to form ketal i.e aldehyde or ketone
Also which reducing agent also reduce this ketal.
And how thioketal formed from thioethyleneglycol different from ketal

Aldehydes are generally more reactive than ketones. According to my copy of "Protective Groups in Organic Synthesis" by Greene & Wuts (John Wiley & Sons, 1991) aldehydes form dimethyl acetals preferentially to ketones, for example in MeOH with dry HCl for 2 mins followed by 12 mins reflux (p178, quoting Cameron et al, J. Chem. Soc., 3864 (1953)).
Acetals are generally resistant to reducing conditions, particularly hydride-based reagents. The only conditions where they are reactive are conditions where they are first hydrolysed by the aq. acid present to reform the carbonyl group which is then reduced e.g. Clemmensen
Thioketals are produced by reaction of carbonyl compounds with thiols under similar acidic conditions to ketals mechanism here. They are different from ketals in that they are resistant hydrolysis by aq. acid and in the case of thioacetals of aldehydes the proton on the carbon bearing the two sulfurs is activated to deprotonation (see Corey-Seebach reaction here.)