text:
Reaction mechanism of 2-amino-4,6-dimethylpyrimidine as a nucleophile
My understanding tells me that 2-amino-4,6-dimethylpyrimidine can be deprotonated quite easily because the conjugated base is stabilized be resonance and act as nucleophile attacking aldehyde carbon. But I found no resource in literature about this reaction:

This source states that the pKa of 2-amino-4,6-dimethylpyrimidine is 4.8 and classifies it as a weak acid, so your mechanism is probably correct.

@AniruddhaDeb, that's probably the pKa of the conjugate acid.

Your mechanism is almost correct; this is essentially an aldol condensation. The only issue is with the protonation states, which (as you've drawn) aren't entirely feasible.
Let's start by establishing how acidic that methyl group actually is. You're right that it is more acidic than a typical methyl group, because the charge can be delocalised onto those two nitrogens. But note that this delocalisation is pretty similar to that in a typical ketone enolate:

so to a very, very rough first approximation, that proton you're removing has a $\mathrm{p}K_\mathrm{a}$ of $\sim 20$, roughly that of a ketone. Yes, there are two nitrogens in the ring which improves the delocalisation, but there is also an amino group sticking off the ring, which is also trying to delocalise electron density onto the ring nitrogens. The value of $4.8$ cited in the comments is far more likely to be the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid (where the amino group is protonated).
Secondly, if anything were to be deprotonated, the $\ce{NH}$ in the amino group is going to be more acidic than the $\ce{CH}$ in the methyl groups. This is exactly analogous to how ammonia is more acidic than methane, and how amide $\ce{NH}$'s are more acidic than ketone $\ce{CH}$'s.
And lastly, note that your reaction is being carried out using acetic acid as a solvent. The chances that a carbanion (and a strongly basic one, as previously established) can exist for any reasonable amount of time under these conditions are... low, shall we say!
Now, you've probably drawn mechanisms for aldol condensations in acid. In this case, you don't deprotonate to form an enolate; you just use the enol tautomer, which is still nucleophilic on carbon. So the strategy here is not to deprotonate the methyl group outright, but to use the tautomer. Basically, use the same mechanism you've drawn, but add a proton to everything:*

That includes the final elimination, too. Note that if you protonate the $\ce{OH}$ group and let it fall off, you can get a benzylic cation that is stabilised by the 4-hydroxy substituent.

* You could delocalise all the way onto the other ring nitrogen if you wanted.

You have a excellent point.

To your question, orthocresol has given an excellent answer, which do not need any more elaboration. My focus here to address your concern of:

My understanding tells me that 2-amino-4,6-dimethylpyrimidine can be deprotonated quite easily because the conjugated base is stabilized by resonance and act as nucleophile attacking aldehyde carbon. But I found no resource in literature about this reaction.

Actually, there are tons of reference there to find. What you have to do is change your key words appropriately. For example, in your case, use a phrase such as reaction of methyl pirimidine with aldehydes.
In general, methyl groups attached to aromatic heterocyclic rings may behave very differently than methyl substituents on aromatic hydrocarbons. For example, methyl groups on $\pi$-deficient six-membered heterocyclic systems such as 2-methylpyridine (substituent is at $\alpha$-position to the hetero atom) are more reactive toward bases and/or electrophiles than are methyl groups attached to a benzene ring (e.g., $\bf{A(1)}$, Ref.1):

As expected, methyl groups on heterocyclic systems where a ring nitrogen is formally positively charged (e.g., pyridinium salts) are more reactive than those in uncharged systems. Thus, acid-catalyzed reactions are also possible, as in $\bf{A(2)}$ (Ref.1).
On the other hand, methyl groups in the 2-, 4-, or 6-position of pyrimidine nucleus are ‘active’ in the sense of that in 2,4-dinitrotoluene as depicted in $\bf{B}$ of above figure. In the reaction with benzaldehyde, 2,4-dinitrotoluene undergoes aldol condensation at 'active' methyl group in the presence of catalytic amounts of DBU to give compound $\bf{2}$ ($\ce{R = Ph}$, $\gt 65\%$ yield; Ref.2). In addition, if a methyl group is in the 5-position of pyrimidine nucleus, that methyl would resemble that of toluene (more aromatic behavior). In general, additional heteroatoms on the nucleus increase the reactivity of the methyl group. For example, methyl groups attached to pyrimidine, pyrazine, and pyridazine are more reactive than those on pyridine (Ref.1).
There are a lot of examples for aldol-type condensation can be found in literature. Early literature have been reviewed in Ref.3 with list of specific literature. Two such references are Ref.4 and Ref.5. The baseline is the conversion of appropriate methyl- into styryl-pyrimidines can be done by condensation with benzaldehyde derivatives in the presence of an acid, Lewis acid, dehydrating agent, or strong organic base as catalyst. For example, 5-bromo-2-methylpyrimidine was reacted with benzaldehyde in the presence of a Lewis acid, $\ce{ZnCl2}$ to yieldobtained 5-bromo-2-styrylpyrimidine ($2\ce{-CH=CH-Ph}$) in good yield (Ref.6). Similarly, 4-methylpyrimidine and benzaldehyde with acetic anhydride has given 4-styrylpyrimidine (Ref. 7).
I think these are good enough evidence to convince you.
About $\mathrm{p}K_\mathrm{a}$ of a methyl group on a heterocycle ($\mathrm{p}K_\mathrm{a}$s of toluene and $\alpha$-$\ce{H}$ of a ketone are given for comparison):
$$
\begin{array}{c|c|c}
\hline
\text{Compound} & \mathrm{p}K_\mathrm{a} & \text{Reference} \\
\hline
\text{Toluene}  & \sim 41 & \text{Ref.8,9} \\
\text{2-Methylpyridine}  & 34 & \text{Ref.8,9} \\
\text{3-Methylpyridine}  & 37.7 & \text{Ref.8,9} \\
\text{4-Methylpyridine}  & 32.2 & \text{Ref.8,9} \\
\text{4-Methylquinoline}  & 27.5 & \text{Ref.8,9} \\
\alpha \!- \! \ce{H}\text{ of a ketone}  & 19\! - \! 20 & \text{Ref.8,9}\\
\hline
\end{array}
$$

References:

Fathi A. Abu-Shanab, Basil J. Wakefield, Mohamed Hilmi Elnagdi, “Chapter 3: Methylpyridines and Other Methylazines as Precursors to Bicycles and Polycycles,” In Advances in Heterocyclic Chemistry – Volume 68; Alan R. Katritzky, Ed.; Academic Press: San Diego, CA, 1997, pp. 181-221 (ISBN: 978-0-12-020768-8).
Robert Bujok, Marcin Wiszniewski, Piotr Cmoch, Zbigniew Wróbel, “A general and convenient method for the synthesis of 2,4-dinitrobenzyl ketones. Almost unlimited access to 2-substituted 6-nitroindoles from 2,4-dinitrotoluene and aldehydes,” New J. Chem. 2018, 42(5), 3260-3269 (https://doi.org/10.1039/C7NJ04919F).
Desmond J. Brown, S. F. Mason, In The Pyrimidines; John Wiley & Sons Inc.: New York, NY, 1962, Chapter IV: Pyrimidines and Its $\ce{C}$-Alkyl- and $\ce{C}$-Aryl Derivatives, pp. 116 - 137.
Reuben G. Jones, Edmund C. Kornfeld, Keith C. McLaughlin, “The Synthesis of Some β-Aminoethyldiazines as Histamine Analogs,” J. Am. Chem. Soc. 1950, 72(8), 3539–3542 (https://doi.org/10.1021/ja01164a061).
C. G. Overberger, Irving C. Kogon, “Monomer Synthesis. X. The Preparation and Polymerization of 4-Vinylpyrimidine and 2-N,N-Dimethylamino-4-vinylpyrimidine,” J. Am. Chem. Soc. 1954, 76(7), 1879–1883 (https://doi.org/10.1021/ja01636a043).
J. F. W. McOmie, I. M. White, “623. Pyrimidines. Part VI. 5-Bromopyrimidine,” J. Chem. Soc. 1953, 3129-3131 (https://doi.org/10.1039/JR9530003129).
C. E. Loader, C. J. Timmons, “Studies in photochemistry. Part III. The photocyclisation of 4-styrylpyrimidine to benzo[f]quinazoline,” J. Chem. Soc. C 1967, 1343-1344 (https://doi.org/10.1039/J39670001343).
Robert R. Fraser, Tarek S. Mansour, Sylvain Savard, “Acidity measurements on pyridines in tetrahydrofuran using lithiated silylamines,” J. Org. Chem. 1985, 50(17), 3232–3234 (https://doi.org/10.1021/jo00217a050).
John A. Joule, Keith Mills, “Chapter 4.1.9: Side chain metallation,” In Heterocyclic Chemistry; Fifth Edition, John Wiley & Sons, Ltd.: Chichester, West Sussex, United Kingdom, 2010, pp. 55 (ISBN: 978-1-405-19365-8).


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In a battery, what happens to LiCoO2 when the lithium leaves? It becomes CoO2?
What chemical compound exists at the cathode of a lithium-ion battery when the lithium is not there?
The cathode is usually described as $\ce{LiCoO2}$, so does it become $\ce{CoO2}$?
$\ce{CoO2}$ puts cobalt in an oxidation state of +4, which is possible but rare and unstable. I think $\ce{CoO2}$ doesn't even have a Wikipedia page.

CoO2 exists: http://cobalt.atomistry.com/cobalt_dioxide.html

Not an expert on this, but I think this is one of the reasons you don't ever want to overcharge a LiCoO2 battery...

It is quite complicated. See https://www.sciencedirect.com/science/article/pii/S0364591613000035 for some of the thermodynamics and crystal structures.

@JonCuster That statement is a bit over the top, as is the paper you link as a first read. The general principles of the Li ion battery have been textbook knowledge for quite a few years now.

@Karl - no, I'd say the layering in the crystal structure and the relevant thermodynamics is fairly complicated. Yes, how it works (intercalated Li) is straightforward. How to get to which crystal structure under which conditions is another question.

@Karl - you have made it clear that what you think is trivial is not what the OP apparently thinks is trivial. Feel free to write up a simple answer for them, but be sure to cover all the multiple phases and order-disorder transitions.

@JonCuster That´d be quite outside my area of expertise. :-)

@Karl - but you did a good job at it!

The general textbook formula for the Li ion battery is
$\ce{Li_{1-a}Ni_{1-x-y}Mn_{x}Co_{y}O2}$
, and the simplest/earliest type has $y=1$, i.e. cobalt-only. These "mixed (lithium transition metal) oxides" have a layered structure (see e.g. wikipedia), where you can relatively easily electrochemically remove (and later reinsert) a part $a$ of the Li atoms, without destroying the whole structure. You cannot increase $a$ to much more than 0.1 - 0.3 or so, depending on the exact formulation/grain size/quality/temperature/etc., before the structure does break down.
Even the "uncharged" $a=0$ Li-cobalt(III) oxide gives off oxygen (!) if heated above 180°C. (It is produced by tempering at ~800°C in a pure oxygen atmosphere.) With $a>0$, this happens earlier already, and basically always leads to a runaway destruction of your whole battery setup. With higher contents of Ni and Mn, this danger becomes less pronounced. I´m not sure the Co-only variant was ever commercialised. (?)
Simply speaking, a cobalt-rich battery blows up when overcharged, and in ones with more nickel and manganese the layered structure still breaks down, and you get somewhat stable Ni/Mn(III/IV) species (and still quite a lot of overheating/general destruction). In any case this happens long, long before you reach the ideal $\ce{CoO2}$ (or generally $\ce{MO2}$) stochiometry.
If you look closely (e.g. via XRD) at the crystal structure during charging, you seemingly find that there are a number of intermediate deformed structures, and even more mixed/disordered states in between. And that of course becomes even more complicated (or washed out) with a (more or less) random arrangement of different transition metals in the structure, which will depend also on the specific synthesis route.

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$\psi$ Represents amplitude of electron wave and $\psi^2$ represents electron density in a given space. How?
The $\psi$ represents amplitude of electron wave. So it's dimension should be $\left[L\right] $. But $\psi^2$ represents density of electrons in a given space. So it's dimension should be $\left[ML^{-3}\right]$.
Now $\left[L\right]^2 \neq \left[ML^{-3}\right] $.
So what's wrong with the concept?

Basically everything is wrong, unfortunately.
$\psi^2$ isn't a mass density, it's a probability density. So there is no point in which you should be getting units of mass. As porphyrin pointed out, the units of $\psi^2$ (and hence $\psi$) will depend on the system which you are looking at.
Next, $\psi$ isn't a length amplitude. In fact, it's not an amplitude at all. When people talk about wavefunctions being amplitudes, what they are usually talking about is the coefficients of the basis states of $\psi$. For example, if you express $\psi$ as a linear combination $c_1\phi_1 + c_2\phi_2 + \cdots$, then the $c_i$'s are called amplitudes. But they aren't length amplitudes of some wave, they're probability amplitudes.
Wikipedia's pages (linked above) describe the concepts in far greater details, and you will also see these terms arise frequently when reading about QM.

The wavefunctions are normal mathematical functions; nothing special. The dimensions of $\psi$ depend on the problem. A particle in a 1D box has the normalised wavefunction  $\displaystyle \psi(x) =\sqrt{\frac{2}{L}}\sin \left( \frac{n\pi x}{L} \right) $ so has dimensions of $[L]^{-1/2}$. In 2 and 3 dimensions the units of $\psi$ change accordingly.
The expression $\psi(x)^*\psi(x)$ is interpreted as the probability of being in the infinitesimally thin region $x\to x+dx$ so the probability is strictly $\psi(x)^*\psi(x)dx$ which is dimensionless (as a probability should be) and has a value $0\to 1$ as the waavefunction is normalised.
This works for other measurements also, for example to find the average of the position in the box, $\displaystyle \langle x \rangle=\int_0^L \psi(x)x\psi(x)dx=\frac{L}{2}$  (for all $n>0$) which you can see by looking at the dimensions before integrating (don't forget to include dx), and similarly the average of the square of the position is $\displaystyle \langle x^2 \rangle=\int_0^L \psi(x)x^2\psi(x)dx$ which you can see has dimensions of $[L]^2$. For the first level, $n=1$ the exact result is $L^2(1/3-1/(2\pi^2))$.

The simplest wave function of the Hydrogen atom in the $\ce{1s}$ state is :  $$\ce{\Psi _{1s} =  \pi^{-1/2}} ·(1/a_o)^{3/2} · e^{-{r/a_o}}$$ It has a dimension  $\ce{L^{-3/2}}$

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text:
What is "ionic-covalent "resonance?
Advanced Inorganic Chemistry: A Comprehensive Text on Pg. 228, Chapter 8 titled 'Boron' talks about three factors which appear to be responsible for the shortness of $\ce{B-X}$ bonds(where $\ce{X}$ is a halogen). The second point was stated as:

Strengthening and hence shortening of the $\ce{B-X}$ bonds by ionic-covalent resonance, especially for the $\ce{B-F}$ and $\ce{B-O}$ bonds because of the large electronegativity differences. Evidence that this is important, in addition to the dative pπ-pπ bonding, is afforded by the fact that even in $\ce{BF3}$ complexes such as $\ce{(CH3)3N^+B^-F3}$ and $\ce{BF4^-}$, where the pπ-pπ bond must be largely or totally absent, the $\ce{B-F}$ bonds are still apparently shortened

What is this "ionic-covalent" resonance exactly? I tried looking it up,but found some unsatisfactory and vague answers here where it seems like they are presenting regular resonance as ionic-covalent. This answer on the other hand seems to be talking about compounds possessing both ionic and covalent factor although which is correct,does not really address the issue at hand.
So is this term really just refering to regular π-bond resonance, or is it an entirely different phenomenon?

I recommend checking rather earlier posts here like https://chemistry.stackexchange.com/questions/27535/percentage-ionic-character-when-electronegativity-is-given https://chemistry.stackexchange.com/questions/32306/can-100-covalent-bonds-exist

@Mithoron I understand the concept being talked about in these posts: no bond is completely ionic or completely covalent. But my question is regarding what ionic-covalent resonance is, and how is it related to this concept

But you didn't bother to read what Dave cited?

Oh sorry,got what you meant. So in the examples written in the question above, they will also form structures like B+ F- and B- F+ with B-F being a minor contributor(if we talk about BF4-)? @Mithoron

I guess covalent component still dominates, but structures with "broken" bond would be also important in depicting it.

I see. Still, I would like a more comprehensive answer pertaining to my case, with evidence supporting it like what has been shown on page 74 here


text:
Aluminum Hydroxide electrolysis
Does molten Aluminum Hydroxide break up to aluminum, water, and oxygen by electrolysis? It's easy to write the equation $$\ce{4Al(OH)3 -> 4Al + 6H2O + O2}$$ but is this actually possible?

Just look here in wikipedia. The aluminum hydroxide would dehydrate to produce alumina, which is used in the H-H process.

Why would alumina be formed instead of splitting into aluminum and hydroxide ions? If this was done in air then I understand but in an inert atmosphere? If the hydroxide ions interact then water would form and the excess oxygen could react with the aluminum

Look at the equation you have in your question: it is not balanced. Two aluminum hydroxides, when heated sufficiently, lose three water molecules and form alumina, i.e., $\ce{Al2O3}$. This is exactly what happens on a large scale as an intermediate step in the production of aluminum via the H-H process. It has nothing to do with the atmosphere. It is simply the reaction that actually happens. This is quite common. For example, magnesium hydroxide, when heated sufficiently, loses water and magnesia, i.e., MgO, is produced.


text:
Is it possible to produce sulfuric acid at large scale by oxidizing and hydrating pyrite?
I was reading The Knowledge by Lewis Dartnell, which is a book about recreating our technology from scratch should we experience a civilization-ending apocalypse. The author thinks that acquiring vanadium catalyst for contact process is too difficult, so his recommendation to produce sulfuric acid is to: (1) bake pyrite, (2) react the sulfur dioxide produced with chlorine gas (electrolyzed from brine), using activated carbon as the catalyst, (3) then react the resulting sulfuryl chloride with water to produce sulfuric acid.
I don't know how hard acquiring vanadium is, but I think it will be easier to replicate the same process happening in acid mine drainage:
$$\ce{2 FeS2 + 7 O2 + 2 H2O → 2 FeSO4 + 2 H2SO4}$$
The iron(II) sulfate can then be baked to produce sulfur trioxide, which can also be processed into sulfuric acid.
I think this method would have been simpler. Am I correct? Is there another simple method of producing sulfuric acid?

Well, you want to produce sulfuric acid which is reasonably highly concentrated. Because as a dilute acid, it´s not so very useful.

Distillation shouldn't be too hard, no? Sulfuric acid also has a much higher boiling point than water, so boiling should be good enough in a pinch.

Have you ever tried to distill even pure water? I guarantee you sulfuric acid is far, far worse. Plus I bet it´s an azeotrope. And definitely badly corrosive. ;-)

A post-apocalyptic world should be littered with broken stainless-steel tools. A lot of the items in my toolbox are embossed with names that spell xxvanadyy.

@Karl Thankfully, I have never needed to distill sulfuric acid, but I'm aware of the danger. Will the method I proposed make a very dilute acid, or at least more dilute than what the book recommended? Also, what's xxvanadyy? Google gave nothing on that.

Hm, acid mine drainage gives very dilute acid, bad enough for the environment, but I guess far less than 1 mol/l. If you do the reaction on purpose, it might be possible to be far better.

@Karl To the best of my knowledge, acid mine drainage is dilute only because it's exposed to the environment eg. rainwater, river. I'm guessing that performing the reaction on a sealed environment would result in a much higher concentration. I've never done anything like that, though, so nothing conclusive.

The xxvanadyy thing was simply referring to the fact that steel tools are often chromium vanadium alloys. See here for one in my own toolbox: https://chemistry.meta.stackexchange.com/a/4757/79678. I will delete this comment in a day or two.

@Karl The H2SO4-H2O azeotrope IS concentrated sulfuric acid (98.3%). H2SO4 has a boiling point that is much higher than water (290°C pure, 338°C for the azeotrope).

If your already using electricity, you could make O3 or H2O2 using electric arcs, and they should be able to oxidize SO2 to SO3 or H2SO4. That's very energy-intensive, though.

Per a source, there is an iron bacteria path acting on FeS2 to create sulfuric acid:

Thiobacillus spp. are participants of active oxidative corrosion. They oxidize a range of sulfur compounds to sulfuric acid, and the acid released from the cells may attack alloys.

My experience, however, suggests that the easiest way to produce H2SO4 is by the action of a concentrated solution of Oxalic acid (used in commercial products) on magnesium sulfate (FeSO4 may also work, but likely leads to an iron impurity).
Per a source (old Wikipedia commentary) to quote:

Relationship to oxalic acid
The dissociation of protons from oxalic acid proceeds in a stepwise manner as for other polyprotic acids. Loss of a single proton results in the monovalent hydrogenoxalate anion HC2O4−. A salt with this anion is sometimes called an acid oxalate, monobasic oxalate, or hydrogen oxalate."

As such expected reactions:
$\ce{MgSO4 + 2 H2C2O4 -> Mg(HC2O4)2 + H2SO4}$
$\ce{MgSO4 + Mg(HC2O4)2 -> 2 MgC2O4 (s) + H2SO4}$
Be careful, the created Sulfuric acid could be very strong depending on the level of hydration.
Warning: Attempting (in the presence of H2C2O4) to produce fuming sulfuric by boiling down the solution can lead to a spontaneous ejection of the acid (reported in an old thread of mine).


text:
Can submicron mould particles go through building structures, non permeable foil paper?
I read that submicron mould particles and vapours related to mould can go through walls, floors and ceilings?  Is this correct?

The fact that a room isn't air tight is the problem. There is a light, a wall switch, plumbing, cracks in the flooring, or air conditioning vent that leaks contaminated air into the room.

You are asking two different questions

I read that submicron mould particles can go through walls, floors and
  ceilings. How so aren't these structures meant to be air tight?

Not true. Mould particles can not seep through concrete walls or thick wood, even though they are porous structures. The interstitial pore size is too small and tortuous to let any mold particle pass through it randomly.

What about the lowest thickness of foil paper which is meant to be non
  permeable, can submicron mould particles go through that?

No, an aluminum foil is impervious to solids no matter what their size is. Yes, probably electrons and X-rays will pass through but nothing particulate.


text:
Can Aluminum Hydroxide (Al(OH)3) be heated to it's melting temperature before decomposing?
On the wikipedia page for aluminum hydroxide the listed melting point is 300 °C (572 °F), but in the same page it states that aluminum hydroxide decomposes at only 180 °C (356 °F). Can aluminum hydroxide be melted without decomposing?

The decomposition of pure aluminium hydroxide to water and oxide takes place above 550 °C. What you were reading in Wikipedia refers to the use in a mixture as a fire retardant.

Thanks for the clear answer! Do you have a source for the decomposition temperature of pure aluminum hydroxide? I am trying to find good resources for chemical properties

TL.DR: Aluminium oxide (sometimes known as alumina) is made by heating the aluminium hydroxide to a temperature of about 1100 - 1200°C. (Chemguide)
$$\ce{Al(OH)3 ->[\Delta] Al2O3 + 3H2O}$$

As @andselisk said, the temperature Wikipedia mentions is aluminium hydroxide mixture in a fire retardant which might have other compounds. Decomposition of pure aluminium hydroxide has been studied in various papers and its decomposition reaction mechanism, temperature, reaction kinetics has been exhaustively studied.
A paper examines$\ce{^{[1]}}$ the decomposition of aluminium hydroxide at a temperature of 973-1123 K(700 °C-850 °C).
Another paper$\ce{^{[2]}}$ shows aluminium hydroxide being heated from room temperature to 1200 K(926.85 °C) where the exact temperature of decomposition products i.e $\ce{Al2O3}$ is noted:

[...] we find the first and the smallest endothermic peak at 519
  K, which is due to the partial dehydroxylation of gibbsite
  ($\ce{Al(OH)3}$) and formation of boehmite ($\ce{AlOOH}$). Endothermic
  peak at 585 K corresponds to two processes: (i) transformation of
  gibbsite to phase $\ce{χ-Al2O3}$ and (ii) additional conversion of
  gibbsite to boehmite. The first process is in accordance with the
  results obtained. Another endothermic peak at 815 K is due to
  decomposition of boehmite and formation of alumina $\ce{γ-Al2O3}$. The
  obtained TG curves show clearly three steps of weight loss. The weight
  loss in the first step (about 5 wt.%) is due to the partial
  transformation of gibbsite to boehmite; the second step (about 25
  wt.%) corresponds to decomposition of gibbsite to boehmite, otherwise
  to $\ce{χ-Al2O3}$. The last step of around 3 wt.% relates to the
  formation of $\ce{γ-Al2O3}$. Further observations show that the total weight
  loss is equal to 33 wt.%. All these transformations of gibbsite and
  the chemical composition are confirmed by the XRD analysis.

References

Chemical kinetics and reaction mechanism of thermal decomposition of aluminum hydroxide and magnesium hydroxide at high temperature (973-1123 K) by Ienwhei Chen, Shuh Kwei Hwang, and Shyan Chen, Ind. Eng. Chem. Res. 1989, 28, 6, 738–742 DOI: https://doi.org/10.1021/ie00090a015
http://przyrbwn.icm.edu.pl/APP/PDF/131/a131z3p62.pdf
Bhattacharya, Indra & Das, S. & Mukherjee, P. & Paul, Subir & Mitra, P.. (2004). Thermal Decomposition of Precipitated Fine Aluminium Trihydroxide. Scandinavian Journal of Metallurgy. 33. 211 - 219. 10.1111/j.1600-0692.2004.00686.x.(DOI) (Here aluminium hydroxide is subjected to dehydration upto a temperature of 1440 °C)


text:
Why do we see complementary colors in UV/Vis Spectroscopy?
From what I read, a compound that absorbs visible light will produce a complementary color (using the color wheel), that can be seen. Why is this the case? Is it because, for example, if a molecule absorbs red, for instance, all the other colors will reflect and combine (?) to give off the color green that we see? I thought if we excite an electron to its excited state, with it absorbing a certain wavelength, when it drops back down to its ground state, it will give off a photon of that same wavelength (and same color). Also, when a molecule absorbs UV radiation and reflects all of the visible wavelengths, I read that it will appear white or colorless (isn't black the absence of all colors?), is this just two ways of saying the substance appears white or does that mean that it can be white or clear?

There are several points

From what I read, a compound that absorbs visible light will produce a
  complementary color (using the color wheel), that can be seen. Why is
  this the case?

This has to do with the physiology of the eye and how it has been designed to precept the colors in the visible range. As you stated, if we remove green from the visible spectrum, the transmitted light appears reddish or purplish to our eye. May be a cat or dog sees it differently. Color is a human perception. This complementary chart is just an approximation. If we take potassium permanganate solution, and if you look at what colors does it allow to pass through, you will see that it absorbs mostly in the green. You can easily do such experiments with a cardboard box and a diffraction grating made from CDs and a white LED.

I thought if we excite an electron to its excited state, with it
  absorbing a certain wavelength, when it drops back down to its ground
  state, it will give off a photon of that same wavelength (and same
  color).

This does not happen in majority of the cases i.e, in solutions and solids. The phenomenon you are referring to is called resonance fluorescence and it only happens to atoms in the gas phase. You shine 589 nm to Na atoms in gas phase, and you will see emission of 589 nm from Na atoms. This is accomplished in a flame. See Atomic Fluorescence Spectroscopy.
Normally, once light is absorbed, the absorbing material does not emit light. Instead, the absorbed energy is dissipated as heat.

Also, when a molecule absorbs UV radiation and reflects all of the
  visible wavelengths, I read that it will appear white or colorless
  (isn't black the absence of all colors?), is this just two ways of
  saying the substance appears white or does that mean that it can be
  white or clear?

Clear and white are two different things. Something clear like water does not absorb much in the visible range, perhaps a very very small portion from the red region. Therefore when you have huge amount of water you start to see the effects of absorbing red portion of the light. Do you recall that ocean water looks bluish? Even liquid oxygen is light blue!
You can say that water transmits most of the visible light. Majority of the substances will also absorb UV. How deep UV do you want? Most of the substance will start absorbing deep UV. Even air does, otherwise we would have been fried by the Sun.
A white piece of plastic or a cotton cloth, is reflecting most of the visible light, hence  it appears white to us. If you look the same piece in red light, it will look red, and in blue light it will appear blue and so on.

"This has to do with the physiology of the eye and how it has been designed to precept the colors in the visible range. As you stated, if we remove green from the visible spectrum, the transmitted light appears reddish or purplish to our eye."   I just had a question about this point. Does our eye interpret the complementary color, because it is perceiving the remaining reflected colors (for absorbed red, the reflected colors approximately would be orange, violet, yellow, blue, and green, which our brain interprets as green) as the complementary color?

Yes, this is a physiology question on the design of the eye. Nothing to do with chemistry or physics in a traditional way. You apparently have several misconception about colors which I have tried to address in the answer.

Eye and color perception is such complex topic that perhaps nobody can tell you the answer "why" this is so.

"Does our eye interpret the complementary color, because it is perceiving the remaining reflected colors (for absorbed red, the reflected colors approximately would be orange, violet, yellow, blue, and green, which our brain interprets as green) as the complementary color?" I think you can observationally or superficially say that but what is the exact mechanism- perhaps nobody knows.

You can mention that even fluorescence taking places, photons are randomly emitted in space. Still a certain colour is effectively reduced, and a complementary tint still emerges. That is why absorption spectropy is generally possible.

Is the colored light (due to absorbing) different in its constituent wavelengths from "pure colored light" i.e. If we see red (due to absorbing green) is this different from light in the wavelength region of red (620-720nm)


text:
Rearrangement of 1,1-Dichloro-1a,9b-dihydro-1H-cyclopropa[l]phenanthrene on heating

Due to angle strain , the three membered ring is not stable. So i think it would break and kick out one of the chlorine atoms resulting in the formation of a carbene. But i don't know how to proceed further and how it finally breaks into 2 products.

As I mentioned in the comments, this is a simple homolytic bond fission to give a carbene and a stable aromatic compound (Phenanthrene). Here's the mechanism:


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Why is this compound named (Z)-but-2-ene-1,1-diyldibenzene and not (Z)-1,1-diphenylbut-2-ene
A recent answer of mine involved the synthesis of the following compound:

I initially assumed the name of this compound to be (Z)-1,1-diphenylbut-2-ene as the butene is the main chain, with two phenyl groups attached to it. However, the preferred IUPAC name seems to be (Z)-but-2-ene-1,1-diyldibenzene, as verified by ChemDraw. Why is this so? I have not encountered the naming of biphenyls in this manner. What rules apply here to give this compound this name?

Close, but the preferred IUPAC name is also not (Z)-but-2-ene-1,1-diyldibenzene. You probably have not encountered the naming of biphenyls in this manner because this is not a biphenyl. Also note that even with only one phenyl group, the parent structure of the resulting [(2Z)-but-2-en-1-yl]benzene would not be the butene chain.

The first rule you should know is about selecting between a ring and a chain as parent structure. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups
cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Thus if you only have butene with a phenyl group, the preferred parent structure is the benzene ring and not the butene chain. Therefore, your initial assumption "I initially assumed the name of this compound to be (Z)-1,1-diphenylbut-2-ene as the butene is the main chain, with two phenyl groups attached to it." can generally be ruled out.
In this case, however, another rule is relevant. Since there are two identical parent structures (the two benzene rings), which are not alkanes, multiplicative nomenclature is preferred to substitutive nomenclature.

P-51.3.1 Preferred IUPAC multiplicative names
For a multiplicative name to be categorized as an IUPAC preferred name, certain restrictive conditions must be met. Multiplicative nomenclature is preferred to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical parent structures, other than alkanes when
(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical and
(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.

Therefore, the preferred IUPAC name for the compound that is given in the question is
1,1'-[(2Z)-but-2-ene-1,1-diyl]dibenzene.
Also note that this is not a biphenyl.


text:
Reaction of multiple diols with sulphuric acid

Find the total number of possible products

I can think of atleast three reactions that will take place in the acidic medium:

Pinacol-Pinacolone rearrangement
Intermolecular dehydration (leading to ether formation)
Intramolecular dehydration (epoxide formation)

Here are the possible products that would be formed:

This gives us a total of $8$ possible products.
Are any different reactions possible? Also, are any of the products that I have listed wrong? (Will any of them not be formed).

You have not got any alkenes in there. I think you might get some ethers of allylic alcohols

Dihydrobenzofurans may also be possible

Epoxides are unlikely. Unfortunately your question is too broad. Its author in textbook vastly underestimated practical aspects of such reaction.

I agree it's a poorly designed question. But anybody who voted to close as "too broad", could you please explain how "this question currently includes multiple questions in one", the description which you presumably read when voting to close?


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What happens to a flat lemon and can it be recharged?
A lemon can be used as a battery, and is often used to teach kids the basics about electricity and chemistry.
This is due to their acidic and electrolytic properties.
What happens to a lemon when it goes flat, does it drain like an ordinary battery?
Once the lemon is flat, can it be recharged like a rechargeable battery?
How does this affect the lemons decomposition?

A lemon does not discharge, or recharge, or go flat; it is metal that is the source of all charge-related phenomena here.

Is this any better? Could you explain this a little, maybe provide an answer if you have one

See this one: https://chemistry.stackexchange.com/a/80252/79678 and search for “lemon battery” here. Minor point: technically, it is a lemon cell, since a battery is two or more cells.

It is the same thing as my answer here: https://chemistry.stackexchange.com/a/116171/79678. All the lemon cell does is inefficiently tap some of the energy resulting from the reaction of the metal anode, e.g., zinc, with the citric acid in the lemon juice. The citric acid gets consumed, the metal gets consumed (oxidized), and the lemon is then contaminated with metal ions. The lemon cell is not rechargeable.

Could also be a dried-out lemon, losing contact. Give it a squeeze.

Please also see this answer: https://chemistry.stackexchange.com/a/9332/79678.

@DrMoishePippik I just about dropped my flask of colored water and dry ice when I saw your “Give it a squeeze.” comment! ;-) I wonder why no one uses grapefruits ...


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Why do we say that a transition state that resembles a stable species in free energy also resembles it in structure?
My textbook states," According to the Hammond-Leffler postulate, the transition- state structure for a step that is uphill in energy should show a strong resemblance to the structure of the product of that step."
Also stating,"One way that the postulate can be stated is to say that structure of a transition state resembles the stable species that is nearest to it in free energy. For example, in a highly endergonic step the transition state lies close to the products in free energy,and we assume,therefore,that it resembles the products of that step in structure as well. "
(Source:Solomons' Organic Chemistry
Book by Craig B. Fryhle, Scott A. Snyder, and T. W. Graham Solomons)

If you quote from a book, please cite it.

That is quite literally what the Hammond postulate says (Wikipedia). So the direct answer to why we say that is because that's what the postulate says. Are you asking about the justification for the postulate?

Yes , I meant is there any logical explanation for it?Or should we just accept it just because it's a postulate..

Because it seems quite reasonable at first. Of course there might very different structures with similar energy, but it is unlikely. And even more so if the structures lie along a reaction path.

Fallon, it is clear to you that the transition state is higher in energy than any other species involved (reactants, metastable intermediates that can be eventually isolated, products) involved?

Yes , I think the answer is really helpful and clears my doubt. Thanks.

EDIT for a different approach:
The energy of a quantum mechanical system of nuclei and electrons is governed by the positions of the pieces. So, we're looking at the energy associated with the wave functions of all the pieces.
For two species that are similar in structure, the idea is that those two structures lead to relatively similar solutions of the wave functions. If the wave functions are similar, the energies of the particles described by those wave functions must be similar as well.

The use of early and late makes this answer very obscure. At least to me.

Early and late are standard terms when discussing the Hammond Postulate. I tried to relate early to "early on the reaction coordinate." Do you have any specific suggestions for improving the answer?

Because isn't clear to which logic you refer to, at the very start. Then what you said reiterate the postulate. Perhaps I am wrong but the "answer" should be that in the comment by @orthocresol, integrated by my comment. The problem is how to explain something intuitive to OP. As I said one can imagine exceptions or violations, of course. But it should be natural to relate similar energy to similar structures, and this is what the postulate does.

I see what you mean. Let me try again.

@Alchimista Better?

I think my comment was not all that helpful! It was meant more to clarify the question...

@orthocresol is what I mean. And I've tried to integrate it to provide what it seems obvious to me.

Yes somehow it is better. It can holds also for classical system like bodies and springs. More to the OP now: perhaps I am the naive one, but it seems a very reasonable postulate if one has the context clear. Perhaps it is not, at the end it is named after Hammond, but it is how I always saw/interpreted it.

@Alchimista In the end, it's just a mathematical statement. For a smooth, single-valued function $f$ of $n$ variables, small perturbations $\delta n$ led to small perturbations of $\delta f$ which can be determined from (and in some sense, capped by) the partial derivatives of $f$.

Right. The movement done is the association to structure.


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Number of Stereoisomers of C(CHClCH3)4
I am having trouble understanding why the structure
$$\ce{C(CHClMe)4}$$
has only five stereo isomers. According to my textbook, the stereo isomers correspond to the following series of absolute configurations:

(R,R,R,R)
(R,R,R,S)
(R,R,S,S)
(R,S,S,S)
(S,S,S,S)

But my question is, does not the order in which these appear change the total? Isn't (R,S,R,S) different from (R,R,S,S)?

Just to make sure, is this the structure of your compound?

Did you try drawing it out or playing with a model?

@andselisk Yes it is

4 1.RRRR
2.RRRS
3.RRSS
4.RSSR SSSR will be same as RRRS on rotating and 3 and 4 will be different .

To answer your question, if you haven't figured it out by now, RRSS is the same as RSRS and it is the only one of the five that is optically inactive (Compare with nonactin). Structure A numbers the positions of chiral chlorine-bearing carbons. Structure B is RSRS and C is RRSS. Clearly, they are superimposable and, therefore, identical. A similar exercise with RRRS and RSSS will give a similar result.


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Home made PCB tinning solution
By reviewing a couple of articles and videos I decided to give some try to develop a PCB tinning solution at home. Based on most of the articles, the easiest way to develop this solution is by using the following chemicals:

Stannous Chloride ($\ce{SnCl2}$)
Thiourea ($\ce{CH4N2S}$)
Sulfuric Acid ($\ce{H2SO4}$)

After checking with a couple of stores, I noticed that Thiourea is not available to buy in any of the stores. Now my questions are:

What Thiourea does in this solution?
Is there any substitute available for Thiourea (for this process)?
Is it possible to prepare Thiourea from any other chemicals? (It seems using Ammonium Thiocyanate ($\ce{NH4SCN}$) can do that job, but it looks like a lengthy and complex process to do at home.)
Is there any possible way to build a tinning solution without using Thiourea?

According to this, thiourea is essential to tinning. Yet, silver polishing contains thiourea so follow them and see what happens. Good luck.

Also keep in mind that thiourea is a human carcinogen. No wonder you cannot find it.

Thiourea ($\ce{(H2N)2C=S}$) is a very important component of tinning solution because tinning is a redox process. That's because thiourea is a powerful reducing agent (Ref.1), chemically and biochemically used in redox reactions in vitro or in vivo:
$$\ce{(H2N)2C=S <=> (H2N)2C-S-S-C(NH2)2 + 2e-} \quad E^\circ = \pu{0.418 V} \tag1$$
The oxidizing agent in this process is $\ce{Sn^2+}$ ion, which reduces it self during the process:
$$\ce{Sn^2+ + 2e- <=> Sn} \quad E^\circ = \pu{-0.1375 V} \tag2$$
As a result, following redox reaction happens:
$$\ce{(H2N)2C=S + Sn^2+ <=> (H2N)2C-S-S-C(NH2)2 + Sn} \quad E_\mathrm{rxn}^\circ = \pu{0.2805 V} \tag3$$
Since $E_\mathrm{rxn}^\circ $ is positive, the reaction is spontanious in standard conditions and no electric current is necessary. When you dip your circuit in the solution (or touch surface to surface of the solution), metallic tin would deposit as a thin layer.
As an alternative to $\ce{(H2N)2C=S}$, dimethylated or diethylated thiourea can be used in those reactions, which require electrode potential within the range of $\pu{0.418 V}$ or higher (Ref.2):
$$\ce{(CH3NH)2C=S <=> (CH3NH)2C-S-S-C(NHCH3)2 + 2e-} \quad E^\circ = \pu{0.460 V} \tag4$$
$$\ce{(C2H5NH)2C=S <=> (C2H5NH)2C-S-S-C(NHC2H5)2 + 2e-} \quad E^\circ = \pu{0.450 V} \tag5$$
In tanning process, theoretically, any component with oxidative potential bigger than  $\pu{-0.1375 V}$ can be used as the reluctant. Thus, best alternative come to my mind is ascorbic acid (Vitamin C). Recently, natural source of ascorbic acid source, extract of fruits of Averrhoa bilimbi (Sinhala name: Billin), has been used to prepare silver and gold nanoparticles from corresponding salts (Ref.3):

In any event ascorbic acid didn't work, there is a way to synthesize thiourea derivatives without using thiocyanate or isocyanate. The scheme is illustrated in the bottom box of above diagram (Ref.4). The $E^\circ$ of these derivatives should be closer to that depicted in equations $(4)$ and  $(4)$. However, any attemp to synthesize these compound should be done in certified chemical laboratory under the observation of an experienced chemist. Keep in mind that thiourea is a potent carcinogen.
Note: It is well known that silver polishing reagents contain thiourea ($50\%$ by weight; Ref.5). You may able to extract thiourea from such solutions.

References:

Paul W. Preisler, Louis Berger, “Oxidation-Reduction Potentials of Thiol-Dithio Systems: Thiourea-Formamidine Disulfide,” J. Am. Chem. Soc. 1947, 69(2), 322–325  (https://doi.org/10.1021/ja01194a048).
Paul W. Preisler, “Oxidation--Reduction Potentials of Thiol--Disulfide Systems. III. Substituted Thioureas and Compounds Containing Two Thioureido Groups,” J. Am. Chem. Soc. 1949, 71(8), 2849–2852  (https://doi.org/10.1021/ja01176a076).
R. S. Rimal Isaac, G. Sakthivel, Ch. Murthy, “Green Synthesis of Gold and Silver Nanoparticles Using Averrhoa bilimbi Fruit Extract,” Journal of Nanotechnology 2013, Article ID 906592, 6 pages (https://doi.org/10.1155/2013/906592).
Wei Tan, Jianpeng Wei, Xuefeng Jiang, “Thiocarbonyl Surrogate via Combination of Sulfur and Chloroform for Thiocarbamide and Oxazolidinethione Construction,” Org. Lett. 2017, 19(8), 2166–2169 (https://doi.org/10.1021/acs.orglett.7b00819).
Mark Laing, Mark Beeston, "Silver polish formulation containing thiourea," US Patent 5,958,854, 1999 (Application No.: 08/869,685 in 1997).


text:
Optical activity in cyclohexane
In case of checking for the optical activity of cyclohexane or its derivative why we check for POS and COS in its cyclic 6 membered planar eclipsed state but not by its real conformation it forms chair, boat,twist boat etc)  .

Are you asking why we determine if the compound is chiral from the simplified planar hexagonal drawing, rather than its true conformation drawing?

Yes I am asking this

Why might you think? Which drawing is easy to show and visualise planes of symmetry? Remember the drawings are saying the same thing - the chair conformation is implied in the drawing of a hexagon.

It is not about being visualised easily but it is about the real structures which should be seen for the molecule.The complete planar form doenot exist in any of its conformation.Check out for finding the plane of symmetry of a simple molecule 1,2 cis dimethyl cyclo hexane which has POS according to planar structure,If there is no problem with this then the actual conformer should also have POS ,but where it is

So with the specific example of cis - 1,2 - dimethylcyclohexane, the molecule is achiral not because it has a plane of symmetry. Each of the conformations of the ring are chiral, but they interconvert rapidly at room temperature and so the overall compound is said to be achiral.

The idea of a compound being achiral doesn't mean it has to contain a plane of symmetry.

Can you explain it further

So each of the conformation is chiral because they have no plane of symmetry like you suggested. However the ring is constant flipping between the two chiral conformations, which are enantiomers of each other. So at room temp you moving between each enantiomer so fast you can call the overall compound achiral. But at very low temp when there isn’t the energy to ring flip it is technically a chiral compound as you stuck in one conformation

You can add this one in the answer section


text:
Is this statement for the First Law of Thermodynamics correct?
My Biochemistry book states the First law of Thermodynamics like this:
"The First Law of Thermodynamics states that in a closed system, energy is conserved".
My book defines Closed System as a system which can exchange energy with its surroundings but not matter.
My doubt is how can energy be remain conserved(constant) if the system can exchange energy with its surroundings.
Please clarify.

You are right. Internal energy can change in a closed system. Your book statements are misleading. The first law of thermodynamics considers the energy balance of your system (internal energy) keeping into account the work exchange and heat exchange between your system and the surrounding environment. Isolated system do not exchange matter nor energy.

Are you sure it does not use "isolated" rather than "closed"? In any case, energy is always conserved for the usual systems of interest. The first law is a restatement of this property.

It states like this: "The first law of thermodynamics states that in a closed system energy is conserved. In other words, although energy can be transferred between the system and the surrounding in different ways,  in chemical process we will consider here, energy can neither be created nor destroyed."

You deleted another question which actually was interesting but somewhat unclear. In essence melting of ice is accompanied by an increase in the entropy ("disorder") of water. However, energy has to be transferred into the ice for this to take place. That transfer results in cooling of the surrounding and thereby in a loss of entropy of the surrounding.

You can think of the first law of thermodynamics as an energy accounting method. It says that all inputs and outputs of energy should add up to the total change in the energy of a system. The exchanged energy is generally classified as work or heat. By "energy can neither be created nor destroyed," it is meant that nuclear reactions, in which the structure of nuclei changes with accompanying transformation of matter into energy and vice-versa, will not be considered. The reactions considered are ordinary chemical reactions involving making and breaking of chemical bonds resulting from interactions between electrons and nuclei.

Your book is misleading. It should have stated "An isolated system", but probably for the sake of convenience, mentioned "closed", ignoring the fact that closed has a completely different terminology.
As an isolated system can exchange neither energy nor matter, there is no form of energy entering or exiting it.
Thus the correct statement would be that "The total energy in an isolated system remains constant".


text:
What is the structure of CF3 radical?
In the class, I was told that $\ce{H3C^.}$ has a trigonal planar structure with the unpaired electron in $\mathrm{2p_z}$ orbital. But $\ce{H3C -}$ has a trigonal pyramidal structure.

But why does this happen?
Why does $\ce{F3C^.}$ also doesn't have a trigonal planar?
Edit : I have seen one possible answer in a different question.  But is there any answer that is explained with the help of VSEPR or MOT or VBT?

Please do not use MathJax formatting in titles.

How could it be trigonal planar if it has an electron in a p orbital?

@Mithoron , Is there any other explanation from VSEPR theory or VBT or MOT?

@JoshMitchell , Carbon in $H_3C•$ is $sp^2 $ hybridised . All the unpaired electrons present in $sp^2$ hybrid orbitals form bonds with Hydrogen.  The remaining unpaired electron is in $p$ orbital

https://chemistry.stackexchange.com/questions/34993/geometries-of-methyl-and-silyl-radicals

CH3 radical is Sp2 hybridised carrying the lone electron in a pure p orbital. This is so because Sp2 hybridised orbitals are lower in energy than Sp3 hybrid orbitals (The more the S character the lesser is the energy of the orbital). When there is a lone electron present, it is more efficient to keep the bond pair electrons in lesser energetic Sp2 hybrid orbitals and lone electron in higher energy pure P orbital than keeping all bonding pairs and the lone electron in Sp3 hybrid orbitals. So most of the radicals are Sp2 hybridised.
But in situations like •CF3 where the atoms/groups attached to C are of high electronegativity things may change. According to Bent's rule more electronegative atoms/groups will favour less electro negative orbitals i.e. orbital s with lesser s character(Sp3 has lesser S character compared to Sp2). This is quite obvious because the more the S character of the C center the more is the tendency of the C center to attract the bond pair towards itself and in cases like electronegative atom, it will also try to attract the bond pair towards itself. This makes some unhappy situation. This is reflected by the Bent's rule. That is why •CF3 radical favours Sp3 hybridisation.
It's not that only such radicals like •CF3 can acquire Sp3 hybridisation. In many other cases like bridge head of adamentum also acquire Sp3 hybridisation becauseof opposition by its rigid structure to acquire planarity on generation of radical though it's not that much easy process to form radical here.
In conclusion, a radical will favour that hybridisation which will be energetically favourable for it.

Sp2 hybridised orbitals are lower in energy than Sp3 hybrid orbitals - this is not true. Hybrid orbitals are no eigenfunctions of the Hamiltonian, thus they cannot be assigned an eigenvalue. It is also not true to say 'electronegative orbitals', because electronegativity is the property of an element. I also cannot get behind your description of Bent's rule. It is very, very important to state that hybridisation is a model, which results from a particular structure. Obviously the structure that is most common will have the lowest possible energy.

What I meant about energy of hybrid orbital is- S orbital is much lower in energy than P orbital and this is because the electrons in S orbital are much closer to nucleus than p orbital. And also if you consider the hybridization, the more the S character in hybrid orbital, the more are the electrons closer to the nucleus and so if you put them in more S charactered hybrid orbital then being much more closer to nucleus it will have lesser energy. Bent's rule is just a rule to deal with elements having different electronegativities while putting them in hybrid orbital model.

I would like to say that electronegativity is not property of an element or anything. It's rather a relative concept. What ever I meant here by electronegativity of hybrid orbital is the attraction force on electrons when they are kept in that hybrid orbital. Please feel free to review my comment. I am even in my undergrad and wrote this answer with the articles or books I ever read.

I'm sticking with my initial assessment: Your answer is wrong. Electronegativity is the power of an atom to attract electrons to itself (https://doi.org/10.1351/goldbook.E01990); it is therefore the property of an atom (or element). It is a relative concept, but not in the way you are describing it. It is still wrong to say that an hybrid orbital can be assigned an energy eigenvalue. Hybridisation is a descriptive concept. Bent's rule, while derived with hybridisation in mind, is independent of it. It is an empirical observation, which remains valid for any wavefunction.

Please answer the question in your view point. I appreciate your knowledge. I wanna know answer to this question too. Yes you are correct that we can't assign eigenvalue to hybrid orbital. I also mentioned in the last comment what I meant to say in my answer. Neither do I deny your logic nor I am obstinate to my answer. I am still a learner. So please feel free to answer this question. (Please don't say again and again 'your answer is wrong'). What happens in chemistry is just because it's energy favourable. So this is not a perfect answer. Please answer the question.


text:
How does overvoltage affect the products of electrolysis?
When you electrolyse an aqueous solution of $\ce{NaCl},$ the product formed at cathode is $\ce{H2}$ gas (preferential discharging), but the product formed at anode is $\ce{Cl2}$ gas. According to standard electrode potential, reduction of $\ce{Cl2}$ is 1.36 V and $\ce{O2}$ is 1.23 V.
Keeping the values in mind, $\ce{O2}$ gas should be formed at anode because it has a smaller electrode potential. But it is given that chloride ions is preferred by anode because of a phenomena called overvoltage.
I don't understand how overvoltage can effect the products of electrolysis.

Basically, there is a kinetic hindrance to the production of oxygen gas at the anode. Overcoming that requires increasing the voltage over what would be required if there was no hindrance. The extra voltage is the over-voltage. The over-voltage depends on factors such as anode composition and surface characteristics.

@EdV Where does this kinetic hindrance come from?

Good question! I have no idea about the exact mechanistic details that are ultimately responsible for the hindrance. It is likely very complicated and hard to study. But the oxidation of chloride is evidently relatively easy: take electrons from a chloride ions and then chlorine atoms form chlorine molecules.

@EdV But I just feel like I can't grasp the entirety of preference of O2 gas over Cl2 gas. I have so many questions! My textbook says the slowness of the electrode reactions create a resistance on the electrode surface which the over voltage needs to overcome? How is this resistance being set up? How do I predict products of electrolysis of other salts like PbBr2 without knowing if it is either preferential discharge or over voltage the priority that the ion is following? Its all very confusing for me.

The resistance is just another way of looking at it: there is a reaction rate “bottleneck”, i.e., the rate of oxidation is hindered. So the current flow, involved in oxygen production, is reduced by this impedance. Over-voltage effects are basically empirical and you just have to deal with not having certainty. Sorry, but that is how it plays out.

Keep in mind the actual redox potential is more important than standard redox potential. Oxygen overvoltage is just additional effect.

When you electrolyse an aqueous solution of NaCl, the product formed at cathode is H2 gas (preferential discharging), but the product formed at anode is Cl2 gas.

There is another complication. The book is missing the fact whether chlorine or oxygen is evolved depends on the concentration of the salt. At low salt concentrations, oxygen will be formed and at high NaCl concentrations $\ce{Cl2}$ will be produced at the anode.
Keep in mind the thermodynamic "competition" is between the oxidation of water and oxidation of chloride ions. Electrochemists have spent their lives and careers in understanding the oxidation reduction of water for hundreds of years. So if you are slightly confused and you have a lot of questions that is a sign of a healthy scientific mind. I am glad that you did not digest what the textbook was saying right away.
Wikipedia hints as how complex is the process of hydrogen and oxygen production at a electrode.

The numbers which you are quoting are like comparing apples and oranges.
This half cell
$$\ce{2H2O(l)} → \ce{O2(g) + 4 H+(aq) + 4e−}$$ $E^o$= +1.23 V is valid for pH of 0, i.e., 1 M H$^+$
ion. Your aqueous NaCl solution is not present in 1 M H$^+$.
You might need to use the Nernst equation to see what is the true potential at a neutral pH.

So the given electrode potentials aren't correct because here NaCl is more in concentration? Then why would my textbook give the reason of over-voltage? I think the experiment not taking place under standard conditions is a better explanation than over-voltage.

No my point is that we cannot ignore concentrations. Read about Nernst equation first. Overpotential is another story but first think about what is the concentration of NaCl.

First one comment on "overvoltage": It doesn´t mean that there is zero reaction below. Even below the theoretical electrochemical potential, electrolysis runs, very slowly. Around the potential plus overvoltage, the slope of the U-I plot has a bend and goes from close to zero to a constant, finite value.
Very simplistic explanations (or rationalisations, rather) for the overvoltage:
To create a diatomic gas, you need to oxidise or reduce two ions in very close proximity. That requires a little extra energy/potential, which turns into heat once the two single atoms recombine. Depends on the mobility of the reduced atom, the adsorbance of which also is an equillibrium process.
There are more possible contributions. The extra "dipolar" charge on the ion by its deformed hydrate hull requires a small extra potential to overcome. In the moment the ion discharges, the ordered water molecules suddenly depolarise, again creating heat.
To adsorb the ion on the surface, you need to push aside other species, which might have a weaker or stronger interaction with the electrode.
Etc.

I am slightly unsure about it "To create a diatomic gas, you need to oxidise or reduce two ions in very close proximity. That requires a little extra energy/potential". The occurrence of two Cl radicals nearby should be a completely random process and it would to be dependent on the concentration of the ions next to the electrode. Why would appearance of two oxidized Cl radicals next to each require an additional energy?

@M.Farooq Good point. Perhaps if you put it this way: The discharge of the ions is an equillibrium process. Either you push it in one direction with a larger potential, or the probability of two radicals occuring next to each other becomes very small. If the radicals are very mobile on the surface, the overvoltage would be smaller, and we know it varies a lot with different electrode materials.


text:
Graph representation of polymer nanocomposites
Graph theory has been used as a means of representing molecules for decades. For a single molecule, a graph representation assigns a vertex for every atom in the molecule and the covalent bonds are represented using edges of the graph, which connect the pairs of vertices. I was wondering if there are established representation for multi-component materials as a single graph, in particular polymer nanocomposites?

Just because Lewis formulas look like graphs doesn´t mean they are, or rather that you can gain much by treating them as such. I´m afraid you are reasoning by analogy. Nanocomposites can (and are) often be characterised as network structures, that´s of course correct. How useful that is depends on what you want to know about them!

@Karl They in fact are frequently used in machine learning. Machine learning algorithms rely on some representation of the molecules. Graphs and SMILES are the frequently used representation methods to train these models.

Hm, I don´t know much about the current state of such methods. You should perhaps ask yourself what machine learning can do (=is good at) when describing molecules with graphs, and if your questions are somewhat similar. You don´t want to be the prototypical  man with a hammer, who sees nothing but nails. ;-)

Maybe you want to provide an example, and I agree with Karl, without specifying what property you are attempting to teach your method to correlate with whatever graph representation you come up with, then it seems you aren't providing enough information. I would say that any sort of abstract representation (molecular formula, dimensions, physical and chemical properties etc) that ultimately correlates in some unique way with the property of interest would provide useful predictive power.

@BuckThorn What you are referring to is supervised learning: you associate input with some value/class and train a model to do prediction. There is another category called unsupervised learning: you learn from the data itself. You can think about face generation networks, you train with many pictures of faces, algorithm learns what does a human face look like and generates new human faces. No value/label is associated with the training data. In chemistry, you can train a model with drug molecules and learn what a drug looks like and produce a new drug molecule.

I'm not quite up to date with AI. Has non-supervised machine learning ever been shown to work? Because even humans are terrible at it. ;-) Basically science and technology evolve because we are a swarm intelligence where each individual is able to learn almost arbitrary things, under supervision. That enables rare strokes of luck/genius to propagate, while they would otherwise flicker and be gone.

Ah, of course! Writing 'ever' in italic won't help with the philosophical argument. It's all statistics. If you learn distribution of an event, you can sample from that distribution and have new samples of the same nature. But let's not get sidetracked.

;-) Putting aside the philosophicals, what do you (intend your computer to) learn nonsupervised? How do you "grade" (t)his learning, or put it to task?

The objective can come in the form of reconstruction of the training data. For instance, VAEs learn to extract features from the input data and then learn to reconstruct the input from those features. They are trained based on how good their reconstruction was.

Why go from drugs to "polymer nanocomposites". Seems quite a leap. What about everything inbetween?

Mind to elaborate?

There are a lot of additional features you'd need to capture moving to a composite. It seems you've had extensive experience with small drug-like molecules. Why not start with a "smaller" problem, say describing a polymer using graphs? In any case, a search of "graph theory polymer composites" actually turned up some hits that could be relevant to you in terms of providing info on features that might be considered, but it seems you are interested only in describing bonding contacts using graph theory. A quick search did not turn up much on this specifically.

@Blade BTW rem to use handle if directing message at someone, otherwise these are likely to go unheard.


text:
Elution and void volumes in a gel chromatography experiment
I was using gel-filtration for separation Hb from xylene cyanol FF with Blue Dextran to determine the void volume. Sephadex G-25 is stationary phase, and mobile phase was buffer $(\pu{0.020 mol dm-3}$ $\ce{Na2HPO4}$ and $\pu{0.150 mol dm-3}$ $\ce{NaCl},$ $\mathrm{pH} = 7).$
Is it possible that my sample has smaller elution volume than void volume?

I took a liberty to correct spelling, formatting, capitalization (use genuine compound names) and wording (don't mix your native language with English) to make the question more understandable, but still it's rather hard to comprehend what is being asked. What is Hb? Hemoglobin? "Is it possible that my sample has smaller elution volume than void volume?" — how did you come up with this conclusion? Feel free to [edit] your question to help us help you.

Blue dextran (molecular weight is about in a range of $\pu{5\!- \! 2,000  kDa}$, according to the manufacturer) is used in affinity chromatography, gel filtration chromatography, protein chromatography and molecular weight markers. Blue dextran is a collection of molecules of dextran to which a blue dye (Cibacron blue F3GA) is covalently attached (Ref.1 & Ref.2):

Most of the molecules of dextran are so large and so irregular in shape that they cannot penetrate the beads (of Sephadex G-25 here) at all, but there may be some of the blue dextran molecules that are smaller and elute at a volume greater than the void volume. The volume at which the first peak of $\mathrm{A_{620}}$ from the blue dextran emerges from the tubing is the void volume, $V_\circ$, of the column plus any volume between the bottom of the stationary phase (Sephadex G-25 here) and the end of the tubing feeding the fraction collector.
According to Wikipedia:

In adult humans, the most common hemoglobin type is a tetramer (which contains four subunit proteins) called hemoglobin A, consisting of two $\alpha$ and two $\beta$ subunits non-covalently bound, each made of 141 and 146 amino acid residues, respectively. This is denoted as $\alpha_2\beta_2$. The subunits are structurally similar and about the same size. Each subunit has a molecular weight of about $\pu{16,000 Da}$, for a total molecular weight of the tetramer of about $\pu{64,000 Da}$ $(\pu{64,458 g/mol})$. Thus, tetramer of hemoglobin-A ($\alpha_2\beta_2$) would have larger elution volume (enter the pores) than the $V_\circ$ measured by blue dextran (which would not enter the pores of Sephadex G-25).

For comparison, $\beta$-amylase (Theoretical Molecular weight: $\pu{223.8 kDa}$)would elute first after Blue dextran with an elution volume little greater than $V_\circ$.
References:

John E.Wilson, "Applications of blue dextran and cibacron blue F3GA in purification and structural studies of nucleotide-requiring enzymes," Biochemical and Biophysical Research Communications 1976, 72(3), 816-823 (https://doi.org/10.1016/S0006-291X(76)80206-9).
David Viet, Stephanie Beck-Candanedo, Derek G. Gray, "Synthesis and characterization of blue dextrans," Carbohydrate Polymers 2008, 74(3), 372–378 (https://doi.org/10.1016/j.carbpol.2008.03.003).


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Is the XRF spectroscopy equally sensitive for every element?
I am curious whether or not every element can be equally well detected by using X-Ray-fluorescence-spectroscopy. Might there be any contitions or circumstances that make it hard/impossible to detect a certain element in a probe?

In a very very broad reasoning, XRF is much more sensitive to the "mid-range" elements.
Elements below Na are generally not detectable.

The x-ray tubes must have a beryllium window to seal the tubs which absorbs the low energy x-rays.
The lower the atomic number the less the fluorescent yield.
Low energy x-rays don't penetrate from the sample efficiently.
Low energy x-rays tend to be absorbed in the detector windows.

For the very heavy elements the detectability is poor too. Say anything about lanthanum.

The K x-rays are so energetic that a high voltage on the x-ray tube is needed to excite them.
The lanthanides and actinides tend to behave similarly chemically so geological samples tend to contain a mixture which creates peak overlaps.
Higher energy detection of x-rays is poor. EDS because the x-rays penetrate the detector. WDS because the available crystals don't diffract high energies well.
The L x-rays are not quite as useful because the greater number of x-rays leads to more peak overlap.


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Why HCl is used in Lucas test?
In Lucas test $\ce{HCl}$  is used for acid-base reaction between $\ce{Zn(OH)Cl}$ to give $\ce{ZnCl2}$ and $\ce{H2O}$ which is kind of good because $\ce{ZnCl2}$ is formed again and used in first step quickening the reaction. But why don't Lucas test use other acids like $\ce{H2SO4}$ (and add $\ce{ZnCl2}$ in bulk) or $\ce{HClO4}$ (stronger acid than $\ce{HCl}$ may faster the reaction). But why $\ce{HCl}$?
Is it had to do with acidic strength of $\ce{ZnCl2}$ and acid (I mean if stronger acid than $\ce{ZnCl2}$ may react with alcohol)?

ZnCl2 can also donate Cl2 (in fact both ZnCl2 and HCl can).That doesn't resolve my question (Why HCl why not strong acid like HCLO4)

HCl is strong and with ZnCl it makes an adduct that's even stronger. Also adding conc. HClO4 may lead to explosion...

ZnCl2 would not form ZnCl4^2-
with other acid but HCl.

The use of concentrated H2SO4 is not amenable. Per a source:
Concentrated sulphuric acid produces messy results. Not only is it an acid, but it is also a strong oxidising agent. It oxidises some of the alcohol to carbon dioxide and at the same time is reduced itself to sulphur dioxide. ... It also reacts with the alcohol to produce a mass of carbon.
On the other hand, per Wikipedia, to quote:

The Lucas test in alcohols is a test to differentiate between primary, secondary, and tertiary alcohols. It is based on the difference in reactivity of the three classes of alcohols with hydrogen halides via an SN1 reaction:[3]...The differing reactivity reflects the differing ease of formation of the corresponding carbocations. Tertiary carbocations are far more stable than secondary carbocations, and primary carbocations are the least stable.

And further:

An equimolar mixture of ZnCl2 and HCl is the reagent. The alcohol is protonated by this mixture, and H2O group attached to carbon is replaced by the nucleophile Cl−, which is present in excess. Tertiary alcohols react immediately with Lucas reagent as evidenced by turbidity owing to the low solubility of the organic chloride in the aqueous mixture. Secondary alcohols react within five or so minutes (depending on their solubility). Primary alcohols do not react appreciably with Lucas reagent at room temperature.[3] Hence, the time taken for turbidity to appear is a measure of the reactivity of the class of alcohol, and this time difference is used to differentiate among the three classes of alcohols:

Now, with respect to HClO4, there is at least one catalytic study of HClO4-SiO2 during the reaction of benzyl alcohol (1 mmol) and thiophenol (1.1mmol) at room temperature under solvent-free conditions (see).


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Using solubility product to determine the mass of the precipitate of a reaction
I mix together $\pu{100 mL}$ of an aqueous $\ce{NaCl}$ solution at $\pu{0.5 M}$ and $\pu{100 mL}$ of an aqueous $\ce{AgNO3}$ solution at $\pu{0.3 M}$. Assuming that the solubility of $\ce{NaCl}$ in water is $\pu{360 g/L}$, the solubility of $\ce{AgNO3}$ in water is $\pu{2340 g/L}$, the solubility of $\ce{NaNO3}$ is $\pu{921 g/L}$ and the $K_\mathrm{sp}$ of $\ce{AgCl}$ is $1.77 \times 10^{-10}$, what is the mass of the precipitate ($\ce{AgCl}$) formed?
I know that the answer to this problem is $\pu{42.1 g}$, I solved it using stoichiometry. However, I want to know how to solve it using the solubility product and the other solubility figures given.

@EdV Apparently there is a way to use the solubility figures given to get to the same answer and I would like to know how.

@EdV well, I was told to use the notions of equilibrium solubility, which is not what I used to find 42.1 g

What is the shaggy dog ?

@EdV: With all the respect, the answer is $\pu{4.3 g}$. I think, you forgot the initial volume of $\ce{AgNO3}$. :-)

@MathewMahindaratne You are correct: I spaced a factor of 10 due to the volume of silver nitrate solution. I will delete some comments as penance and thanks for correcting me! ;-)

This is awfully closed to a homework question. Since you already have an answer (real answer is actually $\pu{4.3 g}$, see following calculations) and still insisting on how to use the solubility product information, I'd like to give some insight. Since all components are in solutions, solubilities of $\ce{AgNO3}$, $\ce{NaCl}$, and $\ce{NaNO3}$ should not be in concern. Solubility of $\ce{AgCl}$ is the only one should be concern:
$$
\begin{array}{lccc}
\ce{&2 AgNO3(aq) & + NaCl(aq) &<=> &AgCl(s) &+ & NaNO3(aq)} \\
\text{Initial}, \pu{mol} & 0.3 \times 0.10 = 0.03 & 0.5 \times 0.10 = 0.05 && 0 && 0 \\
\text{Final}, \pu{mol} & 0 & 0.05 - 0.03 = 0.02 && 0.03 && 0.03 \\
\text{Final}, \pu{mol/L}& 0 & \frac{0.02}{0.2}= 0.1 && \pu{0.03 mol} && \frac{0.03}{0.2}= 0.15 \\
\end{array}
$$
Thus, amount of $\ce{AgCl}$ precipitated $= \pu{0.03 mol} \times \pu{143.3 g/mol} = \pu{4.3 g}$
Since $K_\mathrm{sp}^\ce{AgCl} = \pu{1.77 \times 10^{-10} mol^2L^-2}$ is such a small number, it is safe to say almost all of $\ce{AgCl}$ is in solid form. However, the solubility product come to play a role if there is a question about what is the final $\ce{[Ag+]}$ in solution. Now you have to consider following equilibrium:
$$\ce{AgCl(s) + H2O <=> Ag+ + Cl-}$$
Hence, $K_\mathrm{sp}^\ce{AgCl} = [\ce{Ag+}][\ce{Cl}] = \pu{1.77 \times 10^{-10} mol^2L^-2}  \tag1$
Suppose the solubility of $[\ce{Ag+}]$ is $\alpha \ \pu{ mol\:L-1}$. Then, $[\ce{Cl-}]$ is $\alpha \ \pu{+ 0.1 mol\:L-1}$ (common ion effect). Yet, $K_\mathrm{sp}^\ce{AgCl} = \pu{1.77 \times 10^{-10} mol^2L^-2} $ is such a small number, the $\alpha$ amount from the dissolution of $\ce{AgCl}$ can be considered negligible.
Thus, from equation $(1)$:
$$K_\mathrm{sp}^\ce{AgCl} = [\ce{Ag+}][\pu{0.1 mol L-1}] = \pu{1.77 \times 10^{-10} mol^2L^-2} \\ \Rightarrow [\ce{Ag+}] = \frac{\pu{1.77 \times 10^{-10} mol^2L^-2}}{\pu{0.1 mol L-1}} = \pu{1.77 \times 10^{-9} molL-1} $$


text:
Why is boiling point of carbon tetrachloride greater than silicon tetrachloride?
$\ce{CCl4}$ and $\ce{SiCl4}$ both are non polar molecules thus London forces must be the only forces of interaction.
Then London forces are supposed to be greater in $\ce{SiCl4}$ due to its larger size and its boiling point should be higher than comparatively smaller $\ce{CCl4},$ but $\ce{CCl4}$ has higher boiling point.
The only explanation I got is the greater polarity of the $\ce{Si-Cl}$ bond. I don't understand how polarity of individual bonds in a non-polar molecules increases its boiling point.

The larger Si atom changes the outside shape of the molecule. I´ll guess that the "contact area" between neighbouring molecules in SiCl4 is less.

Interesting! That contrasts with b.p. of CF4 & Si4, −128 °C and −86 °C; and with CI4 and Si4, 171 °C and 287 °C, respectively, where the Si compounds have higher b.p., as opposed to 77 °C and 58 °C for the chlorides.

@karl As Si is larger  than C shouldn't the contact area be greater for sicl4

Carbon tetrachloride is more compact in shape. Try calculating the molar densities, and compare with the molecules' diameters.

@karl I asked the same thing. If ccl4 is smaller and lighter then why does it have a higher boiling point?

That´s not the point I am trying to make. If you scale down SiCl4 so the molecule has the same size as CCl4, how many of them are in the same volume? Or in other words, how good is the packing of both of them in the liquid state? Or you compare the unit cell volumes of the crystalline substances. My guess is that SiCl4 is not quite as "dense" as it looks, in terms of packing.


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Free refined value in Rietveld
I have a question about Rietveld refinement
The author of this paper claimed that the free refined CN ligand is 1.081.
However, they does not show how they got that number. They stated it was based from this table however I am not sure how you can get 1.081 from this table. What is the calculation?

Thank you.
Source:
Reference

Wang, W.; Gang, Y.; Hu, Z.; Yan, Z.; Li, W.; Li, Y.; Gu, Q.-F.; Wang, Z.; Chou, S.-L.; Liu, H.-K.; Dou, S.-X. Reversible Structural Evolution of Sodium-Rich Rhombohedral Prussian Blue for Sodium-Ion Batteries. Nature Communications 2020, 11 (1), 980. DOI: 10.1038/s41467-020-14444-4

"they does not show how they got that number"—Well, they kinda do, it's just that you don't have raw data to follow it: "All PXRD data analysis was done in TOPAS 5 software. The PXRD data was first indexed to get unit cell, lattice parameters, and crystal symmetry information. Then the indexed unit cell was used for Le Bail fitting the PXRD data to derive the suitable peak profile, and lattice parameters. These derived data was fixed and used for further Rietveld refinement."

I guess that 1.081 refers to the distance length (in Angstrom) separating the C and N atoms. You obtain this value by applying by using the values listed in the table; this is quite tedious, takes some time and require quite good knowledge about crystallography. Otherwise (and much more easy), you can use the crystallographic data listed in the table to build the structure using a proper crystallographic software (such as Vesta); then the software will calculate bond distances for you.

I would highly doubt that $d(\ce{C#N})$ can be as small as 1.081 Å (typical nitrile bond is 1.16 Å), that's more likely $d(\ce{C-H}).$ Besides, if you read the paper, it explicitly says that 1.081 is the site occupancy of free refined CN ligand before they fixed occupancies for carbon and nitrogen atoms at 100%. I do agree though regarding your point on VESTA, this is wonderful piece free of software, but here it won't help to get this number. Raw data + TOPAS 5, like authors did, is probably the only way.

I didn't read the paper. In any case bond length ranges around typical values, but is not fixed. it depends on temperature, pressure and even chemical nature of the surrounding atoms. In the present case 1.081 is not so different than 1.16;if you use the crystallographic data reported in the table you will find that the bond lenth results even smaller than what I supposed, about 0.95 Angstrom. In any case a site occupancy equal to 1.081 is meaningless; an atomic sita cannot be more occupied than 100%. for this reason authors fixed occupancy at 1 (100%).


text:
Why does solubility of Carbonate salts of Alkali metals in water increase down the group?
Carbonate salts have a very large anion so hydration energy should dominate over lattice energy. Since hydration energy is inversely proportional to radius of ion, I would expect Lithium to release the largest hydration energy and thus be the most soluble. But my textbook (Concise Inorganic Chemistry by JD LEE) says that the "Solubility of carbonates and fluorides of Alkali metals increase rapidly on descending the group. The reason for this is that their Lattice energies change more than the hydration energies on descending the group." However, it provides no
further explanation as to why that is true. Could someone please explain this better?

Did you find the answer for this?


text:
Distilling milligram quantities of oil like it's late 1970s
In a currently retracted* manuscript by Tomas Hudlicky [1, Retraction Watch] the following has been noted:

Fifty years ago professors took an active part in all laboratory instructions and they taught
  advanced techniques to research undergraduates, graduate students, and even postdocs.
  During my postdoctoral stay in Geneva Professor Oppolzer personally showed me how to distill 3 mg of
  an oil after I commented that it could not be done.
  Such practices are all but absent at present.
  Students are not taught by professors, most of whom have not been at the bench for decades, but
  by more senior personnel in the research laboratory.

Does anyone know or can suggest a plausible method used by Prof. Wolfgang Oppolzer to distill milligram amounts of oil, considering this demonstration has likely took place between 1977 and 1978 (according to this CV)?
I flipped through the distillation handbook of that time [2], especially focusing on section 5.1.1 Micro- and semi-micro-distillation, and I suppose it was possible that some version of an extremely miniaturized distillation apparatus combined with a set of spinning band columns or concentric tube columns operating under reduced pressure could've been used:

The separation of essential oils with spinning band columns at 4 to 13 torr has been reported by Spiegelberg [9].
  Examples of separations of high-molecular organic compounds with a concentric tube column are given by Jantzen and Witgert [3] and Fischer [7].

Bonus question: how relevant mastering this distillation technique would be these days?
Are there better alternatives and what are they? (Gas chromatography, I presume?)

* Thanks to M. Farooq for drawing my attention towards the article and motivating to find and actually read the original paper.
References

Hudlicky, T. “Organic Synthesis—Where Now?” Is Thirty Years Old. A Reflection on the Current State of Affairs. Angew. Chem. 2020, ange.202006717. DOI: 10.1002/ange.202006717. (Retracted 2020-06-05)
Krell, E. Handbook of Laboratory Distillation: With an Introduction into the Pilot Plant Distillation, completely rev. 2nd ed.; Techniques and instrumentation in analytical chemistry; Elsevier Scientific Pub. Co. ; Elsevier North-Holland: Amsterdam; New York, 1982. ISBN 978-0-444-99723-4.

During the 1970's we used a homemade Kugelrohr for non-fractionating distillation of small amounts of liquid under reduced pressure. Today there are hi-tech Kugelrohrs: en.wikipedia.org/wiki/Kugelrohr

Today, in order to purify 3 mg sample, one would do a semi-preparative liquid chromatography or supercritical fluid chromatography (preferred). It is not a big deal today. GC columns have very poor capacity.

@Andselisk, If you search microdistillation in ACS Analytical Chemistry, you will find plenty of apparatus descriptions until the 1960s.


text:
Is it possible to determine if a substance can dissolve in water just by looking at its chemical structure?
Is it possible to determine if a substance can dissolve in water just by looking at its chemical structure? For example, how am I supposed to know if $\ce{NaHCO3}$ and $\ce{Na2CO3}$ can dissolve in water or not? Or should I just memorize all of it?

No, but with experience you will start guessing whether the given compound will dissolve in water or not. At this stage learn the solubility rules. There are not many.

Adding to what @M.Farooq said, keep an eye out for exceptions eg Silver Fluoride, Lead Chloride, Sodium Bismuthate etc

Generally speaking, several thousands sodium and potassium compounds are known. They are all soluble in water. The only exception are rare : sodium bismuthate, sodium uranyl acetate, potassium perchlorate, potassium bitartrate, and may be some odd molecules, which remain exceptions. So please, just memorize that sodium and potassium compounds are soluble.

Yes. At least in practice and, as far organic compounds and different solvents are in scope, you don't have to memorize one by one. Note that this doesn't mean you always can or it is easy.


text:
Classification of Organic Compounds And Definition of certain terms
I am currently studying in class 12, and I realised that I have a couple of fundamental doubts in the classification of organic compounds which was not covered well in class 11. My doubts are not related to nomenclature, which I regard as one of my strong points rather, it is directed at the definitions of certain terms such as saturated, unsaturated, cyclic, aliphatic etc.
In lieu of this, I have a couple of questions which I would be highly appreciative any of you could answer. Thank You so much in advance.(P.S. You can simply respond with a yes if my understanding is correct)
Q.)Is a cyclic compound defined as a compound comprising of at least one cyclic ring? So it may also comprises of multiple open chains as long as it comprises of at least  one cyclic ring.

E.g Will the above compound be considered to be a cyclic compound even though it also has open chains?
Q.) My second question is whether the terms saturated and unsaturated are only defined for hydrocarbons or for all organic compounds in general? This stems from the fact that I read that the definition of a saturated a compound is a compound that only has C-C bonds.

E.g Is the above compound, considered to be a saturated compound since it only has C-C?

Query regarding cyclic compounds- yes your reasoning is correct. Your structure has multiple functional groups (C=C) and a cyclopropane ring. Look at the structure of cholestrol, it has a number of rings and a long chain as well.
Query regarding unsaturation: Saturated is a very old term in organic chemistry, dating back when the atomic structures and electrons did not exist. The meaning of saturated, even today is "chemically combined to the maximum extent, having no ability to combine further." The carbon atoms in cyclopropane are saturated in the sense that all the four valencies of each carbon atom are satisfied by four individual atoms.
You will also encounter another term called the degree of unsaturation, it is a related concept: It tells us how many rings or double bonds are in the molecule. Don't mix these two terms.
You are again right in the sense that saturated and unsaturated terminology is frequently used for carbon chains or cycles.
For example we say, saturated fatty acid and unsaturated fatty acid as shown below:

We are silently ignoring the C=O bond in this usage.

Thanks for your response. So is the second compound I have drawn(1-chloropropane) a saturated or an unsaturated compound? Thus it is also possible for derivatives of hydrocarbons to be termed as saturated compounds right?

Yes, 1-chloropropane will be considered as a saturated compound because there is no double bond in the carbon chain. Chains (linear or cyclic) and their derivatives without double bonds/ triple bonds in C to C will be considered saturated.

Thanks for your response ! An unrelated question which I know you will be able to clarify, alkanes only refer to acyclic saturated hydrocarbons right? Cyclic saturated hydrocarbons belong to the class of cycloalkanes right?

Yes, correct. Accept the answer if you understand it now.


text:
What is fugacity?
When one attachs a manometer to car tires or carbon dioxide fire extinguishers, what is being measured: the pressure or the fugacity?
For ideal gases both are the same, but not for real gases.
As fugacity is used on thermodynamics calculations, I see it as the correct value and would expect it to be the property being measured on the situation above.
However, the Wikipedia page says that the vapour pressure of a liquid is not equal to the fugacity of the vapour. Therefore, it cannot be the property being measured on the situation described.
What is it then? Is it just a value used for making the thermodynamic equations work? Therefore being a correction factor to thermodynamics.

Fugacity is a terrible invention. It causes more confusion than anything. It is the effective pressure, useful when modeling stuff, but not useful in the real world

The original paper of Lewis [1, p. 54], who invented the concept of fugacity explains it in a clearer way than Wikipedia's convoluted explanation:

II.  FUGACITY.
If any phase containing a given molecular species is brought in contact  with any other phase not containing that species, a certain quantity will pass from the first phase to the second. Every molecular species may be considered, therefore, to have a tendency to escape from the phase in which it is. In order to express this tendency quantitatively for any particular state, an infinite number of quantities could be used, such, for example, as the thermodynamic potential of the species, its vapor pressure, its solubility in water, etc. The quantity which we shall choose is one which seems at first sight more abstruse than any of these, but is in fact simpler, more general, and easier to manipulate. It will be called the fugacity,† represented by the symbol ψ and defined by the following conditions:—

The fugacity of a molecular species is the same in two phases when these phases are in equilibrium as regards the distribution of that species.
The fugacity of a gas approaches the gas pressure as a limiting value if the gas is indefinitely rarefied. In other words, the escaping tendency of a perfect gas is equal to its gas pressure.

That these two conditions are sufficient to define a property of every substance which is not a mathematical, fictitious quantity, but a real physical quantity, capable of experimental determination in every case, must now be shown. It is obvious from the above conditions that in any case where our present methods of measurement are unable to show a deviation of the vapor of a substance from the gas law then the vapor pressure is the nearest approximation to the fugacity. In all cases the vapor pressure is an approximation to the fugacity, the approximation being nearer the nearer the vapor is to a perfect gas. […]

†In the earlier paper this quantity was called the escaping tendency and represented by the same symbol. For the sake of brevity I have chosen to substitute the word “fugacity” for “escaping tendency” without the slightest change in the meaning of the function.

More in Lewis's original words [2, p. 262]:

The fugacity of a substance is therefore equal to its vapor pressure if the vapor behaves like a perfect gas. Speaking in terms not very precise, we may say that the fugacity of a substance is equal to the vapor pressure that the substance x could have if its vapor were a perfect gas. It has been shown in the preceding paper that for a given substance in a given state the fugacity is a definite property of which the numerical value can in most cases be readily determined, and which is well suited to serve as an exact measure of the escaping tendency.

References

Lewis, G. N. The Law of Physico-Chemical Change. Proceedings of the American Academy of Arts and Sciences 1901, 37 (3), 49. DOI: 10.2307/20021635. (Free Access)
Lewis, G. N. Outlines of a New System of Thermodynamic Chemistry. Proceedings of the American Academy of Arts and Sciences 1907, 43 (7), 259. DOI: 10.2307/20022322. (Free Access)

Very clear indeed. The manometer measures the pressure then. I thought fugacity could be something like a "real pressure". Once I read on Reddit that the papers that introduce concepts often offer clearer explanations, as no reader has any idea of the concept whatsoever (it was about quantum physics, specifically).

Straight to the source, splendid answer! It's also good to know what time does the usage of em dash after colon stems from:)

I still have trouble understanding fugacity. Lewis's article speaks of "any phase" in contact with "any other phase". So these phases may be two liquids, like water and cyclohexane. And here the fugacity is not related to the gas pressure. It is related to the concentration or to the activity. So what is fugacity ?

@andselisk, Your perfectionism in formatting is unbeatable :-) I didn't even notice that. Your grammar teacher must have been very good.

@NecaNe, Yes, that why a look at original papers is handy and this one is freely available from Hathi Trust (legally)

@M.Farooq I also found this intersting: https://hsm.stackexchange.com/questions/1926/who-named-the-fugacity-who-coined-the-variable-name-and-did-it-already-relate-t

This definition is better than the one Poutnik got from the Wikipedia? : "1. The fugacity of a molecular species is the same in two phases when these phases are in equilibrium as regards the distribution of that species." Can you explain what that means? I find the explanation from the Wikipedia perfectly clear and simple.

Here's my problem with this definition: I can't remember seeing the term fugacity except in the context of gases. So what two phases are we talking about here?

@BuckThorn You probably want to read the paper further, namely pp. 55–56.

@BuckThorn, The problem with Wikipedia "explanation" is that it needs the crutches and jargon of other terms "molar Gibbs free energy", "real gas", "activity", "reference pressure" etc etc. Lewis explained it more elegantly like a master!

Granted it might seem more "jargony", but Lewis' definition also refers to a "perfect gas" (one which follows "the gas law"). And Lewis' comment does not refer to the molar Gibbs energy by name, but the molar Gibbs energy is just the chemical potential for a pure substance (for instance for an ideal gas they are the same), and how exactly do you quantify when "the fugacity of a substance is equal to the vapor pressure that the substance x could have if its vapor were a perfect gas" without bringing up the chemical potential?

GN Lewis was certainly brilliant, but it may be that he did not win the Nobel Prize because physical chemists resented his introducing terminology such as the fugacity into the canon (I jest). Lewis is also responsible for coining "activity".

@BuckThorn, Yes in the same cited papers he coins activity. Oh chemical potential is another can of worms. I would be so glad if anyone can explain me chemical potential without using circular definitions of thermodynamics.

@M.Farooq Chemical potential is the rate of change of the Gibbs free energy with respect to amount of substance when temperature and pressure are held constant. Defining Gibbs free energy requires the introduction of entropy and applying it requires thinking about limiting (reversible) processes which does make it a difficult concept. Still, that doesn't make it any less real or useful.

@BuckThorn Yes indeed. Chemical potential is quite complex and highly abstract. The only good explanation of free energy I every found was in Laitinen's or Lingane's electrochemistry.


text:
Mixture of products in ring opening epoxide reaction with acid
I ran across a question that says:

In the following reactions, the result is sometimes one product, and sometimes a mixture of products. Design the mixture of products, explain why this is happening and predict what the main product is.

Since all of them are ring opening with acid reactions, I know that the nucleophile attacks the most hindered place. What it means by mixture of products? To draw the attack on the most and least hindered place and explain that the major one is the attack on the most hindered place, or there is a racemic mixture? And if so, why is that?

Let's see what you think the answers are. Answers are not free ;)

@user55119 I obviously do know where and why the attack will happen each time, and how the epoxide will open. The part which i don't understand is if they want the racemic mixture or to draw the attack on the most and least hindered place and explain which is the major product.

Structures that appear chiral are assumed to be racemic (a and b) unless indicated otherwise. The issue of mixtures would be regioisomers. You should put your attempts in the original post which will invite comments or answers.