text:
What is an aromatic alcohol?
The following question was given in NCERT Exemplar Problems (Class XII):

According to Wikipedia:

In organic chemistry, the aromatic alcohols or aryl-alcohols are a
  class of chemical compounds containing a hydroxyl group (—OH) bonded
  directly to an aromatic hydrocarbon group, in contrast to the benzyl
  alcohol, where the hydroxyl group is bonded indirectly to an aromatic
  carbon atom.

But, I didn't understood the part: aromatic alcohols or aryl-alcohols (as aromatic alcohols are not always aryl alcohols).
Further, we know that aryl halides are compounds in which halogen group (-X) is directly attached to an aromatic ring. In the same way, aryl alcohols also mean that those compounds in which hydroxyl group (-OH) is directly attached to an aromatic ring.
Hence, according to my understanding, the answer should be (ii), but the answer given is (iii).
Where I am wrong? And, what is an aromatic alcohol?

Sometimes phenols (OH group connected to the aromatic ring) are not considered to be alcohols, basically because they are much more acidic than aliphatic alcohols. When you use such a convention, aromatic alcohols are those in which you have an aromatic ring and a hydroxyl group connected to a sp3 carbon. If NCERT follow this convention, the right answer would be III.

@RaulLuciano seems to make the correct distinction based on current terminology. From IUPAC gold book  alcohols are Compounds in which a hydroxy group, –OH, is attached to a saturated carbon atom $\ce{R3COH}$. Phenols are compounds having one or more hydroxy groups attached to a benzene or other arene ring. // So is "aromatic alcohol"  outdated terminology?

@MaxW aromatic alcohol is still used, but it means a compound which has both the alcohol group and an aromatic ring in its structure.

@RaulLuciano, My question is why Wikipedia has said "aromatic alcohols or aryl alcohols" as aryl alcohols are Ar-OH which is not congruent to your definition of aromatic alcohols ?

@Rahul It should be pointed out that Wikipedia is not to be considered an authoritative source and has many incongruities and errors of fact.

Answer (iii) is the correct one.

I agree that the correct answer should be (ii). My only experience with NCERT questions are from Stack Exchange, but it seems that there are either many errors in those questions or they rely on unusual conventions.

'Aromatic alcohol' is not a term that appears anywhere in the IUPAC literature - while its meaning is evident, it doesn't make sense in the same way that 'aromatic heterocyclic' does so is probably best avoided.
What is defined by IUPAC is the definition of a phenol, and the definition of an aryl group. The definitions below come from an old IUPAC glossary of  'class names' which, although old, is still referenced in the most recent organic nomenclature guidelines and the most recent IUPAC Gold Book.

phenols:
  Compounds having one or more hydroxy groups attached to a benzene or other arene ring
aryl groups: Groups derived from arenes by removal of a hydrogen atom from a ring carbon atom.
Glossary of Class Names of Organic Compounds and Reactive Intermediates Based on Structure.
  Pure App. Chem. 1995, 67, 1307

To actually answer your question, the answer is (II) - as Jerepierre alluded to, the NCERT books / guides are renowned for being less that accurate, and a quick search around chem.SE will reveal just how many errors there are! In this case, its hard to tell if the error is just due to typesetting (someone typed III instead of II), or it it lies with the ambiguity of the term 'aromatic alcohol' (is it referring to phenols, or is it referring to a compound which contains both an aryl ring and a hydroxyl group).
I think its fair to assume that the question is asking you to identify the phenols, in which case the answer is (II) is because A and D are the only structures in which the hydroxyl group is directly attached to a benzene or other arene ring, which is our working definition of a phenol / aromatic alcohol.

So, according to you, aromatic alcohol is not defined anywhere in the world of chemistry. Right?

No. Just according to me its an ambiguous term and not one used by IUPAC (the people who define the Internationally accepted system of nomenclature).

Unfortunately textbook was correct and Wikipedia article had a mistake at that moment. Aromatic alcohol is an alcohol that is aromatic and is not phenol. How otherwise would you call such compound?

I've been working as a researcher in the chemical sciences for 20 years.  I upvoted this answer and would assume without any other context that aromatic alcohols referred to things like phenols, napthols, furanols, pyridinols, pyrazinols, etc etc etc.  I'm sure people do it sometimes but I can't recall people saying "aromatic alcohol" to mean benzylic alcohol or similar compounds.


text:
How do I make a ZMAT or XYZ file from scratch?
From a structure representation like the one below for Bacteriochlorophyll A, how do I make the ZMAT or XYZ file?
The closest similar question asked here was about converting a PDB entry into XYZ, but what I have here is not in PDB format, it is just the image of the molecular structure, reported in a paper.

PDB format per se is not related to proteins, other than historically. (Yes, I know what the "P" stands for.) You may have pretty much any molecule in PDB format. Do you have yours or not? If no, then the referenced question is of no use. In that case start with drawing your structure in some molecular editor, then click "Export as..." and pray.

@IvanNeretin: I do not have any "molecular editor", is there a free one that works well? I downloaded MOLDEN but it is old fashioned and seems hard to use. Do I have to draw this entire ~120 atom structure, or can I just drag and drop the image that I linked to?

Ten years ago the answer would be "no, you have to draw it all by hand". Today I'm not so sure, so let's wait for other opinions.

See if there is a crystal structure (or something similar).  You can get coordinates from that (XYZ) and then tweak by hand in Molden.  It saves a lot of time if you have access to a crystal structure database (CCDC) and an appropriate database program (such as ConQuest).

I want to drag and drop the image and get the XYZ, is it possible?

@LordStryker: Why does CCDC not seem to have "Bacteriochlorophyl a" ?

@user1271772 I doubt that is possible. Its unlikely that there is a program that can determine the correct bond distances, angles, etc, even approximately, just from an image file. But I don't think the structure would be too difficult to make in a free software like Avogadro.

I don't know of any way to drag/drop a ChemDraw image to get XYZ coordinates.  If there is a way, someone let me know ASAP please.

@user1271772 I don't know the steps to go directly from an image to a SMILES string (in your case you know the name of your compound, so you can find the SMILES string without the image), but this site suggests that openbabel can be used to convert a SMILES string to a 3d structure.

Download some PDB entry of some protein with this chromophore and take the coordinates from there. Proteins like that are FMO (pdb 3BSD), LH2 (pdb 1NKZ) etc. As far as I'm aware, there's no way to pass from an image to a 3D representation.

There are some people working on generating molecules from images. Also, search a pdb,xyz or mol2 file of similar molecule and then finish your draw on Avogadro. Where you can export it.

@LordStryker - Avogadro can read ChemDraw files. There are sometimes difficulties with the CDX binary file, but CDXML are text and can be imported easily for refinement.

I know of two three ways to do this, but in either case it's quite a bit of work since I don't know any software that can read molecular structures from pictures. The second and third way use freeware only.

Draw the structure in Chem3D. For whatever stupid reason Chem3D doesn't support xyz files but you can save as *.pdb. If you need *.xyz then:

save as Gaussian input
delete the first 3 lines
Put the total number of atoms in the resulting first line
remove any "atoms" labeled as LP (lone pair, Chem3D uses them for force field optimization)
remove the second column next to the elemental symbol (if you sue Notepad++ you can select that by using ALT)
change file extension

Use openbabel 2.2.0 or higher to convert SMILES into *.xyz or *.pdb using the "gen3d" keyword. Of course for this you need the SMILES so you either have to produce that yourself or you need to draw it in a software that is able to produce SMILES like, again, Chem3D or this free website
It turns out this NIH website is also able to produce 3D structures, also in pdb file format, directly from SMILES or drawing the structure in their editor.

"Bacteriochlorophyl a" has the formula C${55}$H${74}$MgN$_{4}$O$_6$

It's about 140 atoms, I really have to draw this whole thing from scratch in that editor where we just add C-C bonds one by one? Can we at least start from a similar structure that's in a database somewhere and then change the middle atom to Mg, for example? How can no one have made an XYZ file for "bacteriochlorophyl a" before. Everyone has to repeat other people's work every time? The issue is that the molecules I'm working with are hundreds of atoms per molecule, not just O$_3$ or CO$_2$.

Ok, about 70 "heavy" atoms isn't actually that much, an experienced user could draw that within few minutes. My answer was also about how to solve that for any random structure. In this case there are easier ways. You can go to wikipedia and open the page for that compound and copy the SMILES from there.

@user1271772 I tried it. I agree that this one can be tricky due to the coordinated Mg but if you do it without the Mg and insert it afterwards it works very well.  Here's a PM6 optimized structure: http://textuploader.com/dpy6q

55+74+1+4+6 = 140, so why does it say 135 at the top of your XYZ? Also I am very grateful for you spending time on making this XYZ for me. I also want to learn how to do it myself so that in the future I do not need to ask people for help.

In addition to my question 22 minutes ago, what is meant by "copy the SMILES from there". I am at the bacteriochlorophyl wikipedia page, which has "bacteriochlorophyl a" as one of the molecules listed there, but what and where is the SMILES ?

Avogadro (free) does support xyz file types, and so using Avogadro saves you a step. Building your molecule with Avogadro should not take more than 15-20 minutes. Avogadro also offers many different templates for common structures, such as amino acids, vitamines, common organic molecule moieties.
However, if you end up doing anything "manually", then the bond lengths and bond angles probably end up a little wonky. You can perform (in Avogadro) a geometry optimization with real time visual feedback using simple force-fields, but the final geometry will not of course be extremely accurate.

@user1271772 https://pubchem.ncbi.nlm.nih.gov/compound/11953947#section=InChI-Key You can find the SMILES here under identifiers

Building such large structures can be a nasty task, and as it is quite large, you can easily loose track of parts of it. You should try a couple of molecular editors and find something you are comfortable with, then editing molecules becomes something routine and even a large one like your example is less tedious.
Apart from this very general advice, here are some additional tips.
Like DVSA states there are some more or less tedious ways to do it. The most error-prone is obviously trying to build them one atom at the time. That is not only hard work, but it also consumes a lot of time. Using smaller (pre-optimised) fragments can help a lot. Obviously you'll need a molecular editor capable of merging/joining structures.
Probably the best way to built a structure from scratch is to start from something you know, like a literature known structure of a derivative, or a substructure from a database. Unfortunately conversion from InChI or SMILES with open babel does not always work as you might intend it to, especially in those cases, where the SMILES or InChI becomes ambiguous.
Let's take a look at your example: Bacteriochlorophyll A. It is easy enough to find on PubChem, where they provide you with an InChI, and some SMILES.
If you try to convert the following to a 3D structure with openbabel 2.4.1, you will get an error, and I assume this is due to the fact that there are essentially two entities stored in the InChI (the dot . separates those entities).
InChI=1S/C55H75N4O6.Mg/c1-13-39-34(7)41-29-46-48(38(11)60)36(9)43(57-46)27-42-35(8)40(52(58-42)50-51(55(63)64-12)54(62)49-37(10)44(59-53(49)50)28-45(39)56-41)23-24-47(61)65-26-25-33(6)22-16-21-32(5)20-15-19-31(4)18-14-17-30(2)3;/h25,27-32,34-35,39-40,51H,13-24,26H2,1-12H3,(H-,56,57,58,59,60,62);/q-1;+2/p-1/b33-25+;/t31-,32-,34-,35+,39-,40+,51-;/m1./s1

I have executed open babel with the following command, where the above InChI was stored in structure.inchi
obabel -iinchi structure.inchi -oxyz -Ostructure.xyz --gen3d

The result is breathtakingly awful, and I am probably going to file a bug-report later on. The full output is:
*** Open Babel Warning  in InChI code
  For InChI=1S/C55H75N4O6.Mg/c1-13-39-34(7)41-29-46-48(38(11)60)36(9)43(57-46)27-42-35(8)40(52(58-42)50-51(55(63)64-12)54(62)49-37(10)44(59-53(49)50)28-45(39)56-41)23-24-47(61)65-26-25-33(6)22-16-21-32(5)20-15-19-31(4)18-14-17-30(2)3;/h25,27-32,34-35,39-40,51H,13-24,26H2,1-12H3,(H-,56,57,58,59,60,62);/q-1;+2/p-1/b33-25+;/t31-,32-,34-,35+,39-,40+,51-;/m1./s1
  Problems/mismatches: Mobile-H( Hydrogens: Number; Mobile-H groups: Number; Charge(s): Do not match)
1 molecule converted

The magnesium is somewhere, and the porphyrin ring is all crooked; that does not help at all. (I have also tried SMILES, but that did not work any better.) In some cases you might be luckier, so that approach is something to keep in mind.
I tried the NIH website as DVSA suggested, and apart from the manganese being somewhere, that worked reasonably well. (If you try this approach then you can skip the next sections.)
If you have accesses to the ChemOffice suite things may become a little easier. Copy the InChI into the paste buffer, open ChemDraw, choose Edit > Paste Special > InChI. You will hopefully see something like the Lewis-like structure from your post then. You will need to clean it up a little, you should definitely delete the metal. Now copy the cleaned structure into the paste buffer and open Chem3D. On the right side you should have the ChemDraw - LiveLink, where you can paste your structure. This will create a 3D image on the main screen. I have tried it with the example molecule and it worked reasonably well. (I can only recommend this for organic molecules.) You can already use the implemented force field to optimise it (and you should). Depending on what molecule you are doing, there might be parameters missing (mostly metals, very heavy atoms), charges wrong applied, or other problems. It is basically trial and error, it might not always lead to a success.
Safe the resulting file as something another molecular editor can open. Since ChemOffice handles even common file formats badly, I recommend pdb as the best (still bad) option. (I should really try MarvinSketch or similar software.) It still produces plenty of warnings in open babel, which can mysteriously still recover most of it, and you'll have a somewhat reasonable structure to riff off on.
I was unable to open the pdb directly with molden (unrecognised elements, probably bad formatting), but it was no problem reading it with Chemcraft (payware). Since Avogadro (open source) works on top of open babel, I assume it should be able to read the file, too.
If you edit molecules a lot I recommend spending some money on Chemcraft (that is my personal preference, but I am not affiliated with it), in my experience it is one of the best molecular editors. It has a rather extended fragments database (which you can customise), that may even take care of 'clicking together' a rather huge molecule. It comes with some more nice features, but things like that are better discussed in chat. (I have used it occasionally on the site, so you might find some interesting showcases here, too.)
Some people love their GaussView, but I simply cannot see the appeal of that program.
Obviously you can use open babel to transform even the garbage pdb to a xyz file, which you then can view/edit with molden. Unfortunately the zmat interface of molden is quite tedious, but it comes with a few pre-built fragments which makes extending molecules easier. The most tedious thing about that program is, that it does not support drag and drop. In the ZMAT Editor there is an option called 'substitute atom by fragment', I suggest you give that a look. (Chain-like organics can be easily prepared with that starting from methane.)
I believe Avogadro also has a fairly sophisticated molecular editing tool, however I currently have no access to it, so I can't really comment.
There is no harm in trying a few molecular editors and sticking with what works best for you.
Eventually you should have a somewhat good structure to start a calculation on. Keep in mind that whatever you created is only one conformation and likely not the minimum structure. You should use conformation generators to test whether what you have is good or bad.
I have developed a preference for GFN2-xTB from Stefan Grimme's group (semi-empirical program package capable of optimising largen structures) for these tasks.

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How does the 1.5 bond order of ozone contribute to its instability?
I'm confused as to why ozone is so unstable? I've been told it's because it has a fractional bond order. However, this doesn't make sense to me. Don't fractional bond orders mean resonance and doesn't resonance stabilize ozone?

Fractional bond order by itself does not mean anything.

CVers: what is unclear in this question?

It's important to bear in mind stability is always a relative judgment.  Things are stable or not only with respect to that into which they could turn. So we only say ozone is "unstable" because it can and will easily decompose into oxygen (O2) if given a chance.  In some sense it might even be more correct to say what's important here is the stability of O2 (which derives from its strong double bond).  If O2 weren't possible, or were significantly higher in energy itself, then we would say O3 is more "stable" just because it wouldn't have anything into which it could as easily turn, so it would stick around longer.  So "stability" is not just a property of one molecule, but a property of that molecule plus all the other molecules into which it could turn.
A famous example of this is the high stability of N2 (with its very strong triple bond), which means certain compounds (like TNT or ammonium nitrate) with nitrogen atoms in higher oxidation states (e.g. bonded to O atoms) can readily decompose to products including N2 and O2, producing a lot of energy.  That is, they are good explosives, and you can consider that as coming from the strong stability of the O2 and N2 molecules.  (This is somewhat akin to the argument in organic chemistry that the pKa of an acid depends strongly on the stability of its conjugate base -- that into which it can turn.  The more stable the base, the higher the pKa, i.e. less "stable" the acid.)
All that said, fractional bond order does indeed suggest resonance, because it suggests the failure of a single Lewis structure to describe the molecule accurately.  And indeed O3 needs two resonance structures to describe it. It's also correct to say resonance stabilizes a molecule, but, again -- with respect to what?  In general, the answer is "with respect to a similar (perhaps hypothetical) molecule that does not have resonance."  So O3 is more stable (because of resonance) than some hypothetical O3 molecule trapped in one of the O3 resonance structures.  It can still be "unstable" for most purposes, it's just less unstable than this hypothetical non-resonant O3.

A fractional bond order does not mean resonance, it's just a parameter obtained from molecular orbital theory accounting for the difference in population of electrons in bonding and antibonding orbitals. For example, you can use the bond order to explain, in comparison, why oxygen (gas) is more stable than ozone, but you cannot use it to predict ozone stability by itself.

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text:
How to find the percentage composition of halogen using Carius method?
In the formula:
$$\text{Percentage}=\frac{\text{atomic mass of halogen}}{\text{molecular mass of Agx}}\frac{\text{mass of AgX}}{\text{mass of compound}}\cdot100$$
here mass of compound is the mass of the organic compound whose chemical composition is to be found.
What I don't understand is that: $\frac{\text{mass of AgX}}{\text{molecular mass of AgX}}\times\text{atomic mass of x}$ gives the mass of halogens in the product and to get percentage they are dividing the value with mass of compound and multiplying by 100. But we found the mass of halogen in the product then why are we diving it with mass of compound?

Let's break it down.
We start with the mass of AgX.
$$\text{mass of AgX (g)}$$
Then, we divide by the molar mass of AgX to get the number of moles of AgX.
$$\text{moles of AgX(mol)}=\frac{\text{mass of AgX (g)}}{\text{molar mass of AgX (g/mol)}}$$
That's the same as the number of moles of halide (this is the crux of the problem).
What about the total mass of halide?
$$\text{mass of halide (g)} = \frac{\text{mass of AgX (g)}}{\text{molar mass of AgX (g/mol)}} \times \text{molar mass of halide (g/mol)}$$
So, if we divide this by the total mass of the compound, we should get the mass proportion of halide.
$$\text{proportion of halide} = \frac{\text{mass of AgX (g)}}{\text{molar mass of AgX (g/mol)}} \times \frac{\text{molar mass of halide (g/mol)}}{\text{mass of compound (g)}}$$
We just multiply this proportion by $100\%$ as a formality to get a percentage between 0-100 rather than 0-1.


text:
Can someone confirm and/or correct my logic regarding this mechanism?

In the picture, Ethyl 2-oxocyclopentanecarboxylate (I) reacts with ethyl 2-(4-chloromethylphenyl)propionate (II) in the presence of KOH in hot DMF to afford ethyl 2-[4-(1-ethoxycarbonyl-2-oxocyclopentan-1-ylmethyl)phenyl]propionate (III), which is then hydrolyzed and decarboxylated by treatment with 47% HBr in refluxing dioxane. I'd just like to walk my way through this and see if my theories regarding the mechanism are correct.

(I) has an $\alpha$ carbon that is deprotonated (the one beside the ester group). This creates a carbanion that is stabilized both by resonance as an enolate and inductively by electron withdrawing from the ester group. This stability leads to marginal nucleophilicity; chlorine is a marginal leaving group here (although the carbocation is resonance stabilized by the aromatic ring), and no inversion is observed, which leads me to believe an $S_N1$ rxn occurs between (I) and (II) to yield (III). DMF is used as a solvent so as not to inhibit nucleophilicity by using a polar protic solvent.
It's not specified, but I suspect (III) can be  captured via distillation to due boiling point/density differences between it and (I) and (II).
(III) is reacted with HBr in dioxane (I'm not sure why 47% is used; I'm assuming at STP, it's the maximum concentration of HBr(g) in water) is used, as dioxane is relatively stable at extremely high and extremely low pH values. The hydronium acts as an impetus for hydrolysis on the ester groups on both rings. I think the Bromide anion acts as just a counterion here (but I'm not sure why).
The step where (III) is reacted with sodium hydroxide is where I get lost. I'm not sure how the sodium hydroxide spurs decarboxylation and selective decarboxylation, at that (why not the ester on the aromatic ring as well?).

47-49% is the maximum strength of hydrobromic acid that can be made i.e. concentrated HBr. It is a very good reagent for acid hydrolysis of esters

Thank you for confirming that. Do my other hypotheses make sense with respect to the mechanism?

I would not attempt to distill III, column chromatography would be a better way to purify it

That makes sense @ Waylander. Thank you for the contribution. Do you have any comments about my first bullet point?

I doubt it is SN1, benzyl halides are pretty decent electrophiles. DMF is a good solvent because the potassium enolate of I will be soluble in it.

The reaction of III with NaOH has nothing to do with decarboxylation - the HBr step accomplished that. The NaOH neutralises excess HBr and forms the sodium salt of the carboxylic acid.

What mechanism do you think occurs between I and II?

Standard SN2 attack by the enolate

Related: Mechanism for decarboxylation

The decarboxylation occurs specifically at that position, because that is a $\beta$-keto acid. In presence of slight amount of heat only the decarboxylation occurs due to the presence of keto group adjacent to it. Actually, in acidic medium only,it decaboxylates in presence of heat as follows,
But no keto group is present at the carboxylic acid group attached at the benzylic position so it cannot decaboxylate.

Thank you for the answer. Do the other aspects of my explanation for the mechanism seem feasible?


text:
What happens to iron electrodes (attached to a battery) in copper(II) sulphate solution?
After a few hours, the following observations were made:

The solution turns green. My guess: $\ce{FeSO4}$ is formed.
There are reddish-brown and black powder deposits on both the electrodes.
The deposits fall down easily when the set-up is shaken a little.

What are the reactions that are occurring?
According to me, the copper is displaced from the $\ce{CuSO4}$ and is deposited on the iron electrode (the cathode), and $\ce{FeSO4}$ is formed as a green solution. What is getting deposited on the anode? Are there ferrous and ferric oxides/hydroxides being formed?

Is there any visible difference between the electrodes, to begin with?

@mhchem, there was no visible difference between them, initially. Two normal iron nails were used. They were physically cleaned before conducting the experiment.

The reaction is spontaneous, you don't need put a power supply.
Let's see the reaction with the potentials (obtained from the table of standard reduction potentials):
$$
\begin{align}
\ce{Fe^0 &→ Fe^2+ + 2 e-} &\quad E^\circ &= \pu{0.44 V} \\
\ce{Cu^2+ + 2 e- &→ Cu^0} &\quad E^\circ &= \pu{0.337 V} \\
\hline
\ce{Fe^0 + Cu^2+ &→ Fe^2+ + Cu^0} &\quad E^\circ &= \pu{0.777 V}
\end{align}
$$
So, the solid formed in both electrodes is copper. The green solution is related to iron sulfate.

If you actually passed current thru the cell, you would have been cathodically protecting the cathode - that is, preventing the dissolution of iron. On top of this "cathodic protection", however, you have copper deposition. And the copper could have been loosely deposited as black nano particles.
At the anode, you are discharging SO4-- dianions, which then form H2SO4 (H+ ions plus O2 gas). The acid produced will be corrosive to the iron, so Fe++ will go into solution. You also have the iron metal replacing Cu++ in solution, so some copper may be plating out - similar to the situation at the cathode.
Essentially you have two reactions going on: 1) copper deposition at both electrodes because iron is more active than copper, and 2) more copper deposited at the cathode because of electroplating, and more iron dissolving at the anode for the same reason. More iron dissolving at the anode may not inhibit copper plating out there.
The overall effect at both electrodes can be affected seriously by the current density.


text:
How can I balance this reaction using the algebraic method?
How can I balance this reaction using the algebraic method?
$$\ce{K4Fe(CN)6 + KMnO4 + H2SO4 + H2SO4 -> KHSO4 + Fe2(SO4)3 + MnSO4 + HNO3 + CO2 + H2O}$$
My solution:
$$\ce{a K4Fe(CN)6 + b KMnO4 + c H2SO4 -> d KHSO4 + e Fe2(SO4)3 + f MnSO4 + g HNO3 + h CO2 + i H2O}$$
$\ce{K}{:~} 4a + b = d$
$\ce{Fe}{:~} a = 2e$
$\ce{C}{:~} 6a = h$
$\ce{N}{:~} 6a = g$
$\ce{Mn}{:~} b = f$
$\ce{O}{:~} 4b + 4c = 4d + 12e + 4f + 3g + 2h + i$
$\ce{H}{:~} 2c = d + g + 2i$
$\ce{S}{:~} c = d + 3e + f$
I couldn't solve further than that.

I will just point it out - although you require the algebraic method - that the redox method instead is much easier than this.

Now all you have to do is variable substitution, which is kinda tedious since it's such a large equation. Remember that solutions are defined up to multiplying everything by a constant factor, so we might as well pick

$a = 2$ (this guess is inspired by the equation for $\ce{Fe}$)
$\ce{Fe}{:~} e = 1$
$\ce{C}{:~} h = 12$
$\ce{N}{:~} g = 12$
Because $\ce{Mn}{:~} b = f$, we can rewrite $\ce{O}$ as $4c = 4d + 12e + 3g + 2h + i$
Subtract 4 times $\ce{S}$, and we get $0 = -4f + 3g + 2h + i$, so $4f = 60 + i$ (call this $\ce{O}'$)
Now rewrite everything in terms of $b$:

$\ce{K}{:~} 8 + b = d$, so $d = b + 8$
$\ce{Mn}{:~} f = b$
$\ce{O}'{:~} 4b = 60 + i$, so $i = 4b - 60$
$\ce{S}{:~} c = (b + 8) + 3 + b = 2b + 11$
$\ce{H}{:~} 2(2b + 11) = b + 8 + 12 + 2(4b - 60)$, so $4b + 22 = 9b - 100$, so $5b = 122$ and $b = 122/5$. This is a bit unfortunate since later on we have to multiply all variables by 5.

We now have $a=2$, $b=122/5$, $c=299/5$, $d=162/5$, $e=1$, $f=122/5$, $g=12$, $h=12$ and $i=188/5$.
By multiplying everything by 5 we get the final solution:

$a = 10$
$b = 122$
$c = 299$
$d = 162$
$e = 5$
$f = 122$
$g = 60$
$h = 60$
$i = 188$

You can check that these figures will make the equation balanced.

You are lucky. From time to time, I take on a challenge like this, just to see if I can still do it. But normally, we expect that you show some effort yourself. See our homework policy.
We start off with some mathematics.
(1) $4a + b = d$
 (2) $a = 2e$
 (3) $6a = h$
 (4) $6a = g$
 (5) $b = f$
 (6) $4b + 4c = 4d + 12e + 4f + 3g + 2h + i$
 (7) $2c = d + g + 2i$
 (8) $c = d + 3e + f$
and rewrite this in the "equals zero" form.
(1) $0 = 4a + b - d$
 (2) $0 = a - 2e$
 (3) $0 = 6a - h$
 (4) $0 = 6a - g$
 (5) $0 = b - f$
 (6) $0 = 4b + 4c - 4d - 12e - 4f - 3g - 2h - i$
 (7) $0 = 2c - d - g - 2i$
 (8) $0 = c - d - 3e - f$
I fill this in a table. (I could have done this right from the original chemical equation.)
     a   b   c   d   e   f   g   h   i
(1)  4   1      -1
(2)  1              -2
(3)  6                          -1
(4)  6                      -1
(5)      1              -1
(6)      4   4  -4 -12  -4  -3  -2  -1
(7)          2  -1          -1      -2
(8)          1  -1  -3  -1

Now, (2) to (5) look simple enough to do some replacements. I start with (2) where $a = 2e$. In the first line, we replace $4a$ by $8e$ and so forth.
     a   b   c   d   e   f   g   h   i
(1)  0   1      -1  0+8
(2)  0             -2+2
(3)  0              0+12        -1
(4)  0              0+12    -1
(5)      1              -1
(6)      4   4  -4 -12  -4  -3  -2  -1
(7)          2  -1          -1      -2
(8)          1  -1  -3  -1

We got rid of $a$, (2) is cancelled out. (5) allows us to get rid of $b$ via $b=f$.
     a   b   c   d   e   f   g   h   i
(1)      0      -1   8  0+1
(3)                 12          -1
(4)                 12      -1
(5)      0             -1+1
(6)      0   4  -4 -12 -4+4 -3  -2  -1
(7)          2  -1          -1      -2
(8)          1  -1  -3  -1

Let's use (3) and (4) to remove $g=12e$ and $h=12e$.
     a   b   c   d   e   f   g   h   i
(1)             -1   8   1
(3)                12-12         0
(4)                12-12     0
(6)          4 -4 -12-36-24  0   0  -1
(7)          2  -1  0-12     0      -2
(8)          1  -1  -3  -1

i.e.
     a   b   c   d   e   f   g   h   i
(1)             -1   8   1
(6)          4  -4 -72              -1
(7)          2  -1 -12              -2
(8)          1  -1  -3  -1

Now we take (1) with $d = 8e+1f$
     a   b   c   d   e   f   g   h   i
(6)          4     -104 -4          -1
(7)          2      -20 -1          -2
(8)          1      -11 -2

Now we take (8) with $c = 11e+2f$
     a   b   c   d   e   f   g   h   i
(6)                -60   4          -1
(7)                  2   3          -2

One last step with (6) $i=-60e+4f$
     a   b   c   d   e   f   g   h   i
(7)                122  -5

That is $f=122e/5$
We cannot solve that further with mathematics. There is an infinite number of possible results. We need just one solution, so we guess one number. $e=5$ seems good so that we get a nice $f$. Now, from the many equations we had above, we search some that fit.
$e = 5$
$a = 2e = 10$
$f = 122e/5 = 122$
$g = 12e = 60$
$h = 12e = 60$
$b = f = 122$
$c = 11e+2f = 299$
$d = 4a+b = 162$
$i = -60e+4f = 188$


text:
Relative magnitudes of intermolecular forces from phase diagrams
Is it possible to say whether or not a system A or B has stronger intermolecular forces from a phase diagram according to which:

The $T_{\text{b}}$ of A is lower than the $T_{\text{b}}$ of B.
The $T_{\text{melting}}$ of A is higher than the $T_{\text{m}}$ of B.
The triple points of A and B are the same.

Or is it not possible to compare A and B's intermolecular forces from a phase diagram?
I was sure that B has stronger intermolecular forces, as its boiling point is higher than for A at the same pressure.

This is an interesting situation. It would be easier to answer if the mp and bp of A were both either higher or lower than for B.
But your measure of the strength of intermolecular forces will be melting point - and boiling point. And they give different answers, so how do we make a decision?
Consider the boiling point first: the higher boiling point of B suggests higher intermolecular forces, as you have concluded. Heat of vaporization is needed to put a molecule into the vapor phase, and it seems that B needs more energy to go gaseous.
But A melts at a higher temperature, suggesting that its molecules are held together in a solid phase by stronger bonds. Solid A needs more energy to turn it into a liquid; perhaps A can pack into a tighter solid structure. It would then need more heat to melt it, so it has a larger heat of fusion.
Then your answer must have two parts: A is more tightly bound in a solid framework, but after you put enough heat into both A and B to melt them, B has more intermolecular bonding capacity as a liquid than A does. So B boils higher and is more strongly bound as a liquid.

James, first of all thank you for your answer. Though I'm still not sure how to answer this question. Do you think it would be appropriate to say, that as A and B are in different states of matter, therefore it's impossible to compare their intermolecular forces?

Rather, melting and boiling are two different types of strength measurements; each has a heat required, and they are not necessarily related. So you can compare one set of heats (fusion) and decide which of A or B has stronger bonds. Then you compare the heats of evaporation and decide which of A or B seems to show stronger intermolecular strength at the higher temperature. Perhaps strength is a concept that you can measure by the amount of heat required to "break".


text:
Which compound results in the following products when mixed with AgNO3 and Chlorinated water
I'm currently working on attempting to identify a particular substance of which I know the cation is Na, however I am unsure as to what the corresponding anion is based off of these characteristics.
The salt is soluble, and when drops of AgNO3 are added to the dissolved substance the liquid turns a dark brown and a nearly black precipitate forms on the top. The liquid remains relatively transparent however. In addition, I observed that when chlorinated water was added to the solution, the solution slowly turns white with a light tint of yellow, like a very light creamy color, it releases some pungent odor however I am not familiar enough with scents to determine whether or not it is an ammonia or sulphur odor. Of the sodium compounds, the following are options:

Sodium bisulfate
Sodium borate
Sodium citrate
Sodium chloride
Sodium fluoride
Sodium nitrate
Sodium nitrite
Sodium oxalate
Sodium sulfate
Sodium sulfite
Sodium thiosulfate

As far as I'm aware the reaction with chlorinated water is just a reaction with Cl

The black precipitate can either be

metallic silver, formed by reduction of the $\ce{Ag+}$ ions (most likely)
an insoluble silver salt

Possible candidates for the sodium salt:
╔══════════════════╦═══╦═══╗
║      SALT        ║ 1 ║ 2 ║
╠══════════════════╬═══╬═══╣
║Sodium bisulfate  ║   ║   ║
║Sodium borate     ║   ║ ? ║
║Sodium citrate    ║   ║   ║
║Sodium chloride   ║   ║   ║
║Sodium fluoride   ║   ║   ║
║Sodium nitrate    ║   ║   ║
║Sodium nitrite    ║   ║   ║
║Sodium oxalate    ║   ║   ║
║Sodium sulfate    ║   ║   ║
║Sodium sulfite    ║ ✔ ║   ║
║Sodium thiosulfate║ ✔ ║   ║
╚══════════════════╩═══╩═══╝

The reaction with chlorinated water tends to show it is the first case (maybe with sodium thiosulfate).
Regarding the pungent odor, $\ce{H2S}$ has a strong rotten-egg odor.
A specific test for sodium thiosulfate is adding an acid (such as $\ce{HCl}$ to the unknown salt solution). A yellowish-white sulfur colloid should slowly appear.
A specific test for sodium sulfite is the decoloration of fuschine (should be available in a microbiology lab): add a buffer at pH = 7 to 1 mL of your unknown salt solution, then 1 mL of fuschine. After a few seconds, it should turn transparent.
Can you conduct those test to get more insight on this? If the first one is positive, no need to do the second one.


text:
Would this be a reasonable mechanism for the formation of a diazomethene?

I am mostly concerned about the first step. Are there any species in this mechanism that look too reactive for this mechanism to make sense? Any thoughts would be greatly appreciated.

The accepted mechanism is here: 1
KOH attacks the carbonyl of the urea, the tetrahedral intermediate breaks down with Me-N=N-O- leaving. This picks up a proton from solvent to give Me-N=N-OH. The second step is OH- removing a proton from the Me group of Me-N=N-OH which then loses OH- giving diazomethane.

The mechanism cited by @Waylander involves attack of base at the carbonyl group is an "accepted" mechanism but your approach is not unreasonable with some adjustment to arrow pushing.  Your initial step provides no role for base implying that N-nitrosomethylurea is unstable or unstable in water. Instead, deprotonate the amide nitrogen with base is required in your mechanism. The decomposition of isocyanic acid is fine. For the species CH3N=N-OH, use the electron pair on the lefthand nitrogen to eliminate OH-, then deprotonate CH3NN+ with base to get CH2N2.


text:
How can residual entropies be reconciled with the Third Law?
The third law of thermodynamics states that the absolute entropy of a pure crystalline substance is $0$ at $0\,\mathrm{K}$. In the analysis of chemical reactions this absolute scale is sued to calculate the entropy change.  However, it is also noted that some substances like $\ce{CO2}$ have residual entropy at $0\,\mathrm{K}$ because their structure is not perfectly crystalline.
Now my question is, given all of the above why is it that in thermodynamics tables such as in the book "Physical Chemistry" by Bawendi the entropy of $\ce{CO2}$ is listed as $0$ at $0\,\mathrm{K}$ when it should be some finite (albeit small) number? The same goes for other substances like $\ce{H2O}$ whose entropy is listed as $0$ at $0\,\mathrm{K}$ when it should be a finite number.
I understand that different tables use different reference scales and the entropy of $\ce{CO2}$ could be arbitrarily taken as $0$ at $0\,\mathrm{K}$. However the problem is then if the tables give the entropy of a substance relative to the entropy of the same substance at $0\,\mathrm{K}$, then the reference scale for each substance is different. This is fine but in all the example problems in the textbook mentioned above (and others) they use the entropy values listed in the table for each substance without accounting for the different reference states.
For example, if I have the reaction
$$\ce{A + B -> AB}\,,$$
the entropy change is
$$\Delta S ~=~S_{\ce{AB}} - S_{\ce{A}} - S_{\ce{B}}\,,$$where $S_{\ce{A}}$ is the entropy of $\ce{A}$ and so on.
Now in order for this to work the entropies must be taken relative to the same reference state (like when we use heat of formation to calculate enthalpies). However in the Bawendi book they use the values listed in the tables for each substance, which are again relative to the entropy at $0\,\mathrm{K}$ of that particular substance. This would all be fine if the entropy at $0\,\mathrm{K}$ was indeed $0$ for all substances but it is not. Could someone please explain why a lot of books do this? or is there something that I am missing?

First of all, entropy is an absolute value; you don't use different reference scales and reference states for it. Internal energy is another story: it never appears in the equations as an absolute value, but only as a difference of starting and ending values, hence you may define a scale for it and place zero on that scale more or less arbitrarily. Ditto for enthalpy and Gibbs energy. But entropy is not like that.
Now to the point. The only thing that has zero entropy at $0\,\mathrm{K}$ is an ideal crystal, which is also the equilibrium state (whatever that might mean) of crystalline compounds at that temperature. Getting to equilibrium in fluids might be anything from slow to very fast; in solids it is usually rather slow, and in solids near absolute zero it is so slow that the Sun might well become a red giant before the last defects leave your real crystal. See, thermodynamics is all about eternity. Wait until then, and you'll see the Third law at work.
That's what we mean when we are talking of residual entropy.

 I know about the possible degeneracy of the ground state, but this post is already getting too long.

Does the said law have no bearing on earthly things, then? Why, it does. Many crystals just get good enough while growing at reasonable temperature. They have very few defects, hence their entropy is already pretty low and gets lower as we cool them down. On the other hand, many other crystals are not that good.
So why would the book list entropy of all compounds at $0\,\mathrm{K}$ as zero? Well, getting to absolute zero is infinitely hard, and getting close to it is pretty hard too. Maybe the authors just put all those zeros there in a well-intentioned attempt to remind the readers of the Third law? Too bad for those who by that point have heard of residual entropy.
Then again, I don't see why would $\ce{CO2}$ have much of residual entropy in the first place. It is a good rigid molecule with nice polar bonds and without nasty things like nuclear spin. Perhaps its entropy near the absolute zero is really $0.0$ within the given precision. Water ice is another story, with all its hydrogen disorder. But wait, isn't it transforming into a hydrogen-ordered form at low temperatures? So maybe the zeros are true, after all?

The crystals do have to be perfect, any old crystal at $\pu{0 K}$ will not have zero entropy, and so $\ce{CO2}$ and $\ce{H2O}$ are given zero entropy. On the other hand, $\ce{CO}$ has the possibility of having arrangements $\ce{CO...CO}$ or $\ce{OC...CO}$ so should not have zero entropy at $\pu{0 K}$.
The convention usually used in chemistry is therefore to make the entropy  zero when translational, configurational, rotational, vibrational and electronic contributions are zero. Contributions due to nuclear spin and isotope mixing are ignored. This can be justified because nuclei are conserved in reactions and isotopic ratios are effectively constant thus any changes cancel out when taking entropy differences.

There are actually two school of thought on this.
One is that residual entropy (the entropy a substance has as a result of not being a perfectly ordered crystal at $\pu{0 K}$) is a thermodynamic entropy.  Accordingly, according to this school, a substance must be a perfectly ordered crystal in order to have zero entropy at $\pu{0 K}$.
The other is that "residual entropy is only an apparent entropy and not a real thermodynamic entropy" (J. Non-Cryst. Solids 2009, 355 (10-12), 617–623).  Thus, according to this second school, all substances have zero entropy at $\pu{0 K}$.


text:
If dG is a state function, why do reversible and irreversible paths differ?
For a reversible process at constant T,P in a closed system with no non-expansion work dG = 0. For an irreversible process under the same conditions, dG < 0. Since dG is a state function I would expect two process at the same T,P to give the same result just as dS is calculated for reversible processes but can be applied to any process with the same initial and final states. I suspect that I have incorrectly assumed that the states for dG are the same but I'm unsure.

G is not only a function of T and P, but also a function of chemical composition.

Right, but in this case I'm assuming single phase pure substances so that n is constant.

For the same initial and final states of the system, it has to be the same.  If you have a single phase pure substance and the initial and final temperatures and pressure are the same, then $\Delta G$ of the system is zero, irrespective of the path.  Of course, the intermediate states can vary, and the changes in the surroundings may be different.

Can you provide a specific example that we can discuss?

I think I figured it out. For a constant T,P process heat is dq=dH and is no longer a path function so that dG=0 regardless of whether the process is reversible or irreversible. dG is only <0 if there is a change in another variable such as n1,n2... or if we get rid of the constant P condition.

I have a little different perspective for you to consider.  Either no process has taken place at all, or the system has experienced a cyclic process, such that its final state is the same as its initial state.  If the cyclic process was irreversible, then any entropy that was generated within the system during the process (as a result of the irreversibility) has been transferred to the surroundings.

dG = -TdS_total right? So if entropy was transferred to the surroundings, wouldn't dG<0?

Also my question is similar to https://chemistry.stackexchange.com/questions/42758/why-at-constant-pressure-and-temperature-gibbs-energy-change-of-a-process-can-be?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa except that I'm interested in the contradiction when dG<0 for no reaction.

The total $\Delta G$ would be negative, but the $\Delta G$ of the system would be zero.


text:
Which is the cheapest plastic that retains its structure at 85 °C or higher temperature?
In an academic project, I need to protect some electronics components from heat (around 85 °C). I am searching for both a cheap plastic film and a cheap plastic to be used in a injection molding process.
As matt_black referred, the main criteria of the selection is "the point where structural integrity is reduced by repeated use at that temperature or the point where the polymer starts to creep".
By now I am using cork, but I would like to use a more robust material.
Those materials should be both thermal and electrical insulators and should not absorb water. They should also be resistant to the external environment: rain, sun (UV radiation), dust, etc.
Which materials are the best option in this case? From what I have searched, these are the main options in terms of thermal requirements:

PPO/PPE based
Polypropylene
Polyurethane
Vinyls
Polybutylene
Acetals
ABS and SAN
Polystyrene
ABS/Polycarbonate alloy
Acrylics
Cellulosics
Polyethylene and copolymers

When you say "protect from heat" do you mean insulate or something else? Protecting from things happening at that temperature (like water ingress) is different from protecting against reaching that temperature.

@matt_black The main function of the film is to thermally insulate, being the no water absorption also an important factor. If used a container it would be a plus to protect it from external environment.

It is also worth noting that most solid plastics won't do much insulating from heat (except for short periods) though they will act as a barrier to water or air to some extent. Good insulation requires something else like a foamed material (or cork which you already use) possibly plus a heat reflector. You could use cork coated in foil all sealed inside a plastic film.

Would a thermoset plastic be OK (e.g. Bakelite?).

The melting point is not going to give you the answer you want.  To properly select a polymer you need to look at the glass transition temperature $(T_\mathrm{g})$.  This is the temperature at which a material loses crystallinity and starts to more easily flow.  This parameter varies widely with thermal history and internal stress of a polymer material and can be raised by annealing.
Based on UV resistance, I would exclude any olefin or vinyl polymers as well as ABS.
For water resistance, I would exclude cellulose polymers and acetals.
Based on the temperature requirement I would say polystyrene ($T_\mathrm{g} = \pu{100^\circ C}$), acrylics $(T_\mathrm{g} = 85 - \pu{165^\circ C})$, polycarbonate ($T_\mathrm{g} = \pu{135^\circ C}$), or PPE/PPO ($T_\mathrm{g} = \pu{215^\circ C}$) have sufficiently high $T_\mathrm{g}$'s to remain as acceptable for the mentioned specifications.

In terms of prices, can you give me a hint on the prices (the lowest)? Do you think ABS with a UV stabilizer would be an option?

Well, you can get acrylic and polycarbonate at most hardware stores; acrylic being the cheaper of the two, for which, I would go with acrylic then.

The Tg of a polymer depends also on its molecular weight. So it doesn't mean a lot of thing to speak only about the Tg. See Kanig ueberreiter's equation.

I did intend to include molecular weight in my discussion but forgot to include it.

Googling "common plastic melting point" produces this list for me, and taking a quick look at the result it looks like ordinary polyethylene might do OK (mp 140C) and if you go to nylon (mp 220-270C) you would have a pretty comfortable margin.  A well-known issue with nylon, however, is that it tends to absorb water and in damp environments might not keep its dimensions exactly.  Only you know whether that affects your application.

It isn't the melting point you have to worry about: it is the point where the structural integrity is reduced by repeated use at that temperature or the point where the polymer starts to creep. Polyethylene is not usually recommended for use at ~100°C but polypropylene is better (if I remember correctly).

@matt_black Is there a table with that info? Thanks


text:
Why do my equilibrium calculations on this HF/NH4OH buffer system not match those in literature?
I've been trying to reproduce Figure 2 from this research paper (full text available).
The Problem
However I can't seem to get the same values as in the paper. I did the math both by hand and with MATLAB, but neither matched their results. I have the approximate solution (as it's from a graph) but after entering all the data and checking multiple times I still haven't been able to find what is wrong. The source for the graph/equation is the article at this link: The chemistry of co-injected BOE. The graph I want to reproduce is their Figure 2:

Above the figure it says: "distribution of the different species in a solution using $1800\,\mathrm{cc}$ of DI water, $100\,\mathrm{cc}$ of $49\%$ w/w HF and $0$ to $200\,\mathrm{cc}$ of $28\%$ w/w $\ce{NH4OH}$."
Here are the results I get for $10\,\mathrm{cc}$, $40\,\mathrm{cc}$ and $90\,\mathrm{cc}$:
$$
\begin{array}{c|cccc}
\hline
\text{Volume (cc)} & \ce{HF} & \ce{H2F2} & \ce{F-} & \ce{HF2-} \\
\hline
10 & 43\% & 48\% & 3\% & 6\% \\
40 & 35\% & 33\% & 14\% & 18\% \\
90 & 21\% & 12\% & 37\% & 30\% \\
\hline
\end{array}
$$
To calculate the fractions, I'm summing $\ce{HF}$, $\ce{H2F2}$, $\ce{F-}$ and $\ce{HF2-}$ to get the total and then I divide each of them by the total (both concentration and moles should both be valid).
I was wondering if anyone might have an idea of what I might be doing wrong. Or if the graph in the paper might possibly have an error or if there's some other numeric error in the paper.
Originally, I thought something was wrong with MATLAB's solution but I've checked that so many times by now, in more than just the two ways I show down below, that I really feel that that's not the issue. I've been at this problem for a month now, and about a week into it (3 weeks ago) I asked for help with the MATLAB side of things (see post on SO of original problem; I had some friends help out in the end). Honestly I just really want to know what the problem/mistake is.
The Equations
The equations being used from the article are in the image below:

Matlab scripts used
The matlab scripts I used are the following:
First mols.m as a depency of the other two,
%%%% amount of mol, Vol=volume, d=density, pwt=%weight, M=molecularweight
function mol=mols(Vol, d, pwt, M)
mol=(Vol*d*pwt)/M;
end

First script, all matlab calculated
clc;
clear all;
%Units to be used
%Volume is in CC also cm^3, 1 litre is 1000 CC, 1 cc = 1 ml
%density is in g/cm^3
%weigth percentages are in fractions of 0 to 1
%Molecular weight is in g/mol
% pts=10; %number of points for linear spacing

%weight percentages of NH4OH and HF
xhf=0.49;
xnh3=0.28;

%H2O
Vh2o=1800;
dh2o=1.00; %0.997 at 25C when rounded 1
mh2o=18.02;

%HF values
Vhf=100;
dhf49=1.15;
dhf=dh2o+(dhf49-dh2o)*xhf/0.49; %@ 25C
Mhf=20.01;
nhf=mols(Vhf,dhf,xhf,Mhf);

%NH4OH (NH3) values
% Vnh3=linspace(0.1*Vhf,1.9*Vhf,pts);
Vnh3=10;
dnh3=0.9; %for ~20-31% @~20-25C
Mnh3=17.03; %The wt% of NH4OH actually refers to the wt% of NH3 dissolved in H2O
nnh3=mols(Vnh3,dnh3,xnh3,Mnh3);

if max(nnh3)>=nhf
    error(['There are more mols NH4OH,',num2str(max(nnh3)),', than mols HF,',num2str(nhf),'.'])
end

%% Calculations for species
Vt=(Vhf+Vh2o+Vnh3)/1000; %litre
A=nhf/Vt; %mol/l
B=nnh3/Vt; %mol/l
syms HF F H2F2 HF2 NH3 NH4 H OH
eq2= H*F/HF==6.85*10^(-4);
eq3= NH3*H/NH4==6.31*10^(-10);
eq4= H*OH==10^(-14);
eq5= HF2/(HF*F)==3.963;
eq6= H2F2/(HF^2)==2.7;
eq7= H+NH4==OH+F+HF2;
eq8= HF+F+2*H2F2+2*HF2==A;
eq9= NH3+NH4==B;
eqns=[eq2,eq5,eq6,eq8,eq4,eq3,eq9,eq7];
varias=[HF, F, H2F2, HF2, NH3, NH4, H, OH];
assume(HF> 0 & F>= 0 & H2F2>= 0 & HF2>= 0& NH3>= 0 & NH4>= 0 & H>= 0 & OH>= 0)
[HF, F, H2F2, HF2, NH3, NH4, H, OH]=vpasolve(eqns,varias);% [0 max([A,B])])

totalHF=double(HF)+double(F)+double(H2F2)+double(HF2);
HFf=double(HF)/totalHF %fraction of species for HF
H2F2f=double(H2F2)/totalHF %fraction of species for H2F2
Ff=double(F)/totalHF %fraction of species for F-
HF2f=double(HF2)/totalHF %fraction of species for HF2-

Second script, equation were written out and solved by hand until being left with the final equation which was then solved in matlab
clc;
close all;
clear all;
%Units to be used
%Volume is in CC also cm^3, 1 litre is 1000 CC, 1 cc = 1 ml
%density is in g/cm^3
%weigth percentages are in fractions of 0 to 1
%Molecular weight is in g/mol
% pts=10; %number of points for linear spacing

%weight percentages of NH4OH and HF
xhf=0.49;
xnh3=0.28;

%H2O
Vh2o=1800;
dh2o=1.00; %0.997 at 25C when rounded 1
mh2o=18.02;

%HF values
Vhf=100;
dhf49=1.15;
dhf=dh2o+(dhf49-dh2o)*xhf/0.49; %@ 25C
Mhf=20.01;
nhf=mols(Vhf,dhf,xhf,Mhf);

%NH4OH (NH3) values
% Vnh3=linspace(0.1*Vhf,1.9*Vhf,pts);
Vnh3=10;
dnh3=0.9; %for ~20-31% @~20-25C
Mnh3=17.03; %The wt% of NH4OH actually refers to the wt% of NH3 dissolved in H2O
nnh3=mols(Vnh3,dnh3,xnh3,Mnh3);
if max(nnh3)>=nhf
    error(['There are more mols NH4OH,',num2str(max(nnh3)),', than mols HF,',num2str(nhf),'.'])
end

n2=nnh3;
n3=nhf-nnh3;

%% Calculations for species
Vt=(Vhf+Vh2o+Vnh3)/1000; %litre
A=nhf/Vt; %mol/l
B=nnh3/Vt; %mol/l

syms H
yl=H*(1+B/(H+6.31*1e-10));
HF=(sqrt((H+6.85*1e-4)^2+4*A*H*(H*5.4+5.42*1e-3))-(H+6.85*1e-4))/(2*(H*5.4+5.42*1e-3));
yr1=(1/H)*(1e-14+6.85*1e-4*HF*(1+3.963*HF));
hf=matlabFunction(yr1-yl);
H3O=fzero(hf, 6.23e-3)
yval=double(subs(yl,H,H3O))
yvalcheck=double(subs(yr1,H,H3O))

figure
hold on
bnds=[H3O-0.1 H3O+0.1 yval-0.1 yval+0.1];
ezplot(yl,bnds);
ezplot(yr1,bnds);
plot(H3O,yval,'or','MarkerSize',10,'LineWidth',2)

HF=double(subs(HF,H,H3O));
F=6.85*1e-4*HF/H3O;
HF2=3.963*HF*F;
H2F2=2.7*HF^2;

totalHF=HF+F+H2F2+HF2;
HFf=HF/totalHF %fraction of species for HF
H2F2f=H2F2/totalHF %fraction of species for H2F2
Ff=F/totalHF %fraction of species for F-
HF2f=HF2/totalHF %fraction of species for HF2-

Your math and program(s) appear to be correct.
HOWEVER: The discrepancy between your calculation and that from the paper arises from different definitions of the fraction of each fluorine-containing species that is present.
You have calculated the fractions based on the total amount of each substance present; e.g.:
$$
\require{begingroup}\begingroup\newcommand{\conc}[1]{[\ce{#1}]}\newcommand{\Ka}[1]{K_{\mathrm a, \ce{#1}}}\newcommand{\Keq}[1]{K_{\ce{#1}}}
f_\ce{F-} = {\conc{F-}\over \conc{F-} + \conc{HF} + \conc{HF2-} + \conc{(HF)2}} \\
f_\ce{HF2-} = {\conc{HF2-}\over \conc{F-} + \conc{HF} + \conc{HF2-} + \conc{(HF)2}}
$$
In the paper, they have calculated these fractions based on the amount of fluorine atoms embodied by each substance present; e.g.:
$$
\hat f_{\!\ce{F-}} = {\conc{F-}\over \conc{F-} + \conc{HF} + 2 \conc{HF2-} + 2 \conc{(HF)2}} \\
\hat f_{\!\ce{HF2-}} = {2 \conc{HF2-}\over \conc{F-} + \conc{HF} + 2 \conc{HF2-} + 2 \conc{(HF)2}}
$$
As I'll show, the former definition reproduces the results you provided in your question, and the latter reproduces the figure from the paper quite nicely.

I like to use Excel for equilibrium calculations—I find that being able to see the full sweep of the math at a glance helps me keep a grasp of where I'm at in the process of putting the solution together. Also, the Solver add-in makes it really easy to calculate results once the worksheet is assembled.
In the below, I've used the following color convention:

Purple: Reference data and given values from the problem
Gray: Calculated by formula
Orange: Input parameter
Blue: Parameter for Excel Solver
Green: Outputs for the Excel Solver objective function

First, I set up the calculation for the composition and volume of the final mixture, with the volume of 28% ammonium hydroxide as a parameter:

The specific gravities of the stock $\ce{HF}$ and $\ce{NH4OH}$ solutions were appreciably different from unity $(\ce{HF}$ source | $\ce{NH4OH}$ source$)$.  After mixture into the $1.8\,\mathrm L$ of water, though I just assumed all $\ce{SG} = 1$. I tracked the masses of total $\ce{HF}$ and $\ce{NH3}$ $($since the 28% is presumably weight percent of $\ce{NH3})$ as well as the overall mass of the system all the way through.  In combination with the relevant molecular weights, this allowed calculation of the total concentrations of fluorine and nitrogen species as well as the final volume of the system.
From here, I calculated the distribution of species in the final system:

I took the system $\mathrm{pH}$ as an adjustable parameter for the Solver.  From there, the $\ce{H+}$ and $\ce{OH-}$ concentrations are trivial to calculate.  For simple acid-base equilibria, I like to formulate the calculation in terms of the ratio between the acid and base species:
$$
\Ka{NH4+} = {\conc{H+}\conc{NH3} \over \conc{NH4+}}
  \quad \Longrightarrow \quad
  \rho_\ce{NH4+} =
  {\conc{NH4+} \over \conc{NH3}} = {\conc{H+} \over \Ka{NH4+}}
$$
In this case, where there is only the one acid-base equilibrium for the ammonia/-ium system, the concentrations can be calculated as:
$$
\conc{NH3} = {\conc{N_\mathrm{total}} \over 1 + \rho_\ce{NH4+}}
  ~ ; \quad
\conc{NH4+} = \rho_\ce{NH4+} \conc{NH3}
$$
Unlike in the reference linked in the question, I chose not to carry out the substitutions needed to express the concentrations of the fluorine-containing species solely in terms of $\conc{H3O+}$. Instead, I defined them as follows:
$$
{\conc{HF} \over \conc{F-}} = {\conc{H+} \over \Ka{HF}} \\
{\conc{HF2-} \over \conc{F-}} = \Keq{HF2-} \conc{HF} \\
{\conc{(HF)2} \over \conc{HF}} = \Keq{(HF)2} \conc{HF}
$$
I then set up a guess for $\conc{F-}$ as an adjustable parameter for the solver (blue cell), and used the above ratios to calculate trial concentrations for the other three species (gray cells).  The appropriately weighted sum of these concentrations $(1$ for $\conc{F-}$ and $\conc{HF}$; $2$ for $\conc{HF2-}$ and $\conc{(HF)2})$ was then matched against $\conc{F_\mathrm{total}}$ (green cell).
The species fractions in bold italics, which will be used in the charts below, were defined either as $f_i$ or $\hat f_{\!i}$, appropriately. Note that these values are not used in solving the system of equilibrium equations, so it didn't matter which one I had defined when I executed the Solver.
The charge balance was the final equation to set up, with no changes from that shown in the original question, except for pulling all the terms to one side so that the target value was zero.
I then set up the Solver add-in to find the best-fitting values for the $\mathrm{pH}$ and $\conc{F-}$:

I ran through the set of relevant addition volumes of $\ce{NH4OH}$, from $0$ to $200\,\mathrm{mL}$, storing the values of all of the $f_i$ and $\hat f_{\!i}$ values for plotting.  This is what I got for the fractions $f_i$:

The heavy squares are the individual data points you provided in your question. As can be seen, they match the $f_i$ results quite well.
Here is the chart of the $\hat f_{\!i}$ values, which is a spot-on match to the one in the paper:
$\endgroup$

Thank you so much I was wondering what I did wrong. This helps me out a lot as I can finally understand what I did wrong. I'm going to read through this properly when I have a bit more time and after that I'll accept it as the solution, for now I upvoted it. Thank you so much.

@BobvandeVoort Sure! Yeah, the key difference is absolutely between the definitions of $f_i$ and $\hat f_{!i}$.

This explains it perferctly and again thank you. I'm just wondering if you would know one other thing maybe. Why would we look at the concentration based of fluorine atoms? I'm trying to take a good initial guess of the relative etching/dissolving/reaction speed between this chemical mix/soup and another material  ($HfO_2$).

@BobvandeVoort I have no idea. For this application, calculating the distribution of species for a buffered oxide etch to estimate a rate of reaction, I wouldn't use either of the $f_i$ or $\hat f_{!i}$ fractions. I'd just stick with the concentration values themselves.

@BobvandeVoort Though, I should say -- their choice of $\hat f_{!i}$ does make sense in that it properly bookkeeps all of the initial fluoride added to the system with the 49% HF. (IOW, it makes sense to count $\ce{HF2-}$ and $\ce{(HF)_2}$ twice since they each have two fluorine atoms.)

thank you yes I understand  that. Although for me it might be more useful use the total concentration. I know which of the 4 have an effect on my substance (2 of them) and which ones don't ;)


text:
What is the pH of starch?
We need to identify an unknown for a chemistry lab. It's either starch or zinc sulfate.
We know that zinc sulfate is acidic, but what is the pH of starch?

Assuming its a solid powder just burn it in a crucible. Starch will burn completely but zinc sulfate will leave zinc oxide.

I used to teach college chemistry courses. In this case, you would want to use the iodine test for starch. If you get a deep, blue-black/purple color, then you have a positive result for starch. This is far more reliable and less invasive than pH or any other tests mentioned here. Think smarter, not harder!

Compounds don't have a pH, solutions do. So I think what you are asking is: What is the pH of a mixture made by adding starch to water? As you can see from the comments and the answers, there might be better strategies than measuring the pH to figure out the identity of the white solid.

Starch is a weak acid, but a stronger acid than water.
If you make a slurry of starch and add sodium hydroxide the starch will react with the sodium hydroxide, the starch slurry will then change in viscosity. It normally becomes a lot more viscous in my experience. This is part of the method by which starch is alkylated. I have made moderate sized batches of benzyl starch from corn starch by first stirring the starch with sodium hydroxide and then treating the resulting paste with benzyl chloride and then heating it up.
If you were to get a very senstive pH meter you would see that a slurry of starch in water is slightly more acidic than water would be. An alternative would be to use a dilute sodium hydroxide solution to make the measurement, I will warn you that it can get messy as the starch can be transformed into a paste / slime.
MaxW is right in part, a pure starch will give no ash but zinc sulfate will dehydrate to anhydrous zinc sulfate unless you heat it very very hard.
Another method is to add iodine, many starches will form a deep blue / black colour with a small amount of iodine. Zinc salts will not.


text:
How come one of the hydroxides in the phosphate group are more acidic than the other?

Three phsiologically relevant $\mathrm{p}K_\mathrm{a}$ values are associated with psilocybin: 1.3, 6.5 and 10.4. Suggest the appropriate charged species and mark the $\mathrm{p}K_\mathrm{a}$ values next to the corresponding protons in your structure.

How come one of the hydroxides are more acidic than the other in the phosphate group? Is because of electron repulsion? Can the substituents on a phosphorus group not move around? Why are the two hydroxide groups on the phosphorus atom not the same functional group?

Because the rendering is not really accurate.  It would have been better to render the two dissociation constants without trying to associate them with specific hydrogen atoms.
Given the neutral phosphate group with both hydroxyl hydrogens attached, either one has an equal probability of being removed as a proton, corresponding to the lower of the two $\mathrm{p}K_\mathrm{a}$ values.  The drawing simply picked one of the protons at random.  Then with that proton removed and the functional group now negatively charged, the second hydroxyl proton has become harder to remove and that is what the higher $\mathrm{p}K_\mathrm{a}$ represents.


text:
What happens to an acid when it completely dries out?
What happens to an acid when it completely dries out?  Does it stop being acidic?  And if there is a base there also, does it give back it's H's?

When it dries out, it is gone. In other words, it continues to be acidic elsewhere. As for the "base also" situation, that couldn't happen in the first place, since acids and bases do no sit well with each other.

When an aqueous solution of an acid is dried, water evaporates and the $\ce{H+}$ ions crystallize with the anionic part of the acid and form a crystal.
If the acid used to make the solution was not of crystalline form then after evaporation of water the anion and cation would have combined to form a compound of some other type (may be a solid formed by co-ordinate bonds or a liquid or a gas like $\ce{CO2}$)
Initially, because of presence of $\ce{H2O}$ molecules, the $\ce{H+}$ ions were bonded with them to form $\ce{H3O+}$ molecules and the anionic part of the acid were solvated by these $\ce{H3O+}$ ions. This solvation happened because it liberated more energy than consumed in dissociation of the acid molecule (i.e. was thermodynamicaly favorable)
But, now since the water molecules are absent, the $\ce{H+}$ and $\ce{A-}$ (anions) will combine to form $\ce{AH (s)}$ as its formation will be favorable because there is no more water to compensate the energy required in there dissociation.
If there had been a solution of acid and base then the $\ce{H+}$ ions from the acid and $\ce{OH-}$ ions from the base would have already reacted to form $\ce{H2O}$ and hence the scenario will be the same as above instead we will also notice crystallization of the cation $\ce{(B+)}$ from the base $\ce{(BOH)}$ and the anion ($\ce{A-}$) from the acid ($\ce{AH}$) to form $\ce{BA (s)}$.
If it would had been an aqueous solution of only a base then also the water would have evaporated and crystals of the corresponding base would have formed.
To talk a bit more (though not necessary in the answer): Acidity is a measure of concentration of $\ce{H+}$ ions in water and now since there is no solution present but only the crystals, the crystals won't behave like an acidic solution until and unless solvent is added to it.

Your first sentence make it sound like all pure acids are crystalline, which is not quite true.

I actually took it to be crystalline as an example to explain,  I have edited it and have written the condition in a code block at the start of the answer

Now you assume that all pure acids are solid, which is still not quite true.

Yup, I have edited it, I added CO2 as an example, but I guess if it's correct or not because in solution of CO2 in water  first H2CO3 will form then it will disassociate to form CO2

Great answer.  So if there is a protonated alkaline with an acid, will it deprotonate and go back to being a freebase when the solution is removed?

Use HCl as an unambiguous example. @ua57874: With weak acids this may be the case, with strong acids hardly so. To both of you: instead of "acid with a base" (which is quite a bit of an oxymoron), better use the word "salt".

@ua57874 Thank you :)    By protonated alkaline you mean salt ?    I think de-protonation of salt in an acidic solution will depend upon values for Ka and Kb for the acid and base.
The amount of de-protonation will depend upon the strength of that base and also the acid.
Whatever base it may be certain amount will de-protonate in establishment of equilibrium

Yeah, a salt.  I mistyped above, where I said solution, I meant solvent.  So, if let's say a weak base alkaline is added to a solution with a weak acid, and it protonates, will it deprotonate when the solvent is removed?  Is it if the ka of the acid > kb of the base, the acid will take back the H's when the solvent is removed?

Whatever may be added, either salt or base, chemical combinations will take place according to the equilibrium constants and on drying out water, we will get those combinations in there solute form(solid, liquid or gas)

@Ivan Neretin am I correct ?

The solvent doesn't matter much. If the acid itself is volatile (and weak, too), then it will get removed - in other words, the base will deprotonate. Same thing will happen if the base itself is volatile, like ammonia. If both the acid and base are weak, but not volatile, then the salt will not be destroyed when you remove water. On the contrary, in some cases it can be destroyed when you add water.


text:
Safely lowering bleach pH without chlorine gas?
According to Eurekalert, by lowering the pH of household bleach (e.g., by adding vinegar or citric acid), we can increase $\text{[HOCl]}$ and consequently greatly improve the disinfecting capabilities of the bleach itself.
However, doing so could technically liberate chlorine gas, either through hypochlorous decomposition
$$ \ce{4H+(aq) + 4OCl- (aq) -> 2Cl2(g) + O2(g) + 2H2O (\ell)} $$
or mere comproportionation
$$ \ce{2H+(aq) + OCl- (aq) + Cl- (aq) ->  Cl2(g) + H2O (\ell)} $$
(depending on how the bleach was originally made) and so, mixing bleach and acid seems to be contraindicated everywhere else on the internet.
However, the authors of that link also suggest

... first diluting one cup of household bleach in one gallon of water and then adding one cup of white vinegar.

So then does that make it safe? Wouldn't chloroacetic acid formation also be a concern?

There won't be chloroacetic acid to speak about. As for the rest, in effect you'll acidify bleach to make it stronger and then dilute it to make it weaker, much like a man who takes sleeping pills together with wake-up pills.

The oxidizing effect of chlorine is quite powerful even at concentrations of 1% or less. Mixing bleach and vinegar gives a much more concentrated solution. Although chlorine is soluble to the extent of 1.46 grams per 100 mL cold H2O (CRC Handbook), its vapor pressure above the water may be significant. A solution diluted by a factor of 7 or 8 would be in contact with a surface long enough to disinfect with a much reduced odor. Safer for people, but still deadly to bacteria.


text:
How does the "sample type" selection on the MassHunter software for Agilent 7900 ICP-MS affect the quantitative results?
I'm new to using the instrument, and recently I have tried to measure Cr in both the eluent(1% nitric acid) and a method blank (also 1% nitric acid, but after it has gone through the entire analytical procedure).
In the batch settings, I set the "sample type" for the eluent as "bkgd"  (it stands for background), and my method blank as "samples". The result was that I obtained a negative CPS signal for my method blank. On the offline data analysis program, I was able to change the sample type of the the eluent to "samples" and told the program to reprocess the data, and lo and behold, the CPS signal goes back to positive.
What I think happened was that whenever I set something in my batch as the sample type bkgd, the instrument automatically references that cps count as zero, so that the subsequently analysed sample's signal will be referenced from that. However, I do not know exactly how the different sample types will affect my measurements. I have tried looking up the instrument manuals from Agilent, but none of them so far have given me any insight into this.
There are several sample types to choose from, such as blk (for blanks), sqstd(for tuning solution standards), calblk, etc. I don't want to risk selecting any of these until I'm confident I know what they do to my data.
Any response will be appreciated.

Precisely because vendor software is often opaque, I tell it that every one of my samples, whether a blank, a standard, a spike-in, or whatever, is a Sample or an Unknown.  Then I do all data processing, such as background subtraction, regression to find standard curves, etc., in an environment that I control and understand (e.g., R, Excel, or python).


text:
Dissolving of non polar gases in water (liquid)
How does dissolving happen for non polar gases in water? Is it like being in the space between the water molecules or does there occur some kind of interaction occur between gas molecules and water molecules?
Also if the gases dissolve in water, how can their solubility be increased? I have also heard what's called Henry's law accounted for this?
Answers on molecular scale would be more appreciated.

Certainly there are attractive interactions between water molecules and nonpolar solutes, there are always dispersion attractions.  Additionally, water will "layer" against a solute  particle, producing solvation shells, even in the complete absence of attractions, just because of the excluded volume forces packing  everything together.   However, for nonpolar gases the new gas-water interactions are significantly weaker than the water-water interactions they replace, so the enthalpic change on dissolve is strongly positive (unfavorable).  Hence the only reason gases dissolve is because the entropy increases, and that is simply because the volume available to the gas molecules is larger if it includes the volume of the water as well as the volume of the gas above it.
Another way to put that is that molecules of nonpolar gas are found in the water simply because they can be.  Id est, those molecules will zoom around in the gas phase, occasionally strike the surface of the water, plunge in, and wander around randomly for a while before exiting.  Net result: a low concentration of nonpolar gas dissolved in the water.  The concentration will increase with pressure simply because a higher pressure means a higher density of the gas, which means more gas molecules  will strike each cm2 of the water surface per second, and at equilibrium that must be matched by a higher rate of exit from the surface, which requires a higher equilibrium concentration in the bulk water.  For small pressures, the effect is approximately linear in pressure, and the coefficient is called the Henry's Law coefficient.
The dissolved gas particles do not occupy spaces between the  water molecules, if by that you mean preexisting spaces.  They shove the water molecules out of the way, create a hole, and live in that.  The volume of the solution may not increase, however, or may not increase as much as one might think, because as I mentioned above, the water will form solvation shells around the dissolved particle, and will moreover disrupt the hydrogen-bonding network in the water.  Both effects could result in an increase in density in the regions near the hole created by the solute particle.  So predicting whether and by how much the volume of the water expands (or contracts) is in general difficult.


text:
Why tertiary amines can't show chirality?

Why is a tertiary amine with three different substituents not chiral? Doesn't the lone pair represent a fourth, different substituent?

See this https://en.wikipedia.org/wiki/Nitrogen_inversion

@SoumikDas Also addressed in this question...probably close enough to be duplicate, actually.

The other question is not a fully dupe of this question... For one thing, the statement of this question is wrong. Tertiary amines can definitely be chiral. It's just that the epimerize quickly, but they are definitely chiral when there are 3 different substituents. In addition, you can easily create an tertiary amine that does not invert by tying back the substituents.

@Zhe Sir, plz. explain a bit " you can easily create an tertiary amine that does not invert by tying back the substituents  " what is"tying back the substituents"

Constraining the substituents within a rigid structure, possibly a cyclic species

A quinuclidine will do the trick.

Also, see quinine.

Good point, @Zhe, I hadn't considered structural constraints against inversion. Reopen vote cast.

The short answer to your question

Under conditions that inhibit inversion, an amine that has three different groups attached is chiral.

The above applies not just to tertiary but also to secondary amines.

Now, why inhibit inversion? Why does it make a difference?
As mentioned by Zhe (and mentioned in Newton's answer) in the comments:

Tertiary amines can definitely be chiral. It's just that they epimerize quickly, but they are definitely chiral when there are 3 different substituents. In addition, you can easily create a tertiary amine that does not invert by tying back the substituents.

If you notice, they are chiral for very short periods of time due to the inversion that takes place in amines. In regards to the time scale – Ammonia ($\ce{NH3}$) flips $4 \times 10^{10}$ times every second. Converting that into the time for one inversion that translates into an inversion every $\pu{2.5 \times10^{-11} s}$ which is $\pu{25 picoseconds}$. Such a scale of time cannot be discerned by any of our current methods to identify stereochemistry.
From http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch07/ch7-3.html

In this case, even though a chirality center is present in each molecule, the sample is optically inactive since the optical activity of the two extremes of inversion
averages out because they are enantiomeric.

For why umbrella inversion occurs, see this answer by SendersReagent

Now, what if we restrict this inversion by attaching free-rotation restricting groups? What happens then?
This restriction brings (or rather removes) a fact into the mix. There is no more inversion possible.  This is because rearrangement into $\mathrm {sp}^2$ cannot take place because of the groups added since it becomes sterically inhibited in becoming planar. An example of this would be quinuclidine which has no way of forming a planar structure.

I'm not sure that one can go so far as to say what the "net observed" molecule is...

There is actually another way to get a tertiary or secondary amine to be chiral.  Put it into a chiral environment, such as a chiral solvent.  You will get nonequivalent interactions with the two enantiomers of the amines, so the equilibrium between the two enantiomers is nudged off symmetry.  We see this experimentally as a change in optical activity with addition of the amine to the system.

@OscarLanzi Is this a method to find chirality in amines? Also, why is there a difference in interaction?

To follow-up on @OscarLanzi's comment, there are Europium based chiral shift reagents frequently used to cause enatiomeric peaks to shift (by creating disastereomeric complexes).

@Safdar Is the planar structure a transition state? Also why in the electrostatic maps there is electron density above the nitrogen and not below it if the two structures transform into each other?


text:
Would a hydrogen/oxygen fuel cell be a good choice for energy storage and generation in H2 rich atmospheric environments?
So imagine an atmospheric probe sent to enter the atmosphere of a giant gas planet. There would be no need to store hydrogen since a compressor or an inlet with sufficient dynamic pressure would allow sufficient $\ce{H2}$ to enter the system in the upper atmosphere. The output would be basically limited by dynamic pressure and max output of the cells. $\ce{O2}$ would be delivered from a pressure vessel or cryogenic tank (seems possible, it's cold in the shade in space) and valve regulated.
Besides low operation temperature, what would be the drawbacks?

An "inverted" $\ce{H2/O2}$ fuel cell should work well, and be easier to maintain than one in an $\ce{O2}$ atmosphere, because there are a plethora of membranes to pass pure $\ce{H2}$ (e.g. hot platinum, plastics, ceramics...). That means that the cell would be less likely to be poisoned by trace gases.
It would also be possible to use methane, common on gas giants, in a fuel cell.
That said, carrying the oxygen in the probe from Earth to the target would take energy. A fuel cell might be useful for a short-lifetime probe, such as one penetrating deep into a gas giant's atmosphere, only to be crushed in a few days. For a longer lifetime, nuclear thermoelectric power (e.g. MMRTG) is more practical.

Good point. Using answer mode because did not fit in comments.
But when you wrote about trace gases I thought that there are indeed traces gases in the Jovian atmosphere and a descending probe has some probability of hitting a cloud deck made of
NH4, then NH4SH, NH42S, and finally water clouds. Are these sulfur compounds nasty for the majority of membranes?
But I worry too about temperature, surely the hydrogen supply would need to be preheated, and oxygen exiting the tank would lose temperature too due to pressure drop. Or are there any fuel cells that operate well at low temperatures?
FYI the possible design would be a quadcopter assisted by an infrared balloon
(derived from https://www2.jpl.nasa.gov/adv_tech/balloons/outer_jupisat.htm)
The quadcopter would need at least 6kw of power for a 10kg descent module to just hover in free fall conditions due to high g (in the case there is a balloon failure/malfunction). That's quite a lot of power. So the cell stack would need to be lightweight and compact (less or equal to 1kg and 1l volume would be ideal)
A 2L oxygen tank would provide the fuel cell up to 12kwh of energy at 80% efficiency. that would be enough for 2 hours of hovering, so the best would be to have it assisted by a solar infrared heated balloon. (a big one, pumped with hydrogen taken from the atmosphere, and may be preheated before pumping)
The thermodynamic equilibrium temperature could be modeled taking into account, heat gain from the fuel cell, heat gain from joule losses and payload, heat loss from gas input temperature and heat loss from the enclosure. As the probe descends, the problem would be reversed, heat gain would come from Jupiter atmosphere, and H2 inlet temperature and pressure would be significantly higher and I think the efficiency of the cell would be better. At some point however the valve would need to be closed to prevent pressure and temperature damage. Maybe an empty lightweight tank could be filled just before to make some reserves for the late descent.
At this point, the probe is on final reserves and then secondary battery power (presumably not much capacity, only for science payload and critical systems at this point) and is sinking.
With that design, the probe could stay operational in the atmosphere for 72 to 96 hours, maybe.

The problem, however, seems that even with current state of the art technology, fuel cells are notoriously heavy. A 10kW fuel cell weighing 100kg is at least two orders of magnitude too heavy for this kind of application. Maybe when (if) technology matures it would be worth it.


text:
Which will give faster SN2 reaction

In $\ce{H2C=CH-Br}$ and $\ce{H3C-CH2-Br}$, which will react faster towards a $\mathrm{S_N2}$ reaction?

According to me, as double bond exhibit −I effect, hence the 1st should do a faster reaction. Am I right, or is there any other reason?

$\ce{CH3-CH-Br}$ will give faster $\mathrm{S_N2}$ reaction because when a nucleophile will approach $\ce{CH2=CH-Br}$ for $\mathrm{S_N2}$ reaction the double bond between $\ce{CH2=CH}$ will hinder its approach (steric effect), but there is no such hindrance in case of $\ce{CH3-CH2-Br}$.
To support the answer we can add one more point that in case of $\ce{CH3-CH2-Br}$ the charge $δ^+$ on the $\ce{C}$ atom of $\ce{CH2}$ will be greater in magnitude than that at the $\ce{C}$ atom of $\ce{CH}$ in case of $\ce{CH2=CH-Br}$ because the double bond has better −I effect than single bond, hence it will be easier for $\ce{-Br}$ to attract the shared electron pair towards it and develop a greater $δ^+$ charge on $\ce{C}$ in case of $\ce{CH3-CH2-Br}$, which will ultimately support the approach of the nucleophile for the $\mathrm{S_N2}$ reaction.

There is no sense of comparing the reaction rate when one of the compound doesn't show SN2 mechanism. And, that compound is $\ce{CH2=CHBr}$, due to the fact that it is a vinyl halide.
$\ce{CH2=CHBr}$ will hinder the approach of nucleophile due to the presence of pi-electron cloud around double bond.
And, therefore the obvious answer is $\ce{CH3-CH2Br}$

I think you mean that the obvious answer is ethyl bromide, rather than vinyl bromide, also considering your answer. Right?

@GürayHatipoğlu, +1 for pointing out!


text:
Sec-butyl vs Butyl - Alphabetizing and Ordering in Alkane Nomenclature
For the structure below,

if we use trivial name for complex substituent, do we call it
5-butyl-10-sec-butyltetradecane, or
5-sec-butyl-10-butyltetradecane?
Considering that both substituent are placed in a symmetrical way in the parent chain and sec- prefix are not included in alphabetizing, which one is preferred?

Note that the prefix sec-butyl was still contained in the 1993 IUPAC recommendations but is no longer recommended in the current (2013) recommendations.

Related: t-butyl VS s-butyl

This may be resolved by IUPAC Rule A-2.3(ii) which says "The name of a complex radical is considered to begin with the first letter of its complete name," which seems to suggest to me "sec-butyl" is considered to start with "s" and not "b" in this case.  As I interpret Rule A-2.3(i) it is only multiplying prefixes (di-, tri-) which are ignored when alphabetizing substituent names.

but as far as i understand, sec, tert prefixes are not used for alphabetizing as well as the multiplying prefixes. only iso prefix is included in alphabetizing. pls correct me if i am wrong.

this problem wont occur if we change the sec-butyl to 1-metylpropyl, thus the name will become 5-butyl-10-(1-metylpropyl)tetradecane


text:
Radial probability curve
Can somebody give me a clue on how to solve the question in the picture for me given below

What do you want us to do with this?

Whyever not do it, it might be for the simple joy of doing it.

The normalised wave function of $2s$ orbital is $$\Psi_{2s} = \frac{1}{4 \sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{\frac{3}{2}}\left(2 - \frac{Zr}{a_0}\right)e^{-\frac{Zr}{2a_0}}$$ Now the radial probability distribution function will be $$P(r) = |\Psi|^2 4\pi r^2 = \frac{Z^3}{8a_0^3}r^2\left(2-\frac{Zr}{a_0}\right)^2 e^{-\frac{Zr}{a_0}},$$ so that $\int_{0}^{\infty} P(r) ~\mathrm dr =1.$ Now for getting the maxima and minima, solve $\frac{\mathrm d}{\mathrm dr}(P(r)) = 0.$ Simply we have to solve $$\frac{\mathrm d}{\mathrm dr}\left[r^2\left(2-\frac{Zr}{a_0}\right)^2 e^{-\frac{Zr}{a_0}}\right] =0.$$ Solving this one root will be from the equation $(2- \frac{Zr}{a_0})=0$, which will give you the value of $ r$ for which $P(r) =0$, that will be the minima as you can easily verify. The maximas will come from the other quadratic equation $$\left(\frac{Zr}{a_0}\right)^2 -6\left(\frac{Zr}{a_0}\right) + 4 =0$$ which will give you two root as $\frac{Zr}{a_0} = 3+ \sqrt{5} ;\, 3-\sqrt{5}.$
So, the difference between two peak points will be $r_d =\frac{2\sqrt{5} a_0}{Z}.$ If it is Hydrogen atom, the distance will be $2\sqrt{5} \times0.529 A^0 = 2.365 A^0.$


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Why would diclofenac have a higher skin permeability when formulated as diclofenac epolamine rather than diclofenac sodium?
There is a suggestion on the literature that the formation of ion pairs creates a permeation enhancement.
Usually only non-ionised compounds penetrate the skin.
Is it possible that the conjugate species does so?
Some sources claim the epolamine salt has enhanced solubility, others say it is about the same.
Appreciate any help

Where does the literature suggest this?  Would you be able to add a link?

They likely penetrate the skin as a rather unpolar pair of two large ions, where the organic rest shields the charges.

Voltaren is a 1 -2% paste of the ammonium salt of diclofenac (epolamine is hydroxyethyl pyrrolidine) and is used a topical pain reliever. You can smell the ammonia.
It could be that the increased pH of the paste softens skin and makes it more permeable. High pH will peel your skin right off!

Transdermal drug delivery is a complicated process governed not only by chemistry/physics, but also biology.
Limiting ourselves to the physical/chemical aspects:

Different solubility, both kinetic and thermodynamic
Different tonicity - alters the osmotic pressure
Different pH and buffering capacity of the formulation - altering charge state populations of the solutes in the local environment
Surfactant effects of different solutes - changes the surface tension at phase interfaces
Membrane perturbation by different solutes - ability to alter skin/cellular membrane behavior and dynamics

These factors are interrelated with one another to varying degrees
As for primary literature, see: Ex vivo study of transdermal permeation of four diclofenac salts from different vehicles


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IUPAC of heterocyclic compounds terminology
During my studies of heterocyclic compounds I learned that monocyclic compounds has prefix+ stem+ suffix.
the prefix is coming from the heteromolecule name (e.g oxgen --> oxa- )
however the stem and suffix is ambiguous to me. I can link some of the stem (e.g et --> from tETra)  but what about all other stems and suffixes. Where do they came from?

It will be very helpful if the source of terminology is documented here.

This way of naming heterocycles is known as the Extended Hantzsch-Widman system of nomenclature (the above link is a free reproduction of a publication that revised this system : W.H. Powell, Pure Appl. Chem., 1983,55, 409-416).
The stems do actually come mainly from numeric prefixes. " 'ir' from tri, 'et' from tetra, 'ep' from hepta, 'oc' from octa, 'on' from nona, and 'ec' from deca."Source. They are mainly taken from the middle letters of the numeric prefixes. As for "ol" and "in" for 5 and 6, these are holdovers from the intial system from 1887 that was developed for nitrogen containing rings, so my guess is they were kept to maintain the connection to the endings of pyrrole and pyridine.


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Nomenclature of lithiated dithianes
Lithiated dithianes have a few uses in organic synthesis (albeit fairly niche ones) such as the Corey-Seebach alkylation / Corey-Seebach reaction, which uses dithianes as masked acyl anions (d1 synthons) as shown below.

Searching through the 2013 IUPAC recommendations for organic nomenclature for dithian, dithiane didn't come up with any useful results, and the usual technique of looking for the corresponding oxygen analogue seems redundant in this case as the corresponding molecule with oxygen is impossible!
In most papers, the authors avoid explicitally naming the lithiated intermediate (certainly one way to get around the issue):

Lithiation of dithiane X with t-BuLi followed by alkylation with epoxide Y resulted in Z upon quenching.

In others, a derivative name is used based on the 1,3-dithiane itself:

2-lithio-1,3-dithiane derivatives are usually prepared by deprotonation with n-BuLi in THF at low temperatures2 and the anionic species obtained are able to react with many types of electrophiles.
Taken from Tetrahedron 2003, 59, 6147

ChemDraw outputs something more in-keeping with other nomenclature:

The exact nomenclature of alkyllithiums is an inorganic problem (cf. 2005 Red Book) and in there one finds as an example (IR-10.3.2, p 229; repeated in P-69.3, p 1043 of the 2013 Blue Book)

$\ce{LiMe}$ lithium methanide (compositional name)
$\ce{[LiMe]}$ methyllithium
$\ce{[(LiMe)4]}$ tetra-µ3-methyl-tetralithium
$\ce{(LiMe)_n}$ poly(methyllithium)

so, the name you choose will depend on what structural features you want to emphasise. Note that the Red Book does not prescribe "preferred" IUPAC names, and the Blue Book explicitly refuses to assign PINs (p 1036):

Neither preferred IUPAC names or preselected names (see P-12.2) for organometallic compounds involving the transition elements (including the Group 3 elements) and Groups 1 and 2 elements, excpet for 'ocene' compounds, are noted. This determination will await consideration by a task group on inorganic and/or organometallic nomenclature.

Obviously, from a practical point of view, the name "methyllithium" is more commonly used and more easily recognised. In line with this, we will therefore look for a name of the form "_yllithium". The use of "methyl" instead of "methanide" is permissible because (p 1037, 2013 Blue Book)

Because of a long established tradition, names for substituent groups as defined in Section P-29 can also be used as ligand names.

and hence the question becomes: what is the preferred prefix for the following group?

This is a purely organic question, so can be answered with the 2013 Blue Book. The relevant rules are as follows:

P-29.2 [...] (2) The suffixes 'yl', 'ylidene', and 'ylidyne' are added to the name of the parent hydride with elision of the terminal letter 'e', if present, when followed immediately by the letter 'y'. The locants for the atoms of free valences are as low as is consistent with any established numbering of the parent hydride and, except for mononuclear parent hydrides or the suffix 'ylidyne', the locant '1' must be cited. This method is used to generate names of 'alkanyl-type substituent groups' that are simple substituent groups with free valences at positions other than '1'.
P-57.1.5.1 Preferred prefixes derived from cycloalkanes are of the 'cycloalkyl type' (see P-29.2); preferred prefixes derived from cyclic compounds other than cycloalkanes are all of the 'alkanyl type'.

The preferred IUPAC name of the parent hydride, according to Hantzsch–Widman nomenclature (P-22.2.2.1), is 1,3-dithiane. Correspondingly, the preferred prefix is 1,3-dithian-2-yl:

Finally, returning to the Red Book:

IR-2.2.3.2 Parentheses are used in names in the following ways:
(a) To enclose substituent group or ligand names in order to avoid ambiguity, for example if the substituent group or ligand names contain multiplicative prefixes, such as (dioxido) or (triphosphato), or if substitution patterns would otherwise not be unambiguously specified, or if the substituent group or ligand name contains numerical or letter descriptors [...]

we can conclude that a suitable name would be (1,3-dithian-2-yl)lithium.

I'd never thought of looking in the red book – I've always very much put organolithiums in the 'organic' box and never thought to consider them as inorganic based on the metal


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Finding the pH value of HF

Find the pH value of HF.

$pH = -\log[H+]$ for strong acids
$pOH = -\log[OH-]$ for strong bases
$pH = 1/2(pKa - \log C)$ for weak acid
$pOH = 1/2(pKb - \log C)$ for weak base

HF is weak acid so $Ka = 6.6 \times  10^{-4}$
$pH = 1/2(-\log(6.6 \times 10^{-4}) - \log(0.04))$
$\boxed {pH = 2.2892}$
Is my approach correct?

This is complete nonsense. You need concentration to calculate pH.

Have a look at my answer to this question as it outlines the full method and its approximations https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566#95566

Well, yes but barely.  Assuming you forgot to say that $C = 0.04M$ is the molarity of the HF solution, then you have applied your equations correctly.
A concern here is that your shortcut equations rely on what's often called the "small x approximation," which says the amount of HF that actually dissociates is so small we can neglect the difference between the initial and equilibrium molarity of HF.  Is that actually true?  We would need to do the full calculation to see, starting from the Ka equation that describes the equilibrium composition of the solution: $$K_a = \frac{[{\rm H}_3{\rm O}^{+}][{\rm F}^{-}]}{[{\rm HF}]}$$ If we let $x$ stand for the equilibrium molarity of ${\rm H}_3{\rm O}^{+}$ and $C$ for the initial molarity of HF, then this equation turns into $$x^2 = K_a (C - x)$$ which can be solved with the quadratic equation.  Using $C = 0.04$ we find $x = 4.819\times10^{-3}$ which gives pH = 2.317...  You'll notice that's different from your value, and that's because in fact about 12% of the HF dissociates, which isn't that small.
However, a point we haven't addressed yet is that you've given your answer to far more precision that the data justify.  You have only 2 significant digits in your pKa and only 1 in your value for C.  That means your answer should have only one significant digit, and you should give it as pH = 2. In that case, taking the precision of your data into account, your answer is correct.  In fact, your answer is correct even to two significant digits (pH = 2.3).  But that's kind of just lucky.  You might want to study up on when you can use those equations and when you can't.