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Inexact differential-path function
From Wikipedia:

an inexact differential cannot be expressed in terms of its antiderivative for the purpose of integral calculations

What's the mathematical reason that inexact differential cannot be expressed in terms of its antiderivative? Not enough information? Too complicated? $\mathrm{d}U$ is an exact differential. If you integrate $\mathrm{d}U$, what will you get then?

This question would be better off in a part of stack exchange devoted to maths

I think you answer comes from the part of the sentence that you left of the quote from Wikipedia:

... i.e. it's value can't be inferred by looking just at the initial and final states of the given system.

Inexact differential are the result of integrating nonstate functions (functions which aren't path independent). The fundamental theorem of calculus, which allows the integration of a function over a range by determining the antiderivative at the end points of that range, requires path independence in order to be true. So it won't work for inexact differentials that are path dependent.
As for integrating over $dU$, $U$ is a state function and so you can apply the fundamental theorem and solve the integral using the antiderivatives at the end points.


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What is chemical decomposition in the context of a crystal or amorphous solid?
The IUPAC Gold Book definition of chemical decomposition is:

The breakdown of a single entity (normal molecule, reaction intermediate, etc.) into two or more fragments.

That is a very general definition, but I am not sure if it is supposed to be this inclusive. For example, think of a diamond crystal. It is a single entity. Now if chemical decomposition occurs, it can break down into multiple fragments. These fragments are not defined more closely, so it could be that the fragments are single C atoms, polyatomic C molecules... or even diamond fragments again.
In this sense, simply cutting or polishing a diamond would be a special case of chemical decomposition. This seems quite absurd to me. So the question is: Is chemical decomposition really that broad, or is there some better definition that is more commonly used in chemistry? It seems to me that the intuitive definition of chemical decomposition would be that a molecule is broken into smaller fragments, while a crystal or an amorphous solid would need to be broken into non-crystalline (non-amorphous) fragments in order for it to be considered chemical decomposition. Otherwise it's just cutting!
The question is relevant to me because I am writing my PhD thesis on laser-induced breakdown spectroscopy. The "breakdown" in the name of the technique refers to a chemical breakdown, which is the same as chemical decomposition. Usually, solid samples are analyzed, in which case some material from the sample surface is removed by laser ablation. Laser ablation is basically the same as laser-induced breakdown, if you look at the definitions. So there is some confusion on the terminology in general, and I want to unravel it as good as possible in my theory chapter.
Thank you very much for your input!
(Note: By the way, some non-IUPAC definitions I have found for chemical decomposition also confusingly define it as relating to compounds. As far as I know, the definition of a compound excludes crystals and amorphous solids, but it also excludes homonuclear molecules. So this definition seems way too exclusive, because it makes sense that decomposition can also occur for homonuclear molecules.)

It's decomposition if a relevant part of the sample (e.g. the one you actually analyse) is not the same substance as before. Polishing a diamond, you still have a diamond, plus some dust you don't care about much.

When it comes to non-molecular solids such as glass, metal or diamond, "a single entity" is in fact the formula unit, the concept identical to a "molecule" for, well, molecular compounds. Both formula unit and molecule are the smallest portion of a substance that preserves its chemical properties. As long as the formula unit remains the same, there is no chemical decomposition to speak of.
However, this doesn't mean that there is no chemical processes that can be initiated mechanically. An entire branch, mechanochemistry deals with them. As for laser ablation, I think you are basically introducing a secondary phase, a thin film of another substance (for example, metal oxide) on the surface.
One more practical example I can think of: for a single-crystal diffraction experiment one has to use a 3D crystal (in a sense that it shouldn't be a 2D plate for the proper absorpton correction) with linear dimensions not exceeding fractions of a millimeter. For the probe preparation a crystallographer often has to literally cut the crystals with a special scalpel under the microscope. However, the resulting crystal structure doesn't depend on how many cuts and in what direction the crystallographer did. The x-rays though, can alter the structure of some sensitive molecular crystals (often biological samples, proteins), but this is another story.

Thank you, this sounds very reasonable and is something I had not thought of. The definition of a chemical entity is not given by IUPAC, but your interpretation as only relating to the formula unit probably makes the most sense. One unrelated question about the "phase" - I understand it correctly that a mixture of NaCl and KCl, for example, has a NaCl phase and a KCl phase? Or does a phase need to be physically connected so that a homogeneous mixture in fact contains hundreds of small NaCl and KCl phases?

This depends on what mixture you are talking about. A mechanical mix of both salts would indeed have 2 phases. However, a system of molten salts with an arbitrarily chosen composition $\ce{Na_xK_{1-x}Cl}$ ($0\leq x\leq 1$) is homogeneous all right (single phase, see e.g. this NaCl-KCl phase diagram).

Thank you, this is much appreciated! Yes, I was talking about a mechanical mix, because I was wondering whether several independent grains of NaCl would be considered the same "phase" in a mechanical mixture with something else. (From your reply, I get that the answer is "yes".)

Yep, if you take table salt and pour the powder of potassium chloride in, then thoroughly mix them, it's still considered a two-phase system. However, if you increase the temperature enough (say, up to 700 °C, see the phase diagram) and cool the molten salts down, you form a new chemical compound (single phase). This is not always the case and you always have to check the phase diagram first.

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What do the dots mean in HN.CO.CH3 connected to the second carbon of N- Acetylglucosamine? And what do the N and acetyl indicate?

What do the dots mean in HN.CO.CH3 connected to the second carbon of N- Acetylglucosamine? And what do the N and acetyl indicate?

What is glucosamine? What is an acetyl? The acetyl is connected to the amine at the nitrogen atom, hence N-acetyl. These "dots" (they look more like blobs to me) are just a simple covalent bond. Why they look like blobs I have no idea.

These don't look like dots to me, more like en-dashes used for denoting chemical bonds (e.g. $\ce{HN-CO-CH3}$). Probably they were chosen short by the typography in order to fit an entire structure on the page. N- indicate the substituent on the nitrogen, and the task of finding what is acetyl is I leave to you:)

They look more like dashs to me, a dash is a common symbol for a single bond

These are not dots. They actually are single bonds though I do not know why they are printed that way. As such, they should be represented by a half-long dash (called EN DASH in Unicode).
In the name N-acetylglucosamine, acetyl means that a hydrogen atom has been replaced an acetyl group (i.e. –CO–CH3) and that the replacement occurred on the nitrogen (aka "N") atom.
Overall, it means something like: "consider glucosamine but replace one hydrogen on the nitrogen atom by an acetyl group". This is the standard "substitutive naming" defined by the IUPAC.

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Are AcO- and PhO- better described as soft or hard nucleophiles?
I would presume soft nucleophiles as resonance increases the energy of the HOMO and therefore they favour SN2 over E2?

Well, many properties like nucleophilicity are expressed while comparing it to another chemical species'. Creating an absolute scale of nucleophilicity only relies on our current knowledge of the "poorest nucleophiles" and the "best nucleophiles"

Preference of SN2 over E2 is controlled by basicity and solvent. Stronger bases shift preference toward E2. Protic solvents favor E2; polar aprotic solvents favor SN2.


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What is thermal isomerization?
Context: a retinal molecule the isomerizes from all-trans to 13-cis form, then "reisomerizes thermally".
I tried googling it but couldn't find a simple, straightforward explanation.

The term "thermally" is used here to contrast with a photochemical isomerization. In the chemistry of vision, 11-cis retinal absorbs light and is isomerized to the all-trans isomer. This is a photochemical process because it involves the absorption of a photon to promote one electron into an antibonding orbital. In this excited state, the geometry can rearrange to the more stable all-trans form before the electron relaxes back to the ground state.
In a "thermal" process, no photons are absorbed, no electrons are promoted, and no excited states are needed. The molecule takes advantage of ambient or supplied kinetic energy to reorganize.

In photochemical isomerization absorbed light is responsible for conversion of cis to trans where as in thermal isomerization molecule uses its own thermal energy where molecules donot need to absorb photos .In addition no excitation takes place.

That's exactly what the accepted answer says, please elaborate further to make your answer stand out. Welcome to the site!


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How many natural products are there in existence?
Hi I'm doing a project on natural products and I'm wondering if somewhere there is at least an estimated number of how many exist?
If this exists somewhere can you please provide a reference.
Please and Thank you! :)

Depending on what you call a product, the estimates will vary wildly.

What I would do is to search for the capacity of the largest database for the natural products fulfilling your definition of them. For instance, here is an outdated (2015) review covering databases from the structural data to some ancient Chinese recipes. On average, there is probably about 50000 or so chemical substances characterized (I might be entirely wrong, not my field).

@Bianca this article gives the number of known natural products and tries to provide an estimate of the number of possible ones (though keep in mind Ivan's point about the definition of product). They come up with a value on the order of 50 million. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3447393/

Would we count DNA strands from different individuals from different species as different "products"?


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Why does diamond have no molecular formula?
Why does diamond have no molecular formula, while buckminsterfullerene does have it, despite both being allotropes of the same element?

somewhat related to https://chemistry.stackexchange.com/questions/75170/why-is-the-buckminsterfullerene-the-purest-form-of-carbon/75177#75177

To quote my answer to Is a diamond a single molecule?:

Diamond is a covalent network solid, like a number of other common materials (quartz, graphite, glass, and a whole bunch of stuff).
Because they are not discrete molecules - there is no 'diamond' molecule the same way there are molecules of caffeine, benzoic acid, citric acid, N,N-dimethylaminopyridine, etc.

Another way to think about it, since diamond is a covalent network solid, each diamond is a single giant molecule. Since each diamond is one molecule, there is no universal diamond molecule with the exact same number of carbon atoms every time it occurs.
Fullerenes form distinct molecules with the same number of carbon atoms in each species every time. Thus buckminsterfullerene is always comprised of molecules of $\ce{C_{60}}$.


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Nucleophile vs base and Electrophile vs acid
How we will figure out that a given compound will behave as nucleophile or base ? Same problem for electrophile and acid.
By the way I have already seen a similar question though I didn't find it helpful but if someone could correct the answer posted by Vikas Avasthi then I think (no sure) that it might become part of answer to my question.

What do you mean by "acid" and "base"? If you're talking about Lewis acid/base theory, then there is no difference. If you're talking about Brønsted acid/base theory, then you need to learn more or gain more experience to identify which is more likely to be operative. It's not that easy to explain in a few sentences.

I recommend reading the first comment on the question you linked and visiting the link of "SHAB principle".


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Why can the acidic strength between H2S and HCl be explained on the basis of electronegativity but H2S and H2O can't?
Consider simply aqueous solutions of $\ce{H2S}$  and $\ce{HCl}$ . It is pretty much known that $\ce{HCl}$ is stronger than $\ce{H2S}$ in water. The reason behind that first is the electronegativity of $\ce{Cl}$ is higher than $\ce{S}$, so, in a polar medium, it becomes easier for the $\ce{H-Cl}$ bond to break than the other one.  More ever , $\ce{Cl^-}$ ion being smaller in size is extensively hydrated in water than $\ce{HS^-}$ ion, which also favours in the increased acidity of $\ce{HCl}$.
But now consider acidity of $\ce{H2S}$ and $\ce{H2O}$ in water. If we apply the similar reasons as before, we will ultimately conclude water is  a stronger acid than $\ce{H2S}$. But actually it’s the reverse. The only fact supporting the actual phenomenon is the weaker bond energy of $\ce{H-S}$, but if we apply this to the earlier case of $\ce{HCl}$ and $\ce{H2S}$, it will again contradict the actual phenomenon.

Why is this the case ? If my reasoning is incorrect, what is the actual reasoning that will support the correct acidity in both the cases ?

You could make a similar argument for comparison between the 'HX's (X = halogens).

You are right that electronegativity is an important consideration but another factor that is also very important is size of the atom. Although there is a decrease in electronegativity from $\ce {O}$ to $\ce {S}$, the size of the atom which the charge is mainly centred on is now larger. This allows the charge to be distributed over a larger volume of space, decreasing inter-electronic repulsions, providing greater stabilisation of the negative charge.
However, the greater size decreases charge density, decreasing the strength of ion-dipole interactions between the anion and the surrounding water molecules. Thus, the extent of solvation decreases for the larger anion.
Electronegativity is defined as the ability of the atom to attract a bonding pair of electrons. It merely serves as a proxy for the ability of the atom to stabilise negative charge. It should not be viewed as the most important factor that influences the atom's ability to stabilise negative charge.
Also, mentioned in your answer, the strength of the bond formed with the larger atom with more diffuse orbitals decreases. A bond that is more easily cleaved would facilitate bond cleavage and thus, favour the acid with the larger atom bonded to $\ce {H}$ as the stronger acid.
Consolidating the above discussion, there are really two factors on each side of the table below. Perhaps, the distribution of charge increases very significantly and the bond energy also decreases significantly while the decrease in extent of solvation and the electronegativity exhibit a relatively less significant effect. It would be great if someone is able to quantitatively show this...
\begin{array}{|c|c|c|c|} \hline
\text{Favouring acids with larger atom} & \text{Favouring acids with smaller atom} \\ \hline
\ce{Distribution of charge} & \ce{Extent of solvation}\\ \hline
\ce{Bond strength} & \ce{Electronegativity}\\ \hline
\end{array}

But you just restated my arguments only. You didn't proceed much towards the actual logic (if any) supporting both the cases . I don't see much new things in your answer.

I have presented an additional important factor to consider - the greater extent of charge dispersal in conjugate bases with larger atoms having more diffuse orbitals. I have also presented clearly a framework for analysis of the acidity phenomena. I agree that my answer does not provide the most convincing answer to your question but it has provided a  comprehensive framework (i.e. the consolidation of the 4 factors) on which we can work on.

It is not just electronegativity.  You also have to look at how strongly the atom bonds to hydrogen.  If a strong bond forms, it might convince the atom to act as a base even if it would be grumpy based on its electronegativity.  If only a weak bond forms, even an atom with relatively low electronegativity might balk at being protonated.
As a rule, hydrogen forms stronger bonds with itself and with nonmetals in the second row ($\ce{C}$ through $\ce{F}$) than with just about anything else.  It's due to favorable covalent overlap; take account of orbital sizes and nodes.  Respecting this difference, treat $\ce{H,C,N,O,F}$ as one group and heavier nonmetals as another, and just compare electronegativities within each "class".

Acidity/ Basicity is a bulk property - it also depends on the medium where a substance acts as an acid or a base. For example, ammonia, a weak base in aqueous medium, can be a rather strong base in another medium.
Similarly, pure 100% dry $\ce{H2S}$ is theoretically a stronger acid than perfectly dry gaseous $\ce{HCl}$ due to a smaller bond dissociation energy and weaker dipole interaction. But, when it comes to aqueous medium, many factors are taken into account to decide the  strength  of acidity. To behave as a stronger acid, constituent ions need to be hydrated very well to facilitate the dissociation of $\ce{H+}$ ion or proton.
In case of $\ce{H2S}$, it is not so polar, (electronegativity difference  between $\ce{H}$ and $\ce{S}$ is only about $0.4$ just like the case of $\ce{CH4}$), it doesn't dissolve very well in water. Also the dissolved $\ce{H2S}$ feebly dissociates to give proton or $\ce{H+}$ ion since the $\ce{HS-}$ is not well hydrated due to its bigger size. On the other hand, $\ce{HCl}$ dissolves (due to a bigger electronegativity difference, about 1.4) and is solvated very well, since $\ce{Cl-}$ ion is smaller in size when compared to $\ce{HS-}$ or $\ce{S^2-}$.
Hence, $\ce{HCl}$ is a stronger acid than $\ce{H2S}$ due to the environment in which it is present and also note that the electronegativity difference plays a minor role here.

+1 I've seen  your contributions and they look good to me. Just a few tips in writing an answer. 1, Use paragraphs, it makes the whole thing look a lot neater. Two new lines breaks the word wall into paragraphs. 2. Use $\ce{}$ when formatting chemical compounds. It makes them look a lot prettier.

Why is solubility a factor in pKas...? The equation for Ka only involves aqueous species, i.e. molecules that have already dissolved. The question of solubility is quite separate. The case of $\ce{H2CO3}$, or $\ce{CO2}$, is similar: the "pKa of $\ce{H2CO3}$" already assumes that the molecule is formed from $\ce{CO2 + H2O}$, and it's only the "pKa of $\ce{CO2(aq)}$" where solubility / reactivity comes into play. The common confusion in this case only arises because $\ce{H2CO3}$ basically doesn't exist; but such arguments cannot be applied to $\ce{H2S}$.


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How can the stereochemistry of the product of an E1cb reaction be determined?
Say if this is your reagent.

I guess similarly to how you'd determine the stereochemistry of the product of an E1?

Since your example compound has a chirality center, let's start by picking one enantiomer. We'll do the other enantiomer later. In the presence of a suitable base, this compound can be deprotonated to form the enolate anion. The enolate group itself can be E or Z, but for this purpose it does not matter.

In the second step of the mechanism, the electrons in the new pi bond from the enolate pi electrons shift toward the beta carbon. The new pi bond will be formed from the enolate p orbital on the alpha carbon and the C-Cl sigma-antibonding orbital on the beta carbon.

For this to work, the C-Cl must be oriented perpendicular to the plane of the pi bond. We can draw Newman projections for our specific enantiomer of the enolate anion. There are two conformations that put the chlorine in the correct orientation. The second one below has less steric hindrance with the larger group (ethyl) near the H atom and the smaller group (methyl) oriented toward the enolate carbon.

The major product forms from this conformation, with the methyl group cis to the enolate/carbonyl carbon.

Now, the other enantiomer forms the same major product, but I would like you to convince yourself of that.


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Combinatorics of Molecule given the Elements
I have search a lot on Internet and my own researches on a sheet of paper failed to found an analytical closed formula for the following problem:
Considering:

Common ranges of temperature and pressure
A FINITE number of given pure elements with their respective valence

Found a lower-bound of all possible complete/incomplete molecules formed involving ALL the given elements (ignoring valences that are indistinguishable).
Examples:

Given one C, one H. We can form only one "incomplete" molecule: CH (lowerbound=1)
Given two H. We can form only one "complete" molecule H_2 (lowerbound=1)
Given one O and two H. We can form H_2 but also H_2O, but also incomplete HO (lowerbound=3)

Thx for feedback

Not sure what temperature and pressure have to do with this problem.  How would you expect T and P to restrict formulae?

Also, when you say "molecule" I assume you mean "empirical formulae".  You're not expecting this algorithm to tell apart glucose (C6H12O6) and galactose (also C6H12O6), right?  If what you really want is just empirical formulae, OK.  There is no way an algorithm could enumerate all of the possible structures/molecules, though.

As a first answer, ignore valence requirements and just do the combinatorics.  If O is in {0, 1} and H is in {0, 1, 2}, then there are six possible formulae, one of which is the nonsense H0O0.  I guess that leaves five.

Check out the database at https://academic.oup.com/bioinformatics/article/29/2/290/201920


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Why is this named 3-phenyl-2-butanol instead of 2-phenyl-3-butanol?
Why is the following compound named 3-phenyl-2-butanol and not 2-phenyl-3-butanol? Does the alcohol or the phenyl group have priority?

You have already correctly identified the compound as a phenylbutanol. The only remaining problem is the numbering of locants. The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds (first to double bonds if there is a choice)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Clearly, the $\ce{-OH}$ group is the principal characteristic group in this case since you have already used it as principal characteristic group to find the correct suffix ‘ol’.
Therefore, a low locant is assigned first to the $\ce{-OH}$ group according to Rule (c), which yields phenylbutan-2-ol.
After that, a low locant is assigned to the phenyl group according to Rule (f); however, the only possible numbering left is 3-phenylbutan-2-ol.


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Reducing Strength of Phosphorus Oxoacids
My chemistry textbook suggests that "acids which contain a P-H bond have strong reducing properties". Both hypophosphorous acid (H3PO2) and pyrophosphorous acid (H4P2O5) contain two P-H bonds; a P-O-P bond is present additionally in pyrophosphorous acid (as in the figures below).
Which of these two turns out to be a stronger reducing agent, and why? Or do they have equal reducing properties?

IMHO, pyrophosphorous acid has comparable reducting properties as phosphorous acid, not as hypophosphorous acid.

@Poutnik Phosphorous acid has three OH groups, while pyrophosphorous acid has two. Phosphorous acid is a stronger reducing agent in general, hence? (I understand the reducing strength will change in acidic, neutral and basic media)

AFAIK, phosphoric acid had 3 OH groups, phosphorous acids has 2 OH and 1 H.

IMHO, condensation of acids does not affect reduction ability much. What does affect it is number of H on P.

@Poutnik Yeah, 2 OH's (was confused about phosphorous as phosphonic acid). Having one H, you say phosphorous acid doesn't compare well with phosphinic acid as a reducing agent, which has 2 H's (ie phosphinic acid is weaker). Is this somewhat backed by exp. evidence?

I guess phosphinic=hypophosphorous to keep constant naming within the question.In my knowledge, hypophosphorous acid is stronger reducting agens than phosphorous acid.

@Poutnik And the relative reducing strengths of hypophosphorous and pyrophosphorous acids?

the best is to find and compare redox potentials. But they may not be available for Pyro versions.

See https://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29  -0.499V versus -0.276V


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Can anyone tell me what reaction is this?

I came across this reaction, but I have no idea what reaction is this. Can anyone please tell me?
What will be the products?
My first guess is that a carbene will be formed which will attack on the alkene, but I don't know what to do after that.

Diazomethane under the action of light or heat releases nitrogen leaving a  neutral, divalent carbon species with a lone pair and an empty orbital called a carbene. This carbene then reacts with the double bond of alkene breaking it to form a cyclo- species. in your example, cis-2-butene will react with diazonmethane to form cis-1,2-dimethylcyclopropane with no detectable amount of the trans isomer.
However, do note that if the same reaction is carried out via a triplet carbene, generated via triplet sensitization, then a 1.3:1 mixture of trans to cis dimethylcyclopropane is observed. [3]

Source

https://www.masterorganicchemistry.com/2011/11/05/reagent-friday-diazomethane-ch2n2/
[Source for pic. 1]
http://www.ochempal.org/index.php/alphabetical/c-d/diazomethane/ [Source for pic. 2]
http://reag.paperplane.io/00000832.htm
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Diazomethane%2C_Carbenes%2C_and_Cyclopropane_Synthesis


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How to calculate the kinetic order of an enzymatic reaction?
This question is concerning the metabolism of ethanol by an alcohol dehydrogenase enzyme.
Usually people metabolize alcohol equivalent to "one beer per hour". One beer is said to contain 33cL. The ethanol content of a beer is 4.6%(V/V). If this approximation is correct for an average lean person of 75kg after drinking two beers, then what is the kinetic order of reaction of the enzymatic reaction metabolizing the alcohol?
The alcohol content in one beer is:
$(33 \cdot 0.046)cL = 1.518cL$
The ethanol content in two beers is then:
$(2 \cdot 1.518)cL = 3.036cL$
How do I link this to the weight of the person (the 75 kg)? And how to I determine of this follows a zero, first, or second-order reaction?
I know that a first order reaction is given by the following rate expression:
$-\frac{dA}{dt} = k \cdot A$
The second order reaction is given by the following rate expression:
$-\frac{dA}{dt} = k \cdot A^2$

This reaction is zero order (except at very, very low concentrations) and this is why it can be used in breathylisers.

@porphyrin Could you please elaborate on how you determine that it is a zero order reaction?

People do not metabolize one beer per hour, see e.g. here

... and here.

The legal limit in some states of 0.08% BAC corresponds to about 17 mM ethanol in the blood. The $K_M$ value for human liver alcohol dehydrogenase is 0.5 mM according to this paper. So at this particular legal limit, the enzyme works at 97% of its maximal capacity. ($v = v_{max} \frac{[S]}{[S] + K_M}$).
Above or very much above the legal limit, the rate does not change much, making it approximately zero order in ethanol. Because the breakdown of alcohol does not get faster with more intake, it can take past the next morning to metabolize sufficient alcohol to reach values below the legal limit if someone drinks heavily at night.
For ethanol concentrations below the $K_M$, the reaction is approximately first order in ethanol, so it gets slower and slower as the ethanol concentration approaches very small concentration.
While the $K_M$ is the same for different individuals, the amount of enzyme is not (depends on the size and health of the liver, gender, and on genetics). That's where the body mass comes in.


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A doubt on the action of NBS/CCl4
Suppose we have a alkene( ex- 2- butene)  we try to subject it to allylic bromination using NBS/CCl4 and then the mixture is separated using fractional distillation. According to me as the reaction proceeds with a free radical pathway we must get  three  products namely cis , trans  1-bromobut-2-ene and the rearranged  product but the answer is given as that the fractions obtained will be only 2 . Where am I going wrong ?

Is your but-2-ene starting material an E/Z mixture or single isomer?

Not specified and that is the main problem

You are quite correct that this is a radical bromination, know as the Wohl–Ziegler reaction. The mechanism is here Wohl–Ziegler. The key point is that the radical intermediate at the allylic position is stabilised by resonance with the double bond giving a 3-centred intermediate. This can brominate at either end (favouring the more substituted), but it does not isomerise the double bond. So I think the question setters are refering to 3-bromobut-1-ene and 1-bromobut-2-ene as the possible products without concerning themselves with the E/Z isomers.

The OP suggested (E)-"ex"-2-butene as a substrate (bp 0.8 oC). Given that the W-Z NBS reaction is normally done in refluxing CCl4 (bp 77 oC) to enhance the solubility of NBS. There may be difficulty in containing the substrate in the reaction vessel. Dry ice/actone condenser???

A poorly chosen example substrate, perhaps.


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Opening of ethers using acids

I got that following my solution i get the product is 3HCO2H
but the given answer is 3HCHO

Each carbon has two bonds to oxygen and two bonds to hydrogen, there is no oxidant present. How do you get HCOOH as product?

This compound is a special type of ether called an acetal. An acetal is a functional group with two alkoxy groups attached to the same carbon atom. They hydrolyze in aqueous acid to form aldehydes and ketones. The pair of carbon-oxygen single bonds is replaced by a carbon-oxygen double bond. The mechanism can be similar enough to what you drew, though an SN1-like mechanism is also reasonable.

The carboxylic acid cannot form. There are two hydrogen atoms attached at that carbon atom, and they need to remain in the structure of the product. In order for the carboxylic acid to form, you would need to displace a hydrogen atom with a hydroxy group, not a likely step in a mechanism. When in doubt draw in the implicit hydrogen atoms.

For my interested what software did you use to draw the mechanism for this or was it an image you found?

I used ChemDoodle by iChemLabs: https://www.chemdoodle.com/. It's not free, but it is inexpensive compared to ChemDraw, and there is a student discount.


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Inorganic chemistry for condensed matter physicists
I'm a theoretical condensed matter physicist, on a field of magnetism research. I wonder if any of you could recommend a concise and well-written textbook on inorganic chemistry to learn concepts of the ligand field theory, coordination of transition metal complexes and their magnetic states, etc.
During my work, I just look up the periodic table and see the electronic configurations, but the oxidization of the transition metal complexes depends on ligands, and it's frustrating as I don't seem to understand a general principle.
I'm quite familiar with solid state physics, and a bit of quantum chemistry. It would be great if you could recommend me textbooks or materials on inorganic chemistry for "solid state physics" or "material science".

Related: https://chemistry.stackexchange.com/q/37303/41556

Any text book suitable for an advanced undergraduate course in inorganic chemistry should cover these topics. Which text may be best for this purpose will be a matter of opinion, so I am not going to recommend a specific one. Links to specific books/editions will risk being out of date for future readers of this question and answer. However, some author combinations have been active in writing text books in this field for many years. I would look for a recent edition of the book by either:

Meissler and Tarr (as of this writing, the most recent edition is the 5th edition by Meissler, Fischer and Tarr, published by Pearson, 2013)
Shriver and Atkins (as of this writing, the most recent edition is the 6th edition by Shriver, Weller, Overton, Rourke, and Armstrong, published by WH Freeman, 2014)

Just be careful since the phrase "inorganic chemistry" in some uses refers to the introductory course sequence in chemistry, what is more commonly called "general chemistry".

After searching for books in Amazon, I came to realize that Inorganic chemistry actually deals with almost everything that is not organic chemistry. But I think now I can find topics which I am interested in. Thanks for  the advice!


text:
Oxidation state of Mn in MnO₄⁻
I have tried to understand what the oxidation state for Mn in $\ce{MnO4^-}$ is. So far i have understood that O have the oxidation state of -2 so because there is 4 O the number turns to -8. What happens to the minus at O? and what is the oxidation state for Mn?

The sum of the oxidation states of all of the atoms needs to equal the total charge on the ion.
For example, for $\ce{NH4+}$, if each hydrogen atom has an oxidation state of +1, and the overall charge is +1, then we can solve for the oxidation state of nitrogen:
$$\begin{aligned} 4(+1) + OS_\ce{N} &= +1\\
OS_\ce{N} &= +1 -4(+1) \\
OS_\ce{N} &= -3
\end{aligned}$$
Similarly, we can do $\ce{SO4^2-}$. If each oxygen atom has an oxidation state of -2, and the overall charge is -2, we can solve for the oxidation state of sulfur:
$$\begin{aligned} 4(-2) + OS_\ce{S} &= -2\\
OS_\ce{S} &= -2 -4(-2) \\
OS_\ce{S} &= +6
\end{aligned}$$

Seeing as Oxygen has an oxidation state of -2 and you have 4 Oxygen atoms, the total charge contributed by Oxygen = -8. Because the total charge of this compound = -1, then in this case Mn = +7
+7 - 8 = -1

As you said Oxygen has the oxidation state of -2 so if we add up all the oxygens they have a -8 oxidation state.
As the oxidation state of the ion is equal to its charge, the Mn has to "balance" the -8 to get it up to -1. Therefore the oxidation state of Mn in Mno4- is +7.


text:
Dipole moment in aromatic compounds
Compare the dipole moment of the following:

o-nitrophenol
o-dichlorobenzene
o-xylene

What I thought was:
o-dichlorobenzene should have the maximum, due to the acute angle and being more electronegative than the I- effect of $\ce{CH3}$ groups, and that o-nitrophenol would be the least since OH shows M+ and $\ce{NO2}$ is M- effect.
But the data from the internet says:
o-nitrophenol (3.01 D) > o-dichlorobenzene (2.67 D) > o-xylene (0.64 D)
How?

Think of it this way: o-xylene is a virtually undisturbed aromatic ring, o-dichlorobenzene is an aromatic ring with electrons slightly pulled towards chlorine, and o-nitrophenol is an aromatic ring with electrons pulled towards NO2 with great force, and also pushed to it from another direction (should the direction be the same, maybe the effects would mostly negate each other, but this is not the case).

@IvanNeretin any reason why those effects dont cancel or atleast reduce the net moment?

Why, of course they do reduce it. In p-nitrophenol the dipole moment would be even greater. As to why they don't cancel out: why would they? Different directions, like I said.

@IvanNeretin Shouldn't the dipole moments in o-nitrophenol cancel out? I mean, -OH is strong donor, -NO2 is strong acceptor. Now if I try to draw vectors, they seem to slightly opposite to each other (not exactly! But at around 60 degrees from -OH dipole moment's head to -NO2 dipole moments tail.). A reasoning I found from an unreliable source suggest that, "o-nitrophenol has high dipole moment because of its ionic resonating structure". Could you please enlighten me?

Like I said, of course they do partially cancel out.


text:
Aufbau principle in filling Ti
A few weeks ago, I was given this question to complete:

The ground state electronic configurations of Ti is $\ce{[Ar] 4s^2 3d^2}$
 State whether it follows the Aufbau principle or not. If not, explain why and state what the configuration would be if the principle was followed.

Now, I know the Aufbau principle tells us that when filling orbitals with electrons, the lowest energy orbital is filled first.
Looking at this diagram and the periodic table, I thought the filling was right - $\ce{4s}$ is before $\ce{3d}$ so I thought it was reasonable.

However, according to the Lecturer this is apparently wrong.
He says:

The $\ce{3d}$ level is at a lower energy than the $\ce{4s}$ and should be filled first according to the principle. It takes the above configuration to minimise the d-d interelectronic repulsion energies. The alternative configuration of [Ar]3d4 would have a higher energy.

I'm slightly confused here.
How am I wrong?

The 3d/4s thing is much more complicated than textbooks would have you and I believe. The total energy of the system is not simply a sum of individual orbital energies (orbitals are quantum mechanical objects, and the maths doesn't work out that way). So, the fact that the aufbau principle works is a bit of a coincidence (and the "breakdown" of aufbau in some contexts should not come as a surprise). See e.g. https://chemistry.stackexchange.com/a/8426/16683

But, afaik, your lecturer's interpretation is correct. There are some references in the disclaimer here: https://chemistry.stackexchange.com/a/33310/16683 (I suggest not bothering with the rest of the answer.)

The energy level diagram which you give and is to be seen in the standard textbooks is assumed to apply to all elements. This is not the case, nor would we expect it to be so. After calcium the 3d drops below the 4s so fills first. Coulombic repulsions tend to push electrons up into the larger 4s orbital where repulsion is less. This means the 4s electrons are the outer electrons which are lost first in chemical reactions and define the atomic radius of the atom.


text:
Stability of formyl chloride
Why is formyl chloride unstable? I know it is just a combination of CO and HCl. But why it is unstable whereas other acyl chlorides are stable, like acetyl chloride?

It is unlike other acyl chlorides as formyl chloride dissolves in water. At the same time, it reacts vigorously with it, and no other acyl chlorides produce carbon monoxide and hydrochloric acid.

Can you explain the reason for its solubility in water?

Because of its structure (HCOCl), there is hydrogen bonding between the H- atom and water.

authors: ə̷̶̸͇̘̜́̍͗̂̄︣͟


text:
Charge distributions in the transition state of the SN2 reaction
I understand that a similar question has been posed here. However, the scope of my question is framed differently. I will present analysis of two different sources of information and I would like to question if there is a contradiction between the two.
Perspective by Carey & Sundberg (2007)
Carey & Sundberg (2007) provides the following perspective on the effect of substituents on the rates of $\ce {S_N2}$ reactions, on p. 418-419:

The extent of the rate enhancement of adjacent substituents is dependent on the nature of the TS. The most important factor is the nature of the $\pi$-type orbital that develops at the trigonal bipyramidal carbon in the TS. If the carbon is cationic in character, electron donation from adjacent substituents becomes stabilizing. If bond formation at the TS is advanced, resulting in charge buildup at carbon, electron withdrawal is more stabilizing.

It is accompanied by  the following diagram:

What I would like to point out is that the authors claim that it is possible for the carbon atom involved in an $\ce {S_N2}$ reaction to develop a slight positive charge when the cleavage of the $\ce {C-LG}$ bond is more advanced than the $\ce {C-Nu}$ bond. Perhaps, we can see this as more electron density is leaving the carbon atom via the transfer onto the leaving group, compared to the electron density transfer from the nucleophile onto the carbon atom.
Computational study by Pross & Shaik (1982)
Pross & Shaik (1982) seeked to dispel the commonly-held belief that:

the extent of charge development in a transition state is dependent on the earliness or lateness of that transition state. For example, in the $\ce {S_N2}$ reaction of an anionic nucleophile with a neutral substrate, most of the negative charge would be expected to be localised on the nucleophile for an early transition state ... while for a late transition state most of the charge would be localised on the leaving group ...

They used a method they referred to as the "linear combination of valence-bond configurations". They write that the wavefunction describing the transition state will reflect the fact that both configurations shown below are of equal energy at the crossing point of the diagram (Fig. 1) below by having equal weights, where the two configurations refer to $\ce {Nu^- + R-X}$ and $\ce {Nu-R + X^-}$. This would mean that the transition state would have around $\ce {-0.5}$ of the charge localised on $\ce {Nu}$ and the other $\ce {-0.5}$ localised on $\ce {X}$. Does this suggest that the carbon atom is roughly electrically neutral?

Does research presented by Pross & Shaik (1982) contradict what is mentioned by Carey & Sundberg (2007)? And is the theory presented by the textbook in that paragraph incorrect? Personally, I have never seen such an effect described in this way in other texts on the $\ce {S_N2}$ reaction.

References

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.
Pross, A.; Shaik, S. S. Is Charge Development a Measure of SN2 Transition State Structure? Tetrahedron Lett. 1982, 23 (51), 5467–5470. doi:10.1016/0040-4039(82)80159-7


text:
Use of fractional distillation in organic chemistry
Can fractional distillation be used to distinguish between:
1) d/l form of a compound
2) geometrical isomers of a compoun (Cis and Trans)

The boiling point of an (R)-isomer is the same as the one of the enantiomer, the (S)-isomer.   Since the condensor does not offer a chiral environment to the chiral molecules, there is no discern of them and consequently, neither "single" nor "fractional distillation" separates them.   (You may expose the molecules to a chiral surface to adsorb, though, and separate the enantiomers by chiral gas chromatography; a specialized approach in analytical chemistry and -- because of the scale of amount typically accessible -- much less frequently deployed on a preparative one (a review).)
A successful separation of (cis)-(trans)-isomers by distillation of course depends on how different the boiling points of the pure compounds are.  Looking up the data for (trans)-stilbene (bp 305 C reference) and (cis)-stilbene (307 C at 1 atm reference) suggests other separation techniques may be much more suitable since less demanding.  This especially becomes relevant if you need a technique easy to scale, to manage and to be run continously (e.g., distillation).

fraction distillation means seperation of compounds on the basis of specificity of boiling point.so we cannot find the geometrical and stero isomerism using fractional distillation.
But we can find the form of the substance wheather it is gas or solid or liquid using fractional distillation.


text:
Geometry optimisation in Gaussian09
I have looked at various sources regarding this, including the Gaussian official website. However, I have not managed to understand the geometry optimisation method used by Gaussian, a method known as the Berny algorithm. I understand that this algorithm uses the idea of updating the Hessian matrix, a matrix of 2nd order derivatives of the energy with respect to spatial coordinates, with past gradient history, something similar to the conjugate gradient method. My mathematical understanding is not that advanced so I would prefer if the answer uses a step-by-step approach. An overview of the approach in laymen terms would also be appreciated.


text:
Confusion in change in enthalpy relation
In the equation
$$\Delta H = \Delta U + \Delta PV,$$
under what circumstances can we simplify it into
$$\Delta H = \Delta U + P\,\Delta V + V\,\Delta P$$
by product rule?
In this article (pdf slides) it is mentioned that it isn't true. In one question I got a wrong answer using
$$\Delta H = \Delta U + \Delta PV + V\,\Delta P,$$ but the correct answer using
$$\Delta H = \Delta U + \Delta PV = \Delta U + P_\mathrm f V_\mathrm f -P_\mathrm i V_\mathrm i.$$
The question was:

one mole of a non ideal gas undergoes a change of state from ($\pu{2 atm}$, $\pu{3 L}$, $\pu{95 K}$) to ($\pu{4 atm}$, $\pu{5 L}$, $\pu{245 K}$) with a change in internal energy, $\Delta U = \pu{30.0 L atm}$. Calculate $\Delta H$ in the process.

Using
$$\Delta H = \Delta U + P\,\Delta V + V\,\Delta P$$
you get $\Delta H = \pu{40 L atm}$, which is wrong. And using $$\Delta H = \Delta U + \Delta PV$$ you get $\Delta H = \pu{44 L atm}$, which is correct.
Is it something to do with the non ideal gas?

It's nothing to do with ideality or lack thereof. The "product rule" works for infinitesimals, i.e. $\mathrm d(pV) = p,\mathrm dV + V,\mathrm dp$, or $\mathrm d(pV)/\mathrm dx = p(\mathrm dV/\mathrm dx) + V(\mathrm dp/\mathrm dx)$ where $x$ is some other variable. For finite quantities that doesn't make sense, because you have two pressures $p_1, p_2$ and two volumes $V_1, V_2$; so exactly which pressure should $p$ in $p\Delta V$ be, and which volume should $V$ in $V\Delta p$ be?

@orthocresol - the product rule works just fine in this case, if you add back in the term that's dropped out for the infinitesimals. See my answer for details. The answer to your question is that Pi should be in PDV and Vi in VDP.

@Andrew you're correct. This is bordering on splitting hairs, but I think the only thing we're disagreeing on is how one defines a product rule. OP seems to  associate it with the one used in basic calculus: $(fg)' = f'g + g'f$, and so I went along with that; that doesn't work with finite changes, of course. If you add back in the $\Delta p \Delta V$ term, the expansion of $\Delta (pV)$ no longer explicitly has the same form as the 'product rule' $d(pV) = pdV + Vdp$; but if you define $\Delta (pV) = p_i\Delta V + V_i\Delta p + \Delta p \Delta V$ to also be a 'product rule', then yes, it works.

The answer is clear if you pay attention to how the "product rule" comes about.
Let's write out $\Delta (PV)$ as $P_fV_f-P_iV_i$, where subscript $f$ means final and $i$ means initial.
Now substitute $P_f=P_i+\Delta P$ and $V_f=V_i+\Delta V$:
\begin{align}
\Delta (PV) &= (P_i+\Delta P)(V_i+\Delta V)-P_iV_i\\
&=P_iV_i+P_i\Delta V + V_i\Delta P + \Delta P \Delta V -P_iV_i\\
&=P_i\Delta V + V_i\Delta P + \Delta P \Delta V
\end{align}
When we are dealing with infinitesimals in derivatives, that last term is $\mathrm{d}P\mathrm{d}V$ and is assumed to be negligible. But when $\Delta V$ and $\Delta P$ are not infinitesimal, you need to include it.


text:
Cis Orientation of dienophiles orientation of groups maintained in Diels Alder
When I was reading about Diels-Alder reactions, I was confused about the cis and trans rules for dienophiles, as shown below.

The text told me that "a cis dienophile will generate a ring with cis substitution, and trans dienophiles will generate a ring with trans substitution."
I'm a bit confused.
Why is this the case?

To answer you question, imagine (or: build a molecular model) that the newly formed six-membered ring may be simplified as a flat disk.  Then

either you start with a (trans)-configurated diene, as in the first example drawn by you.  Since the Diels-Alder simultaneously forms the new $\sigma$-bonds on the expense of $\pi$-bonds and moves the location of the remaining double bond, the substitutents R are either both below, or both above this reference plane.  Relatively to each other, these substitutents R in the product conserved their (cis)-relationship from the dienophile.
or, if your dienophile is (trans)-configurated, yielding a product were one of the substitutents is above, and the other is below this reference plane; corresponding to the second example drawn by you where are R conserved their (trans)-relationship to each other.

On paper, the observed conservation of sterochemical information is both the synchronicity of the events (pushing the pairs of electrons, absence of [charged] intermediates as e.g. in nucleophilic substitutions), as well as the principle of conserving the symmetry of molecular orbitals envolved here.

You are right, this process and many others that involve 3 pairs of electrons are concertated reacctions, and this happens as before having the final bonds filled, there exist a combination of sigmas and pi wave functions so that the other bonds that have been previous have no time to take other positions. After profesor Eliel, and also my own experience doing this in 500 l reactor using methylpentadiene as diene and acrolein or crotonic aldehyde and having huge GLC showing the different products.


text:
Why gas solubility decreases with temperature?
I found this quote on a Chemistry LibreTexts website:

If the heat given off in the dissolving process is greater than the heat required to break apart the solid, the net dissolving reaction is exothermic (energy given off). The addition of more heat (increases temperature) inhibits the dissolving reaction since excess heat is already being produced by the reaction. This situation is not very common where an increase in temperature produces a decrease in solubility.

I understand that when dissolving releases more energy than is being used in the dissolution, the net reaction is exothermic. Why exactly does the excess heat inhibit the dissolution? I cannot see the reason why. Could somebody explain this in more detail?

On which website? Gah, ninja'd. Also citation is about solids. Dissolving gas decreases entropy, so with rise in temperature, process becomes less favorable.

Water, for example, holds the most dissolved gas just above freezing... but in the transition to ice, most gases are excluded (methane being an exception), leading to bubbles in the ice. Boiling the water before freezing it removes dissolved gases, enabling formation of bubble-free ice.


text:
Why can't you form an anhydride from an ester via nucleophilic attack?
The reaction referred to is the nucleophilic acyl substitution reaction, where the nucleophile is a carboxylate ion and the electrophile is the ester. I am having a difficult time visualizing why this cannot occur.

It may be somewhat plausible in the E conformation of the ester, but certainly not in the Z form

Carboxylate is a poor nucleophile, alkoxide a much better one. If the reaction occured, the alkoxide would then go on to react with the anhydride - a very much better electrophile - so the outcome would be ester plus carboxylate again.

@YUSUFHASAN I don't see how ester conformations are relevant. Even if it made a difference to the fundamental reactivity of esters vs anhydrides as Waylander explained (and it doesn't), it's not as if you could isolate two different conformations of the ester.

The reaction you are asking about is a carboxylate plus an ester reacting to form an anhydride plus an alcohol. The reverse reaction, making an ester from an alcohol and an anhydride, is one commonly performed in teaching labs.
So the equilibrium lies on the side of the ester. The principle of micro-reversibility would suggest that this reaction can go in either direction, but you would have trouble detecting formation of the anhydride and alcohol from the carboxylate and ester because of where the equilibrium lies.


text:
Water vapour pressure in oxygen gas bubble
Suppose we have a source of O$_2$ bubble formation in water at specific nucleation sites, how can we estimate the water vapour pressure inside the O$_2$ bubbles?
I know that:
(1): $p_\mathrm{O_2}+p_\mathrm{H_2O} = p_\mathrm{tot}$
(2): $c_\mathrm{O_2} = K_\mathrm{O_2}\cdot p_\mathrm{O_2}$ (Henry's law)
where $c_\mathrm{O_2}$ is the concentration of $O_2$ in the liquid phase, and $p_x$ are partial pressures. How can I get $p_\mathrm{H_2O}$ probably as a function of $O_2$ gas pressure?

If I understand your question then you're looking in the wrong place for a solution. If the bubbles are 100% oxygen then the question has nothing to do with partial pressure in the gas phase and should instead be about the surface tension of the solvent and the pressure will vary in inverse proportion to the size of the bubble. You could try editing your question so it can be answered here. Or ask on physics.se .

There is still a partial pressure, because there is also a vapor pressure of water right? So it is not 100 % oxygen

I will also post it on physics.se maybe someone can help me there: https://physics.stackexchange.com/questions/451473/water-vapor-pressure-in-oxygen-bubble

Quite right about water partial pressure, but unless you're working at 99 celsius or in a partial vacuum then compared to surface tension it'll be tiny in a tiny bubble.

The oxygen is acting as essentially an inert gas holding water. The equilibrium vapor pressure of water will be less than its saturation vapor pressure due to the concave surface. The more concave (i.e. the smaller the bubble) then the lower will be equilibrium vapor pressure. This is a surface tension effect and is called (by some) as the Kelvin equation:
$$RT\ln\left(\frac{p_\mathrm r}{p_\mathrm o}\right)=2g\frac{V_\mathrm m}{r}$$
Where $R$ is the gas constant, $T$ the absolute temperature, $p_\mathrm r$ is the vapour pressure inside the bubble, $p_\mathrm o$ the vapour pressure on a plane surface,  $p_\mathrm o$ the surface tension, $V_\mathrm m$ the molar volume of the liquid, $r$ the radius of the bubble.
Note that the sign of $r$ is negative for bubbles and positive for drops, which show an increase in vapor pressure to that of a planar surface
Reference

introduction to colloid and surface chemistry by D J Shaw, 2nd edition, page 57, ISBN 0 408 70021 1

I agree with this. This gives however the pressure relative to a planar surface. Is there a way that the oxygen pressure could also change the water vapour pressure for a planar interface?


text:
Hyperconjugation and the stability of alkenes
Clearly, this issue has been touched on here and here. However, I would like to raise a new point on the issue.
I have always thought that stabilisation of the $\ce {C=C}$ $\pi$ bond via hyperconjugation is through an interaction between a $\ce {C-C}$ or $\ce {C-H}$ $\sigma$ MO and the $\pi^*$ MO. This is also the explanation presented by orthocresol and Philipp in the posts linked above. However, the explanation given by Fleming (2009) is somewhat different as can be seen from the diagram below, taken from p. 74

There seems to be an internal contradiction in the text. On the previous page, Fleming does mention about the interaction between the $\sigma$ MO and the $\pi^*$ MO yet the diagram shows an interaction between a $\sigma$ MO and the $\pi$ MO. Not sure if it's a misinterpretation on my part...
On p. 73, he wrote:

One factor is the hyperconjugative stabilisation of the $\ce {C=C}$ $\pi$ bond by the alkyl groups. Fig. 2.11 shows the interaction of the orbitals of a $\sigma$ bond with the orbitals of a $\pi$ bond, which is similar to the interaction of two $\pi$ bonds in butadiene (Fig. 1.32). Although the $\sigma$-bonding orbital and the $\pi^*$-antibonding orbital are further apart than $\psi_2$ and $\psi_3^*$ in butadiene, they are just close enough to mix in a bonding sense effectively to lower the energies of $\psi_1$ and $\psi_2$ in Fig. 2.11, making the drop in energy $\ce {E1}$ a little greater than the rise inenergy $\ce {E2}$.

Reference
Fleming, I. Molecular Orbitals and Organic Chemical Reactions. John Wiley & Sons, Ltd. United Kingdom, 2009.

Ordinarily without the contribution of the π* orbital, we would expect the interaction of two filled orbitals to be destabilising, in the sense that $E_2 > E_1$ (as drawn in the diagram). This is the same rationale as why simple MO theory predicts the helium dimer to be unstable with respect to dissociation, for example.
It might be slightly unclear, but my interpretation of the text is that the empty π* orbital contributes a little bit to both MOs $\psi_1$ and $\psi_2$. This leads to a tiny stabilisation of both of these two MOs, such that the destabilisation $E_2$ is a bit smaller and the stabilisation $E_1$ a bit larger. Although it is not very obvious, you can see this from the figure.
It could probably have been made clearer by drawing extra dotted lines from the π* orbital to those MOs. My guess as to why Fleming hasn't done that is because he wanted to use dotted lines to illustrate the most significant orbital interactions (those between orbitals of similar energy). Note how he emphasises in the text that the σ and π* orbitals are not very close in energy (which implies that the interaction is small).

So the main contribution to the stabilisation is still as what Fleming draws, the interaction between $\pi$ and $\sigma$ MOs?

The interaction of two filled orbitals (e.g. σ/π) is not stabilising in nature. The stabilisation arises from σ/π* interactions, as has been described before, but is just not clearly illustrated in the diagram.

If you read the text again, it is quite clear about this: "Although the σ-bonding orbital and the π*-antibonding orbital [...] they are just close enough to mix".

Hmmm... Then it is puzzling why he showed a different phenomena occurring in the diagram

That's what my third paragraph addressed.

Recently, I have been thinking about what is the physical interpretation of this MO interaction. For example, in this case, is the C=C pi bond strengthened or weakened? The sharing of electron density with the antibonding pi MO of the C=C bond seems to suggest some sort of weakening of the pi bond. Does this C-H sigma - C=C pi* interaction suggest that electron density originating from the C-H bond is being "shared with the C=C bond?


text:
Does the reaction between phosphorus and chlorine produce phosphorus trichloride or phosphorus pentachloride
Phosphorus has two covalent compounds it can form with chlorine, where it is a single, central atom bonded to chlorine, which are phosphorus trichloride and phosphorus pentachloride.  Because of this, I am not sure as to whether phosphorus will react with chlorine to form the trichloride and the pentachloride.
I am thinking that it would probably be more likely for it to form the trichloride, since it is more favorable for phosphorus to not have an expanded octet, but does the amount of chlorine gas in relation to the phosphorus (excess or not) play a role in what compound will be formed?

Yes, it will depend on concentration of chlorine

vide https://en.wikipedia.org/wiki/Phosphorus_trichloride#Preparation & https://en.wikipedia.org/wiki/Phosphorus_trichloride#Preparation

Sir Humphry Davy had mentioned that when he burned phosphorus in chlorine gas, he obtained a clear liquid (phosphorus trichloride) and a white solid (phosphorus pentachloride) - reference 15 of Wikipedia.


text:
RO- as leaving group in saponification process

If RO- is a stronger base than OH- (and leaving groups are weak bases), why is RO- a leaving group here?

It's unproductive step otherwise. It may happen more often then productive one but you just get substrate back, ready for another try.

$\ce{OH-}$ of course can, and will, be a leaving group here. The reaction is in equilibrium, $\ce{OH-}$ adds and leaves very many times, but that is a dead end as it does not drive the reaction forward. $\ce{RO-}$ can also be the leaving group - the difference between the basicity of $\ce{OH-}$ and $\ce{RO-}$ is not so great that it does not happen. Even if it happens one time in ten thousand, when it does the cascade it sets off makes the reaction irreversible so the reaction proceeds, if slowly, and the salt of the carboxylate is produced.


text:
Equilibrium :. Why does a system at equilibrium try to undo the effect of external stress applied on it?
" Le Chatelier principle states that a system at equilibrium when subjected to external stress tries to undo the effect of that stress"
In short, why does system at equilibrium follow Le Chatelier principle?

It is not that a system follows Le Chatelier's Principle, as it is not a law such as Newton's laws of motion,  but rather that this is a shorthand way of working out what happens; a 'rule of thumb' if you wish.
If a system is in stable equilibrium it must have the lowest energy possible under the prevailing conditions. If now a perturbation is made, say by changing temperature or pressure, the system is no longer at equilibrium because some energy has been added or removed. It will now change to lower its energy and again reach equilibrium under the new conditions. Le Chatelier's Principle is then a statement of this.
For example, suppose that as a result of molecular motion, a small part, A, of a larger volume of a gas spontaneously increases by  $\Delta V$. Thus the pressure  decreases, $\Delta p < 0$, and so the net force on A exerted by the surroundings is in a direction that tends to restore equilibrium.

Thank you porphyrin for answering my question. I really appreciate your answer.


text:
Why is concentration of nickel nitrate constant?
Consider a cell with both nickel electrodes and a solution of nickel nitrate ($\ce{Ni(NO3)2}$). Current is passed through the solution. The $ E° $ values predict that  nitrate in the solution should get reduced to nitrogen oxide. But a question regarding this has the answer given that the concentration does not change. Why is the concentration of nickel nitrate constant?

Are both electrodes made up of nickel and immersed in the same solution of nickel nitrate? What type of cell is it? If you can elaborate on the question, then it would be easier to answer.

Yes both are immersed in the same solution of nickel nitrate.

Negatively-charged nitrate ions travelling to the also negatively-charged cathode doesn’t seem to be efficient.

Also, the reduction of nitrate requires acid: $$\ce{NO3- + 2H+ + e- -> NO2 + H2O}$$

Okay then the next possibility is reduction of $ H+ $ . Are there any other reactions taking place?

Don't forget the overpotential on nickel when you look at other reactions.

That means there would be a deviation in the calculated and experimental values of the electrode potential, but the reactions will still be same, won't they? Is reduction of $ H^{+} $ the only reaction taking place?

To plate nickel out of a 1 molar Ni(NO3)2 solution should require 0.25 volts (plus a little more). But the fact that a nickel anode in H2O could generate 0.25 volts suggests that a small current could carry nickel from one electrode to the other at a voltage significantly lower than 0.25 volts. The total effect is a mechanical (physical) shift of the metal, not a change in the chemistry.
The potential for reduction of NO3- is high in 1 molar H+, but this concentration will not be nearly approached in the nickel nitrate solution (pH will probably be about 4-5). In the Nernst equation, a factor of 10000 in the hydrogen concentration and a factor of 4 in the exponent predicts that the reduction of NO3- ion will not take place at voltages which would be enough to move nickel from one electrode to the other.
Now if the current is increased by raising the voltage very high, I think that not only would the nickel be transferred faster, but some nitrate could be reduced to NO.


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Testing for asbestos using readily available chemicals and tools?
Given that it is difficult to determine who is telling the truth in the recent claim that J&J covered up finding asbestos in their powder, is there any chemical test that we can perform on our own to determine with reasonable confidence whether the particular sample of powder we have bought contains asbestos? (I am aware that one test involves viewing the sample under a sufficiently high-powered polarizing light microscope, but not everyone has such a tool.)
I have not much chemistry background beyond high-school, and it seems that there are numerous kinds of asbestos, apparently characterized more by their large aspect ratio than by their chemical composition. For example, chrysotile (which accounts for ≈95% of asbestos in buildings in the US) has an idealized chemical formula $\ce{Mg3(Si2O5)(OH)4}$, but may not even occur as exactly that. Talc according to wikipedia is either $\ce{H2Mg3(SiO3)4}$ or $\ce{Mg3Si4O10(OH)2}$. These break down as follows:

Chrysotile: $\ce{(Mg^2+)3[Si2O5]^2−([OH]^−)4}$
Talc A: $\ce{(H+)2(Mg^2+)3([SiO3]^2+)4}$
Talc B: $\ce{(Mg^2+)3[Si4O10]^4−([OH]^−)2}$

So we cannot test for $\ce{Mg^2+}$ or $\ce{[OH]^−}$. But is there a chemical test for $\ce{[Si2O5]^2−}$ that can be done at home? And is it safe to rule out chrysotile if no $\ce{[Si2O5]^2−}$ can be detected, or can the amount be too low for chemical detection but still dangerous?

The chemical difference between asbestos and talc is very small: the differences are structural. This is why the best test is probably to look at a sample with a microscope.

@matt_black: Thanks, but I'm not sure I understand your comment. Are you saying that talc has Si2O5 as well? Or are you saying that Si2O5 has very similar chemical properties to Si4O10? Or something else?

@user21820 Very different minerals result from small changes to the structure of the silicate units they contain. The differences are caused by the subtly different way the SiO2 units are connected. Those differences are not easy to detect chemically. So, yes, Si2O5 is hard to tell apart from Si4O10 using chemistry (but trivially easy looking at the crystals of the mineral).

@matt_black: I see. It's a bit disappointing to be unable to do it myself, since I've read that some talc sources are contaminated while others are not, so I wouldn't simply trust any testing done on samples provided by the company, and one would need to test a statistically significant number of samples from off the shelves to be relatively confident there is no asbestos. As of now, with many involved parties withholding a lot of information, I would've liked to just ignore them and check for myself.

you can do it yourself, build x-ray diffraction apparatus

Is there a chemical test for $\ce{[Si2O5]^2−}$ that can be done at home?

First $\ce{[Si2O5]^2−}$ isn't a poly atomic ion like sulfate ($\ce{SO4-}$) or nitrate ($\ce{NO3}$) but represents a stoichiometry in the material.
Unfortunately at this time a test without x-ray diffraction is not possible. Asbestos/chrystolite is not just a chemical formula, but a crystal structure. It's not just what it is made of but also how the atoms are structured that makes it potently carcinogenic. If you were to take asbestos and melt it then cool it sufficiently fast to form a glass (amorphous/non-crystalline substance), it would not have similar carcinogenic properties as the crystalline material.

I am aware that one test involves viewing the sample under a sufficiently high-powered polarizing light microscope, but not everyone has such a tool.

It is worth noting that using polarized can be done at home with polarized filters but is a presumtive test not a direct method. Essentially all chrystolite will rotate light but not all minerals that rotate light are chrystolite.

Oh thanks very much for your answer. So am I correct to say that it was wrong for me to break down the formula "Mg3(Si2O5)(OH)4" into ionic 'components'? Also, thanks for explaining that the polarization test yields false positives.

@user21820 Yes, but I will note that the bonding of silicates is somewhat an exception compared to other oxide ions (phosphate, borate, sulfate, carbonate, nitrate, chlroate, etc...) so I would say the intuition was good but silicates are a curve-ball.

I see. I was aware that silicates can be macromolecules, such as SiO2, but I got confused by the metal ions. Based on what you said, I would now guess that you have an Si2O5 macromolecular structure with embedded Mg2+ and OH− ions to balance out the charges. Right?

@user21820 Essentially yes, but the word is network solid not macromolecule.

Ok thanks, though the article you cite says "In a network solid there are no individual molecules, and the entire crystal or amorphous solid may be considered a macromolecule". Haha..


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Electron affinity of nitrogen and exchange energy
When studying the effects of electron filling in chemistry, I was studying why energy is needed to add an electron to a nitrogen atom i.e. electron affinity.
Nitrogen already has an electron structure of $\ce{1s^2 2s^2 2p^3}$ and using my knowledge of exchange energy, the energy released as exchange energy is
$$\frac{3 \times 2}{2} = 3K $$
But adding an electron means there is 1 electron  $\ce{1s^2 2s^2 2p^4}$ with opposite spin to the rest of the other electrons so the total exchange energy is still
$$\frac{3 \times 2}{2} + \frac{1 \times 0}{2} = 3K $$
so why is it unfavourable i.e. endothermic for nitrogen to became a nitrogen anion $N^-$?

$ \rm N^{-}$ has the outer electron structure of $ \rm 2s^2 2p^4$. This means we have coulombic repulsion in one of the p orbitals which means adding a single electron will be endothermic. This accounts for the decrease in ionization energy between N and O.
When you get to $\rm N^{3-} $ we have 6K units of exchange energy which is enough to offset the coulombic repulsion.
You see this effect when considering the electron configurations of the transition elements.


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vapor pressure and elevation clarification
I’m a bit confused on why higher elevations/decreased atmospheric pressure wouldn't affect the vapor pressure of a liquid-or have a negligible affect on vapor pressure.
With decreased atmospheric pressure, boiling point decreases, and a decrease in BP would increase vapor pressure. So isn’t atmospheric pressure changing the vapor pressure?

The only variable that affects the vapor pressure of given substance is its temperature.

Think about what vapour pressure is instead of using analogy

The boiling point of a liquid is the temperature such that the liquid's vapor pressure is equal to the atmospheric pressure. When you go to a higher altitude, the boiling point is lower not because the vapor pressure of the liquid at a given temperature has changed, but because there is less atmospheric pressure.
The vapor pressure of a liquid at a given temperature is an intrinsic property due to the equilibrium between the liquid and gas phase of that substance.

Vapor pressure does affect boiling point, but it doesn't go the other way. BP depends on VP, and VP depends on temperature. It is a one way relationship.


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6-oxoheptanal reaction with LiAlH4
Is it possible?
the reaction below

That isn't 6-oxoheptanal.

You have lost a carbon, did you intend this?

This is not 6-oxoheptanol and the product cannot be as shown in the snapshot.  You have to keep all the carbon-carbon bonds, and the reactant has secen carbons in the chain versus your product having six.
Lithium aluminum hydride can reduce not only ketones but also carboxylic acids (see here).  So you are going to convert both of the functional groups to alcohols.  With that in mind, please review how the LAH reacts, and then you can get the correct answer.


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Eriochrome Black T titration - Concentration of Indicator and endpoint
While not as frustrating an indicator as murexide, Eriochrome Black T's wine red to blue transformations provide significant challenges. The change from "fully wine red" to "fully blue with no trace of red" takes around 1mL. It goes through first from the wine red to a purple, which then becomes are deeper blue after several more drops (with still very slight traces of purple and hence red). Addition of several more drops thereafter produces a dark blue colour which appears to have no purple trace (there may be a slight amount), and adding 10 or so drops converts the darkish blue to a beautiful, light blue. Where is the endpoint here? I assume it is at the deep blue, although the wine red to purple transformation is very sharp and abrupt.
Another challenge that this indicator provides is the amount used. It appears the more indicator I put, the more a pure blue solution appears to have a purple trace. The endpoints differ with a tenfold difference in Erio T amount up to 1mL. I have tested this theory - with a dilute solution I titrate to the "deep blue endpoint", add more Erio T, and then hints of purple become visible etc.
Does anyone know the true endpoint of Erio T and the optimal amount to add?
Note: This was for a Mn (II) titration (ascorbic acid added). The problem manifests itself in magnesium titration as well, and prevents obtaining endpoints to any accuracy.
Edit: The concentration of titrant and analyte are both ~0.01M. This is quite standard for EDTA titrations.


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Equilibrium constant from mole ratio
Chapter 4, problem 13 from the Chemical Priciples [1, p. 170]:

Equilibrium concentrations
Experiments have shown that at $\pu{60 °C}$ and $\pu{1 atm}$ total pressure, the equilibrium ratio of $\ce{NO2}$ to $\ce{N2O4}$ in moles in a closed vessel is exactly $2:1$.
a) Calculate the equilibrium constant, $K_\mathrm{c}$, for the dissociation of $1$ mole of $\ce{N2O4}$ into $2$ moles of $\ce{NO2}$.

My answer:
The chemical equation is $\ce{N2O4 <=> 2 NO2}$ and at equilibrium the ratio is $2:1$ so shouldn't
$$K_\mathrm{c} = \frac{[\ce{NO2}]^2}{[\ce{N2O4}]} = \frac{(\pu{2 mol L-1})^2}{\pu{1 mol L-1}} = \pu{4 mol L-1}?$$
The answer in the back of the text book is $K_\mathrm{c}= \pu{0.0488 mol L-1}$.
I do not understand how this answer was calculated from the information provided.
References

Dickerson, R. E.; Gray, H. B.; Haight, G. P. Chemical Principles, 3d ed.; Benjamin/Cummings Pub. Co: Menlo Park, California, 1979. ISBN-13: 978-0-8053-2398-6.

Where did you get your concentrations of $\pu{2 mol/L}$ and $\pu{1 mol/L}$ from? The question only told you the ratio is 2:1, so in general the concentrations should be represented as $2x$ and $x$, where $x$ is not necessarily equal to $\pu{1 mol/L}$. Therefore $K_c = (2x)^2/x = 4x$. The real task for you is to find $x$. To find the concentration $x$, you need to find (1) the amount of substance, i.e. how many moles of each substance there are; and (2) the total volume. The former was, of course, already given to you, but the latter wasn't.

From the ideal gas law, what is the molar volume of an ideal gas mixture at 60 C and 1 atm pressure?  From this result, what is the moles/liter of the gas?

I suggest to pick one gaseous component, say, $\ce{N2O4}$, and, applying ideal gas law, express $K_\mathrm{c}$ as a function of the partial pressure of the chosen gas. Since at the equilibrium the ratio $n(\ce{N2O4}):n(\ce{NO2})=1:2$ is exact (e.g. molar fractions are $1/3$ and $2/3$, respectively), and the volume doesn't change, you can immediately deduce that
$$[\ce{N2O4}] = 2[\ce{NO2}];$$
$$p(\ce{N2H4}) = \frac{p_\mathrm{tot}}{3}$$
where $p_\mathrm{tot}$ is the known total pressure of the system. Using ideal gas law
$$[\ce{N2O4}] = \frac{p(\ce{N2H4})}{RT} = \frac{p_\mathrm{tot}}{3RT}$$
and the law of mass action
$$K_\mathrm{c} = \frac{[\ce{NO2}]^2}{[\ce{N2O4}]} = \frac{(2[\ce{N2O4}])^2}{[\ce{N2O4}]} = 4[\ce{N2O4}] = \frac{4p_\mathrm{tot}}{3RT},$$
one can plug in the parameters and calculate $K_\mathrm{c}$:
$$K_\mathrm{c} = \frac{4\cdot\pu{101325 Pa}}{3\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{333 K}} = \pu{48.8 mol m-3} = \pu{4.88e-2 mol L-1}$$


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Does the priority go for the longer substituent, the closer one, or one which contains a pi-bond in naming Alkenes and Alkynes?
According to the new IUPAC nomenclature, the longest chain is given the priority over that which contains the most number of double bonds.
In naming this alkene:

From which end would we start numbering? Because of what substituent?

'ethenyl' (= 'vinyl') as a substituent is not fundamentally different from 'ethyl' or 'propyl', there's no special rule for it. Hence, number in the direction which generates the smallest set of locants.

You have already correctly identified the parent structure to be the longest chain, i.e. dodecene. As usual, the most important simplified criteria for the numbering of locants for substituent groups in such compounds are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds (first to double bonds if there is a choice)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
| (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound that is given in the question doesn’t have a principal characteristic group that is expressed as a suffix. Therefore, Rule (c) is not relevant in this case.
A low locant is assigned to the double bond according to Rule (e). In this case, this rule always results in the same locant for the double bound (the ‘ene’ ending); it doesn’t matter at which end of the chain you start counting; i.e. the name of the parent structure is dodec-6-ene.
Finally, low locants are assigned to all the substituents as a set according to Rule (f). Therefore, the systematic name is 8-ethenyl-3-ethyl-5-propyldodec-6-ene rather than ‘5-ethenyl-10-ethyl-8-propyldodec-6-ene’ since the locant set ‘3,5,8’ is lower than ‘5,8,10’.


text:
How did scientists discover the concepts of anode and cathode without knowing about electrons?
Scientists researched electrons and came to know about electrons after the battery was developed as we can see in the JJ Thomson's model that he used the cathode and anode. My question is that that if the scientists have known already that there are positive and negative charge that they have made the battery they should have known about electrons.
I am confused in this. Research tells that electrons were first discovered by JJ Thomson and that there are negative charges in atoms. My question is that that, without knowing about electrons how were the ideas of cathode and anode developed.

The bottom line is that scientific experiments describing electrical phenomena are possible without any good theory about what causes the phenomena. Static electricity has been known since ancient times, batteries were discovered long before any theory that explained them well. Thomson's breakthrough gave us a good theory (or parts of it) not all of which mashed well with previous ideas (electricity flows in the opposite direction to electrons which actually explain the electricity because the flow was defined before a good theory about what was happening).

Electricity is a macroscopic phenomenon. It does not depend on the notion of an electron.
In 1733 Charles François de Cisternay du Fay discovered the existence of positive and negative charge. In 1800, after the invention of the voltaic pile, the first electrolysis experiments that showed the effect of electric current on water have been made.
In 1833 Michael Faraday defined the terms cathode and anode:

The anode is therefore that surface at which the electric current, according to our present expression, enters : it is the negative extremity of the decomposing body; is where oxygen, chlorine, acids, &e., are evolved; and is against or opposite the positive electrode. The cathode is that surface at which the current leaves the decomposing body, and is its positive extremity ; the combustible bodies, metals, alkalies, and bases, are evolved there, and it is in contact with the negative electrode.

His definition does not involve particles of any type. In fact Richard Laming firstly hypothesized the concept of an indivisible quantity of electric charge in 1838.
You may want to study https://en.wikipedia.org/wiki/History_of_electrochemistry for further information.

https://en.wikipedia.org/wiki/Cathode#Etymology This section of Wikipedia also provides some insight from the perspective on the origin of the names "cathode" and "anode."


text:
Vibronic coupling: How do I determine the new point group after vibration?
So I was recently discussing the transitions in Egyptian Blue ($\ce{CaCu[Si4O10]}$) with some of my students, who had to prepare this compound. What I like in particular in this case is how, at least in a simplified view you can show, that the blue is not simply due to one single transition with the complementary color to blue but actually composed of three possible transitions in the visible region that cause an absorption by all colors but blue.
I also looked for a paper so they could give a reference in their report and this paper summarizes the transitions pretty well.
But when I thought about it for a while some questions came up in my mind on the actual transitions. So there are three transitions, either from the $\mathrm{a_{1g}}$, the set of $\mathrm{e_{g}}$, or the $\mathrm{b_{2g}}$ into the $\mathrm{b_{1g}}$. As the square plane has an inversion center we are dealing with the same problem as in octahedral geometries, the odd / even parity rule upon excitation.
Then I found this line in a text on symmetry rules in electronic transitions

In cases where transitions coincide with vibrations of the initial or final state, the electronic transition moment R needs to be replaced by the transition moment Rv. [...] Here Ψv denotes a vibronic wave function. In complete analogy with electronic transitions, we could derive the following set of selection rules for vibronic transitions:
   $$\Gamma(\psi_v') \otimes \Gamma(\psi_v'') = \Gamma(T_x)$$

At first I misread that line and thought that the dipole moment would change to vibrational modes as well, so I could choose from a larger set of possible symmetry elements to do the transitions, but then going through the examples at the bottom of the page it seems more like the vibration would cause a lower symmetry and change the point group into one where the transition may be allowed. It further says:

The total symmetry of a system can be expressed as a conjunction of the symmetry of the electronic states Γ(Ψ) and the symmetry of the vibration Γ(Ψv).
   $$\Gamma(\psi_{ev}) = \Gamma(\psi) \otimes \Gamma(\psi_v)$$

And this is the line that I don't understand anymore. So I looked at the examples And for the one with the point group $\mathrm{c_{2v}}$ for the transition between $\mathrm{b_{1}}$ and $\mathrm{b_{2}}$ it says:

The picture changes if we account for vibrational modes too. An asymmetric stretching causes the molecule to get from $\mathrm{c_{2v}}$ to $\mathrm{c_{s}}$ and only the molecular plane remains as symmetry element. State $\mathrm{B_{2}}$ becomes $\mathrm{A'}$ and state $\mathrm{B_{1}}$ $\mathrm{A''}$. Consequently, the transition dipole moment is of symmetry $\mathrm{A''}$ and perpendicular to the plane of the molecule. Obviously, reduced symmetry increases the number of possible transitions.

I also found a correlation table for the point group $\mathrm{c_{2v}}$ that links it to it's sub-groups. For $\mathrm{c_{2v}}$ the irreducible representation for the vibrational modes should give:
 $$\Gamma_{vib} = 2 A_1 + B_2$$
So for $\mathrm{B_{2}}$ we have this case, mentioned in the link, where we have to change to a vibrational mode. But how do I determine what sub-group is created by this vibration? In their example they showed that $\mathrm{B_{2}}$ becomes $\mathrm{A'}$. In the correlation table this would be $\mathrm{c_{s} (σ_{yz})}$. And if I then go back to the character table for $\mathrm{c_{2v}}$ the only thing that I could find would be that for this entry, $\mathrm{c_{s} (σ_{yz})}$, $\mathrm{B_{1}}$ would return -1 while $\mathrm{B_{2}}$ remains unchanged.
So does this give me any hint on how to transform my point group to its sub-group by a vibrational mode? As I have never had any lecture on point groups and symmetry I can unfortunately only look for tables, texts or similarities since I have no clue about point groups at all. This means the above-stated ideas could be terribly wrong. I tried to understand the text I quoted above as well as possible and this would be my interpretation.
This means that in my Egyptian Blue case, I would determine the symmetry element of the transition dipole moment. If the transition is not allowed I consider the vibrational normal modes and if either the initial and or final state has the same symmetry element, then I need to find out (using a correlation table) how the point group can be changed to a sub-group, where the transition would be allowed.
Therefore, I finally remain with the question: How do I determine which sub-group is chosen by the actional of a vibrational mode in this specific case?

@ Tyberius has give you the answer, but for more detail have a look at chapter 7 of 'Modern Spectroscopy' by Hollas, (4th ed, publ Wiley.), in particular 7.3.2 to 7.3.5.

Thank you, I will try to get a copy of it when I return to university tomorrow.

I don't believe it is strictly necessary to determine the subgroup induced by a vibration in order to determine if a vibronic transition is allowed. You can simply use the equation from your 2nd quoted block of text.
For example, consider whether an electronic transition would be allowed between a $b_1$ and $b_2$ state of water. Without including vibrational contributions,
$\langle b_1|\overset{a_1}{\underset{b_2}{b_1}}|b_2\rangle\neq a_1$, so the transition would seemingly not occur. However, if we instead consider the vibronic state $b_1 \otimes B_2=a_2$ (formed from the $b_1$ electronic state and the $B_2$ asymmetric stretch), the transition can occur, as $\langle a_2|b_1|b_2\rangle= a_1$.
The $b_2$ state is also a vibronic state, in this case the original electronic state paired with a totally symmetric vibrational state. We could also obtain other transitions by considering the $b_2$ electronic state paired with other vibrational states.

Thank so, so that was what they meant by multiplying it with the vibrational state. I didn't do it explicitly so I just assumed it would not change anything if both symmetry elements were the same, to begin with. Now that I see the product it makes sense to me now. Thank you!

@Justanotherchemist yeah some of the notation of point groups and direct products can be confusing if it isn't clearly explained. Glad I could help!

Can I combine it with any possible vibrational mode or does it have to be the one that has the same irreducible representation as the initial or final state?

@Justanotherchemist you can combine any of the molecules vibrational modes, not just those that match the electronic state.

I don't know if it's okay to give an additional question like that but I feared that editing the original post would lead to confusions in the continuity of the answers and it's probably not advisable to make it an extra question.
So I tried it with combining only one state with a vibrational mode and it worked. But can I really treat it that way? What I'm basically saying is that I need a distortion to make the transition happen. And as we know from Franck-Condon principle will the electronic transitions be much faster than the vibrational ones. So I guess this means that we need to have the vibration first, followed by an electronic transition. Which means I am starting from a distorted initial state, one where the geometry has to be different but there is apparently no change in the point group or overall symmetry and just this one initial state changes for some reason. But the change in electronic states should be almost vertical and therefore there wouldn't be any changes in the configuration. Sure the transition is into the excited vibrational state with the biggest overlap integral, but does that mean that it will automatically have this new vibrational quantum state? To my understanding, those are separated steps. The absorption at one molecular geometry, followed by relaxation and Stokes-shift into the lowest vibrational electronically excited state. So for the resonance condition of both initial and final state, don't we automatically have to assume the same vibrational distortion? Why would that vibronic coupling only act on one state and not on the other, because if it did to both it wouldn't change much about the transition conditions? So a description of it being a change in the point group as a distortion by some vibrational mode that lowers the overall symmetry and therefore allows more electronic transitions to happen seems probably a lot more plausible to me.
I'm okay if it can be solved by just having one of the two states being coupled to some vibration but what is the reason that the other state doesn't do the same thing?

I think it may be better to post this as another question and link back to this one saying it is a continuation. I can post something on that new question, but I believe the issue you are having is that while the vibrational state and the geometry/configuration are related, they are not the same thing.


text:
Does an acetal need to have one hydrogen attached to its carbon?

I was asked to identify the functional groups in this molecule above.
So far I have found an amide ($\ce{-NHAc}$), an ether ($\ce{-OBn}$), and a conjugated diene, but I was also told that there is an acetal ($\ce{RHC(OR')2}$).
However, in the only part of the molecule where I can see a carbon atom attached to two other oxygen atoms, the carbon atom is not attached to a hydrogen atom.
Where is the acetal group in this molecule?

According to the IUPAC definition, an acetal is any compound with a formula $\ce{R2C(OR')2}$ (R' ≠ H). In the past, "acetal" referred only to compounds of this type, which were derived from aldehydes (one R = H), however, it now broadly encompasses all compounds of this type, including those derived from ketones, which were formerly only referred to as ketals. Interestingly, there was a period of time in which use of the term "ketal" was discouraged (IUPAC says 'abandoned', but many chemists kept using it anyway), but it has now been re-embraced as a term, for the subset of acetals that are derived from ketones.
One last note - the molecule shown in this question is a "spiroacetal" - a type of acetal where the acetal carbon is the only atom joining two rings.

IUPAC notwithstanding, the acetal/ketal distinction is more meaningful than acetal itself.


text:
What could be causing noisy pH measurements?
I designed and built a hydroponics system with pH logging and I am trying to understand why pH measurements in the nutrient reservoir vary ~ ±0.2 pH while measurements in a separate bottle of probe storage solution only vary ~ ±0.01 pH.  The nutrient solution is in a 3.5 gallon plastic pail, as shown below.  The pH probe is the one with a blue top.  The black probe measures EC and removing it has no effect on the pH noise in the reservoir.

The plot below shows the  logged pH. The abrupt change is when the probe is moved from the nutrient solution to the storage solution.  Note the relatively large noise in the nutrient reservoir and the nearly imperceptible noise in the storage solution.
In the plot,the pH is measured every 5 minutes at the start of a spray cycle, when the pump is off and has not been running for the last ~4.5 minutes.  Note that similar noise occurs when measuring at a rate of 1 Hz.  During the storage solution measurements, the probe tip is in a separate O-ring sealed bottle.  The bottle is sitting at the spot where the pH probe would normally be inserted into the reservoir.
What are potential causes of the pH noise in the nutrient solution?

First of all, very cool project - I've bookmarked your page! For sources of error, you might be seeing ambient carbon dioxide levels affecting pH - that's the first thing that comes to mind. Your storage solution is likely buffered, and your nutrient solution less so. Others with more relevant experience with hydroponics will likely get things cleared up. Again, very cool!

Carbon dioxide would be the change of pH rather than noise. Noise might be some vibration including that from near machines (refrigerator, conditioning). Vibration should be less in the storage bottle for this to hold (less volume, different material, different place....).

Several factors may influence your sensor readings:

Electromagnetic noise. However, since EMF should be directly proportional to pH, it seems to be not your case. If it were the main reason, the same amount of noise would show up in both the nutrient and the storage solutions.
Temperature. I've included this one because it is unclear from your post whether you are measuring the actual solution temperature when you calculate pH from EMF, or just assume that the temperature is some constant value. The Nernst equation may be used to estimate the amount of "noise" (i.e. the difference in EMF) one may expect from the temperature fluctuations: $\Delta E=2.303\frac{R\Delta T}{F}pH$. That is, if we can neglect the temperature dependence of pH itself, i.e. when $\Delta T$ is relatively small (when $\Delta T$ is large we can no longer neglect the temperature dependences of equilibrium constants). If, for some reason, the daily temperature change in the nutrient bath is different than that in the storage solution, you may expect to get slightly different readings.
Math. We should not forget that the value measured (pH) is logarithmic, and that it is the concentration (well, activity) of protons that makes the difference. For example, to change pH from 6.45 to 6.05 (±0.2), you need to add just around $5.4\cdot 10^{-7}$ mol/l of protons, but to change it from 4.26 to 4.24 (±0.01) - $2.6\cdot 10^{-6}$, or almost 5 times as much. In other words, the same actual $[\textrm{H}^{+}]$ fluctuations cause different pH changes around different pH values. In turn, these $[\textrm{H}^{+}]$ fluctuations may be due to slight inhomogeneities of the nutrient solution, differences in dissolved carbon dioxide concentration, etc. The more neutral your solution is, the more sensitive, with respect to $\Delta [\textrm{H}^{+}]$, your pH electrode becomes. In this regard, you may want to prepare a few solutions with different pH and try to investigate the noise vs. pH dependence. Or do something even more simple - as a reference system, instead of the storage solution (which, in addition, probably possesses some buffer characteristics), try to use something more neutral, like water, and see if the noise is as large as in the nutrient solution.

Since my analytical chemistry is a bit rusty, no other potential cause comes to mind.


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Nucleophilicity in Polar Aprotic Solvent
Why is it that fluoride is more nucleophilic than iodide in an aprotic solvent? The Clayden textbook suggests that lower-row elements of the periodic table are more nucleophilic since their valence electrons are in higher energy levels and are therefore more reactive. However according to the slides from my lecture class, the order of nucleophilicity for halides in an aprotic solvent goes from F- > Cl- > Br- > I- which is opposite from what is stated in the Clayden textbook. Can someone please explain this?
Thank you

It's because of the anion stability. Fluoride is less stable than iodide, therefore more likely to donate electrons.
In protic solvents, the fluoride was stabilised by the partially positive hydrogen end. Therefore, it is less nucleophilic then iodide.