text:
Impact of flow rate on peak areas in Mass Spectrometry
As a follow up to the discussion in comments: it's now clear that UV peaks will have larger area if flow rate is decreased, because the same portion of molecules are measured again and again.
With Mass Spec it's different because it's a destructive detector. So does the flow impact peak areas in Mass Spec? And how much does this impact differ from UV?

If the same molecules are recorded «again and again» (sorting the column, passing the UV-Vis detector), than why shouldn't the TIC integrated over time when these molecules pass the MS detector equally rise?

@Buttonwood, since I've never actually worked with the instrument let alone looked under its hood, I may easily mistake here: I thought that the whole flow that comes out of the column is directed to MS. Then it's ionized and injected into the mass analyzer. If we decrease the flow, we don't change the number of particles passing to the analyzer. Since MS just counts the number of particles (which didn't change), we should end up with the same area. And after the particles passed the detector, they're "destroyed" and never return to the detector which means they aren't counted again.

@Buttonwood, oh, or I may have misunderstood your question. With UV-Vis detector we have a flow cell and the detector is shining light through that cell. The longer the molecules pass through that cell the more they're exposed to the detector. With Mass Spec there's no flow cell - it just takes the particles from the chromatographic column and "counts" them. This process means that the particles fall onto to the mass detector (or are filtered out) and get out of the system. So they have only one shot at being measured.

The flow rate of the mobile phase (like a carrier) affects when the molecules sort the column, broad or narrow peak shaped, and (van Deemter equation) the quality of separation of a mixture passing the column. UV, MS, FID; these all are merely observers at the exit gate (cf. e.g., HPLC simulation on ChemCompute). Different to UV and FID, for HPLC-MS, only a tiny split portion sorting the column actually enters the MS detector. In total (i.e., over time), the same number of molecules pass column and (fixed portion) detector, high, or low flow rate.

If the molecules of interest sort the column like a flat broad peak, then the recorded signal for a brief increment of time (say $\pu{10 s}$) might be well different to an increment of recording the TIC for a $\pu{10 s}$ interval about a narrow, sharp and high band of molecules sorting all at once the column.  And true, once hitting the sensor, the molecules are «dead», and not accumulated.  The Faraday beaker isn't about collecting the molecules to fill a vial.

@Buttonwood, I agree that flow rate impacts height & width of peaks. But the area should still be the same, right? Also, when you say "sort" the column do you mean "leave"? Haven't heard of this term before..

Yes, with «sort», there was an accidental code-switching with the French verb to leave something/someone (which is sortir, in the infinitive).

As we discussed previously in the comments given in your link, in the analytical chemistry literature, you will see two types of detector classifications especially in the field of chromatography, namely, concentration sensitive and mass flow rate sensitive detectors. This terminology was introduced by Istvan Halasz in the 1960s.
When we wish to classify detectors as concentration sensitive vs. mass flow rate sensitive detectors, we have to eliminate the column discussion in between. This is the stumbling point and therfore the concept is taught or discussed quite incorrectly even by highly experienced chromatographers. So how are we supposed to call HPLC-UV detector as concentration sensitive vs. HPLC-MS as mass flow sensitive detector?
Do a thought experiment. Fill a HPLC eluent bottle with a solvent and add a few crystals of benzoic acid and dissolve. Connect the pump to the detector directly and monitor the absorbance at 254 nm as a function of flow rate.
At 0.05 mL/min the absorbance is $A$, 0.25 mL/min absorbance is still $A$, and 5 mL/min, it is still $A$. This is because Beer's law does not have a flow rate term in it. Now dilute that solvent containing benzoic acid by half and repeat the same experiment. The signal is now $SA/2$, and same observation is to be made at high flow rates.
Thus, HPLC-UV detector is a concentration sensitive detector.
Come to a GC-MS system (I am avoiding HPLC-MS, see below). Use an empty silica column as before. Inject the mobile phase containing a small amount of benzoic acid as before. At 0.05 mL/min, the TIC signal is $S_1$, increase the flow rate to 0.15 mL/min, the signal is $S_2$. It turns out that $S_2$> $S_1$. Mass spec it therefore mass flow rate sensitive detector, because by increasing the flow rate more molecules are reaching that detector and the response will increase. In this classification, the column has no role and we are only interested in the signal magnitude not the peak areas , because there is no column.
The story changes, when you are interested in peaks areas and when there is a column. This classification, like every human classification, breaks down.
Let us reconnect the column and use HPLC-UV detector. Use a clean solvent, and inject a solution of benzoic acid. You should see one peak with area $X_1$ at 0.25 mL/min, and for the same injection, at 1 mL/min, the peak area would be $X_2$. It will be found that the peak areas $X_1$>$X_2$, the reason is not that the same portion of the analytes trapped in the flow cell are measured again and again, the reason is that the residence time of the analytes is larger at low flow rates. Peak area is the integral of the Signal x Time! More time means larger peak area.
The story with mass spectrometer is slightly challenging with flow rates and much less predictable, because now we have two coupled effects
(i) Mass flow rate effects, that the signal should increase with flow rate because more molecules are reaching.
(ii) Peak area should not be affected too much from flow rates because the residence time of the analyte in the mass spectrometer is too short regardless of flow rate. It is a destructive detector unlike the UV. So residence time does not matter much,
So, it is hard to predict what will happen to the peak areas in MS because a third complicating factor is the ionization efficiency may change as a function of flow rate.
To show  you what a mess HPLC-MS is, look at the following graph. The peak area is non-linear function for a very common HPLC mass spectrometer, the ESI.

Figure taken from Combined Electrospray Ionization−Atmospheric Pressure Chemical Ionization Source for Use in High-Throughput LC−MS Applications; Richard T. Gallagher et al. in Anal. Chem. 2003, 75, 4, 973–977 (doi 10.1021/ac0205457).
In short, Halasz was an excellent separation scientist but his classification breaks down in real life, yet he did a lot of good work. I heard that he commanded good respect in his circles in the 1960-70s.


text:
What is the major product of the reaction between 1‐methylcyclohex‐1‐ene and BrCl?
Would the bromine add to the more substituted carbon due to selectivity?

This is not a radical process. $\ce{BrCl}$ is a source of $\ce{Br+},$ so this is the addition of $\ce{Br+}$ to the alkene followed by capture of the cation (bromonium ion) by $\ce{Cl-}$ [1].
The product is that of $\ce{Cl-}$ addition to the most stable cation centre i.e. the tertiary centre, so 1-methyl-1-chloro-2-bromocyclohexane is the product.
Reference

Buckles, R. E.; Long, J. W. The Addition of Bromine Chloride to Carbon—Carbon Double Bonds. J. Am. Chem. Soc. 1951, 73 (3), 998–1000. DOI: 10.1021/ja01147a033.

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sort: votes


text:
What is meant by "analogous reactions" in the Bell-Evans Polanyi Principle?
The BEP principle says that for analogous reactions, enthalpy change and activation energy are linearly related. What is meant by analogous reactions?
Eg. For electrophilic addition of HBr to pent-1-ene, two products (2 or 1-bromopentane) are possible. But carbocation for formation of 2-bromo-pentane is more stable, so 2-bromopentane is also more stable than 1-bromopentane. Here, BEP applies.
But then for E2 elimination of HBr from 2-bromobutane, the transition state leading to formation of but-1-ene is more stable than that for formation of but-2-ene, although the product but-1-ene itself is less stable. Here, it seems to contradict BEP principle? Are these not considered analogous reactions? If so, why?

The 'analogous' would refer to, for example, HBr + pent-1-ene and the analogous reaction HBr + hex-1-ene.  These two analogous reactions are in the same family of reactions, HBr addition to alkenes.

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text:
Energy of reactants and products
In some reactions, energy is released and when I want to understand the reason for the release of energy, I find that the energy of the products is lower than that of the reactants.
But What is this energy difference caused by?
It is written in various sources that this energy difference is due to the potential energy difference, But what is potential energy?
Other sources make it clear that the high energy of the reactants causes instability, What exactly does this high energy describe? When we say that the energy of the reactants is higher, does it mean that the bond energy is higher? Well, if so, why is it necessary to break strong bonds in the reactants and weak bonds are formed in the products?
I would be grateful if you could explain the matter to me completely and clearly.

Related:  https://chemistry.stackexchange.com/questions/157655/why-does-the-pe-becomes-negative-in-potential-energy-versus-intermolecular-dista?noredirect=1#comment322805_157655

Unfortunately this is much too broad for this site. Please consult a general chemistry book. If you have trouble with any concepts you encounter, you are welcome to ask more detailed questions here. Essentially you are asking here: What is Chemistry and why and how does it work?

Creation of a chemical bond releases energy ( and vice versa ).
This released energy is different for different bonds.
If breaking bonds of reactants takes less total energy than there is released by forming of product bonds, the reaction releases energy ( and vice versa ).
The source of energy is combined kinetic energy and electrostatic potential energy of valence electrons.
Potential energy is energy needed to move a free object ( away of external forces ) to the place of the interest ( is negative when energy is released ).
When reaction energy is released, valence electrons gain in average more negative potential energy. The means more energy is needed to release such electrons from a molecule.


text:
Reaction of 1-Bromo-1-phenylethene with two equivalents of LDA
I was confronted with the reaction of 1-bromo-1-phenylethene with two equivalents of LDA, then the product is hydrolyzed.
As it is a bulky base, it will surely deprotonate and form the phenylacetylene, and form phenylacetyl-lithium in a second step, but what happens when it is hydrolyzed?
Will the product stay as it is or what happens?

The phenyllithium acetylide is a reasonably stable species (pKa mid 20s). Protonation will give phenylacetylene.


text:
How displacement reaction actually taking place?
"When a zinc plate placed in copper sulphate solutionfor some time,after sometime we can see holes in plate , why?"
So this was a question in my book , I don't want answer of this one.
But I Want answer of this one:-
Does displacement reaction only take place on surface of any metal plate  ( high reactive)? Beacuse we heard 'deposit' word , like copper deposited on zinc. So I zinc from surface of plate started dissolving and in place of zinc , copper deposit on that plate. So I think reaction should stop after all surface zinc has been displaced by copper, beacuse now it is covered by copper and solution has ZnSO4 .
Than why hole get build in plate , does inner zinc also taking part in displacement reaction?
Why if copper is already deposit on surface(surface zinc should have start dissolving first and copper simultaneously cover the surface of plate)
Also according to above , some CuSO4 should still left in solution along with ZnSO4 , as inner (zinc under copper deposit on zinc plate ) hasn't displaced.
This make sense because if whole zinc plate has been dissolved than copper from solution just don't get place to deposit and settle like sand at bottom?
Please clear my doubts,
Does only surface of a metal plate(high reactive) take part in displacement reaction ? (And remaining metal plate  under deposited layer of less reactive metal remain undissolved) If yes than why holes build in plate?why does layer deposit on plate protect inner metal from further dissolving.
If you want any clarification, please let me know but don't mark my question negative If u think deeply it is a good question.

Copper deposited on the zinc surface does not protect the remaining zinc to be dissolved. Copper does not form a real layer on the zinc surface. As long as some metallic zinc remains in contact of the copper sulfate solution, the reaction will proceed. It may be irregular and produce holes in the zinc plate. The reaction will stop when and only when the zinc metal or the copper solution is missing

Thank u....so much   .    So u mean even after copper layer is formed on zinc , it will still cause dissolving of inner  Zinc   ? And what do u mean "when zinc metal missing" , do u mean when it is fully covered with Copper or do u mean all zinc got dissolve? And why that holes appear in plate of zinc?

U said "as long as some metallic zinc remain in contact with cuso4" , that i am asking ....suppose outer surface zinc will start displacing copper and after some time  cu  will form a covering on zinc metal. Than why it is said we will see holes in zinc plate after some time . It is like saying " zinc is keep dissolving until it created hole in plate " where does cu went , it should deposit on plate after "Surface" zinc has replaced Cu from CuSO4 , and protect inner zinc .....just like galvanisation( just analogy i know they are 2 different things but zinc protect inner metal from corrosion )

Copper will never form a covering on zinc. The copper deposit does not adhere on the zinc plate.  It looks like moss or foam. Copper and zinc do not have the same density. The copper layer does not hold on the zinc. As long as the reaction proceeds, the amount of metallic zinc decreases. The copper does not protect the zinc. The zinc plate becomes thinner and thinner. The amount of copper becomes more and more important.  After some time, there is no metallic zinc any more. But before, holes may appear in the zinc plates, depending on the currents in the solution.

So u mean covering is not like a sticky ....I mean eg. Paint on a iron rod, so covering is not like that, instead it is foam like? And zinc will keep decreasing?

Yes. When a new $\ce{Zn^{2+}}$ is created it needs room to be dissolved in water. It repells the newly created copper atom away from the metallic surface. And zinc keeps decreasing


text:
How can a catalyst be selective if it does not change the equilibrium constants?
\begin{align}
\ce{CO(g) + 3H2(g) &->[Ni]  CH4(g)  +  H2O(g)}\\
\ce{CO(g) + 2H2(g)  &->[Cu/ZnO-Cr2O3]  CH3OH(g) }\\
\ce{CO(g) + H2(g) &->[Cu]  HCHO(g) }
\end{align}
As we see here, using different catalysts in the reaction between Carbon monoxide and hydrogen yields different products. Is this in contradiction to the following description of the properties of catalysts?

However, it is very important to keep in mind that the addition of a catalyst has no effect whatsoever on the final equilibrium position of the reaction. It simply gets it there faster.
https://courses.lumenlearning.com/introchem/chapter/the-effect-of-a-catalyst/

Adding a catalyst makes absolutely no difference to the position of equilibrium
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Le_Chateliers_Principle/Le_Chatelier%27s_Principle_Fundamentals

If a catalyst is not supposed to affect the reaction's final equilibrium position how do we explain the catalyst selectivity seen here?
I saw a similar question (Selectivity of catalysts) but it wasn't addressed directly at this principle (and unanswered still).

The quotes you have quoted are in contradiction with the title. Catalysts do not affect the position of equilibrium, they accelerate the respective ( and possibly selected ) forward and backward reactions. Acceleration just some reactions ( in both ways) is not against any principle.

The question itself is probably not a bad one, but it needs a lot of work reformatting to make it more attractive.

If a catalyst is not supposed to affect the reaction's final equilibrium position how do we explain the catalyst selectivity seen here?

If you wait long enough so that all three reactions attain equilibrium, the presence or absences of catalysts have no effect on the product mixtures.
In the examples, however, the reactions without catalysts are all slow. If you add one catalyst but not the other two, one reaction will reach equilibrium and you can work up the product mixtures, which will contain little or none of the other possible products.
The fancy way of describing this is to say that the product ratio is under kinetic control (rather than thermodynamic control). Here are two specific scenario where reaction conditions determine product identity: Kinetic and thermodynamic control of sulphonation of toluene and Thermodynamic vs Kinetic Sulphonation of Naphthalene

Catalysts very much do affect end products because they may act differently on competing reactions. For instance, given ethylene and oxygen a suitable catalyst may promote formation of ethylene oxide and not as strongly promote oxidizing the ethylene to carbon dioxide and water. (In this particular case, a silver catalyst with carefully controlled temperature is used to maximize ethylene oxide yield.)
What a catalyst does not affect is the equilibrium product mixture for a particular reaction.


text:
Spin contamination and expansion of unrestrict determinants
I was studying with the book "Modern Quantum Chemistry" of Sazbo and Oustland. In the chapter 2, they talk about unrestrict (UR) determiants. It's explaneid that UR determinants are not eigenfunctions of $S^{2}$, but we can expand them as linear combinations of singlet $|1 \rangle$, doublet $|2 \rangle$, triplet $|3 \rangle$ states and so on. For exemple, UR determinants used as approximatons to a system that is a singlet (with $N^{\alpha} = N^{\beta}$) is
$$
|\Psi_{S} \rangle = a_{1} |1 \rangle + a_{2} | 3 \rangle + a_{3} |5 \rangle + ...
$$
To UR determinants that represents a system with $N^{\alpha} = N^{\beta} + 1$ (a doublet) and $N^{\alpha} = N^{\beta} + 2$ (a triplet), we have
$$
|\Psi_{D} \rangle = b_{1} |2 \rangle + b_{2} | 4 \rangle + a_{3} |6 \rangle + ...
$$
$$
|\Psi_{T} \rangle = c_{1} |3 \rangle + c_{2} | 5 \rangle + c_{3} |7 \rangle + ...
$$
and so on. That is, in the expansion we use either even or odd eigenstates and the series begins in the eigenstate with the same spin of the system. I would like to know why this is so. I understand that, since $H, S_{z}$ and $S^{2}$ commute, we can construct states that are eigenstates of both of them, and that this set of states is complete, so any function can be write as a linear combination of these states. But I can't see why in the above expansions we exclude some of the eigenstates.

Forgive me, as my specialty is not in electronic structure theory. But the root cause of this is angular momentum coupling and that's a topic that I feel qualified enough to comment on.
In this answer, small letters ($s$, $m$) refer to quantum numbers and big letters are the corresponding quantum mechanical operators.
When you have two particles with spin quantum numbers $s_1$ and $s_2$, the spin quantum number $s$ of the resulting two-particle system has the possible values
$$s = |s_1 - s_2|, |s_1 - s_2| + 1, \ldots, s_1 + s_2 \tag{1}$$
(the absolute values can be left out if you assume, without loss of generality, that $s_1 \geq s_2$.) This fact is stated in pretty much any textbook that covers quantum mechanical angular momentum. Since the allowed values for $s$ increase by $1$ each 'step', the multiplicity, $2s + 1$, increases by $2$ each 'step'.
The simplest example is two spin-1/2 particles, e.g. electrons. In this case $|s_1 - s_2| = 0$ and $s_1 + s_2 = 1$, so the resulting coupled states are triplet ($s = 1$) and singlet ($s = 0$) states. Note that $m$ here refers to the projection quantum number (i.e. the $m$ in $\hat{S}_z|\psi\rangle = m\hbar|\psi\rangle$), not the multiplicity $2s + 1$.
$$\begin{align}
|s = 1, m = 1 \rangle &= |\alpha\alpha\rangle \\
|s = 1, m = 0 \rangle &= 2^{-1/2}(|\alpha\beta\rangle + |\beta\alpha\rangle) \\
|s = 1, m = -1\rangle &= |\beta\beta\rangle \\
|s = 0, m = 0 \rangle &= 2^{-1/2}(|\alpha\beta\rangle - |\beta\alpha\rangle)
\end{align}$$
In particular, note that it is not possible to construct a doublet state from two spin-1/2 particles. In general, it is not possible to construct a system with half-integer spin from two half-integer spin subsystems. We can generalise this quite easily:

Spin of system 1 ($s_1$)
Spin of system 2 ($s_2$)
Spin of overall system

Half-integer
Half-integer
Integer

Half-integer
Integer
Half-integer

Integer
Half-integer
Half-integer

Integer
Integer
Integer

Larger systems are constructed by repeatedly coupling together smaller systems. However, this table applies at every stage: so if you couple together system A which consists of two electrons (integer spin), with system B which consists of a third electron (half-integer spin), you get a three-electron system which must have half-integer spin.
From this it is not too hard to see that if you have an odd number of electrons, the resulting overall spin must be half-integer (i.e. multiplicities of $2, 4, 6, \ldots$); and if you have an even number of electrons, the resulting overall spin must be integer (i.e. multiplicities of $1, 3, 5, \ldots$).

Bonus
An extra mathematical proof of the table above is reproduced from Sakurai's Modern Quantum Mechanics, 3rd ed. It follows from the relationship between the 'uncoupled' and 'coupled' representation, which differ in the choice of the basis sets. In the 'uncoupled' form, we have simultaneous eigenstates of $\hat{S}_1^2$, $\hat{S}_{1z}$, $\hat{S}_2^2$, and $\hat{S}_{2z}$, i.e. quantum numbers of $s_1, m_1, s_2, m_2$. In the 'coupled' form we have simultaneous eigenstates of $\hat{S}^2$, $\hat{S}_1^2$, $\hat{S}_2^2$, and $\hat{S}_z$, i.e. quantum numbers of $s, s_1, s_2, m$. (In both cases, these are the maximal number of operators that mutually commute, so it's not possible to add any more quantum numbers to either set.)
Notice that $s_1$ and $s_2$ are 'shared' between both representations. So in order to cut down on the notation, we denote the 'uncoupled' basis states by $|m_1;m_2\rangle$; and the 'coupled' basis states by $|s;m\rangle$. Essentially, we cut out the quantum numbers that aren't unique.
The transformation between the two bases is given by
$$|s;m\rangle = \sum_{m_1,m_2} |m_1;m_2\rangle \langle m_1;m_2 | s;m\rangle,$$
which is an expansion of the coupled state $|s;m\rangle$ as a linear combination of uncoupled states $|m_1;m_2\rangle$. The scalar quantity $\langle m_1;m_2 | s;m\rangle$ is a Clebsch–Gordan coefficient. It can be shown that these coefficients vanish unless $m_1 + m_2 = m$; and this relies on the fact that the corresponding quantum mechanical operators add up:
$$\hat{S}_z = \hat{S}_{1z} + \hat{S}_{2z}.$$
Thus, for any arbitrary coupled state $|s;m\rangle$,
$$(\hat{S}_z - \hat{S}_{1z} - \hat{S}_{2z}) |s;m\rangle = 0.$$
Note that this is purely because the operators cancel each other out. It does not depend on the state $|s;m\rangle$ being an eigenstate of all three operators: in fact, in general it will not be, because this is an eigenstate of $\hat{S}^2$ which does not commute with $\hat{S}_{1z}$ or $\hat{S}_{2z}$. Left-multiplying by some arbitrary uncoupled basis bra $\langle m_1m_2|$, we get
$$\langle m_1m_2|(\hat{S}_z - \hat{S}_{1z} - \hat{S}_{2z}) |s;m\rangle = 0.$$
The trick here is now to apply the relevant operator to the relevant side of the braket. In particular, we want to apply $\hat{S}_z$ to the ket, and apply $\hat{S}_{1z}$ and $\hat{S}_{2z}$ to the bra, because these components are eigenstates of the respective operators. We don't need to worry about complex conjugates because the operators are Hermitian and thus have real eigenvalues.
$$\begin{align}
\langle m_1m_2|(\hat{S}_z - \hat{S}_{1z} - \hat{S}_{2z}) |s;m\rangle  &= \langle m_1m_2|\hat{S}_z|s;m\rangle - \langle m_1m_2|\hat{S}_{1z}|s;m\rangle- \langle m_1m_2|\hat{S}_{1z}|s;m\rangle \\
&= m\hbar \langle m_1m_2|s;m\rangle - m_1\hbar \langle m_1m_2|s;m\rangle - m_2\hbar \langle m_1m_2|s;m\rangle \\
&= (m - m_1 - m_2)\hbar \langle m_1m_2|s;m\rangle = 0 \\
\end{align}$$
which means that either $m = m_1 + m_2$, or $\langle m_1m_2|s;m\rangle = 0$. So essentially, all the terms in the sum
$$|s;m\rangle = \sum_{m_1,m_2} |m_1;m_2\rangle \langle m_1;m_2 | s;m\rangle$$
vanish, unless $m = m_1 + m_2$. This means that it's not possible to get a coupled state $|s;m\rangle$ if $m$ cannot be expressed as a sum of some $m_1$ and some $m_2$ (because all the terms would go to zero, and we'd get a zero state, which is unphysical).
Now, consider two particles which have half-integer spin, i.e. half-integer $s_1$ and $s_2$. Because $m_1$ is allowed to take values $-s_1, -s_1 + 1, \ldots, s_1$, it follows that $m_1$ also has half-integer values. The same is true of $m_2$.
But we showed above that for the coupled states, the only valid states are those where $m$ can be expressed as a sum of $m_1$ and $m_2$: this means that $m$ must be an integer. Furthermore, the quantum number $m$ is allowed to take values $-s, -s+1, \ldots, s$. Consider the case where $s$ were to be a half-integer: this would mean that $m$ had to be a half-integer as well, a contradiction. So $s$ must be an integer.
Altogether, this completes a proof that if $s_1$ and $s_2$ are half-integer, then $s$ must be an integer (i.e. the first row of the table above). The other cases follow in a similar fashion.


text:
Conversion from Spectral Line Intensity to Absorbance (HITRAN)
The HITRAN spectral database gives access to the spectral absorption lines of various gaseous species. These lines are given as $S$ values in units of $cm^{-1}/(molecule·cm^{-2})$ (or $cm.molecule$).
Given a simple Beer-Lambert law of absorption:
$I = I_0 \cdot exp(-l \cdot \varepsilon_{gas} \cdot \mathcal{C})$
Wherein:

$I_0$ and $I$ are the incident and emerging light intensities on a given gaseous sample
$l$ is the light pathlength through the sample (in $m$)
$\varepsilon_{gas}$ is the molar extinction coefficient of the gas (in $m^{-1}/(mol.L^{-1})$)
$\mathcal{C}$ is the gas concentration (in $mol.L^{-1}$)

How can we convert this $S$ value into an $\varepsilon_{gas}$ value (or similar) usable in the Beer-Lambert law?

I apologize for my definition of the Beer-Lambert law may be more appropriate to species in solution. If one prefers, they may write it with $\mathcal{N}$ (in $molecule/L$) instead of $\mathcal{C}$ or play around with pressure instead of concentration. My question remains essentially the same : converting from spectral intensity into absorbance.
I had a look into the HITRAN documentation and especially:
1 - The FAQ which-on the topic of spectral line unit-leads to this article. Appendix A1 details units and appendix A.2.4. From what I understood (the following may be completely wrong), I could get the optical depth (absorbance) as:
$\tau = u \cdot S \cdot f$
with:

(I guess) $u = \mathcal{N} \cdot l$ with $\mathcal{N}$ in $molecule/m^3$, $l$ in $m$
$S$ the line intensity as given by the HITRAN database
$f$ which I have no idea how to find / with which value to compute it.

In other words, the Beer-Lambert law would write: $I = I_0 \cdot exp(-l \cdot \mathcal{N} \cdot S \cdot f)$. Is it correct?
2 - The definitions and units page which reproduces the content of the article.
I also found some resources here.

Related topics: here.

What is the unit of $S$ ?

@Maurice −1/(·−2), I wrote it at the top or may find it under several of the HITRAN pages.

I followed the HITRAN link and, under “cross section data definitions”, it gives equation 1. The tau value is the transmittance as a function of wavenumber. So you basically have it ready to go.

@EdV, thank you for the answer, I did find the equation that you mention. However, cross-section data are only available for complex molecules here: https://hitran.org/xsc/ Simpler ones such as CO2, N2 or O2 are unfortunately not available in this format... That being said, it comforts my initial guess that tau = uk = uSf with u=NL, they call it rho instead of N...

TL;DR: HITRAN offers a Python API called Hapi which gives the possibility to fetch spectral data from the database directly, and convert it into absorbance spectra within tens of line of code. It is useful and quite user-friendly.

Disclaimer: I am no expert in infrared spectroscopy or molecular energetic transition. The following answer is mainly an attempt at explaining what I understood from online reads and discussions with other scientists. I tried to justify my allegations with appropriate citations but do not hesitate to correct me in the comment if I misinterpreted / misreported something.

Line Broadening Phenomenon
The absorption of electromagnetic radiations is caused by the presence of resonance modes in the structures of chemical species (atoms or molecules)[1]. In theory, if we take a single atom or molecule at very low pressure and temperature, these resonance modes appears at very specific frequencies (or wavelengths) causing very sharp lines in their absorption spectra.
However, when the molecules or atoms are in the presence of other species, collisions can occur between an excited molecule and its surrounding, interfering with the absorption process and broadening the shape of the lines[2]. Conceptually, it is as if the original spectral lines—which can be seen as Dirac impulsions—were convolved by a line-broadening profile, as depicted below.

 Convolution of two spectral lines by a broadening function of profile f(x). Image source.
This broadening profile can be modelled using three different functions, depending on the broadening effects taken into account[3, 4]:

Lorentz profile, used to characterize pressure broadening i.e. the interactions with other molecules. The higher the total pressure, the more Lorentzian broadening.
Doppler, or Gaussian profile, defined in the absence of collision effects. It accounts merely for the effect of temperature (the molecules are moving towards and away from the observer, hence the Doppler shift.
Voigt profile, which combines the two above.

Link With the HITRAN Values
For a given spectral line in the HITRAN database, which is characterised by:

its amplitude $S$, given in $cm^{-1}/(molecule\cdot cm^{-2})$
its wavenumber $\nu$, given in $cm^{-1}$

The contribution to the overall absorbance of the gas of a single spectral line, itself corresponding to a resonance mode $t$ (of resonnance wavenumber $\nu_t$), observed at a given wavenumber $\nu$, can be computed as $\tau$ (dimensionless optical depth of the sample)[5]:
$\tau = u \cdot S \cdot f$
with:

$u = \mathcal{N} \cdot l$ with $\mathcal{N}$ in $molecule/m^3$, $l$ in $m$, the density of absorbing molecules per unit path length ($molecule/m^2$)
$S$ the line intensity as given by the HITRAN database
$f = f(\nu,\nu_t,T,p)$, the normalized line shape function

$f$ can be computed as:
\begin{equation}
f(\nu,\nu_t,T,p) = \frac{1}{\pi}\cdot \frac{\gamma(p,T)}{\gamma(p,T)^2 + [\nu - (\nu_t + \delta(p_{ref})\cdot p]^2}
\end{equation}
Wherein:

$p$ is the pressure (atm)
$T$ the temperature (K)
$\gamma$ the pressure broadened line half-width ($cm^{-1}$)
$\delta$ the pressure shift ($cm^{-1}.atm^{-1}$)

With appropriate calculation formulae for $\gamma$ and $\delta$ also given in [5] (equations A12-A16). Here we can see that to obtain the full absorption spectra of a given gas, we must add the contribution of the absorbance of all the spectral lines broadened by the above presented function, modelled by f. In other words, the optical depth at a given wavelength can be expressed as:
\begin{equation}
\tau(\nu) = u \cdot \sum_{t} S_t \cdot f(\nu,\nu_t,T,p)
\end{equation}
Although such computations could be done by hand, HITRAN luckily developed a Python package called Hapi, which makes such calculations automatically (see below for an example with $CO_2$).

 $CO_2$ absorption spectrum using the Hapi interface. In blue are the raw transition lines, while in red is the broadened absorption spectum using Lorentzian coefficients computed for a dilution in air.
WARNING: be sure tu use the Hapi functions appropriately. In particular, the broadening of the spectral lines changes with temperature, total pressure, but also with the nature of the dilution medium at hand. In other words, 5% $CO_2$ in air will not have the same absorption spectra as 5% $CO_2$ in argon or hydrogen, for instance.

References

Donald A. Burns, Emil W. Ciurczak, Handbook of Near-Infrared Analysis, 2007. ISBN: 9780849373930.
Wikipedia, Spectral Line: Line Broadening and Shift, Archive 20/09/2021
Irina N. Sokolik, Basics of gaseous absorption/emission and line shapes, EAS 8803, Study notes for Environmental Science, Link
Wikipedia, Spectral Line Shape, Archive 20/09/2021
Rothman et al., THE HITRAN MOLECULAR SPECTROSCOPIC DATABASE AND HAWKS (HITRAN ATMOSPHERIC WORKSTATION): 1996 EDITION, p.46, 1998, DOI: S0022-4073(98)00078-8


text:
Acidity of triphenylmethane
The question is : Why does triphenylmethane have such a high pKa value (going up to $33$)
My reasoning suggests that the conjugate base is highly stabilized and hence triphenylmethane should be a really good $\ce{H+}$ donor, but clearly isn't.
Where am I missing out?

"conjugate base is highly stabilized" - Why? Just because of resonance with a benzene ring? I suggest looking up the pKa of toluene to get an idea of how much stabilisation this actually provides.

Its the triphenylmethyl cation that is very stable, not the anion

It's like 20 units less then methane - quite a lot of "stabilisation".

It is a very good H+ donor, as far as hydrocarbons go.

Interestingly, diphenylmethane has about the same pKa as triphenylmethane.

I guess it would be a great H+ donor if all the benzene ring were to lay on the same plane. The fact that they are all twisted one from each other makes this hypotethical very well resonance-stabilized negative charge, just quite enough stabilized.


text:
Why doesn't the protonated propyne rearrange?
In the reaction of propyne with $\ce{HBr}$ this cation (prop-1-en-2-ylium) was mentioned by my teacher:
$$\ce{H3C-\overset{+}{C}=CH2}$$
He told me that the carbocation will not rearrange as it will have no change in stability. When I pointed out that the carbocation could shift away from the double bond, as this will provide resonance stability to the carbocation, he just got silent for two minutes and then said it does not rearrange.
Can someone please explain why it cannot rearrange?

This seems quite relevant: Why are vinylic and arylic carbocations highly unstable?

You changed the structure. Is this what you really mean? Is it a triple or double bond?

This link may be helpful. https://chemistry.stackexchange.com/questions/2774/resonance-in-carbocation-derived-from-allenes?rq=1 @Martin

You can refer the book David Klein's Organic Chemistry. It is discussed in there

Apologies, I have put the wedding structure.


text:
How could we synthesize tert-butyl-alcohol, starting with just methane as only carbon source?
would it be possible to synthesize tert-butyl-alcohol with methane as the only carbon source? I mean, I almost figured it out, so basically methane is radically substituted by bromine under UV, then it is given to a suspension of Li in dimethyl ether, and in the last step, it is combined with CO2 (dry ice or by burning CH4) to yield the lithium propane-2,2-bis(olate), but I don't get the last step where we would get the tert-butyl-alcohol.

How would only adding water give the tert-butyl alcohol?

Carbon dioxide is also a carbon source in this scheme. Presumably the last step is to add more MeLi to the acetone, and something got skipped in the scheme.

If my memory serves me correctly, the addition of MeLi to lithium carboxylate to form the dilithioalkoxide, requires carboxylic acid salts of 6 carbons or greater. There is solubility problem. As an academic exercise, this example is fine. Follow @orthocresol’s lead. If you have access to Organic Reactions, you can read Margaret Jorgenson’s review here: https://onlinelibrary.wiley.com/doi/full/10.1002/0471264180.or018.01


text:
Quantifying location and volume of "lone pairs" for VSEPR
Of course molecular shape and the VSEPR model come up frequently.
As discussed previously in a valence bond picture like VSEPR, the lone pair electrons are closer to the atom than bonding pairs, and "fatter."
I can think of a few ways of quantifying this, but I'm trying to add more data when discussing molecular shape:

Are there examples where the lone pair distance is estimated (e.g., for ozone or $\ce{NO2}$)? This could be compared to a typical bond length.
Is there an example where the lone pair volume is quantified (e.g., through a polarizability volume, or an electron density isosurface)?

VSEPR is typically used to give qualitative valence bond understanding of shapes, but the lone pair effects are often confusing. So I'm wondering if data on volume / position can help clarify these effects for students.

There have been a few efforts to experimentally quantify the polarizability volume of lone pairs, including:
"Polarizability of Non-Bonding Electron Pairs" Australian Journal of Chemistry (1965) 18 (3) , pp.253-.
They estimate lone-pair polarizabilities from a series of molar refraction measurements (per lone pair).

The refractivity of a molecule can be regarded as the sum of the refractivities of the component bonds and of the unshared electron pairs.

Polarizabilities of Non-Bonding Electron Pairs in $Å^3$
$$
\begin{array}{ll|ll|ll}
\hline {\text { Group V }} & & {\text { Group VI }} & & {\text { Group VII }} \\
\hline \mathrm{N} & 1.0 & \mathrm{O} & 0.5 & \mathrm{~F} & 0.2 \\
\mathrm{P} & 2.13 & \mathrm{S} & 1.5 & \mathrm{Cl} & 0.8 \\
\mathrm{As} & 3.3 & \mathrm{Se} & 2.2 & \mathrm{Br} & 1.2 \\
\mathrm{Sb} & 3.9 & \mathrm{Te} & 2.8 & \mathrm{I} & 1.8 \\
\mathrm{Bi} & 4.6 & & & & \\
\hline
\end{array}$$
A related follow-up "Polarities and anisotropic electron polarisabilities of neutral ligands" J. Chem. Soc. A, (1969), 2735-2739 used dipole moment and molecular Kerr measurements in dioxane and benzene of a bunch of N, P, and As compounds. The polarizabilities of $\ce{N:}$ and $\ce{NO}$ compounds were remarkably similar, suggesting for N, P, and As, that the lone pair had similar polarizability volume to an oxygen atom.


text:
Method to remove impurities from acetaminophen synthesis experiment
I'd like to know about impurities from acetaminophen synthesis experiment. Are there any impurities produced by oxidation of p-aminophenol? If there is any impurity produced by oxidation of p-aminophenol, I would like to know the formula and impurity removal method. Also, it is said that acetic acid and water wash away when they react, and I want to know more about this.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

Which acetylation conditions are you using?

After the synthesis, your crude solid acetaminophen contains dark impurities, which have been formed from oxidation of the starting material, 4-aminophenol that carried along during the synthesis (if you have used 4-aminophenol as it is without purification). These impurities are intensely colored dyes of unknown structures with highly conjugated chromophore, hence the color. Although the amount of the dye impurities is very small, their intense color is bright enough to impact the color of the crude acetaminophen, which supposed to be colorless. Most of these colored impurities is destroyed by heating the crude product with sodium dithionite $\ce{(Na2S2O3)}$, a reducing agent. The heating with an aqueous solution of dithionite reduces double bonds in the colored dye to produce colorless substances. Thus, decolorized acetaminophen can be collected on a Büchner funnel by filtration. Following is an example for micro-scale decolorization procedure:

Dissolve $\pu{0.2 g}$ of sodium dithionite $\ce{(Na2S2O3)}$ in  $\pu{1.5 mL}$ of water in a  $5$-$\pu{mL}$ conical vial. Add your crude acetaminophen ($\approx \pu{200 g}$) to the vial. Heat the mixture at about  $\pu{100 ^\circ C}$ for  $\pu{15 min}$, with occasional stirring with a micro-spatula. Some of the acetaminophen will dissolve during the decolorization process: Cool the mixture thoroughly in an ice bath for about  $\pu{10 min}$ to reprecipitate the decolorized acetaminophen (scratch the inside of the vial, if necessary to induce crystallization). Collect the purified material by vacuum filtration on a Hirsch funnel using small portions (about  $\pu{0.5 mL}$ total) of ice water to aid the transfer. Dry the crystals for $5$-$\pu{10 min}$ by allowing air to be drawn through them while they remain on the Hirsch funnel. Weigh the purified acetaminophen and compare the color of the purified material to that obtained above.

It is better if you further purified your decolorized materials using a microscale crystallization technique.

Note: Usually, your limiting reagent in this experiment is 4-aminophemol. The excess acetic anhydride would decompose to acetic acid during the reaction. Acetic acid is also a byproduct of the reaction. The product, acetaminophen (N-(2-hydroxyphenyl) acetamide) is insoluble in cold water and would precipitate. Thus, your first filtration and washing would remove all soluble byproducts including acetic acid. The crude product would be colored if there are impurities present in the initial 4-aninophenol starting material.

Are you sure of your units in the above procedure - 200g of crude acetaminophen in a 5ml vial??

@Waylander: No, I wasn't sure about the measurement. But, it was from a college procedure. Acetaminophen is not supposed to dissolve but it is enough to dissolve reducing agent so would react with the color pigment.

@Waylander: I presume the text should read ~200 mg given that the experiment appears to be on a microscale. p-Aminophenol has a solubility in water of 1.5g/100mL at unspecified temperature. Activated charcoal decolorization may be applicable.


text:
Oxidative cleavage of alkenes by Chromic Oxide
Does CrO3 (hot concentrated or acidified like Jones Reagent) cleave alkenes and alkynes like hot conc KMnO4 would, like this reaction?

If not, why not? And do any other reagents other than Ozone capable of oxidative cleavage of $\pi$ bonds?

Osmium Tetroxide + Sodium Periodate will cleave alkenes to dicarbonyls - search Johnson Lemieux oxidation.

thanks! but what about the chromic oxide?

Chromic oxide does not typically cleave alkenes


text:
Number of gas molecules in activated state vs normal state
I have a simple bimolecular reaction
$$\ce{A + M <=>[k_2][k_1] A^* + M}\tag{1}$$
where $\ce{A}$ is a molecule which collides with another molecule $\ce{M}$. $\ce{A^*}$ is the $\ce{A}$ molecule in the activated state. $k_2$ is the reaction rate constant in the forward direction and $k_1$ is the reaction rate constant in the reverse direction.
The activated complex then forms the product by the following reaction where $k_3$ is the reaction rate constant.
$$\ce{A^* ->[k_3] P}\tag{2}$$
For this mechanism it's said that at low pressures $k_3 > k_2$ and vice versa at high pressures. Now according to Le Chatelier's principle at low pressures the system moves to produce more gas molecules and vice versa.
Now at low pressures, by low pressures I mean lowering the pressure from a set pressure, $P$, reaction step 1 will move to produce more gas molecules. But when we are comparing  $\ce{A}$  and  $\ce{A^*}$, which one will have more gas molecules i.e how to deduce in which direction will the step 1 move in response to stresses in terms of more pressure and less pressure?

A*/A will probably follow the Boltzman distribution E=E0.exp(-Ea/kT), eventually shifted due reaction entropy change. As both (1) and (2) are supposed to be in equilibrium, (2) would have zero net effect on (1). If (2) has not reach equilibrium by P production yet, the ration A*/A would be even smaller.

Sorry Poutnik step 2 is not supposed to be in equilibrium. I have edited the question.

So IMHO [A*]/[A] < [A*]/A < 1 // BTW are you sure you mean by K1,K2,K3  equilibrium constants and not kinetic rate constants ?

Instead of `K3` K3, use should rather use $K_3$  $K_3$ to follow the proper typography. // You can find useful  Notation basics and How can I format math/chemistry expressions on Chemistry SE.

Yes I meant kinetic reaction rate constant. I have edited in the question. What do you mean by IMHO [A*]/[A] < [A*]/A < 1? Can you please clarify?

I assume you are familiar with basics of algebra. [A] means molar concentration of A, but it can be replaced by partial pressure, as I have forgotten you probably speak about gaseous phase.

Yeah sorry I didn't what IMHO meant ,I got that you are saying in equilibrium [A*]/[A] will be more than [A*]/[A], but how did they assume at low pressures K3>K2 and at high pressures K3>K2? If you could clarify that it would be very helpful for me to understand the mechanism going on

Let us continue this discussion in chat.

This is the Lindemann mechanism for unimolecular reactions, second order at low pressure and first order at high pressures. There is no need to think about Le Chatelier as this can be analysed exactly to answer your questions. Any decent phys. chem. text book will describe this .


text:
Determining reaction order from absorbance

A reaction is initiated in a solution which contained a dissolved component. The transmittance was measured at $\pu{240 nm}$ in a $\pu{2.54 cm}$ long cell. At $t = 0$ the transmittance was $1$. $\varepsilon{}$ for the product is $\pu{470 dm^3 mol^{-1} mm^{-1}}$.
$$
\begin{array}{l|cccccccc}
\hline
t \text{[min]} & 4 & 8 & 12 & 16 & 20 &  \infty \\
\hline
\text{transmittance} & 0.507 & 0.442 &  0.417 & 0.403 & 0.395 & 0.360  \\
\hline
\end{array}
$$
Determine the reaction order and rate constant

I am unsure if the way that I am solving this problem is the correct way of doing it. I begin by calculating the transmittance for the reactant(s) by taking the transmittance for the different times $t$ minus the transmittance for $t = \infty{}$. The way I am thinking about this is that at $t = \infty{}$ then all the reactants have become product which means that the transmittance at $t = \infty{}$ is the initial transmittance for the reactants (correct me if I'm wrong).
From this I get:
$$
\begin{array}{l|cccccccc}
\hline
%
t \text{[min]} & 4 & 8 & 12 & 16 & 20 &  \infty \\
\hline
%
\text{transmittance} & 0.147 & 0.082 &  0.057 & 0.043 & 0.035 & 0  \\
\hline
\end{array}
$$
Then I used Beer-Lambert's law to convert the transmittance to absorbance and then to concentration, getting:
$$
\begin{array}{l|cccccccc}
\hline
%
t \text{[min]} & 4 & 8 & 12 & 16 & 20 &\\
\hline
%
\text{concentration} & 4.68*10^{-4} & 6.11*10^{-4} & 6.99*10^{-4} & 7.69*10^{-4} & 8.19*10^{-4}  \\
\hline
\end{array}
$$
Plotting $t$ in seconds either against $\ln(c)$, or $1/c$ gives me that the reaction is of first order with a rate constant $k = \pu{5.62 \times 10^{-4} s^{-1}}$.
However, I also tried first converting the transmittance to absorbance, then calculating the absorbance for the reactants (by taking the last measured data minus the rest as for the transmittance) and then from that calculate the concentration. This then gives me that the reaction is of second order with a rate constant $k = \pu{32.9 M^{-1} s^{-1}}$.
Which way is the correct way of solving the problem? Or is both wrong? I find it confusing that the two methods give different answers depending on whether I use the transmittance or absorbance first..
Appreciate all the help!
EDIT NEW ATTEMPT
So, I think that this might be the way to do it and would appreciate if someone could correct me if I'm wrong.
I being by converting all the transmittance for the given product to the concentration of the product using Beer-Lambert's law.
$$
\begin{array}{l|cccccccc}
\hline
%
t \text{[min]} & 4 & 8 & 12 & 16 & 20 &  \infty \\
\hline
%
\text{Concentration products} & 0.000166 & 0.000199 & 0.000214 & 0.000222 & 0.000227 & 0.00025  \\
\hline
\end{array}
$$
Now, I calculate the concentration for the reactants by subtracting the concentration for t=$∞$ by the remaining concentrations to get:
$$
\begin{array}{l|cccccccc}
\hline
%
t \text{[min]} & 4 & 8 & 12 & 16 & 20 &  \infty \\
\hline
%
\text{Concentration reactants} & 0.000084 & 0.000050 & 0.000036 & 0.000028 & 0.000023 & 0  \\
\hline
\end{array}
$$
I then plot the concentration as a first order and second order and get that the second order has a better fit. So the answer is that the reaction is of second order and the rate constant is $33$ M$^{-1}$s$^{-1}$.
Does this make more sense than the first two methods?

I don't know if you know mhchem including one of the later implemented (perhaps specific to chemistry.se) short hand of \pu{} as in $t = \pu{32.5 s}$ (to yield $t = \pu{32.5 s}$).  It isn't siunitx (whose scope is not needed here on chemistry.se), but helpful to prevent line breaks in formulae, to retain formatting, and easing input in the body of questions, answers, comments.  (Not safe for titles of questions, though.)

If you have these plots at hand, why do you don't substantiate your question and add them to your question?

There are no unique answers to your question. It depends of some informations that are missing in the text.
First : Is the initial solution made of only one colored substance ? In other words, is the end product the result of the reaction or is it due to some other compound already present initially ?
Second : Do you know the extinction coefficients epsilon of the initial and of the final products ? Which one did you use to calculate the concentration ?

@Maurice I only know epsilon for the final products. Perhaps that justifies that I should convert the transmittance of the products to concentration before calculating the concentration of the reactants. Since I can't use epsilon later for the reactants.. if you understand what I mean

@Katara. Some information is still missing. 1) Is the initial mixture containing only one colored substance ? 2) How can you explain that the transmittance goes from 0.147 to zero, and simultaneously the concentration goes form $4.68$ to $8.19$ ? These data are contradictory !

@Maurice I wrote the question as given to me so I don't know more than what is said. 2) I didn't think of that.. but how would I solve this problem then, disregarding my two attempts.

@Maurice I have edited my question with another method that I think is more correct than the other two.. would truly appreciate it if you could have a look at it :)


text:
stoichiometric crystal
I wonder if there is something like a "stoichiometric crystal"? In other words: Is it possible to have a crystal that is non-stoichiometric?
I'm not interested in crystal defects but don't want to ignore them.

Well, $\ce{Fe_{0.95}O}$ is definitely a thing, and crystalline at that. As for the defects, unless you are using some unconventional definition of a crystal, they are inevitable.

Thanks, I guess, what I actually wanted to know is: Are there crystals which are not stoichiometric?

Crystals of alums  $\ce{KAl(SO4)2.12 H2O}$  and  $\ce{KCr(SO4)2.12 H2O}$   have ions $\ce{Cr^3+}$ and $\ce{Al^3+}$ freely interchangeable in any ratio, making possible the non stoichiometric $\ce{K(Al,Cr)(SO4)2.12 H2O}$, if crystalized in mixed solutions - once done at home as a young student.

There are many binary phases that have broad phase fields. One example is titanium nitride. But, being off-stoichiometry means that there are vacancies on the nitrogen sub-lattice.

Can it be said, that a pure/ideal crystal has to be stochiometric? In case it consists of several different elements.

Extending the first part of your question, i.e. are there stoichiometric crystals as in crystals of stoichiometric composition, the answer is yes.  These are co-crystals IUCr defines as

«Solid consisting of a crystalline single-phase material composed of two or more different molecular and/or ionic compounds generally in a stoichiometric ratio which are neither solvates nor simple salts.»
source: IUCr Online Dictionary of Crystallography

While this excludes crystals of e.g., $\ce{CuSO4 * 5 H2O}$, there are plenty examples consisting of organic molecules such as the 1:1 mixture of quinole and pyridine (reference), or the 2:1 co-crystal of meloxicam and acetylendicarboxylic acid (reference) to mention just two.
Since the simultaneous presence of two components in one crystal typically alters the physical properties, e.g., of the rate of dissolution (which may translate into a change of rate their constituents become accessible for an organism [ADMET data]), this is a large field of research of interest for pharmaceuticals and material science.   See, for example, the reviews by Buddhadev, Nugrahani, or Ross. Again, it is a large field of research; don't drown in the literature.
References:
Buddhadev, S. S.; Garala, K. C. Pharmaceutical Cocrystals—A Review. Proceedings 2020, 62, 14; doi 10.3390/proceedings2020062014, open access.
Nugrahani, I.; Jessica, M. A. Amino Acids as the Potential Co-Former for Co-Crystal Development: A Review. Molecules 2021, 26, 3279; doi 10.3390/molecules26113279 , open access.
Ross, S. A.; Lamprou D. A., Douroumis D. Engineering and manufacturing of pharmaceutical co-crystals: a review of solvent-free manufacturing technologies. Chem. Commun. 2016, 52, 8772-8786; doi 10.1039/C6CC01289B.

Thanks a lot for the detailed answer! A crystal is not necessarily stoichiometric, or? I mean a pure crystal would not fall within such a description(?).

It is not clear if "stoichimetric" in your question about a "stoichiometric crystal" refers to a) compounds (like above quinole:pyridine), or b) like in Ivan's example $\ce{Fe_{0.95}O}$.  Chemistry "to the letter of the book in high school" expects/only allows either $\ce{FeO}$ (only Fe(II)), or $\ce{Fe2O3}$ (only Fe(III)), or $\ce{Fe3O4}$ (with Fe(II) and (Fe(III)) to keep the subject accessible enough without entering into study of chemistry at an university.

@Buttonwood I intuitively understand the OP meaning "stoichiometric crystals" as crystals where constituent molar ratios are ratios of small integers, similarly as ratios in chemical equations. the question may be, what deviation is just a corruption and what would be "nonstoichiometric crystal".

@Poutnik Yes, what Poutnik said :) At least this is how I'd understand it but I'm not a chemist(?).

Then «non-stoichimetric» crystals were (iron oxides/sulfides appear to more prone to this, than other metal oxides/sulfides) e.g., with Frenkel .or./.and. Schottky defects (an atom in excess [definitely too many]/an atom missing [a vacancy] where anticipated on the lattice).  Right of the tip of my hat, I don't know of an (equivalent) example for a purely organic crystal («organic» as antonym to «inorganic»).


text:
How to convert benzyl alcohol to 2-phenylethanoic acid?

How can we convert benzyl alcohol to 2-phenylethanoic acid?

This is what I have tried.

Can someone tell me whether this is correct because I am not sure about the second reaction.
In my textbook it is given that the following reaction takes place.

But I am not sure whether the reaction will take place if R' is replaced by H.
Also other methods to convert benzene to 2-phenylethanoic acid are welcome.

The reaction of a silver carboxylate salt with an alkyl halide gives a ester, not really useful in this context

One easy way is to do the first step as you have drawn making benzyl chloride, then treat that with KCN (in DMSO  or use water/CH2Cl2 with a phase transfer catalyst) to give 2-phenyl acetonitrile then hydrolyse that with aqueous sulfuric acid as described in this Organic Syntheses procedure here
To convert benzene to 2-phenylethanoic acid: benzene reacts with ethylene oxide under Friedel-Crafts conditions to give 2-phenylethanol (see this SE answer here). This is readily oxidised to 2-phenylethanoic acid.

Is the method which I did correct?

Yes, that will work. Goes even better with a trace of DMF to catalyse it.

Certainly the formation of benzyl chloride will work but, as @Waylander has pointed out, silver formate will give an ester, benzyl formate. The use of a silver salt here is overkill because potassium or sodium formate would suffice.


text:
Units of collisional cross-section and reactive cross-section
I haven't been able to find the units of the reactive cross-section (σ* = Pσ) online but when I do a unit analysis from the equation:
$$ k = PσN_A √(\frac{8k_BT}{πμ}) e^{-E_a/RT}$$
then when plotting lnk against $1$/T the intercept is equal to:
$$intercept = lnPσN_A √(\frac{8k_BT}{πμ})$$
solving Pσ then gives that the units should be [m$^2$s$^{-1}$mol$^{-1}$]. However, the units of the collisional cross-section is [m$^2$] and the reactive cross-section is by definition the collisional cross-section times a steric factor P which is a constant, so how come the units change? Shouldn't the units for the reactive cross-section also be [m$^2$]?

what is the dimension of P ?

@porphyrin I believe it is unitless, in our lecture notes it only says its a steric factor

ok, then the units of rate constant $k$ are $\mathrm{m^3/mol/s}$ which are correct.


text:
Current density from overpotential

A current of $190$ mA is measure for a $2.5$ cm$^2$ electrode in contact with a solution containing  M$^{2+}$/M$^{3+}$ metal ions at room temperature at a overpotential of +$180$mV. The transfer coefficient is $0.41$.
Determine:
$1$. The exchange current density for the electrode
$2$.
The current density if the overpotential is +$240$ mV and +$50$mV

I know how to solve the first part of the problem, for which I get:
$$ j_0 = 1.22  mA/cm^2$$
But I don't understand how to solve the second part. How can I calculate the current density from the overpotential if I don't have the exchange current density j$_0$? Or can I use the exchange current density from the first part? I would assume that it can't be used since the overpotential changes but maybe I'm wrong

I think the ECD is inherent to the charge transfer reaction, doesn't depend on overpotential.

@MarkWolfman Rather, the OP value depends kind of reciprocally on the ECD value, right ?

I suppose so. The net current depends on the exchange current combined with the overpotential.


text:
Transparent material used in the movie Metropolis
This is probably a rather unusual question for this forum. The review of the movie Metropolis (made in 1926) by Roger Ebert shows what I assume is a still frame from the movie:

The shield looks very thin and has the typical small scratches that I would expect from a transparent thermoplastic.
My question: What material is this? If this is a silica-based glas and not a thermoplastic (did that exist in 1926?), what characteristics does it have? Is it flexible?

Possibly still ordinary glass.  If you think about e.g., the canopy of a Spitfire, this likely was acrylate which by Wikipedia was developed by 1928 and really entered production in the 1930s to cast large panels ready to mold after heating.  Perhaps a cheaper soda-lime glass (less high melting temperature than modern borosilicate glass, easier to bring into this shape).

Perhaps cellophane?

When we still had a laboratory glass shop, they could make amazing glass structures.

Most synthetic plastics were commercialized after 1930s, I think. But semi-synthetic cellulose derivatives like cellulose acetate and especially cellulose nitrates were around from the middle of 19nth century. Both were used at some point to produce thin film, in particular photographic film.
My first bet would be that it is a celluloid plate, made from mixture of nitrocellulose and camphor. It was one of the first thermoplastics and can be made transparent.

Indeed, if it is a movie, then it must have been made on a photographic film, so it surely was available.


text:
Materials for McBride's Freshman Organic Chemistry at Yale University
I have been looking over the online Organic Chemistry course taught by McBride at Yale and named "Freshman Organic Chemistry", 125A, which is on Itunes University.  Only the lectures are online; and McBride regularly refers to his website as the source of additional materials. Looking at the exams it turns out that much of the course material must be on his webpages, which are no longer available on the specified website since he retired.  Can I find similar materials somewhere?

Most materials are available for the fall 2010 at
http://chem125.webspace.yale.edu/indexFall10.html
and for Spring 2011 (the last time the course was given)
http://chem125.webspace.yale.edu
the author is at j.mcbride@yale.edu

It appears that Yale University Prof. Emeritus McBride himself has provided the requested lecture files, etc.! Excellent! ;-)

@Ed V: Yes he has after my e-mail alerting him.

There is a webpage called Freshman Organic Chemistry I described as CHEM 125a. On the bottom of the page, there is a zip of all course pages $(\pu{10 MB})$. However, there are Terms of Use described in a separate page so you may need to donate money or pay for the usage. There is an age restriction as well:

Use of Open Yale Courses website is restricted to those individuals over 18 years of age, emancipated minors, or those who possess legal parental or guardian consent, and are fully able and competent to enter into these terms, conditions, affirmations, representations and warranties herein. If you are under the age of 13, you may not use the Open Yale Courses website without consent or approval from your legal parent or guardian.

Yet, there is a section called "Sessions" just above this image:

If you click on the "Sessions", there are 37 lectures with appropriate 37 videos. There are also few exams and final exam. Enjoy the learning.


text:
Determining ppm concentrations from food label information for ICP-OES analysis?
Overview: I am looking to determine what ppm values of trace metals are present in a particular formula while considering digestion, dilution, and detection limits for ICP-OES. For example, one serving is 15 g of powdered formula which is then dissolved in 150 mL of solution, according to the label. The label reads 15 micrograms of Mn.
Goal: Find concentration of a metal from analysis of 1 g of formula digested and then diluted to 100 mL.
Attempt: 15 micrograms of Mn per 15 g of powdered formula. As that microgram/g is equivalent to ppm (15ug/15g = 1.0 ppm), there should be 1 ppm of Mn per 15 g of dry formula.
Taking 1 g of the formula, digesting it, and diluting to 100 mL: 100mL/1g x (X read) = 1 ug/g (ppm)
The AAS or ICP-OES read value should then be 0.01 ppm.
This would be well within the detection limit for ICP-OES. However, I am not sure if my above approach is correct. Is there something obvious that I am missing? Any insight would be appreciated.

You should not have a single sample preparation for all the elements. The sample preparation will depend on element by element. Your calculations have subtle misconceptions. Don't convert to concentrations but work with masses for such problems.
So, you have 15 $\mu$g Mn in 15 g sample. This means you have 1 $\mu$g Mn in 1 g sample (see, I am avoiding calculating ppm in the solid as yet). So, if you dissolve 1 $\mu$g of Mn in 100 mL of water, your final solution Mn concentration is 0.001 mg Mn/ 0.1 L or 0.01 mg/L Mn. So our numerical values match but your statement is slightly problematic.

there should be 1 ppm of Mn per 15 g of dry formula.

No, even if you take one ton of sample, it will still be 1 ppm Mn (w/w), because concentration does not depend on sample mass or volume. Analyte masses do, so if you take a larger sample mass, analyte mass will increase too, but its concentration in the sample will remain the same. This is reason to work with masses and then calculate concentration at the end.
As you already know AAS is less sensitive than ICP-OES in general, but it is less expensive. The critical info is called characteristic concentration in AAS for a each element, which is defined as the concentration that will produce 0.0044 absorbance units under optimized concentrations in air acetylene flame on a standard 10 cm burner head. Clearly, 0.01 ppm Mn is out of question on FAAS, because even under the best conditions, 0.05 ppm will produce an absorbance 0.0044. Such a value is prone to tons of errors. As a result you need a concentration sample, which is 10x fold more concentrated. I will leave it as an exercise.
This is a screenshot of a so-called Cookbook of Perkin Elmer's AAS.

The AAS or ICP-OES read value should then be 0.01 ppm.

Don't trust machine calculated concentrations. Make a calibration curve youself. No instrument can tell you the concentration without a calibration curve.


text:
Height of FCC arrangement
I have read that height of HCP arrangement is given by $\mathrm{4r\sqrt{\frac23}}$ as distance between consecutive layers is $\mathrm{2r\sqrt{\frac23}}$(in both FCC and HCP). If that is the case, then what is the height of FCC arrangement? Can it be $\mathrm{3r\sqrt{\frac23}}$? I highly doubt that FCC is an "abc" arrangement and not "ab" arrangement.
(r=radius of sphere)

My first instinct was to say that FCC has no height, because that's a cubic cell and it only has one parameter.
Then again, you didn't ask about the cell. The question was about the FCC arrangement, and as such, it makes perfect sense (if maybe in imperfect wording, as "height" is not quite the word to be used here). What is the "vertical" distance from a layer to the next layer with identical placement, really?
Well, it is as simple as that: like you said, HCP is a two-layer packing, hence its height is twice the distance between the layers. On the other hand, FCC (when thought of in terms of hexagonal layers) is a three-layer packing, hence its "height" is three times the distance between the layers. You also said that the distance between the adjacent layers is $2\sqrt{2\over3}r$, which is correct.
See where this is going?


text:
Side-chain Oxidation vs Oxidative Cleavage

When the above compound is subjected to acidified $\ce{KMnO4}$, can both side chain oxidation and oxidative cleavage of $\ce{C=C}$ bond occur? Or will one have priority over the other?

It will depend on the reaction conditions used. I think it would be possible to find conditions that give majority oxidative cleavage.

@Waylander what kind of conditions would favor oxidative cleavage over side-chain oxidation? I thought the reactants and conditions are largely similar for the two?

Lower concentrations and reaction temperature. But if you want selective oxidative cleavage there are better reagents to use.


text:
Prepairing a Sodium Acetate buffer - Meaning of molarity
We have instructions to prepare a "50 mM Sodium Acetate, pH 5 solution". I am not entirely sure what does the molarity refer to in this case. Does it mean:

Add Sodium Acetate salt up to 50 mM, and adjust with Acetic Acid until pH 5 - Na+ remains at 50 mM, Acetate- higher
Add Acetic Acid up to 50 mM, and adjust with Acetic Acid until pH 5 - Na+ remains at 50 mM, Acetate- higher
Add Sodium Acetate salt up to X mM, and add Y mM Acetic Acid, so that X + Y = 50 mM Acetate ions - Acetate- ends up at 50 mM, Na+ lower

It is the molarity of total acetic acid/acetate content, so the 3rd option.
It is quite general principle applied in the pH buffer context. You can have a wide rage of $\ce{pH}$ values with the particular buffer type, with the same total molarity of "active substance", like citrate, phosphate or mixed buffers, being adjusted by variable amount of strong acid/base.
Note that calculation of $\ce{pH}$ from concentrations is not exact, because it is just approximation of activities. The key is, you have to know in advance the ratio, e.g. from the handy $\mathrm{pH}$ buffer lab tables or other sources. The alternative is starting with pure acetic acid, or better from the approximately precalculated ratio, and adjusting final $\mathrm{pH}$ by $\ce{NaOH}$ or $\ce{HCl}$.
Sometimes, preparation of buffers is combined with the constant target ionic strength by addition of indifferent salt. In our case, it would be the less of added salt ( like e.g. NaCl ) for higher acetate content.

Thanks for the answer. So the goal ion is always the one that has a dissociation constant, right? e.g. in a 50 mM Tris/HCl, the goal compond for the 50 mM is the Tri

Rather conjugate(d) acid/base system, like HA <=> H+ + A-, or B + H+ <=> BH+, where A,,B may or may not be neutral.


text:
At equilibrium, which one of the following is always Zero?
At equilibrium, which one of the following is always zero?
(A)∆H
(B)∆S(Total)
(C)∆S(System)
(D)∆G°
I know that at equilibrium ∆G=0, and was facing difficulty in connecting the above mentioned thermodynamical parameters to equilibrium. Can someone help me regarding this? (I am an undergraduate student and don't want to dig deep into this. Please keep things simple for me.)

If you know that at equilibrium ∆G=0, isn't that enough to answer this question?

Here is a hint:  For an ideal constant temperature reservoir as the surroundings,  what is $\Delta S$ of the surroundings if the system undergoes a change at constant pressure?

Be aware that in contrary to $\Delta G$, $\Delta G^{\circ}$ is independent on being in equilibrium and is zero just accidentally.

This question is potentially misleading. It would be better to use differential (d) rather than finite (∆) differences.

See other posts, eg https://chemistry.stackexchange.com/questions/124412/what-is-wrong-in-this-argument-that-dg-must-always-be-zero/ and https://chemistry.stackexchange.com/questions/51732/what-is-the-difference-between-the-equilibrium-position-and-the-equilibrium-cons/51735#51735 and many more....

See also: https://chemistry.stackexchange.com/questions/41862/what-is-the-difference-between-%e2%88%86g-and-%e2%88%86g


text:
Major product in the reaction of alcohol with unsymmetrical acid anhydride
When alcohol reacts with symmetrical acid anhydride to form ester. We have the following reaction,

But if the acid anhydride used is unsymmetrical how can we identify the major product? For example in the following reaction which is ester is the major product?

I tried searching the internet for the mechanism of this reaction but was able to find none. Any help would be appreciated.

My feeling is that it will be determined by which R group confers more hindrance to the carbonyl group. If there isn't much difference then you will get close to a 1:1 mixture

@Waylander Do you know which bond of the alcohol will break, C-O or O-H? Is it similar to the reaction of alcohols with carboxylic acid?

O-H will break. It is similar to the reaction of an alcohol with an acyl chloride or activated ester. The other product is the carboxylic acid


text:
Does a larger bond energy in reactants mean a more exothermic reaction?
In the book "Combustion Physics" for C.K Law, page 42, it is stated that:

The parameter that has the next strongest influence on [adiabatic flame temperature] $T_{ad}$ is the $C/H$ ratio of the fuel molecule, with $T_{ad}$ increasing with increasing $C/H$ ratio. There are three factors that need to be considered to explain such a behavior. First, species with large $C/H$ ratios often contain more double and triple $C-C$ bonds, which can hold more potential energy than the single bond. For example, we have shown that the enthalpy of the reaction $C_2H_6 → C_2H_4 + H_2$ is endothermic by $32.7$ kcal/mol. Since the enthalpy of formation of $H_2$ is zero, one can then conclude that ethylene stores more energy than ethane and thereby contributes to the higher $T_{ad}$ for ethylene.

It is implied from the above that a large bond energy for the reactants means there is more energy to be liberated by combustion. However, my understanding is that we need to give energy to the reactants in order to break their bonds, and energy is then liberated by the formation of bonds in the products. Thus a large bond energy for the reactants would mean a more endothermic reaction. Is my view point correct? If so then what does the author mean exactly?

You are correct. The interpretation and wording depends on the point of view.  It is like jumping from a shallow hole to a deep hole. One POV evaluates the absolute hole deepness with the all potential energy negative. Another POV evaluates their relative deepness, taking as the reference not the flat surface above, but the target depth of the deeper hole.

@Poutnik but this still doesn't explain why contradiction in the deductions. Would the reaction become more exothermic or endothermic when reactants have higher bonding energy?

It does. It depends what is meant by higher bonding energy as it is an unclear term. Higher energy because of weak bonding ( leads to exothermic reaction ), or higher energy released by existing bond formation ( leads to endothermic reaction)  ?

@Poutnik Hmm, my understanding is that double and triple bonds are stronger than a single bond and require more energy to break.

$\ce{C#C}$ is very strong bond, but $\ce{C2H2}$ is an endothermic compound and pressurerized can explosively form carbon and hydrogen. ( that is why it is stored in pressurized cylinders dissolved in acetone ).

All this is just semantic

I would not excuse this away by saying it's semantic or simply POV-dependent.  What Law wrote is completely wrong-headed, and shows a fundamental lack of understaning of the details of chemistry.  He writes "double and triple − bonds...hold more potential energy than the single bond", cleary indicating he thinks double/triple bonds have higher PE than single bonds.  No, they have lower PE.   The reason ethylene has higher E than ethane isn't because the C=C double bond is higher in PE than a C-C single bond. Instead, it's because, the presence of the C=C bond in place of a C-C bond means..

you have fewer bonds total.  That's what gives you the higher energy.  I.e., the higher energy of the ethene is in spite of the lower PE of the C=C bond, not becaue of a fallacious higher PE of that bond.  Specifically, in going from ethene to ethane, you are going from one C=C + four C-H + one H-H (6 bonds), to one C-C plus six C-H (7 bonds).  And the former has higher energy than the latter.  That's because, while the C-H and H-H bonds have about the same PE, and C=C has a lower PE than C-C, the combined PE of one C-C + one C-H bond is lower (more negative) than that of C=C.


text:
Determine anion radius in FCC cell

The edge length of an FCC cell is $\pu{508 pm}$. If radius of cation is $\pu{110 pm}$, the radius of anion is?

My approach:
$$(2R_{\text{cation}}+2R_{\text{anion}})^2=2a^2$$
Which on simplification gives:
$$\sqrt 2(R_{\text{cation}}+R_{\text{anion}})=a$$
On putting the values, the answer comes out to be $\pu{249.21 pm}$ which seems correct but here the answer is something else
Please help me about it.

I think your calculation is correct.

Based on the figure, which shows a square of a = 508 pm length on a side, the diagonal, from point A to point B, is a times the square root of 2. This equals the sum of the two ion diameters. And this is what your second equation gives, if you simply multiply both sides by the square root of 2.

To nit-pick: It is actually not fcc, since fcc means that the center of the face is equivalent to the corner. Since at the corner is an anion and at the center of the face is a cation, the cell could be bcc or primitive cubic lattice but not fcc. But with this said. I also think that the answer from the link is wrong.

The lattice is not fcc. And the drawing is not perfect. The rectangle should be a square, as both edge lengths are equal.

@Maurice The lattice is indeed fcc but the problem was that I m considering the cations to be placed at the face centers but the truth is the anions make the whole fcc and cations are present at the octahedral voids

@Samyak Marathe. The horizontal and the vertical sides of the rectangle have both the same length $a$. But the horizontal edge is 7.5 cm long. The vertical edge is 6.5 cm high. So the drawing is not well done.


text:
Which amines show N−N coupling reaction?

We add 1 equivalent of $\ce{KCN}$ followed by catalytic reduction via Lindlar's reagent $(\ce{H2}/\ce{Pt})$ to 4-chlorobenzyl chloride (1).

Verify whether the product (P) obtained will undergo an N−N coupling reaction with a diazonium salt or not.

According to me, it should because I find no valid reason as to why it shouldn't. The product P that I obtained was 4-chlorobenzylamine (2):

But the book mentions that it won't. That made me wonder if there are some amines that don't participate in coupling reactions, or is it because of something else.

Please add a complete reference to the book you are referring to, and Please do not add hyphens after colons.

Shouldn't your first line read "We add 1 equivalent of KCN to 4-chlorobenzyl chloride (1) followed by catalytic reduction via Lindlar's reagent (H2/Pt)". If so, structure 2 is missing a methylene group.

@user55119 Yes.I was trying to say that. I think after the “editing process” there might be some issues(with regards to question’s language) here and there.

I think your book is wrong (what book is it?)
This statement from the Wikipedia page on Azo coupling here seems definitive enough:

In alkaline media, diazonium salt can react with most primary and secondary amines that exist as a free base and produce triazene

It also includes a reaction scheme showing a secondary amine (pyrrolidine) coupling with a phenyl diazonium salt.


text:
Apparent inversion of oxidation state of sulfate ion using minimal basis set
It is well known that the sulfate ion has essentially no 3d-2p bonding; hence I used the minimal basis set(STO-3G) for the NBO calculation I did via Gaussian, just to check it.
However, I got a surprising result- all four S-O bonds had higher coefficients on the sulfur than oxygen, which contradicts the basic fact that the oxidation state of sulfur in the sulfate ion is +VI; the four S-O bonds must all be "assigned to the oxygen", i.e. have higher coefficients on the oxygen.
At first I thought this was because I used the Hartree-Fock method, which is often too rough for precise calculations. So I tried using B3LYP and the density=current command. Same results still.
Now comes the main question- what would be the reason for this contradiction, and what methods can be used to solve this?
(All S-O bonds were set to be 150pm with perfect tetrahedral symmetry)

The basis set is too small, it is probably missing flexibility.

I thought that way until I realised that larger basis sets give sulfur an NBO charge of about +2.5,  which is clearly less similar to the known empirical value(+1.7) than is the value from my calculation(+1.3).

The chosen structure and level of theory is not remotely close to a reasonable approximation.
You are calculating an anion with charge -2, but there are no polarisation or diffuse functions in STO-3G. Hartree Fock behaves quite poorly, especially for those systems.
This is quite obvious when you look at the bond lengths. For lack of time I'll refer to values from Wikipedia (sulfate): $d(\ce{S-O}) = \pu{149 pm}$.
At the HF/STO-3G level of theory that value is $d_\mathrm{HF}(\ce{S-O}) = \pu{176 pm}$.
Running a bonding analysis on an approximate wave-function for a structure, which is not even close to experimental values will not produce any meaningful results.
Gaussian even warns you about this:
Warning!  S  atom    1 may be hypervalent but has no d functions.

So while there is little involvement of the d-orbitals, they still make a huge difference. Remember the correlation energy is only less than 1% of the total energy, but it is exactly that tiny bit you have to get right to make reasonable approximations.
With that in mind it is easy to spot that the NBO analysis is horrible, too. The charge on sulfur is way too low, as is the same for oxygen.
                                     Natural Population
             Natural    ---------------------------------------------
  Atom No    Charge        Core      Valence    Rydberg      Total
 --------------------------------------------------------------------
    S  1    0.50743     10.00000     5.49257    0.00000    15.49257
    O  2   -0.62686      2.00000     6.62686    0.00000     8.62686
    O  3   -0.62686      2.00000     6.62686    0.00000     8.62686
    O  4   -0.62686      2.00000     6.62686    0.00000     8.62686
    O  5   -0.62686      2.00000     6.62686    0.00000     8.62686
 ====================================================================
 * Total * -2.00000     18.00000    32.00000    0.00000    50.00000

If you force the structure, you'll get a wave-function, which won't tell you anything remotely approximate. So you might get all sorts of artifacts, even such that apparently lead to a contradiction.
For the above the bond orbital then is quite within reason, meaning that it is obviously also wrong:
  22. (1.97736) BD ( 1) S  1- O  2
               ( 68.31%)   0.8265* S  1 s( 25.00%)p 3.00( 75.00%)
                                         0.0000  0.0000  0.5000  0.0000  0.8660
                                         0.0000  0.0000  0.0000  0.0000
               ( 31.69%)   0.5629* O  2 s(  3.15%)p30.79( 96.85%)
                                         0.0000  0.1774 -0.9841  0.0000  0.0000

You can see how terrible this approximation is as oxygen uses an almost pure p-orbital for bonding.
If you increase the level of theory to DF-BP86/def2-SVP, you'll find $d_\mathrm{BP86}(\ce{S-O}) = \pu{154 pm}$, which is still not good, but at least closer.
The charges make a lot more sense:
                                     Natural Population
             Natural    ---------------------------------------------
  Atom No    Charge        Core      Valence    Rydberg      Total
 --------------------------------------------------------------------
    S  1    2.34734      9.99995     3.43215    0.22057    13.65266
    O  2   -1.08684      1.99997     7.07959    0.00727     9.08684
    O  3   -1.08683      1.99997     7.07959    0.00727     9.08683
    O  4   -1.08683      1.99997     7.07959    0.00727     9.08683
    O  5   -1.08683      1.99997     7.07959    0.00727     9.08683
 ====================================================================
 * Total * -2.00000     17.99984    31.75051    0.24965    50.00000

So does the bonding orbital:
  22. (1.97871) BD ( 1) S  1- O  2
               ( 34.07%)   0.5837* S  1 s( 25.00%)p 2.95( 73.81%)d 0.05(  1.19%)
                                         0.0000  0.0000  0.5000 -0.0018  0.0000
                                         0.8557  0.0772  0.0000  0.0000  0.0000
                                         0.0000  0.0000  0.0000  0.0000  0.0000
                                         0.0000  0.0944 -0.0545
               ( 65.93%)   0.8119* O  2 s( 23.20%)p 3.30( 76.64%)d 0.01(  0.16%)
                                         0.0000  0.4812  0.0210 -0.8754 -0.0102
                                         0.0000  0.0000  0.0000  0.0000  0.0000
                                         0.0000  0.0000  0.0348 -0.0201

With all this keep in mind: This is a calculation for an anion at virtually $\pu{0 K}$ in vacuum. There are no counter charges. That in itself is a poor approximation. You can easily see this in the eigenvalues for the molecular orbitals $\varepsilon_\mathrm{gas}(\mathrm{HOMO}) = \color{red}{+}\pu{7 eV}$.
This might be circumvented by using a solvent model (PCM, water), at least the energy is now more reasonable $\varepsilon(\mathrm{HOMO}) = \pu{-3 eV}$. The structure is virtually unchanged and only slightly shortened bonds.
As a general takeaway: Anions need at least polarisation functions, better are diffuse functions. Split valence basis sets are for heavy duty, work horse calculations only. For analysis and energies always use at least triple zeta basis sets. Always calibrate your DFA calculations (DFT Functional Selection Criteria). The smaller the molecule the better the level of theory needs to be.

"As a general takeaway: Anions need at least polarisation functions, better are diffuse functions. Split valence basis sets are for heavy duty, work horse calculations only."  Does this hold for cations as well, may I ask?

@KanghunKim For cations you will not have the problem with the orbital energies. The rest applies to every calculation. Furthermore, split valence basis sets usually have polarisation functions already. Benchmarking is always the only way to get a reasonable approximation.


text:
What does it mean by non-bonding p pi orbital?
In a paper by Peng et. al.,[1] I read about the valence band maximum of $\ce{TiO2}$ consists of non-bonding O pπ orbitals

... the valence band maximum (VBM) consists of non-bonding O pπ states

What does non-bonding pπ orbital mean? If there is not a chemical bond, why did they call it a π orbital?
References:

Peng, H.; Li, J.; Li, S.; Xia, J. First-principles study of the electronic structures and magnetic properties of 3d transition metal-doped anatase TiO2. J. Phys.: Condens. Matter 2008, 20 (12), 125207. DOI: 10.1088/0953-8984/20/12/125207.

Besides the fact that orbital as functions do exist independently of their occupancy, the fact that those orbitals are involved in the valence band formation means that are occupied.

Well, you have that planar triangle around O, and a $p$ orbital perpendicular to it, so it feels kinda natural to call that orbital $\pi$.

And, if you are concerned by non bonding orbitals in general, then Wikipedia can be a starting point https://en.m.wikipedia.org/wiki/Non-bonding_orbital

$\pi$ is a symmetry label, not a label that indicates whether there is a bond or not. These concepts are orthogonal to one another.

When you have atoms bonded all in one plane, there will be $p$ orbitals oriented perpendicular to the plane which may not interact significantly with adjacent atoms. Such orbitals would then be called nonbonding.
We may compare water with carbon dioxide. Introductory textbooks often describe the oxygen in water as having a distorted $sp^3$ hybridization, but in reality only two oxygen $p$ orbitals mix with the hydrogen and oxygen $s$ orbitals to form the bonds. The third, perpendicularly oriented $p$ orbital is a nonbonding $p$ orbital. We can see this orbital (HOMO) in the middle of this diagram from flickr.com:

In contrast, carbon dioxide does have the $p$ orbitals on the central carbon atom interacting strongly with out-of-plane orbitals on the oxygen atoms forming the familiar $\pi$ bonds.
Now let's look at titanium dioxide. Titanium dioxide may come in different phases depending on temperature and pressure, but the phase we generally see under ambient conditions is the rutile structure by WP user Ben Mills:

The red atoms are the oxygen atoms in titanium dioxide, and these are seen to have a distorted, but planar, triangular coordination with the gray titanium atoms (an equilateral triangular coordination does not fit with the octahedral coordination of the titanium). The anatase form described in the OP's reference has a more complex atomic arrangement but similar triangular coordination around each oxygen atom. So, there are perpendicularly oriented $p$ orbitals. The titanium atoms also have out-of-plane orbitals, but unlike the carbon in carbon dioxide these do not interact strongly with the out-of-plane oxygen $p$ orbitals. The difference in orbital energy and size is too great. So the out-of-plane oxygen $p$ orbitals are more like their counterparts in water, essentially nonbonding.


text:
Symbol to denote a group which is either an oxygen atom or NH group
I'm drawing the generic structure of several different organic molecules in my thesis. They share some features, one of which is a carbonyl carbon attached to either an oxygen atom (i.e. ester) or an $\ce{-NH}$ (i.e. amide).
My first instinct was to draw it like this†: $\ce{R-X-C(=O)-R'}$
But I wonder if X wrongly implies that it's a halogen? In which case, what would be the correct letter or symbol to use? I've seen Z used sometimes when X is used elsewhere, would that be more appropriate? Or perhaps Q for heteroatom, as in Reaxys? Or A for 'any'?
† I'm actually drawing in ChemDraw, this is simplified for the sake of the question
Edit: corrected mistake (nitrogen atom -> $\ce{-NH}$ group)

Oxygen and nitrogen have different valencies, so you can't use the same letter to denote literally an $\ce{O}$ or an $\ce{N}$ atom, as they can't be directly substituted for one another.
You should use the same letter to denote $\ce{O}$ or $\ce{NH}$, for example. It's perfectly permissible to write $\ce{R-X-C(O)-R'}$ and say $\ce{X} = \ce{O}, \ce{NH}$. This approach is commonly used in the literature. If your amide is tertiary then write something like $\ce{X} = \ce{O}, \ce{NH}, \ce{NR}$. The same applies to structures drawn in ChemDraw.
Beyond that, the choice of letter is arbitrary (as long as you define it!) so $\ce{X}$ is perfectly fine, although you should obviously avoid letters which already represent a chemical element (e.g. $\ce{B}$, $\ce{C}$, ...).
What one might consider the "official rules" are given in: Brecher, J. Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008). Pure and Applied Chemistry 2008, 80 (2), 277–410. DOI: 10.1351/pac200880020277 (free version provided here). See especially section GR-2.2. However, there isn't really much in there that constrains you; a lot of it is "common sense".

Of course I meant NH and feel very silly now... Edited my question for the sake of future readers. Thank you for the excellent answer!


text:
Calculation of the mass of N2 in the container
Two vessels of equal volumes are connected to each other by a valve of negligible volume. One vessel containing $0.1$ mole $\ce {N_2}(g)$ and $0.05$ mole $\ce {I_2}(s)$ at temperature $T_1$. The other vessel is completely evacuated. The vessel containing gases is heated to temperature $T_2$ whereas the other vessel is kept at $\dfrac{T_2}{3}(\lt T_1)$. Calculate the mass of $\ce {N_2}$ in container at temperature $\dfrac{T_2}{3}$ when equilibrium is attained.

My attempt: Assuming the final pressures to be $P$ in both the container at equilibrium and using the respective ideal gas equations for both the containers,
$$PV=(0.1-x +0.05-y)RT_2$$
$$PV=(x+y)R \frac{T_2}{3}$$
Where $x$ and $y$ are the final moles in the right container. Using this we get $$x+y=\frac{0.45}{4}$$
But now I have no clue for finding another equation for $x$ and $y$. In these type of questions, I always get stuck for the $2^{\text{nd}}$ equation.

Confusions:

Another equation for $x$ and $y$

Is it valid to assume final temperatures of both the containers same as initial?

And why they have given the information about $T_1$, means how can we use it?
Thank You

(1) There is none. (2) Yes. (3) You use $T_1$ to deduce the phase state of iodine in the second vessel.

@IvanNeretin,then how will we solve for x and Y?

There is no $y$, just $x$.

Let's assume the problem can be solved using only nitrogen data. The amount of iodine is independent of the amount of nitrogen. With a little bit of luck, the final temperature in the cold container is so low (= $T_2/3$) that the iodine is entirely condensed.
In the final state, there is $n_1$ moles in the hot container, and $n_2$ moles nitrogen in the cold container. The pressures are respectively $p_1$ and $p_2$ in both containers. The temperatures are : $T_2$ and $T_2/3$. Equalizing the pressures, one gets : $p_1 = n_1RT_2/V = p_2 = n_2RT_2/3V$, which gives : $n_1 = n_2/3$.
Now the total number of moles is $n = n_1 + n_2 = 4 n_2/3 = 0.1 $. As a consequence, the total amount of nitrogen in the cold volume is : $n_2 = 0.1·3/4 = 0.075$ mole.

The condensed state of iodine is not left to luck, but stated explicitly.


text:
Do coordination complexes with stronger ligands form more stable complexes?
I read that chelate complexes generally are more stable than monodentate complexes. However, I noticed that in the case of $\ce{[Fe(CN)6]^3-}$ and $\ce{[Fe(en)3]^3+}$, this is not observed as the former is more stable. Why is this so?

See this post: https://chemistry.stackexchange.com/questions/43297/which-complexes-are-stable-ammine-complexes-or-carbonyl-complexes

@NilayGhosh Thanks, that helped. By the way, I have to ask: are you Nilay Ghosh from IIT Kharagpur?

Chelate complexes are more stable IF EVERYTHING ELSE IS THE SAME.
In your case, the anionic nature(cf. ethylenediamine is neutral) of cyanide makes cyanide a better sigma-donor than is ethylenediamine, while the pi-accepting nature(cf. ethylenediamine is pi-neutral, if not (hyperconjugative) pi-donating) of cyanide makes cyanide a better pi-acceptor than is ethylenediamine.


text:
Calculate concentration of product after time passed
If I have two parallel reactions:
A --> P$1$   (k$_1$ = $0.00010$  s$^{-1}$)
A --> P$2$   (k$_2$ =  $0.00050$  s$^{-1}$)
and know that the initial concentration of A is $0.0417$ M. How can I calculate the concentrations of P$1$ and P$2$ after $30$ min?
Thank you!
EDIT
I believe the downvote is because I haven't tried to solve the problem myself. This is because I have no idea how to do it. I can get a relationship between the two products:
dP$1$/dt = k$_1$[A]
dP$2$/dt = k$_2$[A]
=> dP$1$/dP$2$= k$_1$/k$_2$
Integrating gives:
k$_2$P$1$ = k$_1$P$2$ => $0.0005$P$1$ =$0.0001$P$2$ => P$1$ = $0.2$P$2$
But what I can do with this relationship I don't know since I don't know how to calculate the products.

Hint: You know from the two rate constants that the second product, P2, is produced at 5 times the rate of the first product, P1, so at any time you will have a 5 = P2/P1. So add the two rate constants, solve the rate equation, and use the ratio.

The kinetics is given by :
$\pu{d[A]/dt}$ = $\pu{- k_{1}[A] - k_{2}[A] = - k_t[A]}$
Integration gives : $\pu{[A] = [A]_{0} e^{-k_tt}}$
At any time, the proportion of $\pu{P_{1}}$ and $\pu{P_{2}}$ are $\pu{k_{1}/k_t}$ and $\pu{k_{2}/k_t}$
As a consequence, the wanted concentrations of $\pu{P_{1}}$  and $\pu{P_{2}}$ at time $t$ are the following :
$$\pu{[P_{1}] = \frac{k_{1}}{k_t} [A]_{0} ( 1 - e^{-k_t t} )}$$
$$\pu{[P_{2}] = \frac{k_{2}}{k_t} [A]_{0} ( 1 - e^{-k_t t} )}$$
In the present problem, $\pu{k_t = 0.00060 s^{-1}}$, $\pu{[A]_{0} = 0.0417 M }$, and $\pu{t = 1800 s}$, so that  $\pu{k_tt = 1.08}$
$$\pu{[P_{1}] = ( 1/6) · 0.0417 M · ( 1 - e^{-1.08}) = \frac{0.0417}{6} ( 1 - 0.340) = 0.0046 M }$$
$$\pu{[P_{2}] = ( 5/6) · 0.0417 M · ( 1 - e^{-1.08}) = \frac{5·0.0417}{6} ( 1 - 0.340) = 0.0023 M }$$


text:
Does C8H2N4 (in pentagon shape) exist and what is it called?
My primary school released a folder which contains some sketches. As you can see, they included two molecules. One of them is isopropanol. What's the name of the other molecule (bottom)? Does it exist? The formula would be $\ce{C8H2N4}$. It needs to be in a pentagon shape. I guess it has something "tri cyano cyclo penta" in the name.

The  formula reported in the first and in the second picture are not finished. The nitrogen atom down at left should be attached to two H atoms. These two H atoms are correctly reported in the third picture, with spherical atoms. The total number of H atoms is $4$, and not $2$.

I suspect the scribble you are interpreting as as N at the bottom is not related to the structure and that carbon is supposed to be a CH2 group so the core structure is cyclopentadiene and the overall molecule is tricycanocyclopentadiene.

Then again, at this level of precision N might well be a part of the structure, and even a part of the ring (as in pyrrole). Who cares. All suggested molecules exist, but are not particularly famous or important.

Aside apparent possibilities of a made-up structure or structure being a fragment ripped off from a larger heterocycle by the designer, there is a third one.
Let's pretend the black dash next to the label ‘N’ on the bottom is a minus sign.
Then we have 2,3,4‐tricyano‐1H‐pyrrol‐1‐ide $\ce{C7HN4-}$:

I haven't found this exact structure, but a more symmetrical tetracyanopyrrolide anion $\ce{C8N5-}$ (2,3,4,5-tetracyanopyrrol-1-ide, TCP) is well-known and has been structurally characterized with single crystal diffraction [1] (see also supplementary structures with CSD IDs PEMTOS, PEMTUY and PEMVAG).

Figure 1. Fragment of the crystal structure of tetramethylammonium 2,3,4,5-tetracyanopyrrol-1-ide (CSD ID: PEMTOS). Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H};$ $\color{#909090}{\Large\bullet}~\ce{C};$ $\color{#3050F8}{\Large\bullet}~\ce{N}.$
Reference

Becker, M.; Harloff, J.; Jantz, T.; Schulz, A.; Villinger, A. Structure and Bonding of Tetracyanopyrrolides. Eur. J. Inorg. Chem. 2012, 2012 (34), 5658–5667. DOI: 10.1002/ejic.201200790.

What an unpleasantly high N:C ratio. This molecule looks like a vicious accident wanting to happen.

@StianYttervik See nitrogen enriched energetic materials e.g., by the Klapötke group. The military is interested in these.

@StianYttervik 5:8 ? That's a piece of cake ;) Something like 5:1 and things are interesting :D