text:
Are Anomers Epimers?
epimers are the class of stereoisomers which differ along configuration at any carbon in a carbohydrate(generally),
And anomers are the kind of stereo  isomers that differ at the carbonylic carbon, which becomes achiral only after hemiacetal (hemiketal) formation
Websites on search give that Epimers are bigger group where Anomers are a part of them, but in certain books on biochemistry like that of Lippincott(d.ferrier) writers define Epimers as change in configuration at any carbon except anomeric carbon..
So are Anomers Epimers?

Related: Difference between Epimers and Diastereomers

IUPAC basic terminology of stereochemistry defines anomers and epimers:

Anomers
Diastereomers of glycosides, hemiacetals or related cyclic forms of sugars, or related molecules differing
in configuration only at C-1 of an aldose, C-2 of a 2-ketose, etc.
Epimers
Diastereomers that have opposite configuration at only one of the two or more stereogenic centers present in the respective molecular entities.

So yes, anomers are a subset of epimers.
(Interestingly, on the contrary, ‘anomer’ term is exceptionally used in much broader meaning than ‘epimer’, in the term ‘anomeric effect’, which became generalized to e.g. conformations of molecules completely unrelated to saccharides.)


text:
How many samples would be worth when following a desorption? How to relate with limit of detection?
I am following the desorption of a compound from a polymer. Let's imagine that it occurs at a certain rate and most of the compound is desorbed after 5 days (as in the figure). Then, I could use many analytical techniques to follow the compound desorption by measuring at, e.g., t = 1 day and t = 4 d. The concentration difference between these two points is large enough to be detected.

However, if I the desorption is much slower (as in the figure below) and I can only take samples for 10 days, the only way to characterise the desorption rate would be with a much more sensitive analytical technique, as I would need to detect changes from, e.g. 0.98 to 0.96

My question is, supposing that I have a good guess of the concentration differences that I will encounter, how can I choose an analytical technique beforehand based on reported limits of detection and/or quantification? How can I relate limits of quantification with the ability to differentiate a sample at 0.98 and another one at 0.96?

I think you are confusing two concepts: the limit of quantification (LOQ) and the sensitivity of the technique.
LOQ is the lowest concentration of the analyte that can be quantified at an acceptable level of uncertainty and accuracy, and is determined (by most methods) as the calculated concentration of the blank + 10*sd of the blank measurements.
Sensitivity, on the other hand, is the method's ability to distinguish between different amounts of analyte.
There are some guidelines that do not make a differentiation between the two (such as in the FDA's Bioanalytical Method Validation guidance document), but this is generally not the case in other situations.
In your case, you will need to find an analytical method that is sensitive enough to differentiate between 0.96 (a.u.) and 0.98(a.u), that is, 0.02 units.
As for the limit of quantification, you need to ask yourself what is the lowest concentration of analyte that you want to quantify. Based on the decay plot you have shown, it seems like the lowest concentration you would encounter is above 0.8 a.u. In that case, a method with an LOQ of 0.8 a.u should be sufficient. If you want to be even more rigorous, your LOQ should be at a level that is low enough that even if it is extended by the expanded standard uncertainty of your method, it should still be low enough to quantify the lowest level of analyte you expect to encounter.

I know the difference between both. My question is whether I can relate the sensitivity and the LOQ somehow, even as a rough approximation

Right. I'm sorry, I had to read that again to understand better what you mean. I doubt that LOQ can be related to sensitivity since the former is a characteristic of your entire analytical procedure (including sample preparation steps, etc), whereas sensitivity, if I remember correctly, is an instrumental characteristic. I suppose your goal is to find an instrument with that sensitivity, but since it is not a parameter that is often quoted by manufacturers, you choose to look st LOQ instead?

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text:
Basification of acid solution. pH calculation. How?
This is my first post.
In order to understand basification I've created a pretty simple example as I currently understand it mainly because.. I really don't understand all of it.
Example:
Chemicals = Lye & Vinegar

Reaction  = NaOH + C2H4O2 = H2O + NaC2H3O2

Since the molar ratio of the equation is 1:1...

                Molar Mass          It takes...
Lye     NaOH    39.99710928  = 40   Grams of Lye to NEUTRALIZE,
Acetic  C2H402  60.05196     = 60   Grams of Acetic Acid.
                               *....a Weight Ratio of 2:3

Now to the part I don't understand...
1 liter = 1000 grams

At 5% standard acidity, a liter of vinegar has exactly
1000 Grams x 0.05 = 50   Grams of acetic acid in it.

Thus, at the above calculated Weight Ratio of 2:3...
50 Grams of acetic acid x 2/3 = 33.33   Grams of Lye to NEUTRALIZE,
                                50      Grams of Acetic Acid.
                                *....proportional Weight Ratio of 2:3

So far...
We have neutralized the 50 grams of acetic acid in the solution to pH7.
Question:
What would be the pH of the remainder 950ml of water in the bottle of vinegar? Is it the obvious? pH7?
Is my goal only to neutralize the acetic acid in the solution because the remainder is obviously just water and naturally at a of pH7?

Your neutralized solution will not have pH 7, because the reaction of sodium hydroxide and acetic acid produces sodium acetate, which is a weak base.

You've made some assumptions that we can deal with, and some that need adjustment.
1) 1 liter of water weighs 1000 grams. Correct. However, 1 liter of 5% acetic acid weighs 1005 grams. No big deal; 1 liter will still contain 50 grams of acetic acid.
2) As the commenter noted, neutralizing the acetic acid gives sodium acetate, a weak base with pH ~9. You can use phenolphthalein as an indicator because it changes color at this pH.
3) The liter of neutralized acid will be homogeneous, so the 50 grams of acid and 33.33 grams of sodium hydroxide will be distributed throughout; any sample of the solution will have the same pH.
(Although it is interesting to do an acid/base titration with phenolphthalein. As you approach the endpoint, a drop of base will splash red, then fade to colorless. The next drop will splash a larger red cloud, then fade; etc. Finally, you might get a huge red cloud and think you are done, but you are just shy of the last bit of neutralization and the solution turns colorless again. The next drop flashes totally red.  The reason I mention this is that you could say that for those few milliseconds when the solution is partially red and partially colorless, the pH is different in different locations, but this is not a steady state, so you mix it till you can get a repeatable measurement.)
4) What you measure when you stick a pH meter into the solution is the concentration of the hydrogen ion (hydronium ion) that results from the water molecules reacting with everything else all over the solution. The acetic acid is a reactant in the original solution that increases the hydrogen ion quantity, but the sodium hydroxide reduces it, so you have made a solution of sodium acetate, pH ~9.

Just a final thought. If the water I had started with was Deionized water at a starting pH7, would the  solution of sodium acetate then basify to pH7 or pH9 is just what it is?

Sodium acetate does not give pH = 7 when dissolved in pure water. Sodium hydroxide is a strong base (completely dissociated); water molecules will attach to a sodium ion about as readily as hydroxide ions. Acetic acid is a weak acid; the acetate ion tends to hold onto its proton in acetic acid, and in water, acetate ions (from sodium acetate) will even pick up some protons, so the pH rises to 9.

Part A
why the final pH of the solution resulted from the neutralization reaction of a weak acetic acid with a strong sodium hydroxide base in the bottle of vinegar>$7$?

I can infer from the post that you maybe have a misconception about the neutralization reaction and neutrality which hinder the scientific understanding to what happened in the bottle of vinegar when the acetic acid neutralized. You need to differentiate between neutralization and neutrality and enhance your scientific knowledge about the  concepts related to neutralization and neutrality as :

•   Neutralization and neutral $\mathrm{pH}$  are often different things. The neutral $\mathrm{pH}$  is when$\ce{ [H3O +] = [OH-]}$ while the neutralization reaction is when acid and base react to form water and salt, but the salt is not necessarily neutral. The students must success in realizing of the central role of water in neutralization reactions to accept that the neutralization of acid and base is not always given a neutral product.
•   The equivalence point is the point at which the moles of acid are equivalent to the moles of base, according to the equation.

When you  have neutralized the$ 50\text{ grams} (50/60 = 0.83\text{mol})$of weak acetic acid by a$ 33.33\text{ grams} (33.33/40 = 0.83 \text{mol})$of strong base sodium hydroxide means that you completely have converted the acid ($\ce{CH3COOH}$) to its corresponding base($\ce{CH3COO-}$), so you will get $0.83\text {moles}$ of sodium acetate ($\ce{CH3COONa}$).
The neutralization is fairly straightforward:
$$\ce{CH3COOH+NaOH⟶CH3COONa+H2O}$$
Hence, at the equivalent point, the acid in the vinegar bottle has been fully neutralized by the base (no unreacted acid or base present anymore) the bottle will only contain $\ce{H2O}$ and the basic salt ($\ce{ CH3COONa}$). The result, at the equivalence point, will be the same as dissolving($\ce{ CH3COONa}$) in water at the same concentration as you have at the equivalence point.
The salt ($\ce{ CH3COONa}$) is  electrically neutral substance formed by cation $\ce{Na+}$ (an acid)and  anion$\ce{ CH3COO−}$ (a base), completely dissociated or ionized in an aqueous solution as :
$$\ce{CH3COONa   -> Na^+_{(aq)}   + CH3COO^−{(aq)}}$$
a)  Since $\ce{Na+}$  is the conjugate acid of a strong base, it won't be strong enough to react with water; $\ce{Na+}$  actually spectator ion.

b)  Meanwhile, since $\ce{ CH3COO−}$ is the conjugate base of a weak acid, and  therefore strong enough to be able to hydrolyze and accept ions$\ce{ H+}$ from water, so water act as an acid leaving a hydroxide ion $\ce{OH−}$ as :
$$\ce{CH3COO^− +H2O <=>CH3COOH + OH^−}$$
Therefore, at the equivalence point, even though there is no acid or base present,$\ce{ [OH−]>[H3O+]}$, hence the$\mathrm {pH}$ will be >$7$.
Part B
Calculating the exact and approximate pH of $\pu{0.83 mol L-1} \ce{CH3COONa}$ in the bottle of vinegar?
The main reactions which take place are
$$
\begin{array}{ll}
\ce{CH3COONa->CH3COO-  +  Na+}&\quad\left (\mbox{Full Dissociation}\right)\\
\ce{CH3COO- + H2O <=> CH3COOH + OH-} &\quad\left(K_\mathrm{{b}\ce{(CH3COO^-)}}=\frac{K_\mathrm{w}}{K_\mathrm{{a}\ce{(CH3COOH)}}}=\frac{10^{-14}}{1.8\times10^{-5}}=5.56 \times10^{-10}\right) \\
\ce{ H2O <=> H+ + OH-} &\quad\left(K_\mathrm{w}= 1\times 10^{-14}\right) \\
\end{array}
$$
The salt  ($\ce{ CH3COONa}$)  is  an ionic salt,and most sodium salts fully dissociate in aqueous solution, and so this "equilibrium" doesn't exist and you can then consider all $\ce{ CH3COONa}$ to be converted to $\ce{CH3COO-}$   ions and $\ce{ Na+}$ ions into the water.
$\ce{ Na+}$  is weak acid it won't undergo hydrolysis.
$\ce{CH3COO-}$   is a relatively strong base it will undergo hydrolysis forming weak acetic acid, which dissociates partially.
The autoionization of water produces very few hydrogen and oxygen ions.so;
The resulting solution in the bottle contains the ions $\ce{  Na+, H3O+, OH−, CH3COO- }$ and undissociated acid  $\ce{CH3COOH}$. In order to specify the concentrations of the five  species present in an aqueous solution of $\ce{ CH3COONa}$ salt, we need five independent equations between them. These are :
1- Equilibria. We have two equilibrium equations:
The hydrolysis of the conjugate base $\ce{CH3COO-}$  :
$$ K_\mathrm{b} =\frac{[\ce{OH-}][\ce{CH3COOH}]}{[\ce{CH3COO-}]}\tag{1}$$
The dissociation equilibrium of water:
$$ K_\mathrm{w} =\ce{[OH-][H+]}\tag{2}$$
2- Mass balance. The mass balance equation that relates the concentrations of the various dissociation products of the substance to its initial concentration, which we designate here as $ [\ce{CH3COONa}]_0$.
For the solution of $\ce{ CH3COONa} $, these equations would be
a- The mass balance equation for the cation  ($\ce{Na+}$), but it is trivial:
$$ \ce{[CH3COONa]_0 =[Na+]} \tag{3}$$
b- The mass balance equation for the anion($\ce{CH3COO-}$)  ; the "total concentration of acetate". This concentration is the sum of all concentrations of species that are related to or derived from acetate. In this solution the relevant species are the acetate ion and acetic acid, so
$$\ce{ [CH3COONa]_0 =[CH3COO-]_T=[CH3COO-] +[CH3COOH]} \tag{4}$$
3-Charge balance. In an ionic solution, the sum of the positive and negative electric charges must be zero; in other words, all solutions are electrically neutral. This is known as the electroneutrality principle.
$$\ce{ [Na+] + [H+] =[CH3COO-] +[OH-]} \tag{5}$$
Substituting $\text{Eq }(3)$ into $\text{Eq }(5)$ yields an equation for$ [\ce{CH3COO−}]$:
$$\ce{ [CH3COO-] =[CH3COONa]_0 + [H+] -[OH-]} \tag{6}$$
Inserting $\text{Eq }(6)$  into $\text{Eq }(5)$ and solving for $ [\ce{CH3COOH}]$,
$$\ce{[CH3COOH]  = [OH-] –[H+]}\tag{7}$$
Substitute these last two equations ($6$ and$7$) into the equilibrium constant $\text{Eq}(1 )$:
$$ K_\mathrm{b} =\frac{[\ce{OH-}]([\ce{OH-}] –[\ce{H+}])} {[\ce{CH3COONa}]_0 – [\ce{OH-}] – [\ce{H+}]} \tag{8}$$
substitute these two expressions (${[\ce{OH-}]=\frac{K_\mathrm{w}}{[\ce{H+}]}}$ ) and ($K_\mathrm{b} =\frac{K_\mathrm{w}}{K_\mathrm{a}}$) into the $\text{Eq}(8 )$:
$$K_\mathrm{a}=\frac{\mathrm{[H^+]}{([\ce{CH3COONa}]_0 + \mathrm{[H^+]} - K_\mathrm{w}/\mathrm{[H^+])} }}{( K_\mathrm{w}/\mathrm{[H^+]} - \mathrm{[H^+])}} \tag{9}$$
which is cubic equation in $\ce[{H^+}]$  that is solved to get exact value of $\ce[{H^+}]$
; it becomes as
$$\
 [\ce{H+}]^3
+( K_\mathrm{a}+[\ce{CH3COONa}]_0) [\ce{H+}]^2
  -K_\mathrm{w}[\ce{H+}] -    K_\mathrm{w} K_\mathrm{a}
  \tag{10}
$$
"This is exact. Now to the approximations."
The $\text{Eq}( 9)$  can be simplified If we assume that $\ce{[OH−] >> [H+]}$ in basic solution, then $\text{Eq} (9)$ can be simplified to
$$K_\mathrm{a}=\frac{\mathrm[{H^+}]([\ce{CH3COONa}]_0  - K_\mathrm{w}/\mathrm{[H^+])} }{ K_\mathrm{w}/\mathrm{[H^+]} } \tag{11}$$
which is quadratic equation in$\ce[{H^+}]$  that is solved to get approximate value of $\ce[{H^+}]$ :
; it becomes as
$$\ce[{CH3COONa}]_0 [\ce{H+}]^2
    -K_\mathrm{w}[\ce{H+}] -    K_\mathrm{w} K_\mathrm{a}
  \tag{12}=0
$$
The $\text{Eq} (11)$  can be simplified further  If we assume that $\ce{[OH−] << [CH3COONa]_0}$ , then $\text{Eq} (11 )$ can be reduced to
$$K_\mathrm{a}\approx\frac{\mathrm[{H^+}] [\ce{CH3COONa}]_0    }{ K_\mathrm{w}/\mathrm{[H^+]} } \tag{13}$$
Or
$$[\ce{H+}]\approx\sqrt{\frac{K_\mathrm{a}K_\mathrm{w}}{[\ce{CH3COONa}]_0}}\tag{14}$$
- Comparing of the $ [\ce{H+}]$ values resulted from solution the three derived equations when: $$[\ce{CH3COONa}]_0=\frac{50\text{grams}}{(60\text{grams\mol)}\times{1\text{Liter}}}=\frac{33.33\text{grams}}{(40\text{grams\mol)}\times{1\text{Liter}}}={\color{red}{0.83325}\mathrm{M}}$$
and
$K_\mathrm{a}={\color{green}{{1.8\times10^{-5}}}}$
1- The completely exact answer solution using cubic equation ($10$)bywolframalpha
$[\ce{H+}] =4.6478\times{10^{-10}}\text{M}$ , $\mathrm{pH}=9.3327526$
2- The approximate answer solution using quadratic equation ($12$)wolframalpha
$[\ce{H+}] =4.64782\times{10^{-10}}\text{M}$ , $\mathrm{pH}=9.3327507$
3- The further approximate answer solution using the reduced equation ($14$) $[\ce{H+}] =4.657\times{10^{-10}}\text{M}$,$\mathrm{pH}=9.3318$

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text:
Hybridization and excitation of electrons, how is it possible?
When studying the structures of molecules by hybridization why we take into consideration of excitation of electron if the excited electron only stays in the upper shell by absorbing for only 10^-8 seconds{approximately }?

We don't.$\mathstrut$

Hybridization is a rather simple concept used for educational purposes mostly. The description of electronic excitations goes way beyond the scope of the model. Nobody (that I'm aware of) really quantitatively studies the structure of molecules using  hybridization.

We don't really consider the excitation of electrons. If it appears like this, then this is probably due to an incorrect understanding of what an orbital actually is: The attempt to describe each single electron by its own wave function which is independent of all the other electrons (which it is not).
An orbital is defined as single electron wave function $\varphi(\vec r_i)$ where $\vec r_i$ are the coordinates of electron $i$. We then try to approximate the many electron wave function as an (anti-symmetrized) product of orbitals
\begin{equation}
   \Psi(\vec r_1, \vec r_2, \dots \vec r_N) \approx \prod_{i=1}^N \varphi(\vec r_i)
\end{equation}
But this is strictly speaking not true, because the electrons are correlated with each other, due to the Coulomb repulsion $V\propto \frac{1}{|r_i-r_j|}$. The product ansatz does not work for the kind of differential equation we need to solve. The Hartree-Fock method tries to do so anyway, and gives therefore wrong (or at least only approximate) result.
This means the whole concept of orbitals is an approximation, which may give a qualitative understanding, but are failing when trying to get quantitative results.
One approach to correct this approximation is expanding the many electron wave function in some basis and solve the resulting eigenvalue problem. This basis is not unique, but one possible choice is to use the Hartree-Fock determinant and the excited configurations you can generate from it. But these excited configurations are more of an abstract mathematical thing and have limited physical meaning, because they are based on the above approximation. The real physics/chemistry is in the solutions of such methods. And those solution cannot be describe by the above ansatz for $\Psi$.

Indeed, one of the surprising things about the universe is how good the single electron solutions are even for atoms with many electrons. One might argue on 'good', but the fact we can identify noble gases all the way down the periodic table means they are fantastically good at some level.

@JonCuster Good point. This is nothing one would expect from a purely mathematical perspective. I think it is consequence of nature preferring locality.


text:
When are transition state's energy barrier "reasonable" at a certain temperature?
Every computational chemist who does a little bit of mechanistic studies has asked himself this question. When you obtain your first transition state and look at the activation barrier needed to access it, when would you tell that this required energy is reasonable and can be reached at a certain temperature ?
I have heard (never found a proof yet) that a biological process can be performed under standard conditions ($\pu{37 ^\circ C}$) with energy barrier as high as $\pu{15 kcal/mol}$.
Does anyone has a guideline for this?

Why not just plug the activation barrier and temperature into the Arrhenius equation and check if the resulting rate constant is reasonable?

Your question is really about what you consider to be a reasonable rate constant and so is somewhat subjective. The fastest a bimolecular reaction can be in solution is given by how fast the reactants can collide and then react on first collision. i.e reaction is diffusion controlled. In this case as reaction is at first meeting the activation energy must be small, $\le \pu{5  kJ mol-1}$. In water the diffusion controlled rate constant is in the region of a few times $\pu{10^10 dm3 mol-1 s-1}$. Most reactions are far slower than this.
The simplest, but very approximate, approach for other reactions is to use the Arrhenius equation to estimate the rate constant. The equation is
$$ k = A \exp(-E_a/RT),$$
where $E_a$ can be taken as your activation energy and you have then to estimate $A$. This can  be done for simple reactions using statistical mechanics and calculating the partition functions, but for anything complex you are left only with guessing values based on experimental data from similar reactions. However, because of the exponential dependence of the rate constant on activation energy even a small change in this can have a dramatic influence on the rate constant, far bigger than any similar change in $A$.
If you do have a potential energy surface then you can calculate trajectories along this surface by solving the (classical) equations of motion, repeatedly using energy sampled from a Boltzmann energy distribution,  and so work out a theoretical rate constant.

If you don't have any idea what $k$ value would be appropriate, arguably the best option is to identify a catalyst that is experimentally verified to do the reaction of interest and is assumed to involve the same mechanistic step you computed. Identify the lowest temperature the experiments are successfully run at. Determine the rate constant for the elementary step with known catalyst, either from previously published work or by running the calculation yourself (preferably at the same level of theory as your predicted catalyst).  With $k$ known for the experimentally tested catalyst, you can back out the required $T$ for your catalyst by plugging in your catalyst's $E_{\text{a}}$ and $A$ into the Arrhenius equation (or you can just assume the prefactors are identical since this is all an estimate anyway).
This process is a very rough approximation though for many reasons. It assumes the mechanism is identical to the reference material, it assumes the DFT-computed energetics are fairly accurate (even small inaccuracies can greatly change the required $T$), and it neglects the effect of concentrations on the rate, among other things. Depending on your needs, it may be okay for a ballpark estimate though. For instance, if you are studying a biological system and get that $T$ must be greater than 400 °C, you probably need to look for a new catalyst.
I also feel obligated to point out the following two limitations to this kind of estimate. The first is that it tells you little about the thermodynamic feasibility of the reaction. It would be worth making sure that the reaction is not predicted to be highly endothermic (although a low barrier necessary implies this won't be the case). In addition, if you are studying a mechanism where the reactant for the elementary step involves a bound adsorbate, it's always possible the energetically favorable binding mode is different between the reference catalyst and your own, which would further complicate matters.

Using this website, input the activation energy and temperature and it gives half life. A reasonable half-life is up to the user, but more than 1 day is going to be super slow. Usually assume the transmission coefficient is less than 1 (e.g. 0.5).
For example, an upper bound at room temperature is usually less than 25 kcal/mol $\approx$ 105 kJ/mol. With a 0.5 transmission coefficient the half-life is 4.2 days.

The other answers have given you approaches to estimating the overall rate constant as a function of temperature, which in turn requires you get an estimate of the prefactor, $A$. Knowing rate as a function of temperature is of course useful.
However (and I write my answer not just for you, but for others who might be searching for this topic), your question wasn't about estimating the overall rate; rather it was about estimating the temperature at which the activation barrier could be 'reasonably' overcome by thermal energy:  "When are transition state's energy barrier “reasonable” at a certain temperature?"
In response, let me offer a very simple general guideline commonly used by chemists and physicists.  The activation barrier is comparable in height to the amount of available thermal energy when:
$$RT \approx E_a$$
I.e., when:
$$T\approx E_a/R$$
At this temperature, the exponential term in the Arrhenius rate expression is simply $e^{(-E_a/RT)}=e^{-1} \approx 0.37 $.
A good number to memorize is that, at room temperature (298 K), $RT\approx 2.5\,\, \text{kJ/mol.}$  This tells you that the average thermal energy at room temperature is within the range needed to break water-water or water-protein hydrogen bonds. It also says you're in the regime where changes in temperature have a significant effect on the stabilty of these bonds.  [The stability of stronger (e.g., covalent) bonds isn't significantly affected by temperature changes around room temp, and weaker associations (those due to van der Walls forces) are thoroughly disrupted at much lower temperatures.]
[This isn't to say VDW forces are unimportant at room temp — they are!  Rather, it's saying these bonds are in a continuous state of dissociation and re-association at room temp.]


text:
Entropy as the driving force for osmosis
How is entropy responsible for osmosis and is movement of solvent possible from its higher concentration to lower concentration?
By concentration, I will be referring to the concentration of solvent molecules below.
I understand that the entropy of the universe tends to increase. But I do not understand how this principle applies to osmosis. Basically, in osmosis the solvent moves from its higher concentration  to lower concentration. Here, it appears that the entropy of the lower concentration side is increasing and the entropy of the higher concentration of solvent side is decreasing. How can we claim that the net change is increase in entropy?
Also, is the reverse process i.e. movement of solvent molecules from  lower concentration side to their higher concentration side ever possible (= thermodynamically favorable) without the presence of external driving forces?

(a) The tendency for the solute undergoing diffusion to occupy as large a volume as possible is similar to that of a gas filling the volume available to it and in each case the driving cause is the increase in entropy. In any  distribution of particles there are more ways of distributing them in a larger volume than in a smaller one.
Recall that as the solvent enters the concentrated solution the solution volume increases and this continues until the chemical potential on either side of the semi-permeable membrane is equalised. The pressure is also increased as solution is pushed up into a column, the difference in pressure is the osmotic pressure.
(b) From the statements above you can see that reverse osmosis cannot occur without applying external energy.
Notes: The change in entropy of a gas with volume is $\displaystyle \frac{\partial S}{\partial V}=\frac{R}{V}$. In a dilute solution the mole fraction $x_s \sim 1/V$ where $V$ is the  volume of the solvent (or solution)  with one mole of solute. As $G-G^0=-T S = RT\ln(x_s)$ then $\displaystyle \frac{\partial S}{\partial V}=\frac{R}{V}$. Thus the change in entropy is the same for the dilution of a gas as for the solute in a dilute solution.

" Recall that as the solvent enters the concentrated solution the solution volume increases and this continues until the chemical potential on either side of the semi-permeable membrane is equalised. "  What about the decrease of the volume on the other side?

Any change in the pure solvent's entropy is balanced by the opposite change of the solvent's entropy in the solution side of the semi-permeable membrane plus the increase in entropy due to the presence of the solute as the volume increases.  Nothing would happen if there was no solute.

Were the pure solvent side in a rigid container the chemical potential of the pure solvent would lower until it matched that in the solution.

Consider two decks of 52 playing cards. One deck has blue backs and one has yellow backs. A group plays poker with the yellow deck. If blue cards slowly get mixed into the yellow deck, the possibilities of types of five card hands increases. The faces in five card hands will have increased variations (a player could get two cards with the same face, like two aces of spades). Additionally, the variations of card back colors adds to the possible unique types of five card hands (four yellow-backed and one blue-backed). If the yellow-backed cards is the solute and the blue-backed cards is the solvent (and only the blue-backed cards can enter or exit the game) then the blue-backed cards are getting diluted from their higher "concentration" to a lower "concentration".
A pure solvent which can make five hydrogen bonds with neighboring solvent molecules will have less entropy than a solvent that has the opportunity to make any combination of five ion-dipole attractions and hydrogen bonds. The solvent is effectively getting diluted and entropy increases.


text:
Is the energy of the orbital an electron "resides" in, considered to be a factor of that electrons ionization energy?
The answers to a Quora question mainly refer to the factors as generally being, the electrostatic forces of attraction between the electron and the positively charged nucleus and all other events, by which this force is affected, i.e. atomic size (distance of the electron from the nucleus), the penetration effect, the shielding effect (by inner electrons) and the spin-pair repulsion phenomenon.
My textbooks also mention that electrons in the same shell have a "reasonably constant" distance from the nucleus, leading to the key factor of the observed progressive pattern down a period being due to the increase in atomic charge.
I get all that, but what I fail to understand is why none of these sources mention the energy of the electron's resident orbitals. It's my understanding that when an electron is "plucked" from the atom by means of the ionization energy value, or put into orbit by that value, they do so from their orbitals/subshells.
Now, each of these subshells have a different integer value of energy which increases up the count. I also concur that these orbitals have differing forces of attraction acting on them from the nucleus due to distance, inner electrons and the electrons proximity to the nucleus leading to different values of Ionization energy for the electrons "in" them, but I find it difficult to grasp that all electrons in the same shell have similar "distances". In one shell, for instance $n = 3$, there are 3 subshells each of which have a different three dimensional geometry and hence distance from the nucleus. The electrons in these subshells also have different amounts of energy which is what I'm asking about.
Don't these different energy values play a part in their ionization energy? i.e. can't the lower ionization energy of electrons in further subshells be explained by the higher potential energy of that subshell, as the higher the energy the lower the value of ionization energy?
Can't the energy of the orbital an electron resides in, be considered to be a factor of its ionization energy?

In short, you ask: "Does A affect B?" The answer is: A is B.

You mean to say the ORBITAL POTENTIAL ENERGY IS THE IONIZATION ENERGY?

Orbitals are an artificial construct. I think you are basically talking about this: https://en.wikipedia.org/wiki/Koopmans'_theorem . This of course is a very crude approximation, but the quasiparticle formalism in GW can give you ionization potentials that way.

OP, please don't shout. We are all reasonably well educated in reading, we will probably understand your point when you use ordinary spelling and capitalisation.

https://chemistry.stackexchange.com/questions/76111/energy-of-electrons-in-the-same-subshell

@Martin-マーチン sorry didnt mean to come off like that, was just trynna emphasise


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Do the dimensions of the cathode and anode matter in electrochemistry?
I'm curious whether the dimensions of the cathode and anode matter in an electrochemical process. Let's take copper electrorefining as an example. For instance, should the cathode and anode be of the same length and be in alignment with each other or does the process not care?
From my perspective, the cathode doesn't 'see' the anode, it only sees the $\text{Cu}^{2+}$ ions and therefore the anodes dimensions and placement relative to the cathode don't have an effect. Note, let's ignore the increase in resistance as the cathode and anode are placed further away and other similar effects. The essence of what I'm trying to understand is how their placement/alignment effects the occurrence of the reaction.

It depends $ $.

Dimensions - no; curvature - yes.

The dimensions of your electrodes should not effect the open circuit voltage, since this is determined solely by the identity and concentrations of the redox species.
Once the cell is running, the dimensions may affect how much the voltage drops off as you draw current, known as "overpotential". If your cell is limited by its ability to deliver ions to the surface of the electrode, or to transfer charge into the electrode, then the higher surface area of a large electrode is likely to result in a more efficient battery.

If you examine the Nernst Equation, there is no indication that the cathode or anode have any effect. What does effect the current is the resistivity of the solution. The further apart the electrode are, the more resistance the solution offers to the current. If the electrodes get too close, a short can occur.
Also, a high current between the electrodes can deplete the carriers due to a lack of diffusion or electrostatic repulsion.
At lower currents, you will see the same amount of current for any electrodes with the same surface area at the same distance.
If the electrodes are not aligned, then the electrodes look like tapered cones due to the increasing resistance from the closest point to the ends.


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Why don't buffer solutions return to equilibrium?
Suppose I want to make a buffer solution in any of the three ways possible with dihydrogen phosphate as my weak acid. In the first way, a certain ratio of the weak acid and its conjugate base are added at once to form the buffer. Why does this not return to the equilibrium described by its Ka value instead of forming a stable buffer? With a pka of 7.2, shouldn't the mixture shift to the weak acid to reestablish equilibrium.
The other two ways involve adding a strong acid or base to a weak base or weak acid respectively, but I still see the same issue. Say a strong base is added to a solution of the dihydrogen phosphate and the same moles added react and form an equal number of moles of the conjugate base. After this is complete, why doesn't equilibrium get restored. Is the weak acid/base equilibrium out-competed by the seperate strong base/weak acid to weak acid/weak base equilibrium?
Hope this made sense. Thanks in advance!

Can you explain why you think the buffered solutions are not at equilibrium?

@a-cyclohexane-molecule  Well say you're using a weak acid (not so weak that the conjugate base is especially strong). Only very few molecules of this acid should dissociate. In preparing a buffer with a weak acid and its conjugate weak base at the same time, isn't the solution suddenly not in equilibrium because there are too many molecules of base for every one of the weak acid? Why doesn't the base just convert itself to back to the acid to restore the equilibrium ratio that would exist if it were just the weak acid isolated in the solution?

In fact, equilibrium does restore, and that pretty fast.

At equilibrium, the ratio $\ce{[HA]}/\ce{[A-]}$ is not necessarily constant. The ratio $\ce{[HA]}/(\ce{[H+]}\ce{[A-]})$, however, is.

@a-cyclohexane-molecule Gotcha. So how is it that it can be relied on that the [H+]  will rise to the necessary amount for the desired pH after the weak acid and weak base are added? Shouldn't [H+] lower itself after it's added since the amounts of the WA and WB don't change. That way the ratio of the Ka value remains small. I feel like I must be missing something big here

The amounts of the WA and WB do change.

@a-cyclohexane-molecule Your answer is exactly what I was looking for. Thanks!

@Elmer, glad to help! Took a bit of questioning to figure out what you were looking for, sorry.

Let us consider the standard acetic acid/sodium acetate buffer, with, say, 0.1M of acetic acid and 0.1M of sodium acetate. These are not the equilibrium concentrations of acetic acid and sodium acetate; the buffer solution will equilibrate. In this case, the pKa of acetic acid being less than 7, we know that some acetic acid will dissociate.
Quantitatively, the equilibrium expression can be rearranged into the Henderson-Hasselbalch equation $$\text{pH} = \text{pKa} + \log\left(\frac{\ce{[A-]}}{\ce{[HA]}}\right) \approx \text{pKa} + \log\left(\frac{0.1+x}{0.1-x}\right) \approx \text{pKa} + \log\left(1+20x\right) \approx \text{pKa} + 20x \approx \text{pKa},$$ where $x$ represents the concentration of HA that has dissociated, assumed small. The third equality follows from a Taylor expansion of $0.1-x$, and the fourth from a Taylor expansion of $\log(1+20x)$. We can check whether this quantity is indeed small by solving the equilibrium expression for $x$ exactly; intuitively, it should be small because the difference in proton concentration at a pH of 5 and at a pH of 7 is small relative to the concentration of the acid.
Crucially, the extent of dissociation is small enough to be neglected as a first approximation.


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What is 'power law' kinetics?
I've come across a lot of instances where 'power law' description of the kinetics was mentioned, but I cannot think of any integrated rate law that follows something like:
$$A(t)=bt^n+c$$
I'm pretty sure the 'power law' refers to an integrated rate law instead of rate equations like:
$$r = k\prod_i [A_i]^i$$
So what is power law kinetics anyway? Does this 'power law' have a physical meaning or just a polynomial approximation that tells nothing about the mechanism?

Usually these type of complex kinetics arise when there is an inhomogeneous distribution of properties in a system. For example reaction over barriers which have a range of sizes, this can occur in proteins or in quantum dots but not in 'normal' kinetics. Fluorescence quenching in micelles can follow non-single exponential decays because there is usually a distribution of quenchers  solubilised in micelles. The fluorescence has the form $\ln(I)\sim -At+B\exp(-Ct)+D$ where , $t$ is time and $A,B,C,D$ constants.
In diffusion controlled quenching or energy transfer quenching in solution the rate constant will depend on time as $k=k_0+k_1/\sqrt{At}$. So there are many situations in which 'strange' kinetics can be observed.
However, vey often 'power law' kinetics is just a name given to an empirical method of fitting data to a function, say of the form $\exp(-kt^n)$ where $k, n$ are variables used to fit the data. In this form the decay function has no physical meaning.

Addendum. The first equation above was calculated as shown below.
As a model we suppose that the fluorophore and quencher are both in a micelle, as example of a restricted geometry, and that effectively no fluorophore exists outside. There is always an excess of quencher over fluorophore whose concentration is small so that no more than one will exist in any micelle. It is possible that zero, one or  several quenchers can be in any micelle.
We suppose that the rate constant for quenching is $k_i=k_0+ik_q$  where $k_0$ is the fluorescence decay rate constant value without quencher and $i$ the number of quenchers in any micelle, $i=0,\,1,\,2\cdots$ and $k_q$ the constant for quenching of the fuorophore.
The measured fluorescence signal is the sum of contributions from all micelles and is therefore
$$I(t)=I_0\sum_{i=0}^\infty p_i\mathrm e^{-k_it}$$
where $p_i$ is the probability of there being $i$ quenchers in a micelle. It seems reasonable to suppose that this has a Poisson distribution, and in that case $\displaystyle p_i= \frac{m^i\mathrm e^{-m}}{i!} $ where $m$ is the mean number of quenchers in a  micelle. Substituting and simplifying the intensity then becomes
$$ I(t)=I_0\mathrm e^{-k_0t-m}\sum_{i=0}\frac{(m\mathrm e^{-k_qt})^i}{i!}$$
The summation term has the form $\sum x^i/i!$ which is the expansion of the exponential thus
$$ I(t)=I_0\mathrm e^{-k_0t-m}e^{m\mathrm e^{-k_qt}} = I_0\mathrm e^{-k_0t +m(\mathrm e^{-k_qt} -1)      }    $$
which when logs are taken is the expression above.

very well explained ! could you give me a link that would lead to the derivation of your first equation ?

I have added the derivation as I could not find the reference.


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Atoms in Solar Cells
An electric field in solar cells is created through photons knocking electrons off of atoms. What happens to the positively charged atoms once the electrons have been knocked off? Furthermore, are the atoms the photons collide with in order to knock the electron off of the atom any specific kind of atom, and are there any chemical equations modelling this?

What happens to the positively charged atoms once the electrons have been knocked off?

The atom grabs the next closest electron, usually from its neighbor; which grabs an electron from its neighbor, etc. Eventually the electron comes from the terminal that completes the circuit (usually) or from the leakage current (when no current is flowing).

Furthermore, are the atoms the photons collide with in order to knock the electron off of the atom any specific kind of atom, and are there any chemical equations modelling this?

Yes, it depends on the material making up the photo cell. Look up photoelectric effect. But instead of the electrons being kicked free from the metal, the silicon layer captures the electron and transfers it to the anode to make  the electrical circuit.


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Can I get feedback on assignment? (chromatography)
HW question asks me why samples can be separated into their components by chromatography. My answer is that each component in the mixture will have different affinity for the solvent (mobile) phase as it moves up the paper, because each component can be more or less polar/nonpolar. The technique exploits the fact that polar components will move further distance with a polar solvent, while less polar components move less distance. Is this basically correct? Thanks

You missed an important interaction.

also an interaction with the water on the paper (polarity again?) is that what you mean?

Not just the water, but the components in the mixture with the paper. If the components have zero affinity for the paper, they would move with the solvent front (even if they have very low affinity for the solvent).

You really missed something about chromatography and you assume that the stationery phase is neutral.
In fact, chromatography is based on the balance between the interactions between a component and BOTH the mobile phase (aka the eluent) and the stationary phase. Depending on your choice of mobile and stationery phases, results can be entirely different.


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Do bond lengths change upon protonation of a molecule such as an H2O molecule?
Wouldn't the hydrogen atom in a hydronium cation feel a greater proton repulsion force (nucleus-nucleus repulsions) from the oxygen atom and orient away from the oxygen atom to maximize the attractive force it has for the shared pair of electrons by minimizing proton repulsions?

The bond length is definitely different, but what is a “proton repulsion force”?

For a water molecule, the shared pair of electrons between the hydrogen and oxygen atoms tend to spend more time towards the oxygen atom resulting in a slight negative charge for oxygen atom and a slight positive charge for the hydrogen atom. Now, this means that electron that is bound to hydrogen is now farther away from the hydrogen atom than it would be if it wasn't shared with oxygen. From this perspective, I thought that there must be some other force preventing the hydrogen atom from moving closer to the shared pair of electrons and I thought that possible force was proton repulsions.


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Signs of electrode potentials
Why is the oxidation potential given a positive sign and the reduction potential a negative one?
Also, in my book a bunch of formulas are given without any explanation:
$E^{o}_{cell}= $ Oxidation potential of anode + Reduction potential of cathode
$E^o_{cell}=$ Standard reduction potential of cathode - Standard reduction potential of anode
$E^o_{cell} =$ Oxidation potential of anode - Oxidation potential of cathode
I am unable to get any intuition for this and the physics behind these signs. Could someone please provide an intuitive explanation of what's going on?

Why, that's simple: negative sign means that the reaction is running in reverse direction.

We use a convention for the sign of potentials. The standard reduction potentials are referenced wrt  hydrogen reaction $\ce{2H+ + 2e^- <=>H2_{(gas)}}$ which is given the value $0.0$ V. Oxidising agents have positive values, for example $\ce{Ag^+ + e^- <=> Ag_{solid}}$ $E^\mathrm{o}=+0.799$ V, or $\ce{Cl2_{(gas)} + 2e^- <=> 2Cl^-}$, $E^\mathrm{o}=+ 1.36$ V, and reducing agents negative values, e.g. $\ce{Cr^{2+} + 2e^- <=>  Cr_{solid}}$, $E^\mathrm{o}=-0.91$ V.
The oxidation potentials of these reactions are just of the same magnitude but of opposite sign. Thus for the same species  the oxidation potential + reduction potential add to zero. Normally we only deal with reduction potential so if you have an oxidation potential just reverse the sign and it becomes a reduction potential.
Note that the same species can be an oxidising agent or a reducing agent depending what it is being reacted with.
A reducing agent is an electron donor, and oxidising agent an electron acceptor. If a species is oxidised it looses electrons but gains electrons if reduced. Thus a reducing agent is itself oxidised;  confusing, yes  but its just how it is.
The cell potential is defined in your second definition as this uses the standard reduction potentials. The others are just variants by messing with reduction vs oxidation and changing signs as appropriate.
( Note that in some papers and books you may come across the redox potential values quoted wrt the saturated calomel electrode, not hydrogen, and this has a value $+0.246 $ V so its always worth a quick check. )


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Why does aluminum in zeolites form tetrahedras?
It forms a bond with four oxygen atoms even though it has only there electrons in his outter shell?

One should understand that the rules to complete outer shells apply to individual molecules (although they don't always get followed). Solids are a different matter - look at any number of binary phase diagrams and you can find multiple phases with different ratios of one element to the other. The extended nature of bonding in solids is just different.


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Is (Zn(OH)4)2- acidic or alkaline?
My chemistry teacher told me that $\ce{Al(OH)4^-}$, $\ce{Pb(OH)4^2-}$, $\ce{Zn(OH)4^2-}$ are called complex ions. So will they break down in some conditions and give out $\ce{OH-}$ to show alkaline property or just remain neutral?

Yes they are alkaline, and pretty strongly so.

These ions exist only in presence of excess of hydroxide ions, otherwise they easily liberate few of their own and precipitate as a hydroxide.

On the other hand, $\ce{Al(OH)3}$, $\ce{Pb(OH)2}$, and $\ce{Zn(OH)2}$ are amphoteric, because they can react both as acids and bases.

@permeakra So these ions can be broken down back to ppt and release hydroxide ions to prove their alkalinity? Thank you.

Let's first imagine that one of these complexes is placed in neutral (pH 7) water. If the resulting solution possesses a pH of > 7, we can conclude that the anions show an "alkaline character" as you describe. If the resulting pH is < 7, we'll conclude that the material is contributing an "acidic character."
The way one might go about answering this question without dunking one of these salts in water is by starting with the a more well know complex such as the hexa-aqua complex of each metal and observing the behavior.
$\ce{AlCl3 + 6H2O -> Al^3+.6H2O + 3Cl^-}$ Chloride becomes a spectator anion
This $\ce{Al}$ water complex ($\ce{Al^3+.6H2O}$) possesses acidic protons, just like any polyprotic acid (eg. phosphoric acid.) Thus, each proton also has an acid dissociation constant or pKa. The first pKa for $\ce{Al^3+.6H2O}$ is approximately 5 and would thus acidify the solution. The pKa's of each of the subsequent protons donated to solution is higher than the last and eventually, no more protons would be donated to the solution.
With this basic understanding out of the way, let's look at the approximate pKa's of each of the neutral hydroxide salts of each of these metals.
For $\ce{Al(OH)3}$ the pKa is > 7. This means that by adding another hydroxide to the complex to make it a tetrakis hydroxide complex and would result in an even higher pKa and a more alkaline solution.
$\ce{Pb(OH)2}$ and $\ce{Zn(OH)2}$ are both weakly basic (pKa > 7) and possessing low solubilites. Adding more hydroxide ligands will make each of them more alkaline and should also make each of them more soluble. The zincate anion $\ce{Zn(OH)4^2-}$ is often formed by dissolving Zinc metal in a strongly alkaline solution.$\ce{^1}$
I hope this helps a little.
ref 1. http://depts.washington.edu/chem/facilserv/lecturedemo/BrassPenny-UWDept.ofChemistry.html


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Substituent effects on acidity
It is said that 4-methylbenzoic acid will be a weaker acid than benzoic acid. 4-nitrobenzoic acid will be a stronger acid than benzoic acid.
I don't understand why the methyl group on 4-methylbenzoic acid would affect the acidity. I know that it has an inductive effect (hyperconjugation) but I thought induction only had an effect over a one to two carbon distance. The carboxylic acid is separated from the methyl group by four carbons.

Never undermine the power of conjugation, they can go further than you might think.
Chlorophyll has 24 p-orbitals in conjugation and Hemoglobin has 26!!

(1) It is true that the inductive effect is weak across several bonds, BUT conjugated bonds (i.e. alternating simple and multiple bonds) propagate this effect much farther, even more for mesomeric effects. You can think of "conjugated bonds" like a tunnel for electrons.
(2) In your case, the electronic effects will stabilize the salt more/less than the acid. As they are in equilibrium, the salt is more/less favored than its acid counterpart.
I think that the most important thing here is that the more electric charges are "diluted" (i.e distributed among more atoms), the more stable a molecule is. This is an effect of entropy, which is too often poorly described as disorder.


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Hydrogen bonds - why not in HCl?
In a textbook1, I found the following clear definition of hydrogen bonds:

The strongest secondary bonding type, the hydrogen bond, is a special case of polar molecule bonding. It occurs between molecules in which hydrogen is covalently bonded to fluorine (as in $\ce{HF}$), oxygen (as in $\ce{H_2O}$), and nitrogen (as in $\ce{NH_3}$). For each $\ce{H-F}$, $\ce{H-O}$, or $\ce{H-N}$ bond, the single hydrogen electron is shared with the other atom. Thus, the hydrogen end of the bond is essentially a positively charged bare proton that is unscreened by any electrons.

This is not an "etc." definition. It is made clear that we only call it a hydrogen bond when we have either of the three bonds: $\ce{H-F}$, $\ce{H-O}$ or $\ce{H-N}$.
My question is why this definition excludes compounds such as $\ce{HCl}$? $\ce{HCl}$ and $\ce{HF}$ are very similar with one covalent bond that pulls the hydrogen electron towards their "shared" region, exposing the hydrogen atom as a positive end of this permanent dipole. Both can create secondary dipole bonds to adjacent molecules of the same type.
Why is $\ce{HF}$ called a hydrogen bond while $\ce{HCl}$ is not?
1 Materials Science and Engineering by W. D. Callister & D. G. Rethwisch, Wiley, 8th edition

HCl and HF are less similar than you think.

@IvanNeretin Please enlighten me.

As you surely know, HF is more polar. Now consider this: it is really, really much more polar.

Very very related.
Short answer:
Chlorine is just too large for a hydrogen bond to form.
One thing you need to realize is that Hydrogen Bonding has a fair bit of covalent character as well which makes it considerably more powerful than van Der Waals forces. The Chlorine molecule has its valence electrons in 3p orbitals. Hydrogen has valence electrons in 1s orbital. For covalent character, the orbitals need to overlap. The overlap between the 1s and 3p is very weak and this is the reason why chlorine lacks that covalent character required for hydrogen bonding.

Hydrogen bonds to chlorine atoms are known. In the gas phase HCl dimers are hydrogen bonded but the bond has only about half the dissociation energy of $\ce{(HF)2}$ which has a larger electrostatic attraction energy. The adduct $\ce{HCl\cdots OH2}$ is also known and has a similar hydrogen bond energy to that of  $\ce{HOH\cdots OH2}$.
Using neutron diffraction hydrogen bonds have been studied in many crystals. From a study of many types of amino acid crystals (and derivatives thereof) the normal $\ce{HOH\cdots O}$ hydrogen bond is found to have  a range of lengths, from about 0.175 to 0.2 nm with most lengths at around 0.185 nm. The $\ce{OH\cdots Cl}$ hydrogen bond peaks about 0.2 nm with an range of $\approx \pm$ 0.05 nm. The $\ce{NH^+\cdots Cl}$ bond is longer peaking at 0.21 nm.
Calculations (Singh & Kollman, J. Chem. Phys. v83, p4033, 1985) show that the linear hydrogen bond, X-H-X, has an energy profile that is not that dissimilar to any chemical bond when energy is plotted against X to X separation. There are electrostatic attraction terms and exchange repulsion terms. These form a potential well with repulsion dominating (positive energy) at close separation of the two X atoms, a minimum energy (negative) at about 0.18 nm for water dimers, and a gradual increase in energy as the distance X - X increases reaching zero at large separation. However, there are also polarisation terms and those due to dispersion and charge transfer to take into account. These terms are all attractive and become important only at small separation of the X atoms. However, they are not insignificant and  contribute  a few tens of percentage to the total energy. The result of these terms is to move the minimum of the potential energy to smaller values that that due to exchange and electrostatic terms alone. Thus one can imagine the H atom residing in the potential energy between the two X atoms and being close to the minimum energy. Having described this ideal case,  IR spectroscopy does  indicate that in some cases the potential may have two minima rather than just one as described so the nature of these bonds is still a subject of debate.


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States of matter.Liquid
In a book a read that a simple defination of liquid i.e."it is a material that assumes the shape of a container without filling it completely" but i cannot able to understand  "without filling it completely".
please eleborate this ?

Presumably this is meant to distinguish liquids from gases, which both assumes the shape of the container and expands to fill the entire shape.

Along the same lines, the definition of gas would be a 'material which assumes the shape of container and fills it completely'.
These definitions are very much intuitive as a gas always fills the container in which it is put.
The liquid on the other hand, does not fill the container.

@ShahMHasan As in my mind filling a container not depend the quantity of liquid that we put into container?

If a glass of water has 100 ml capacity and you pour in 50 ml, only half of the glass will be filled. The liquid does not expand to occupy the 100 ml.

@ShahMHasan  If the cpacity of glass 100ml .we pour also 100ml of liquid in it.then container is fully filled.then this situation not satisfy defination of liquid.

@user77330 The point is not whether the liquid actually fills the container if you add the exact amount: it is whether the liquid fills the container if you don't. Liquids have a defined volume: if you add less than the volume of the container, the container will not be full with liquid. The point being the volume of liquid doesn't depend on the volume of the container.

Both liquid and gas will assume the shape of the container.
It's also intrinsic for the gas to  fill up the volume of the container but such a feature is not intrinsic to the liquid. Therefore the book is implying that if the second feature does not imply, the phase of the matter in question is liquid.
please note:
Though we may fill up a container to it's volume by a liquid, this is us filling the container not the liquid intrinsically. Any volume of gas can fill up a whole container Intrinsically (Imagine the space).
Hope this helps abit.


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Connection between Gibbs Energy of Formation and Gibbs Energy of Mixing
Recently I found the following formula, which I cannot verify to be correct. Since I found this formula in several publications I assume that it is correct but I don't know why.
First there is the Gibbs Energy of Formation $$\Delta_fG$$ and second the Gibbs Energy of Mixing $$\Delta_mG$$
The formula I found says, that $$\Delta_mG = \frac{\Delta_fG^{\circ}}{N}$$ where N is the number of particles. I am not sure if this is really N or n (amount of substance). Have anyone seen this formula and/or can explain why it is valid/how it can be derived?
Thanks!
Edit:
Indeed, my problem is also in the frame of binary alloys. Consider the reaction $$2A + B = A_2B$$ which is performed in an electrochemical cell (emf measurement). Then $$\Delta_fG^\circ = -zFE$$ because the reactants are elements and A is the active species. So far this should be correct. In this context the formula in question reads as follows $$\Delta_mG = \frac{\Delta_fG^\circ}{3}$$
The full definition of the Gibbs Energy of Mixing is $$\Delta_mG = x_A RT\ln{a_A}+x_B RT\ln{a_B}$$
Unfortunately, I cannot give more detail, because I don't understand the formula. :)

Perhaps you could edit in more context. Where I play most with thermodynamics (binary alloy phase diagrams), that relationship would definitely not hold.

Nobody? What information should I add to make answering easier?

Can you give some links to the places where you have seen this formula? You mention several publications.  It might help give more background to show where it came from.

here are two documents (they are not open access): DoIs: 10.1149/1.2127590 and 10.1016/j.jct.2004.07.033

Thanks for the links in the comments above. The second one J Chem Thermo, 37, 97 (2005) seemed to be clearer. Eqn (12) reads
$$\Delta_M G=\Delta_f G^\circ /4
$$
but is preceded by the sentence

According to the definition of the molar free energy of mixing,
  $\Delta_M G$ for Ni$_3$Al would refer to the formula
  Ni$_{3/4}$Al$_{1/4}$, and thus the equation ... was obtained.

In the first link J Electrochem Soc, 128, 1181 (1981) it is harder to spot your equation. However, I think I see it, in comparing eqns [7] and [10] which relate $\Delta G_m$ and $\Delta G_f^\circ$ to a Coulombic titration expression involving an integral over the compositional variable $y$. This is for the alloy Li$_y$Sn, and the expressions look identical except for a factor $1/(1+y)$. So I guess that's it. However there's no helpful explanation.
Bearing in mind the sentence I quoted from the (2005) paper, I suspect that this factor is simply a question of how these quantities are defined, in relation to the formula of the alloy under consideration (i.e. per atom or per formula unit), and not intimately connected with the expression for the free energy of mixing in terms of mole fractions and activities (except in that it is written per atom, or more accurately, per mole of atoms).
[Edit following OP comment]
I think that this is less complicated than it may have looked. $\Delta G$ is just the difference in Gibbs energy between two well defined states. I see, for instance, on p43 of Statistical Thermodynamics of Alloys by NA Gokcen (1986)
the definition

The molar Gibbs energy of mixing refers to the process prescribed by
$x_1$ (pure component 1) $+$ $x_2$ (pure component 2) $\rightarrow$ $1$
  gram atom of alloy

where, of course, "gram atom" is old terminology; but the meaning is clear. If the original metals are in their standard states, this is essentially the same as the definition of the standard Gibbs free energy of formation of the alloy, except that the latter is usually expressed in terms of one mole of the compound. I think this difference is what is intended by the numerical factor, irrespective of any particular ways of writing the Gibbs energy of mixing in terms of other variables (for example, one can sometimes use regular solution theory to estimate $\Delta S$ and $\Delta H$ of mixing, but I don't think that's relevant here).

You are looking at the right equations. But how can this be a matter of definition, when therms like $\Delta_fG$ and $\Delta_{mix}G$ have their own fixed definitions. There is even more: $\Delta_fG$ is a differential value $(\partial G/\partial \xi)$ and $\Delta_{mix}G$ an integral value. This is really confusing...

I think the two quantities are defined to be essentially the same (apart from this numerical factor), and I've edited my answer accordingly.

Sorry for the late response, I start thinking that you might be right. Therefore, I will accept your answer until I found a better explanation (if there is one :) ). Thank you!


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Can there be any substance whose number of electrons and protons are different at neutral state
I want to ask that if there any substance that has the different number of electrons and protons at its neutral state

Something neutral must be uncharged.

It's certainly possible, you just need other charged particles that aren't electrons or protons.
For example, a positron is positively charged, so something that's made out of an electron plus a positron would work. See e.g. https://en.wikipedia.org/wiki/Onium, or this answer on SE, for a few examples.

It is worth noting that such substances will not last long.

Well your question itself is sort of wrong because you assume that individual atoms would have a definite number of electrons... which is usually not the case. In a molecule, electrons can be spread around several nuclei depending on several factors like the atoms' electronegativity or the molecule structure.
In fact, it is not really a matter of how many electrons are around a nucleus but more "how likely" they will be around that nucleus compared to other surrounding nuclei.

I want to know if there a way in which the different number of electrons can neutralize the charge different number of protons. Suppose there is a neutral atom with 4 electrons and 3 protons.

@Nick Charge is a fundamental quantity. In normal matter all the charges come from either protons or electrons. The net charge is what you get when you add up all the positive charges from protons and all the negative charges from electrons. So there is no way that 3 electrons can balance the charges from 4 protons.

@Nick — No, there cannot be a neutral atom with 4 electrons and 3 protons


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Melting of ionic compound hydrates
My professor asked us the reason of the melting point of potash alum being so low, despite it being an ionic salt. He later went on to answer it by saying that melting point is a misnomer for the phenomenon that is happening here.

What's happening here is that the alum is losing its water of hydration and is then dissolving in that water itself. So it is actually a dissolution point rather than melting point.

I couldn't find any sources for what he said. Though what he said makes logical sense, I would like to read a bit more about it. Can you please provide some sources confirming what my professor said?

A misnomer is a word that causes confusion by being used incorrectly. "Melting" in application to this process is something different. It is a word that causes confusion by being used correctly. What we have here is melting all right, but not of the sort we are used to seeing.

Your professor's premise of dissolution is only true if the process is incongruent melting which by the way is still melting. Whereby the process proceeds as:
$$\ce{KAl(SO4)2.12H2O_{(s)} ->[\Delta][T1] KAl(SO4)2~_{(s)} + 12H2O_{(l)}\tag{$\beta$ to $\alpha$ + liquid}}$$
Then
$$\ce{KAl(SO4)2~_{(s)} ->[\Delta][T_2 > T_1] KAl(SO4)2~_{(aq.)}\tag{$\alpha$ to liquid}}$$
This concept is outlined in the generic binary phase diagram below where $\alpha$ is anhydrous potash alum, $\beta$ is the hydrated form, and liquid is potash alum solution. Above a certain temperature $\beta$ will decompose into $\alpha$ and liquid. As the temperature rises more potash alum will dissolve until eventually it will all be liquid again.

Figure 1. Binary Phase Diagram with Peritectic melting
However if it is a direct conversion from liquid to solid then it is congruent melting and your professor is wrong. This can occur either as melting an eutectic (shown in Figure 2) or congruent melting of a compound (shown in Figure 2) and proceeds directly as follows.
$$\ce{KAl(SO4)2.12H2O_{(s)} ->[\Delta][92.5^\circ C] KAl(SO4)2~_{(aq.)} + 12H2O_{(l)}\tag{Interionic Melting}}$$
$$\ce{Sn_{(s)} + Pb_{(s)} ->[\Delta][183^\circ C] Sn_{(l)} + Pb_{(l)}\tag{Eutectic}}$$
$$\ce{Ti5Si3_{(s)} ->[\Delta][2130^\circ C] 5Ti_{(l)} + 3Si_{(l)}\tag{Intermetallic Melting}}$$

Figure 2. Binary Pahse Diagram with Eutectic.

Figure 3. $\ce{Si\! -Ti}$ Binary Bhase Diagram with Compound at $\ce{Ti5Si3}$ That Undergoes Congruent Melting.
Usually differential scanning calorimetry is used to determine which melting regime you are in indicated by a range of temperature causing a spike in heating rate corresponding to incongruent melting or a single temperature where heating spikes which would indicate congruent melting.
In this case, however, we can confidently assume that $\ce{KAl(SO4)2.12H2O}$ is under going incongruent melting since it melts above the temperature of water and below the temperature of the anhydrate and there are no other hydrates that it forms, the process is the peritectic melting thus the process your professor describes is correct, but it is still considered a melting process.

Excellent answer +1. But I have a query. Shouldn't the potash alum decompose into more discrete species on further heating or will it remain an aqueous solution? See this question for context.

Yes, it will form the separate ions in solution, My point was the difference between a two step and one step melting

Also, you want to add the concept of "burnt alum" to your answer. Wikipedia states: When heated to nearly a red heat, it gives a porous, friable mass, which is known as "burnt alum".

I could, but I felt like breaking out phase diagrams was far enough into the weeds already.


text:
Solid structures related question
In which of the following compounds are all tetrahedral voids vacant?

$\ce{NaCl(s)}$
$\ce{CaF2(s)}$
$\ce{ZnS(s)}$
$\ce{Na2O(s)}$

Kindly explain what this question asks from me and provide a detailed answer. Also, please explain tetrahedral voids.

So in the NaCl lattice you have either $\ce{Na^+}$ or $\ce{Cl^-}$ forming the ccp while the other one occupies all of the octahedral voids. This means there are no tetrahedral voids occupied. You can see this if you look for the $\ce{NaCl}$ structure on the net where they drew in the polyhedra. Both of them are octahedra only.
$\ce{CaF2}$ is a structure where the $\ce{Ca^2+}$ cations form the ccp and the $\ce{2 F^- }$ anions occupy all of the tetrahedral voids. You can see this as well when you look at the structures, there should be a tetrahedron in there.
Now $\ce{CaF2}$ marks the beginning of a whole series of cubic and hexagonal structures by something we call 'thinning' (at least in German). What happens here is that we start to occupy fewer and fewer tetrahedral voids. The first series, where half of the tetrahedral voids are filled only are:
$\ce{ZnS}$, $\ce{PtS}$, $\ce{PbO}$ and the hexagonal $\ce{ZnS}$ and the hexagonal $\ce{BeO}$.
Therefore $\ce{ZnS}$ being cubic and hexagonal (wurtzite) has half of the tetrahedral voids filled only.
Finally, there is the $\ce{Na2O}$. $\ce{Na2O}$ or $\ce{Li2O}$ is what we call an anti-type to $\ce{CaF2}$. It features the same setting but the atoms are switched. So this means still all tetrahedral voids filled.
Therefore only the very first one, $\ce{NaCl}$ has no tetrahedral voids filled up. In your case vacant means that there are no tetrahedral voids filled.
Other structures would include all of the once that fill certain amounts of octahedral voids instead of tetrahedral ones, so $\ce{NaCl}$, $\ce{NiAs}$, $\ce{Al2O3}$, $\ce{CdCl2}$, $\ce{CdI2}$, ... (note those are all structure types and no random compounds as it might look like at first glance).
Additional information:
As I was asked to edit my post to also include some picture I did the following here. The crystal structures were taken from the ICSD. The ionic radii were chosen from the Shannon table.
There is a problem with structures on the net. They often forget a proper scaling of atoms. You cannot simply take the ionic radii you find on the net. The ionic radius is dependend on the coordination geometry.
So it depends whether your anion or cation sits in an octahedron or a tetrahedron. I scaled all of them to a common size by deviding them by 2.5.
In the first structure you can see the $\ce{NaCl}$ with both polyhedra added. As you can see they are all octahedrally coordinated. So there is no tetrahedral void filled here.

Next is the $\ce{CaF2}$ and the $\ce{Na2O}$ structure. Both $\ce{CaF2}$ and $\ce{Na2O}$ crystallize in the same space group Fm-3m. For $\ce{CaF2}$ the $\ce{Ca^2+}$ occupy the 4a Wyckoff-position and the $\ce{F^-}$ occupy the 8c Wyckoff-position. If you look at $\ce{Na2O}$ it has the same space group and $\ce{O^2-}$ has the 4a and $\ce{Na^+}$ has the 8c so metal and non metal or cation and anion swapped places. This is what we call the anti-type.
Therefore both have the same amount of voids be it octahedral or tetrahedral filled. As you can see there is a cubic coordination around the 4a position and a tetrahedral around the 8c position.

Finally to your $\ce{ZnS}$. Both the cubic and the hexagonal phase can be derived from the $\ce{CaF2}$ lattice which results in tetrahedra around both ions.

EDIT 2:
Just in case it might not be visible from the structures I highlighted the tetrahedral voids that match in both structures. From this you should be able to see that half of the tetrahedral voids are now vacant. The structure which forms is a diamond-type structure due to what we call Grimm-Sommerfeld rule. If you look where $\ce{Zn}$ and where $\ce{S}$ are in the PSE you can see that in the middle of them we find Carbon or Silicon. So it adopts this structure. For $\ce{PtS}$, which is also a half filled $\ce{CaF2}$-type the atoms arrange so the $\ce{d^8}$ on $\ce{Pt(II)}$ get's his desired square plane coordination. For $\ce{PbO}$  we have quite large gaps because $\ce{PbO}$ has a lone pair and those lone pairs can then point into the empty space.

Could you please upload an image to enhance your answer.


text:
Artificial nose sensors
I'm am doing research in creating an artificial nose. I need to know how this process works by someone with domain expertise; essentially, what technologies are available currently to detect a broader range of particles in a closed system. My understanding so far goes like this on a high-level overview: particle comes into contact with the olfactory cells and particles from the environment have a custom signature which we associate as smell of a banana or orange, now how to do this artificially or how is this done artificially?

Welcome to Chemistry.SE!  Please take a minute to look over the help center and tour page to better understand our guidelines and question policies.

"What technologies are available currently to detect a broader range of particles in a closed system" a LOT of technologies allow this: spectrophotometry, NMR, GC-MS, etc. These technologies give a lot of information about the molecular structure (but not about the molecular  function).
The hard part in your idea is trying to replicate the sensivity of a nose, which is, at the moment, impossible as far as I know.
Trying to deduce via a robotic nose the smell of a novel compound is a task that could, theorically, be made if that "nose" had exactly the same receptors and "circuits" of our nose, since receptors interact in a unique way with molecules, and that is the source of all those unique smells that we sense.
A more approximate approach, doable with current technologies, could be a database-based approach: a lot of "smelly" molecules could be categorized and described in terms of smell.
The analyzer (one of the machines mentioned earlier, for instance) could identify the molecule(s) involved, and the database could provide information about the smell in that environment.
The problem? A sistem like this doesn't work like a nose at all! Any unknown molecule couldn't be "smelled".
Maybe in the future, with the improvement of technology, a molecular docking study will be performed in seconds, and that, when all the proteins associated with smell will be known, could allow us to know the exact way a molecule affects our receptors, our brain and our perception, hence creating a working "digital nose".
And finally, let me add a note: you said "particles from the environment have a custom signature which we associate as smell of a banana or orange". One might argue wether that "signature" is in the molecule or in the nose itself. While many compounds are sensed in a similar way within a population, there are genetic mutations between species and even individuals. Puppies are known to be appealed by cat feces, for instance: is that just a behavioural thing, or their perception of the smell of cat's feces is different from ours?
Some people like the smell of truffles, others hate it: it's (at least partially) because of the individual ability of detecting molecules.
The point is not that of detecting a "signature", but in detecting molecules and interpreting them via our specific "signature".
If olfactory proteins didn't exist, smell wouldn't exist: it's not a physical property of the molecule, but the result of an interaction with who "smells".
Don't be fooled: that's not a philosophical sentence: light wouldn't cease to exist if we had no visual receptors. But for sure colours wouldn't exist, since also colours exist only because we are evolved to detect light as a mixture of three different chemical responses (each associated with one of the RGB colours).

That was a response I was really hoping to hear, thank you. I'm going to go through your points and attempt to address and add some thoughts to your statements.. I agree with you about the database approach it would be more efficent then having all these stored on the device but something like you have the system and it's sensors that can detect the molecules and send that via RF to the other computer to be processed. Essentially these molecular structures would have a dictionary associated with the structure in theory.

--- combine the use of Machine Learning we could detect these molecular structures and associate them with their "smell"  the hard part or rather time consuming part would be the collections of all these unique smells; this part is my background but i'm more or less void of knowledge in biological systems and biochem. I think smell(s) are subjective for sure and there are mutations among different species that have no doubt had a enviromental motivator for survival or perhaps no explanation like the puppies but leaving of the subjectivness of the smells and subcouncious biases or experiences

But you're confident these sensors could react with a more broad range or particles ? not just a specific target? again, I'm new to some of these technolgies you meantioned i'm just trying to get a better understanding if the current technology can be manipulated in a way that could detect broad range or particles and have a unique reaction producing dynamic results given the input of the particles.

I missed a very important point as well. "replicating the sensitivity of the nose" may be a naive statement but couldn't this be solved by a greater density of sensors in the system? the problem that comes to my mind is the physical space of the device but this isn't an issue for a prototype given it's not > 100 lbs or X of course the technology in the future or a better understanding of the system would directly contribute to more efficient sensors kind of like the transistor perhaps

Basically any liquid or gas can be analyzed, so there are no limits to the analyzed compounds.
What I meant by "replicating the sensitivity of the nose" was intended in the distinction of smells.
The machines I mentioned (expecially NMR and MS) are possibly more sensitive than our nose. They distinguish perfectly between extra-virgin oil and regular oil, or ethyl benzoate and a real banana. And in some applications, they find compounds at a much lower concentration than the one we can smell.

The problem is in giving the machine an "olfactory" sensivity, since the machines have no olfactory receptors.

Since the receptors are proteins, you might want to google something about:
-protein crystallography (and take a look at PDB)
-molecular docking and molecular dynamics
And maybe, since you care about the analytical part of the project, any book about analytical chemistry.

great thank you, i'll sit on this for now and any new questions I have in the next few days i'll be sure to post them here again, thank you for the excellent responses.

I think the major gas sensing technologies are NDIR(cheap), semiconducter type(cheapest), fuel cell type(cheap) and conductive polymers(expensive)


text:
Why does it require so much pressure to create diamonds?
Why does it require so much energy/pressure to create diamonds, yet with silicon it is very easy and (I believe) naturally occurs in a "diamond" structure, i.e. where each atom is covalently bonded to four adjacent atoms? I guess it has something to do with the fact that the outer shell is further away from the nuclear, since that is the only significant difference between the two atoms, but I can't work out why that should matter here.

In short, carbon loves to form $\pi$ bonds, but silicon doesn't.

If you squeeze hard enough on silicon, it transforms to an hcp structure. Why is that so hard, when titanium is hcp at STP?

Secondarily, silicon does not occur naturally with the diamond structure. It does not occur naturally as the pure element at all, only combined with oxygen.

There are two allotropes of carbon crystal, graphite and diamond.   Both
are stable, so the formation of diamond structure is competitive with
the other structure (or structures, if you include buckyballs
and tubes).   The important consideration, therefore, is crystal growth rate,
which will depend on solvent and seed material, as well
as temperature and
pressure.
For manufacture of large artificial diamonds, liquid iron 'solvent' is
used, which is why high temperature is important; carbon has to diffuse through the liquid solvent. At high temperature,
the (relative) instability of diamond means that
a pressurized environment is required: diamond has to expand (diamond is 3.53 g/cc, but graphite is 2.23 g/cc) in order to become graphite, and pressure prevents that.   Pressure alone might
become so high that graphite recrystallizes as denser diamond, in a shock wave (as long as you only want to make small diamonds) made with explosives.  At elevated temperatures and low pressure, diamond
recrystallizes as graphite, however.
It is
possible to grow diamond films (depending on a substrate that
seeds the crystal form of interest) at low pressures as well, from
vapor.   Pyrolyzing CH4 in the presence of a cold surface that doesn't
seed the graphite form is the general approach.
To summarize: in growth of a carbon crystal, the environment determines
the dominant crystal structure in a number of ways.  The Diamond Makers by Robert Hazen, is a good treatment of this subject.


text:
What is Nesbet's theorem?
In a talk by Frank Neese there was a slide with title "Nesbet's theorem" and an equation indicating something like "if your wavefunction has the double excitation amplitudes correct, then you can get the correct FCI energy for free". The caveat being that to get the doubles exactly correct you need the triples and quadruples, and to get the quadruples you need the pentuples and hexuples, etc.
Unfortunately I am not able to find this theorem stated precisely anywhere with a Google search. I have been told that the theorem can be found in the book Molecular Electronic Structure Theory by Helgaker, Jorgensen, and Olsen, however I do not have the book and need the precise statement of this theorem.
Does anyone know the precise statement of Nesbet's theorem and the original source?

Section 4.1.1 (p 237) of Szabo and Ostlund's Modern Quantum Chemistry also describes this issue.
Using intermediate normalisation the correlation energy can be written as
$$\begin{align}
\langle \Psi_0|\hat{H}-E_0| \Phi_0\rangle &= \langle \Psi_0|\hat{H}-E_0 \left(| \Psi_0\rangle + \sum_{ar}c_a^r |\Psi_a^r \rangle + \sum_{abrs}c_{ab}^{rs}|\Psi_{ab}^{rs} \rangle +.... \right) \\
&= \sum_{abrs}c_{ab}^{rs}\langle \Psi_0|\hat{H}| \Psi_{ab}^{rs}\rangle
\end{align}$$
$\langle \Psi_0|\hat{H}| \Psi_{a}^{r}\rangle = 0$  according to Brillouin's theorem.
$$\langle \Psi_0|\hat{H}| \Psi_{abc}^{rst}\rangle = \langle \Psi_0|\hat{H}| \Psi_{ab}^{rs}\rangle \langle \phi_c | \phi_t \rangle = 0$$
and similarly for higher excitations, because $\hat{H}$ is a 2-electron operator.

It's Nesbet's theorem, and you can find it in equation (41) of

Nesbet, R. K. Electronic Correlation in Atoms and Molecules. In
Advances in Chemical Physics; Prigogine, I., Ed.; Wiley: New York, 1965; Vol. 9, pp 321–363. DOI: 10.1002/9780470143551.ch4.

$$E_\text{corr} = \frac{1}{4}\sum_{ijab} c^{ab}_{ij}\langle ij || ab \rangle$$
Note Nesbet uses the more general two-body operator $R$ and does not have the prefactor. See slide 7 on this talk by Neese (PDF).

The other answers were very helpful, especially in telling me the correct spelling of Nesbet! With the correct spelling a quick search gave this paper: ChemPhysChem 2017, 18 (23), 3478 which I found to be the most helpful (for me) since it also defines all the notation:

Now let us build an analogous definition for $e_\mathrm c(\mathbf{r})$ in the case of a CI wavefunction $\Psi$ represented by a linear combination of the ground-state determinant $\Phi_0$ and excited determinants $\Phi_i^a$, $\Phi_{ij}^{ab}$, etc.:
$$\Psi = C_0\Phi_0 + \sum_i^{\text{occ}}\sum_a^{\text{virt}}C_i^a\Phi_i^a + \sum_{i<j}^{\text{occ}}\sum_{a<b}^{\text{virt}}C_i^{ab}\Phi_{ij}^{ab} + \cdots$$
In this case, the correlation energy is conveniently provided by Nesbet's theorem:
$$E_c^{\text{CI}} = \sum_{i<j}^{\text{occ}}\sum_{a<b}^{\text{virt}} c_{ij}^{ab} \left< \Phi_0 \middle| \hat{H} \middle| \Phi_{ij}^{ab} \right>$$
where intermediate normalisation is assumed: $c_{ij}^{ab} = C_{ij}^{ab}/C_0$.

So if you have the correct doubles amplitudes of an FCI wavefunction, you can calculate the correlation energy easily using the integrals corresponding to the Hartree–Fock determinant and the double-excitation determinants.


text:
How are the LCAO ansatz and Slater determinant ansatz for the wavefunction connected?
Say we have a system of 2 electrons, an ansatz for the wavefunction is then a Slater determinant consisting of two orbitals.
$$
\left|\Psi_{Slater}\right>= \frac{1}{\sqrt{2!}}\left(\left|\phi_1\phi_2\right>-\left|\phi_2\phi_1\right>\right)
$$
How is this related to the basis expansion ansatz:
$$
\left|\Psi\right>= \sum_i c_i \left|\phi_i\right>
$$
The basis expansion formula would give me a function of 1 coordinate x_1 if i projected it on the position space, while the slater determinant ansatz would give me a function of two coordinates x_1, x_2. x_i stands for a set of x,y,z
coordinates. Spin is neglected in this example.
I understand that for the basis set expansion in an orthonormal basis, one can obtain the coefficients by the inner product
$$
\left<\phi_i|\Psi \right>=c_i
$$
but how can i use this on a slater determinant ? I only understand how this works if everything relies on one coordinate x_i, then i can calculate this:
$$
\left<\phi_i|\Psi \right>=\int\phi_i(x)\Psi(x)dx=c_i
$$
But how does this work if the wavefunction is multidimensional like the slater determinant wavefunction ?
$$
\left<\phi_i|\Psi_{Slater} \right>=\int\phi_i(x_1)\Psi_{Slater}(x_1,x_2)dx= c_i(x_2)?
$$
This expression makes no sense, since the remaining object would still be a function of $$x_2$$
To my knowledge only basis functions of one coordinate are used in calculations, like 3D Gaussians. Assuming that i have a wavefunction given in slater determinant form, how would i find the coefficients of the basis functions ?
I am afraid i am mixing up things that don't belong together, clarifications would be welcome.

I think I see where you are getting tripped up. LCAO is a way of constructing MOs, not multielectron wavefunctions. The Slater determinant is a way of forming a wavefunction from MOs, which may have themselves been formed using LCAO. @HansWurst

To put @Tyberius comment into equations: your second equation should be: $\phi_i=\sum_\nu c_{\nu,i} \varphi_\nu$, where $\phi_i$ are the molecular orbitals from your first equations, and $\varphi_\nu$ are atomic orbital basis functions of the LCAO ansatz.

That makes sense. But i am still not clear on some other equations then.It is often  assumed that you can write $$\left|\Psi \right> = \sum_i c_i \left|\phi_i\right>$$ The small phi's are then not molecular orbitals but some eigenfunctions of an operator for the given space ? I.e. i could use the Hamilton Eigenfunctions as the basis functions  and they would dependent for the example above, each on two coordinates x_1 and x_2 ?

It is hard to tell what you are asking about. It is unclear where this equation is coming from. What is its physical context, other than being some linear combination? Alternatively we would need to know what exactly $|\phi\rangle$ is here. But without knowing either, I can only make wild guesses: it could be the Hartree-Product,  CI,  or something completely different. Please clarify your question with some more context.

I found the question somewhat confusing. I think it is because mixing up some concepts.

I understand that for the basis set expansion in an orthonormal basis,
  one can obtain the coefficients by the inner product  $$
 \left<\phi_i|\Psi \right>=c_i $$

Please, think about what you mean by the coefficients and where the functions live, before continue reading.
As you thought this expression as an inner product (which is right), both functions must live in the same space so they have the same number of coordinates. What you are doing with that inner product is projecting $\Psi$ over the $\phi_i$. If $\{\phi_i\}$ is an orthogonal set, then you can make the expansion you wrote. But, if the are not, then such expansion is meaning less. You still can make expansions that recover $\Psi$  (in fact, an infinite number of them). So, I get confused about which coefficients are you referring to. As for example, $\phi_1$, appears in both: many Slater's det. and in many terms in one (or more) Slater's det., to ask  which is its coefficient in the Slater det. makes not sense, except if you want to expand the Slater det. into a LCAO which is not possible (but you still can project it).
See below.

But how does this work if the wavefunction is multidimensional like the slater determinant wavefunction ?

First, note that even for 1-electron problems the problem is not mono-dimensional. So, the case of Slater's determinants is exactly the same. You only need to integrate over every coordinate.

To my knowledge only basis functions of one coordinate are used in
  calculations, like 3D Gaussians.

3D-Gaussians have three coordinates. It is obvious using Cartesian coordinates. Under spherical coordinates they only depends on $r$, but they still are functions of $\theta$ and $\phi$. Although you can take only one coordinate ($r$) for its evaluation, it is not true for computing an arbitrary inner product.

Assuming that i have a wavefunction given in slater determinant form,
  how would i find the coefficients of the basis functions ?

At first, take in mind that those coefficients are for expanding the Slater det. into the subspace spanned by your $\{\phi_i\}$. That is, it they are not an complete set, you only will get an approximation. You just have to perform those integrations. In your example, the $c_i$ will not be a function of $x_2$, because you have to integrate over every coordinate.
I am not sure if I understand correctly your doubts. Please, do not hesitate in ask for clarification. Maybe this answer could be useful.

I was initially confused by the fact that the explicit variables are suppressed and was unsure about the dimensionality of the involved objects, but that has been cleared up by now.

@Hans Wurst , books often do not characterize rigorously the mathematical objects that they are using. That leads to an extra effort for the reader to realize exactly what those objects are. Normally it is skipped leaving some holes.


text:
Feasibility of Pyrolytic Rearrangement

I've been asked whether or not the above reaction happens - and if it does, what's the mechanism?
I'm really confused, I think it shouldn't happen; but I'd like to confirm with Chem SE, just to be sure.
If it doesn't happen, then what are the possible methods of accomplishing the conversion?

Seems unlikely as it would involve acetate anion attacking the phenyl ring. Think it more likely you would get benzonitrile and acetic acid

Maybe oxime/methylketone might form in similar rearrangement.

How is this off topic? It's not a homework question, and I have no other way of finding out whether or not the reaction is feasible.

If you describe why you think it's unlikely to happen there's a good chance it will be reopened.


text:
Chair conformations of cyclohexane with two substituents
I have been studying the chair conformations of cyclohexane but I get confused when it comes to two substituents.
I came across an example, in which the one substituent is the ethyl group (Et) and the other one is methyl (Me).  The ethyl is drawn like a solid wedge, whereas the methyl is drawn like a dashed wedge. As far as I  know, I have to draw the chair and then the equatorial positions and the axial ones. I thought that since the bigger substituent is the ethyl group, that it would be better in a equatorial position and the methyl would be better in a axial position. But the example, draws the ethyl group in an axial position and the methyl in an equatorial position
Why is that?I really don't understand.
Thank you.

Welcome to SE! What you have described is a trans isomer. Since one group appears to be axial and one equatorial, it sounds like the 1,3-isomer. With the ethyl axial it is the less stable of the two chair conformers.

Here are some structures to help you. You noted that one group was a solid wedge and the other a dashed wedge without mentioning the substitution pattern. I presume that this was a flat, 2-D structure. You also note that in the chair structure, one group is axial, the other equatorial. Only the 1,3-isomer 2 fits these conditions. The conformation 2a is more stable than 2b.
ADDENDUM: The 1,1-disubstituted isomer would also fit the conditions of the question. The questioner did not designate the structure.


text:
How do I get Mass spectrometry graph, history, and Infra-red spectroscopy graph for Secnidazole?
My chemistry teacher wants me to create an academic poster for a recently-discovered biologically important organic molecule with chirality that is relatively small and is also recent.
Secnidazole (also called Solosec) seems to fulfill these requirements fairly well, however I have no clue on how I would be able to obtain the mass spec or infra-red spec of it, as they doesn't seem to be listed online anywhere. As well as this, there doesn't seem to be a listing for what organisation created it or anything either.
How do I either get a rough approximation of the results, or find a legitimate set? Bonus points for directing me to the history of the molecule.

https://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi (or search sdbs database, it should be the first link) is a useful database that contains a variety of spectral data for a large number of compounds. If you search for Secnidazole, they appear to have MS and IR data for this compound.
If you have access to journal articles, you can also do a literature search for this compound. It is likely that spectral data for this compound will be present in any paper where this molecule was synthesized (either in the main article or its supporting information).
Edit: I just did a brief literature search through Reaxys - most of the papers seemed centered around clinical trials rather than organic synthesis, although I will admit I didn't look over any of them in any great detail.  I'll leave that to you.
Chemical companies such as Sigma Aldrich or Fisher Scientific also sometimes supply spectroscopic data for the compounds they sell.
If you have access to ChemDraw, you can simulate $^1$H and $^{13}$C NMR data for any compound you draw.  I'm sure other spectral simulation programs exist, but don't have much experience in that area.

Hey, I went to SDBS.com, and it says the domain name is for sale

Edited. Sorry about that!


text:
Conditions for the reduction of carboxylic acids by LiAlH4
From my textbook, the reduction of carboxylic acids by LiAlH4 "has to be in anhydrous conditions, such as dry ether, followed by aqueous acid." I read from the internet that the reduction reaction has to be in anhydrous conditions since LiAlH4 reacts violently with H2O. But why does the reduction reaction need to be 'followed by aqueous acid'? Isn't that contradictory to the fact that LiAlH4 needs to be in anhydrous conditions?

When the reaction is done, you need to quench it to (1) destroy the remaining LiAlH4 (2) protonate the lithium alkoxide to give you the alcohol. The aluminium salts formed can be separated from the alcohol. If you don't quench the reaction it makes separation essentially impossible.
Of course, if you add the water before the reaction is done, then that defeats the purpose of using an anhydrous environment.

Saturated aqueous Na2SO4 is better than aqueous acid for decomposition of excess LiAlH4 in that all of the aluminum salts are coagulated and a clear ether solution of the alcohol is formed.

@user55119 indeed, one can do much better than just adding something like HCl (the Fieser workup seems to be fairly popular, too, although I've no personal experience with it).


text:
Is water vapour formation possible at -10 degree celcius?
Under any pressure conditions is it possible ? If not then is there no water vapour present in places where temperature drops to -10 degree celcius ?

Yes it is possible, albeit to a little extent.

Welcome to Chemistry.SE! Questions are the central focus of this site, but we ask that questions demonstrate an attempt to solve/understand the question to maintain quality.  Please take a minute to look over the help center and tour page to better understand our guidelines and question policies.

As @Ivan Neretin comments just a little. At $-10$ C the vapour pressure is $\approx 260 $ Pa (Pascal)  where $1\text{ atm} = 101325$ Pa. Even at $-100$ C there is some absolutely minute amount, $0.0014$ Pa.
The reason is that there is still some energy around even at these low temperatures. The chance (probability) that there is enough energy to propel a molecule into the gas phase, decreases, of course, as the temperature falls. It does so very rapidly as it is given by the Boltzmann distribution which decrease exponentially with energy at any fixed temperature.; $prob(E)\sim \exp(-E/(k_BT))$ where $E$ is the energy/ molecule needed to put the molecule into the gas phase and $k_B$ the Boltzmann constant.

Snowflakes are ice crystals that are formed directly from water vapor present in the air when the ambient temperature is below 0 Celsius. So it is quite common to have water vapor present in the air at below-freezing temperatures. Snow formation occurs on a gigantic scale every day, somewhere in the world.
Sometimes you can observe ice and snow slowly disappear even during a cold snap in winter when temperatures remain below freezing over a long period of time. In this case, the ice sublimes directly into the vapor phase. You could also observe this process in the laboratory on a small scale by placing ice in a vacuum chamber and pumping on it to a high vacuum.


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Are there chiral carbons in PVC polymers?
PVC polymers are formed from the monomer CH2=CHCl.
Below picture is a section of PVC molecule.
Is there any chiral carbon centre in this polymer molecule. If yes, then how?

I have a source of CAIE examination paper implying carbon bonded to Cl group is chiral, but as in the picture the group on both sides of that carbon is same.
(Source, Qno:38)

See: Tacticity

Yes, all polymer chains of the form - (CHR - CH2 -) are in principle chiral, the problem is that in the left and right side of the backbone in  BBL - CHCl - BBR is in effect indistinguishable.
Even in a PVC molecule with perfect tacticity, only the end groups (they usually are different) make the whole thing chiral, and they are too far away. In a structure with a defect tacticity, the next defects in the chain create chirality at a C atom, but they (usually) occur with the same probability distribution on both sides of the local carbon atom and the whole molecule.
That's why in polymers, optical activity is not observed, instead, the relative occurrence of pairs, triplets, etc. of neighbouring C* atoms with identical stereochemistry can be specified (and detected via NMR). That way, you can differentiate e.g. chains which go RRRRRSRRRRRSRRRRRS or RRRRRSSSSSSRRRRRRSSSSS, which will show remarkably different mechanical properties due to their crystalline structure.
A different thing are of course polymers with chiral side groups, e.g. proteins. A solution of those can (or rather does, unless the side group is racemic) show optical activity.