text:
How is 1 mole hypochlorite ion equivalent to 1 mole chlorine gas?
According to my book, 1 mole ClO- is equivalent to 1 mole Cl2. How is that even possible? 1 mole ClO- contains 1 mole Cl atoms. On the contrary, 1 mole Cl2 gas contains 2 mole Cl atoms. So, how are they equivalent?
Let me put it in this way. If I have 257.5g (5 mole) of ClO- ion, will it be correct to say that I have 355g (5 mole) Cl2 gas?
I have provided some context to my question. It's not required to be read in order to answer this question.
Context (not required to be read):
3.04g bleaching powder is dissolved in water to make a 400mL solution. To titrate 25mL of this solution, 40mL 0.075M sodium thiosulphate solution is needed.
Q) Calculate the amount of available active Cl2 from the mentioned bleaching powder.

My book started by finding the molarity of liberated I2 from KI (0.06M). Then my book wrote the following equation:
$$\ce{ClO− + 2I− + 2 H+ → I2 + H2O + Cl−}$$
Then my book said, "Here, two electrons from iodide ion (I-) are accepted by a single hypochlorite ion (ClO-), and oxidation number of Cl (+1) in ClO- is reduced to -1 in Cl- ion. Hence, the oxidation number of Cl is reduced by 2 units. [$\ce{Cl2 + 2e- -> 2Cl-}$] (I couldn't write the oxidation numbers in latex as my book did for this reaction). So, one mole ClO- ion is equivalent to 1 mole Cl2. The molarity of liberated I2 = Molarity of active Cl2 = 0.06M." Then my book went on to find that supplied bleaching powder contains 1.704g of active Cl2, which they claimed was the answer to the given question.

In terms of redox reactions in aqueous solutions, they are equivalent. In fact, chlorine in basic solutions turns quantitatively into hypochlorite while retaining its reducing power: $$\ce{Cl2 + 2OH- -> OCl- + Cl- + H2O}$$ When you acidify the solution, some of the hypochlorite and chloride turn into chlorine again.

@KarstenTheis, Did you mean "chlorine in basic solutions turns quantitatively into hypochlorite while retaining its oxidizing power"? Only the hypochlorite ion releases chlorine upon acidification. The other chloride is just a spectator.

@AChem Yes, oops, that is what I meant to say. In a reaction, either the two chlorines in dichlorine change oxydation state from zero to -1, or the chlorine in hypochlorite changes by two steps, from +1 to -1. In either case, two electrons are transferred

You will find the "equivalency" relations in many classical volumetric or gravimetric analysis. Active chlorine and active oxygen are such representative terms; these terms are academically old but they are still used in industrial analysis around the world.
When we say that a certain bleach contains a certain amount of active chlorine (old word for hypochlorite), all it means that this bleach will also liberate the same amount of iodine from potassium iodide if the same amount of chlorine gas were used.
I think your confusion stems from counting the atoms in chlorine and hypochlorite. This equivalency is not based on atom counting, it is based on the causing the chemical reactions as Poutnik showed in the equations.
For example do not be surprised if someone says,
1 mol $\ce{KMnO4}$ $\equiv$ 5 mol $\ce{Fe^{2+}}$. Now there is no atomic equivalency.

Your answer makes a great deal of sense to me. This is what I thought initially too. However, my book muddied the concept for me: when asked to find the amount of active Cl2 in a given amount of Ca(OCl)Cl, they found that 1.704g of Cl2 is present. They should've made it clear that 1.704g of Cl2 gas is not actually present in the given amount of bleaching powder, but rather the given amount of bleaching powder only behaves (liberates the same amount of Iodine from potassium iodide) as if 1.704g of Cl2 gas was present instead of the given amount of bleaching powder. Am I making sense?

@tryingtobeastoic, Yes you are right.

If we consider reaction $$\ce{ClO- + Cl- + 2 H+ -> H2O + Cl2},$$
there is 1:1 ratio, 1 $\ce{ClO-}$ is equivalent to 1 $\ce{Cl2}$.
The equivalence does not mean the atom counts must match, as they can come from other sources (from $\ce{Cl-}$ in this case).
Similarly
\begin{align}
\ce{ClO- + 2 I- + 2 H+ &-> Cl- + I2 + H2O}\\
\ce{Cl2 + 2 I- &-> 2 Cl- + I2 }
\end{align}
The equivalent molar amounts or masses are those, where one form can be in respective quantity actually or formally converted to the other, with eventual participation of other components. Or, equivalent molar amounts or masses may in an applicable context substitute each other.
Note that the below clarifies the very frequent confusion of mole vs mol.

quantity
unit name
unit symbol

mass
kilogram
kg

molar amount
mole
mol

Working with Bleaches like NaOCl and Ca(OCl2) and some organic reagents etc, I can tell you that Cl2 has very low free chlorine. Cl2 in water makes ClO- as the active specie. So, its fair to say 1 mol of Cl2 has same free chlorine as 1 mol of ClO- though its highly inaccurate. Because bleaching happens in the solution and not air, I'd say 1 mol of ClO- has more free chlorine compared to 1 mol of Cl2


text:
Why is polystyrene (PS) an electrical insulator
The reason behind why graphite is electrically conductive is as ancient as the age of the Queen: that, electrons inside the benzene ring are delocalised.
However, if one looks the repeating unit of polystyrene:

we see the familiar benzene (phenyl?) ring, which logically should also contain delocalised electrons. But as all things turn out to be disappointing; polystyrene is an insulator (both thermal and electrical), and a pretty good one at that.
Why is this the case?

The delocalised electrons are not continuous. If you consider the 3-D structure of the polymer they are sticking out from the chain so there is no overlap from one delocalised system to the next

https://chemistry.stackexchange.com/questions/65021/why-is-fullerene-60-an-insulator-while-graphite-is-a-conductor

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
The major product of radical halogenation: Why is it a halogen attached to a primary carbon and not a tertiary carbon in the given example?
There's this problem that I saw in my Organic Chemistry book:

The reaction between $\ce{(CH3)3CH}$ and $\ce{Cl2}$ forms two products: $\ce{(CH3)2CHCH2Cl}$ (63%) and $\ce{(CH3)3CCl}$ (37%). Why is the major product formed by the cleavage of the stronger primary $\ce{C-H}$ bond?

I understand that chlorine is more reactive than bromine, so it is unselective and its reaction would yield a mixture of products. However, from what I have understood, radicals are more stable when they are more substituted, so why is the major product the less substituted compound and not the more substituted one?

Notice it is 37% what is significantly more than if purely statistical.

Related: What is the major product on chlorination of 2-methylbutane?

The experimental relative rates of chlorination to primary, secondary, and tertiary positions are $1$, $3.8$, and $5$, respectively. In the given compound $\ce{(CH3)3CH}$, there are $9$ identical primary hydrogens and $1$ tertiary hydrogen.
Roughly speaking, the rate of chlorination at primary carbon is proportional to the product of relative rate of chlorination at primary position and the number of identical primary hydrogens and similarly to the rate of chlorination at tertiary carbon.
So theoretical yield of $\ce{(CH3)2CHCH2Cl}$ is nearly $64.3\%$ $\left(=\dfrac{1\times9}{1\times9 + 5\times 1} \times 100 \right)$ and theoretical yield of $\ce{(CH3)3CCl}$ is nearly $35.7\%$ $\left(=\dfrac{5\times1}{1\times9 + 5\times 1}\times 100 \right)$. These percentages are very close to the experimental results. So the yield of monochlorinated compound also depends on the number of hydrogen atoms besides the bond energy and other factors.

I have corrected the formula of monochlorinated product at tertiary carbon.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Water Vapor content at equilibrium in vacuum
Given a hard bodied vacuum sealed container $C$. $C$ itself is internally divided into two sealed sub-containers $C_{lower}$ and $C_{upper}$ (each having volume $V$ $m^3$).
$C_{lower}$ is completely filled with water, whereas $C_{upper}$ contains (assume perfect) vacuum. Assuming the environmental temperature of $T$ degrees Celsius, when we open the separation between both how do we (approximately) calculate the maximum amount of water vapor ($kg/m^3$ at temperature $T$) (at equilibrium)? If needed, we can assume standard atmospheric pressure to keep things simple.
I am aware about the Absolute Humidity values for air (https://www.engineeringtoolbox.com/maximum-moisture-content-air-d_1403.html) but unclear about this vacuum scenario.

Related: https://physics.stackexchange.com/q/440703

@KarstenTheis Thank you. Yes agreed its related but that seems to be just a conceptual question. I am unsure how to calculate the actual vapor content from it..

You’d get the equilibrium vapor pressure for temp T which you can read off of an H2O phase diagram

As I understand you are referring to this (https://en.wikipedia.org/wiki/Vapor_pressure_of_water). Just a followup aren't these values for an open system (air at standard 1 atm). Or can I assume these values are for a closed vacuum system (as in Q) and I can use them to calculate the vapor density of above system?

The vapor pressure determines the temperature T, or vice versa. When you say you can impose "standard atmospheric pressure" the effect is probably not what you think. The relevant pressure is the vapor pressure. See for instance https://chemistry.stackexchange.com/questions/138256/change-in-water-vapor-with-change-in-pressure-and-volume?rq=1

The presence or absence of air does not affect the equilibrium  vapor pressure of H2O. It is solely a function of temp

The vapor pressure is a function of T and inert gas pressure. this adds a second component to the phase rule.


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How useful is thymol blue (acid) as an indicator?
The pH range of thymol blue (acid) is 1.2-2.8.. This pH range seems pretty useless. Generally, the sudden change in pH for SA-SB titration occurs in the range of 4-10; the sudden change in pH for SA-WB titration occurs in the range of 4-7, and the sudden change in pH for WA-SB titration occurs in the range of 8-10.
None of the ranges overlap with the pH range of thymol blue (acid), which is 1.2-2.8. When I found this indicator listed in my book, naturally I thought why this indicator is even listed here as its pH range seems pretty useless.
Are there any acid-base titrations/reactions where thymol blue (acid) is used as an indicator?

Being pH indicator and being pH indicator usable for acidobasic titrations are 2 different things.

I guess you didn't check https://en.wikipedia.org/wiki/Thymol_blue ?

@Mithoron thanks. I assume the comment was mainly for the OP. I have remembered there is an indicator with 2 transitions, but I have forgotten the one is thymol blue. Red 1.2->Yellow 2.8 +  Yellow 8.0->Blue 9.6

@Mithoron Thymol blue is indeed a notable and useful indicator as it has a useful pH range of 8-9.6. However, I was wondering whether its other pH range of 1.2-2.8 is used in any acid-base titrations?


text:
Is the relation [K] = [C] + 273.15 exact?
In reading Atkins' Physical Chemistry, the first section of the first Chapter, the author says that the conversion equation between the absolute temperature and celsius is exact and at the same time the author explains that the Kelvin is defined by setting the triple point as 273.16 [K], and it happens to be the melting point of water is 0.01[K] under this point. How can this be exact? At the first glance ,if Celcius is defined with the melting point and boiling point setting as 0 and 100, it should be empirical.

As Kelvin scale was redefined via the Planck constant and as Celsius scale was redefined to keep following the former by 273.15 offset, the standard water melting and boiling point does not exactly match 0 and 100 deg C. But it does for very most practical purposes. See more at wikipedia - Kelvin

https://chemistry.stackexchange.com/questions/122828/is-there-an-uncertainty-associated-with-the-value-0-k-for-absolute-zero


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Why is this equation true for a salt in acid?
I previously asked a question about a calculation (below)

Calculate the molar solubility of $\ce{SrF2}$ in a solution buffered at pH = 2.00. ($K_{a}$ for HF is $7.2 \times 10^{-4}$). The $K_{sp}$ of $\ce{SrF2}$ is $K_{sp} = 7.9 \times 10^{-10}$

The answers used the fact that:
$$\ce{[HF] + [F-] = 2[Sr^2+]}$$
But I don't understand this. I can understand why this is true in a pure water solution:
$$\ce{[F-] = 2[Sr^2+]}$$
This must be true because if $\ce{SrF2}$ dissolves, it must release $\ce{2F-}$ ions for every $\ce{Sr^{+2}}$ ion it releases. This makes sense.
But for the case where there is a HF buffer, I don't get why you add on the $\ce{[HF]}$. What I personally expect is that because the $\ce{HF}$ will undergo this equilibrium:
$$\ce{HF <=> H+ + F-}$$
then it will decrease the solubility of $\ce{SrF2}$ due to the $\ce{F-}$ being a common ion. But I really don't see how this translates to adding a $\ce{[HF]}$ onto the expression above.

The link is broken, this should be OK. But AFAIK, there is nothing said about HF buffer. In fact, nothing explicit was said about the pH buffer but pH value.

The expression  in a solution buffered at pH = 2.00. (Ka for HF is 7.2×10−4). says that 1/ there is a pH buffer with pH=2.00  AND 2/ the acid HF has acidity constant K_a=7.2×10−4.  It does not say the pH buffer is based on HF/F-.

It is an example for parallel reaction. The reactions taking place are $\ce{SrF2 <=> Sr^2+ +2F−}$ and $\ce{H+ +F− <=> HF}$. The $\ce{F−}$ ions formed from the first reaction reacts simultaneously with $\ce{H+}$ ions in the medium forming
$\ce{HF}$.
For your better understanding let us say that the second reaction is freezed for sometime. So $2[\ce{Sr^2+}]_i = [\ce{F-}]_i$. Now let us say that second reaction is unfreezed. So some quantity of fluoride ions gets converted to $\ce{HF}$. And from second reaction the lost amount of fluoride ions should be equal to the amount of $\ce{HF}$ formed. So $[\ce{F-}]_i =[\ce{F-}]_f + [\ce{HF}]$ and $[\ce{Sr^2+}]_i =[\ce{Sr^2+}]_f$. Here subscripts denote initial and final concentrations. So finally the relation between these concentrations is
$$2[\ce{Sr^2+}]=[\ce{F-}] + [\ce{HF}]$$

(1 of 2) Sorry but I don't get it. If there is some $\ce{F-}$ being turned into HF, then shouldn't there be a decrease in the amount of $\ce{F-}$? Thus we should instead write: $[\ce{F-}]_f =[\ce{F-}]_i - [\ce{HF}]$

(2 of 2) Unless you mean that the SrF2 would disassociate more to make up for it? But then this line wouldn't be true: $[\ce{Sr^2+}]_i =[\ce{Sr^2+}]_f$ since we have changed the amount of F-. Hence the $[\ce{Sr^2+}]$ will be different too as Ksp is constant

Sorry now I have edited

@JohnHon Let us say that quantity of $\ce{Sr^2+}$ formed is more than the solubility obtained from $\ce{K_{sp}}$ equation.  Now the flouride ion concentration decreases in such a way that $\ce{K_{sp}=[\ce{Sr^2+}][\ce{F-}]^2}$

It is just a way of interpretation. In reality no reaction freezes for sometime and proceeds after sometime.

@JohnHon I do not see any big deal here.  Both dissolution and dissociation processes are linked, coming to their equilibrium together, according to the respective equation.


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Why is PCl3 acidic?
Why is it that when adding PCl3 into distilled water, the solution becomes acidic? What exactly is the reaction occurring? Must it not be neutral as the PCl3 is molecular, or then P has a lone pair of electrons, so is it not a Lewis base?

Answers in here may give some insights.

Can you prepare an aqueous solution  of $\ce{PCl3}$?

According to this source  here

Phosphorus trichloride (phosphorus chloride, PCl3) is a colorless fuming liquid that reacts violently with water to liberate phosphorous acid (H3PO3) and hydrogen chloride (HCL) gas.

The oxygen of water attacks the P atom, which is positively polarised because of the 3 electronegative chlorines attached to it.
The HCl gas dissolves in the water to give hydrochloric acid so the resulting solution contains 2 strong acids.

Hey thanks loads! But why does it give HCl and phosphoric acid?

The citation says phosphorous acid, not phosphoric acid.

If you go to the source linked it incorrectly says phosphoric acid


text:
Two wavelength photo-initiators?
Not a chemist so I hope you'll forgive any amateur mistakes. I found an overview of 3D printing resin photo-initiators here:
https://pubs.acs.org/doi/10.1021/acsapm.8b00165#tbl1
All these activate when subjected to light with a compatible wavelength.
Are there photo-initiator (blends) which require two different wavelengths to be present at the same time to activate?
If so, where can I find more information about them? If not, is there a specific reason something like this can't ever be created?

Usually if a molecule absorbs at, say, 400 nm it will also absorb at a shorter wavelength, but not the other way round. Usually it is easier to get visible light sources than  UV i.e. 400-600 nm is easier/cheaper than 350 nm and below.  (Also if you use  light shorter than 300 nm ish there is a risk to your skin and cornea so great caution is needed, screening etc.) There seems no reason to make blends.

Thank you, that explains why there aren't such products.

There are two-photon processes, requiring two separate: https://analyticalscience.wiley.com/do/10.1002/imaging.3042/full/ . This can, indeed, cause the material (photoactivatable GFP, in the citation), to accept differing energy photons. Another example is an IR viewing card, containing a phosphor activated by visible light, that then emits visible light when struck by IR photons: https://www.newport.com/f/near-infrared-nir-sensor-cards


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Difference between Protoporphyrin and Porphyrin?
I understand that Porphyrin is defined as 4 pyrole rings linked by methenyl bridges. When this porphyrin ring is associated with a Ferrous ion (Fe 2+) we get the Heme molecule. However my book mentions that Heme is a complex of Protoporhyrin IX and Ferrous ion. So what is the difference between a Porphyrin ring and a Protoporphyrin ring?

There's no difference, protoporphyrin IX is one of a large number of porphyrins. What you think of as porphyrin is actually porphine, a parent compound of the group, a "core" of protoporphyrin IX.

From here:

Protoporphyrin is a derivative of porphyrin. The key difference between porphyrin and protoporphyrin is that porphyrin is a group of aromatic chemicals which has four modified pyrrole subunits interconnected to each other, whereas protoporphyrin is a derivative of porphyrin which has propionic acid groups.

Protoporphyrine contains porphine as core which is the simplest of the porphyrin chain. Picture from Wikipedia:


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Is There a Problem with the Assignment of Stereochemical Descriptors by ChemDraw in Select Examples?
While using ChemDraw Professional 21.0.0.28 to construct some cyclic allenes and rotationally restricted biphenyls, I came across anomalous assignments of stereochemistry. [For the latest version of the 2013 IUPAC BlueBook go here and for errata look here.]
The major issue is the assignment of a stereodescriptor (R/S) to the hydroxyl carbons in the allenes 1/2 and biphenyls 3/4. By the CD analysis allenes 1 and 2 are diastereomers as are biphenyls 3 and 4 when, in fact, allenes 1 and 2 are identical as are biphenyls 3 and 4! Because the pairs are not diastereomers  but identical, a stereochemical assignment [non-stereogenic] is not warranted.
Rotate the structures 180° about the horizontal axis. Allene 1 superimposes on 2; biphenyl 3 superimposes on 4. These two structures are no different from the case of cycloheptanols 5/6 rotated about the horizontal axis or ᴅ-arabitols 7/8 about a vertical axis. Structures 5 through 8 are handled properly; structures 1 through 4 are not. Is it just me or is there something wrong with the CD algorithm that doesn't address this issue?

A minor issue is the use of the terms R/S when the intent is Ra/Sa. Indeed, IUPAC prefers the descriptors M/P. In this instance, the allenic component of 1/2 and the biphenyl entity in 3/4 would be designated as ‘P’ rather than ‘S’.
Rule P-92.1.2.2.1 (PDF):

The helicity rule can also be applied to the stereogenic axis in allenes and biphenyls. Looking along the chirality axis the ligands are arranged in pairs. When proceeding from the nearer ligand having priority in the pair to the further away atom or group having priority in the pair, the chirality is described by the symbols ‘M’ if the path is anticlockwise; the symbol is ‘P’ if the path is clockwise. Stereodescriptors ‘M’ and ‘P’ are used in preferred IUPAC names. The lowest locant on the axis is cited before the stereodescriptor.

It looks inconsistent. For (5P)-cyclodeca-5,6-dien-1-ol 1/2, "Show Stereochemistry" shows a label at the -OH group, but "Analysis -> Name" only gives "(6S)-cyclodeca-5,6-dien-1-ol".

@Loong: Well, at least one section of CD gets it right. I have contacted Perkin-Elmer about this issue. Hopefully, they will take corrective action.

@Loong: Perkin-Elmer (ChemDraw) has acknowledged the error. Presumably it will be corrected in a future version.


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Mathematical models of Vaporization-Condensation dynamics
I am a mathematician now studying an introductory chemistry course.
Consider a liquid in a closed container, at (say) room temperature.
Then some of the liquid will vaporize. Then some of the resulting gas would condensate, and vice-versa, until a dynamic equilibrium will be achieved.
Let $L(t),g(t)$ be the mass of the liquid and the gas substances respectively, at time $t$.
Assuming that a fraction $c_1$ of the liquid is converted to gas, and a fraction $c_2$ of the gas is converted to liquid, we arrive at the following system of differential equations:
$$
\dot g =c_1L-c_2g, \dot L =c_2g-c_1L,
$$
or in matrix form
$$
\begin{pmatrix} \dot{g} \\\dot{L} \end{pmatrix}=\begin{pmatrix} -c_2 & c_1 \\\ c_2 & -c_1 \end{pmatrix} \begin{pmatrix} g \\\ L \end{pmatrix}.
$$
(We also have the initial conditions $L(0)=L_0, g(0)=0$ where $L_0$ is the initial mass of the liquid. The initial amount of gas is zero).
Question: Is this (admittedly very naive) model considered in the literature? Is it a reasonable model for describing the phenomena?

Short answer is yes. These are standard reaction rate equations that are used for any chemical kinetics description. Read the kinetics section of any intro chem text for more info

I fail to see how this explains anything about the equilibrium between liquid and vapor. At equilibrium the free energy change between gas and liquid is zero and is independent of the amount of material as long as both phases exist. If the chemical potential of the vapor [fugacity]exceeds that of the liquid, vapor will condense, if that of the liquid exceeds the vapor, liquid will evaporate.

What you're trying to develop is a kinetic model of a gas-liquid equilibrium. There are  are such models, but I'm afraid yours won't work.
The reason your model won't work is that it says the rate of evaporation is proportional to the mass of the liquid, and the rate of condensation is proportional to the mass of the gas.
But the rate of evaporation is instead proportional to the surface area of the liquid-gas interface and the tendency of the liquid molecules to escape into the gas phase from that surface.  The latter is in turn determined by the nature of the liquid and the temperature (and, to a much lesser degree, the pressure, since changing the pressure changes the chemical potential of the liquid).
For instance, if you compare two sealed cylinders of the same diameter containing a liquid-gas mixture at equilibrium (same substance in both cylinders), increasing the mass of the liquid simply raises the position of the liquid-gas interface in the cylinder. It will have no effect on the rate of evaporation at that interface.
And the rate of condensation depends on the rate at which gas molecules are hitting the surface and the fraction of those striking the surface that condense.  The former will depend on the the surface area of the liquid-gas interface, the temperature and molecular mass (temperature and molecular mass together determine molecular speed), and the concentration of the gas.  The latter (the fraction hitting the surface that condense) is knowns as the condensation coefficient and, surprisingly, there remains disagreement about its value. See: D. T. Jamieson, Condensation Coefficient of Water, Nature, v. 202, p. 583 (1964).
I do applaud your effort to apply your mathematical background to what is a new field for you.  But to make useful mathematical models, it is first necessary to understand the phenomenon physically.  So I would focus on developing your physical intution for the field.
Indeed, if you want to explore this more deeply, you might want to take a look through the articles that cited that 1964 Nature paper.  That list is easily obtained by searching for the paper in Google Scholar (there's about 200, but you should be able to filter that down quite a bit):
https://scholar.google.com/scholar?cites=13593954657415698794&as_sdt=2005&sciodt=0,5&hl=en
Note that, while you're considering an equilibrium system, most of the above appear to be for non-equilibrium systems.  This makes sense, since being able to model rates of evaporation and condensation is important for commerical and industrial applications, and is also a key part of atmospheric chemistry and the study of water cycles.
Also, just a note about terms: You wrote "Then some of the resulting gas would condensate". Condensate is a noun—it's the substance that condenses.  What you want is the verb, which is "condense".

Thanks, that is a perfect answer. It clarifies that I used wrong physical assumptions. In particular , I really liked your comparison between two cylinders containing different masses. I guess I was too hasty to apply math without waiting for more advanced stages in the course, where kinetic models are discussed. I now look forward to studying them. Thanks for your patience and clear explanations.


text:
How does the second hydrolysis of phthalimide work (hydrolysis of amide)?
In the hydrolysis of Phthalimides we can see two successive hydrolysis of "amides" see ChemLibre for a nice mechanism
This is interesting because normally (also according to ChemLibre) amides are very stable (the amine is a poor leaving group) and don't react.
Nevertheless the first hydrolysis can be rationalized because the nitrogen electrons are delocalized and there is ring strain. Similar to how beta-lactams are more reactive than normal amides.
However, the second reaction appears to be just a normal amide hydrolysis. How does this happen?

Generally under forcing conditions. Elevated temperature, concentrated reagents and long reaction times.

It is normal amide hydrolysis, so I don't get why you're asking.

@Mithoron I was under the impression that amide hydrolysis wasn’t possible based on the ChemLibre website. However, I guess it can be forced.


text:
Why probability for an atom is fraction for population?
In the canonical ensemble the probability of finding an atom (or molecule) in its $i$-th energy state is given by the Boltzmann factor.
This probability is interpreted as the number of microstates that we will find the atom in its $i$-th energy state (neglecting degeneracy) divided by the total number of microstates.
Why we can convert this number to the fraction of atoms (or molecules) that are in this energy state Wikipedia?

The Boltzmann distribution is often used to describe the distribution of particles, such as atoms or molecules, over bound states accessible to them. If we have a system consisting of many particles, the probability of a particle being in state i is practically the probability that, if we pick a random particle from that system and check what state it is in, we will find it is in state i. This probability is equal to the number of particles in state i divided by the total number of particles in the system, that is the fraction of particles that occupy state i.

Is this fraction constant at every time or is the expectation value of the fraction that equals $p_i$?
Intuitively we could say, assuming particles behave independently, that each behaves like a toss coin. If we think of each microstate as flipping all the atoms at the same time, then a microstate is like flipping a coin $N$ times, and because $N$ is so large, then the fraction equals $p_i$.
Also one can view this flipping from a different perspective. Each microstate, is like flipping a number of $N$ coins at the same time, but each flip of the $N$ coins is a single random experiment. In this case how can we equate the probability with the fraction of population?
Is there any formal derivation which shows why $p_i$ equals:
$$ \left\langle \frac{n_i}{N} \right\rangle $$
?

I don't have enough time for a proper answer right now, but to cut a long story short: it's an expectation value, just like everything in QM, and I suggest reading about density matrices, which is the formalism used for quantum stat mech (i.e. using the QM of one particle to describe the QM of many (independent) particles) and will likely clarify many of your questions. I don't think there is any contradiction between your two perspectives; it is a random experiment, but the expectation value is exactly $n_i/N$.

That should not be taken to mean that exactly $p_i$ of the particles are in state $i$, but rather, that if you were to measure their state, then you would expect $p_i$ of them to collapse to state $i$. (replace 'collapse' with QM interpretation of choice if desired)


text:
Determining the average relative mass of an element from the percentage of isotopes

Suppose we have an element $\ce{A}$. Its relative mass is $16$ and it has three isotopes: $\ce{^16A}$ ,$\ce{^17A}$ and $\ce{^18A}$. The available percentage of $\ce{^17A}$ is $0.037\,\%$. What is the available percentage of $\ce{^16A}$ and $\ce{^18A}$?

Let the percentage of $\ce{^16A}$ be $x\,\%$. Then the percentage of $\ce{^18A}$ is $(100 - (x + 0.037))\,\%$ or $(99.963 - x)\%$. From this we can write
$$\frac{17\times 0.037 + 16x + 18\times (99.963-x)}{100} = 16$$
If I solve this I will get $x = 99.9815$ and the available percentage of $\ce{^18A}$ as $(100 - (99.9815 + 0.037))\,\%$, which is a negative number.
What is wrong in my calculations?

What you have done is correct but there is something mistake in average mass.

@Infinite , The formula I used to calculate average relative mass, does that work for 3 or more isotopes? I mean is something like this correct? $$\frac{((p * a) + (q * b) + (r * c) + (s * d) )}{100} = relative-atomic-mass$$. where p,q,r, and s are the atomic masses of isotopes and a,b,c, and d are there's percentages of abundance?

Yeah of course, it doesn't matter on the number of isotopes.

If you use a relative atomic mass of 15.9980, does it work?

Your calculations are fine, but the scenario is impossible.

Yeah, what @ScottCookson said.  If the average mass of an element is 16, and the isotopes have weights of 16, 17, and 18, then there can't be any 17-A or 18-A at all—their abundances have to be 0%.  I.e., the only way you can get an average mass of 16 is if the sample is 100% 16-A.  Any amount of 17-A or 18-A will increase the average mass above 16.   So if you actually do have some 17-A then, mathematically, the only way you can get an average mass of 16 is to have a negative amount of 18-A!  That's why you're getting that result.  Bottom line: The problem wasn't set up correctly.

The problem setting ( and your trying) has messed up isotope nucleon numbers, isotope molar masses and element molar mass.
It is rather:
$$\frac{M(\ce{^{17}A}) \times 0.037 + M(\ce{^{16}A}) \times x + M(\ce{^{18}A})\times (99.963-x)}{100} = M(\ce{A})$$
but molar isotope masses are not available in the task.
It is obvious the element is oxygen and the molar mass data can be found, but the task is not about that. Such a task is a shame.


text:
What is the black sticky residue after heating copper hypophosphite?
I am experimenting with double sided PCB (printed circuit board) via (connections from one side to the other) holes activation.
The activation is the process of making the walls of  holes electrically conductive enough to then electro plate copper. Forming a copper tube with good mechanical and electrical properties.
One activation technique consists of coating the surfaces to activate with copper hypophosphite mixed with amonia water. Then when heating at 150°C, the solution burns and leaves behind a conductive surface.
I understand the reaction produces nano particules of metallic copper. Annoyingly, this also leaves a sticky residue. Which is relatively hard to clean. But successfully cleaning it also takes most of not all of the conductive layer away. And of course it's pretty much impossible to clean within small (less than 2mm) holes anyways. Within holes this residue can leave too rough of a surface, which electroplates badly.
I know there are other techniques for via holes activation. This one has the advantage of being pretty fast with minimal steps. Which is why I am exploring it. Dip in activator making sure it gets within the holes. Cook past 150°C until black. It's now conductive. Sometimes as low as a few ohms.
The activator solution was made with calcium hypophosphite, copper sulphate, amonium hydroxide, and water. There is a bit of excess calcium hypophosphite compared to the sulfate. I filtered the calcium precipitation out. And I diluted the solution to about 15% of amonia. Finally I added 1% of non ionic surfactant (triton X).
What is the sticky residue made of? What could I try to reduce the amount of residue?
PS: I am no chemist.

Have you tried to wash your plates with some basic solution, like a sodium carbonate solution ? I guess this sticky residue may be a polymeric phosphoric acid, which should be soluble in a base, like $\ce{Na2CO3}$. If it does not dissolve this sticky residue, try with some diluted $\ce{NaOH}$ solution, cold or warm. At least, copper metal should resist to this operation.

I only used water, with and without surfactant. Copper does get attacked a tiny bit with caustic solutions. Its a good way to clean it in fact. I wonder if the nanoparticules of copper will resist. Only one way to find out. I will report here.

I tried up to a 3% solution of NaOH. For up to 10 minutes. It didn't seem to do anything to the black residue. Nothing different than just water as far as I can tell. When wet, the sticky residue can be wiped, taking with it the conductive layer. Maybe what I need, is a way to make it extremely thin. I tried diluting the activator in water. But this made for a patchy conductive layer. I will experiment with vacuuming the excess before heating.

My leading suspect is decomposed TritonX.  TritonX is an organic compound used as a surfactant, or synthetic "soap".
You might try to clean your substrate first with TritonX (only) solution as a separate step first, then rinse with distilled water.
Make your activator solution without TritonX, and apply it after cleaning.
After heating, there should be less undesirable residue.

That is really good information thank you. I did try with a "pure" solution. And sadly,  it wasn't noticably cleaner after heating (the amount of triton X is very minute). But it has to be better.  I also did try: ultrasonic cleaning with water and a bit of tritonX. Rinse. Dip in cationic surfactant (sold as moss killer). Not sure if I should rinse after that. And then dip in clean activator and heat. I need to do more testing. I haven't touched it in a while.

Then, I would wonder what the PCB boards were made of, and maybe try a lower drying temperature, maybe 100 - 120 C.  The other thing that comes to mind is electrodeposition of copper from solution.  Also, make sure your ammonia is pure and maybe go with 2-5 % rather than 15%.  Some "ammonia" sold at stores also has other ingredients.

This was a big jug of industrial 30% ammonia. I will try with a different brand and seek lab grade purity. I did try different temperatures. In fact you know there is copper when the solution turns black. Before that, its not conductive. Thank you for the recommendations. I will report here when I get back to it and do more testing. I am busy on other projects right now.

Three months later I hope you are still interested and testing. The black residue is one (if any) of the disadvantages of this method of activation. From the literature and internet info available, there is no clear states about the need (or not) to rinse/blow/dry this residue to a minimum.
I also never trusted about the use of soap (no matter how pure), because is for sure that there are some extra ingredients added. I always believed that the use of soap is only as a surfactant and to ease the surface tension of the liquid activator, so I will be using ethoxylated nonyl phenol instead of soap (warning: like you, I am not chemist, so DO NOT trust what I am saying).
Your industrial 30% ammonia is enough and very good for this application. You do not need reactive lab grade purity for this process. TritonX seems to me very overloaded for this, so I prefer a purest surfactant like ethoxylated nonyl phenol (hope to test it in a few days or weeks).
Another not very clarified step is the oven process. When I was into PCB shop manufacturing floor, all the oven processes were carried through batch racks, so it is uncommon in this process to enter the PCBs into the oven with the activator still wet (a no-no).
You stated that you post-diluted the ammonia to 15% ¿why? ¿it was because of irritating odor? ¿another reason? Let me ask you some questions so maybe we could find some causes... (also please, clarify, at which stage, if any, did you post-diluted ammonia)

Of course, at first, when adding ammonia water to the solution you must be at outdoors and with some respirator mask if possible, because of the pungent odor of ammonia.

Then, after the complete activator solution is ready to use, and when you were hole-activating a PCB, how strong was the ammonia odor? This is a good indicator if the solution was prepared correctly. Did you was indoors? Did you make the activation process without using a respirator mask? It was odor tolerable? Did you perceive ammonia odor at this stage? If so, how much, none, low, medium, high odor, not-tolerable?

At the oven process, again, how pungent was the ammonia odor, if any? Did you had the oven indoors, so you can perceive the heating process odors? Take in consideration that a very precise monitoring of the oven temperature is needed since you do not want to overheat the ammonia and enter some decomposition (for health issues and for the activation process itself).

A convection oven is mandatory so you can place your PCBs vertically (NOT horizontally), and again, this is influenced if some drying is needed (or not) into the activated boards (more testing needed at this stage to observe results).
Regards.

By diluting to 15% ammonia, I meant the concentration in the final solution. For example for 1 liter of final solution, it would mean 0.5l of 30% ammonia. I did try few concentrations, ranging from 5% to almost 24%, and besides the annoying stink, I am not sure it made a significant difference.
For the smell, 15% in the final solution indeed stinks. 5% is very reasonable on the other hand. While cooking the board, the most noticable smell is of white phosphorus. But we are talking about incredibly minute amount here.

As for drying, as long as water had time to evaporate without boiling it did not seem to matter.
I found drying the PCB horizontal made for a cleaner result, simply because the activator seems to spread more evenly instead of pooling at the edges of the holes and make a thicker reside at the lowest edge of the PCB. For cooking the PCB I did it both ways unintentionally and I don't remember noticing a difference.

For the surfactant, I found positive improvement to dip into a 0.1% solution of benzalkonium chloride (found in commercial products like moss remover). Shaking the excess off, then dipping into the activator. This is a cationic surfactant and it should help break the surface tension and hold the activator on the surface to coat (thanks Applied Science on YouTube for this).

Also tried micro etching the surfaces of the hole walls with a very hot solution of permanganate + lye. It did not seem to improve anything. And it's just too dangerous and time consuming for my liking anyways. My understanding is this helps when your have those 100k RPM drill that make holes that are too smooth because it melts the epoxy a bit. Not a problem I have.

The issue with the black residue is how it often makes an uneven surface. Sometimes it clogs the holes a bit. We are talking 0.4mm to 0.8mm holes here. If one side is partially clogged, the electroplating won't be able to make it through.

After reading the original patent again, I can see that the authors clearly state that the black residue are the metallic copper particles (so, if black, then you have copper formed). Also, I can see that the common article widespreaded (from instructables) differs in many aspects from the original patent, obviously in an effort to simplify the process, but I think that in some concepts it adds complexity as well.
You can avoid desmearing (you named it microetching the holes) with permanganate if you are only making 2-layers PCBs. Only when making innerlayers is mandatory.

I recommend that you read the whole patent document carefully. My efforts now are to reduce the black residue, not in the holes, but over all the whole substrate, since it would be useful to have a batch rack of activated boards ready to enter the batch oven witthout dipping liquid solution. Also, I will be experimenting replacing ammonia with glycerol, like in the patent document examples.

One additional step may be laminating, exposing, and developing dry film over both sides, with via artwork exposing, similar to LPI soldermasking. Then you will have a board with dry film over almost all the surfaces, with the vias exposed in plain copper. Only then, you could dip the PCB into the activator (the concentrations and ingredients of the activator solution may no pose a problem for dry films, but testing is needed for Ca(H2PO2)2, but I think it is pretty safe for dry films).

With the use of previous dry film laminating, your PCB only should be receiving the activator into the holes. Of course, into the oven, you must do a pre-heating step to reach only a safe temperature before the dry film starts to melt or damage. And with hope that at this temperature, hopefully most of the water in the activator started to evaporate so maybe you could strip the dry film to continue with the heating process. More testing here is needed, since dry films tend to increase bonding to the surface after heating, but in some cases with overheating, the dry film starts to crack.

Again, these are only vague ideas. More testing and experimentation is needed, but I think that a reliable method only could be assured individually. That is, the equipment, consumables, etc. each person have available influences all of the process, and a practical solution serving all the needs for everyone (like a tutorial) it is not practical. So more testing is needed.

I have read the patent a few times. As well as a few others that talk about copper hypophosphite. If I knew what I was doing, I wouldn't be asking here. The black residue contains the nanoparticles of copper. But it is also sticky and clog holes.


text:
Displacement Reaction
In a displacement reaction I understand that the more reactive metal essentially takes the place of the less reactive metal. But I can't seem to understand how this more reactive metal can break the bonds of this compound. Why would this more reactive metal want to take the place of the less reactive metal in the compound, if the compounds is already very stable?

This can be easily understood if one consider the electrochemical series. I can explain this concept with a simple and famous illustration. Consider the following reaction:
$$ \ce{Zn + Cu^2+ -> Zn^2+ + Cu}\quad E^\circ=+1.1\ \mathrm V$$
Where, $E^\circ$ is the standard cell potential for a cell reaction. From thermodynamics we have a relation between $\Delta G^\circ$ and $E^\circ$ which is given by, $$\Delta G^\circ=nFE^\circ$$ where, $n$ is the number of electrons transferred and $F$ is known as Faraday and it is the magnitude of charge on one mole of electrons. So from this relation we will get $\Delta G^\circ\lt0$ for the above reaction. And this is the criterion for spontaneity or feasibility of the reaction. So this reaction would happen in forward direction. This illustration explains why a metal would displace another metal in such kind of reactions.

User124053's explanation does not correspond to the reality. He or she says that
"the more reactive metal essentially takes the place of the less reactive metal. But I can't seem to understand how this more reactive metal can break the bonds of this compound"
No ! No metal "takes the place" of another one. No bonds are broken : the reaction takes place between ions and a metal in aqueous solution, as explained by Infinite. The only effect is a charge transfer between $\ce{Zn}$ and $\ce{Cu^{2+}}$.


text:
Does nickel precipetate gold and copper?
I dissolved some CPU s and RAM s  in aqua regia the day after I found a green solution with black powder in the bottom. So what just happened is it nickel or tin from solder?

The chart implies sharp boundaries where in fact they are indistinct. Magnesium does react with water, but very slowly (and aluminum and zinc can react with bases as well as acids), and the metals below hydrogen ca still be reacted with some acids.

CPU and RAM ard made of millions of transistors. Each transistor is made of silicon or germanium doped with neighbors in the periodic table, like gallium. arsenic, or plenty of other elements in traces like indium, antimony, etc. And these transistors are interconnected by golden copper wires. Capacitors and resistors are made of still different metals or semi-metals. Your solution must have a complex composition. It is worth being analyzed by standard techniques of qualitative, then quantitative analysis

As you said silicon, which doesn't dissolve in aqua regia and even so the amount of those doping elements are decimals percent.


text:
Calculate frequency dose from different light intensities

The quantum yield for homolytic O-O cleavage of (CH$_3$)$_3$COOC(CH$_3$)$_3$ at $266$ nm is $0.21$. In one experiment, a pulsed laser of this wavelength was used to induce the formation of t-butoxyl radical from (CH$_3$)$_3$COOC(CH$_3$)$_3$. The experiment is to study the kinetics of recombination (dimerization) of phenoxyl radicals. These are formed by the t-butoxyl radical abstracting a hydrogen atom from phenol in solution (inert solvent). To get good statistics, the laser is allowed to shoot light pulses with a certain frequency (ie with a certain time between the pulses) and then follow the disappearance of phenoxyl radical by measuring its absorbance as a function of the time after each pulse. Then the results from each pulse are superimposed to obtain a statistically reliable mean value. You can regulate both pulse frequency and energy per pulse. To obtain a good mean value, at least $95$% of the phenoxyl radicals from the previous pulse must have been consumed before the next light pulse hits the solution. Calculate the maximum pulse frequency that can be used if the incident light intensity is $10^{-7}$, $10^{-5}$ and $10^{-3}$ J cm$^{-2}$, respectively.
The illuminated area is $0.5$ cm$^2$ and the volume is $0.5$ mL. The absorbance at $266$ nm (which is entirely attributed to (CH$_3$)$_3$COOC(CH$_3$)$_3$) is $0.14$. We can assume that the length of the laser pulse is negligibly short in this context and that the formation of phenoxyl radical from the reaction between t-butoxyl radical and phenol is quantitative and immediate. The rate constant for recombination of phenoxyl radicals is $3.5 *10^7$ M$^{-1}$ s$^{-1}$.

I am completely lost when it comes to this problem. To begin with I have never heard of pulse frequency nor can I find any equation related to it. The only equation I can find with the frequency of radiation is: $T = I*t*h*v$, where $v$ is the frequency and $T$ is the radiation dose (which I don't really know how to calculate).
I would be truly thankful if someone could push me in the direction of how to solve this. I have never encountered a similar problem and there is so much information given that I don't know what is useful and what is not.
Thank you!
EDIT: Attempted answer
$1$. Calculate the absorbed energy, in J.
E = I$_0$ * A = $10^{-7}Jcm$^{-2}$ * $0.5$cm$^2$ = $5*10^{-8}$J
$2$. Calculate the energy of one 266 nm photon, in J.
E = hc/λ = $\frac{6.62*10^{-34}Js * 3 * 10^8m/s}{266*10^{-9}m}$ = $7.46*10^{-19}$ J
3. Divide to get number of photons absorbed.
Here I am not sure which energy should be in the numerator and which in the denominator and depending on it I either get a very high number ($6.675*10^{10}$) or a very small number ($1.49*10^{-11}$).
4. Multiply by quantum efficiency to get number of effective photons.
Continuing with both gives me: $0.21$ * $6.675*10^{10}$ = $1.4*10^{10}$
and, $0.21$ * $1.49*10^{-11}$ = $3.13*10^{-12}$
5. Multiply by 2 to get number of radicals.
$3.13*10^{-12}$ * $2$ = $6.258*10^{-12}$
$1.4*10^{10}$ * $2$ = $2.8*10^{10}$
6. Divide by volume and Avogadro’s number to get molar concentration of radicals.
[R] = $\frac{6.258*10^{-12}}{5*10^{-4}L * 6.022*10^{23}mol^{-1}}$ = $2.08*10^{-32}$ mol/L
[R] = $\frac{2.8*10^{10}}{5*10^{-4}L * 6.022*10^{23}mol^{-1}}$ = $9.3*10^{-11}$ mol/L
7. Use the second order integrated rate expression to find the time required for 95% of the radicals to recombine.
Here I am also a bit unsure if what I am doing is correct. So we have a second order reaction so the rate equation is:
$\frac{1}{a-x}-\frac{1}{a}$ = k$_2$t
We have that at least $95$% has to be consumed so $a-x$ is the $5$% that's left.
$a-x$ = $2.08*10^{-32}$ mol/L * $0.05$ = $1.04*10^{-33}$ mol/L
$a-x$ = $9.3*10^{-11}$ mol/L * $0.05$ = $4.65*10^{-12}$ mol/L
We now solve for $t$:
t =$\frac{\frac{1}{1.04*10^{-33}}-\frac{1}{2.08*10^{-32}}}{3.5*10^{7}}$ = $2.61*10^{25}$s
t = $\frac{\frac{1}{4.65*10^{-12}}-\frac{1}{9.3*10^{-11}}}{3.5*10^{7}}$ = $5837$ s
8. Take the reciprocal to find the max laser repetition rate.
max laser repetition rate = $\frac{1}{t}$ = $\frac{1}{2.61*10^{25}}$ = $3.83*10^{-26}$ s$^{-1}$
max laser repetition rate = $\frac{1}{t}$ = $\frac{1}{5837}$ = $1.7*10^{-4}$s$^{-1}$
Which is not the same answer that @Ed V stated that he got. Did I understand the help given in the comments correctly or where am I going wrong?
Thank you!

Laser pulse frequency is confusing here, it means 1/time between laser pulses, so 10 Hz is 0.1 seconds apart.   Normally laser frequency it means the frequency of the laser's light.  The fact that the laser is at 266 does not matter here.

@EdV I'm sorry but I still find it a bit confusing.. what exactly am I calculating? Am I trying to calculate 1/time? I don't have any equations that involve laser pulse energy nor frequency so following what you are saying is a bit difficult.

@EdV Thank you for your help, I have tried to follow the steps that you wrote but I don't get the same answer as you stated that you got. I have edited my question and included my calculations if you have the time to take a look at it :)


text:
How does the redox reaction in a voltaic battery pile start?
From the Wikipedia article, the redox reaction for a voltaic pile is:
\begin{align}
\text{Oxidation:} && \ce{Zn &-> Zn^{2+} + 2e^-}\\
\text{Reduction:} && \ce{2H^+ + 2e^- &-> H2}\\
\text{Total:}     && \ce{Zn + 2H^+ &-> Zn^{2+} + H2}
\end{align}
What is causing the reaction to take place?  It seems that the zinc anode must be oxidized and lose electrons, which travel through the anode/cathode wire to the copper cathode.  Is the reduction of the ionized hydrogen at the cathode causing this?  Also, how is the hydrogen ionized?

Don't abbreviate Wikipedia as Wiki!

Water already has hydrogen ions in it: water auto-ionizes. Acidic solutions, e.g., vinegar, lemon juice, hydrochloric acid, have significantly higher concentrations of hydrogen ions. Place a solid piece of zinc metal in vinegar or lemon juice, and it will spontaneously react (get oxidized, i.e., corrode away) as shown in your equations. You would see little hydrogen bubbles on the zinc. The voltaic pile just lets you get electrical energy rather than useless heat energy.

You are right ! It is difficult to believe that in the voltaic cell (zinc and copper plates separated by a neutral $\ce{NaCl}$ solution), the cathodic reduction may be due to the reduction of $\ce{H+}$ ions into $\ce{H2}$, because a neutral solution has practically no $\ce{H+}$ ions : $\ce{[H+] = 10^{-7} M}$. The cathodic reduction is not $$\ce{2 H+ + 2 e- -> H2}$$ It must be
$$\ce{2 H2O + 2e− -> H2 + 2 OH^− }$$ As a consequence, $\ce{H2}$ bubbles do appear on the copper plate. But the $\ce{OH-}$ ions created at the cathode react with the $\ce{Zn^{2+}}$ ions created at the anode according to : $$\ce{Zn^{2+} + 2 OH-  ->  Zn(OH)2 (s)}$$ so that a precipitate of $\ce{Zn(OH)2}$ appears in the solution. To go back to Nick's question, the final equation is not the reaction $$\ce{Zn + 2 H+ -> Zn^{2+} + H2}$$ as proposed in the text, but it is $$\ce{Zn + 2H2O -> Zn(OH)2 + H2}$$

$2e^-$ is an antecedent to the product.  Where are those electrons coming from such that the reaction is started?

The two electrons are coming from the metallic $\ce{Zn}$ atom. $\ce{Zn}$ is more stable as an ion than as a neutral atom. In other words, it looks as if the neutral zinc atom "feels like" losing two electrons. Pardon me this image. But zinc can do this only if some other ion, atom or molecule does accept its electrons, here $\ce{H2O}$

Thanks. Why is it that the two electrons leave the anode and enter the cathode, vs simply leaving the anode and reacting with H2O?

Probably the electrons are repelling one another in the anode. And they prefer leaving the anodic zone, rather than reacting with $\ce{H2O}$near the anode. I am not sure. I would be pleased to hear another device.

This answer is misleading.  It's true that there are not a lot of H+ ions in water at equilibrium, however this ignores that if any of these ions are consumed, they will be immediately replaced to maintain equilibrium, so there is effectively an "unlimited supply" of them even if they're not all present initially.  Electrons cannot "react" with (neutral) H2O directly.  H2O naturally dissociates into H+ and OH- and then H+ combines with electrons.  However, Zn++ ions being produced at the anode will repel H+ ions, so there will be none at the anode for the electrons to come in contact with.

It is important to remember that both electrode reactions:
\begin{align}
\ce{Zn &<=> Zn^{2+} + 2e^-}\\
\ce{2H^+ + 2e^- &<=> H2}\\
\end{align}
are reversible and are ongoing even without galvanic connection of the anode and the cathode.
Rates of electrode oxidation reactions increase with the electrode potential and rates of electrode reduction reactions decrease with potential. Each redox system has some particular equilibrium potential value where the rate of the net reaction is zero.
When anodes and cathodes are galvanically connected, they force each to other different then equilibrium potential, what leads to non-zero net reaction, free electron production/consuption and the current.
Zinc anode forces lower than equilibrium potential on the hydrogen cathode and $\ce{H+(aq)}$ reduction reaction takes the upper hand.
Hydrogen cathode forces higher than equilibrium potential on the zinc anode and $\ce{Zn(s)}$ oxidation reaction takes the upper hand.


text:
Information about liquefied carbon factories from carbon dioxide
I am looking for some information about the factories that turn CO2 gas into the liquid CO2. I want to know the methods that they are using. Can anyone help me?
Thanks

See https://pubs.acs.org/doi/pdf/10.1021/ie300431z or just do a search on "carbon dioxide liquefaction"

If you compress a gas in a syringe at $20°$C, its temperature increases a bit (say $+ 10$ °C). If you now maintain the pressure and let the compressed gas cool down, its temperature goes slowly back to its original value ($20$°C). If you now release the pressure, the gas temperature will decrease down to say +$10$°C : temperature drop is equal to minus the heating step during compression.
If you now repeat this same experiment with another syringe containing another compressed gas cooled down to $20°$C, the first cold syringe ($10°$C) can be used to cool down the second one down to say $15°$C. By releasing the pressure of this compressed and cold gas, this second gas goes down to +$5°$C.
Repeating this process over and over with a lot of syringes,  you can go down to as low temperatures as you want it. Of course this phenomena stops when the gas is cold enough to become a liquid or a solid.

I think you mean dry ice; solid carbon dioxide, not carbon. It is made very cold and freezes. At atmospheric pressure carbon dioxide sublimes, no liquid; gas to solid. Look up a CO2 phase diagram. Not as cold as natural gas separation or air separation.

I wouldn’t assume the question is about dry ice. CO2 is transported as a liquid in bulk

Not at atmospheric pressure.  Actually CO2 liquid pipelines are a very special case requiring crack arresters every few miles.

Liquid CO2 is also transported by ship. Obviously not at atmospheric pressure.

Ships seem unlikely., it would require 100 psi pressure vessels or insulation. Both expensive for a cheap commodity like CO2. Pipeline applications are CO2 wells which can be pressured into pipeline and shipped to oil producing areas where it is used for third stage recovery by injecting into wells .

The volume shipped is small, but it happens. Typically food grade CO2. Ships are similar to LPG tankers. See for example https://gcaptain.com/mol-invests-in-liquid-co2-shipping-line/

On a side note, there's been a lot of interest in CO2 shipping recently because it is assumed that it would be necessary for large scale carbon capture and storage systems. I suspect that is the motivation for the OPs question given the use of "carbon" rather than "carbon dioxide" in the question


text:
Comparison of boiling points between 2‐butanol and isobutanol
Which has higher boiling points among butan‐2‐ol (1) and 2‐methylpropan‐1‐ol (2)?

I know that branching decreases the boiling points, but I want to know how position of substituents affects the melting (m.p.) and boiling points (b.p.). Google search only resulted in the b.p. of isobutyl alcohol listed in Wikipedia: 107.9 °C.

2-butanol should be written instead of 1-methyl propanol as it is an incorrect IUPAC name.

2-butanol boils at $99°$C - $100°$C

https://chemistry.stackexchange.com/questions/132765/why-is-the-boiling-point-of-homologous-primary-secondary-tertiary-alcohols-w

@Infinite The correct IUPAC name is butan-2-ol


text:
Acidification of a vicinal diol
The following is a question from Advanced problems in organic chemistry by MS Chouhan:

The options given are:

My initial thought was that this could be a pinacol–pinacolone rearrangement. However the aryl shift would lead to an absurd amount of crowding, and there is no option that corresponds to this.
Secondly, it could form an epoxide, but I'm unsure of that as well.
Any explanation/reference to this reaction is appreciated!

b and d can be discarded as products immediately. Which aryl group is the more electron-rich and thus the more likely to capture a cation?

@Waylander it would be the anisole ;-)

So, I think we're agreed that the answer is either (a) or (c). I got most of the way to the mechanism by trying to map the product onto the starting material (the connectivity really changed!) to determine a likely intermediate. Due to the almost-symmetry, there are two mechanisms at play, but the more likely situation now depends on evaluating the two-step mechanism for what the RDS is...

@Zhe what were the two mechanisms that you tried? We could try building on it!

The diol is nearly symmetric. Since the first step is almost certainly protonation of one of the alcohol groups, that's the dichotomy. Then you can have cation formation followed by cyclopropane ring opening + ketone formation; or, doing all of that in a concerted manner. Finally vinylogous attack on the ketone with one of the aryl rings.

It's a neat mechanism. I wouldn't necessarily have predicted a cyclopentane as a product a priori, but that's what you get when you put a string-loaded cyclopropane in your starting material.

@Zhe: This question is a duplicate: https://chemistry.stackexchange.com/questions/124940/is-pinacol-rearrangement-the-source-for-the-ring-expansion/124955#124955

Steo 1 should be cyclopropane ring collapse, after protonation to give alkene and ketone. Step 2:  Protonation of alkene keading to stable tertiary carbocation. Step 3 : carbocation as electrophile attacks benzene ring(ekectrophillic aromatic substitution).

This reaction is based on cyclopropane ring collapse forming allyl carbocation.

@user55119 Excellent answer and a very very interesting mechanism, especially, a pi-bond formation by pinacol rearrangement! I suppose the instability of the ring is the driving force.

@ChakravarthyKalyan your step 2 does not go along with user55119's mechanism, but I suppose that tautomerizing the ketone would lead to a similar product anyway.

Step2 is collapse of cyclopropane ring to enol vs allyl alcohol carbocation. In either cases ketone is formed.


text:
Understanding the molar free energy to simulate phase separations (Na2O-SiO2)
I am a mathematician and I want to understand the molar free energy model given in Kim and Sanders (2020) Equation (1), that is,
\begin{equation}
f(c,T) = RT \big(c \log(c) + (1-c) \log(1-c)\big) + \big(A_0(T) + A_1(T) (1-2c)\big) \, c(1-c),
\end{equation}
where $T$ is the temperature, $c$ is the concentration (of $\text{SiO}_2$), $A_0$ and $A_1$ are temperature dependent coefficients of the Redlich-Kister interaction model which are  given in $\text{kJ}\,\text{mol}^{-1}$ by
\begin{equation}
A_0(T) = 186.0575 - 0.3654\, T\quad \text{and}\quad A_1(T) = 43.7207 - 0.1401\, T
\end{equation}
(see also Kim and Sanders (1991)) and, as usual, $R = 0.0083144\ \text{kJ}\,(\text{mol K})^{-1}$ is the gas constant.
Problem: According to Kim and Sanders (2020), Figure 1, I would expect to get a double well potential for a temperature T = 900 K with two minima at approximately $c_1=0.82$ and $c_2=0.97$. However, plotting $f$ at $T = 900\ \text{K}$ yields:
$f(\cdot,900)$" />
Questions: Did I misunderstand something concerning the energy model? Is there a transformation missing, especially concerning the composition $c_0 = 0.88$ mol of $\text{SiO}_2$ (and $0.12$ mol of $\text{Na}_2\text{O}$). Can someone give me good references to basics of similar problems (scaling etc., discussions of energy models ...)?

In an Ideal solution, based on how many ways one can randomly arrange the constituents, one gets:
$\Delta S_{mix} = -R(X_{A} \ln{X_{A}} + X_{B} \ln{X_{B}})$ with $X_{A}+X_{B} = 1$ being the relative molar concentrations.
so
$\Delta G_{mix} = RT(X_{A} \ln{X_{A}} + X_{B} \ln{X_{B}})$
that is your first bit.
For a Regular solution,
$\Delta H_{mix} = \Omega X_{A} X_{B}$
for some $\Omega$ that may well depend on temperature and composition. So in your equation, $A_{0}(T)$ is the part of $\Omega$ that depends only on $T$. Composition dependence is represented by a power series in $(X_{A}-X_{B})$, or equivalently $(2X_{A}-1)$ or $(1-2X_{B})$. The subscripts on the $A$ terms are the power of $(X_{A}-X_{B})$ for that term.
This representation is quite commonly used in the Calphad method for computing binary (and higher) phase diagrams.
So, in your particular case, what does this look like? Plugging those values into my homebrew Calphad program I get a miscibility gap that looks like:

And, stopping at 600K, a free energy curve that looks like:

so I'm not quite sure what went wrong for your calculation. One thing to double check would be that your units (J/mol vs kJ/mol) for the entropy and enthalpies match.
Note also that you are not necessarily looking for minima in the curve - you are looking for common tangents to the curve which indicate that the Gibbs free energies of the two components are the same at the two compositions. The red line on the second plot shows the common tangent at that temperature (600K in this case). (That said, you can freely choose a component reference free energy that would make the curve at a particular temperature show those points as minima, but in general you will just get a common tangent.)

Thanks for your explanations regarding the model building. My problem is that the concrete energy $f$ with values $A_0$ and $A_1$ etc. as given above does not yield a double well potential (cp. figure in the question). However, I would expect a double well potential, that is, I would expect that $f$ has two minima. (Setting e.g. $\Omega = 20$ in your notation (example) would yield a double well potential at $T = 900\ K$). That's why I cannot accept your answer as an answer.

@Henning - I added my calculations based on those papers.

Actually, your free energy curve does not look very different from my one in the question (up to a scaling factor). I am wondering if this is reasonable. I think that due to the miscibility there should be two minima, not just one minimum.

@Henning - again - not minima, places of common tangent to the curve. Perhaps that is the confusion? You are free to scale one or both free energies of the constituents, which is just adding or subtracting a straight line to the whole free energy calculation. The common tangent remains the same regardless of that.

I missed somehow your last paragraph of your extended answer, sorry and thank you! In fact, that could be the confusion. I was obsessed with the fact that the system moves in two minima. OK, I have to think about it. Many thanks!

@Henning - I would highly recommend Porter and Easterling's Phase Transformations in Metals and Alloys. 2nd edition is best in my opinion, but see if you library has it. The first 40 pages or so cover a lot of ground on binary phase diagrams.

As an extension of the above answer of Jon Custer, I would like to draw attention to the paper of Charles (1967). On page 634 Charles is concerned with a similar unfortunate fact that "the changes in mixing free energies, which might denote immiscibility, would be too slight to be observed on the scale necessary to present all the calculated free energy values." He therefore considered an additional linear term -- with arbitrary coefficient -- which enables a detection of a miscibility by a visual inspection of the corresponding (double well) potential.
Correspondingly, for the above system (cf. question) we may consider
\begin{equation}
f(c,T,B) = RT \big(c \log(c) + Bc + (1-c) \log(1-c)\big) + \big(A_0(T) + A_1(T) (1-2c)\big) \, c(1-c).
\end{equation}
This additional linear term $Bc$ does obviously not affect the conditions for the common tangent
$$
\begin{gather}
0 =& f'(x,T) - f'(y,T)\\
0 =& f(x,T) + f'(x,T) (y-x) - f(y,T).
\end{gather}
$$
Hence, the miscibility gap is not affected. However, the visual presentation of the common tangent seems to be more appealing. For the above Gibbs free energy at $T = 923.15 \text{K}$ I made the following graphs in MATLAB.


text:
How does coenzyme A produce proton from pyruvate ions?
Coenzyme A (CoA) is a complex molecule consisting of mercaptamine, β-alanine, pantoic acid, diphosphate, pentose, phosphate and adenine moieties:

Which part of it actually plays role in the formation of proton and formate from pyruvate ion, and what is the mechanism? I know that the mercaptamine part has the SH group (thiol group) and there are many acidic H atoms in the molecule.
Plus, how does the lyase catalyst play its role here? Is it a SN1 or SN2 reaction? Please provide some proper references.

I think you will get better information just searching the literature on pyruvate formate lyase enzyme mechanism.

The reversible conversion of pyruvate ion and coenzyme A (CoA) into $\ce{acetyl-CoA}$ and formate ion (Figure 1) is not a one step reaction but a complex multistep radical reaction (Ref.1), thus it is neither $\mathrm{S_N1}$ nor $\mathrm{S_N2}$:

Two research groups (Ref.2 & 3) have proposed two mechanisms both involving radical initiation (Figures are from Ref.1):

Both mechanisms have taken similar path for the radical assisted chevage. The theoretical studies (Ref. 1) and crystal structure of an intermediate (Ref.4) have confirmed the validity of mechanism(s):

References:

Fahmi Himo and Leif A. Eriksson, "Catalytic Mechanism of Pyruvate Formate-Lyase (PFL). A Theoretical Study," J. Am. Chem. Soc. 1998, 120(44), 11449–11455 (DOI: https://doi.org/10.1021/ja9820947).
J. Knappe, S. Elbert, M. Frey, and A. F. V. Wagner, "Pyruvate formate-lyase mechanism involving the protein-based glycyl radical," Biochem. Soc. Trans. 1993, 21(3), 731–734 (DOI: https://doi.org/10.1042/bst0210731).
Camran V. Parast, Kenny K. Wong, Sandra A. Lewisch, John W. Kozarich, Jack Peisach, and Richard S. Magliozzo, "Hydrogen Exchange of the Glycyl Radical of Pyruvate Formate-Lyase Is Catalyzed by Cysteine 419," Biochemistry 1995, 34(8), 2393–2399 (DOI: https://doi.org/10.1021/bi00008a001).
Andreas Becker and Wolfgang Kabsch, "X-ray Structure of Pyruvate Formate-Lyase in Complex with Pyruvate and CoA: How the Enzyme Uses the Cys-418 Thiyl Radical for Pyruate Clevage," J. Biol. Chem 2002, 277(42), 40036-40042 (DOI: https://doi.org/10.1074/jbc.M205821200).


text:
How many hydrogen environments are there in phenyl acetic acid?
How does the molecule below have only three hydrogen environments? Should it not have 5 different hydrogen environments according to the labelled diagram?

See also: Phenyl group in proton NMR.

You are correct that there are 5 different hydrogen environments in phenylacetic acid. But whether or not the aromatic hydrogens are resolvable at some particular magnetic field is a different issue. The lower the field; the lower the chances.


text:
Cubic nature of van der Waals' gas equation

$V_m^3-\dfrac{RT+bP}{P}V_m^2+\dfrac{a}{P}V_m-\dfrac{ab}{P}=0$

So, At $T<T_c$ the above equation has three real roots say $V_1,V_2,V_3$, my doubt is what does this mean physically because for some particular pressure and temperature gas must have a fixed volume , so what does these three roots mean.

1 of 3 roots for T < Tc is just the artefact of the mathematical model, using the cubic equation, to archive 2 V values for the gas and the liquid phase. Be aware the equation is not much usable near the critical point. Note that multiple phenomena use models where only some of solutions have physical significance.

Cross posted on Physics SE.

It is discussed at Wikipedia:  van der Waals' equation: Maxwell equal area rule

The van der Waals equation describes the isotherms of a gas with respect to pressure and volume. The first van der Waals root (the first solution of the cubic equation) will give you the dew point, which is the point where the first drop of liquid forms from the gas, the other root of the equation is the so-called bubble point, which is when the first liquid drop of the compound is formed. The intermediate root will always be found in a region where liquid and vapor coexist, but just as it is a region and not a point, that root itself has no physical meaning.


text:
Rate of polymerization for copolymers with two monomers
I've found in textbooks that for a chain growth homopolymerization the rate of polymerization $r_\mathrm{p}$ is
$$r_\mathrm{p} = k_\mathrm{p}\sqrt{\frac{f_\mathrm{kd}}{k_\mathrm{t}}}c_\ce{A}^ac_\ce{B}^b,$$
where $k_\mathrm{p}$ is the rate constant of propagation; $k_\mathrm{t}$ is the rate constant of termination; $f_\mathrm{kd}$ is the constant for chain initiation.
What is the rate equation for chain growth copolymerization $r_\mathrm{p}$ such as styrene-butadiene rubber (SBR) synthesis? In many polymer textbooks it only shows the Mayo–Lewis equation which I understand is not the $r_\mathrm{p}$ equation required for volumes calculations for reactors.


text:
Why does band heads arise in vibrational-rotational spectra?
I learn that a band head is a region of maximum intensity due to many lines falling together. It can be determined by differentiating the expression for the R branch.
To no avail, I cannot find any explanation on why band heads arise after going through a few textbooks and Internet resources.
My guess: It might be due to the breakdown of Born-Oppenheimer Approximation which explain the unequal spacing in vib-rot spectra. The rotational constants decreases at higher vibrational states, to a limit?
Why does the absorption lines of R branch seem to 'U-turn' at higher J'' value?
Attached is the vib-rot spectra. This is one of the materials I searched from the Internet.(https://cefrc.princeton.edu/sites/g/files/toruqf1071/files/Files/2015%20Lecture%20Notes/Hanson/pLecture2.pdf , slide 28)
FYI, I am at the entry-level of learning molecular spectroscopy. Any explanation is really appreciated.

It's been a long time since I did this, but from my old notes... this is caused by the rotational constant $B$ being different in the two vibrational states $v = 0$ and $v = 1$. Generally we have that $B_1 < B_0$ (where $B_n$ denotes $B$ in the vibrational state $v = n$). This makes sense because $B = h/8\pi^2cI$ (expressed in wavenumber units) is inversely proportional to the moment of inertia $I$, and we expect $I$ to increase with vibrational excitation.
So, if $v_0$ is the vibrational frequency, then the R-branch transition frequencies (for the $(J, v) \to (J + 1, v+1)$ transition) are given by
$$\begin{align}
R(J) &= v_0 + B_1(J+1)(J+2) - B_0J(J+1) \\
     &= v_0 + J^2(B_1 - B_0) + J(3B_1 - B_0) + 2B_1
\end{align}$$
and the spacing between successive transitions is
$$\begin{align}
R(J + 1) - R(J) &= [(J+1)^2 - J^2](B_1 - B_0) + [(J+1) - J](3B_1 - B_0) \\
                &= (2J + 1)(B_1 - B_0) + 3B_1 - B_0 \\
                &= 2J(B_1 - B_0) + 4B_1 - 2B_0
\end{align}$$
Since $B_1 - B_0 < 0$, this separation decreases with increasing $J$ as observed. (Notice that if we let $B_1 = B_0 = B$ then we recover the "simple" behaviour where the spacing between lines is $2B$.) Naturally, exactly the opposite scenario is seen in the P-branch; I'll leave the maths to the reader.
A very thorough book for spectroscopy is Hollas, Modern Spectroscopy (4th ed.), which explains all of this in detail.


text:
Are there any types of adhesive which are designed to be easily soluble in strong acid (HCl with pH 1-4)?
I am a mechanical engineer so forgive my ignorance in chemistry. I know there are water soluble adhesives like polyvinyl alcohol but I am wondering if there are glues which easily dissolve in low pH HCl without the need for scrubbing or much perturbation.

Welcome to Chemistry! I don't quite get the need why it has to be acidic and there are probably quite a number of water soluble glues out there. There are probably various degrees of adhesion, so I guess there can't be a general answer to this question. If you could add more detail and context, you'll probably get a more meaningful answer.

Rather than use strong acids, chlorinated solvents will be better e.g.  dichloromethane.

You might look up the chemical compatibility for a number of adhesives that are otherwise suitable for your purposes. For example, https://www.norlandprod.com/techrpts/chemresit.html shows NaOH (a strong alkali), attacks its product. That said, if an adhesive is sandwiched between to impermeable surfaces, chemicals may not be able to infiltrate to dissolve the adhesive.


text:
Why does the equation for chemical potential apply to liquids and aqueous solutes?
The equation for chemical potential of a gas can be derived as such:
At constant temperature,
$$\mathrm dG = V\,\mathrm dP \label{eqn:1}\tag{1}$$
Substituting with the ideal gas law $PV = nRT$,
$$\int_{p^\circ}^p\mathrm dG = \int_{p^\circ}^p \frac{nRT}{p}\mathrm dP \label{eqn:2}\tag{2}$$
Then
$$\mu = \mu^\circ + RT\ln\frac{p\phantom{^\circ}}{p^\circ} \label{eqn:3}\tag{3}$$
What I don't get is how this equation, which was derived using the ideal gas law, then becomes applicable to liquids as well as aqueous solutes, where:
$$\mu = \mu^\circ + RT\ln\frac{[\ce{A}]\phantom{^\circ}}{[\ce{A}]^\circ} \label{eqn:4}\tag{4}$$
where $[\ce{A}]$ is either $\ce{A(l)}$ or $\ce{A(aq)}$.
Note: I am aware that for liquids, at equilibrium the chemical potential of the liquid phase is equal to the vapor phase, and by considering Raoult's law:
$$\mu = \mu^\circ + RT\ln\chi_A \label{eqn:5}\tag{5}$$
But how would I go from \eqref{eqn:5} to \eqref{eqn:4}?
I know these are probably quite trivial but I would appreciate any help.

What is A ? Is it the generic name of any chemical ? Is it the absorbance of a colored stuff ? Why don't you call $p$ as $p$(A) or ${p_A}$ ?
What is $\pu{\chi}_A$ ? Is it the molar fraction ? Anyway you can go from $(3)$ to $(4)$ by admitting that the concentration [$\pu{A}$] of a gaseous stuff $\pu{A}$, once dissolved in a liquid is proportional to its pressure $p$ in the gas phase (Henry's law)

In general you would use activities or fugacities instead of concentrations or mole fractions.

The kinetic theory of matter says that matter consists of individual particles be they atoms, molecules, ions, nanoparticles, even elephants; the state of aggregation can have an effect. Let's restrict to molecules and ions and worry about the elephants when they recover in number. The second factor is that these molecules, ions etc. have thermal energy or heat content that is caused by the random velocity of the molecules [tempered by changes in quantum mechanical states, rotational, vibrational electronic]. The total energy of something is a constant under constant conditions and is determined by the temperature[or vice versa]. It has been noticed that reactions take place only under certain conditions. You must use your sense of observation, imagination, education etc to see what these conditions might be. One factor is actual physical state, another is distance between molecules, another is the identity of adjacent molecules [ie concentration or dispersion]. The way of describing all these factors is the nebulous chemical potential or activity of a chemical moiety in a mix. For ideal behavior these are the pressures, concentrations[molarity, molality], mole fractions. In real live they are chemical activities or chemical potentials. It takes a while to accept this. For now use it and try to think about what is going on and discard preconceived notions and what you learned in high school or college or even grad school It was all incomplete and possibly wrong, but still a fair start.


text:
Could mixing naphtha-based cleaner and industrial floor cleaner produce hazardous substances?
I work in a car dealership, and the guy who cleans the workshop uses an industrial floor scrubber / vac on the floor tiles. On particularly stubborn oil / soot marks, he uses naphtha-based brake cleaner in a spray dispenser as almost a "pre-wash", and then goes over it with the scrubber.
If the machine has been left for a while, perhaps for over a week during the holidays, the machine produces an acrid, pungent smell that burns the eyes and back of the throat. Almost like bad drain-type smell, and he thinks its just stale water, and we're being wusses — but the odour goes around the whole building, and my proximity to the workshop means I get a good dose of the terrible smell.
I have a theory that the brake cleaner is having a reaction with the detergent in the machines waste tank, and when he starts it after it has been festering. It's really bad, and what's more, I think it is hazardous.
The two chemicals are: a naphtha-based brake cleaner, and an industrial floor cleaner containing things such as 2-(2-butoxyethoxy)ethanol, isotridecanol ethoxylate etc. I can provide more details from the safety data sheets if wanted, but is the above enough to make a forward conclusion?

These situations are pretty tough to get to the bottom of. Most of the substances probably don't have long term health studies. On the other hand, the burning sensation is telling. I suspect you're not going to get a lot of traction on this topic, but maybe you can just circumvent the whole issue by checking to see if they can improve the ventilation in the workshop and just flush everything out.

The floor-cleaner is meant for large-scale application, and likely has been tested to meet OSHA and other standards, but brake-cleaner is usually used in only a limited area, and in well-ventilated spaces. Check its ingredients, particularly for more toxic chemicals than naphtha, e.g., chlorinated solvents or aromatic compounds such as toluene or xylene.

Normally, surfactans (like the ethoxylated alcohols you referenced) are relatively stable under various conditions. I'd worry a bit about the hydrocarbon exposure by itself; in the long run, your brain is not going to be all too happy if exposed over longer periods of time in a badly ventilated setting.

bad drain-type smell

That would trigger me towards dihydrogensulfide formation. Even with the MSDS at hand, it can be hard to tell if you run that risk, e.g. because of him accidentally mixing an acidic cleaning product with a sulfur compound-containing product.


text:
How to synthesize anhydrous zinc(II) nitrate?
I'm interested in making anhydrous zinc nitrate. I know it decomposes fairly rapidly if you heat it. It is also soluble in ethanol.
Would another desiccant or a molecular sieve do the trick? Or maybe it can be made directly somehow? There is surprisingly little information on ranking desiccate materials.

Reaction with dinitrogen tetroxide* $\ce{N2O4}$ is probably the most prominent and convenient way of obtaining anhydrous zinc(II) nitrate, as well as alkali metal and mercury(II) nitrates according to the scheme (the reaction with zinc goes to completion and is therefore studied in more detail [1]):
$$\ce{M + x N2O4 -> M(NO3)_x + x NO} \quad \ce{M} = \ce{Li, Na, K, Rb, Zn, Hg}$$
A piece of zinc (5 g) with a freshly cleaned surface is immersed in liquid nitrogen tetroxide (10 ml).
Within 10 minutes the metal is covered with a thin layer of white fine-crystalline substance, nitrogen oxide is released, and the liquid in direct contact with the metal turns green due to release of nitrogen trioxide $\ce{N2O3}$.
As the product layer becomes thicker, $\ce{Zn(NO3)2·2N2O4}$ solvate crystals begin to form, which are easily separated from the metal surface with gentle stirring.
In 24 h, about 1–2 g of the product is formed, but its formation in the future is increasingly difficult to control.
After removal of excess metal, nitrogen tetroxide is distilled off by evaporation.
Here, a free-flowing fine-crystalline powder is formed.
Nitrogen tetroxide is weakly bound to zinc nitrate and is released when the powder is heated.
When about $10\,\%$ of the attached tetroxide is removed, the solid turns into a liquid.
The change in the phase state is caused by a decrease in the melting temperature of the initial solvate as a result of the formation of zinc nitrate.
Since the reaction
$$\ce{Zn(NO3)2·2N2O4 <=> Zn(NO3)2 + 2 N2O4}$$
is reversible, a complete removal of the tetroxide occurs by heating for 6 h at 100 °C under reduced pressure.
The resulting anhydrous zinc nitrate is a free-flowing white powder.
Nitrogen trioxide formed in solution significantly affects on the reaction rate. [2]
If the piece of metal is stationary, the reaction rate gradually increases, but if the solution is stirred by the metal itself, the rate decreases.
This behavior distinguishes this system from the usual heterogeneous processes of solid-liquid interaction when a concentration gradient of active forms is established near the surface, and stirring increases the reaction rate.
This phenomenon is explained by the presence of nitrogen trioxide at the metal surface.
The molecules of this substance are more polar than the molecules of tetroxide, and therefore they increase the dielectric constant of the medium and the concentration of the active form of $\ce{NO+}.$
The presence of the trioxide at the surface facilitates the reaction, and with stirring, the substance is distributed in solution.
The trioxide does not directly participate in the reaction; otherwise, the formation of metal nitrite would inevitably occur as a result of the ionic dissociation
$$\ce{N2O3 <=> NO+ + NO2-}.$$
The rate of dissolution of zinc is proportional to $\sqrt{[\ce{N2O3}]}$.
At 0 °C the reaction rate in a solution containing $40\,\%$ nitrogen trioxide is eight times greater than in pure nitrogen tetroxide, but the product does not contain traces of nitrites.

For an extensive overview I suggest reading chapter Anhydrous Metal Nitrates, section II Preparative Methods in [3, pp. 73–136].
Below are some relevant points and considerations taken from that review.

Thermal dehydration as well as dehydration of the solid hydrate over phosphoric oxide in a vacuum desiccator at room temperature yields insoluble basic salt.

Desiccants and molecular sieves won't be effective due to the fact that water molecules in crystallohydrates usually belong to inner coordination sphere and are not merely held by hydrogen or van der Waals bonding or host-guest interactions.

Displacement in nonaqueous solvents is not a viable option due to poor solubility of nitrates (silver(I) nitrate is typically used) in organic solvents and highly problematic or impossible solvent adduct removal from obtained  solvated metal nitrate.

Anhydrous nitric acid, as straightforward as it seems, usually doesn't react completely with the dried salt, and only a partial replacement occurs, often yielding in double salts.
Additionally, there is always some water present due to equilibrium
$$\ce{2HNO3 <=> N2O5 + H2O}.$$

Addition of dinitrogen pentoxide to nitric acid assists in binding water and provides an additional source of water-free nitrate anion due to ionic dissociation in $\ce{HNO3(conc)}$:
$$\ce{N2O5 <=> NO2+ + NO3-},$$
although conversion of zinc(II) salts into anhydrous nitrate using either this method or chlorine/acetyl nitrates $(\ce{ClNO3},$ $\ce{CH3C(O)ONO2})$ hasn't been reported at the time.
Pure $\ce{N2O5}$ reacts directly with many metals, albeit with low selectivity giving mixed products.

* In a lab pure dinitrogen tetroxide is easily obtained by thermal decomposition of lead nitrate.
Under normal conditions, $\ce{N2O4}$ is a liquid substance (m.p. −11.2 °C, b.p. 21.15 °C), and the reactions are almost always carried out at room temperature.
Traces of moisture in $\ce{N2O4}$ can be detected simply by freezing the liquid, and this is one of the most sensitive methods of testing.
Pure $\ce{N2O4}$ freezes to form a colorless glassy substance, but in the presence of traces of moisture nitrogen trioxide $\ce{N2O3}$ is formed, which turns the frozen tetroxide green.
Usually it is recommended to cool the liquid to about 10 °C prior to synthesis, otherwise the heat released during the reaction may cause vigorous boiling and the release of toxic fumes.
References

Addison, C. C.; Lewis, J.; Thompson, R. 628. The Liquid Dinitrogen Tetroxide Solvent System. Part VII. Products of Reaction of Zinc with Liquid Dinitrogen Tetroxide. J. Chem. Soc. 1951, 2829. DOI: 10.1039/jr9510002829.
Addison, C. C.; Lewis, J.; Thompson, R. 630. The Liquid Dinitrogen Tetroxide Solvent System. Part IX. Products, Rates, and Possible Mechanisms of the Reaction of Zinc with Liquid Dinitrogen Trioxide–Tetroxide Mixtures. J. Chem. Soc. 1951, 2838. DOI: 10.1039/JR9510002838.
Advances in Inorganic Chemistry and Radiochemistry; Emeléus, H. J., Sharpe, A. G., Eds.; Academic Press: New York, 1964; Vol. 6.


text:
Are substituted sugar/carbohydrate derivatives such as sucralose still sugars/carbohydrates?
Sucralose is made from sucralose, with chlorination of 3 of the hydroxyl groups. If broken up into the 2 saccharide-type units, you'll get 1,6-dichloro-1,6-dideoxyfructose and  4-chloro-4-deoxy-D- galactose. Firstly, are those 2 compounds considered to be monosaccharides/sugars? And if so, is sucralose considered a disaccharide/sugar?
They wouldn't conform to the regular general formula for monosaccharides, (CH2O)x, but IUPAC Gold Book (https://goldbook.iupac.org/terms/view/M04021)  says "for Monosaccharides - A term which includes aldoses, ketoses and a wide variety of derivatives. Derivation includes oxidation, deoxygenation, introduction of other substituents, alkylation and acylation of hydroxy groups, and chain branching." At what point would a derivative not be considered a monosaccharide anymore?

I think as long as the ring structure is intact with the ether bond one may still call it a carbohydrate derivative. For instance cylcodextrins are heavily derivatized for chromatographic purposes such as permethylated cyclodextrin. It is a derivative but most of its hydroxyls have been replaced by methyl groups.


text:
How should I evaluate $\left(\frac{\partial U}{\partial V}\right)_T$ with van der waals equation: $P = \frac{nRT}{V-nb}-a(\frac{n}{V})^2$?
So in my homework I was able to get this-
$\left(\frac{\partial U}{\partial V}\right)_T = T(\frac{\partial P}{\partial T})_V - P$
but now I'm trying to plug in $P = \frac{nRT}{V-nb}-a(\frac{n}{V})^2$ in the above equation.  I'm struggling here, I got zero for
$\left(\frac{\partial U}{\partial V}\right)_T$ but I'm then asked to evaluate $\left(\frac{\partial U}{\partial V}\right)_T$ in an integral from infinity to $V$ with respect to $V$ (i.e. $dV$).

Chemistry SE site strongly recommends plain text titles for index/search reasons and due possible displaying issues in question lists.

For $(\frac{\partial p}{\partial T})_V$, just take the derivative of your entire expression for p ($p=\frac{nRT}{V-nb}-a(\frac{n}{V})^2$ )with respect to T at constant V.  Then multiply by T, and subtract  the expression for p....exactly as your equation says.  If you're getting zero, I suspect what you are doing is subtracting only the first of the two terms in the expression for p, rather than the whole thing.  General suggestion: When you can't get the right answer, instead of trying to find the error in what you've done, start over from scratch; that way you're less likely to repeat it.


text:
Purpose of measuring chloroform soluble extractives
I noticed that tests such as the FDA ones are measuring "chloroform soluble extractives". I understand that chloroform is toxic and its effect to human if exposed to it. But does that mean those "chloroform soluble extractives" are also as bad? i.e. if chemical-A is soluble in chloroform, does that mean chemical-A is toxic too? Otherwise, why are we measuring the "chloroform soluble extractives"?
For reference: http://www.stq-cert.com/en/news_show.php?article_id=206

There is no correlation between chloroform solubility and toxicity

@Waylander then what's the point of measuring "chloroform soluble extractives"?

@xcoder  ask the FDA not me

Why should be solvent toxicity and solute toxicity related? Potassium cyanide is water soluble. As water is not toxic, does it mean the cyanide is not toxic either ?

Really? Chloroform is a carcinogen...it should be avoided whenever possible.

@Achem  Chloroform is not a known carcinogen. The MSDS states " Possible cancer hazard based on tests with laboratory animals." Big difference.

Difference perspectives ...synthetic organic chemists are quite comfortable with hazardous chemicals whereas many analytical chemists are not. EPA has classified chloroform as a Group B2, probable human carcinogen. Some 30-40 years ago it was okay to have benzene in wash bottles in student organic labs and it was also okay to do qualitative analysis with mercury salts. Today it will be a nightmare in a teaching lab.

From the link you can see an example "Closure with sealing gaskets for food containers  - U.S. FDA CFR 21 177.1210：Net chloroform soluble extractives."
For analytical testing of leacheates in samples, chloroform is not the only solvent- it is one of the solvents which is useful for representing the capability to  "extract" fat-like molecules (long chain molecules), polymers, and a wide range of other molecules particularly plasticizers. In this particluar case, the purpose is to assess if the sealing gasket in food container is able leach potentially toxic molecules or not over a long term in food or not.
Suppose your food in that container has oils, fats and water (like gravy or tuna fish etc.). As a chemist, one would be concerned what type molecules can leach into food after a long time  storage from the container. In that case, chloroform extraction test will suggest what kind of molecules might partition into the food fats from the container.
Chloroform extract does not necessarily represent toxic molecules but potentially harmful stuff. Secondly chloroform extraction is analytical chemist friendly- after extraction the analysis can be directly carried out in a gas or a liquid chromatograph for further analysis.


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Gibbs free energy for an irreversible process
In an irreversible process taking place at constant T and P in which only pressure-volume work is being done, the change in  Gibbs free energy (dG) and change in entropy (dS) are?
The answer to this question was given as dS > 0, dG < 0. I know that ∆G is negative and ∆S is positive for an irreversible process but ∆G gives the maximum amount of "non-expansion" work that can be extracted from a system, whereas in this question only pressure volume work is being done which is non-useful (expansion) work. So how can ∆G be negative here?  Please help me to clear this confusion. Thanks!

For a process at constant T and P in which no non-expansion work is done, $$\Delta U=Q-P\Delta V$$ and $$\Delta S=\frac{Q}{T}+\sigma$$where $\sigma$ is the irreversibly generated entropy.  Combining these equations gives $$\Delta U=T\Delta S-P\Delta V-T\sigma$$or$$\Delta G=-T\sigma$$which is negative.  On the other hand, the sign of $\Delta S$ can be positive only if $$Q>-T\sigma$$


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If you are doing an ONIOM calculation in Gaussian, how do you add both custom MM parameters and a custom basis set using GEN and Amber=SoftFirst?
I am trying to calculate a biological system using AMBER forcefields as my low-layer in ONIOM3. I also want to use a non-native basis set (aug-pc-1 for opt+freq, aug-pc-2 for SPE) that requires custom input.
Using ONIOM(M062X/gen:amber=softfirst) gives me syntax errors and errors in RdPar.
How do I resolve this?
Gaussian documentation provides no resolution, and google provides no answer either!

I suspect that many Gaussian experts hang out over at Matter Modeling SE (https://mattermodeling.stackexchange.com/)

I've decided to troubleshoot the issue presented earlier, and feel it might benefit others to spare them the headaches through an easily google-able problem.
If using JUST the gen keyword...
After the connectivity table, leave exactly 1 empty line before pasting the specification from Basis set exchange. End the input file with exactly 2 empty lines.
If using JUST the Amber=SoftFirst keyword,
After the connectivity table, leave exactly 2 empty line before pasting your custom AMBER parameters. If you have custom non-bonding parameters, they must be flush with the rest of the parameter specification. Any line within the parameter specification that is preceded by an extra space will cause syntax errors.
If using BOTH, (ergo: ONIOM(m062x/gen:amber) amber=softfirst
After the connectivity table, leave EXACTLY 2 empty lines, THEN FIRST input the molecule mechanics parameters as above.
AFTER the last line of the molecule mechanics parameter specification, leave EXACTLY 1 empty line, THEN input the gaussian formatted basis set from Basis Set Exchange, and leave EXACTLY 2 empty lines at the end of the input file.
I hope it helps others avoid headaches!