text:
Use CaCO3 powder as antacid agent
A well known strategy for paper maintenance is to store it in a antacid enviroment
(special boxes, envelopes etc.).
There are also special preservation acid-free papers and archival material (envelopes etc.) "buffered" with a base material (usually $\ce{CaCO3})$ to help keep the pH of the paper to be preserved in the $>7$ pH zone for as much as possible.
I was wondering if we could simply use a $\ce{CaCO3}$ powder (talk or chalk powder) to spray the paper to be preserved instead of using a $\ce{CaCO3}$ infused paper?

Instead of what? You said yourself that we already do that, i.e., use paper with CaCO3. If you are referring to spraying the old documents, then no, we'd rather keep them as they are.

If the paper is unavailiable wouldn't a CaCO3 powder have the same result? Edited the question to be more precise.

To an extent, it would, though it probably won't penetrate quite as deep. Then again, what is it that we want to keep? If it is a new document, just use the modern paper (which in these days is nearly always CaCO3 infused). If it is a historical document, the archivists are very reluctant to alter them in any way.

Preserving paper by deacidification is not a simple process as you have to infiltrate the paper with the antacid component
The basic strategy of putting some sort of buffer against acids to preserve paper against degradation is sound. But it isn't easy to achieve. Using carbonate powder won't work as the powder won't infiltrate the substance of the paper where the degradation occurs (and chalky books will be messy to handle).
But there has been a lot of work of various processes to overcome this limitation. This Wikipedia page summarises some of the known processes but also their limitations (one example originally promoted by NASA used diethyl zinc to infiltrate paper but it turned out to be a very dangerous process to manage as diethyl zinc is dangerous and spontaneously flammable with water and air). But a variety of other commercial processes exist. Some of them seem to work better than others as all have some side effects on the preserved materials (eg there is no point preserving the paper if the process damages the ink!)
But some processes seem to work well. The one described here, using aminoalkylakoxysilanes even seems to improve the mechanical strength of the paper with few side effects.

The vast majority of books aren't historically significant for their book form, but rather for their content. So another approach is to computer scan the books and reduce them to a computer file.

To answer your question, yes on calcium carbonate layer but its surface action is just limited to pH control. Also, consider baking soda, which may add more than surface pH control due to its solubility.
Some other ideas to address other concerns as well, which may be present in more tropical locales, including threats from things like fungus, bacteria, worms,.. may I suggest adding a layer of basic copper carbonate ($\ce{Cu2(OH)2CO3}$) which I would create as a nano-suspension. Per Wikipedia on basic copper carbonate, to quote

It also has been used for many years as an effective algaecide in farm ponds and in aquaculture operations.

And, it can also address acidic pollution as would CaCO3:

Basic copper carbonate is decomposed by acids...

Only one concern is that upon significant acidic dissolution over time into an aqueous copper salt, a possible pigment development.
Interestingly, per a source, to quote:

Bacteria, yeasts, and viruses are rapidly killed on metallic copper surfaces, and the term “contact killing” has been coined for this process.

So, one may also consider just adding a layer of copper metal (in the form of a copper nanoparticle suspension by fast methods, which may be transformed into the basic copper carbonate with time (but still potentially, but now more minor, staining issue). However, copper metal is but one choice along with other metals (and their oxides, like nano-zinc oxide) and even precious metals like nano-silver (see this source, Silver nanoparticles as an effective disinfectant: A review) with related properties (and less potential staining).


text:
CO2 dynamics-vector correlations
I would like to ask if anyone has any information (schematic diagram)of vector correlations for CO2 molecule. (Is it bending in the excited state with μ in plane? How CO fragment can be ejected after the break?)
Thank you in advance

What's μ? Also, there is more than one excited state, and more than one mode of oscillations.

μ is dipole moment. I am studying photodissociation of CO2 so I was wondering what is happening if I give a specific amount of energy and break the bond.

You will need a potential energy surface for the (excited?) state you intend to dissociate from. I would imagine that some of these have been calculated although I can't give you a reference.

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text:
What can we learn from the energies of molecular orbitals?
Molecular orbital theory is successful in calculating the structure of molecules (minimizing the total energy with respect to atomic positions). The solutions of an MO calculation are useful in deriving molecular properties such as dipole moment or vibration frequencies. These do not rely on energies of electrons in distinct orbitals, i.e. even though the theory is based on combining atomic orbitals to molecular orbitals, the somewhat arbitrary separation into orbitals is not used in these calculations.
On the other hand, many conceptual explanations of chemical properties and reactivity are based on orbital energies, such as HOMO-LUMO considerations, rationalization of electronic spectra, nucleophilicity etc. This answer suggests that the orbital energies of occupied orbitals are somewhat arbitrary and those of empty orbitals are almost meaningless. In the answer to a different question, there is a claim that you can apply orbital energies in just one way, and there is no guarantee how good the answer will be:

Basically the only interesting quantity that you can get from an orbital energy is the first ionization energy based on Koopmans' theorem, which also is just approximately true.

So, what can you do with the orbital energies (either the numeric values or perhaps the ranking by energy)?

There is nothing arbitrary about the orbital energies. It is reference point that is arbitrary, but that's not a big deal.

Not a full answer, but some general thoughts:
In general, what we can learn from individual orbitals (and their energies) is very limited, since orbitals are just an abstract tool to represent the many particle electronic wave function. This is manifested by the need for methods beyond Hartree-Fock (DFT, post-HF, etc) and the fact that the orbitals in these methods become even more abstract: Kohn-Sham DFT considers the orbitals to be non-interacting and moves the electron-electron into an external potential. In post-HF the ansatz for wave functions is extended in different (yet similar) ways, but ultimately still based on orbitals. Thus, an exact theory does not consider individual orbitals, but always combines them. In the context of such a theory, it is thus very hard to relate a molecular property to orbital energies only.
However, there still seems to be some rough correspondence, as MO diagrams are widely used to explain things like stability and reactivities. But I think this is all just empirical. Also, the fact that orbitals are an approximation means it does work on some scale. For example, we can distinguish between core and valence electrons based on the orbital energies, as those are typically well separated. (the cutoff energy may however depend on the system). But arguing which valence electron is more strongly bound is more difficult, as their orbital energies are more similar. But this is not really helpful, since we can separate core and valence electrons without applying quantum mechanics.
But we can extend that idea a little bit: Looking at differences in orbital energies, we can estimate the distribution of the electrons and predict a high-spin or low-spin situation of unpaired electrons (e.g. in transition metal compounds). Or the stability of radicals. Or the stability of bonds, e.g. in $\ce{C=C}$ bonds, the $\sigma$ bond is more stable than the $\pi$ bonds. We can even explain the more stable bonding $\ce{N2}$ compared to $\ce{O2}$, although we are comparing two systems with different nuclear charges and number of electrons here. But these arguments mostly work without looking at the actual numbers for orbital energies. It is mostly just qualitative.
Ultimately, we still know something about the orbital energies that we can use. The values are just very fuzzy (high uncertainty, large error bars), therefore we have to be careful with any conclusion. This may actually be the reason for how a large part of chemistry actually works: Often there are concepts which kind of work, but are not applicable in every situation, e.g. the HSAB concept. There may be other competing theories which cover other situations, but a full exact theory (other than elaborate quantum mechanics) rarely exists. In chemistry everything is a bit fuzzy and often it comes down to just know or test the properties of a certain compound. Physics on the other hand can be much more systematic.

When is an answer ever complete? Thanks for your thoughts.

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Covalency of metal oxides in high oxidation states
So, metal oxides are generally basic, right? But in higher oxidation states their covalent character is predominant. For example in
Mn2O7 the Mn - O bond is covalent. Why is that?

Metal oxides are not always basic.  In many cases where the bonding is strongly ionic, the oxide ions do act as a base.  However, if the metal is in a high oxidation state an ionic bond would require a prohibitively high charge on the metal ion; the ionization energy required to put seven positive charges on one manganese atom far exceeds what can be reclaimed from the electron affinity of oxygen and electrostatic attraction alone.  The energetically preferred option for $\ce{Mn2O7}$ is to form polar covalent bonds instead.  Such bonds are similar to those in $\ce{CO2, SO3, P4O10,}$, etc., and thus $\ce{Mn2O7}$ will act similarly to those oxides -- as an acid anhydride.


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What exchange timescales are needed to cause decoupling in NMR?
I'll try to explain the source of my confusion as clear as possible:
(*) The methyl protons in toluene are decoupled because the C-C bond between the methyl group and benzene allows very fast rotation. Hence, on average, these protons will see the same environment.
It is said that the time scale for the rotation must be faster than the time scale of the instrument. For a 250 MHz apparatus this would be (1/250MHz)=4 nanoseconds.
(*) In impure solvent the methylene protons of ethanol don't couple to -OH,  since fast exchange of the OH protons results in decoupling and the protons will see an average environment.
Here it is said that this occurs if the timescale of exchange must be faster than the coupling constant J for the coupling. This is typically around 5-10 Hz. So in this case, if the exchange is in the 100 millisecond timescale decoupling already occurs.
This utterly confuses me, in the first case averaging requires nanosecond timescales (related to the frequency of the instrument) and in the second case almost second timescales (related to the small difference in frequency between coupled peaks)? Why?
ALSO and maybe related to this, why do we see in the spectrum of ethanol peak broadening for the OH proton, but not for the methylene group?

The first statement ist imo wrong. With a good enough spectrometer, you should be able to see the $^4J$ coupling between ortho and methyl. Unless the rotation would average the coupling to zero. In ethanol, the methyl group rotates quite as fast, and you get a fine $^3J$ coupling.

If a teacher says this averages on the NMR time scale, he either means bear with me for the moment, i´ll come back to it later or you all know what I mean, let´s not split that hair again, and/or he hasn´t got the foggiest idea why exactly. :-))

You need to differentiate what gets averaged, a CS difference or a scalar coupling.
Methyl protons have a different chemical shift depending on the current conformation, but they rotate so fast this gets averaged out perfectly, and they have a very sharp signal. They show no coupling between themselves because it´s an $AB_2$ system with CS difference zero, so $A_3$ actually.
$J$ coupling to a methyl group (e.g. the ethyl group in ethanol) gives sharp lines and clear splitting, because the coupling varies very rapidly (with the methyl rotation) around a stable and finite value, and it only depends on the occupation of the spin states in both groups.
The coupling methylene to OH is inverted every time the proton gets exchanged for one with different spin state, so it averages to zero. The exchange happens so often that the methylene group still gives a sharp quartett from the methyl.
The OH group coalesces with the water peak. Because the exchange happens randomly and differently for each molecule, the coherence in your spin ensemble deteriorates, hence the line broadening. Note: It´s not a splitting that get´s averaged out, but a change in chemical shift.
If the proton exchange was even faster, the line would sharpen again at some point I guess. Just like the methyl protons themselves. If the exchange was even slower (milliseconds), the line would broaden so much it became invisible. If you have a really dry sample, the exchange becomes so slow it does not happen on the NMR time scale, the line becomes sharp, and you even see the $^3J$ coupling to the methylene group.

Dear Karl, I see what you mean between the difference between CS and coupling in these two examples. Nonetheless I think my main confusion is, what is meant with the NMR timescale? Quite often this term is used, it is fast 'compared to the NMR timescale'. But in the case of rotation it is claimed in a text I've read the timescale is nanoseconds, for exchange milliseconds. That's 10^6 difference!

In the mean while I've found this blog: http://u-of-o-nmr-facility.blogspot.com/2008/08/nmr-time-scale.html. Here the time scale is also linked to the difference in frequency between the two peaks (ie in Hz), and not to the frequency of the spectrometer (ie in MHz). I conclude thus that my first statement was wrong..?

The timescale of one NMR experiment is between the dwell time (time distance between two points in your FID) and the length of the FID. Generally, it can be between two times the Lamor frequency (for double quantum NMR) and five times the longest longitudinal relaxation time, i.e. hours in extreme cases. The term is not exactly well defined, per se. ;)

The maximal resolution (frequency difference between two points in the Fourier transform) is the inverse of the length of the FID recorded, no matter how many points you sample in that time. The spectrum just gets broader.


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What does it mean to accept or donate a proton?
I've been trying to find out the answer to this question but can't seem to find it anywhere. I would like a very simple explanation.

Give me 100$, and I will explain you.

Think of it kind of like mugging: the mugger (base) “accepts” a wallet (proton) from a victim (who had a relatively easily swiped wallet). So the victim “donates” the wallet to the mugger.

@Ed Great minds think alike, they say.

Just to be clear, something acting as an acid in water cannot just have a bond break, spontaneously and without an agent, resulting in a proton being tossed out into the solvent! Breaking a bond takes energy and the separation of the positively charged proton, and the negatively charged remainder, is not energenically favored! The moiety that takes the proton, i.e., the base, actively takes the proton from the acid moiety! Strong acids are easy prey: even water, as pathetic as it is as a base, is almost 100% effective in taking protons from strong acids in water.

Weak acids actually are hard to steal from: a weak acid in water is mostly intact. At any given instant in time, only a small percentage of weak acid in water exists as the conjugate base, i.e., the remainders after protons have been taken by water molecules. Most of the weak acid exists as molecules of the weak acid, just like when it was added to water. So weak acids are actually bad mugging victims: the large majority, at any given time, hold on to their protons when they are in water. Bases take protons from acids and acids do not just toss protons to the solvent crowd!

Donate a proton (acting as an acid)
A covalent bond with hydrogen breaks, hydrogen leaves without electrons (as a hydrogen ion, sometimes called proton as a shorthand), and the two electrons from the bond remain with the remainder of the substance (typically in form of a lone pair).
Accept a proton (acting as a base)
A substance makes a new covalent bond with a hydrogen ion, providing both electrons (typically, a lone pair turns into the bond with hydrogen).


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Why does the electron affinity increase become more exothermic down group 2 and group 5?
It is generally true that the electron affinity becomes less exothermic down a group, because of the increase in atomic radius.
There is a well-known exception that the electron affinity of Cl is more exothermic than that of F, because fluorine has a high electron density; adding electrons would be unfavorable because of the electron-electron repulsions.
In Group 2 and Group 5, this is more extreme: the electron affinity becomes more exothermic down the group. Why is this so?

I can tell why it's positive for Be, Mg and N, which is quite clear. I can't explain the "trend"...

There is no clear tendency anywhere. The only column where the electron affinity is regularly modified the way you say when going down the column is the first one.


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Acidic and basic salts with ph indicator
Is magnesium sulfate an acidic, neutral, or basic salt? Which color will it show with the universal ph indicator?
In my opinion it needs to be slightly acidic hence turns indicator yellow

Which pH indicator? There are dozens of them...

Universal indicator

From the Merck Index, I see that magnesium sulphate is neutral in water : pH 6 - 7.

It's a pity nobody else can see this. Could you please provide an URL or a complete reference, if you are using a paperback version?

Here is some info in Pubchem


text:
What exactly are the frontier orbitals in diradical species like O2?
I read about frontier orbitals, namely HOMO, LUMO. But what's exactly HOMO/LUMO? I know HOMO stands for highest occupied molecular orbital and LUMO stands for lowest unoccupied molecular orbital.
But I am not getting which orbital is to be tagged as HOMO, and which as LUMO. Consider the example of $\ce{O2}$, highest occupied MO is $\pi^*_\mathrm{2p}$ and lowest unoccupied MO is also $\pi^*_\mathrm{2p}$. Which orbital is HOMO/LUMO? Is $\pi_\mathrm{2p}$ the HOMO and $\pi^*_\mathrm{2p}$ the LUMO? If so, why? There is a similar case with $\ce{B2}$.

O2 is a bad example to consider. As for the bonding/nonbonding part: typically yes, but not always.

@IvanNeretin what about $\ce{B2}$?

That's better. In this case I expect $\sigma2p$ for HOMO and $\pi2p$ for LUMO. Note that the LUMO is bonding.

@IvanNeretin is it $\sigma$2p or $\sigma$2s or $\sigma^\ast$2s?

OK, looks like I got it wrong. π2p sits below σ2p and hence is half-filled, which technically makes it both HOMO and LUMO.

It's important, that these labels only work within the framework of MO theory. There they are straight forward. The orbital highest in energy which is occupied is the HOMO and it does not matter how many fold the degeneracy is. The immediately following orbital would be the LUMO, which is unfortunately not always well defined. The case of bonding and antibonding is another story, when you venture out to more than two atoms one orbital can be both. Frontier Molecular Orbital Theory is a very helpful approximation, but it is quite crude.

For O2 and other radicals, it may be wise to use the term SOMO (singly-occupied molecular orbital), instead of HOMO/LUMO

Truthfully, there are two questions here which are not very closely related and could stand as separate Q + A's. The first is how are the HOMO/LUMO defined in the case of a diradical like O2. The second is their bonding / antibonding character.

@orthocresol so, what should I do?

Since the majority of what you have written pertains to the first question, I suggest removing the second and asking it separately. But I also would not be surprised if it has been already been asked and answered. You could do a quick search.

@orthocresol I searched the second part, couldn't find one on SE, if you could tell the link.

at first glance: Bonding and Antibonding Why do bonding and anti-bonding states split in energy (mathematically)? Maybe How can two orbitals constructively and destructively interfere simultaneously? This: How can a molecular orbital with a nodal plane still be considered “bonding”? (!!)

@Martin - マーチン none of them answer particularly relation between HOMO/LUMO with BMO/ABMO. Also this post ain't getting reopening votes (2 till now, one of them is mine), what should I do?

That isn't really that surprising. The concept of bonding and anti-bonding orbitals is independent from the concept of HOMO and LUMO. It does - especially in Organic Chemistry - often get mixed up and non-bonding orbitals are introduced, too, but that's too much for the comments. I'll have to think about the Question, and we should continue conversations in [chat].

@Martin - マーチン I am working on vector analysis right now, I would be grateful if you could spare some time from your busy schedule, tomorrow and (or in) succeeding days. Also, in which room will you be most comfortable?


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How to solve for the amount the equilibrium was disturbed with an ICE table?
On a recent quiz, we were asked the following question:

$$\ce{A(aq) + B(aq) <=> C(aq)}$$
  At $\pu{298 K}$, the equilibrium amounts of each of the component are as follows: $[\ce{A}] = \pu{0.125 M}$, $[\ce{B}] = \pu{0.400 M}$, $[\ce{C}] = \pu{1.200 M}$. At $\pu{298 K}$, the system is perturbed by increasing the concentration of $\ce{B}$ by adding $X$ amount. When equilibrium is re-established, the concentration of $[\ce{C}] = \pu{1.250 M}$. The volume of the system remains unchanged. How much $X$ (in $\pu{M}$) was added to the system?

The correct answer for this is $\pu{0.344 M}$, but I'm struggling to understand the setup for how the ICE table would work, since my original construction would mean the $X$ value is just $\pu{0.05 M}$:
\begin{array}{lccc}
\text{Stage}       & \ce{A (aq)}      & \ce{B (aq)}   & \ce{C (aq)} \\\hline
\text{Initial}     & \pu{0.125 M}     & \pu{0.4 M}    & \pu{1.2 M}  \\
\text{Change}      & -X               & +X            & +X          \\
\text{Equilibrium} & \pu{0.125 M} - X & \pu{0.4 M} +X & \pu{1.25 M} \\
\end{array}
But this does not seem to be right. Have I set up my initial values incorrectly? Any help would be appreciated.

The change for B should be -X, just like for A, the other reactant. And, as given in one of the answers, you know what X is from the last column.

This exercise is horrible, it uses terminology which is incorrect. You have been given equilibrium concentrations not amounts. Then the equilibrium is perturbed by adding $X$ amount which is again supposed to be a concentration, but the volume stays the same. That question was clearly posed by someone who has never done (and/or studied) chemistry. Please add a source (citation, not just a screenshot) for this garbage exercise.

The key insight here is that the amount added, "$X$", is NOT the typical "$x$" in a standard ICE table. The typical $x$ represents the changes in concentration toward equilibrium. The disturbance $X$ is a perturbation away from equilibrium. The initial state of your ICE table should be the state at which you will begin moving to equilibrium, i.e., with $X$ already added to [B]. Try starting with [B]+$X$ as your initial concentration for B in the ICE table and proceed as usual (I encourage you to actually try it this way). And to find the value of $X$ just use the information given about how [C] increased, as I'm sure you had no trouble doing.

Ah, I suspected something like that may have been at play. Thank you for explaining the theory and clearing up that ICE table is only once you start moving towards equilibrium. I'll keep this in mind for the exam!

Calculation of $K_c$ using initial concentrations
$K_c = \dfrac{1.2}{0.125 \times 0.4} = 24l mol^{-1}$
+-------------+-------------+--------------+----------+
|    Stage    |    A(aq)    |  B(aq)       | C(aq)    |
+-------------+-------------+--------------+----------+
| Initial     | 0.125M      | 0.4M + X     | 1.2M     |
| Change      |    -y       |    -y        |   +y     |
| Equilibrium | 0.125M  - y | 0.4M + X - y | 1.2M + y |
+-------------+-------------+--------------+----------+

Final concentration of C = 1.25 M, $\therefore$   y= 0.05 M. Since $K_c$ won't change, thus
24 = $\dfrac{1.25}{0.075 \times (0.35 + X)}$
$\implies$ X = 0.3444 M

The way the change row is usually interpreted, it should reflect changes due to the chemical reaction, so every species should change according to stoichiometry. You could have the initial B at 0.4M + x, and have -0.05, -0.05, +0.05 as change. The equilibrium row is correct, so you get the correct answer.

This is not really an answer, it's just a table without much of an explanation; I can't follow what is presented here. We have given up the stance on only providing hint answers for homework-like questions. It is overall better to have a worked example with no questions left to ask, than a hint which produces even more questions. (This was flagged NAA, so the above was a service announcement.)

@Martin - マーチン I updated my answer, is this format correct?

It's better, but still not good. Please never ever omit units. Also, while some people think and speak in formulas, quite a few don't. Those who do are usually not asking such kind of questions. Therefore, adding explanations would be a good idea.

@Martin - マーチン questions like this have little to explain, I don't know how to explain the working. If you could help me by editing the answer.....

I find questions like these (I mean the original exercise, not the one by OP) awful, because they don't actually make any sense. Please see my comment to the question. In general, there is much, much to explain. You should start with writing the reaction equation again, then start deriving the general formula for the equilibrium constant, then explain the variables used (and never ever, not in a single instant omit any unit) and then insert them and do the calculation. Explain how to set up the table. And please, write actual sentences and nut just phrases.


text:
1H NMR splitting of 3-methyl-1-butanol
I have a question about 1H NMR splitting of 3-methyl-1-butanol.

The spectrum looks as the following.

The assignment of the NMR spectrum is the following.
Assign.     Shift(ppm)

  A             3.673
  B             1.66
  C             1.57
  D             1.49
  E             0.922

So based on the results, A hydrogen is splitted into triplet. Is this splitted by D hydrogen? If so, why not C hydrogen on the alcohol group?
The results are retrieved from the following link: https://www.chemicalbook.com/SpectrumEN_123-51-3_1HNMR.htm

The integration shows that the most upshifted signal has two hydrogens and the NMR table also agrees to show that H-C-OH shows a signal near 3.4ppm.

There is no OH signal visible here. And I´m guessing this is a fake spectrum, unless they took D2O as solvent.

Well, it's possible that OH is in this multiplet if solvent is non-polar like CCl3D. If the solvent was D2O there still would be peak from HDO.

There is no residual chloroform peak so it is not in CDCl3, the quoted reference does not say what the solvent was.

In this example, HA is indeed split by HD, but you have quite rightly identified that we don't see coupling to the alcohol proton, HC. This is pretty normal that we don't observe couplings to protons that are termed labile; -OH, -NH2, -SH, for example. Whenever samples contain trace amounts of an acid or base, there will be very fast exchange between the alcohol proton and that other species (water is often the culprit). In solvents like chloroform (which this spectrum was run in), the exchange is very fast, and you just observe a single peak. In other solvents, such as dmso, you might observe two broadened peaks. But, the exchange means that any coupling to that nucleus is lost
We do sometimes observe coupling to alcohol protons. If you can eliminate exchange, or slow it sufficiently, you will observe 3J couplings as you would for any other nuclear coupling pathways. 3J couplings to alcohol protons are in the order of 3-6 Hz, commonly.


text:
Is equilibrium constant at a given temperature derived from Gibbs free energy of reaction valid for doing simple kinetic modelling?
I intend to do a kinetic study of simple alcohol catalytic dehydrogenation reactions in the gas phase. I want to start with simple power law kinetics using $K_\mathrm{eq}$ to account for the reversibility of the reactions. Would the equation below, knowing the Gibbs free energy of reaction, be valid?
$$Δ_\mathrm{r}G^\circ = -RT\ln K_\mathrm{eq}$$
I came across a paper that used this equation to calculate $K_\mathrm{eq}$ to build a kinetic model of the ethanol-to-butadiene reaction. However, other papers I read, i.e. on the methanol-to-diethyl ether reaction, have used complex equations derived from empirical observations to calculate $K_\mathrm{eq}.$ Why use one approach over the other?

In general I would say 'no', thermodynamics only tells us about starting and ending points, thus you know the $\Delta G^\text{o}$ for the reaction but nothing about its actual mechanism, (since time does not come into thermodynamics). As $K_\mathrm{eq}$ is a ratio of rate constants these can take on many different values and still have the same ratio. To study the mechanism you will need to monitor species and intermediates vs time as the reaction proceeds.
Before doing any experiments you can only make informed guesses as to what the mechanism might be, even apparently simple reactions, such as between hydrogen and oxygen, is not simple but by experiment found to be a chain reaction.

May I ask you for the reason for reverting my edit? \circ in ΔG^\circ is not a typo, it's one of the two standardized ways of typesetting standard state. \Delta G^\text{o} is semantically wrong and its usage can only be justified if there is no technical means to input proper symbols.

Nothing in particular you can change it back if you wish. Whatever the standard way is, using \circ in the font used makes a rather small symbol that does seem awkward when reading.

I don't see how it would be incorrect to do what the asker posted, assuming the reaction is at the standard temperature. By definition, it would yield exactly  Keq. The real issue at hand is that Keq is not a good metric of the kinetics of the system.

At equilibrium, the forward and reverse rates have to be equal. For mass action kinetics, this tells you that the reverse rate constant must be equal to the forward rate constant divided by the equilibrium constant.


text:
Acidity order checking for α-furoic acid and acetic acid
α-Furoic acid's conjugate base is aromatic and hence is more stable. Then why acetic acid is more acidic than α-furoic acid?

The conjugate base of 2-furoic acid is aromatic, but so is the acid.
'
The aromaticity does not contribute to the stability of the conjugate base because the negative charge from the carboxylate anion cannot delocalize into the ring. The major resonance contributors for the carboxylate anion are:

However, because of the oxygen atom, furan is an electron-rich arene, and one position that has higher than expected electron density is the 2-position:

Furan is therefor electron-donating, and electron-donating groups destabilize anions. Consider the following resonance contributor of 2-furoic acid and the close proximity of the formal negative charges:

Can I say that cross conjugation due to O wins over aromaticity in this case?


text:
What is the largest, noncrystalline, non-polymer inorganic molecule?
A question came up in class from one of my chemistry students. He asked what the largest, non-crystal, non carbon-based, molecule is. We ruled out complex proteins, DNA, and other polymers, as well as crystalline structures and metals from the list. I didn't have an answer for him, so I looked it up but could not find a definite answer. I found two threads on Reddit, listed below. Polyoxometalates come up, but without any specifics. I know "largest" can be subjective, so please specify if the molecule is "largest" because of its physical size or because of its mass.
https://www.reddit.com/r/chemistry/comments/2phl4s/whats_the_largest_inorganic_molecule_you_know/
https://www.reddit.com/r/askscience/comments/i6v07/what_is_the_biggest_noncrystalline_inorganic/

I noticed someone down-voted my question. Can someone tell me why?

You didn't exclude amorphous substances... Another thing is what you call a polymer? Polyoxometalates already have quite repetitive structure. If you link together some big fullerenes, at which point it would become "a polymer"?

I agree with the redditors that the answer is likely a polyoxometalate.
$\ce{Na48[H_xMo368O1032(H2O)240(SO4)48] · \text{ca.}~1000 H2O}$ aka
$\ce{\{Mo368\}}$ (core: $M \approx\pu{60.7 kDa},$ dimensions $\approx \pu{2.5 nm} × \pu{4.0 nm})$ is the primary candidate [1]:

Figure 2. Structure of 1a in crystals of 1 in polyhedral (a) and ball and stick (b) representation perpendicular to the $C_4$ axis and along the $C_4$ axis ((c) and (d); building units $\ce{\{Mo1\}}$ yellow, $\ce{\{Mo2\}}$ red, $\ce{\{Mo(Mo5)\}}$ blue with blue-turquoise pentagonal bipyramids; $\ce{O}$ atoms small red spheres, $\ce{S}$ atoms gray spheres).
Note that one capping part is turned against the other by $360/16^\circ.$
The different building blocks and the arrays of different local symmetry are assigned.

A recent review of mass-spectrometry studies of large polyoxometalates is also available [2].
References

Müller, A.; Beckmann, E.; Bögge, H.; Schmidtmann, M.; Dress, A. Inorganic Chemistry Goes Protein Size: A $\ce{Mo368}$ Nano-Hedgehog Initiating Nanochemistry by Symmetry Breaking. Angewandte Chemie International Edition 2002, 41 (7), 1162–1167. DOI: 10.1002/1521-3773(20020402)41:7<1162::AID-ANIE1162>3.0.CO;2-8.
Surman, A. J.; Robbins, P. J.; Ujma, J.; Zheng, Q.; Barran, Perdita. E.; Cronin, L. Sizing and Discovery of Nanosized Polyoxometalate Clusters by Mass Spectrometry. J. Am. Chem. Soc. 2016, 138 (11), 3824–3830. DOI: 10.1021/jacs.6b00070.

Are these things really molecules?

@Alchimista. Yes, they are. I'm not sure about this one, but majority of polyoxometalates is also stable in appropriate solution, and some even in gas phase.

Isn't the prefix "poly" a give away vs the question?

@Beerhunter "Poly-" in polyoxometalates (POMs) usually refers to the presence of more than three metal centers linked via oxo-bridges. Under given conditions, one usually doesn't get all sorts of molecules/chains/blobs etc. of POMs, only one type of species. There is no molecular mass distribution to speak of, as one would, for example, in case of traditional organic polymers.

@andselisk so the nomenclature states that if its three or more,  tri, tetra etc aren't used and the reference is simply "poly"?

@Beerhunter Why? POM is a collective term, and "poly" preserves original meaning (many). For example, you can name cyclic anion $\ce{[V4O12]^4-}$ equally as tetravanadate(V) or polyoxovanadate in general. If you are interested in inorganic nomenclature of POMs, take a look at the IUPAC Red Book, namely Nomenclature of Inorganic Chemistry II: Recommendations 2000 (ISBN 978-0-85404-487-0). This and many other cases are covered in there. Don't try to superimpose organic polymers with POMs, it doesn't work like that. If you are looking for the organic analogy, then it's proteins.


text:
Why does Tyndall scaterring produce conical shape?
When light is passed through a collidal solution, due to Tyndall scattering, the path of light appears to be a bright cone. Why does it have to form only a conical shape?
The conditions for Tyndall effect are that the diameter of dispersed particles should not be much smaller than wavelength of light used and refractive index of dispersed phase and dispersed medium should differ greatly.
This is perhaps because colloidal particles scatter light in all directions. Normal refraction shouldn't cause such a shape.
Is there any logical reson for the conical shape?

Source: NCERT class 12

Any chance you could post a photo or reasonable figure to illustrate the effect?

I have added a picture.

The answer is fine, but the reason your figure shows a conical shape is that the double convex lens focusses the beam in that fashion so that the incident light intensity is high in the region being observed with the microscope. This just increases the Tyndall scattering light intensity.

One last thing: the illustration should not have mistakenly labelled the light cone as “Tyndall cone”. That is just nonsense.

It is simple.
Light always travels in a straight line and if a lens is present it would diverge or converge ( in a straight line ) accordingly.
The question is why no other shape like maybe a hemisphere or a paraboloid ?
Now let me ask the question, have you seen such thing?
Check out this diagram to make everything clear.
If the beam of light increases in width ( diverging beam) linearly with distance. It is because all the particles scatter equally in all directions and some of them reach your eyes and this reveals the shape of the beam. The adjacent particles does the same and so on.
Therefore you can also see the light getting more and more diffused linearly as more and more light is scattered.
It also depends on the refractive index of the medium where the incident light ray will either move closer to the normal or move further. Again this would lead to a conically diverging beam.
EDIT

This is to be noted that the colloidal solution does not affect the shape of the beam of light ( converging or diverging ).

This is also evident from your diagram.
From your question I feel that you think that the colloidal solution is responsible for the cone shape, it is not so. It is simply the geometry of the light rays
What the solution does is scatter the light in all directions and makes the "beam" visible.  A straight beam of light in perfect vacuum would not be visible as a stalk of light ( that we see when light enters a dark room thought a window )

Check the Edit @user600016


text:
Is it correct to put a methyl group on each end of a skeletal formula?
On my organic chemistry exam there was a skeletal formula that had, supposedly one methyl group at each end of the skeletal formula and one somewhere in the middle.

But is it correct if you draw a formula like that and then put those methyl groups on each end? I feel like it's very misleading because I thought that those were just the ends of the main chain, so not methyl groups. I lost a lot of marks on that question because of this confusion and I just really want to know if it's even correct to write it that way.

Please just answer this question, because if this is an incorrect way of using the skeletal formula then I could have the possibility to pass this exam.

Do you think that the drawing shows something different than 3-Methylhexane?

But the ends of the main chain are methyl groups.

Here, all three terminal carbons are also methyl groups.
Carbon labels may be omitted or displayed, e.g. for the straight angles or implicit hydrogen atoms.
Both structures 1 and 2 below are identical and represent 3‐methylhexane:

From IUPAC Recommendations 2008 [1, p. 290]:

GR-1.4 Terminal single bonds
As discussed in GR-2.1.2, unlabeled atoms are assumed to be carbon atoms, and so terminal single bonds are assumed to represent methyl groups. Unlabeled bonds should not be used to represent unspecified or variable attachment points (see GR-9), as such diagrams are extremely prone to misinterpretation.

Further [1, pp. 300–302]:

GR-2.1.2 Labeling of carbon atoms
[…]
It is acceptable to add labels for terminal carbon atoms connected to unlabeled carbon atoms, but only when it is possible to do so without overlapping other portions of the diagram.

When ethane, ethene, ethyne, and related molecules are drawn with only one explicit bond, both terminal carbon atoms must be labeled explicitly to prevent the molecule from being interpreted as a stray line or set of lines.

References

Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). Pure and Applied Chemistry 2009, 80 (2), 277–410. DOI: 10.1351/pac200880020277. (Free Access)


text:
Computing the commutator of the Fock and density matrices in AO basis (Hartree-Fock theory)
I am making a simple restricted HF code using the Python interface of Psi4. I am now evaluating convergence by tracking the change in the sum of orbital energies, but I want to do this in a better way. It is common to use the fact that at self-consistency, the Fock and density matrices commute
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} - \mathbf{DF} = \mathbf{0}
$$
However, the above expression is only valid in MO basis, while F and D in my code are computed in AO basis. So I need to derive an equivalent expression in AO basis. I am quite sure the correct expression is
$$
[\mathbf{F}, \mathbf{D}]^{\text{AO}} = \mathbf{FDS} - \mathbf{SDF}
$$
as this is equal to zero to within $1\times10^{-14}$. But how to derive this?
Derivation
An arbitrary molecular orbital $\phi_i$ is expanded in atomic orbital basis functions
$$
\phi_i = \sum_\alpha C_{\alpha i} \chi _{\alpha}
$$
Acting the commutator on $\phi_i$ and expanding it to AO basis yields
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha}  - \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$
Since we know the solution contains the overlap matrix $\mathbf{S}$, lets look at the definition
$$
\mathbf{S}_{ij} = \langle \chi_i(\mathbf{r}) \vert \chi_j(\mathbf{r}) \rangle = \int d\mathbf{r} \chi^*(\mathbf{r})\chi(\mathbf{r})
$$
Since this must be part of our expression, it seems to me a good approach is to multiply from the left by $\sum_\beta C_{\beta i}^* \chi_\beta^*$ (dropping the $\mathbf{r}$ dependence from now on) and integrating over $\mathbf{r}$
$$
[\mathbf{F}, \mathbf{D}] = \int \sum_\beta C_{\beta i}^* \chi_\beta^* \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha}  - \int \sum_\beta C_{\beta i}^* \chi_\beta^* \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$
from in braket notation becomes
$$
[\mathbf{F}, \mathbf{D}] = \langle \sum_\beta C_{\beta i} \chi_\beta \vert \mathbf{FD} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle  - \langle \sum_\beta C_{\beta i} \chi_\beta \vert \mathbf{DF} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle
$$
At this point I am not sure what to do - or if am I even on the right track. I can see that we have the "pieces" that make up the overlap matrix, but I don't know how to put them together. Further, due to the orthonormality of the MOs, then I can imagine that the summation terms only survive when $\alpha = \beta$. But I'm not sure how to derive this properly.

Slightly related: https://chemistry.stackexchange.com/questions/102144/is-the-lowdin-orthogonalization-used-in-diagonalizing-the-atomic-orbitals-really

In the context of a DIIS error function, Jensen (2nd ed., ISBN 1118825993) references your AO expression. He gives the following references: https://doi.org/10.1002/jcc.540030413 https://doi.org/10.1063/1.1470195

Start from the AO Hartree-Fock equation and its adjoint
$$\mathbf{F}^{AO}\mathbf{T}=\mathbf{S}\mathbf{T}\epsilon \text{ and } \mathbf{T^\dagger}\mathbf{F}^{AO}=\epsilon\mathbf{T^\dagger}\mathbf{S}$$
where $\mathbf{T}$ is an $N\times n$ matrix that is essentially the occupied block of $\mathbf{C}$ ($n$ is occupied, $N$ is total orbitals). We use this $\mathbf{T}$ matrix because it has the convenient property that
$$\mathbf{T}\mathbf{T}^\dagger=\mathbf{D}^{AO}$$
Now, we can multiply the HF equation by $\mathbf{T^\dagger}\mathbf{S}$ on the right and multiply its adjoint by $\mathbf{S}\mathbf{T}$ on the left, which gives
$$\mathbf{F}^{AO}\mathbf{T}\mathbf{T^\dagger}\mathbf{S}=\mathbf{S}\mathbf{T}\epsilon\mathbf{T^\dagger}\mathbf{S}$$
$$\mathbf{S}\mathbf{T}\mathbf{T^\dagger}\mathbf{F}^{AO}=\mathbf{S}\mathbf{T}\epsilon\mathbf{T^\dagger}\mathbf{S}$$
Subtracting the first equation from the second and yields the desired commutator relationship.
$$\mathbf{F}^{AO}\mathbf{D}^{AO}\mathbf{S}-\mathbf{S}\mathbf{D}^{AO}\mathbf{F}^{AO}=\mathbf{0}$$
The notation I use here is based on a similar derivation given in Chapter 6 of McWeeny's Methods of Molecular Quantum Mechanics, 2nd edition.

If $\mathbf{B}$ is the "normal" coefficient matrix in the Roothan-Hall matrix equation, then the density matrix we use in our calculations is $\mathbf{C}\mathbf{C}^{\dagger}$Is $\mathbf{T}$ only over the occupied molecular orbitals. Is this what is referred to as the "reduced density matrix"? The $\mathbf{T}$ matrix in your answer is not what we get when we diagonalize the fock matrix, right?

The density is just built from the occupied MO coefficients (see this tutorial). While you can solve the HF equation with C instead of T, the Fock matrix itself only depends on the occupied orbitals.

Related discussion on density matrices: https://chemistry.stackexchange.com/q/80351/41556

Starting from statement that the Fock matrix and the density matrix commute in an orthonormal basis.
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} - \mathbf{DF} = \mathbf{0}
$$
The orthonormal basis matrices can be substituted for their equivalents in an atomic orbital basis
\begin{align}
\mathbf{F} = {} & \mathbf{X}^\dagger \mathbf{F}^{AO} \mathbf{X} \\
\implies \mathbf{F}^{AO} = {} & \left[\mathbf{X}^\dagger\right]^{-1} \mathbf{F} \mathbf{X}^{-1}\\
\mathbf{D}^{AO} = {} & \mathbf{X} \mathbf{D} \mathbf{X}^\dagger \\
\implies \mathbf{D} = {} &  \mathbf{X}^{-1} \mathbf{D}^{AO} \left[\mathbf{X}^\dagger\right]^{-1}
\end{align}
where $\mathbf{X}$ is an orthogonalisation matrix.  :
\begin{align}
[\mathbf{F}, \mathbf{D}] = {} & \mathbf{FD} - \mathbf{DF} = \mathbf{0} \\
= {} & \mathbf{X}^\dagger \mathbf{F}^{AO} \mathbf{X}
\mathbf{X}^{-1} \mathbf{D}^{AO} \left[\mathbf{X}^\dagger\right]^{-1}
- \mathbf{X}^{-1} \mathbf{D}^{AO} \left[\mathbf{X}^\dagger\right]^{-1}
\mathbf{X}^\dagger \mathbf{F}^{AO} \mathbf{X} \\
= {} & \mathbf{X}^\dagger \mathbf{F}^{AO} \mathbf{D}^{AO} \left[\mathbf{X}^\dagger\right]^{-1}
- \mathbf{X}^{-1} \mathbf{D}^{AO} \mathbf{F}^{AO} \mathbf{X}
\end{align}
When $\mathbf{X} = \mathbf{S}^{-\frac{1}{2}}$, pre and postmultiplying by $\mathbf{X}^{-1} = \mathbf{S}^{\frac{1}{2}}$:
\begin{align}
\mathbf{X}^{-1} \mathbf{0} \mathbf{X}^{-1} = {} &
\mathbf{X}^{-1} \mathbf{X}^\dagger \mathbf{F}^{AO} \mathbf{D}^{AO} \left[\mathbf{X}^\dagger\right]^{-1} \mathbf{X}^{-1}
- \mathbf{X}^{-1}
\mathbf{X}^{-1} \mathbf{D}^{AO} \mathbf{F}^{AO} \mathbf{X}
\mathbf{X}^{-1} \\
\mathbf{0} = {} & \mathbf{F}^{AO} \mathbf{D}^{AO} \mathbf{S} - \mathbf{S} \mathbf{D}^{AO} \mathbf{F}^{AO}
\end{align}
as you have suggested.

While your derivation makes sense, the choice of $X = S^{-1/2}$ is not the only possible orthogonalization, and I think that the final equation does not depend on the choice.

The transformation of the Fock matrix into an orthonormal AO basis is fine. But it is not clear to me why a very similar transformation of the density matrix changes the basis from MO to AO (and vice versa), and not just between AO and orthogonalized AO basis.

Just noticed that you specifically asked about the MO basis rather any general orthogonalised basis.  Hence why I started in the orthogonal AO basis.  Will update to explain connection to MO basis shortly

@user213305 Did you update your answer?


text:
Aromatic Stretches C=C in Salicylamide
I want to ask a question about the C=C streching modes of Salicylamide.
I was presented with the following spectra and asked to comment on the bonds causing the peaks.

I identified the molecule from this and its NMR spectra to be salicylamide.

Here is my understanding of the IR spectra (solid state, ATR).

$3391.66 cm ^{-1}$ is the asymmetric stretch of the $\ce{N-H}$ in the amide.
$3184.80 cm^{-1}$ is the symmetric stretch of the $\ce{N-H}$ in the amide.

From previous understanding, I was aware that the asymmetric stretch requires more energy, and hence a higher wavenumber than the symmetric stretch.

$2736.27 cm^{-1}$ is most possibly the $\ce{O-H}$ peak of the phenol. I attributed this to a few reasons, possibly due to hydrogen bonding or due to the dissociation of the $\ce{O-H}$ bond, which is driven to the right hand side of the equilibrium by the presence of the EWG amide group.
$1672.98 cm^{-1}$ is due to the $\ce{C=O}$ stretch of the amide.
$1628.09$ and $1589.49 cm^{-1}$ are due to $\ce{C=C}$ aromatic stretches.

I tried to find information to help me describe the $\ce{C=C}$ stretches (i.e. what orientation, any dipole arrows) but I failed to find any information to support my argument.
I recall in ferrocene and acetalferrocene, the ring breathing effect is observed, whereby ferrocene has fewer IR peaks than acetalferrocene due to a lack of a net dipole moment being non-zero, and I wondered whether there was a similar vibrational mode to apply for these two $\ce{C=C}$ peaks, but I was not sure.
What are the directions of the C=C stretches in the molecule of salicylamide in the range $1580 - 1630 cm^{-1}$?

$\pu{2736 cm-1}$ is more like noise to me. Where do you expect peak for $\ce{C-H}$ (aromatic) stretching would appear?


text:
Major product of mononitration of 3-methylphenol
From NCERT Class Chemistry Part 2:

Example 11.5
Write the structures of the major products expected from the following reactions:
(a) Mononitration of 3-methylphenol
(b) Dinitration of 3-methylphenol
(c) Mononitration of phenyl methanoate.
Solution
The combined influence of $\ce{-OH}$ and $\ce{-CH3}$ groups determine the position of the incoming group.

Since $\ce{-OH}$ and $\ce{-CH3}$ are o,p-directing groups, shouldn’t the major product be 3-methyl-2-nitrophenol? Why not?

Weigh inductive, mesomeric and steric effects in the balance.

@SirArthur7 I can’t seem to get the answer right,can you explain it in detail?

In order for a nitro- group to be attached to a phenyl group, you must have an H atom available next to this carbon atom. If NO2 wants to attack the Carbon number 2, there is no H atom on Carbon 1 or 3. So the reaction is not possible.

@Maurice why is this necessary

@Jashanjot : The nitration reaction is due to the ion $NO_2^+$ attrcte

The product you are suggesting is very unstable as compared to the ones given as answers due to steric factors. Thus it would not be formed.

Since −OH and −CH3 are o,p-directing groups, shouldn’t the major product be 3-methyl-2-nitrophenol? Why not?

The reasoning that hydroxyl and methyl are o,p directing groups is not at supportive as to why 3-methyl-2-nitrophenol should be formed in preference to the ones given as answers. The o,p directing effect just rules out the meta product and steric factor decides among the remaining possibilities here.


text:
Can ICE tables give two solutions? What does this mean physically?
The following question comes from General Chemistry: Principles and Modern Applications, 11th ed. by Pettruci.

Exercise 15-33. A mixture consisting of 0.150 mol H2 and 0.150 mol I2 is brought to equilibrium at 445 °C, in a 3.25 L flask. What are the equilibrium amounts of H2, I2, and HI?
  $$\ce{H2(g) + I2(g) <=> 2HI(g)}\qquad K_c=50.2\text{ at }445\,\mathrm{^\circ C}$$

First, I found the concentrations of H2 and I2:
$$
\begin{aligned}
\phantom{}[\ce{H2}]=[\ce{I2}]&=\frac{0.150\text{ mol}}{3.25\text{ L}} \\
&=0.0461538\text{ M}
\end{aligned}
$$
Setting up the ICE table:
$$
\begin{array}{|l|c c c c c|}
\hline
&\ce{H2}&+&\ce{I2}&\rightleftharpoons&\ce{2HI} \\
\hline
\text{Initial}&0.0461538&&0.0461538&&0 \\
\text{Change}&-x&&-x&&+2x \\
\text{Equilibrium}&0.0461538-x&&0.0461538-x&&2x \\
\hline
\end{array}
$$
Then,
$$
K_c=50.2=\frac{(2x)^2}{(0.0461538-x)(0.0461538-x)}.
$$
Most solutions I've seen for similar problems involve taking the square root of both sides of this equation. However, they only take the principal root, effectively ignoring a second valid solution. Here, I'll do the proper solution for finding $x$:
$$
\begin{aligned}
 \\
50.2&=\left(\frac{2x}{0.0461538-x}\right)^2 \\
\sqrt{50.2}&=\pm\frac{2x}{0.0461538-x}
\end{aligned} \\
\begin{aligned}
7.0852&=\frac{2x}{0.0461538-x}&&\text{or}&7.0852&=-\frac{2x}{0.0461538-x} \\
0.327009-7.0852x&=2x&&&0.327009-7.0852x&=-2x \\
0.327009&=9.0852x&&&0.327009&=5.0852x \\
x&=0.0359936&&&x&=0.064306
\end{aligned}
$$
As you can see, there are two positive (and therefore, valid) solutions for $x$. However, only one of the solutions results in a positive $[\ce{H2}]_\mathrm{eq}$:
$$
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}=0.0461538-x
\end{aligned}
\\
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}&=0.0461538-0.0359936&&\text{or}&[\ce{H2}]_\mathrm{eq}&=0.0461538-0.064306 \\
&=0.0101602\text{ M}&&&&=-0.0181522\text{ M}
\end{aligned}
$$
And $x=0.0359936$ does indeed lead to the correct answer according to the textbook.
My question is: was it just a coincidence that I was able to eliminate one of the possible solutions for $x$? In other words, is there a scenario where I could end up with two possible $[\ce{H2}]_\mathrm{eq}$s? If not, what forbids this?
Edit: To reiterate, I am aware that one of the solutions is to be eliminated for being "non-physical". My question is whether it is possible for both solutions to be physical.

See this which is almost the same problem in reverse, and it's not a dupe of what you're asking. In short, in any case when you get a non-physical answer (such as a negative molarity) when solving an equation for more than one root, discarding it as you have is the correct thing to do.

I'm not sure how to prove it, but I think that a system with more than one equilibrium condition would imply very bad things, possibly all the way up to a perpetual motion device. Having two equilibrium concentrations effectively means that even after infinite time, a system still hasn't decided on its ground state.

In other words, is there a scenario where I could end up with two possible [solutions]?

No, there is always a single solution. The reaction quotient Q assumes values from zero (no products) to infinity (no reactants). It varies monotonically with the extent of reaction (the x in the ICE table). So there is always a solution (because Q covers the full range of possible equilibrium constant), and it is always unique (because for a given set of starting values, every value of Q is unique - follows from the monotonic variation).
The only exception is a process where none of the concentration vary with x (when all species are pure solids or liquids). In this case, Q takes a single value (except for running out of reactant or product). If that value is equal to K, x can take a range of values. If Q is different from K, there is no solution, and there won't be any equilibrium (i.e. the reaction would go to completion).

An ICE table is moderately complicated, so it ends up seeming a bit mysterious when you get two solutions.  But it's really not that mysterious at all.  In fact, this sort of thing is pretty common, since it happens anytime you have two variables related to each other through a square.  Let's use a very simple example to illustrate this.
Suppose you are given the following problem:
The area of a square is $\pu{25 m2}$.  What is the length of each side?
To solve this, we use $A = l^2$, where $A$ is the area, and $l$ is the length of each side. Hence:
$$l = \sqrt{\pu{25 m2}} = \pu{\pm 5 m}$$
Now clearly a side can't have a negative length — it's non-physical.  So we choose $l = \pu{+5 m}$ instead of $l = \pu{-5 m}.$  Nothing mysterious about that.  I.e., what you were encountering was merely a property of squares, not a property specific to ICE tables.

I understand that the second solution can be eliminated for being non-physical. I was wondering if there was a situation where both answers are physical and neither can be eliminated.

@Brandon I was mostly responding to your question "was it just a coincidence that I was able to eliminate one of the possible solutions for x?", since the question seemed to imply that there was something unusual (a "coincidence") going on to give you just one solution, and I was trying to explain that it wasn't a coincidence, but rather a natural outcome of having squares. Having said that, I will try to give some thought to addressing your other question.


text:
When an electron is excited to a singlet state, must it flip its spin state to enter the triplet state and flip once more to the ground state?
I understand the Jablonski diagram in that it has intersystem crossing from the singlet state to the triplet state, but how many spin switches are necessary to complete to phosphores? When an electron is excited to a singlet state, must it flip its spin state to enter the triplet state and flip once more to the ground state?

Yes. At least it ends up being reversed again. "Where that happens" escape my competence.

Yes two spin flips, one to form triplet and then again to reform the singlet ground state. See the answer here https://chemistry.stackexchange.com/questions/112814/how-fast-is-the-interconversion-triplet-%e2%86%92-singlet-in-organic-molecules


text:
Differences in efficacy of oxime reactivators in organophosphate poisoning
Compounds containing an oxime functional group, such as 2-PAM and obidoxime, are used in the treatment of poisoning with organophosphorus compounds. These oximes reactivate the enzyme acetylcholinesterase, which has been inhibited by the OPC.
However, some organophosphorus compounds are resistant to re-activation by oximes such as 2-PAM and obidoxime for reasons not related to denaturation of the enzyme (aging), but are responsive to some other oximes. For instance, methylfluorophosphonylcholine and related compounds such as GV are more susceptible to treatment by HI-6 than to obidoxime. Why?
Both of these compounds have fluorine act as a leaving group and have an amino-chain that would occupy the anionic site of the enzyme, so I presume that their resistance to reactivation by older oximes has to do with steric effects, though I have not been able to find research confirming or denying this. Nevertheless, why are some oxime compounds, especially the larger ones such as HI-6 and Hlo-7, better reactivators of AChE than others, especially if some OPCs' resistance to reactivation actually is due to steric effects?
References

Kassa, J.; Bajgar, J. Therapeutic Efficacy of Obidoxime or HI-6 with Atropine against Intoxication with Some Nerve Agents in Mice. Acta Medica (Hradec Kralove) 1996, 39 (1), 27–30. PMID: 9106387.
Worek, F.; Thiermann, H.; Szinicz, L. Reactivation and Aging Kinetics of Human Acetylcholinesterase Inhibited by Organophosphonylcholines. Arch Toxicol 2004, 78 (4), 212–217. DOI: 10.1007/s00204-003-0533-0.
Fredriksson, T.; Tibbling, G. Reversal of Effects on the Rat Nerve-Diaphragm Preparation Produced by Methylfluorophosphorylcholines. Biochemical Pharmacology 1959, 2 (1), 63–64. DOI: 10.1016/0006-2952(59)90058-9.


text:
Do all anhydrous organic matter burn?
All organic matter seems to be inflammable when dried up. Petroleum, Alcohol, Wood, and Coal burns readily. Can I call it a general rule? Are there known exceptions? What makes them the exceptions?

Freons don't burn.

@IvanNeretin - Freons aren't really "organic."

Loosely related: Are there substances that cannot undergo combustion?

@MaxW What's really organic, then? Does it have to be retrieved from a living body? Well, bones don't burn.

Yes all organic compounds can burn. Flammability is another issue. But in proper (hard) conditions everything containing carbon-carbon in principle undergo combustion. Still, a piece of Teflon is not easily lit.

Burning an organic substance is the standard way of analyzing it. The substance is mixed with Copper oxide in a tube heated in a currant of O2. The gas getting out of the tube is analyzed for its CO2 and H2O content. This is the standard way all organic substances have been analyzed.

@Maurice It was standard, like a century ago. Not all was, or can be analysed like this. For example trying to do this with teflon wouldn't be good idea (not to say it wouldn't work). https://chemistry.stackexchange.com/questions/34190/is-there-any-chemical-that-can-destroy-ptfe-or-teflon

https://chemistry.stackexchange.com/questions/119887/are-all-volatile-liquids-combustible

Burning is a non-scientific term, it implies appearance of a luminous chemical reaction in  a gaseous phase. It is commonly called a flame. It does not necessarily imply the presence of carbon or even oxygen. Boron and hydrogen compounds can burn in air, similarly, sodium burns in the presence of chlorine.
You can rather ask, can I oxidize carbon containing materials to carbon dioxide or carbon monoxide in the presence of oxygen? The general answer is yes. You already mentioned the examples of wood, sugar, coal etc. which are carbon rich materials. They will burn with a luminous flame. The carbon in inorganic materials such as refractory carbides, like silicon carbide, can also be oxidized that but only under extreme conditions.
IUPAC acknowledges that the "The boundaries between ‘organic’ and ‘inorganic’ compounds are blurred." So there is no need to create an arbitrary division.


text:
How to search substructures with explicit hydrogen in RDKit?
For a substructure search I would like to search for structures containing unfused benzyl.
The idea was to explicitly add hydrogen. But apparently this does not give the expected results.
So, I assume either my understanding or expectations are wrong or I'm using RDKit not properly.
I just learnt (Substructure search with RDKit) that I better use SMARTS expressions with Chem.MolToSmarts() where I get the expected results.
However, if I have much more complex search structures, how can I get from the SMILES to the SMARTS? Do I have always to figure out the SMARTS code myself? Is there maybe a way just drawing a molecule and add explicit hydrogen and search with this somehow? I guess it should be pretty clear (at least to a human) what I want to find (or not find) with structure 2. Are there maybe any tools or code which can assist here?

Code:
from rdkit import Chem

smiles_strings = '''CC1=CC=CC=C1
CC1=C([H])C([H])=C([H])C([H])=C1[H]
C1(C2=CC=CC=C2)=CC=CC=C1
C12=C(CC3=CC=CC=C3C2)C=CC=C1
C12=CC=CC=C1C=C3C(C=CC=C3)=C2
C12=C(C(C(C=C3)=CC=C4)=C4C=C2)C3=CC=C1
'''
smiles_list = smiles_strings.splitlines()

def search_structure(pattern):
    found = []
    for idx,smiles in enumerate(smiles_list):
        m = Chem.MolFromSmiles(smiles)
        if m.HasSubstructMatch(pattern):
            found.append(idx+1)
    print("Structures found: {}".format(found))

smiles_1  = smiles_list[0]
pattern_1 = Chem.MolFromSmiles(smiles_1)
smiles_2a  = smiles_list[1]
pattern_2a = Chem.MolFromSmiles(smiles_2a)
smiles_2b  = Chem.MolToSmiles(pattern_2a)
pattern_2b = Chem.MolFromSmiles(smiles_2b)
smiles_2c = smiles_2b
pattern_2c = Chem.MolFromSmarts(smiles_2c)
smarts_3 = '[cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]'
pattern_3 = Chem.MolFromSmarts(smarts_3)

print("\nSMILES 1 : {}\n# my expectation: [1, 2, 3, 4, 5, 6]".format(smiles_1))
search_structure(pattern_1)
print("\nSMILES 2a: {}\n# my expectation: [1, 2, 3]".format(smiles_2a))
search_structure(pattern_2a)
print("\nSMILES 2b: {} (Mol from SMILES)\n# my expectation: [1, 2, 3]".format(smiles_2b))
search_structure(pattern_2b)
print("\nSMILES 2c: {} (Mol from SMARTS)\n# my expectation: [1, 2, 3]".format(smiles_2c))
search_structure(pattern_2c)
print("\nSMARTS 3 : {}\n# my expectation: [1, 2, 3]".format(smarts_3))
search_structure(pattern_3)

Result:
SMILES 1 : CC1=CC=CC=C1
# my expectation: [1, 2, 3, 4, 5, 6]
Structures found: [1, 2, 3, 4]

SMILES 2a: CC1=C([H])C([H])=C([H])C([H])=C1[H]
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3, 4]

SMILES 2b: Cc1ccccc1 (Mol from SMILES)
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3, 4]

SMILES 2c: Cc1ccccc1 (Mol from SMARTS)
# my expectation: [1, 2, 3]
Structures found: [1, 2, 4]

SMARTS 3 : [cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3]

If you work in a Jupyter Notebook you can visualize your substructure patterns and the search results, so you can see how the patterns work.
from rdkit import Chem
from rdkit.Chem import Draw

smiles = ['CC1=CC=CC=C1','CC1=C([H])C([H])=C([H])C([H])=C1[H]',
          'C1(C2=CC=CC=C2)=CC=CC=C1','C12=C(CC3=CC=CC=C3C2)C=CC=C1',
          'C12=CC=CC=C1C=C3C(C=CC=C3)=C2','C12=C(C(C(C=C3)=CC=C4)=C4C=C2)C3=CC=C1']

params = Chem.SmilesParserParams()
params.removeHs=False # draw and work with explicit Hs
mols = [Chem.MolFromSmiles(s, params) for s in smiles]
Draw.MolsToGridImage(mols, molsPerRow=3)

This will give you the compounds and especially the one with the explicit Hs.

Now we get the patterns and especially the one with the explicit Hs.
patt1 = Chem.MolFromSmiles('CC1=CC=CC=C1')
patt2 = Chem.MolFromSmiles('CC1=C([H])C([H])=C([H])C([H])=C1[H]', params) # search with explicit Hs
patt3 = Chem.MolFromSmiles('Cc1ccccc1')
patt4 = Chem.MolFromSmarts('Cc1ccccc1')
patt5 = Chem.MolFromSmarts('[cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]')

patts = [patt1,patt2,patt3,patt4,patt5]
leg = ['CC1=CC=CC=C1','CC1=C([H])C([H])=C([H])C([H])=C1[H]','Cc1ccccc1 smiles',
       'Cc1ccccc1 smarts','[cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]']
Draw.MolsToGridImage(patts, molsPerRow=3, legends=leg)

Search and highlight the result.
ms = []
patt = []
allsubs = []
for m in range(len(mols)):
    for p in range(len(patts)):
        sub = mols[m].GetSubstructMatches(patts[p])
        if len(sub) > 0:
            ms.append(mols[m])
            patt.append(leg[p])
            allsubs.append(sub[0]) # substructures could be find multiple time - just take the first

Draw.MolsToGridImage(ms, molsPerRow=3, legends=patt, highlightAtomLists=allsubs)

Compounds 5 and 6 have no nonaromatic bonds, so there is nothing to find.

thank you very much for your code and explanations. It looks like params.removeHs=False is the bit I was missing. I can only test next week and will let you know.

If I'm using params.removeHs=False for searching with structure 2, then HasSubstructMatch(pattern) only finds structure 2 (as in your example).
Why are structures 1 and 3 not found as well? Maybe because of the explicit non-aromatic extra carbon in search structure 2? What do I have to change that by drawing structure 2 with explicit hydrogen and unspecified (is it or not?) extra carbon will also find structures 1 and 3?

@theozh It is not the extra carbon, it is just that 1 and 3 do not have explicit Hs. If I add explicit Hs to 3 it is found.
I thought C[c;H,h]1[c;H,h][c;H,h][c;H,h][c;H,h][c;H,h]1 (see https://smartsview.zbh.uni-hamburg.de/) will work, but it doesn't.

thank you, yes, today I also found out that if you add Hs to all the structures you're searching in, via mol = Chem.AddHs(mol), you will find what is expected. However, there are other cases which I still do not understand. I need more testing.

It took me a while, but you can find 1, 2 and 3 with *[c]1[c;H,h][c;H,h][c;H,h][c;H,h][c;H,h]1
and 1,2,3,4,5 with *[c]1[c;H,h,*][c;H,h][c;H,h][c;H,h][c;H,h]1


text:
Why ZnS can be seen as the solution is made alkaline? Explain with reactions
There is an extract in my book which i am not able to get. It says:
Place strip of metallic zinc in aqueous solution of copper nitrate for about one hour. You may notice that the strip becomes coated with reddish metallic copper and the blue colour of the solution disappears. Formation of Zn2+ ions among the products can easily be judged when the blue colour of the solution due to Cu2+ has disappeared. If H2S gas is passed through the colorless solution containing Zn2+ ions, appearance of white zinc sulphide, ZnS can be seen on making solution alkaline with ammonia.

The final solution contains $\ce{Zn^{2+}}$ ions, which reacts with $\ce{H_2S}$ to produce white $\ce{ZnS}$. But you may not avoid that maybe some of the initial $\ce{Cu^{2+}}$ ions still remains in the solution. And these rare $\ce{Cu^{2+}}$ ions do react with $\ce{H_2S}$ to produce Copper sulphide $\ce{CuS}$ which is dark black.
$$\ce{Cu^{2+}  +  H_2S  ->  CuS  +  2 H^+}$$
So the obtained final precipitate is probably a mixture $\ce{ZnS}$ and some $\ce{CuS}$ ; it is probably not perfectly white, but grey or even black. To be able to see the white color of $\ce{ZnS}$, you add some drops of ammonia $\ce{NH3}$ in the dark mixture containing $\ce{ZnS}$. This will dissolve the copper sulphide, producing a solution containing $\ce{[Cu(NH3)4]^{2+}}$.
$$\ce{CuS  +  4 NH3 -> [Cu(NH_3)_4]^{2+}  + S^{2-}}$$
So the black component of the precipitate will disappear, and you will be able to see the white color of insoluble $\ce{ZnS}$.

You are seeing the difference between a metal that can be precipitated by hydrogen sulfide in acidic solution and one that requires the sulfide bearing solution to be alkaline or at least not as acidic.  In my student days (1970s-1980s) they were called Group II and Group III metals in qualitative analysis, although that may have changed.
Both copper and zinc react as follows:
$\ce{M^{2+} + H2S <=> MS(s) + 2H^+}$
With copper the sulfide has such a low dissociation constant that the equilibrium constant for the precipitation is driven up to very high values, and the copper sulfide precipitates more or less completely even in in strongly acidic solutions.
In the second stage you have just the solution, having decanted it from the solid copper.  For zinc instead of copper the equilibrium constant of the above reaction is not as high and you need to consume the hydrogen ions to drive the precipitation of zinc sulfide forward.  Hence make the solution alkaline with ammonia.  (A strong base might set off redissolution of the zinc sulfide by a different reaction.)

My lab manual states that Group II contains $\ce{Pb^2+}$, $\ce{Cu^2+}$ and $\ce{As^3+}$ and all of these are precipitated by hydrogen sulfide in dilute hydrochloric acid, which Group IV contains $\ce{Co^2+}$, $\ce{Ni^2+}$, $\ce{Mn^2+}$ and $\ce{Zn^2+}$ which are precipitated by hydrogen sulfide in alkaline medium ($\ce{NH4Cl} + \ce{NH4OH}$).


text:
Normal modes of vibration
Why Normal modes of vibration are important when compared to translational and rotation modes. And why the frequencies for rotational and translational modes are close to zero?

what do you mean by important?

Rotational frequencies are not close to zero. They are only smaller than vibration frequencies. You may know that frequencies and energies are proportional at the atomic scale. It does not require much energy to make a molecule move or rotate. More energy is needed if you want to deform it and try to stretch its bonds.


text:
Values for heat of reaction example
I am currently studying the textbook Mass Spectrometry, third edition, by Jürgen H. Gross. Chapter 2.4.3 Bond Dissociation Energies and Heats of Formation says the following:

Energetics of $\ce{H}^\bullet$ loss from $\ce{CH_4^{+ \bullet}}$ The minimum energy needed to form a $\ce{CH_3^+}$ ion and a hydrogen radical from the methane molecular ion can be estimated from the heat of reaction, $\ce{\Delta H_r}$, of this process. According to Fig 2.6, $\ce{\Delta H_r} = \ce{AE_{(CH_3^+)}} - \ce{IE_{(CH_4)}}$. In order to calculate the missing $\ce{AE_{(CH_3^+)}}$ we use the tabulated values of $\ce{\Delta H_{f(H^\bullet)}} = 218.0 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_3^+)}} = 1093 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_4)}} = -74.9 \ \text{kJ mol}^{-1}$, and $\ce{IE_{(CH_4)}} = 12.6 \ \text{eV} = 1216 \ \text{kJ mol}^{-1}$. First, the heat of formation of the methane molecular ion is determined based on the experimental value of $\ce{IE_{(CH_4)}}$:
$$\ce{\Delta H}_{f(CH_4^{+\bullet})} = \ce{\Delta H_{f(CH_4)}} + \ce{IE_{(CH_4)}} \tag{2.14}$$
$$\ce{\Delta H}_{f(CH_4^{+\bullet})} = -74.9 \ \text{kJ mol}^{-1} + 1216 \ \text{kJ mol}^{-1} = 1141.1 \ \text{kJ mol}^{-1}$$
Then, the heat of formation of the products is calculated from:
$$\ce{\Delta H_{f(prod)}} = \ce{\Delta H_{f(CH_3^+)}} + \ce{\Delta H_{f(H^\bullet)}}  \tag{2.15}$$
$$\ce{\Delta H_{f(prod)}} = 1093 \ \text{kJ mol}^{-1} + 218 \ \text{kJ mol}^{-1} = 1311 \ \text{kJ mol}^{-1}$$
Now, the heat of reaction is obtained from the difference
$$\ce{\Delta H_r} = \ce{\Delta H_{f(prod)}} - \ce{\Delta H}_{f(CH_4^{+\bullet})} \tag{2.16}$$
$$\ce{\Delta H_r} = 1311 \ \text{kJ mol}^{-1} - 1141.1 \ \text{kJ mol}^{-1} = 169.9 \ \text{kJ mol}^{-1}$$
The value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{(CH_3^+)}} = 14.35 \ \text{eV}$, which is in good agreement with published values of about $14.3 \ \text{eV}$ (Fig. 2.7).

Note: The amount of energy needed to be transferred to the neutral $\ce{M}$ to allow for the detection of the fragment ion $m_1^+$ is called the appearance energy ($AE$) of that fragment ion.

Taking the values from Fig. 2.7 and using this calculator, we get that $1311 \ \text{kJ mol}^{-1} = 13.588 \ \text{eV}$ and $1386 \ \text{kJ mol}^{-1} = 14.365 \ \text{eV}$.
Unless I have made a mistake or am misunderstanding something, these values do not agree with what the author has presented. It seems that, if we take the values from Fig 2.7, the value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 13.588 \ \text{eV}$, which implies that the published value should be around $14.365 \ \text{eV}$.
So has the author made a small mistake here? Or am I misunderstanding something?
I would greatly appreciate it if people would please take the time to review this.

There is a huge amount of data in this message. But it would be interesting to know exactly what the question is. The title is "Value for heat of reaction". Well ! Which reaction ? There are plenty in the text.

@Maurice I am checking (at the end of the post) the author's work here: "The value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 14.35 \ \text{eV}$, which is in good agreement with published values of about $14.3 \ \text{eV}$ (Fig. 2.7)." As you can see, the point is that the values that I get are different from those of the author.

@ The Pointer. This is exactly my demand. What is the meaning of this $169.9 kJ mol^{-1}$ ? It corresponds to which $\Delta H$ ? Same question for $1311 kJ mol^{-1}$. This $1311 kJ mol^{-1}$  correspond to which $\Delta H $ ?

To answer your first question, it corresponds to $\ce{\Delta H_r} = 1311 \ \text{kJ mol}^{-1} - 1141.1 \ \text{kJ mol}^{-1} = 169.9 \ \text{kJ mol}^{-1}$, which I presumed is written as $\Delta H_r = 170$ in Fig. 2.7. To answer your second question, it corresponds to $\ce{\Delta H_{f(prod)}} = 1093 \ \text{kJ mol}^{-1} + 218 \ \text{kJ mol}^{-1} = 1311 \ \text{kJ mol}^{-1}$. Reading this again, I admit that I am even more confused now about what I did. Do you think I'm interpreting this incorrectly? I'm curious how you interpret it? Sorry for the confusion.

It looks like you’re overlooking that the methane is at -75, so you need to add 75 to 1311 to get the total change in energy that corresponds to AE

@Andrew I already did that (see immediately below the quotes, where I said $1386 \ \text{kJ mol}^{-1} = 14.365 \ \text{eV}$). The author is claiming to have $\ce{AE_{CH_3^+}} = 14.35 \ \text{eV}$, which does obviously not agree with my value of $14.365 \ \text{eV}$. Do you see what I mean? Have I made an error somewhere here, or is it the author who has made an error?

So you're worried about the difference between 14.365 and 14.35??? That's very good agreement in science. Remember that experimental values have errors associated and the calculated value has multiple approximations embedded in it. That they agree to 0.1 % is actually quite impressive.

@Andrew But aren't these values all calculated empirically? As I understand it, none of these values are mathematically derived, as some values are in physics. I was under the impression that the author is just copy and pasting values here into the calculations.  So why would there be any disagreement? That's the point: It seems to me that I am using the same values as the author to perform the same calculations, and yet I am getting different answers than what they claim, which leads me to think that I either am misunderstanding something or making an error, or the author has made an error.

@Andrew Have you read the excerpts yourself and attempted to follow/verify the author's calculations? Do your values agree with theirs (have I made an error)?

I see I misunderstood. I thought you were worried about the sentence "in good agreement with published values."  I see now.

The exercise in question is a prediction of the appearance energy of the methyl cation fragment $\ce{CH3+}$. The authors predict that value by estimating the enthalpy for the reaction of methane ionization followed by loss of a hydrogen atom. The enthalpy change estimation is based on published values for heats of formation and bond dissociation energies that have been determined experimentally.
The values presented, however, are rounded to the nearest integer or tenth, and the authors continue to round values during the explanation. For example, 12.6 eV is converted to the closest integer value of 1216 kJ/mol. All values in Fig 2.7 are presented as integers.
There seem to be two possible explanations for the final value being given as 14.35 rather than your calculated value of 14.365. Most likely, the authors have chosen to round to the nearest multiple of 0.05 rather than the nearest tenth or hundredth. Alternatively, they could have used unrounded values throughout the calculation rather than the rounded numbers presented in the step-by-step description. Either way, the difference is insignificant relative to the (im)precision with which the value can be predicted or experimentally determined.

Thanks for the answer. So there doesn't seem to be any errors in how the author did the calculation? That was what I was mostly worried about, since I don't want to learn something that is actually incorrect.


text:
Trifluoroiodomethane in basic medium

$$\ce{CF_3I + NaOH(aq) -> Major Product}$$

The Major product, according to me, should be ditrifluoromethyl ether $\ce{CF3-O-CF3}$. The hydrogen on the $\ce{-OH}$ group is quite acidic and hence reacts with $\ce{CF3I}$ to give a stable compound according to the following reaction mechanism:

However, the textbook gives the following reaction mechanism and claims that the product is trifluoromethane $\ce{CHF_3}$

Which one of these is correct?

The hydrogen atome on the -OH group cannot be "very acidic" as you say.

Actually it kinda is, but that means the anion wouldn't be that good nucleophile, also F3CO- decomposes, losing F-.

https://en.wikipedia.org/wiki/Trifluoromethanol https://en.wikipedia.org/wiki/Nonafluoro-tert-butyl_alcohol https://en.wikipedia.org/wiki/Trifluoroiodomethane

You didn't give a mechanism for your expected product, so I can only speculate as to how you intended to form the final product.
If you thought to use an Sn2 mechanism, I'd say that the reaction is highly improbable as the halide is attached to a trisubstituted carbon, i.e. a carbon with a lot steric hinderance, and therefore, it would be rather difficult for an Sn2 reaction to take place.
If you chose to use an Sn1 mechanism: with the first step, when iodine leaves, you would create a primary carbocation, which is unstable. Additionally, being that fluorine is an electron withdrawing group, the three fluorines would likely serve to further retract electrons from the already electron-poor carbon, and not serve to stabilize the primary carbocation. Therefore, I doubt that this reaction would occur as well.
In the textbook provided mechanism, the oxygen attacks the accesible iodine, which is next to a very psuedopositive carbon. And although this places a  negative charge  on the carbon atom, the electron-withdrawing fluorines serve to help stabilize the carbanion via their inductive properties. In the second step, the carbons rids itself of its negative charge, by taking a hydrogen atom from a nearby hypoiodous acid (pKa 11).
References:
McMurry, J. (2015). Organic chemistry: with biological applications. Stamford, CT: Cengage Learning.

This is quite thoroughly incorrect unfortunately. Fluorines would stabilise cation mesomerically, and anion is actually difficult to make because of stereoelectronic effect.

@Mithoron flourine has a poor mesomeric effect and its inductive effect is more dominating compared to its mesomeric effect.

Not at all young padawan ;) Check out for example https://chemistry.stackexchange.com/questions/40076/stability-of-carbocations-cf3-vs-ch3#comment63589_40076 https://chemistry.stackexchange.com/questions/67106/which-molecule-is-more-acidic-chf3-or-chbr3

The second pathway looks to be correct because :
1. The carbanion formed is considerably stable
2. $\ce{S_{N^2}}$ dosen't look favourable in the first pathway(in the second step in the first reaction)
The first pathway on the other hand dosen't look to be energetically favorable neither by $\ce{S_{N^1}}$ nor by $\ce{S_{N^2}}$ as the carbon is way too hindered and the supposed carbocation that would form in the $\ce{S_{N^1}}$ mechanism would not be stable in the slightest due to 3 highly electronegative $\ce{F}$ atoms  attached to it


text:
Esterification of Glycine
I came upon the following statement in a true/false question:

Glycine can be esterified by an alcohol in the presence of inorganic acid

Is this statement correct?
Glycine contains the basic $\ce{-NH2}$ group, which would get protonated in the presence of inorganic acid. As a consequence, would the esterification process still be carried out on the $\ce{-COOH}$ group? This paper says that the reaction needs trimethylchlorosilane as a catalyst and not an inorganic acid.

Me3SiCl is not a catalyst, its a reagent used in excess.

The statement is correct.
The most common method for the esterification of amino acids is with anhydrous acid in alcohol. The most convenient way of doing this is with thionyl chloride example procedure or chlorotrimethylsilane example procedure as a source of anhydrous HCl.
The original procedures with HCl gas or conc. sulfuric acid are slower and lower yielding recent example


text:
In pig iron, from where does carbon in it comes from?
I've read that:

"During extraction of iron another impurity appear which is carbon from coke in furnace."

However, I don't understand how there will be enough unreacted carbon that would blend with iron. As reaction of carbon and oxygen is very exothermic so there must be enough energy for all carbon atoms to react with oxygen.

Partly related: Required temperature for the reduction of iron oxide using carbon. Note that for the eutectic of iron containing 4.3 % carbon, the melting point is lowered from 1540 °C to 1150 °C.

@Loong Good point. The blast furnace would be screwed if it weren´t for carbon, in more than one way.

A blast furnace is made of successive layers of charcoal and iron oxide mineral. In the beginning, the lower layer of charcoal is lit and the fire is maintained by air being pressed through the burning matter. The combustion produces first $CO_2$. But as the temperature is high and the charcoal is in excess, $CO_2$ soon reacts with $C$  producing $2$ $CO$  according to the equations :$$\ce{C + O_2 -> CO_2}$$ $$\ce{CO_2 + C -> 2 CO}$$ Then the CO gas touches the bits or pieces of mineral, which produces the following equation :$$\ce{3 CO + Fe_2O_3 -> 2 Fe + 3 CO_2}$$ This produces some metallic iron which is liquid if the temperature is high enough. This liquid flows between the bits of burning charcoal, and dissolves some charcoal during this flow. The obtained pig iron contains about 5% carbon. which must be later on burned in special furnaces, to get steel (containing about 1% dissolved carbon)

No need to dissolve solid coke. $\ce{CO + Fe <-> FeO + C}$.

There is no charcoal in a blast furnace.

So coke and iron are added in the same time but in layers right? I thought each one was added separately that's why I found it weird that there was enough coke left to react with iron that would be added later.

And limestone.....

But if limestone is added since the very beginning why doesn't sulphur phosphorus and silicon react with the CaO produced from limestone thermal decomposition, why is it only silica that reacts while the rest of impurities are getten rid of in the steel-making furnace?

Thinking about it now, it doesn't even matter when exactly it was put it should have neutralised all of the impurities in one time..

The carbon in the pig iron comes from the coke ; although modern blast furnaces may inject hydrocarbon gas and oil. The oxygen is controlled so carbon containing gas exiting is CO ( monoxide).

Modern blast furnaces still use coke!

They need the physical properties of the coke , but anthracite coke is getting more costly so they supplement it with oil and gas in a modern operation. They also use oxygen enrichment so there is less useless nitrogen getting heated in the furnace. I expect Wiki has the info.


text:
How do I convert purity percentage to grams?
I'm trying to find out how many grams of a substance, in this case propylene glycol, are in a 5 litre container that is advertised as being 99.8% pure. Any help would be greatly appreciated.

I'll do the "full problem".
First the volume of a liquid typical increases with a rise in temperature. Since no temperature was specified, one must be assumed. The 88th CRC Handbook of Chemistry and Physics gives 1.0361 g/cu cm at 20 °C. So to work the problem you must assume:
(1) The density of 99.8% propylene glycol is 1.0361 g/ml @ 20 °C

Note: For a catalog the specification is typically given as "at least 99.8%" and if reagent grade then an analysis for the particular batch would be given for the "fresh" bottle. Being hygroscopic, it is quite possible to pick up water from the air once the bottle was opened.

(2) The volume is given for 20 °C
(3) Since the OP is buying $\pu{5 L}$, I find it hard to believe that the volume would be 10% off. So let's assume that the quoted volume is good to $\pu{\pm10 ml}$, which means that the OP can expect 5.00 L.
Thus:
$$\pu{5.00 L}\cdot \pu{1.0361 kg/L} \cdot \pu{0.998}\text{ pure} = \pu{5.170139 kg} \ce{->[rounded]} \pu{5.17 kg}$$
So $\pu{5.17\pm 0.01 kg}$ would be my answer.

+1 Nice! I will delete my other comments, but one question: when 5 L containers are filled, what kind of filling tolerance might be typical? Your 10 mL is reasonable, but I would guess that a slightly higher fill, i.e., nominal plus a filling process error margin, would be used, if only to avoid legal issues. Your take?

@EV - It depends, the @#$%^& companies will try and see how close to the lower limit that they can get. So getting 1001 "5 L" bottles out of a 5000 L tank is 1 bottle of pure profit.

In Wikipedia, propylene glycol has a density $1.04$. So $5$ litre propylene glycol weighs $5200$ g, and contains $5200·0.998 = 5190$ g propylene glycol.


text:
What are the products of these 3 reactants? Acetic Acid + K2CrO4(aq) + Ba(NO3)2(aq)
Acetic Acid + K2CrO4(aq) + Ba(NO3)2 (aq) —> ?

The equation is
$$\ce{K2CrO4 + Ba(NO3)2 -> BaCrO4 + 2 KNO3}$$
The acetic acid does not intervene. $\ce{BaCrO4}$ is insoluble in the solution and make a pale yellow precipitate, that can be filtrated.

What’s the rationale behind acetic acid not having any effect on the reaction?

Acetic acid acid is probably here to be sure that the solution contains no carbonate ions from tap water. Carbonate ions may also reacts with Baryum nitrate, producing some Baryum carbonate which is also insoluble in water. Acetic acid reacts with carbonates which are transformed into $CO_2$ and acetate ions, and acetate ions are not perturbing the reaction.

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text:
I need explanation for reaction rate
That's a sample reaction:
$$2N_2+3O_2->2N_2O_3\ \ \ r=k_1[N_2]^2[O_2]^3$$
Why can't I write this as
$$N_2+\frac{3}{2}O_2->N_2O_3\ \ \ r=k_2[N_2][O_2]^{\frac{3}{2}}$$
Don't bother the elements. Assume it's elementary not mechanism.

First kinetic eq. is incorrect, second probably too.

Furthermore, $N_2$ and $O_2$ have the greatest difficulty to make any reaction at all. And if they react, they will produce $NO$ or $NO_2$, and never $N_2O_3$.

The rate law doesn't change when multiply equation of the  reaction

I'd rather make a sample out of a reaction that actually can occur.

https://chemistry.stackexchange.com/questions/38167/do-coefficients-matter-in-rate-law/38169#38169

A rate law is never deduced from a theoretical equation. NEVER. It is always obtained form experimental measurements. It may happen that the order of the reaction is equal to the stoichiometric coefficient.

This is a bit over the top. I agree that a reaction which such a high reaction order is unlikely to be an elementary reaction which book problems often assume.


text:
How to perform IR spectroscopy on silica particles at various relative humidities when they are passed through a tube using pressurized air?
Surface bonds of silica particles are susceptible to humidity and frequently change between silanol and siloxane with changing RH. I want to measure how these surface bonds change with the humidity using IR spectroscopy when the particles are flown using pressurized air through a tube.

Despite your question is a bit vague about the sample, I assume «when they are in motion» refers to particles passing a tube, are close to each other -- in contrast to particles floating in air.
If so, than you should consider to couple infrared spectroscopy (about the energy band) with a probe based on attenuated total reflection.  Roughly speaking, classical optics predicts you a threshold incident angle where light of a certain frequency no longer is able to leave an optical dense material (high refractive index, for IR for example Silicon, Germanium, or a small diamond) to enter a less dense material (for IR, for example air, most of organic materials).  Recognizing light equally has character of a wave however «allows» that light at this interface still interacts sufficiently enough that one may extract information to calculate an IR spectrum.
What you see in the lab as an ATR-FT-IR spectrometer typically is a gate-like construction

(source)
but there are ATR-IR probes which you may immerse into fluids to monitor the advancement of reactions (e.g., Mettler's reactir)

(source)
where the probe window is at the side or on the tip of the probe itself:

(source)
Potentially -- compared to classical transmission IR -- you truncate any sample preparation (no KBr pill, nor nujol) and may record data of not-too wet samples, too.  Don't forget an ATR correction, as the small penetration depth $d_p$ of light across the phase boundery between window material and sample depends both on $n_c$ (window / crystal material), $n_s$ (sample), the incident angle $\theta$ and wavelength $\lambda$ applied.
$$ d_p = \frac{\lambda}{2\pi n_c \sqrt{\sin^2 \theta - \left(\frac{n_s}{n_c}\right)^2 }}$$

Yes, by motion I mean flow through a pipe. I added that to the question as well. Thank you for your suggestion it is really helpful.


text:
Amide fragmentation of salicylamide
I want to ask a question about the fragmentation of the $\ce{M_17}$ peak of Salicyclamide to a phenolic fragment.
I was reading the following paper giving a profile about salicyclamide.
I want to determine a mechanism that gives the following product from the fragmentation scheme:

I am however, unable to so far produce a mechanism that shows the scheme above.
The figure below shows the closest attempt I have made so far.

I feel my attempt is incorrect since there is a net charge change.
What mistake am I making that can rectify my result and provide a reasonable suggestion for the mechanism involved?
I don't understand how this question can be voted for closure as it is "not relevant to chemistry". This is a clear phenomena given in a research paper and I wanted to discuss the mechanism(s) behind such findings that have been observed in the paper - as noted in the help centre under Questions relating to observed chemical phenomena.

I realised that for an $\ce{m/z = 92}$ you would need a $\ce{C6H4}$ ring with only an oxygen group attached. If a positive charge is present, this implies that there is a radical at position 2 ortho to the $\ce{C=O}$.
I revised my own mechanism below to suggest one possible route:


text:
Why does the test tube face downwards in gas production?
In the preparation of $\ce{NH3}$ from $\ce{NH4Cl}$ and $\ce{Ba(OH)2}$, my teacher said that when heating the test tube, the test tube faces downwards as opposed to upwards. Why is this so? Wouldn't the solid just fall out?

This is from the Royal Society of Chemistry in 2016: https://edu.rsc.org/resources/making-and-testing-ammonia/433.article . The figures show the reaction test tube as horizontal and a suggestion is to put some CaO in that test tube to trap (some of) the water produced.

"As adverbs the difference between downward and downwards
is that downward is toward a lower level, whether in physical space, in a hierarchy, or in amount or value while downwards is towards a lower place; towards what is below." and "In British English, downward (with no final -s) is normally only used as an adjective." Not sure which one is appropriate here.

If you mean the test tube containing the solids (reaction tube), then that tube faces downwards to promote the flow of gas into the receiving tube.  The idea is to let the gas sink away from the reaction zone and towards the receiving tube as it cools and becomes more dense.
The receiving tube, of course, faces down so that the ammonia gas, lighter than air, can float to the top after it meets the air.  So you have the ammonia flowing down with cooling through the reaction tube and then flowing up with buoyancy into the receiving tube.
Now for my bonus question:  Had you been producing carbon dioxide by heating magnesium carbonate, would you change the reaction tube?  How about the receiving tube?

Since CO2 is heavier than air, the receiving tube would be right-side up. Something like this? https://imgur.com/jJh9Ann

Yes, the CO2 would just keep sinking in both tubes.


text:
Biogas temperature
What would be the temperature of Biogas when it is delivered to a boiler through a pipeline by a gas blower. Inlet pressure at gas blower is zero and at delivery it is 3.0 psi.

Initial temperature of Biogas is 40 degree C

What is the gas  mass flow rate and the amount of power being supplied by the blower?

The compression by the blower can be used as an example of adiabatic compression.
$$pV^\gamma=\text{constant}$$
or
$$p_1V_1^\gamma=p_2V_2^\gamma$$
where $\gamma$ is the ratio of the heat capacities $\gamma=C_p/C_V=C_{\mathrm m,p}/C_{\mathrm m,V}=c_p/c_V$. You need to find a value of $\gamma$ suitable for your biogas, which is probably mainly a mixture of $\ce{CH4}$ and $\ce{CO2}$.
According to the ideal gas law,
$$p_1V_1=nRT_1$$
and
$$p_2V_2=nRT_2$$
Therefore,
$$\begin{align}p_1\left(\frac{nRT_1}{p_1}\right)^\gamma&=p_2\left(\frac{nRT_2}{p_2}\right)^\gamma\\[6pt]
p_1^{1-\gamma}T_1^\gamma&=p_2^{1-\gamma}T_2^\gamma\\[6pt]
\frac{T_1}{T_2}&=\left(\frac{p_1}{p_2}^{\frac{1-\gamma}{\gamma}}\right)\end{align}$$
The given initial temperature of $40\ \mathrm{^\circ C}$ needs to be converted to the corresponding thermodynamic temperature $T_1=313.15\ \mathrm K$.
Note that, unfortunately, the given pressures are actually gauge pressures (also note that the use of the unit $\mathrm{psi}$ is deprecated). Assuming an ambient pressure of about $1\ \mathrm{bar}=14.5\ \mathrm{psi}$, the corresponding pressures are $p_1=0\ \mathrm{psi}+14.5\ \mathrm{psi}=14.5\ \mathrm{psi}$ and $p_2=3.0\ \mathrm{psi}+14.5\ \mathrm{psi}=17.5\ \mathrm{psi}$.