text:
Chemical time scales
In calculating UV/VIS spectra with computational procedures (say, TD-DFT), it is commonly assumed that the geometry of the ground-state is still valid also for the excited state due to the short time scale (I more or less took this statement from here).
So I asked myself, are there good references on time scales for chemical processes?
Some phenomena may be obviously faster than others, but others (say, two molecular excitations) may be not that obvious to compare.
I am looking for something simple, like a table (or tables) with examples.

The CH stretch vibration of a CH bond (a very fast vibration) is ~3000 cm-1 which means one vibration takes around 10 fs. Rotations of small molecules are around 10 ps to 1 ns. Going through a transition state without any barrier is around 60fs ( 1/(kB*T/h)). So for basically everything that occurs faster than some fs the atoms (or better the cores) will lag behind.

Thanks for your comment @DSVA!
What about excitations/emissions? They surely span a lot in terms of time.
When can I safely assume the excited state is not going to rearrange itself?

As a rule of thumb, excited state lifetimes are not often less than a picosecond, several hundreds of ps to many  nanoseconds is more typical. e.g. trans-stilbene 70 ps , pyrene 100's ns. Of course there are exceptions particularly when quenching by electron transfer occurs. So most excited states do have time to respond to their new potential before they decay.

If you want to simulate emission spectroscopy, you should optimize your excited states... (hint: stokes shift etc)

According to the Franck-Condon principle which you are implicitly referring to, absorption is indeed very fast compared to all other processes.
If you are looking for time scales of photophysical processes, I can suggest you for example B. Valeur's book "Molecular Fluorescence: Principles and Applications", chapter 3 (Characteristics of fluorescence emission). You'll find this list of characteristic times:

absorption: $10^{-15}$ s
vibrational relaxation: $10^{-12} - 10^{-10}$ s
$\ce{S_1}$ lifetime (fluorescence): $10^{-10} - 10^{-7}$ s
intersystem crossing: $10^{-10} - 10^{-8}$ s
internal conversion: $10^{-11} - 10^{-9}$ s
$\ce{T_1}$ lifetime (phosphorescence): $10^{-6} - 1$ s

as well as examples of molecules with their photophysical data.
For chemical processes, it depends on many things, such as concentration, diffusion rates... and you would need to precise your question (type of photochemical reaction, type of quenching, etc.).

True, thanks for sharing.
That's what I was looking for, I will accept this answer.


text:
Why do we do electrolysis and electroplating using warm electrolyte?
This is a pretty basic question, and I know it has something to do with conductivity, but I'm not quite sure how they are related.

Blake, please see my edits -- if you're unhappy with anything I changed, please feel free to change it back, or roll back my edit entirely.

The reasoning here is two-fold. The solubility of most electrolytes increases with temperature, and water's ionization constant also increases with temperature. On the whole this means more ions, and thereby better conductivity.

Can you clarify on the ionization constant because I thought it decreased when temperature increased.

@jakekimds The value of $\mathrm{p}K_\mathrm{a}$ does decrease with increasing temperature, yes. But, this means that the proper ionization constant, $K_\mathrm{a}$, increases with increasing temperature.

ringo makes good points in his answer. Additionally, though, the increased temperature enhances mass transfer of ions to/from the electrode surfaces by at least two mechanisms:

Higher temperature results in lower electrolyte viscosity, leading to a thinner fluid dynamic boundary layer and concomitant greater mass transfer to/from the electrode surfaces.
Higher temperature also increases the diffusivities of the electrolyte solutes, which also contributes to increased mass transfer to/from the electrodes.

very good additional points.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
What is the solubility order of isomeric alcohols in water?
The order for the solubility of isomeric alcohols in a book was given as:
Primary alcohol >Secondary alcohol >Tertiary alcohol
After this I found a question:

(1) Which is more soluble in water?
(a) 3-ethyl-3-hexanol
(b) 2-octanol

Ans: (a)
The reason was that (a) has a more compact alkyl portion than (b). But as (a) is a tertiary alcohol while (b) is a secondary one, so according to the order given above the answer should be the reverse.
Also, while comparing 2-methyl-2-propanol and 1-butanol, 1-butanol had higher solubility which matched the given order.
I am not understanding what's wrong about the first case. Please explain.

The reasons that you cite are both important for the solubility of an alcohol. Generally speaking, if the molecule is branched its (hydrophobic) surface exposed to water is less than in a linear molecule. This effect is more pronounced when the number of atoms is bigger. On the other hand the effect of the degree of substitution on the solubility is less pronounced when the molecular size rise.
This would partially justify your second question, but the problem here is that your second question is wrong, 2-methyl-2-propanol is soluble with water in every ratio, while n-butanol is not.
The statement primary > secondary > tertiary is valid when comparing ethanol > isopropanol > t-butanol or similar series. The hydrophobic surface in the series stays relatively unchanged this is why we can say that the effect on the solubility is caused by the degree of substitution of the alcohol. In the comparison between alcohols with the same number of carbons branched > linear is a much more important factor.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Mirror reaction - avoiding silver nitride
I am interested in experimenting with depositing silver on glass (mirror). I've read several papers and watched youtube videos but wish to be better informed about the risk of explosion. The mixture is silver nitrate solution, sodium hydroxide solution, ammonium hydroxide and dextrose solution. Is there a combination, or concentration of any of these that forms silver nitride in less than 3 hours? One video suggested heating the mixture to enhance the deposition onto microscope slides, but neglected to mention a temperature, or whether it hastens the formation of the explosive silver nitride. Any comments on whether $\mathrm{50^oC}$ for 1 hour would have ample margin of safety? Thanks!

Very related:  http://chemistry.stackexchange.com/questions/43441/is-tollens-test-safe

Also keep in mind that even though not all errors will not be explosive, the reaction demands very careful handling, of more than 40 people only 2 of them were able to deposit silver as mirror in my organic chemistry laboratory, others just gave precipitation.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Bonding in MOF-5
I came up with this problem today and I can't find the answer elsewhere:
How does the oxygen atom bond to the four zinc atoms? To be more specific, what would be the molecular orbital diagram for the Zn4O cluster
Does this have an obvious answer?

I guess my question to you is why not? An oxygen in liquid water can be coordinated simultaneously to up to four hydrogens. Isn't this case exactly analogous?

Does it have an obvious answer? I have no experience of these types of clusters but I strongly suspect the answer is no.

As Zhe suggested -- it's just a tetrahedral bonding-interaction of oxygen. In the case of $\ce{H2O}$, oxygen has two lone pairs and two bonds to $\ce{H}$; in $\ce{[OZn4]^{6+}}$, those lone pairs are interacting with Zn atoms (strongly, because the $\ce{Zn}$ atoms bear a +2 charge).
Per the MOF-5 $\ce{H2}$ storage paper:

...inorganic $\ce{[OZn4]^{6+}}$ groups are joined to an octahedral array of
  $\ce{[O2C-C6H4-CO2]^{2–}}$ (1,4-benzenedicarboxylate, BDC) groups to form a
  robust and highly porous cubic framework.

Per the original synthesis:

A small amount of hydrogen peroxide was added to the reaction mixture
  in order to facilitate the formation of $\ce{O^{2-}}$ expected at the centre of
  the secondary building unit.

Thus the central oxygen has a full octet, able to supply 2 electrons to each $\ce{Zn^{2+}}$ in the coordination complex.


text:
Is there a mathematical formula for electronegativity?
If we lost all the numerical values we have for the electronegativity of elements, is there any mathematical formula that can be used to re-calculate them based on proton (and neutron) numbers alone? For example, if I had no way to look up the electronegativity of Carbon, could I use some existing formula to calculate it from scratch? Or would I have to go actually measure it?

There is Pauling's method using fitted data, then a bunch of other methods using various atomic properties. https://en.wikipedia.org/wiki/Electronegativity#Methods_of_calculation // You should realize that electronegativity values don't show the % error. There really isn't an "exact" number. It sort of like measuring an atom's size.

Electronegativity cannot be calculated from the number of subatomic particles in an atom. However, using Ionization Energy and Electron Affinity, one can calculate the Electronegativity of an atom using the Mulliken Electronegativity Equation.

Ionization Energy : The amount of energy required to remove the
electron that is least attracted to the nucleus in a gaseous state.
Electron Affinity : The amount of energy an atom gains when an
electron is added.

To find the Ionization and Electron Affinity values of an atom, one must reference a chart.  In the Mulliken Electronegativity Equation, Ei stands for Ionization Energy and Eea stands for Electron Affinity.  Plug in your values and solve.

Equation for Energies in Kilojoules/Mole :
ENMulliken = (1.97×10^−3)(Ei+Eea) + 0.19
Equation for Energies in Electronvolts :
ENMulliken = 0.187(Ei+Eea) + 0.17

Lithium Example in kJ/mol :

The Ionization Energy for Lithium is 520 kJ/mol
The Electron Affinity for Lithium is 60 kJ mol^-1
ENMulliken = (1.97×10^−3)(Ei+Eea) + 0.19
ENMulliken = (1.97×10^−3)(520 + 60) + 0.19
ENMulliken = 1.143 + 0.19 = 1.333

NOTE : There are multiple scales for measuring Electronegativity, including the Pauling Scale, Sanderson, Allen, and more.  This example measures Electronegativity using the Mulliken scale, and will compute an answer that will differ from other models.

@GiantSpruce ionization energy and electron affinity can in principle be calculated by solving Schrödinger's equation, but this is quite hard to do precisely for multi-electron atoms.


text:
Can glass be made with anorthite?
The lunar highlands are made mostly of anorthosite (90% anorthite, the rest oxides of magnesium, iron, a little titanium, and traces of others), and pretty pure anorthite ($\ce{CaAl2Si2O8}$) is common. No minerals that are more than about half silica are known. The use of typical batch reactions to purify chemicals is (mostly) not practical on the Moon, as there is no water*, no halogens, and no carbon, nitrogen, or hydrogen.
Human habitats are going to need clear glass, the more the better. If the Moon was being settled, a way to make clear glass locally would be a priority. There are of course many things to be considered about the proper use of glass, but that is a separate issue. This question is about an advanced colony, not the first few habitats. You can assume the infrastructure for an involved process is available.
Could anorthite be processed into decent clear glass? How much impurity could it have? How can I look into the properties it might have and things to consider about the process?
Below is a table from the Lunar Sourcebook, Ch. 5 pg 129, showing
percentages of pure anorthite in various minerals mixed in igneous rocks
from the highlands:

*Actually there is a lot of frozen water and other volatiles at the poles, but getting it in quantity is a big challenge, and it isn't found anywhere else.

Clear glass? No. // Most such techniques I've read discuss sintering the lunar soil into "bricks." The brick structures would have to be fairly thick for radiation shielding and thus they would be reasonably air tight as well.

Once you get a good lunar smelter set up anorthite seems like a possibility at least.  The keys are what wavelength of light you need for it to pass, required physical characteristic  (hardness, flexibility, etc.).  Is it for habitat windows or solar cells, etc.  In other words all I have to offer are more questions and complications.  But honestly, given your statement about assuming there is an established civilization already there trying to do this, I wouldn't write it off.  Sorry, those are really the only thoughts I can offer.

I guess i know too little about this to frame it in a sufficiently narrow fashion. I'm trying to establish the best way to portray the manufacture of glass for the beginnings of industry on the Moon. My chief interest is glass that can be used for windows and other glazing. Loong said in chat probably it would be better to work with olivine, though that requires finding and excavating pure ore, while anorthite is common on the surface. I'll try to look into it more and come back to the question. Reading suggestions are also welcome, in that case.

@kimholder, incredibly interesting work!  I do think that since the anorthosite is 90% anorthite which can be made into clear, transparent forms, I might start with a focus on anorthite.  Again, it depends on what technological resources have been made available to the colonists.  Best of luck!

In theory, yes it is possible. In practice, not really.
First of all, the anorthite on the moon is never pure anorthite ($\ce{CaAl2Si2O8}$) but rather an anorthite-albite ($\ce{NaAlSi3O8}$) solid solution. That is, the single crystal contains both components. The diagram you added there, where it says "mole %"? The other stuff is albite.
Now, back to glass making. Just about any crystallite solid silicate can be made into a glass. The process is simple: heat it up above the melting point, and then cool it down fast enough so it quenches to glass instead of crystallising. Let's look at these points in details.
Melting the stuff: Anorthite has a very high melting point: 1553 °C. Forsterite (the Mg-rich end member of the olivine group) has an even higher melting point: 1890 °C. To melt the stuff you're going to need a furnace that can go that high, and your energy bill is going to be extremely high. Here's a liquidus phase diagram of the $\ce{CaAl2Si2O8-Mg2SiO4-CaMgSiO6}$ system so you can get an idea of the temperatures required:

Luckily, your material is not pure anorthite. It also has albite! That's good because it takes the melting point down:

However, 1100 °C is still pretty high. The glasses that are commonly used in industry today have melting points of below 1000 °C. This is achieved by having high contents of sodium in the glass, which is perfect in taking the melting point of the material down. However, materials on the lunar surface are not that rich in sodium, and that's a problem.
Adding iron to forsterite (i.e. the fayalite component of the olivine group) will take the melting point of it down, but pure fayalite melts at around 1200 °C which is no good, and the iron will colour the glass green to dark brown (seen as black if thick enough), depending on how much oxygen is around. More about iron later.
Cooling the stuff down: Silicate glasses that we know and love contain abundant silica, commonly above 70%. This makes the glass very viscous because of Si-O polymerisation. A viscous glass can be easily left to cool in air (by the way - not readily available on the moon) and it will remain glass. Cooling a non-viscous glass that has below ~50% silica in air will cause crystals to form in it (anorthite in your case) and it will not be perfectly clear anymore. Therefore, you need to cool it very rapidly, by quenching in water. The problem now is that the glass is very brittle and will most likely crack during cooling or very soon after (this is from my own personal experience in trying to make glasses of these compositions).
So far we have two problems: melting it, and cooling it. There are other problems. You were worried about impurities and rightfully so. Iron is your biggest enemy. Even trace amounts of iron (< 1%) can make your glass dark green or brown. Not only that, if there is oxygen around you might even have magnetite crystals in your glass and it's going to be completely opaque. You need a way to get rid of the iron. Nickel is also a minor problem. As for other "impurities", such as Al, Na, K, Ca, Mg, that's not a problem. They don't give any colour to the glass. It's the transition elements that you should fear.
Yet another problem is that you are not going to get pure minerals. The lunar surface does not contain blocks of anorthite or forsterite or anything else. The regolith is a powder made by weathering down of the minerals, and it contains a horrible mixture of everything. If you want to separate the anorthite or transition-metal-free stuff, you will need to figure out a way to do it.

This provides almost all the information I need. The melting issue is less problematic than it seems - if the furnace is open the vacuum lowers the melting point, usually by more than 200 K, while also reducing heat loss as there is less conduction and no convection. During the day strong constant sun is also quite plentiful. The cooling process I currently lean towards is an enclosure with a very thin, chilled nitrogen atmosphere. Complex infrastructure, but this is a long-term approach. The site has many large anorthosite boulders as it is a large young crater.


text:
How to convert Sodium Carbonate to Sodium Bicarbonate at home (CO2 scrubber)
Is it possible for me at home to react atmospheric $\ce{CO2}$ with sodium carbonate to make sodium bicarbonate.
I am seeing if I can take $\ce{CO2}$ out of the air.

A process commonly used to teach $\ce{CO2}$ scrubbing to chemical engineering students involves not sodium bicarbonate, but sodium hydroxide ($\ce{NaOH}$).
The reaction is exothermic and is as follows:
$$\ce{CO2 (g) + 2 NaOH (aq) -> Na2CO3 (aq) + H2O (l)}$$
The aqueous sodium carbonate is subsequently treated with calcium hydroxide to precipitate out calcium carbonate:
$$\ce{Na2CO3 (aq) + Ca(OH)2 -> 2NaOH (aq) + CaCO3 (s)}$$
Disclaimer:
This is usually done with much higher concentrations of $\ce{CO2}$ (e.g. >10%) than you would find in the atmosphere (~400 ppm). The point of scrubbing in a chemical plant is to reduce emissions to a level that is still much higher than atmospheric $\ce{CO2}$ levels but lower than governmental standards. I suspect that you would not see a noticeable change in ambient $\ce{CO2}$ levels from putting something like this in a place with such (relatively) low $\ce{CO2}$ to begin with. There is a very low driving force to push gaseous $\ce{CO2}$ into the liquid and solvate it when the partial pressure of $\ce{CO2}$ is low.
There exists a unique relationship between the partial pressure of $\ce{CO2}$ above a solution, the dissolved carbon in the solution, and the pH of the solution, shown here (own work):

Note that the equations on the right-hand side are in solution.
So if you wanted to see what was going on in a controlled environment with no other species in the solution / that could enter the solution, you could just use a pH probe to see how much carbon was pushed into the solution!

I am also trying to get the carbon dioxide out of the scrubber as well.

@Catprog see my edit, I expanded on my answer

Or I could just use Ca(Oh)2 + CO2 -> CaCO3

@Catprog added some more info. The CO2 needs to enter the solution first, and this is very favorable w/ NaOH (aq). This could then be precipitated out with, as you say, Ca(OH)2 (s). But a direct reaction involves a solid and a gas, and will likely be slower/achieve a lower conversion as a result (also the reaction you suggest is not possible, try balancing it to see why).

https://en.wikipedia.org/wiki/Limewater#Chemistry (I forgot the other part of the reaction)

The first reaction, Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l),  is pretty similar to my answer, except for the fact that CaCO3 is insoluble unlike Na2CO3. Only when excess CO2 is added does the carbon go back into a soluble form. On another note, biological systems are capable of effectively removing low concentrations of CO2 from air. I also suggest you look into the NaOH-based method for atmospheric carbon capture used by Carbon Engineering in Calgary, discussed in this news article: http://www.nytimes.com/2013/01/06/business/pilot-plant-in-the-works-for-carbon-dioxide-cleansing.html


text:
Calculate [Cr^3+] in equilibrium with Cr(OH4)- when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0
I tried setting up a RICE chart with the initial conditions being:
$$\begin{array}{c|c} \hline
\text{Species} & \text{Concentration / M} \\
\hline
\ce{Cr^3+} & 0.010 \\
\ce{OH-} & 10^{-4.0} \\
\ce{Cr(OH)4-} & 0 \\ \hline
\end{array}$$
I'm assuming I should use the given $K_\mathrm{f}$ value for $\ce{Cr(OH)4-}$ ($8 \times 10^{29}$).
However, when I solve for the change I end up with a negative value for the concentration of $\ce{Cr^{3+}}$? The correct answer is supposed to be $10^{-16}$.

You probably mean $\ce{Cr^{3+}}$ and not $\ce{Cr3+}$. Here is a link on how to properly typeset chemical equations and other formulae using MathJax. Also, please write down the balanced reaction and we can proceed from there.

I forgot to mention that you do not have nitrate in your query. You'll need to account for that in the overall reaction.

You don't have to worry about the nitrate. The answer is the same as if you threw some $\ce{Cr(OH)3}$ into a solution buffered at pH 10. Until you know what you are doing don't ski[p steps by doing some of this in your head. Write out reaction, then the equilibrium equation, solve the equilibrium equation for $\ce{[Cr^{3+}]}$, and then do the RICE chart.

I've written out the full RICE chart and tried setting it up multiple ways, with no luck.

Could you please add the source of the question? (Book title, page number or question number)

I know this problem and the answer the book gives, and unfortunately the book is in error. Let me explain. Adding 0.010 mol of Cr(NO3)3 will also give you 0.010 mol of $\ce{Cr^3+}$, as we can assume the $\ce{Cr^3+}$ will completely dissociate from the $\ce{NO^3-}$. The $\ce{OH^-}$ concentration can be found from the pH, and it is 0.0001 $\frac{moles}{L}$.
The "before reaction starts" gets set up accordingly:
$\ce{Cr^3+     +     4OH-       <=>      Cr(OH)4-}$
0.010 M        .0001 M                      0 M
Notice, however, that there is a 4:1 ratio of $\ce{OH^-}$ to $\ce{Cr^3+}$. It takes 4 $\ce{OH^-}$ for every 1 $\ce{Cr^3+}$, and $\ce{OH^-}$ is quite obviously the limiting reactant. So after the reaction, what is the concentration of $\ce{Cr^3+}$? Let's make it 1:1, then.
$\frac{.0001 M}{4}$ = $\ce{2.5 x 10^{-5}}$. And this is essentially the [$\ce{Cr^3+}$] and [$\ce{Cr(OH)4-}$]. However, this is not what the book says. They meant to put the concentration of $\ce{Cr(OH)4-}$ as .010 M, not $\ce{Cr^3+}$.  Let's try it that way, and see what we get!
Cr(OH)4- <=> Cr3+ + 4OH-
Here, in the equation, the concentrations are:
$\ce{Cr(OH)4-}$ = .010 M
$\ce{Cr^3+}$ = 0
4$\ce{OH^-}$ = .0001 M
-x           +x     +4x
.010 - x     +x   .0001 + 4x
Because .010 - x and .0001 + 4x are so small, let .010 - x = .010 and .0001 + 4x = .0001.
The $\ce{K_f}$ = 8 x $10^{29}$, but that is for the equation when it looks like:
$$\ce{Cr(OH)4- <=> Cr3+ + 4OH-}$$
We switched it around to find out how much $\ce{Cr(OH)4-}$ would dissolve to make $\ce{Cr^3+}$, so we have to use $\frac{1}{Kf}$, which = 1.25 x $10^{-30}$.
$\ce{K_f}$ = $\frac{[{Cr^3+}][{OH-}]^4}{[\ce{Cr(OH)4-}]}$
$\ce{K_f}$ = $\frac{(x)(.0001)^4}{(.010)}$
1.25 x $10^{-30}$ = $\frac{1 x 10^{-16}x}{.010}$
You can take it from there. The answer will be 1.25 x $10^{-16}$ M.


text:
In a neutralization reaction between a strong acid and weak base, can I assume the weak base is totally ionized?
Here's a Chemistry question from the infamously terrible Nelson textbook.

10.0 mL of $\ce{NH3(aq)}$, with a concentration of $\mathrm{5.70\dot\ 10^{-2}}$ mol/L, is titrated to the endpoint (by which I assume they mean equivalence point) with $\mathrm{2.85}$ mL of $\ce{HBr}$ solution. What is the concentration of the $\ce{HBr}$ solution?

The answer they give is $\mathrm{0.200}$ mol/L which assumes that all of the moles of $\ce{NH3}$ reacts with the acid. I am unsure of this because I thought weak bases only ionized partially, and the amount they ionize has to be gleaned from the $\mathrm{K_b}$ table.
Why, if it is at all, correct to assume all of the base reacts?

The magic word is equilibrium.

You're never going to puzzle this out without some guidance so here goes...

Why, if it is at all, correct to assume all of the base reacts?

All of the base will never react no matter how much acid you add. If you setup the equilibrium equation you can solve for $\ce{[NH3]}$ for any given pH. Thus there will always be some small amount of $\ce{NH3}$ in an acidic solution.
The gist is that there is $5.70\cdot10^{-4}$ moles of base. If you add "exactly" that much acid then the necessary assumption is that the acid balances the base and the solution is neutralized.
One other thing to consider in all of this is the significant figures. So +/- $1\cdot10^{-6}$ moles of acid doesn't make any real difference. That is enough to swing the final pH from 6 to 8.
Just like for the acid if $\ce{[NH3]} < 1\cdot10^{-6} $ then the free ammonia is negligible to the amount of $\ce{[NH4+]}$. Thus, considering the significant figures, the assumption that all of the ammonia was converted is reasonable, even though it is not actually absolutely true.

Here follows an alternative answer, but only in words.
Dissolving ammonia in water will establish an equilibrium between ammonia and ammonium ions determined by the concentration and the equilibrium constant. The pH of such a solution will be alkaline. Suppose the pH is 10.5.
By the titration with the strong acid HBr, all ammonia should be considered to be transformed to ammonium ions at the equivalence point. Since we have added a strong acid to the ammonia solution, the pH will drop to about 5 - 6.
The interesting part is what you get at the equivalence point.
The equivalence point will be exactly equal to dissolving pure ammonium bromide in water in the same concentration as you have at the equivalence point. What will the outcome be of doing such a thing?
Well, the ammonium ions of ammonium bromide will react with water and equilibrate. Ammonia will be formed in equilibrium with ammonium ions This is similar to what we had at the starting point, when ammonia reacted with water. The big difference is that the pH is not the same. A new equilibrium will be established. At the new (lower) pH the relative amount of ammonia will be much lower than at the starting point, i.e. before the titration was initiated.
So, even if you have transformed all initial ammonia to ammonium ions by the titration, there will be ammonia present at the equivalence point because of the new equilibrium established at the lower pH.

Yes. Assume that at the equivalent point, all of the base and all of the acid are "neutralized". That is there is an equivalent amount of acid and base present at the equivalence point, by definition. If you have not worked through the logic of the weak/strong acid/base reactions (WA+WB, WA+SB, SA+WB, & SA+SB ) you should. here's a fairly simple (actually somewhat oversimplified --  NH3 not NH4OH is the principle electrically neutral species) http://www.tutorvista.com/content/chemistry/chemistry-iii/ionic-equilibrium/strong-acid-and-weak-base.php.
For a WA+SB salt, its solution (in water) will be basic (pH > 7). For a SA+WB salt such as ammonium hydroxide + hydrobromic acid, the pH of salt solutions will be acidic (pH < 7). Meaning that the titration of NH3 with HBr will have a pH < 7 at the equivalence point. (For a volatile salt such as this there are other potential issues such as does the solid salt decompose? does either the Br- or the NH3 volatilize from solution? BUT ignore these fine points, they're not really relevant to your question) In fact you can calculate the pH, when given the pKa of both (or pKb). Look at it this way. Some of the NH3 is ionized to NH4 and OH- but adding H+ will neutralize all of the OH (in excess of pH 7) allowing more NH3 to ionize. (Pushing the equilibrium toward the right.) The acid pushes the pH down. Meaning the solution will be "more acidic" than what it would be with a strong base. Even at high pH (> 7) there is more H+ than expected (but this is not saying that there's more H+ than OH- above pH 7(!), just there's more than "expected" (for a stong base)). (This is also not saying that at a specific basic pH, that there is more H+ than what is being measure, rather it is saying that the pH is lower than what would be expected given the concentration of Base+ ions present, since WB exists both as B and B+ in basic solution.


text:
Determining the coupling on a substituted benzene ring
I can't seem to find any clear explanation of how benzene ring protons couple with each other when mono and disubstituted with the same constituent.
My professor claims that if a benzene ring is para disubstituted with X, the remaining 4 protons exhibit a doublet interference pattern. This seems strange to me because they are equivalent, and my last two organic chemistry professors said the information isn't useful diagnostically, but never explained why.
Also, he claims that if a benzene ring is monosubstituted the proton in the para position will display a triplet interference pattern. Is this right? Symmetry seems to say that the proton would be a doublet.
Do benzene rings have their own rules for coupling?

Benzene rings follows the same general principles as other spin systems for coupling.
A symmetrically para-substituted benzene, such as 1,4-dichlorobenzene, will have a 1H spectrum which will contain just a singlet. All four proton environments are identical. An XY para-disubstituted benzene will have the aromatic peaks that appear somewhat like two sets of doublets. The appearance is actually incredibly useful diagnostically,as it gives a useful indication of the type of ring substitution you have. They are, however, not true doublets, and consist of second order coupling arising from coupling between magnetically non-equivalent sites. That is, the splitting of the apparent doublets is not a true reflection of the coupling constant.
The para proton on a mono-substituted benzene ring does appear as a triplet. The ortho coupling giving rise to the ~7Hz splitting comes from coupling from two chemically equivalent protons at the meta position. Coupling to 2 equivalent protons will give rise to a 3 line splitting (from your 2nI+1 rule). Strictly speaking, a mono-substituted benzene ring is just a very special subset of a para-disubstituted benzene ring system, but does not typically show the classic splitting expected in a para system, because the 5J coupling is essentially zero.


text:
How can I learn Qualitative Analysis and all its equations?
Qualitative Analysis, as all may know, has so many equations and each element has its own specific properties - colour, reaction, test etc. How can I learn all this?

There is not really an easy way out. You'll have to study the compounds, the trends within periods and groups, specific reactions and properties, and comparative reasoning. There is no shortcut. After you have obtained a decent basic knowledge you'll see the links between theories and behaviours of atoms and molecules.

Well, I'm quite sure I have obtained the decent basic knowledge.

Spend more time with each reaction. Try to identify the reaction types-redox reaction, acid-base, hydrolysis, etc. During this process, you will realize not all the reactions are completely independent. Now, there are a lot of facts that you are supposed to remember, e.g. Try to group the compounds that you find similar. For eg. color of the compounds or specific complexes. Note patterns, e.g. All sulphides that you will encounter in your course will be black except CdS(yellow), MnS(pink), SnS2(yellow),SnS(brown), ZnS(white). Most chromates are yellow and phosphates and sulphates are white.

As Martin has pointed out in his comment, there is no way to get around learning, but you might want to look for a system that interconnects the seemingly overwhelming amount of facts.
In German chemistry education, the Kationentrennungsgang is a systematic workflow to separate and qualitatively detect cations from a complex mixture.
In the course of this workflow,

groups of cations are subsequently precipitated from the mixture and then
analyzed separately by bringing them into solution again, one by one.

In the simple Kationentrennungsgang, the cations are separated into five groups, namely

hydrochloric acid group
hydrogen sulfide group
ammonium sulfide group
ammonium carbonate group
soluble group

Typically, you start by adding concentrated concentrated hydrochloric acid to your sample. This will precipitate  $\ce{Hg2Cl2}$, $\ce{HgCl2}$, $\ce{PbCl2}$, $\ce{AgCl}$, and $\ce{TlCl}$, if present. Cations of the groups 2-5 will remain in solution.
To repeat the sad news: There is now way around memorizing a lot of facts, but using the Kationentrennungsgang, you will learn a lot about periodic trends.


text:
Difference between polarity and polarizability
How do the terms polarity and polarizability differ from each other?
In my opinion, polarity is the degree of ionic character in covalent compounds and polarizability implies towards degree of covalent character in ionic compounds. Am I correct?

Polarizability is the ability to get polar when subjected to electric field. This hardly has anything to do with covalent/ionic distinction.

Polarity and polarisability are almost entirely unrelated.
Polarity refers to the distribution of partial electric charge in molecules (or even materials). Or, more technically, the degree to which a molecule has a dipole moment (which relates to its inherent electric field). Some molecules, like benzene, are non-polar as they have no net dipole moment; others, like nitromethane, have a strong dipole moment. Polarity is a fixed property of the molecule that doesn't depend on the external field.
Polarisability refers to the degree to which the electron clouds in a molecule or atom can be influenced by an external electric field. Everything, polar or not, has a polarisability. Crudely (i.e. simplifying a lot) you can think of it as the degree to which electrons are held tightly in the atom or molecule. So xenon atoms are fairly polarisable compared to helium atoms as their electron cloud is more spread out and less tightly bound. Water is very polar but a lot less polarisable than hexane which is non-polar. Bulk substances have complicated polarisability that depends on the crystal structure and may be inhomogeneous.
Polarity is about the ground state electron distribution; polarisability is about how that distribution changes when subject to an external electric field.


text:
What is the molar mass of a compound with 0.5 atoms in its molecular formula?
I'm trying to calculate the theoretical specific capacity of the $\ce{LiFePO4}$ cathode for a fully discharged state, let's say when the cathode has the following notation $\ce{Li_{0.5}FePO4}$. And I'm using the following formula:
$Q=\frac{nF}{3.6M}$ where F is the Faraday constant
The question is slightly related to this thread. The question is, what is the molar mass (M) of $\ce{Li_{0.5}FePO4}$.
Which one is the correct one for finding the specific capacity:

M = 157.751 g/mol, as if Lithium would be taken as a whole?
or
M = 154.285 g/mol, as if only half of Lithium would be considered?

Can anyone help me with this?

Counting the Li in full is of course totally wrong. Point is, as I already told you with your earlier question, this is a nonstochiomtric compound (the 0.5 is an approximate). It makes little sense to calculate its molar mass. What do you want it for?

Thank you for your answer! I would like to calculate the theoretical specific capacity of LiFePO4 when the cathode is fully discharged (Li0.5FePO4).

So? 0.5 mol of Li are 3.5 g.

Well, that was the question. So I need to consider half of the mass, right?

Please change the question so it actually describes a problem, including the math you intend to do. At the moment, you seem to be just guessing, and Im guessing at your guesses. Otherwise this question should get closed.

You still don't say what the actual question is, how you want to use that formula, etc. Tip: If a question that seems simple proves hard to answer, then it usually just badly put.

Sorry, Karl. I admit I have a problem describing my problems. Especially in front of people that have experience with things.  But thank you for explaining. I will work on my skills to formulate questions. Now I understood that I need to consider 0.5M

You should use 0.5 lithium atoms in the molecular mass. Alternatively you can multiply the formula indices by two to obtain $\ce{LiFe_2(PO_4)_2}$. This formula describes exactly the same compound but it's much more readable, for instance is now clear that the substance contains one $\ce{Li(I)}$, one $\ce{Fe(II)}$, one $\ce{Fe(III)}$ and two phosphate anions per $2n$ electrons transfered.
Here the value of $n$ is multiplied by 2 because you also have your stoichometry multiplied by 2. If you use $\ce{Li_{0.5}Fe(PO_4)}$ to calculate $M$ then the value of $n$ stays unchanged.

Thank you for clarification. Multiplying it by 2 is a nice way to understand it.


text:
Could the difference between absorption time of Diclofenac Sodium and Diclofenac Potassium be related to the cation?
Diclofenac is a common non-steroidal anti-inflammatory drugs (NSAID) that comes in a variety of formulations. Two of the most common forms are diclofenac sodium and diclofenac potassium (the links refer to the FDA prescribing information of both forms). I have heard several times that the Potassium salt starts to act faster than the sodium salt.
In the second page of both leaflets above there is a table with pharmacokinetic parameters:

For diclofenac potassium, Tmax (the time it takes to reach maximum concentration after one dose) is 1.0 hour, and the half-life (the time it takes the drug concentration to go down by 50%) is 1.9 hours.
For diclofenac sodium, Tmax is 2.3 hours, and the half-life (the time it takes the drug concentration to go down by 50%) is 2.3 hours too. This means that for the sodium salt is being "processed" (absorbed, metabolized, eliminated) by the body more slowly than the potassium salt.

One more important point of difference to note is that the sodium salt comes in enteric-coated tablets, while the potassium salt comes in immediate-release tablets.
My question is: Can the difference in "processing time" of both salts by the body stem from the different salts, or is it due to the pharmaceutical formulation (EC tablets compared to IR tablets)?
It might seem at first glance that the difference in formulation should be the answer, but if it is so, why is the cation different between formulations?
I hope anyone can shed light on the mechanism, I was unable to find any useful information that actually addressed this issue.

I'd guess that the difference is due to different $\text{T}_\text{max}$ values. So the peak concentration of diclofenac itself would be greater for the potassium salt, and lower for the sodium enteric-coated salt. So at the peak of the sodium enteric-coated form some of the salt is still dissolving.

Under physiological conditions, the counter-ion is almost entirely dissociated from the drug, meaning that the identity of the counter-ion isn't directly responsible for how well a drug performs. This has been looked at extensively by medicinal/formulation chemists, who often screen many salt forms looking for desirable properties.
From the graph below, you can see that the plasma concentration curves are essentially identical for sodium, potassium, and calcium salts (slight variation in Cmax and Tmax but within experimental error):

Source: Drug Like Properties: concepts, structure, design and methods, Chapter 7
One case in which the identify of the cation does matter is where there is a common ion in the body. This 'common ion' effect is most common when amine drugs are presented as their hydrochloride salts. The stomach is acidic (hydrochloric acid), and therefore the presence of a high concentration of chloride ions can prevent the salt from turning into the free drug.
What is likely to be different between different cations the form that the salt takes. Different salt forms have different packings (some might be amorphous, some might be highly crystalline and ordered). The formulation may also differ (powder coated tables vs gelatine capsules), which will change the time it takes for the drug to be released/absorbed.

Can you link the reference for this graph? Particularly, was this a mouse study?

The reference is below the graph already, you'd need to look up the book

This is the original paper: http://onlinelibrary.wiley.com/doi/10.1002/jps.2600630512/full

My guess is that this would due almost entirely to the formulation. The sodium and potassium are ions and bound ionically to the organic portion of the molecule. That is to say, these compound are "salts". When in the aqueous tissue of your body the Na and K ions likely dissociate to a large extent leaving identical anions which offer the physiological effect.


text:
Reaction of Boron with Alkali
In one of my books it is written that boron does not react with alkali but in another book it is written that it reacts with alkali.
Can any one help in clearing this confusion?

Could you show us those reactions?

When boron reacts with alkali the reaction is this.           3B+NaOH -> 2Na3BO3+3H2O

That equation shouldn't make sense to you. Boron is being oxidised from 0 to +3 but nothing is being reduced.

Boron react with $\ce{NaOH}$ but under certain conditions. Boron exist in two form (1) crystalline form (2) amorphous form. Both forms are generally inert to chemical reactions and is unaffected by acids and alkali at normal conditions but at elevated temperature, concentrated oxidizing acids like nitric or sulfuric acids react with boron to yield boric acid. On the other hand, amorphous boron does react with fused alkali at high temperature to obtain trisodium borate:
$$\ce{2B(s) + 6NaOH(s) ->[t > 773K] 2Na3BO3(s) + 3H2(g) ^}$$
High purity, hydrated sodium metaborate(sodium tetrahydroxoborate, $\ce{Na[B(OH)4]}$) can also be produced from boron in a $\ce{NaOH-H2O}$ system. The process applied was called pressure alkaline leaching.
When fused in the presence of air at $\pu{350-400 ^\circ C}$, sodium metaborate is formed. You can find the reaction in this answer by @andselisk.
References

https://onlinelibrary.wiley.com/doi/abs/10.1002/0471743984.vse1217
Yin, B.W & Qin, S.Y & Zhang, Y.F & Zhang, Y (2014). Preparation of hydrated sodium metaborate from boron concentrate via pressure alkaline leaching. Guocheng Gongcheng Xuebao/The Chinese Journal of Process Engineering. 14. 258-265, (link)

Could you provide other references for the reactions (not chemiday)?

@AshishAhuja see andselisk's answer: https://chemistry.stackexchange.com/questions/118485/reactivity-of-boron-towards-acids

The first link appears not to involve Chemiday, but attempting to click on it gave me an antivirus firewall.  I am unable to sustain an upvote until references more reliablexand secure are found.

@OscarLanzi answer reconstructed, removed the dead link, removed all of chemiday BS.

Your reaction is far from balanced but what you want seems to be trisodium borate.
According to this source (emphasis mine),

Boron reacts with fused sodium hydroxide to form sodium borate and hydrogen.

The cited reaction (already balanced) is the following:
$$\ce{2 B + 6 NaOH -> 2 Na3BO3 + 3 H2 ^}$$
This other source says something similar (again, emphasis mine),

It can also dissolve in alkalies and evolve hydrogen.

So, as it seems, boron reacts not with sodium hydroxide solutions, but with the pure, molten compound.
This may be the source of confusion.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Newer ionic radii data than Shannon's 1976 compilation?
Shannon, R. D., 1976, Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides: Acta Crystallographica Section A, v. 32, p. 751–767. DOI: 10.1107/S0567739476001551 provides a very detailed list of ionic radii for just about any element, for various oxidation states and coordination numbers.
This paper was published in 1976. Is there a newer compilation with new data?

The 82nd edition of the CRC (2001-2002) has tables with data from the Shannon paper you sited plus data from Jia that was published in J. Solid State Chem, 95, 184, 1991.  I have not seen the paper, and that is still pretty old, but a bit newer than Shannon anyway.

@airhuff Of course, the CRC handbook. I'll go and hand a look.

And consultation of this very page equally states that by today (2022-02-02), the publisher is aware of 50411 publications citing this work.  Though there may be some overlap among the 417 in IUCr's journals, 44350 indexed by cross-ref, 50425 indexed by Web of Science, and 50295 indexed by Scopus (reference page).

tldr; it's very difficult to determine consistent additive ionic radii
For various reasons, I've gone back to look at covalent radii recently (e.g., for estimating bond lengths).
In that case, there seem to be fairly useful recent work, e.g. by Pekka Pyykkö:
"Additive Covalent Radii for Single-, Double-, and Triple-Bonded Molecules and Tetrahedrally Bonded Crystals: A Summary"
J. Phys. Chem. A 2015, 119, 11, 2326–2337
For covalent radii, it's clear that one can fit experimental and theoretical interatomic distance r(A-B) to a sum of two covalent radii with systematic accuracy. Notably, Pyykkö defines not only single-bond but double-bond and triple-bond radii for almost all elements.
(N.B.: I'm not sure it makes sense to consider a triple bond to $\ce{Br}$ or $\ce{I}$ but if you want them, it's in the article.)
There are also recent systematic van der Waals radii, e.g., Cramer and Truhlar:
"Consistent van der Waals Radii for the Whole Main Group"
J. Phys. Chem. A 2009, 113, 19, 5806–5812
So what about ionic radii?
There's a really nice article by Liu, Schwartz, and Li, cited by Pyykkö about the problems with ionic radii.
"On Two Different Objectives of the Concepts of Ionic Radii"
Chem (Eur. J.) 2013, 19, 44, 14758-14767
The Liu, Schwartz, and Li point out the competing problems with ionic radii:

Since “bonded atoms” are evidently neither rigid nor spherical, quantum
computational chemists apply “valence multiple zeta and
polarized” basis sets in their molecular calculations. Conversely, X-ray ED researchers use the procrystal model of independent rigid spherical atoms for starting and screening purposes in their software for the processing and analysis of the observed diffraction data.

In other words, we like to consider atoms as having consistent, rigid spherical shapes. For ionic compounds, that's hard to determine, because of charge transfer. By definition, $\ce{A+B-}$ indicates cation $\ce{A+}$ has given electron density to $\ce{B-}$ .. but how much? Is there any covalent character and shared electron density? How do you divide?
Pyykkö also mentions the problem in his covalent radii work. His atomic radii work:

… provided that the A–B bond is not too ionic, or the coordination numbers of the two atoms too different from the original input data, used in the fit.

In short, covalent radii are easy to define because most of the electron density is shared and covalent bonds are thus fairly predictable. In contrast, almost no compound is purely ionic so it's hard to perform the division for ionic species.
Here are two examples from Liu, Schwartz, and Li:

Consider interatomic distances from alkali halide crystals. (The portion is performed by Bader's AIM analysis.)

Note that there's no "vertical" or "horizontal" line for well-defined anion or cation radii. Maybe you can come up with a correction based on electronegativity difference .. but the cation radii particularly depend on the anion.

The same problem shows up for monomeric alkali halide molecules in the gas phase.

It makes some sense. It's difficult to rigorously know how much charge transfers from anion to cation (vs. shared), which is why calculation methods define multiple algorithms to determine it. (Atomic partial charges are something of an eternal debate.)
I wouldn't be surprised to find someone creating a correction using electronegativity differences or covalent character. And it's possible someone will sweep this away with a compelling article - that's the point of science. But it won't be easy, because we expect ionic radii to serve a lot of different purposes.

I'll mention that there's a 2022 article by Hendon's group in Chem. Sci. defining volumes and radii based on electric fields. It's an interesting method, but the radii don't compare directly to Shannon radii. I think it's nevertheless important as a method to define ionic radii in a consistent manner.


text:
Removal of nickel plating on MgB2 without damaging the boride?
I need to remove a nickel plating of a MgB2 wire without causing damage to the MgB2. I felt that electrolysis is the best method but I do not know what electrolyte to use for this. I also need to coat it with a thin layer of copper and would like to know how I can do this without oxidation to the filament.

Could you just plate Cu over the Ni?

As you said you can use electrolysis to strip nickel. Use sulphuric acid diluted in distilled water with concentration of 10-20 % as electrolyte. Submerge the nickel-plated MgB2 after attaching it to the negative terminal of power source in the electrolyte, then submerge the cathode electrode in the electrolyte.
You can use a stirrer to speed up the process and make sure that electrodes don't touch each other.
Now switch on the power source and apply a direct current(DC). Voltage around 10 would do and make sure that you increase the  voltage slowly from the start to get to 10V.
To deposit a thin layer of Cu again use the same process. Use electrolyte as CuSO4 solution (150-250 gm dissolved per litre water), go with the range around 150 cause you want to deposit a thin layer.
Submerge the MgB2 wire as cathode and a copper electrode as anode in the electrolyte.
You can use a stirrer to speed up the process and make sure that electrodes don't touch each other.
Now turn on the power source and slowly increase the DC form 1V and reach upto 3V.
FOR REFERENCE:
https://www.smokstak.com/forum/threads/removing-nickel-plating.61941/
and NCERT

I did not downvote, but I think the answer is more credible if you cite sources for the procedures you recommend.

sure will sir .

is it any good now?

@Anjankumar Not the downvoter either, but I don't see that you've added any citations. You just made it longer and more detailed (which is GREAT), but please cite your sources.

@Anjankumar I've upvoted your answer, but please cite your sources. // From where you have researched and given your answer, that is.


text:
Why does 3-bromopropanoic acid have least pKb value when compared to 3-chloropropanoic acid, 3-fluropropanoic acid and fluoroacetic acid?
Shouldn't 3-bromopropanoic acid have the least pKb due to the reduced inductive effect and electronegativity, leaving the carboxylate anion with more electric charge on it?
Also, what will be the decreasing order of pKb?

Least pKb implies that it is the most basic of all four, which can be predicted using the inductive effect logic as you figured out.

The correct order of PKb is
fluoroacetic acid> 3-fluoropropanoic acid>3-chloropropanoic acid>3-bromopropanoic acid.
To find the order of pKb, first find the order of pKa which can be found by checking the stability of negative charge on carboxylate anion once it donated proton.
Halogens are electron withdrawing groups. The halogen which will withdraw the most electron density will stabilise the anion the best. Fluorine, being the most electronegative withdraws the most electron density thus reducing the charge on oxygen and stabilising the anion, therefore having the least pKa and maximum pKb.

RE: "To find the order of pKb, first find the order of pKa which can be found by checking the stability of negative charge on carboxylate anion once it donated proton." This seems backwards. "The stability of negative charge on carboxylate anion" is essentially the pKb. I think rather you need the notion that pKa + pKb = 14 for aqueous solutions.


text:
If some particles of liquid in a solution convert to vapour then why can't some liquid particles convert to solid?
I was reading about the vaporisation of liquid when I came across the following statement.

The average energy of the particles in a liquid is governed by the
  temperature. The higher the temperature, the higher the average
  energy. But within that average, some particles have energies higher
  than the average, and others have energies lower than the average.Some
  of the more energetic particles on the surface of the liquid can be
  moving fast enough to escape from the attractive forces holding the
  liquid together. They evaporate.

Now if some particles have energies lower than the average, then why don't they convert into solids in the same way the particles with higher energies convert to gases.
If we cover a vessel filled with water with a plate, after some time, we observe droplets of water accumulated at the plate because of vaporisation of water. Why don't we observe any changes that show liquid particles converting to solid?

If water molecules go into the gas phase from the liquid then they are in a different phase. But water molecules go into the gas phase one at a time. If there are the same number of water molecules leaving the liquid as coming into it then the atmosphere above the liquid would have 100% humidity and be in equilibrium with the liquid.
Now if some small number molecules of water happen to line up to form an ice crystal within the water itself what happens? Such an ice crystal is thermodynamically unstable and then "melts" within the liquid phase. Since the whole process takes place in the liquid phase it is virtually undetectable.
Now you also have to realize that form ice crystals big enough to detect visually requires more than a handful of molecules. You'd need billions of billions of molecules. That is just a statistical impossibility.

why is the crystal thermodynamically unstable?

Thermodynamics says that you shouldn't have ice in water that is above 0 C.

Thermodynamics also says that you shouldn't have liquid water that is below 0 degrees C.  Kinetic barriers allow it to happen anyway.  Just a comment on your comment, not relevant to the question ;)

@airhuff - Understand, but where do you try to cut such an answer off?

Not sure I understand the statement.  I +1'd the excellent answer and made it clear that my commentary was only directed at your previous comment.   Just highlighting the point that thermodynamics alone doesn't rule all physical / chemical processes. For further example, you can also have, with great difficulty, superheated ice. In one experiment ice was superheated to 17 degrees C (albeit for a few hundred picoseconds).  I think these things would be of interest to the OP even if his question has been appropriately answered.  I hope I haven't misunderstood your question.

@airhuff - Running joke in my family is that the Dad answer is a 5 minute oration. Mom answers "yes" or "no."

Ohhh, I get it now.  Sorry for the misunderstanding.

The situation you describe, where both solid and liquid exist, would be possible around the freezing point. Coexistence of vapour, solid and liquid is possible if there is a the triple point (for water: 273.16 K; 611.657 Pa).
But in standard conditions, MaxW is right the crystal melts, because the entropy gained by spatial randomization of the particles compensates the energy lost by breaking their packing.


text:
Verifying data acquired from a molecular dynamics simulation
This question is about verifying a system simulated with MD. I am currently building a model of a two phase gas-liquid system, where I initially have $\ce{N2}$, $\ce{CO2}$, $\ce{Ar}$, $\ce{He}$ and $\ce{CH4}$ in the gaseous phase and $\ce{Na+}$ with $\ce{Cl-}$ in an aqueous phase, which is modeled with TIP3P water. The most abundant component of the gases is $\ce{N2}$, which comprises over 95% of the gas.
I have a certain, non-ambient $p/T$ setting with a relatively high pressure (30-40 kbar). My idea is to study partition coefficients using molecular dynamics in multiple different $p/T$ settings. I'm currently using OPLS/AA force field which models intramolecular potentials with harmonic potentials and intermolecular potentials through Ewald-summation and LJ-612 -potential.
Currently I have just equilibrated the system, so that $p,T,V$ and $E$ do not fluctuate much. Before moving onwards with this system, I would like to verify that it produces reasonable data which is on par with other classical computational simulations.
I have noticed that there are not many papers which have focused on a system similar to mine, and I'm probably not going to be able to find an exact match in which to compare to. So, while I've been looking through the literature, I've looked at the following parameters to get a feeling of how well the system is behaving:

Radial Distribution Functions (RDF).
Partition Coefficients of the gases.
Long term behaviour of the system (no odd behaviour such as crystal forming etc.).
Densities of phases ($\rho_{\text{gas}} < \rho_{\text{liquid}}$)

So far these have all been quite reasonable. Are there any other standard methods and/or values that I should be looking at? I am obviously constantly reading through literature to find other results for comparison.

You could try simulating a special case/known system that is close to your system using the same methodology and comparing the results with experiment.

Thanks for the input Felipe! That's exactly what I've been doing. The problem (at the time I was writing the original question) was that I could easily find an abundance of information regarding to the behavior of single compounds (such as methane gas) in TIP3P water, but the amount of information about mixtures such as mine was sparse. I'm fully aware that I'm probably not going to find a 1:1 match from literature. I'm currently looking in to getting hydration free energies for the gaseous compounds and comparing those to experiment.

OPLA has been parametrized based on density and heat of vaporization of liquids. So, I'm afraid of using that for the gas phase. You'd better seek for another FF. Goodluck

Also include Boltzmann distributions of bond lengths and angles? I know that much of this behaviour is captured in the RDF but I find that the distributions are easier to compare and also mean values are widespread in the literature

With the simple molecules in your system, using the Gibbs Ensemble Monte Carlo would be the way to go. Each phase gets its own simulation box, and particles are swapped back and forth until equilibrium is reached.
Cassandra, DL_MONTE and other programs can do this easily for you. Once equilibrium is reached you can sample each box in the NVT or NPT ensemble at that composition.


text:
If copper is insoluble in excess NaOH, does it mean it is soluble in NaOH when not in excess?
I don't understand, does it mean when i add a little bit of NaOH, there will be completely no precipitate/ it will dissolve?

No. Some compounds say $\ce{X(OH)2}$ will form an anion complex in strong (excess) NaOH such as $\ce{X(OH)4^{2-}}$

On the other hand, I do not quite understand this question :). Please edit to clarify.

I don't see why this question is unclear to many. Although the question has not been nicely framed, it is still clear that the OP is talking about the specific precipitation reactions where adding excess of reagent dissolves the precipitate.

When less amount of $\ce{NaOH}$ is added initially, $\ce{Cu(OH)2}$ will be formed and leave the solution as a precipitate. But when excess $\ce{NaOH}$ is added, $\ce{Cu(OH)2}$ will form soluble complex with excess $\ce{NaOH}$.
Complex formation increased the solubility of $\ce{Cu(OH)2}$.
Now, this happened only because copper hydroxide had a tendency to form complex with ex $\ce{NaOH}$ and complex formation was favourable in this case.  Adding excess reagent will only lead to solubility of precipitate if the precipitate has a tendency to form a complex with excess of that reagent.
If "X" is insoluble in excess reagent "Y", then it WILL NOT BE  soluble in "Y" when not in excess.
If "X" is soluble in excess reagent "Y", then it will only be soluble in "Y" if "X" has a tendency to form complex with "Y" and complex formation is favourable.
You will come across a lot such reactions in inorganic chemistry.
e.g $\ce{AgCl}$ dissolves in excess ammonia due to formation of silver-ammonia complex.


text:
Do different vibrational modes correspond to vibrational energy levels?
Do different vibrational modes correspond to vibrational energy levels? Or all modes can be found in ground state of the molecule? Or each vibration for each bond has its own vibrational energy levels? I don't understand, please help.

As a generality different vibrational modes will have different energies. However it is possible that two different modes have the same energy. // In the ground state there would be no molecular vibrations. // Sometimes the molecular vibration can be considered as only one bond vibrating. // Look at Infrared Spectroscopy for more details https://en.wikipedia.org/wiki/Infrared_spectroscopy

@MaxW Are you claiming that "in the ground state there would be no molecular vibrations"?

There are molecular vibrations in the ground state. I've searched a bit and apparently each vibrational mode has its own ground frequency and energy and different vibrational energy levels.

Each vibrational mode has its characteristic frequency $\nu$ (usually reported as the wavenumber $\bar{\nu} = \nu/c$, $c$ being the speed of light). That's all, really; the rest follows from this, since the ground-state energy (zero-point energy) is $h\nu/2$, the vibrational spacings are $h\nu$ and the energy levels are $(n+\frac{1}{2})h\nu$ where $h$ is Planck's constant and $n$ the vibrational quantum number.

Hmmm... It has been a long time. I was thinking the "ground state" was at absolute zero. If I'm wrong please correct me...

@MaxW there exists a zero-point energy (as an example, consider the quantum harmonic oscillator)

@orthocresol I only don't get one thing. So the frequency of vibration upon absorbing energy hν stays the same or is it changed, because now is energy (n+1/2)hν?

No, you're confusing a classical frequency with a quantum frequency.

yes and no: all vibrational normal modes have a different energy (with the proviso that some may be accidentally degenerate) and hence different absorption frequency. (Fundamental frequencies have vibrational transitions that occur with selection rule $\Delta n=\pm 1$.) All normal modes exist in their ground state with zero point energy (quantum number $n=0$) as well as in other levels based on the same mode ($n>0$) which are populated according the Boltzmann distribution at temperature T. Thus each basic vibrational mode i has its own stack of energy levels, with quantum numbers $n= 0, 1, 2 \cdot\cdot$ and (harmonic oscillator) energy $E_i=\nu_i(n_i+1/2)$ with quantum numbers $n_i$ and frequency $\nu_i$ in  $\pu {cm^{-1}}$. The total zero point energy of the molecule is the sum of all the individual zero point energies above the zero of energy.
The number of vibrational levels populated at a given temperature depends on the vibrational level spacing, thus low frequency modes, say, $500 \pu{cm^{-1}}$ have far more levels populated than a high frequency, say $1500 \pu{cm^{-1}}$ mode.
As the energy increases the number of vibrational energy levels / unit energy can become very large indeed in polyatomic molecules. There are overtones where only one vibrational mode is excited, but with more than one quantum of energy, and combinations where two or more different vibrations of different frequency are excited and so each have non zero quantum numbers.
In water molecules there are 3 vibrations (of frequency approx $3650,1595, 3756 \pu{cm^{-1}}$) if we label them in order with quantum numbers then [000] is the zero point energy $(3650+1595+3756)/2=4500 \pu{cm^{-1}}$,  [010] a fundamental in the second frequency and [020] an overtone in this mode and [110] a combination of modes 1 and 2 with energy above zero of $4500+3650+1595 \pu{cm^{-1}}$

Answer to your question. Yes and no. Different vibrational modes may correspond to different energies and may not (it all depends on the molecule and the vibrations concerned). The energies may come to be very close sometimes (degeneracy).
Explanation:
Any molecule has certain number of vibrations possible. These are given by, $(3N-5)$ for linear molecules and $(3N-6)$ for non-linear molecules. These vibrations are called vibrational modes and correspond to distinct types of vibrations (careful, we have not discussed energies yet).
As a simple approximation, the vibration can be modeled as harmonic oscillator. Then each of the mode is an individual harmonic oscillator. (In a real molecules these are not entirely independent but we are working under approximation). Then, if we consider energies corresponding to the vibrational states (regardless of the mode it comes from) in increasing order, we get a table of energies (which are observables either in IR or Raman spectra or in both IR and Raman). In literature these energies are expressed in units of wavenumbers ($cm^{-1}$).

In experimental vibrational spectroscopy, we measure energy differences between the initial and final state and hence we have measurement of energy of first state, second state and so on. The ground state's energy is measured by other techniques. Due to zero point energy it is non zero and has some value in appropriate energy units.
To be more clear, I give example:

i) water: The following data is taken from this paper (http://www.sciencedirect.com/science/article/pii/S0022407312004311) from page 45.

From the above image you can see that, the energy levels are arranged in increasing order. $\nu_{1}\nu_{2}\nu_{3}$ are the vibrational modes (or individual oscillators), and the states occupied are mentioned using numbers (000), (010) and so on. ($\nu_{1}$=symmetric stretch; $\nu_{2}$=anti-symmetric stretch ; $\nu_{3}$=bending).


text:
What is an NOx level
What is an $\ce{NO_x}$ level? I have read some online - but I'm confused as what is used to calculate $\ce{NO_x}$. Reported $\ce{NO_x}$ levels by the government (where I live - Canada) have an $\ce{NO_x}$ level with the attached "On an $\ce{NO2}$ mass basis" What does that mean? I said $\ce{NO_x}$ doesn't exist.

http://www.icopal-noxite.co.uk/nox-problem/nox-pollution.aspx

Concentration level of various oxides of nitrogen in atmosphere.

The atmospheric $\ce{NO_x}$ (usually pronounced as "nox" or "knocks") concentration is the sum of the $\ce{NO}$ and $\ce{NO2}$ species.  These get combined into a single term as they each play a roll in photochemical ozone formation. $\ce{NO_x}$ can form as a byproduct of the combustion of fuels containing nitrogen, like coal, or from atmospheric nitrogen in the high-temperature environment of an internal combustion engine.
Because photochemical ozone production requires $\ce{NO_x}$, volatile organic carbon (VOC) and sunlight, the $\ce{NO_x}$ concentration is really most relevant for environments already rich in VOC that get plenty sunlight.  Modern automobiles have emission systems that have significantly reduced VOC emissions.
Depending on just where you live in Canada, $\ce{NO_x}$ levels are likely very important for photochemical ozone production during the summer, when you have plenty of natural VOC's available from local vegetation (be it forest, tundra, or any other of your rich and diverse ecosystems).
During the wintertime in Canada, or summer in places like the desert Southwest US, the atmospheric VOC concentrations are small.  In these cases it is the VOC concentration rather than that of $\ce{NO_x}$ that limits ozone production.  Of course the deserts also get plenty of the other ingredient, sunlight.
Depending on your latitude, there may be enough sunlight to produce some ozone, and due to temperature inversions the lower rate of ozone production can still lead to the buildup of ozone; so long as you have some $\ce{NO_x}$ and VOC's present.

If eighteen hundred tons of NOx was released - does that mean nine hundred was NO2 ? ( I understand they add the sum together ) do you "not" add all the other combination possible ( N2O etc etc ) ?  Is the NO2 level the most important ?

NOx is the sum of the two mono-nitrogen oxides $\ce{NO}$ and $\ce{NO2}$ only, ie. not $\ce{N2O}$, etc.  Regarding your first sentence, no, that's not quire right. If eighteen hundred tons of NOx were released, there could technically be any amount of $\ce{NO2}$ released up to eighteen hundred pounds, with NO making up the rest of it so that the total equals eighteen hundred pounds.  Let me know if that doesn't make sense :)

All of this is great btw / thank-you . Is NOx a science ? Or .... is it a generic term that represents ( what ever proportion  may be ) levels . If this is true , do we know what the actual NO and NO2 levels are ? Is NOx only relevant in climate change ? NO2 is poison right ..... ? ( I'm being straight up - just a working class guy with a question ) Thank you so much

@Sam, great that this is making some sense. Atmospheric NOx is most important in it's roll in ground-level ozone production (that's bad, not like high in the stratosphere where we need an ozone layer).  Our air is mostly made of nitrogen and oxygen, but under "normal" conditions they don't react to form NOx. Under the heat of an internal combustion engine, the do. The reason for lumping the 2 together is that they are part of a cycle; during daytime, NO2 is photochemically converted to NO producing ozone along the way. At night, the ozone converts the NO2 back to NO again. I hope that helps ;)


text:
Acid and Silicon
What kind of acid would I need to dissolve a silicone compound? Or perhaps silicon in its purest form? I am writing a novel so a layman's level answer would be appreciated.

First off, welcome to SE.Chemistry! This might help. The components of antifreeze (ethylene glycol, propylene glycol) might be the most realistic compounds that someone would have at their home.  The sited article also says some silicone is somewhat soluble in ethyl alcohol and isopropyl alcohol, so maybe that would work. Those aren't acids though. "The butler, holding the murder weapon, spilled the Homemaster's evening shot of everclear..." I think I'll stick to chemistry ;)

You should tell what exactly you have in mind. Pure silicon isn't common outside of electronics, and there's a lot of compounds containing this element.

Silicone and silicon are very different. Which one are you asking about? This is two separate questions...

$\ce{HF}$ (Hydrofluoric acid) is the only one that I know is capable of dissolving silicates.
Other acids, even strong oxidizing mixtures like aqua regia ($\ce{HNO3 + HCl}$) don't cut the mustard.
This is why chemical glassware is typically made of silicon compounds. Not easy to break the $\ce{Si-O}$ bond. Common sand is composed of silicon: $\ce{SiO2}$. Most other compounds are either in rocks or man made (e.g. tetramethylsilane (TMS) used in Nuclear Magnetic Resonance experiments) or reduced (pure) silicon in computer parts.
The common sand-etching reaction is:
$$\ce{SiO2 (s) + 4HF (aq) -> SiF4 (g) + 2H2O (l)} $$

@ToddMinehardt what?

think - VTC = Vote to close, NAA = not an answer

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. || @Todd You can't VTC on an answer. I am not sure if that answers the question of the OP, but that is mainly because I think the question itself is unclear. I'm leaving it as is.

I think the question is unclear because the OP isn't a chemist. "Silicone" was written but Si-compounds (in general) was implied. Especially given the title/context.

@khaverim - Sorry, impedance mismatch on my part. Good answer, unclear question. Upvoting.


text:
Synthesis of Colchicine
Recently one of my relatives have fallen ill and was prescribed with some colchicine. Looking at the structure of the molecule, and with nothing much to do, I decided to put my retrosynthetic skills to the test. Here is a picture of my thought process:

Is there a better way to design a synthesis for this compound using the disconnection method.
From 11b, a Birch reduction is carried out to give the qunione 10b. A rearrangement of the ketone with methanediazonium gives 9b. A dihydroxylation with a peroxy acid and subsequent addition of water gives 8b. A double dehydration reaction with sulfuric acid, coupled with the protection of the ketone with propan-1,3-diol gives the seven-membered quinone 7b. A Heck reaction (or Ullmann reaction) with 7a with a palladium catalyst yields 6. (The protection group is thereafter labelled "PG") Friedel-Crafts acylation with ethanoyl chloride yields 5 (although on second thoughts, I should have done the acylation from 7a from the start). A Michael addition is then carried out with $\ce{BuLi}$ to lithiate the ketone to give the terminal imine 4. Since this terminal imine is unstable, a mild reducing agent converts the imine to the amine 3. The ketone is then removed by addition of dithiol and subsequently reduced by Raney nickel to form 2. Finally, a simple condensation reaction between the amine and acetic anhydride, followed by deprotection of the ketone using an acid, yields the final product colchicine, 1.

While this is an interesting question I have to agree that it is too broad. Maybe focus on a specific aspect of the synthesis or head over to the chat to discuss your question in more detail.


text:
Stability of transition and post-transition metal alkoxides
Many transition metal alkoxides are seen in chemistry, for instance those of $\ce{Ti, Sn}$ and $\ce{Al}$, and in particular one of those of aluminium has its own role in the MPV reduction reaction.
I was wondering: how stable these $\ce{M(OR)_n}$ structures are?
To simplify the question:

Are they susceptible to acid/basic attacks (which could possibly hydrolyze them, maybe)?
Being them metal-alkoxide compounds, do they present similar reactivity as alkali metal oxides (eg: use in Williamson's synthesis of ethers)?
Would it be possible to obtain an alkali metal alkoxide from a transition metal alkoxide, hence synthesizing those compounds without using alkali metals?
When used in a reaction as a catalyst (eg: MPV or Oppenauer reactions), a metal alkoxide can be recovered and reused?

Regarding point 3, you don't necessarily have to use the metal itself - e.g. adding NaH to an alcohol will get you the sodium alkoxide. But I get what you mean.

You definitely can't do MX + M'(OR) -> M(OR) + M'X, where M=alkali metal and M'=TM, because the equilibrium would lie in the opposite direction, since the highly nucleophilic alkali metal alkoxide M(OR) will attack the electrophilic transition metal M'X (especially if it's an early-TM).
I will answer some of these questions in line after I read through some references.

Alkali metal, main group, and transition metal (TM) alkoxides all have significantly different chemistry. Early- and Late-TM alkoxides also have different behaviour, so we should split this family into at least four categories. All of them are basic and will decompose via hydrolysis if dissolved in water; however, this type of chemistry is not usually carried out in presence of water.
In TM chemistry, alkoxide or phenoxide are mostly used as ancillary ligands, meaning they do not usually participate in the chemistry. Although they are liable to be hydrolyzed under protic conditions, alkoxides are actually known for being especially stable TM ligands because they do not undergo β-hydride elimination (BHE) (1), which is one of the most common decomposition pathways for TM alkyls. For instance; you often cannot make a stable t-butyl TM complex because the β-hydride will be transferred to the transition metal, liberating 2-methylpropene unless the metal is totally coordinatively saturated. An important example of a BHE-stable TM t-butoxide is the following Mo(IV) Schrock catalyst, which is stable:
$\hskip2.4in$
It should be noted, though, that the Schrock catalyst doesn't actually have any β-hydrides. Later, I'll show an example of a TM-alkoxide that does.
An important feature of alkoxides as ligands for TM chemistry is that they are electronically malleable. Since an alkoxide has 5 electrons it can behave either as an σ-donating X ligand or as a σ- and π-donating LX ligand (and formally as L2X). Presumably, this electronic malleability has something to do with their stability to BHE. An alkoxide's π-donor character might stop the metal center from liberating a coordination site to accept the β-hydride. For more information on BHE-stability, see (2), as cited by Schrock in (1).
Since early-TMs are more electropositive, the bonding character of their alkoxides is more ionic, although not to the same degree as that of alkali metal alkoxides. Alkoxides are very good bridging ligands, and if they are not sufficiently bulky, alkoxide complexes often polymerize. With a sufficiently bulky monomeric early-TM alkoxide, the main decomposition pathway is formation of the oxo complex. For example; the decomposition of a metal diolate shown in the next image was observed by Gable et al. (3) Their Hartree-Frock calculations implicated an empty metal MO at rhenium in the M=O bond formation, as shown in the MO picture in the next image.
$\hskip0.2in$
Late-TM alkoxides are expected to have weaker M-O bonds; and not as many have been isolated, but a new study (4) has examined these M-X (where X = O, N) bonds.
A really nice example illustrating a few important TM-alkoxide concepts is tantalum(V) ethoxide, Ta2OEt10. See wikipedia: https://en.wikipedia.org/wiki/Tantalum(V)_ethoxide
$\hskip1.5in$
Ta2OEt10 is a stable TM-alkoxide with true β-hydrides, bridging oxygens, and it's readily hydrolyzed when exposed to protons. It can be synthesized via salt metathesis with sodium ethoxide, a reaction that has equilibirium lying in the opposite direction of the alkali metal alkoxide formation you proposed.
In MPV reduction (and OPP oxidation), Al(O-i-Pr)3 is usually used as the catalyst. Aluminium is a metal, but it is a main group metal, and its chemistry is somewhere between that of boron, a semimetal, and that of early transition metals, except, of course, that it does not have a partially filled d shell. Al(O-i-Pr)3 is an electron deficient Lewis acid, comparable in chemistry to boronate esters, but much more reactive. The proposed mechanism for MPV/OPP [R]/[O] goes something like this:
$\hskip0.2in$
It is a true catalytic reaction, so the Al(O-i-Pr)3 is indeed regenerated, but I don't know if it's a routine part of the procedure to recover it. Al(OR)3's are not, however, soluble in IPA, but alcohols and ketones usually are, so you could probably precipitate Al(O-i-Pr)3 in IPA providing you have a way to seperate your ketone and your secondary alcohol afterwards.
Interestingly, Ti(O-i-Pr)4, is analogous to Al(OR)3, but its reactivity totally different, since titanium is a true TM and not just a Lewis acid main group metal like aluminium. Ti(O-i-Pr)4 is the catalyst for sharpless epoxidation, so this might be a good reaction to look at as an example of early-TM alkoxide reactivity.
To answer your questions in list form:

Pretty much all alkoxides are susceptible to acidic attacks, but, being fairly strong bases themselves, they are probably stable to weaker bases
Although they will react with acids, they will not necessarily react with electrophiles unless they are stronger electrophiles than the metal center; however, early-TMs are fairly strong electrophiles. I don't think people use TM alkoxides for ether synthesis though, and you always have to keep in mind d orbitals and the unique rules of TM chemistry when you're discussing TM complexes.
As mentioned above, and in my comment; "You can't have MX + M'(OR) -> M(OR) + M'X, where M=alkali metal and M'=TM, because the equilibrium would lie in the opposite direction, since the highly nucleophilic alkali metal alkoxide M(OR) will attack the electrophilic transition metal M'X (especially if it's an early-TM). As Schrock says in (1), "the relatively high electrophilicity of an early transition metal makes dissociation of alkoxide ion, at least in a neutral species, unlikely".
If you can find a suitable means of seperation like a solvent that will precipiate your metal alkoxide, you should be able to seperate it, although some catalysts will be deactivated after a certain number of cycles.

(1) R. R. Schrock, Polyhedron, Vol. 14., No. 22, pp. 3177-3195, 1995
(2) D. C. Bradley, R. C. Mehrotra and D. P. Gaur, Metal Alkoxides. Academic Press, New York, 1978
(3) K. P. Gable, J. J. J. Juliette. J. Am. Chem. soc, 1996, 118, 2625-2633
(4) J. R. Fulton, A. W. Holland, D. J. Fox, R. G. Bergman. Acc. Chem. Res., 2002, Jan; 35(1): 44-56


text:
Aldolisation reaction and chalcone formation
I realized a reaction of mixed aldolisation between acetone and 2-nitrobenzaldehyde, using a solution of 10% NaOH in ethanol.
I obtained a mixture of products at the end of the reaction. The problem is when I added the ethyl acetate I saw a blue liquid mixed with brown solid.  After a night of this reaction I found a lot of white crystals mixed with the blue liquid.
Can someone explain the presence of this blue liquid with the brown solid, knowing that the crystals are soluble in ethanol?

Well, actually you've made a reaction called the "Baeyer–Drewson indigo synthesis" (original article here), which was route developed in the nineteen century to obtain the indigo dye, nowadays replaced by other methods. I'm not writing the mechanism for this reaction because it is quite long and well-documented elsewhere.
Your blue compound is therefore the indigo dye represented here. As for the brown compound, it's difficult to say without more data (NMR spectra for instance), but the synthesis can give many colored intermediates or by-products such as isatin (bright red).

I hope this helps.

In the aldol reaction, we would expect to come up with the following product:

However, this reaction is bound to happen poorly. This is due to the presence of the nitro group, which has so much electron-withdrawing power that you can consider it to be twice as powerful as a carbonyl group (although the nitro group is not as powerful here due to its attachment to the benzene ring). Hence the nitro group then subsequently attacks the ketone in many intramolecular reactions, an example is shown here:

Since there are so many aromatic compounds that are fully conjugated forming, these should all contribute to the blue colour that you are getting.
I believe the brown solid is the result of some polymerization of the ketone with either the aldol or the nitro group.

I also have to add that the product I drew happens to be an activated form of pyridine, which is susceptible to electrophilic substitution as well. Hence, electrophiles such as another nitro group can easily attack the pyridine to form polymers.


text:
Does aluminium oxide ionize after melting point?
My chem is very basic but I would like to know is there a temperature after melting point where aluminium oxide can ionize enough to pass electrical current through? Is there a temperature where the resistance of aluminium oxide deteriorates?

I'm going to try and answer a bit of confusion in the original question "Does aluminium oxide ionize after melting point?". Aluminium oxide is already ionised in the solid, or more pedantically the ionic model fits aluminium oxide sufficiently well that we can think of it comprised of $\ce{Al^{3+}}$ and $\ce{O^{2-}}$ ions and make reasonable predictions. The point is that in most solids, ionic or not, the constituent atoms and molecules are energetically constrained to vibrate about their "normal", so called equilibrium, position, and so never move very far, whereas in liquids they can move about reasonably freely. Now as conductivity involves the nett movement of charges most ionic solids are not conducting while the liquids are, as in the liquid the charged ions are free to move toward the appropriate electrode while in the solid they are not.
However it's not that quite that simple, but the above is probably good enough to university level. But for completeness I'll add that there are things called fast ionic conductors. These are "ionic" solids that conduct electricity well above a certain temperature. A well known example is silver iodide, which above 419 K start to conduct electricity pretty well. A simple model here is that at that temperature the iodide sub-lattice stay rigid, so providing a solid like framework, but the silver sub-lattice melts, and so the silver ions can move, and thus nett charge migration is possible, and so the solid conducts. For a bit more of an introduction, see Wikipedia article of fast ion conductor. Interestingly one form of aluminium oxide can act as a fast ion conductor, but in this case the charges that are moving are "impurities".

"419.15 K"? Are all those digits significant?

@OscarLanzi Ask Nilay Ghosh, it was changed from 146C by their edit.

It is matter of implicit error propagation during addition of 2 values.  146 + 273.15 = 419.15 mathematically, but 419 statistically.

The Celsius values convert to Kelvin with the 0.15 added on it. vanishes in the difference between measurements not a question of significant figures. that depends on the precision of the thermometer used.

Yes. After molten, every ionic compound (including aluminium oxide, of course) turns into ion. This is how production of aluminium by Hall-Héroult process works.
However, it is very hard to reach aluminium oxide melting point (it is at 2,072 °C). Therefore, we need to put cryolite to decrease its melting point.

Thank you Huy. Does the cryolite just lower the melting point or does it add to the conductivity?

I'm not sure. I looked up on Wikipedia and it said: " ...Besides having a relatively low melting point, cryolite is used as an electrolyte because among other things it also solves alumina well, conducts electricity, dissociates electrolytically at higher voltage than alumina and has a lighter density than aluminum at the temperatures required by the electrolysis...."


text:
Evaluation of a synthesis from aspirin to paracetamol from this Nile Red video
Recently the YouTube chemist Nile Red performed an interesting conversion from aspirin to paracetamol Aspirin to Acetaminophen - Part 1 of 6: Extracting Aspirin from Pills using the following steps:

He performed the following steps:
1) Extracting aspirin from pills (95%)
2) Hydrolysis of the ester (86%)
3) Decarboxylation (51%)
4) Nitration of phenol (6%)
5) Reduction of nitro group (58%)
6) Condensation reaction with acetic anhydride (18%)
Final yield: 0.26%
As you can see, the yield is quite discouraging, particularly the nitration of phenol. The synthesis that he planned was quite linear, first removing functional groups and then adding them. This unfortunately did not take full advantage of the functional groups present.
I feel that this is a good opportunity to reflect how we can carry out the reaction efficiently, and I will illustrate it with the picture below:

1) I feel that nitrating the aspirin directly would have been more effective.

Since the carboxylic acid is 3-directing and the ester 2,4-directing, this would lead to two nitration positions as shown by the asterisks. However, the site to the left would be more ideal as the site to the right is blocked by steric hindrance from the bulky ester group. The combined efforts of both functional groups would also increase the reaction yield quite considerably. Since this is done in the presence of concentrated nitric acid and sulfuric acid, it is not surprising that a hydrolysis can occur here when we add water, thus allowing us to do a one-pot reaction.
2) Now, decarboxylation should be carried out. Due to the powerful electron-withdrawing nature of the nitro group, the $\ce{C\bond{-}C}$ bond should be weaker than the original synthesis, giving us a better yield as well. This is important as the close proximity of the two groups allows for intramolecular hydrogen bonding, thus making decarboxylation more difficult than usual.
I am wondering how much the yield would be affected if it were synthesised in this method. Of course, it is impossible to theoretically come up with a value and can only be done experimentally, but it would be good if I can come up with some estimates. Let me know what you guys think. Thanks!

Ad 1: I would expect the carboxylic acid to be deactivating the aromatic ring for electrophilic aromatic substitution, which is the mechanism for the nitration. Thus, one may need even harsher reaction conditions.

^ True, but there are other factors to consider as well: the activating ester group along with the steric hindrance due to the bulky ester group. That said, it may be prudent to experiment with reagents such as $\ce{Hg(NO3)2}$  to determine if harsher conditions are indeed required.

Two things. 1. You don't know why the yield was low in the nitration. Was it side products or just a crappy reaction? Need more data. 2. This is a really silly pathway. 4-aminophenol is really cheap, so there's no reason to start the synthesis from somewhere else...

I think you're overlooking the big picture here. Of course I would start with the best material if available. This is just a post to reflect upon how the efficiency of the synthesis can be improved.

I'm a big fan of Nile Red Estimating maximum size of magnetite particles in this Nile Red ferrofluid video?

You are absolutely correct that $6~\%$ yield is not acceptable for a synthesis — especially if it is a reaction as simple as the nitration of phenol. A quick search in SciFinder gives numerous reaction conditions that allow for up to $98~\%$ yield of the para-substituted product, depending on which catalysts and sets of conditions you want to use. One of the simplest sets of reagents is probably Ducry’s and Roberge’s Angewandte paper which, however, uses a microreactor. While their best selectivity is $1.2:1$, their yields are way above $12~\%$ (which would be the corresponding yield for $6~\%$ of the para-product in a $1:1$ mixture).[1]
Having an acetate group rather than a hydroxy group greatly modifies the properties of that particular oxygen atom. I synthesised a phenol and the corresponding acetate and the chemical shift of the ortho carbon changes by $7.5~\mathrm{ppm}$ upon acetylation while the meta carbon remains essentially unshifted. Therefore, I wish to challenge your assertion that acetate be ortho and para-directing — please back this with appropriate references. However, I will acknowledge that an acetate group is less meta-directing that a carboxylic acid; therefore the preferred reaction site should still be ortho/para to the acetate.
Whatever the selectivity, I do not agree that nitration of acetylsalicylic acid or salicylic acid would be more efficient than nitration of phenol. Remember that the reaction you are trying to perform is an electrophilic aromatic substitution $\mathrm{S_EAr}$. For an aromatic species to be attacked electrophilicly, it should have a high electron density. The carboxyl and acetate groups are both electron-withdrawing and therefore hinder the reaction. It should be harder and require harsher conditions to nitrate a benzoic acid than a phenol. The choice of steps is therefore better in the originally proposed way, in my opinion.
I will agree that decarboxylation of a meta-nitrobenzoic acid should be easier than of salicylic acid. Since the yield of carboxylation is already rather acceptable already, I do not think that switching the order would provide any significant advantage here.
Finally, it should also be rather easy to increase the yield of the final-step acetylation. $18~\%$ needs a lot of work.

Reference:
[1]: L. Ducry, D. M. Roberge, Angew. Chem. Int. Ed. 2005, 44, 7972. DOI: 10.1002/anie.200502387.

Interesting perspective! Although as mentioned I guess there is no way to predict with certainty whether the yield will be improved or not, we can come up with some theories. Thanks for your input! And yes, apparently for the 18%, he did 3 recrystallisations which further decreased the yield.

@LinusChoy Recrystallisations shouldn’t go from acceptable yields down to $18~%$ unless he needs a course in recrystallising, though ;)


text:
How can difference in neutron number cause a difference in ionisation enthalpies?
It’s a well known fact that protium,deuterium and tritium have the same number of electrons and protons but different number of neutrons. It is expected that their ionisation enthalpies should be same as in all the cases only one proton attracts the only electron they have but their enthalpies are different.
What causes this?

Because the presence of the neutrons changes the nuclear spin state and required wave function symmetry.

there are probably a handful of corrections needed to calculate the exact difference. But the biggest factor is just the mass of the nucleus. Not only is the electron attracted to the nucleus, the nucleus is attracted to the electron. So not only does the electron "move", so does the nucleus. A heavier nucleus moves less. This in turn alters the wave function  of the electron.

There are a handful of additional corrections needed to calculate the exact differences. But the biggest factor is just the mass of the nucleus. Not only is the electron attracted to the nucleus, the nucleus is attracted to the electron. So not only does the electron "move", so does the nucleus. A heavier nucleus moves less. This in turn alters the wave function of the electron.
The difference between protium, deuterium and tritium is reasonably explained by the Bohr model of the atom. The gist is that the electron does not travel around a nucleus of infinite mass, but rather that the electron and nucleus travel around their common center of mass.
For one electron traveling around the nucleus the energy levels of the Bohr system are:
$\text{E}_n = \dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} \cdot \dfrac{\text{Z}^2}{\text{n}^2}\cdot \text{E}_R$
$\text{E}_n$ Energy of the $n$th orbital of the one electron system
$\text{M}_e$ Mass of the electron = $5.489\cdot10^{−4}$ atomic mass units
$\text{M}$ Mass of the nucleus
$\text{Z}$ Charge on the nucleus
$n$ orbital = principal quantum number $n$ = 1, 2, 3, 4, ...
$\text{E}_R$ Rydberg energy = 13.605693 eV
For the $1s^1$ electron of protium, deuterium and tritium, Z=1 and $n$=1.
protium
$\text{M} = 1.007825 - 0.0005489 = 1.007276$ amu
$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{1.007276}} = 0.999455$
ionisation enthalpy = $0.999455 \cdot 13.605693 = 13.598$ eV
deuterium
$\text{M} = 2.01410178 - 0.0005489 = 2.0135529 $ amu
$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{2.0135529}} = 0.999727$
ionisation enthalpy = $0.999727 \cdot 13.605693 = 13.602$ eV
tritium
$\text{M} = 3.0160492 - 0.0005489 = 3.0155003$ amu
$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{3.0155003}} = 0.999818$
ionisation enthalpy = $0.999818 \cdot 13.605693 = 13.603$ eV
Ionization energies From NIST webbook
hydrogen = 13.59844 eV
deuterium = 13.603 eV
tritium = 13.603 eV
Therefore protium < deuterium < tritium

I'll add that when you're doing this sort of thing other factors creep in the calculations.
(1) For a hydrogen atom the electrons aren't relativistic. If you get to uranium though that wouldn't be true.
(2) The 13.6 eV binding energy also reduces the mass of the hydrogen atom. So subtracting the rest mass of the electron isn't absolutely correct, but good enough for the precision used.
13.6 ev $=\quad\quad 0.000 000 014 5$ amu
mass e used $= 0.000 548 9$ amu