text:
Spin quantum number of electron
Why is the spin quantum number of an electron +1\2 or -1\2 in an atom? Is this an experimental value or theoretical value?

Actually, only the projection of the spin of an electron has a sign and is given in unints of $\hbar$. This fact was initially put forward by experiments but could later be explained using relativistic quantum mechanics.

An electron has two quantum numbers the spin which is $ 1/2 $ and the projection (also called magnetic or azimuthal) quantum number onto an axis which has values $\pm 1/2$.

Here is the corresponding question (with answer!) on Physics.SE.

It is both experimental and theoretical. Electrons have angular momentum, and sometimes we want to discuss whether it lines up with another direction (ie typically the direction an electromagnetic field, since electrons are charged their angular momentum makes them magnetism and that interacts with electromagnetic field).
How should we talk about the direction of the angular momentum? We can say "up" and "down" or "forwards" and "backwards", and sometimes we do that. But if we want to use it in calculations, it is easier to give them numbers. So what numbers should we give? They are opposite directions so really it makes sense to use opposite numbers. So what opposite numbers should we use? We can use any, so why not use a gap of "1" between them? So what opposite numbers have a gap of 1 between them? 1/2 and -1/2.
There are some other detailed reasons why 1/2 is a convenient number, but it is in my view fair to say we just want to represent two opposite directions in calculations, and the numbers 1/2 and -1/2 suit this well.

-1. The spin of a particle is fundamental and certainly cannot be reduced to an arbitrary choice of opposite numbers.

@a-cyclohexane-molecule yes, I agree that if we want the absolute value of the spin quantum number to match particle's spin property it is not an arbitrary choice. This is what I was alluding to with my mention of "other detailed reasons". However to understand what this means requires a large amount of theory. So I was trying to focus on what the spin quantum number practically represents: fundamentally it is just a direction. I accept my answer is not ideal and welcome any edits to it or specific suggestions. I think/hope it currenly carries more benefit than no answer.

@a-cyclohexane-molecule thank you though for explaining your downvote.

@a-cyclohexane-molecule Although I am not certain, I believe the historical arrival at the 1/2 labelling for electron spin followed the reasoning I used in my answer, and the labelling of the particle property was chosen to match this, and not the other way round, as I believe you are implying. It is not explicitly stated (or a perfect source) but I believe this is suggested in https://en.wikipedia.org/wiki/Spin_quantum_number


text:
Nickel plating with nickel acetate, nickel chloride, sulfur, saccharin and coumarin
I am a hobbyist trying to learn/apply a nickel electroplating process that I can safely do at my shop.  My substrate is a machinable non-stainless steel with unknown carbon content, and I'm looking to do a close variation of the "Watts Bright nickel" formula.  I have a 99.9% pure nickel anode, nickel acetate, nickel chloride, boric acid, sulfur powder, saccharin and coumadin powders.
The formula calls for nickel sulfate; is it possible to simply add the powdered Sulfur to the solution, or do I have to use a different process to synthesize the nickel sulfate? If so,  can I do so with the materials that I have?
Also, the Watts formula does not include the saccharin or coumarin, but I know that these chemical additives will increase the odds of a smooth mirror like finish.
Also, is there a recommended quantity range, or at least a starting point, for these two chemicals for this application? Any insights here from an experienced chemist/electroplater would be greatly appreciated.

The formula calls for nickel sulfate; is it possible to simply add the powdered sulfur to the solution, or do I have to use a different process to synthesize the nickel sulfate?

This will not work. Sulfur ($\ce{S}$) and sulfate ($\ce{SO4^{2-}}$) are two pairs of shoes. The oxidation state is different and there is no way that sulfur powder will turn into sulfate during the electroplating.

If so, can I do so with the materials that I have?

It might be possible to heat another nickel salt in sulfuric acid, cool down and wait for the nickel sulfate to precipitate, but I wouldn't do it!
Given that the nickel salts are highly toxic, might cause cancer on inhalation and are harmful for aquatic organisms, I wouldn't mess around with crystallisation and the remaining solution for which you would have to find a safe way for disposal.
It might be much easier to aquire nickel sulfate and set up the bath for electroplating according to the given specifications.

That is very helpful.  Thanks.  So, I was now looking at doing the Conventional Sulfamate method.  How is Nickel Sulfamate different from Nickel Sulfate?  And I am curious if this chemical is easier/safer, or more challenging/dangerous to synthesize.  Also, can I do this synthesis with Nickel Acetate, Sulfur, and/or Nickel Sulfate?

I just asked a similar question about the possibility of using NiMH as an anode for Nickel plating. Please help.

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text:
"No variables in Z-matrix" error in scan calculation of Gaussian
I asked a similar question earlier this year but I'm still stuck with a similar problem.
Basically I'm trying to implement a scan calculation using both a z-matrix and Cartesian coordinates. I have the optimized Cartesian coordinates for a nanotube and z-matrix for benzene; I would like to set up the calculation to calculate the potential energy of the system when the benzene molecule is a distance R away from the NT along the z-axis. I'm trying to scan from 2.5 to 5.5 angstroms in 0.1 A intervals.
I've read the Gaussian documentation and it seems like mixing Cartesian and z-matrix coordinates in a scan function is possible but I've tried many different inputs and I still get a "No variables in Z-matrix error."
Here is a my current input file
%mem=40GB
%nproc=4
#P BP86/6-31G scan

Scan using 1 unit nanotube with benzene at 2.5 A

0 1
[Nanotube Cartesian coordinates]

X
X 1 R1
C 2 XC 1 A1
C 2 XC 1 A1 3 60.0
C 2 XC 1 A1 4 60.0
C 2 XC 1 A1 5 60.0
C 2 XC 1 A1 6 60.0
C 2 XC 1 A1 7 60.0
X 3 R1 2 A1 1 0.0
H 3 HC 9 A1 2 180.0
H 4 HC 3 A2 2 180.0
H 5 HC 4 A2 2 180.0
H 6 HC 5 A2 2 180.0
H 7 HC 6 A2 2 180.0
H 8 HC 7 A2 2 180.0

Variables:
R1 2.5 25 0.1

Constants:
XC=1.39893
HC=1.09100
A1=90.0
A2=120.0

System in consideration

Visualization of the z-matrix
Some questions:

Is there an issue with the route section? I tried replacing "scan" with "opt=modredundant" but that introduces its own errors and its clear I don't know what I'm doing.
I suspect this is a syntax issue since the documentation for "z-matrix" and "scan" suggests it is possible to keep the nanotube coordinates frozen and just scan the benzene molecule. Is there an issue with my variables and constants section? I've looked at many sample calculations but I didn't find any that consider moving one molecule toward another. If someone can link me to relevant literature regarding (reaction coordinate calculations?) I'm more than willing to look into it; as for now this is the method my professor suggested.

hm i was never good in the whole z-matrix stuff, but what do you actually want to do here? Why are you using that dummy atom? And if I understand it correctly you are putting the Dummy atom in a distance of R1 to the dummy atom itself?

Sorry if I was unclear, I'm trying to calculate the distance (R1) between the benzene and nanotube to which the potential energy of the system is minimum; like this. I used this z-matrix as I read somewhere that the first atom in the z-matrix can't have any parameters associated with it. I originally had the first entry as X 0 0.0 0.0 R1 (where R1 is the Cartesian z-coordinate) but I could not get that to work for the same reason.

ok, do you want to do a relaxed scan or not?
z-matrix is usually relaxed (but you fixed everything in benzene and you want to have the NT fixed) and you need to use opt=z-matrix. If you want rigid scans, so only one coordinate changed, you need to use the scan keyword.

I believe I need a rigid scan but I might go ahead and try both approaches. I used the scan function but I'm still getting a "No variables in z-matrix" error. I also tried removing all the constants and replacing them with the equivalent values but my syntax apparently still doesn't request the scan of R1 from 2.5 to 5.5 in 0.1A increments.

Ok, so I guess I'm on the right way now. Trying to fix such a complex system is always very difficult so what I did was using a similar system, I got one sodium atom instead of your NT and I'm moving the benzene away.
First of all: you got two different dummy atoms in there, atom 1 and atom 9, which I guess makes some problems. Also the extra lines between the sections are a problem.
So here's the input for this, for a first try just from 2.5 in 5 0.5 A steps which at least runs without any errors:
# opt=z-matrix hf/3-21g

Title Card Required

0 2
 Na                 0.00000000    0.00000000    0.00000000
X 1 R1
C 2 XC 1 A1
C 2 XC 1 A1 3 60.0
C 2 XC 1 A1 4 60.0
C 2 XC 1 A1 5 60.0
C 2 XC 1 A1 6 60.0
C 2 XC 1 A1 7 60.0
H 3 HC 8 A2 2 180.0
H 4 HC 3 A2 2 180.0
H 5 HC 4 A2 2 180.0
H 6 HC 5 A2 2 180.0
H 7 HC 6 A2 2 180.0
H 8 HC 7 A2 2 180.0

XC=1.39893
HC=1.09100
A1=90.0
A2=120.0
R1 2.5 S 5 0.5

So here's the point. The format basically needs to be exact like this. Everything else will confuse Gaussian. You can substitute the "Na" line with your cartesian coordinates, but then the numbering in the z-matrix section changes! you have to adopt those numbers. Oh and of course if you got a more complex system than one atom you want to not only define the distance of that dummy atom but also the angles, so it's in the right position. You might also need to adept the whole matrix so the benzene has the right orientation in relation to the NT, so at least three atoms in there need to be somehow defined by atoms in the NT.
If you want to have fixed coordinates (what you labeled as constants) in the z-matrix, you can add them after a blank line just below the last section.

However, after a test run I'm not sure if it actually scanned that one
  coordinate, I just get one structure right now, I have to check how to
  do the scan correctly.

It works after adding that one 'S' in the coordinate which needs to be scanned.

So to avoid any problems substituting the Cartesian NT coordinates, can I enter the NT coordinates and just append a dummy atom at X 0.0 0.0 0.0? I forgot to mention that I constructed both molecules so that they are centered on the x-y plane and a distance R1 apart on the z-axis; I wouldn't have to worry about reorienting the z-matrix as far as I can tell. I'll try this out.

I don't think that will work, since all atoms are counted from the top to the bottom. So for your input the first hydrogen atom is oriented at atom 3, 9 and 2. But after adding in the NT those atoms will be part of the NT. I also tried it, you need to get the number up by whatever amount of atoms you add before that. You also need to add the additional angles. I also did just an optimization and it gives the same result. So I'm pretty sure right now that scanning a coordinate which involves a dummy atom doesn't work. I guess that it get's deleted after gaussian build up it's coordinates

Ok, I did the input wrong. For scanning the bond length it's 'R1 = 2.5 S 5 0.5'

I attempted this but I'm still having issues. I updated the OP for clarity. If I can't use a dummy atom as part of the Cartesian NT coordinates, I don't think there is a way to build my z-matrix for benzene so that the benzene molecules moves away from the NT in the direction of the z-axis. My previous attempt involved just using Cartesian coordinates for the entire system and manually editing the z- axis coordinates for the benzene from 2.5 to 5.5 A but I have many other NT+arene combinations to consider so that would have taken a very long time.

but, do you even need those dummy atoms? I mean yes, you want to have your benzene centeres above the "benzene unit" in the NT and twisted by 60° from what I can see in the picture. But is this the lowest energy configuration? If it is you could just use the length between one of the carbons of the NT and one of the carbons of benzene and scan this, and the benzene will still stay centered above the NT unit. You won't get exact 0.1 A steps from center to center, but that doesn't really matter.

I agree with you in that there is no evidence so far that this is the lowest energy configuration; I think my professor was just looking for a standard way to calculate the minimum potential energy as we were going to mess around with various other arene+NTs (as well as orientations) after we found a reasonable distance apart to which they have the minimum potential energy. I was also interested in finding out what the minimum NT length would be so that energy contributions from NT atoms near the ends are negligible compared to the NT atoms closest to the arenes.

This is a tedious job as a rigid scan as I am unaware of any GUI program that is able to handle a mixed style coordinate input and modification.
DVSA's answer certainly goes into the right direction, but if I'm not mistaken it should perform a (moderedundant) optimisation instead of a scan.
As a start you need to prepare your Cartesian coordinates from your nanotube and insert dummy atoms in strategic places. I am using ChemCraft because it comes with extremely convenient tools. Most importantly in the edit menu there is an entry that lets you insert a dummy atom into the centre of a selected few atoms. Furthermore, you can resymmetrize your molecule later to ensure correct placement.

I placed X (49) in the centre of C (20), C (22), C (33), C (35) and directly above X (50). In the graphic I marked the coordinates which become most important later in construction. The Cartesian coordinates of X (50) will later be substituted and become the first line of the z-matrix.
You will need a very good 3D imagination to build the z-matrix from there, or you built the complete model first and measure angles and distances from there.
My anchor atoms will be X (49), C (22) and C (35) on the nanotube and X (50) for benzene. To get the alignment right, I am hard-coding the placement of the first carbon into the z-matrix, too. It will be directly above C (35).
It is important, that the geometry description block has no blank lines in it. When using the scan keyword all variables become constants. You will therefore need another block after the geometry specification without any blank lines. If you need to make comments, use ! at the beginning of the line.
The following input will probably do what you want. I have used pm6 as a semiempirical level of theory, because it needed to be quick.

#P pm6
scf(xqc)
scan

benzene nanotube scan

0 1
! 49 Fragment C48X nanotube
C        0.000000000      2.427540000      3.559669000
C        1.196378000      2.072187000      2.878637000
C        2.102311000      1.213770000      3.559669000
C        2.392756000      0.000000000      2.878637000
C        2.102311000     -1.213770000      3.559669000
C        1.196378000     -2.072187000      2.878637000
C        0.000000000     -2.427540000      3.559669000
C       -1.196378000     -2.072187000      2.878637000
C       -2.102311000     -1.213770000      3.559669000
C       -2.392756000      0.000000000      2.878637000
C       -2.102311000      1.213770000      3.559669000
C       -1.196378000      2.072187000      2.878637000
C       -1.196582000      2.072541000      1.430870000
C        0.000000000      2.404111000      0.720134000
C        1.196582000      2.072541000      1.430870000
C        2.082021000      1.202055000      0.720134000
C        2.393165000      0.000000000      1.430870000
C        2.082021000     -1.202055000      0.720134000
C        1.196582000     -2.072541000      1.430870000
C        0.000000000     -2.404111000      0.720134000
C       -1.196582000     -2.072541000      1.430870000
C       -2.082021000     -1.202055000      0.720134000
C       -2.393165000      0.000000000      1.430870000
C       -2.082021000      1.202055000      0.720134000
C       -2.082021000      1.202055000     -0.720134000
C       -1.196582000      2.072541000     -1.430870000
C        0.000000000      2.404111000     -0.720134000
C        1.196582000      2.072541000     -1.430870000
C        2.082021000      1.202055000     -0.720134000
C        2.393165000      0.000000000     -1.430870000
C        2.082021000     -1.202055000     -0.720134000
C        1.196582000     -2.072541000     -1.430870000
C        0.000000000     -2.404111000     -0.720134000
C       -1.196582000     -2.072541000     -1.430870000
C       -2.082021000     -1.202055000     -0.720134000
C       -2.393165000      0.000000000     -1.430870000
C       -2.392756000      0.000000000     -2.878637000
C       -2.102311000      1.213770000     -3.559669000
C       -1.196378000      2.072187000     -2.878637000
C        0.000000000      2.427540000     -3.559669000
C        1.196378000      2.072187000     -2.878637000
C        2.102311000      1.213770000     -3.559669000
C        2.392756000      0.000000000     -2.878637000
C        2.102311000     -1.213770000     -3.559669000
C        1.196378000     -2.072187000     -2.878637000
C        0.000000000     -2.427540000     -3.559669000
C       -1.196378000     -2.072187000     -2.878637000
C       -2.102311000     -1.213770000     -3.559669000
X       -1.041010500     -1.803083000      0.000000000
! 13 Fragment C6H6X benzene
X        49   scan        35   90.000      22   90.000
C        50   xcdist      49   90.000      35   00.000
C        50   xcdist      51   sixty       49   ninety
C        50   xcdist      52   sixty       49   ninety
C        50   xcdist      53   sixty       49   ninety
C        50   xcdist      54   sixty       49   ninety
C        50   xcdist      55   sixty       49   ninety
H        51   chdist      52  onetwen      53  oneeigh
H        52   chdist      53  onetwen      54  oneeigh
H        53   chdist      54  onetwen      55  oneeigh
H        54   chdist      55  onetwen      56  oneeigh
H        55   chdist      56  onetwen      51  oneeigh
H        56   chdist      51  onetwen      52  oneeigh
! 12 Fragment H12 nanotube cappings
H        0.000000000      2.646957000      4.621016000
H        2.292332000      1.323479000      4.621016000
H        2.292332000     -1.323479000      4.621016000
H        0.000000000     -2.646957000      4.621016000
H       -2.292332000     -1.323479000      4.621016000
H       -2.292332000      1.323479000      4.621016000
H       -2.292332000      1.323479000     -4.621016000
H        0.000000000      2.646957000     -4.621016000
H        2.292332000      1.323479000     -4.621016000
H        2.292332000     -1.323479000     -4.621016000
H        0.000000000     -2.646957000     -4.621016000
H       -2.292332000     -1.323479000     -4.621016000

scan           2.500      25    0.100
xcdist         1.396
chdist         1.086
sixty         60.000
ninety        90.000
onetwen      120.000
oneeigh      180.000

In my first attempt I forgot the hydrogens so I added them later on. Note that if you change the lines within the Cartesian blocks it does not matter, als long as you don't change the atom numbers of the anchors.
Here is the scan animation of my first attempt (without the hydrogens).

Note that the dummy atoms will be gone in the course of the calculation. Gaussian just needs them to unambiguously define the redundant coordinates.
If you want to run the calculation yourself, you should obtain this table:

 Summary of the potential surface scan:
   N     scan         SCF
 ----  ---------  -----------
    1     2.5000      1.14872
    2     2.6000      1.12058
    3     2.7000      1.10061
    4     2.8000      1.08668
    5     2.9000      1.07712
    6     3.0000      1.07067
    7     3.1000      1.06641
    8     3.2000      1.06365
    9     3.3000      1.06192
   10     3.4000      1.06088
   11     3.5000      1.06030
   12     3.6000      1.06001
   13     3.7000      1.05990
   14     3.8000      1.05990
   15     3.9000      1.05996
   16     4.0000      1.06004
   17     4.1000      1.06013
   18     4.2000      1.06022
   19     4.3000      1.06030
   20     4.4000      1.06037
   21     4.5000      1.06043
   22     4.6000      1.06048
   23     4.7000      1.06052
   24     4.8000      1.06055
   25     4.9000      1.06057
   26     5.0000      1.06059
 ----  ---------  -----------

This whole project would probably be easier solved generating the different geometries by hand.

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text:
Equation or formula for calculating molecular orbital energies
Is there an equation, similar to the Rydberg formula, for calculating the energies of molecular orbitals?
For example, with the Rydberg formula, I can work out that the energy of a $1s$ orbital in hydrogen is $-13.6\ \mathrm{eV}$. Is there any way that I can calculate (or approximate) the energies of the bonding and anti-bonding orbitals in $\ce{H2}$?

It doesn't exist. Hydrogen atom can be solved exactly, multielectron systems can't.

Ok, thanks! I believe multi electron systems can be approximated by calculating Zeff (for example with Slater's rules) - are there any things like this for the MOs?

Well, for conjugated systems you can use the Hückel method for getting relative energies, but that's about it. Since this was tagged as computational chemistry: it's very easy to use computational chemistry to predict MO energies.

How would you go about doing it computationally? Are there any free software packages which are able to do it? Thank you.

I recommend Psi4. You can use Avogadro to help write input files and look at results.

For future googlers, the Avogadro link is broken, but https://avogadro.cc/ will get you there just fine.


text:
Why isn't HCl called hydrogen monochloride?
Since HCl is a covalent compound, then it would be expected to be called hydrogen ***mono***chloride following the rules for naming. Instead it is named as if it were an ionic compound, without any prefixes. Why?

Why not call it monohydrogen monochloride? If it's unambiguous you leave away anything that could describe "mono-".


text:
Why is bismuthine (BiH3, or bismuth trihydride) unstable if bismuth is considered to be the most stable heavy element?
Bismuth-209 is considered to be the most stable heavy element, though it is weakly radioactive.
Given that, why does bismuthine ($\ce{BiH3}$, or bismuth trihydride) have a half-life of only 20 minutes and is the least stable hydride in its group?

The reactivity of bismuthane is not related to the stability of isotopes. Compare the reactivity of $\ce{CH4}$ with that of $\ce{GeH4}$.

I'm voting to close this question as off-topic because it relates stabilities of isotopes with reactivities of hydrogen compounds.

I object. A confusion between two ideas, one of which belongs to chemistry, is by extension related to chemistry as well.

I think it should stay open. While the basis of the question is wrong, I could imagine other students having a similar question and those who find the SE, could learn from it.

By putting words in incorrect order it is quite easy to arrive at nonsense. Bismuth is by far not the most stable heavy element; indeed, it is not particularly stable at all. Instead, it is the most heavy stable element (if we disregard its radioactivity, that is). Think of these two definitions for a while. Think how different they are.
Moreover, this is about nuclear stability, which has absolutely nothing to do with chemical stability of the element, which in turn (besides not being well defined) has nothing to do whatsoever with stability of its hydride.
I sincerely recommend to all chemistry students to abandon using the word "stability" altogether, for it seems to cause a great deal of confusion every single time someone uses it.

$\ce{BiH3}$ is unstable due to the large size of central atom & therefore its tendency to form stable covalent bond with small hydrogen atom decreases, as a result the bond strength decreases


text:
How much sodium fluoride is needed to precipitate a calcium fluoride solution?
The question was:

If you are given a $\pu{1 L}$ solution of $\ce{CaF_2}$ where there is $\pu{2.0E5 M}$ $\ce{Ca^{2+}}$. How many milliliters of $\pu{0.1 M}$ $\ce{NaF}$ would you need to precipitate $\ce{CaF_2}$? The solubility product of $\ce{CaF2}$ is $1.7\times10^{-10}$ .

The answer given is

$\pu{956 ml}$

What I attempted:
I thought that if $x$ Litres of  $\ce{NaF}$ is necessary to precipitate $\ce{CaF_2}$; then after the mix, the ionic product($K_{ip}$) of $\ce{CaF_2}$ will exceed the solubity product($K_{sp}$). $$K_\mathrm {ip} > K_\mathrm {sp}$$
$$\ce{[Ca^2+][F-]^2} > K_\mathrm{sp}$$  and from this equation I tried to solve the value of $x$.
Here's what I tried in details:-
step 1) Write down the dissociation  reaction of $\ce{CaF_2}$
$$\ce{CaF_2 <=> Ca^{2+} + 2F-}$$
and from the solubility product formula $K_{sp} = \ce{[Ca^2+][F-]^2}$ I calculated the concentration of $\ce{[F-]}$ which I got was $2.91\times10^{-3}$.
step 2) Then I again calculated the concentration of $\ce{Ca^2+}$ and $\ce{F-}$ after adding  $x$ Litres, using the formula $\ce{S_1\times V_1 = S_2\times V_2}$, because the volume now was not $1$ Litre, but $(1+x)$ Litres.
I got $$\ce{[Ca^2+] = \frac{2\times 10^{-5}}{(1+x)}}M$$ $$\ce{[F-] = \left( \frac{0.1x}{1+x} + \frac{2.9\times 10^{-3}}{1+x}\right)}M$$
step 3) Then as I stated before, the condition of precipitation would be as follows
$$K_\mathrm {ip} > K_\mathrm {sp}$$
$$\ce{[Ca^2+][F-]^2} > K_\mathrm {sp}$$
So, from the equation above, I calculated the value of $x$ which I got was $1.8$ Litres (approximately) which was nowhere near the given answer in the book. How to solve this problem?

The question is weird. $\ce{CaF_2}$ is virtually insoluble. If the solution was a saturated solution made by dissolving  $\ce{CaF_2}$ then the $\ce{[F^-] = 2[Ca^{2+}]}$. Also if the solution were saturated then adding any $\ce{F^-}$ would ppt $\ce{CaF2}$. Finally you'd never be able to add enough $\ce{F^-}$ to ppt all the $\ce{Ca^{2+}}$.

But no where in the question they said it was a saturated solution, did they?

As I said the problem statement is weird. The problem as you stated it has the phrase a "solution of $\ce{CaF2}$" which is odd. It sort of makes sense if you assume a saturated solution with a non-stoichometric quantities. But then it really isn't just a solution of $\ce{CaF2}$ is it?!?

Also how can it be a solution of $\ce{CaF2}$ if there is no $\ce{F^-}$ in solution? So there is no way to determine the initial $[\ce{F^-}]$ unless it is $\sqrt{\frac{\ce{K_{sp}}}{[Ca^{2+}]}}$.

Like you said, if the solution was saturated then adding any amount of $\ce{F-}$ will result in precipitating the $\ce{CaF_2}$, so shouldn't we assume that the solution wouldn't be saturated and calculate the initial concentration of $\ce{[F-] = 2[Ca^{2+}]}$ like @Zhe said in his answer? Though doing that also didn't help.

So back to the point I wass trying to make, the question doesn't really make sense. "How many mililitres of $0.1$ M $\ce{NaF}$ would you need to precipitate $\ce{CaF_2}$?" You need a lot less than 956 ml to precipitate some $\ce{CaF2}$, and you could never add enough $\ce{F^-}$ to ppt all the $\ce{CaF_2}$.

" You need a lot less than 956 ml to precipitate some $\ce{CaF2}$, and you could never add enough $\ce{F^-}$ to ppt all the $\ce{CaF_2}$." Could you please explain how you deduced that? @MaxW

First the question doesn't really make sense. $\ce{CaF_2}$ is virtually insoluble, so a "solution of $\ce{CaF_2}$" is a really odd phrase.
Skipping this nit-picking, the reactants are 1 L of a $2.0 \times 10^{-5}$ M $\ce{Ca^{2+}}$ and a $0.1$ M $\ce{F^-}$ solution  and $\ce{CaF_2}$ is to be precipitated.
This is a sloppy use of significant figures since volume of 1L and the molarity of the $\ce{F^-}$ solution are only given to one significant figure.
Ignoring the significant figure problem too, let's work backwards from the given answer of 946 ml.
For the initial solution we can solve for the moles of $\ce{Ca^{2+}}$ present.
$\text{moles }\ce{Ca^{2+}} = 1 \rm{L} * 2.0 \times 10^{-5}\frac{moles}{L} = 2.0 \times 10^{-5}\text{ moles}$
Now if we add 946 ml of 0.1 M NaF, then the moles of $\ce{F^-}$ in solution is 0.0946 moles which vastly more than the calcium. So we can assume the precipitated $\ce{CaF2}$ has a negligible effect on the $[\ce{F^-}]$, and calculate the final concentration of $[\ce{F^-}]$ as:
$\ce{[F^-]}_{\text{final}} = \dfrac{0.0946}{1.946} = 0.0486 \text{M}$
Now from the $\rm{K}_{\text{sp}}$ we have
$\ce{[Ca^{2+}]}_{\text{final}} = \dfrac{\rm{K}_{\text{sp}}} {[\ce{F^-}]^2}= \dfrac{1.7 \times 10^{-10}}{0.0486^2} = 7.20 \times 10^{-8}$
$\text{moles }\ce{Ca^{2+}} \text{ final}= (7.20 \times 10^{-8})(1.946) = 1.40 \times 10^{-7}$
The initial moles of $\ce{Ca^{2+}}$ is $2.0 \times 10^{-5}$ which is +/- $0.05 \times 10^{-5}$. Thus so far as I am concerned the solution is a "magic number" and there is no way to get from the problem statement to that value. In other words why not $2.00 \times 10^{-7}$ or $3.00 \times 10^{-8}$ for the final $\ce{Ca^{2+}}$ concentration?!?
Thus you have to assume that you're going to precipitate some fraction of the $\ce{Ca^{2+}}$. So are you going to precipitate 99%, 99.9%, 99.95% or what? There just isn't anything in the problem statement on which you can pin a value.
Just out of curiosity the % $\ce{Ca^{2+}}$ recovered is:
% $\ce{Ca^{2+}}$ recovered = $(1-\dfrac{1.40 \times 10^{-7}}{2\times 10^{-5}}) \times 100\% = 99.3 \%$

Note that we can also assume that the original solution is saturated in respect to $\ce{CaF2}$. As shown in OP's post, this leads to a concentration of $2.91 \times 10^{-3} \rm{M}$ for $\ce{F^-}$ and given the 1L volume, $2.91 \times 10^{-3} \rm{moles}$ of $\ce{F^-}$. Adding that to the number of moles from the 946 ml of $\ce{F^-}$ solution gives a slightly different value for the total moles of $\ce{F^-}$:
total moles of $\ce{F^-}$ = $0.0946 + 0.00291 = 0.0975$
$\ce{[F^-]}_{\text{final}} = \dfrac{0.0975}{1.946} = 0.0501 \text{M}$
So the magic value for the textbook solution is to assume that: $\ce{[F^-]}_{\text{final}}= 0.0500$
$\ce{[Ca^{2+}]}_{\text{final}} = \dfrac{\rm{K}_{\text{sp}}} {[\ce{F^-}]^2}= \dfrac{1.7 \times 10^{-10}}{0.0501^2} = 6.77 \times 10^{-8}$
$\text{moles }\ce{Ca^{2+}} \text{ final}= (6.77 \times 10^{-8})(1.946) = 1.32 \times 10^{-7}$
% $\ce{Ca^{2+}}$ recovered = $(1-\dfrac{1.32 \times 10^{-7}}{2\times 10^{-5}}) \times 100\% = 99.34 \%$
So we get a slightly different value, but it still seems like a magic number of some sort.

Now if I were to write the problem I would have stated:

You are given $1.00$ Litre of solution which is saturated with  $\ce{CaF_2}$ and has a $\ce{Ca^{2+}}$ concentration of $2.00 \times 10^{-5}$ M. How many mililitres of $0.100$ M $\ce{NaF}$ would you need to recover 99% of the $\ce{Ca^{2+}}$ as a precipitate of $\ce{CaF_2}$? The solubility product of $\ce{CaF_2}$ is $1.7\times10^{-10}$ .

Curious I played with this some more using the assumption of the $\ce{F^-}$ saturated starting solution. Obviously if you add vast amounts of $\ce{F^-}$ solution then any $\ce{CaF_2}$ which formed would dissolve agin. So there is some amount of $\ce{F^-}$ solution which precipitates the maximum amount of $\ce{CaF_2}$.
The max % $\ce{Ca^{2+}}$ recovery = 99.44% with 1913 ml of $\ce{F^-}$ solution. To +/- 0.01% any volume between about 1400 ml and 2500 ml would yield effectively the same result.

I think your $2.91 \cdot 10^{-3}$ value for fluoride concentration doesn't make sense. You are providing the concentration whereby the solution is immediately saturated. Instead, I think you just want double $2.0\cdot 10^{-5}$ for stoichiometric reasons, since both the fluoride and calcium presumably come from dissolution of calcium fluoride.

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text:
Comparison of alkali and alkaline earth melting points - MO theory
If you compare the melting points of alkali and alkaline earth metals, you see that the latter have higher melting points. However, if you look at MO theory, then alkaline earth metals should have quite a considerable proportion of the antibonding part of the bands occupied (even if there is an overlap with the p-band). So, why should the bond in alkaline earth metals be stronger?
(I realise that according to the electron sea model you just say that alkaline earth metals contribute two electrons, not just one, wich solves the problem. MO theory should also be applicable, so what am I overlooking?)

From "The Nature of the Interatomic Forces in Metals" by Linus Pauling

This qualitative description of the interactions
  in the metal is compatible with quantum mechanical treatments which have been given the
  of
  problem, and it leads to an understanding
  such properties as the ratio of about 1. 5 of
  crystal energy of alkali metals to bond energy of
  their diatomic molecules (the increase being the
  contribution of the resonance energy), and the
  increase in interatomic distance by about 15
  percent from the diatomic molecule to the
  crystal.
  The alkaline earth metals, by assuming the
  configuration nsnp, are able to form twice as
  many bonds as the alkalis. Similarly the suc-
  ceeding elements in the periodic table can form
  bonds in increasing number.

You are right on the fact that alkaline earth metals should not bond if we only considered their highest atomic energy levels. The dissociation energy for diatomic clusters of alkaline earth metals is low.
However, the disassociation energy gets higher for tetra-atomic as shown in this computational study: http://www.sciencedirect.com/science/article/pii/0039602885902328
Now, the reason why alkaline earth metals have higher melting points to be due to the p orbitals. The previous study relates p hybridization with bond strength. And so does this study: http://scitation.aip.org/content/aip/journal/jcp/77/8/10.1063/1.444313
Both journals state that p orbitals are involved in bonding. And the more involved the p orbitals in bonding, the stronger the bond. This trend is shown for alkaline earth metals. Beryllium, which has the highest melting point, also has more p orbital character. Also, take a look at the density of states of alkali metals and alkaline earth metals:
http://www.colorado.edu/engineering/MCEN/MCEN5024/
You can see that not all p orbitals are entirely above the s orbitals.
So, to answer your question. The reason why alkaline earth metals have higher melting points is due to p orbitals. As Pauli states in his review, the np orbitals "hybridize" with the s orbitals. This allows for more bonds to form.
I think that p orbitals participate in bonding in alkali metals. But I couldn't find information that compares p orbital "hybridization" of alkali and alkaline earth metals directly.

Thanks for the answer. Actually, I believe quite a number of chemistry textbooks introduce metallic bonds by just expanding normal LCAO MO theory (which says you get as many MOs as you have contributing AOs) to the (almost) infinite metal crystal. And they also talk about antibonding orbitals there. (see http://xbeams.chem.yale.edu/~batista/vaa/band.html for an english example of that description)  -- So I was wondering where the problem comes from and whether this formal derivation from MO theory was missing some essential points.

@Andrea My mistake. I thought of that as band theory. I realize that it also falls as a "subcategory" of MO theory. I'll go read some journals and post my findings.


text:
What is the difference between Sodium Nitrate and NaNH3
I have almost no experience in chemistry so please excuse my ignorance. I am looking to purchase some Sodium Nitrate but the vendor I am dealing with has the chemical formula NaNH3 listed beside the Sodium Nitrate they sell. Just wondering what the difference is?

are you sure it's not NaNO3?

Maybe NaNH2 sodamide not NaNH3? But a simple typo as @costrom suggests NaNO3 makes more sense

Heres this listing I have. https://www.glengarrycheesemaking.on.ca/rennet-and-ingredients/ingredients/sodium-nitrate-en/sodium-nitrate-1-kg.html

So If I am understanding correctly, there is no chemical with the formula NaNH3 ? And its more than likely a typo.

Correct there is no such chemical as $\ce{NaNH3}$. A typo on the webpage is the only thing that makes sense. Since this is from website for making cheese, I doubt that there is much chemistry knowledge there.

Yes it's a typo, that is repeated a few times in the link. It would suggest the website author/updater is not of a chemical persuasion! NaNO3 is sodium nitrate.

$\ce{NaNO3}$ (which we established is the de-typo’d version of what your source calls $\ce{NaNH3}$) is simply the chemical formula of sodium nitrate.

It looks like this is just a typo on the part of the web developer. There is no such thing as $\ce{NaNH3}$. They probably meant to type $\ce{NaNO3}$.


text:
Why are weak acids not considered good leaving groups?
So today in my first year lecture my professor said that weak bases are good leaving groups. I understand why they are good leaving groups in comparison to unstable strong bases but why are weak acids not included? Acids are electron pair acceptors, shouldn't they be better able to take on the donated electron pair? Furthermore, when we say weak bases, weak in comparison to what? To be substrate the solvent or to other potential leaving groups?

You should think Brønsted acids/bases rather than Lewis acids/bases in this context.

I figured, that's the way that it would work according to the weak bases good leaving groups rule, but my instinct was Lewis because of the electrons moving toward the leaving group. Why is Lewis incorrect here?

Well, the thing that’s leaving will always be some kind of base — it’s taking its electron pair with it. The only thing you can still decide is whether it is the base of a weak or a strong Brønsted acid. The Lewis theory would require you to look at the basic properties only (The Lewis acid being the electrophile that is attacked nucleophilicly) so you can’t argue with acidic strength.

My original thought was that it should be a weak acid because they can handle the extra electrons as they are the electron acceptors. However the thought that the base is taking them back also makes sense.

Leaving group ability is tied to the stability of the leaving group. Acidity is directly related to the stability of the conjugate base. If you're a strong acid, the conjugate base is relatively stable in solution. Less so, if you're a weak acid. So the stability of the conjugate base is directly proportional to both acidity and leaving group ability.
Edit:
$$\ce{HA->H+ + A-}$$
$\ce{A-}$ is the species we're evaluating as the leaving group. Basically, it's a good leaving group if it's stabilized. If that's the case, the acid dissociation equilibrium also shifts toward products, making $\ce{HA}$ a stronger acid. You should be careful to differentiate between $\ce{HA}$ and $\ce{A-}$ when you're thinking about these concepts.

So, essentially, acids are not good leaving groups because they have unstable conjugate bases, but bases are good leaving groups because they do have stable conjugate acids?

I'm not sure how you got to that conclusion from my answer, so I guess my answer was bad... Let me edit it...

Oh!!! Okay thank you yes you were right I was thinking about HA and not A- I think I understand now!


text:
Fehling's test arrow pushing mechanism
I am currently attempting to generate an arrow pushing mechanism for the oxidation of glucose to gluconic acid using Fehling's solution. My original thought was to go the route of alcohol oxidation using acidified dichromate where the alcohol attacks in and loses the H using a E2 mechanism.
In the aldehyde oxidation, I would have a hydroxide ion attack prior to the H bond collapse leaving an extra electron on the Cu from the $\ce{C-\mathbf{O-Cu}}$ bond resulting in reduction of $\ce{Cu^2+ -> Cu+}$. Which would satisfy the observed empirical data. After an attempt I googled the mechanism finding very little real information.
A post on this very site suggested the mechanism proceeds via enolate formation followed by single electron transfer with $\ce{Cu^2+}$. Unfortunately, there is no elaboration on this or a source, I also can't find a source online which corroborates this mechanism. Does anyone have a good source for this?

The exact nature of the mechanism is unknown
Be aware that some organometallic chemists consider arrow-pushing mechanisms inadequate/inappropriate for a variety of reasons (including the depiction of single electron transfer). For example, the arrow-pushing mechanisms we draw with alkyllithiums are incomplete in that they do not account for the oligomeric nature of the reagent.[2]
The elements which are known:

The aldehyde or α-hydroxyketone must be enolizable. The enolate tautomerization of reducing sugars is discussed in McMurray[3]. This is common to the Benedict’s test mechanism as well.
The rate of enolization determines the rate of oxidation J. Am. Chem. Soc. 1970, 92, 537
Cu(II) has long been known to mediate oxidative coupling of enolates by a radical mechanism J. Am. Chem. Soc. 1971, 93, 4605. Mechanistic studies point to a metal-chelated single-electron-transfer process.5
The active species is the metastable [Cu(tartrate)2]6- ( Eur. J. Inorg. Chem. 2016, 1798).

Drawing one plausible mechanism

Convince yourself that the Cu(II) complex  is a d9 17-electon complex.  That suggests you could plausibly draw an inner-sphere or outer-sphere mechanism.
For the inner-sphere mechanism: coordinate the aldehyde, and generate the enolate (considering C-vs-O coordination and necessity of ligand-slip).
Form the hydrate of the C-bound enolate, then homolyze the bond to form an organic radical and Cu(I).
A second molecule of Cu(II) oxidizes the radical to the cation, which gets quenched the elimination of H from the hydrate to generate the enol form of the carboxylic acid.

For the basics of an outer-sphere single electron transfer see Mark Vander Wal’s Basics of Electron Transfer
[2]: Advanced Organic Chemistry: Part A: Structure and Mechanisms, Carey & Sundberg, p. 413.
[3]: Organic Chemistry: With Biological Applications,  McMurry, p. 758

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text:
When an expansion of a system does work on its surroundings, is the sign for work negative?
This question comes to me from trying to explain the answer $1.67\ \mathrm{L}$ of this question:
A cylinder with a moving piston expands from an initial volume of $0.250\ \mathrm{L}$ against an external pressure of $2\ \mathrm{atm}$. The expansion does $288\ \mathrm{J}$ of work on the surroundings. What is the final volume?

Work done on the surroundings is the force exerted on the surroundings dotted with the displacement of the surroundings.  Are the force exerted on the surroundings and displacement of the surroundings in the same direction in this problem, or in opposite directions?

I can't tell. How does one tell that?

The work being done is in the same direction , I would imagine. The piston only has in place to move.

Related: Principles of Thermodynamics (closed)

It means the change in internal energy is negative, so you are on the right track. Because the system is doing work on the surroundings, that $288\ \mathrm{J}$ is a loss for the system and a gain for the surroundings.
One way to intuitively work your way through this is to think about what the signs on the terms must be in order for the problem to make sense.
You know the final volume is greater than the initial (it's an expansion of the system into the surroundings), so $\Delta\mathrm{V}$ has to  be positive. You know that $w = -\mathrm{p_{ext}}\Delta\mathrm{V}$ is negative, because $\mathrm{p_{ext}}$ is positive. Thus, the sign of $\Delta\mathrm{E}$ is negative.
You want to solve
$$\Delta\mathrm{E} = -\mathrm{p_{ext}}\Delta\mathrm{V}$$
for $\Delta\mathrm{V}$ and add that to your initial volume of $0.25\ \mathrm{L}$.
Solving for $\Delta\mathrm{V}$:
$$-288\,\mathrm{J} = -2\ \mathrm{atm}\times\Delta\mathrm{V}$$
$$\Delta\mathrm{V} = 144\ {\mathrm{J\over atm}}$$
and then converting units
$$\Delta\mathrm{V} =
\left(144\ {\mathrm{J\over atm}}\right)\cdot
\left({\mathrm{L\cdot atm}\over 101.325\ \mathrm{J}}\right) =
1.42\ \mathrm{L}$$
we now solve for the final volume as (spoiler alert):

$$\mathrm{V_{final}} = 0.25\ \mathrm{L} + 1.42\ \mathrm{L} = 1.67\ \mathrm{L}$$


text:
formaldehyde strong odor neutralization
One of our customers used a very high dose of formaldehyde 37% solution (60 liter formaldehyde  in 40 liter water) for disinfection in  his chicken farm (saloon). Now he says the odor of formaldehyde is very strong and can`t start breeding  season.  How can I help her?
Can he  neutralize  formaldehyde odor with some chemical treatment? Maybe he can use sodium hypochlorite (bleach). Do you think chlorine can oxidize formaldehyde?

Just use water and a fan. Bleach will react exothermically and may start a fire.

No, you don't want to add more water unless you can wash it away. Dampness will "hold" the formaldehyde. This seems to be inside, so you want to use a fan and increase the temperature to dry the place out.

Many tanks I suggest her washing Evey think with water ,and increasing temperature ,when I have report inform you @quantum American

Ammonia water, AKA ammonium hydroxide ($\ce{NH4OH}$), reacts with formaldehyde to form hexamine, a comparatively innocuous solid (though it is flammable when dry).
Safer might be Neutralization of Formaldehyde Gas by Ammonium Bicarbonate and Ammonium Carbonate, which would greatly reduce the flammability of the product.


text:
How to correctly write "7 apples" according to the international system of units (SI)
According to international system of units (SI), we can write "7 kg of apples" to refer to the mass of these apples.
However, if we want refer to the amount of apples, that is, the number of entities, the unit should be the mole. So, what's the correct way to denote "7 apples" in accordance with the SI convention?
Additionally, is it correct (according to SI) to say "7 atoms of hydrogen"? Or must we use mole?

Please have a look at the definition of the mole in the IUPAC Gold  Book.

@KlausWarzecha: done, but not solved the doubt.

Could you [edit] to reflect why the answers didn't clear your confusion? I disagree with the close voters that this is unclear, but you need to be more specific if you believe we didn't sufficiently answer you.

@M.A.R: Answers are very acurate, in particular the one from Loong. I´m waitibg only one expansion or new answer that includes the concept of mole to accept and close the issue.

In SI the unit for kilogram is Kg not kg. And , contrary to a spread habit in organic chemistry, millilitre is ml, not Ml. Just to be pedantic

@Alchimista: according wiki and others, kilo is k, lowercase

@pasaba por aqui.. opps! I see use of capitals starts just at M . Ok I let it here because the part relevant to volumes might be interesting in SE Chemistry. :)

In accordance with the International System of Units (SI) [Brochure in English, 8th edition, 2006; updated in 2014] and the corresponding International System of Quantities (ISQ) [ISO/IEC 80000 Quantities and units (14 parts)], you can define a suitable new quantity, for example with the quantity name “number of apples” and the quantity symbol “$N_\text{apples}$”.
The number of apples $N_\text{apples}$ is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):
$$\dim N_\text{apples} = 1$$
A quantity of dimension one acquires the unit one, (symbol: $1$); i.e. the coherent SI unit for the number of apples is the unit one.
Generally, the unit one is an SI derived unit; for example, the derived SI unit for friction factor is newton per newton equal to one, (symbol: $N/N = 1$). However, the unit one for counting numbers, e.g. number of protons in an atom or number of apples, is considered as a base quantity because it cannot be expressed in terms of any other base quantities. Hence, in this case, the unit one is usually considered as a base unit, although the CGPM has not yet adopted it as an SI base unit.
The name and symbol of the measurement unit one are generally not indicated. Therefore, you may write: “The number of apples is $N_\text{apples}=7$.”
The unit one or its symbol $1$ may not be combined with SI prefixes. For example, if you have 2000 apples, you must not write “$N_\text{apples}=2\ \mathrm k$” for $N_\text{apples}=2000$. (And by the way, when you see something like “10K reputation” mentioned on any stackexchange site, you are looking at at least three nonconformities at the same time.)
Any attachment to a unit symbol as a means of giving information about the special nature of the quantity or context of measurement under consideration is not permitted. Expressions for units shall contain nothing else than unit symbols and mathematical symbols. Therefore, write

“the maximum electric potential difference is $U_\text{max}=1000\ \mathrm V$”, not “$U=1000\ \mathrm V_\text{max}$”
“the gauge pressure is $p_\mathrm e=0.5\ \text{bar}$”, not “$p=0.5\ \text{bar(g)}$”
“the electric power is $P_\text{el}=1300\ \mathrm{MW}$”, not “$P=1300\ \mathrm{MW_{el}}$”
“the water content is $170\ \mathrm{g/l}$”, not “$170\ \mathrm{g\ \ce{H2O}/l}$”

and also

“the number of apples is $N_\text{apples}=7$”, not “$N=7\ \text{apples}$”

The OP wanted SI units so this seems correct, but rationally any paper will be a mixture of "plain" language and scientific language. The phrase “the number of apples is $\rm{N}_\text{apples} = 7$” as compared to “7 apples” seems to be overly complex, and far less effective for good communication.

@Loong: Excellent answer. Could you add any reference to the "mole"? It seems related to the subject. If my understanding of the answer is correct, when finally "unit one" was accepted as base unit, we will have two different base units for same concept, the "unit one" and the mole?

@pasaba_por_aqui - This has nothing to do with the unit moles. You use moles to count atoms and molecules, not apples.

@MaxW: the obvious question is now: why I need a different unit to count atoms than the one to count apples? It is not valid say "because they are smaller" or "because there are more" ;-) Recall we have "yotta" for 10**24

Just get over it. You can also count apples by the dozen and dozen isn't a SI unit.

10K reputation is wrong for only three reasons?

Since the number of particles/molecules in a mole of one chemical is the same as the number in a mole of a different chemical, regardless of the volume, mass, etc, using moles as units makes for simple ratios in equations.

@pasabaporaqui If you have 12 apples, you could say that you have 19.92646848(26) yoctomoles of apples, but I rather think "12 apples" is clearer.

Mole is just a scale factor
I find this description very intuitive:

A mole is the amount of pure substance containing the same number of
chemical units as there are atoms in exactly 12 grams of carbon-12
(i.e., 6.023 X 1023)

I think this should clear out the main part of your confusion. To go into the side questions:
It is definitely acceptable to talk about elementary particles without using mole
You will not often say there are XXX patricles in this jar, because you would need to use huge numbers. But this is not wrong.
A typical example where you would not use the mole scale factors because it would not be convenient:

A typical H2O molecule is made up of two hydrogen atoms and one oxygen
atom.

It is allowed, but inconvenient to talk about bigger things while using mole
If one were to choose apple, or stars as the relevant elementary entity, one could correctly say:

There are XXX mole apples globally.

However, as XXX would be inconveniently small, (and the expression only understood by a very limited audience) there is no reason reason to use mole.

The existing answer already specifies how to define and use a unit according to SI standards. For apples this may not be that neccesary (as 1 apple is a simple enough unit), but if you were to use it in formulas, or if you needed to define something more complicated (like apples that are more yellow than red and more red than green) please follow the given advice.

The first statement says "mole is just a scale factor". According to my information, the scale factors in SI are the prefixes, but mole is not a prefix, is a base unit. Moreover, all prefixes are powers of ten.

@pasabaporaqui You're correct that SI does not define it as a scale factor in the same way as the prefixes, but you can find the definition in the link in Loong's answer, SI's definition does effectively make it a scale factor, and I agree with this answer, it's easiest to think of it as such.

@hvd: It is true that define mole as an scale factor made easier to understand and tech it. However, SI defines it as base unit of a physical property called "amount of substance". That means it is a property of the physical world, like it is the mass, the size, ... . Avogadro constant is the way to convert from/to this unit to unit one. For these reasons, I disagree with the phrase "mole is just a scale factor".

@pasabaporaqui Since you're going full pedant, the mole is a base unit for "amount of substance", which is pedantically distinct from the count number of things. Certainly you can convert a given number of carbon atoms into moles of carbon, but to do so you need a non-SI-defined conversion factor (Avogadro's number), just like you can convert a volume of water into grams using a conversion factor (the density). -- While the count/moles conversion factor is the same for most everything (whereas the density differs widely), to the pedant the distinction still stands.

@R.M.: do you mean SI is pedantic? I'm sure they have good reasons to define "mole" as a base unit instead of use unit one.

@pasabaporaqui SI is not necessarily pedantic, it's just that the point of view you are taking is a very pedantic one, which necessitates distinguishing between "amount of" and "count". The reason mole is a base unit is a practical one, because Avogadro's number isn't exactly known. It's the same reason the ampere is a base unit and not derived: an ampere could be defined as a certain number of electrons per second, but it's not, because that definition was hard to realize experimentally. Likewise, a mole can't be realized as a count (yet).

The mole is defined for "elementary entities" (atoms, ions, molecules, clusters, ...), not for macroscopic objects. Therefore, 7 apples are 7 apples.

Thus, we need define a new SI base unit for microscopic objects (the mole) but not for macroscopic? It is not consistent. If the objective was allow large amount, Mega, Tera, preffixes was enough.

I'm not really sure what confuses you about this, but the amount of things just doesn't have an unit. It's just an amount. We got 1 thing, 5 things, 5 million things or 10^152638 things. There's also no need for a prefix.
Besides that it would also be terrible unpractical to use the mole or a similar unit for the amount of things. Just imaging calling Dominos and ordering 3.3210786e-24 large peperroni pizzas...

If the amount of things doesn't needs a unit, why the amount of microscopic things (elementary entities) needs it? And yes, your example about pizzas is very accurate to show another paradox of this unit.

It doesn't need it, which is one of the things which is criticised about it.  It's just extremly convenient to use it.

Thanks a lot for your patient answers.  This convenience can not be reached in a easier why defining "mole" as a unit prefix like "kilo", "mega", .. instead of define it as a base unit?

-1 this answer is completely wrong. Consider deleting it, to prevent misinformation. It is just not convenient to use for macroscopic objects.

@mcocdawc: if not convenient (I suppose you refer to "mole"), what is the alternative? No other base unit available.

@pasabaporaqui: 1. read the very good answer of Loong in detail. 2. Do you have a usecase where you are missing something like a mole as counting unit?You can always multiply with ten to the power of your desired order of magnitude. There are other counting units in daily life: "a dozen eggs" for example. As a dozen directly translates to 12, a mole directly translates to 6.e23 or N_a. Of course you could write as well 7/12 dozen apples or 7 * (1/6.e-23 mol apples. But why would you want that? Contrary to what DVSA said  7 * (1/6.e-23) mol apples is well defined.

Note that N_a has experimental uncertainty and there are more digits known, so it should be 6.002...=N_a

No, moles of apples are not well defined! The SI system explicitely states that mole is a unit for "elementary entities", apples are not elementary entities.

Obligatory XKCD reference...

7 apples is just 7 apples. Apples are objects, they don't need units, nor does it make sense to give them units.
7 atoms is just 7 atoms.
The Avogadro Constant is defined as the same number of atoms as are found in 12g of carbon 12. That is, 6.022 x 10^23 mol^-1.
Note the units of mol^-1. The number of dimensionless objects per mol.
Hydrogen atoms are all the same. Carbon atoms are all the same. Apples are not like atoms or molecules because they are all slighty different. So in order to talk about a mol of apples, we need to introduce the concept of a standard Apple with a defined chemical formula. I won't propose what that should be, but let's say a standard Apple weighs 100g.
Therefore we can say

1 Apple = 100g of Apples
10 Apples = 1000g of Apples

1 Apple = 100g = 1/(6.022 x 10^23 mol-1) =  1.66 x 10^-22 mol of Apples
Note this also holds for atoms or molecules, should you wish to express a quantity of entities in mol.

1 mol of Apples = 1 mol x 6.023 x 10^23 mol-1 = 6.02 x 10^23 Apples. Which, given that an Apple weighs 100g, weighs 6.022 x 10^25 grams.

Hopefully this illustrates why talking about moles of apples is a strange and rather inconvenient thing to do. If we are discussing hydrogen atoms, which weigh 1.66 x 10^-22 g each, talking about vast quantities of them in terms of moles makes a lot more sense.

Statement "1 Apple = 100g = 1/(6.022 x 10^23 mol-1) = 1.66 x 10^-22 mol of Apples" equals magnitudes that are dimensionality different (grams equal to moles). Is that correct?

@pasabaporaqui indeed, moles and grams are not dimensionally equivalent.  What I said is 1 Apple = 100g and 1 Apple  = 1.66 x 10^-22 mol. These statements allow us to derive a third relation 100g = 1.66 x 10^-22 mol (of Apples.) From this we can find the molar mass of Apples 6.022 x 10^25 g mol^-1 which relates the two.

The unit mole is qualitatively different from the dimensionless unit $1$ because a mole represents an imprecise range of numbers rather than an exact number. It is imprecise since a mole is defined in terms of the number of carbon-12 atoms that together have a mass of $12 \cdot 10^{-3}$ kg, and that number is unfortunately not really a constant since the kg unit is defined by a particular physical metal object that has a mass that varies slightly over time.
So specifying the number of apples as fractions of moles instead of whole numbers adds an unnecessary extra element of uncertainty that is not present if you just specify the number of apples as integer multiples of the unit $1$, assuming your apple-counting device is reliable.
The uncertainty inherent in the definition of the mole could be removed by changing the definition of the kg unit in a way that fixes Avogadro's constant to be an exact value. But that hasn't happened yet.

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A mole isn't really a unit, it is a quantity and (although someone beat me to it in a comment) you can have a mole of moles (at least in theory). Although SI defines it as a base unit this does imply a specific context.
The point here is that the SI system is a practical standard not an axiom of scientific philosophy and no sane person would expect Avogadro's number to be a base unit for macro scale quantities of some arbitrary item.
Indeed it is a ratio of carbon atoms per gram so it is not independently defined as it depends on the definition of the kg.
Moles come about from the fact that in chemistry you are often interested in the specific number of molecules which are taking part in an reaction but at the same time you need to be able to relate this to  a measurable quantity as it is not normally convenient to weight out individual molecules. For example in combustion one molecule  of methane reacts with two of oxygen to form one molecule carbon dioxide and two of water. Molar mass allows us to translate the ration of molecules to mass as long as we know the molar masses of all the elements involved which we do because the value of a mole is chosen to easily relate atomic mass to kg (or more usually grams). E.g. as carbon has an atomic mass of 12 (ish) we know that 1 mole of carbon atoms has a mass of 12 g
Units tell you the specific property which is being described. As far as the pure units are concerned 7 kg is 7 kg whether it is apples, oranges or plutonium.
7 kg describes the quantity of mass, of course apples have many other properties which may or may not be described by SI units so 'apples' needs to be stated to tell the greengrocer what it is that we want 7 kg of. If you are less picky you could say you want 7 kg of edible biomass.
You can also have entirely generic quantities of a unit, physics textbooks can talk entirely legitimately of a mass of 10 kg travelling at 10 metres per second.
This also brings up the concept of dimensional analysis in mathematics which is the principle that for an equality to be valid it must have the same base units on both sides of the equation.
Here it is also useful to introduce the concept of base units these are essentially properties of matter/space  which cannot be defined in any terms other than themselves and are the philosophical core of the SI system.
In  summary SI units can describe both property and scale, quantity is just a number. Since as far as I am aware there is no SI unit for 'applyness' 7 apples is just a quantity of an ad-hoc unit and not (nor can or should it be) an SI quantity.
Also the Mole is only really useful when you are talking about molecules and atoms which have well defined molecular/atomic masses.

Well it is a dimensionless unit, which for the propose of explanation in this context is a quantity, although I do take you point.  the crucial thing is that a mole does not define any fundamental property and is not defined by any SI base units.


text:
Electrolysis of Anhydrous Sodium Hypochlorite
Considering liquid anhydrous sodium hypochlorite:
$$\ce{NaClO (l)}$$
Assuming the ions are free, What would be produced if this were electrolysed?
As I've found lots of information for it's hydrated pair (a.k.a. bleach). And while I know the half equation for sodium:
$$\ce{Na+  + e- -> Na}$$
I cannot find any half equation for the hypochlorite ion: $\ce{ClO-}$, apart from in acid, or even what would be the product (maybe $\ce{Cl2O}$ or $\ce{Cl2O2}$?)

Obviously, sodium does not have a choice other than getting reduced. Oxidising it to sodium(II) is not an easy feat considering that the electron would have to be extracted from a core orbital. That means, the hypochlorite must be oxidised.
Usually, an oxidation of hypochlorite will yield chlorate, $\ce{ClO3-}$. If we attempt to form a redox half-reaction out of that, we can only use $\ce{ClO-}$ as a charge balancing agent:
$$\begin{align}\ce{ClO- \phantom{\ce{+ 4 ClO-}} &-> ClO3- + 4e-}\tag{Ox1}\\
\ce{ClO- + 4 ClO- &-> ClO3- + 4 e-}\tag{Ox2}\end{align}$$
If we do this, we have four extraneous chlorine atoms and two extraneous oxygen atoms. Thus, it makes most sense to do the mass balance by adding $\ce{Cl2O}$:
$$\ce{ClO- + 4 ClO- -> ClO3- + 4 e- + 2 Cl2O}\tag{Ox3}$$
However, it will be difficult confirming this. Sodium hypochlorite is not stable, especially not at elevated temperatures, especially not in pure form, especially not anhydrous sodium hypochlorite (sensing a pattern?). You would need a certain amount of recklessness to attempt this and sufficient luck to be able to report the result. Which is possibly why I cannot find any scientific publication on the topic; so take the equation with two grains (or more) of salt.


text:
Acetophenone reacted with LDA and diethyl carbonate. Why do I need HCl?
I'm working on a practice problem that asks for the product of acetophenone and LDA reacted with diethyl carbonate. After this, it says step 2 is HCl. After I react the first step, I get the correct answer to the problem - ethyl benzoylacetate. I'm asking why the problem says that HCl is added even when everything is neutral in the product I got and it's correct.

If you draw what you have carefully, without omitting any charges or adding magically appearing protons, notice that the actual product you initially get is either the ethanolate addition product of ethyl benzoylacetate, a mixture of ethyl benzoylacetate and lithium ethanolate or the lithium enolate anion of ethyl benzoylacetate and ethanol. In any case, you still have an anionic product.
For isolation and purification, you want to get rid of that negative charge (initially introduced by the strongest base you added: LDA). This is called quenching and typically performed with weak acids. I personally would use ammonium chloride for this type of reaction, but carefully adding dilute $\ce{HCl}$ will do the trick just as well. You should use $\ce{HCl}$’s proton to remove any remaining negative charges and eliminate $\ce{LiCl}$.

So is it a matter of quenching the negative charges on the minor products that aren't ethyl benzoylacetate?

@5antoro I wouldn’t call them minor. 1) They are stoichiometric; 2) not too safe on the $\mathrm{p}K_\mathrm{a}$ values, but I could well assume ethyl benzoylacetate to be equally or more acidic than ethanol.


text:
Would distilling a sodium fluoride solution generate hydrogen fluoride to any noticeable extent?
If  you have a solution of  NaF,  and  you boil  it (to  get  pure water by distillation)  can the  fluoride ions in water  separate  from  sodium ions and  combine with  hydrogen ions to form HF?    If yes,  how much of the molecules vaporized  is pure water and how  much is  HF?

The solubility of sodium fluoride in water is about 40 g/L. Distill off enough water and sodium fluoride will start to precipitate. That's it.

For all intents and purposes, you can consider a solution of sodium fluoride that of a weak base in water. Remember the formula to calculate an acid constant $K_\mathrm{a}$:
$$\begin{gather}\ce{HF + H2O <=> F- + H3O+}\tag{1}\\
K_a = \frac{[\ce{F-}][\ce{H3O+}]}{[\ce{HF}]}\tag{2}\end{gather}$$
For convenience, the acid constant is typically given in logarithmic form as $\mathrm{p}K_\mathrm{a}$; Wikipedia gives us the following value for $\ce{HF}$:
$$\mathrm{p}K_\mathrm{a} (\ce{HF}) = 3.17\tag{3}$$
If we use equations $(2)$ and $(4)$ (the ion product of water) plus a little maths, that allows us to calculate the concentration of $\ce{HF}$ under the assumption of a certain initial amount of $\ce{NaF}$ (e.g. $1~\mathrm{mol\cdot l^{-1}}$). $c_0$ is my notation for the inital concentration (i.e. before any changes).
$$\begin{align}K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}] &= 10^{-14}~\mathrm{mol^2 \cdot l^{-2}}\tag{4}\\
\Longrightarrow c_0(\ce{H3O+}) &= 10^{-7}~\mathrm{mol\cdot l^{-1}}\tag{4'}\end{align}$$
From equation $(1)$, we can derive equations $(5)$ and $(6)$, linking together our unknowns to give us the big picture:
$$\begin{align}[\ce{H3O+}] &= c_0(\ce{H3O+}) - [\ce{HF}]\tag{5}\\
[\ce{HF}] &= c_0(\ce{F-}) - [\ce{F-}]\tag{6}\\
[\ce{H3O+}] &= c_0(\ce{H3O+}) - c_0(\ce{F-}) + [\ce{F-}]\tag{5'}\end{align}$$
$$\begin{align}K_\mathrm{a} &= \frac{[\ce{H3O+}][\ce{F-}]}{[\ce{HF}]}\tag{7.1}\\
K_\mathrm{a} [\ce{HF}] &= [\ce{H3O+}][\ce{F-}]\tag{7.2}\\
K_\mathrm{a} \left(c_0(\ce{F-}) - [\ce{F-}]\right) &= \left(c_0(\ce{H3O+}) - c_0(\ce{F-}) + [\ce{F-}]\right) [\ce{F-}]\tag{7.3}
\end{align}$$
Henceforth, any $c_0$ appearing will be $c_0(\ce{F-})$ and $c_0(\ce{H3O+})$ will be replaced by $10^{-7}$ — ignoring the unit.
$$\begin{align}K_\mathrm{a}c_0 - K_\mathrm{a}[\ce{F-}] &= 10^{-7}[\ce{F-}] - c_0[\ce{F-}] + [\ce{F-}]^2\tag{7.4}\\
0 &= [\ce{F-}]^2 + \left(10^{-7} - c_0 + K_\mathrm{a}\right)[\ce{F-}] - K_\mathrm{a}c_0\tag{7.5}\end{align}$$
Thankfully, the equation is quadratic and we can solve it for $[\ce{F-}]_{1/2}$:
$$\begin{align}[\ce{F-}]_{1/2} &= \frac{- \left(10^{-7} - c_0 + K_\mathrm{a}\right) \pm \sqrt{\left(10^{-7} - c_0 + K_\mathrm{a}\right)^2 + 4 K_\mathrm{a}c_0}}{2}\tag{8.1}\\
[\ce{F-}]_{1/2} &= \frac{1 - 10^{-7} - 10^{-3.17} \pm \sqrt{\left(10^{-7} - 1 + 10^{-3.17}\right)^2 + 4 \times 10^{-3.17}}}{2}\tag{8.2}\\
[\ce{F-}]_{1/2} &= \frac{ 0.9993 \pm \sqrt{0.9986 +0.0027}}{2}\tag{8.3}\\
[\ce{F-}]_{1/2} &= \frac{ 0.9993 \pm \sqrt{1.0013}}{2}\tag{8.4}\\
[\ce{F-}]_{1/2} &= \frac{ 0.9993 \pm 1.0006}{2}\tag{8.5}\end{align}$$
As always, we need to perform a reality check. Subtracting is obviously not an option, as it would lead to a negative concentration. Thus we must add.
$$\begin{align}[\ce{F-}] &= \frac{ 0.9993 + 1.0006}{2}\tag{8.6}\\
[\ce{F-}] &= \frac{1.9999998}{2}\tag{8.7}\\
[\ce{F-}] &= 0.9999999\tag{8.8}\end{align}$$
So of the $1~\mathrm{mol \cdot l^{-1}}$ solution of sodium fluoride we have, a $10^{-7}~\mathrm{mol \cdot l^{-1}}$ solution of $\ce{HF}$ will result. (By the way, the units check out; we do indeed get $\mathrm{mol \cdot l^{-1}}$ at the end.)
You get similar results at other concentrations; if your initial fluoride concentration was $10^{-3}~\mathrm{mol \cdot l^{-1}}$, the resulting $\ce{HF}$ concentration will be $5.966 \times 10^{-8}~\mathrm{mol \cdot l^{-1}}$. There is no need to fear (or hope for) any $\ce{HF}$ development during distillation.


text:
Volume Element in the Integral for Variational Method
I'm learning how to use variational method to calculate the energy of ground state helium. I understand how we can set up a trial wavefunction that is a product of two hydrogen-like wavefunctions; however, when finding the integral $\int_{a}^{b} \phi H \phi dr_1dr_2$ (for $\phi=\phi_1(r_1)\phi_2(r_2)$), what would the volume element be? Would it be $16\pi^2 r_1^2 r_2^2$? I'm assuming the denominator is one for normalized wavefunction of hydrogen-like atom.

If you are working in spherical coordinates, and if the integrand has no dependence on the angular coordinates, then that volume element would be correct ($4\pi$ from each integration over $\theta_i \phi_i$ and $r_i^2$ for each spherical shell); however, the interelectronic potential $\frac{1}{r_{12}}$ depends not only on the radial coordinates, but also on the angular coordinates of each electron, so you can't just integrate those out.

I see. So the $r^2$ terms would be correct, but the $4\pi$ term wouldn't be?

The volume element $\mathrm{d}V$ (or commonly denoted $\mathrm{d}\tau$ in quantum chemistry) is equal to $r^2 \sin\theta ,\mathrm{d}r,\mathrm{d}\theta,\mathrm{d}\phi$. This is the most general form and the one you should use. In the special case where your integrand is only dependent on $r$, then the multiple integral can be separated out, and the integrals over $\theta$ and $\phi$ work out to be equal to $4\pi$.


text:
Help with finding intermediates in the synthesis of Ritalin

Here I have a Chemistry Olympiad question I'm doing that outlines the synthesis of Ritalin. So far I've worked out A, Anion B -, and compound C. You can see my workings, as I've attached them here too.

Nucleophilic substitution with the nucleophile $\ce{C#N-}$ produces A:

A proton is taken away from the $\alpha-\ce C$ to deprotonate it to form Anion B -:

Anion $\ce{B-}$ acts as a nucleophile and substitutes $\ce{Cl}$ to produce C:

But how would I work out the structures of D, E and F?
I'm not familiar with all the mechanisms/reactions of nitriles and am unsure how to proceed.

In A and B, you should probably draw the C-C single bond in the nitrile longer, it's a bit difficult to see at the moment.
The nitrile is sequentially hydrolysed to an amide and then to an acid.
Treatment with $\ce{ROH/H+}$ is a rather obvious Fischer esterification.

You can arguably work all of this out from first principles: the nitrile carbon is electrophilic, and is therefore susceptible to nucleophilic attack by water. The reaction mechanism is given here. On top of this, the IR peaks strongly support the presence of amide and acid functional groups in D and E respectively.
With that said, I'd argue that there is always going to be an element of "I just know this because I read about it/was taught it before". The very fact that you said "I'm not familiar with the mechanisms/reactions of nitriles" indicates this - if you had happened to read a textbook chapter on nitriles, then you would probably have seen this, since it is one of the most fundamental reactions of nitriles. You would also have to read up about the IR wavenumbers - otherwise it's just gibberish.

Thank you very much. In my A-level syllabus the reactions of nitriles are very limited and are not found in my textbooks - this is why I'm working through Olympiad papers, so I can familiarise myself with a wider range of Chemistry and ways to solve questions like this, as I've received an interview for Chemistry at Oxford. Ive seen you're an undergraduate and would very much appreciate some advice on the interview, I'm just doing Olympiad papers right now to improve.

IR is gibberish, of course. */obligatory IR rant ;)*

@MichaelHarding feel free to pop into chat, I'm there often.

@Jan re: IR BOOM!

Just as a point of interest, pyridine's pKa is 5.25, acetic acid's is 4.75. So, in E, that acidic proton should probably be more closely associated with the nitrogen than with the oxygen (I'd assume no matter where the proton is, there's going to be some cyclic intramolecular H-bonding).


text:
How did early chemists measure concentrations and purity?
My chemistry teacher loves going back through the history of famous chemists. This got me wondering how these chemists would first determine the concentration of a sample before they had any other chemicals of known concentration to compare against?
What modern methods are there for determining concentrations and purity without other chemical standards to compare against?
Let me explain my question a little more clearly with an example.
If a historical scientist starts their own laboratory and wanted to do a titration, how would they start? They might produce some hydrochloric acid, or some barium sulfate, and want to confirm its concentration/purity. However, this would require they already know the concentration of something else, which they don't. So it seems a little paradoxical. How did they ever start performing analytical experiments?

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Not everything comes as a solution. Rather, the historical chemist would have started of with solid substances. These were purified according to ‘established procedures’ (distillation, recrystallisation, dissolution and precipitation) which worked quite well. Then, one of these substances would have been used to make a stock solution — and because it is solid you can simply weigh it.
Now you might assume that the historical chemist had no knowledge on how two substances would react — which is likely true. However, once he knew the composition of one type of solution, he was able to express the other in equivalents: ‘this solution is able to reduce two equivalents of the permanganate solution’ (using the first example that jumped to my head; not sure whether that is actually a likely thought).
Going via the diversion of analytical chemistry (is this a compound or an element?), it was established that a law of multiple proportions existed: if two elements form more than one compound, the mass ratio of one of those elements is always a simple integer ratio. This was easily extended to reactions: if something reacted with 2 equivalents of A but 1.5 equivalents of B, then maybe it wasn’t the concentration I assumed but I should multiply by two?
Over time and more data points, the current system would have been established.


text:
Serious introduction to the chemistry of explosives recommendation
I'm looking for a serious introduction to the chemistry (and perhaps to a lesser extend the physics) of explosives. Presupposition of basic university courses in Organic Chemistry and Thermodynamics is no problem, though it would be helpful if some of the specifics needed to understand subsequent material would be briefly outlined as a reminder (e.g. in an introductory chapter). Sadly this search is quite hard because of all the Uncle Fester and like trash clogging up search results. To be clear, that is absolutely not what I'm after. I want an academically oriented introduction, not some "recipe" book.
A book that more or less resembles (I guess) what I'm after is The Chemistry of Powder and Explosives by Tenney L. Davis. However, it is over 70 years old and might be a little to advanced for me (might, I just don't know without some guidance).
Every help is appreciated!
PS. I actually found an online copy of the book I mentioned. It does not look really theoretical?

To be clear, that is absolutely not what I'm after. I want an academically oriented introduction, not some "recipe" book.

I'm glad to read that!
I'm aware of Chemistry of Energetic Materials (Eds. George A. Olah and David R. Squire, Academic Press, 1991) which seem to cover

The Structural Investigation of Energetic Materials.
Studies of Initial Dissociation Processes in 1,3,3-Trinitroazetidine By Photofragmentation Translational Spectroscopy.
Studies of Molecular Dissociation By Means of Ultrafast Absorption and Emission Spectroscopy and Picosecond X-Ray Diffraction.
Computer-Aided Design of Monopropellants.
Polycyclic Amine Chemistry.
Metallacarboranes of the Lanthanides and Alkaline-Earth Metals: Potential High-Energy Fuel Additives.
Methods for Preparing Energetic Nitro-Compounds: Nitration with Superacidic Systems, Nitronium Salts, And Related Complexes.

On a side note, if you are interested in current research on high-energy materials free of nitro groups, have a look at the work of Thomas M. Klapötke.

For a starting point, I'd recommend this book. It covers the basics fairly well, and has a bit of history. It is often used as the introductory text in the few universities that have specific programs in energetic materials.
In addition, this book gives more of a synthetic chemistry perspective, including some very exotic stuff. Some are of practical value, some are only of academic interest.
In addition to that, depending on what you are interested in, I found that this book is quite a useful reference for modern propellants and explosives research (which naturally parrallel each other).

Thank-you so much for your answer! Just curious, what is your expertise in these matters? If you could add some references/details (for example hw do you know such and such book is used in a few universities) I will award you the bounty. At any rate +1. Also it might be valuable to change the hyperlinks "this" to the actual name of the book, so that it is more informative, but that is just a formatting issue. :)

I will give you the benefit of the doubt, because your answer is more what I was looking for (but thanks to Klaus too of course!). So you will get the bounty! Still I would appreciate if you could edit your answer at some later point to include the things I mentioned above.


text:
Why does bromide not react with water when methyl ammonium bromide is dissolved in water?
In the following reaction, the salt $\ce{CH3NH3Br}$ is formed:
$$\ce{CH3NH2 + HBr -> CH3NH3Br} $$
In water this dissociates completely into $\ce{CH3NH3+}$ and $\ce{Br-}$
In my chemistry book, it says that the $\ce{CH3NH3+}$ dissociates further in water into $\ce{CH3NH2}$ and $\ce{H3O+}$ (meaning that $\ce{CH3NH3Br}$ an acidic salt) according to the following:
$$\ce{CH3NH3+ + H2O <=> CH3NH2 + H3O+}$$
Fair enough, but why does the $\ce{Br-}$ not do the same and dissociate in water?
$$\ce{Br- + H2O <=> HBr + OH-}$$
In this way, the dissociation of the $\ce{CH3NH3Br}$ salt would not have so much of an effect on the pH, because the quantity $\ce{OH-}$ ions has also increased?

Somehow related: http://chemistry.stackexchange.com/questions/42696/how-can-bromine-water-be-called-a-solution

Please have a look at the acid dissociation constant of $\ce{HBr}$. Is hydrobromic acid a strong acid? If so, what does that mean for bromide, its corresponding base?

@KlausWarzecha I see now; because $\ce{HBr}$ is a strong acid, the $\ce{Br-}$ won't dissociate in water because that would be effectively undoing the dissociation of HBr. Thanks for that.

The key is comparing $\mathrm{p}K_\mathrm{a}$ values. A good resource for this is the Evans $\mathrm{p}K_\mathrm{a}$ table. Unfortunately, the table does not contain $\ce{CH3NH3+}$, but in the section protonated nitrogen there is a value quoted for $\ce{C2H5NH3+}$ — which is chemically similar enough to methyl ammonium to assume the $\mathrm{p}K_\mathrm{a}$ values to be essentially identical.
\begin{array}{lr}\hline
\text{compound} & \mathrm{p}K_\mathrm{a}\\ \hline
\ce{EtNH3+} & 10.6\phantom{0}\\
\ce{HBr} & -9.00 \\
\ce{H3O+} & -1.7\phantom{0} \\ \hline\end{array}
Notice how the $\mathrm{p}K_\mathrm{a}$ values differ by almost $20$ logarithmic units. $\ce{HBr}$’s $\mathrm{p}K_\mathrm{a}$ value is even lower than water’s by $7$ to $9$ logarithmic units depending on the source used. Thus, while the amine exhibits extensive acid/base chemistry in water, bromide will not be protonated to any noticeable extent. Or you could say that your final equation will be strongly favouring the reactant side.
I’ll also point you to a recent answer of mine wherein I calculated the concentration of $\ce{HF}\ (\mathrm{p}K_\mathrm{a} \approx 3.17$, so notably weaker than $\ce{HBr}$) in a 1 molar solution of $\ce{NaF}$. The case is very comparable to your’s.

So, when we say that a strong acid dissociates completely, and we say that it is not an equilibrium reaction, it actually is protonated as well, but the equilibrium lies so far to the right that the reverse reaction is negligible?

@Adrian For all practical intents and purposes, a strong acid is fully dissociated. Taking a fully correct look at it, the equilibrium is only shifted very far to one side.


text:
Example of an endothermic reaction that can sustain itself for at least an hour?
Recall that in physical chemistry, the spontaneity of a reaction at constant pressure and a given temperature $T$ can be quantified by the Gibbs free energy
$$\Delta G = \Delta  H - T \Delta S$$
A reaction is spontaneous if $\Delta G < 0$. For an endothermic reaction to be spontaneous, $\Delta H < 0$, $\Delta S > 0$ and $|\Delta H|< |T\Delta S|$
Now suppose we monitor the rate of change of $\Delta G$ as the reaction progressed. This means
\begin{align}
\frac{\mathrm d\Delta G}{\mathrm dt} & = \frac{\mathrm d\Delta H}{\mathrm dt} - \frac{\mathrm d(T \Delta S)}{\mathrm dt}\\
\frac{\mathrm d\Delta G}{\mathrm dt} & = \frac{\mathrm dT}{\mathrm dt}\left(\frac{\partial \Delta H}{\partial T} - T\frac{\partial \Delta S}{\partial T}- \Delta S\right)
\end{align}
Now suppose for simplicity $\Delta H$ is roughly constant with temperature. Then:
$$\frac{\mathrm d\Delta G}{\mathrm dt} = \frac{\mathrm dT}{\mathrm dt}\left( - T\frac{\partial \Delta S}{\partial T}- \Delta S\right)$$
Thus for $\frac{\mathrm dT}{\mathrm dt} < 0$, a reaction to be self sustaining indefinitely if the following holds for all $t > 0$
\begin{align}
- T\frac{\partial \Delta S}{\partial T}- \Delta S & > 0\\
- T\frac{\partial \Delta S}{\partial T}  & > \Delta S
\end{align}
In all closed systems, entropy will eventually maximises according to the second law, thus there will be a point where $\frac{\partial\Delta S}{\partial T} < 0$ (i.e. the increase in entropy as the reaction progressed will slow down over time) until the reaction stops (thus explaining why both exothermic and endothermic reactions will eventually go to completion)
Therefore, my question on self sustaining endothermic reaction that last for hours is boiled down to finding a time $t>0$ such that the above inequality holds in the order of hours, given a $\Delta H$ as positive as physically realisable

Any example of an  endothermic reactions that has $\Delta H > 50{-}100\ \mathrm{kJ\ mol^{-1}}$ and can sustain itself in the order of hours?

Thermodynamics does not inform us about the time-scales of processes but only about the starting and ending conditions. Reaction rates are controlled by the activation energies between reactants and products for both  endothermic and exothermic reactions. see answer at http://chemistry.stackexchange.com/questions/10115/rate-constant-units-and-eyring-equation/58739#58739

How about a large block of ice melting? Latent heat is $80,\mathrm{kcal, mol^{-1}}$ if memory serves.

@Zhe A large block of ice melting is not a chemical reaction.  It is a phase change that is controlled time-wise by the heat transfer rate (not reaction kinetics).

@ChesterMiller True, but thermodynamics clearly governs this process in the same way as a chemical reaction. I agree though that I didn't quite read the question right. This isn't quite "self-sustaining" as the OP wanted...

It depends on what you mean by "in the same way."  Conductive heat transfer is certainly fundamentally different from molecules interacting to form different molecules.


text:
How can there be decimal subscripts in a molecular formula?
While learning about how batteries work I have encountered the following notation for a Li-ion cathode: $\ce{Li_{0.5}FePO4}$.[1]
According to Wikipedia, the subscript number in a reaction equation represents the number of atoms in a molecule. But what does the $0.5$ represent? Is it a fraction of atoms in a molecule?
Also, $\ce{LiFePO4}$ is a cathode. I would expect that for a fully charged battery, the anode is populated with all the $\ce{Li}$, therefore the cathode should be $\ce{FePO4}$, right? I can also understand that some $\ce{Li}$ gets stuck in the cathode, but then I read somewhere that a fully charged battery, the cathode is $\ce{Li_{0.5}FePO4}$.
I have added an exaggerated sketch for what I imagine the system would be.

I know that the structure of $\ce{Li_{0.5}FePO4}$ is more complex than this, but this is only a hypothetical sketch. Also, please note that I don't have a background in chemistry, so any additional material that can help me understand the concept better is greatly appreciated.
References:

Tan, H.; Fultz, B. Rapid Electron Dynamics at Fe Atoms in Nanocrystalline Li0.5FePO4 Studied by Mössbauer Spectrometry. J. Phys. Chem. C 2011, 115 (15), 7787–7792 DOI: 10.1021/jp200127b.

Have you understood how Li-Ion batteries work? The Li content is variable, and in practice always substoichiometric.

I think I have an okay understanding of the batteries. I understand that there are structural changes, diffusion and potential differences. And yes, I know that the Li content is variable, that's why I would have expected that for a fully charged battery, all the Lithium ions would be in the anode, leaving the cathode "empty". But I also understand that there are some residual Li-ions that stay in the cathode. I'm trying to link this to chemical reactions. And as I said, I expected that the cathode would be FeFO4 as seen here: http://esl.ecsdl.org/content/9/3/A110/F2.large.jpg

The 0.5 is an approximate value. If you charge the battery significantly beyond that number, the electrode turns into stable iron phosphate, and is thereby destroyed. There must NOT be any structural change during charging/discharging.

Okay! Fair enough. But why 0.5? There should be something with this number... And can one find a formula for getting such a number, even if it is approximated (i.e. such as molar fraction etc.)?

Also, can one sketch such a "formula" Li0.5FePO4, if we break the system at atomic level?

The principle of the Li-Ion battery is that you can take some fraction of the Li ions out of the cathode without structural changes, so they can also get back IN during discharging. You cannot calculate that fraction from first principles. Not with any computer from this century.

Hmm. Interesting information. It's contrary to what I had in mind, but please give me some time to read about it.

For beginners: https://en.wikipedia.org/wiki/Lithium_cobalt_oxide is much easier to understand.

First of all, in chemistry two types of formulas exist: structural and empirical. A structural formula shows the way atoms are connected. An empirical formula only summarizes atoms and their ratios. While often an empirical formula references a molecule, there are cases when the compound in question is not molecular, or it is unknown if it is molecular. In that case, since the total number of atoms in the molecule is undefined, the empirical formula may be scaled by any number, and as long as all indexes are scaled by the same number, it will denote exactly the same information.
So, formulations like $\ce{Ti_{0.9}O}$, and $\ce{Ti9O10}$ denote the same thing: a Ti+O phase with Ti/O ratio 0.9. Usually in such cases, if ratios are fixed, corresponding indexes are kept integral. So, it is  $\ce{Al2O3}$ because the Al/O ratio is fixed here, but $\ce{Li_{0.2}MoS2}$ because the Li/Mo ratio here is not.
Returning to the $\ce{Li_xFePO4}$ issue: in this phase $\ce{FePO4}$ part forms the host network similar to quartz and thus has a fixed composition. However, Li may and does migrate in and out of the host network, so its amount in the bulk is not fixed. This is possible because Fe ions have a lot of free orbitals and some of them have rather low energy, so Li electrons are passed to iron and the small Li ions may travel through the host network. Thus the composition of the electrode changes through the charge/discharge process.
In accumulator chemistry electrodes are rarely forced towards full transfer of traveling ions, because it may lead to irreversible degradation in the electrodes. For example, the $\ce{LiCoO2}$ phase would collapse if all or even a major part of Li left, so never is all or even a major share of the Li allowed to leave it. Consequently, Li-ion battery electrodes typically contain a significant amount of Li both in the fully discharged and fully charged states.
Similarly, it's usually technically impossible to achieve complete transfer even in more traditional batteries because the voltage and consequently the travel speed of ions between electrons falls down with discharge.

Can one say that Li0.5 means Li/FePO4 = 0.5? Or, as Karl said in the previous comments, it is an approximate (or maybe empirical) number?

@Paul One can say that by result of an analysis (or by other consideration) the apparent brutto composition of an electrode is $\ce{Li_{0.5}PO4}$ Since the compound is non-stochiometric (it allows significant change in composition without change of the phase) there is no 'ideal' composition, so any value would be a result of some form of analysis, which by definition gives approximate value (i.e. has some uncertainty)

To add here: sometimes you will even find something like "x" and "1-x" instead of the numbers, which just means that this certain substance can have those two atoms in different ratios. Or sometimes you got something like "0.1-0.5" for a certain atoms.

Thank you. Yes, I recall seeing "x" and "1-x", but I can see that only when the cathode gives away ions. But the formula of the cathode still stays "LiFePO4" (e.g. http://electronicsgate.com/Images/li-ion%20charger/Lithium_Ion_Lithium_Polymer_Battery_Charger_Using_MCP73831_2.jpg) Shouldn't it be LixFePO4 instead of LiFePO4?

In English the counterpoint to a structural formula is typically termed the "empirical formula" (Wikipedia) - I have never encountered the term "brutto formula" before, and it doesn't seem to have an entry in the IUPAC Gold Book.

@R.M. "Brutto formula" seems to be rare, and of mysterious origin. Maybe it means "empirical" in some other language? Google says "brutto" means "gross" in Polish, if it's "gross" in the sense of total/whole/aggregate maybe that lines up somehow.


text:
High voltage AC water electrolysis not working
I was trying to use a $500~\mathrm V$ AC power supply to split water, but nothing happened. I  tested everything with a $9~\mathrm V$ battery, and it all worked fine.
Why did it not work with the high voltage? (Do not worry, I was taking the proper precautions, such as insulation and three pairs of gloves, and I did not touch any live wires.)

Why in the name of all that is good and sensible are you trying to split water at $500~\mathrm V$??? That is a preposterously high voltage for the purpose.

No duplicate, but related: http://chemistry.stackexchange.com/questions/8642/how-does-high-frequency-water-electrolysis-work

Alternating current (AC) is your problem. AC spends half its time pushing electrons forward through the circuit, and half the time pushing them in reverse through the circuit.  Thus, on average the net current pushed through the cell is zero.
So, leaving aside any metal dissolution/deposition, you were forming tiny, tiny amounts of electrolysis gases at the surface of your electrodes while there was current moving through the cell in one direction, but immediately consuming them again when the current reversed. The total number of coulombs passed in each direction was presumably so small that no gas bubble nucleation occurred before the polarity reversed.
Regardless: stop using 500 V for electrochemistry experiments. It's way more potential than you need, and terrifically unsafe. In most cases, anything above 40 V is overkill, and for most experiments that 9 V battery you mentioned will more than suffice.

It has been quite a while since this answer was posted, but I must point out that products are formed with AC electrolysis but mixed, at least if you just use your mains supply(50Hz) (pro: easy to test). The mix is explosive (incase of water electrolysis), but it is formed

@VaradMahashabde Agreed! When I was a kid, I built a simple carbon arc contraption in my basement chemistry lab. I controlled the current using a saltwater rheostat: a jar of saltwater with copper wires stuck in the saltwater. I used 120 V AC line voltage. The carbon arc was blindingly bright, the fuses kept blowing until I got the salt concentration right and chlorine gas was released. I know that smell very well: it is how I was introduced to chemistry and became a chemist.