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Equilibrium Constant & Rate-Determining Step
My first language is not English, so there might be some parts where it's hard to read. Sorry in advance.

I have some questions regarding the derivation of the equilibrium constant and the rate-determining step.
For a hypothetical reaction
$$\ce{aA + bB <=> cC + dD}$$
the equilibrium constant ${K}$ would be
$${K=\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}}$$
However, my textbook states that equilibrium is when the forward reaction rate ${R_{f}=k_{f}\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}$ and the reverse reaction rate ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}$ becomes the same, so that
$${R_{f}=R_{r}}$$
$${k_{f}\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}$$
$${\frac{k_{f}}{k_{r}}=\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}=K}$$
which is a bit unsound because there is no guarantee that
$$\ce{aA + bB -> cC + dD}$$
and
$$\ce{cC + dD -> aA + bB}$$
are both elementary reactions, so ${R_{f}}$ and ${R_{r}}$ might not equal ${k_{f}\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}$ and ${k_{r}\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}$.
So I started searching, and found this:

Kinetic Derivation of the Equilibrium Constant
The usual method of finding the rate equation of a multistep reaction is that it equals the rate equation of the rate-determining step with the caveat that if it contains any reactant which is a product in a previous step, one must substitute the rate factor (e.g., ${\left[\ce{A}\right]^{a}}$, ${\left[\ce{B}\right]^{b}}$) of this reactant with the rate factors of the reactants of this prior step. The idea of this proof is to write this method in a form more amenable to algebraic manipulation, and then find use it to determine the rate equation for the forward reaction, ${R_{f}}$, and then for the reverse reaction, ${R_{r}}$. Once this has done, we can equate ${R_{f}}$ and ${R_{r}}$ and hence derive the equilibrium constant expression.
Consider a general multi-step reaction, but instead of written the usual way, written so that all products appear on the left-hand side of the equation with negative coefficients:
$${x_{1}\ce{X}_{1}+x_{2}\ce{X}_{2}+x_{3}\ce{X}_{3}+⋯+x_{n}\ce{X}_{n}\ce{<=>}0}$$
Imagine that there are ${s}$ steps making up this multi-step reaction, and each elementary step is written the same way as above. Then, for an elementary step ${j}$, define ${x_{ij}}$ so that it is the coefficient of ${X_{i}}$ in the
  ${j}$th step. Since, as stated above, the coefficients of the elementary steps must add up to the overall coefficients,
$${\displaystyle\sum_{j=1}^{s}x_{ij}=x_{i}}\tag{1}$$
Now multiply together the terms of each elementary step, to the power of their corresponding coefficients, up to but excluding the rate-determining step (step ${r}$).
$${\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}}$$
The overall forward rate is proportional to this quantity multiplied by ${A}$, which is the reactants of the rate-determining step with their corresponding coefficients as exponents.
$${R_{f}=k_{f}A\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}}$$
Now assume that we do the same as above, except in reverse. That is, we use the negative of each coefficient, and we start from elementary step ${s}$ (that is, the last step) and go until (but exclude) the rate-determining step:
$${\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}$$
The overall reverse rate is proportional to this quantity divided by ${B}$, which is the concentrations of the products of the rate-determining step with their corresponding coefficients as exponents.
$${R_{r}=k_{r}\frac{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}{B}}$$
Since ${R_{f}=R_{r}}$ in an equilibrium,
$${k_{f}A\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}=k_{r}\frac{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}{B}}$$
Rearranging this to get ${K}$,
$${\frac{k_{f}}{k_{r}}=\frac{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}-x_{ij}}}{AB\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}}}$$
Because there is a negative sign in the exponent in the numerator, we can move the numerator to the denominator if we switch the negative sign to a positive sign.
$${\frac{k_{f}}{k_{r}}=\frac{1}{\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}\right)\left(AB\right)\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}x_{ij}}\right)}}$$
Recalling that ${A}$ is the product of the concentrations of the reactants of the rate-determining step with their coefficients as exponents, and that ${B}$ is the same for the products of the rate-determining step, we can write ${AB}$ as
$${AB=\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{x_{ir}}}$$
Thus,
$${\frac{k_{f}}{k_{r}}=\frac{1}{\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{r-1}x_{ij}}\right)\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{x_{ir}}\right)\left(\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=r+1}^{s}x_{ij}}\right)}}$$
But then this can be written simply as,
$${\frac{k_{f}}{k_{r}}=\frac{1}{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{\sum_{j=1}^{s}x_{ij}}}}$$
Recalling Equation 1, this is simply
$${\frac{k_{f}}{k_{r}}=\frac{1}{\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{x_{i}}}}$$
$${\frac{k_{f}}{k_{r}}=\displaystyle\prod_{i=1}^{n}\left[\ce{X}_{i}\right]^{-x_{i}}=K}$$

I did understand this, but new questions popped up.
First of all, this derivation of the equilibrium constant assumes that the rate-determining step for the forward reaction and reverse reaction are always identical (the ${r}$th step). Is this always true?
For example, in the reaction shown below, the 2nd step is the rate-determining step for both forward reaction and reverse reaction.

However, in this reaction below, the 2nd step is the rate-determining step for the forward reaction, while the 1st step is the rate-determining step for the reverse reaction.

I made the second graph myself, and I do not know if such a reaction is thermodynamically possible. But if reactions like this exist, then the derivation process written above would be flawed.

And here's my second question. Let's bring back the reaction
$$\ce{aA + bB <=> cC + dD}$$
and suppose that this reaction is a 2-step reaction:
$$\ce{aA <=> cC + eE}\tag{Step 1}$$
$$\ce{bB + eE <=> dD}\tag{Step 2}$$
If the 1st step is the rate-determining step, then the forward reaction rate would be ${R_{f}=k_{f}\left[\ce{A}\right]^{a}}$ and the reverse reaction rate would be... uh... ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}$? I tried to derive ${K}$ from ${R_{f}}$ and ${R_{r}}$, but I got stuck in the middle:

The equilibrium constant of Step 2 of the reverse reaction is
$${K_{r2}=\frac{\left[\ce{B}\right]^{b}\left[\ce{E}\right]^{e}}{\left[\ce{D}\right]^{d}}}$$
so
$${\left[\ce{E}\right]^{e}=\frac{K_{r2}\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}}}$$
Equilibrium is when the forward reaction rate ${R_{f}=k_{f}\left[\ce{A}\right]^{a}}$ and the reverse reaction rate ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}$ becomes the same, so
$${R_{f}=R_{r}}$$
$${k_{f}\left[\ce{A}\right]^{a}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}=k_{r}\left[\ce{C}\right]^{c}\frac{K_{r2}\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}}}$$
Thus
$${K=\frac{k_{f}}{k_{r}}=K_{r2}\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}}\tag{???????}$$

What mistake did I make here? I have been trying to solve this for a few days but I can't seem to obtain a clear answer.
(Sorry for such a long question, and thanks in advance.)

EDIT
So about the second question... Thanks to @porphyrin, I've found my mistake. I shouldn't have assumed that the reverse reaction rate is ${R_{r}=k_{r}\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}$. It should be
$${R_{r}=k_{r}\left(\frac{\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}\left[\ce{E}\right]^{e}}×\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}\right)=k_{r}\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}}}$$
Knowing this and solving again, I was able to get
$${K=\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}}$$

EDIT 2
Thanks to @thewitness, I think I got the answer for the first question. So apparently, it is possible for the rate-determining step of the reverse reaction to be different from that of the forward reaction.
Though I got the explanations I needed, feel free to add an answer and give me more insight into these matters. Thanks.

The equation you seek is only true for each barrier, not two as you have in your pictures. The equilibrium applies to each barrier. Secondly you have to assume that $a$ A molecules and $b$ B molecules react in a single step, which is extremely unlikely (i.e. does not happen to any extent)  unless $a=b=1$, same for C and D. So you have to apply chemical knowledge as well as using the maths. If $a, b $ etc are not one there is invariably an intermediate step involved.

@porphyrin Of course, the true values of ${a}$, ${b}$, ${c}$, and ${d}$ would be 1, but why would they matter herein just deriving ${K}$ in a hypothetical example? If the values of ${a}$, ${b}$, ${c}$, and ${d}$ becomes an actual problem in the derivation, let me know. Moreover, could you indicate my mistake more clearly? Which equation is only true for each barrier, and how should I revise it?

Well it does not matter in a theoretical calculation because you can assume as many molecules as you like react in a single step.  K will relate to starting and ending species only and will give no information about mechanism in just the same sense as $aA+bB\ cdots $ does not tell you anything about what actually happens in a reaction. In multiple equilibria the equilibrium constants generally multiply. The forward/reverse rate constants should refer only to each barrier, i.e each step not the whole reaction.

@porphyrin I'm sorry I might be lacking of understanding, what do you mean that the forward/reverse rate constants ${k_{f}}$ and ${k_{r}}$ should refer only to each barrier and not the whole reaction? In general cases, doesn't the rate constant of the RDS become the rate constant of the whole reaction? (Except for some cases, of course.)

Yes just each barrier, of course you could do this for the whole of a complex reaction but such a thing does not make sense. Same for RDS as you don't know what this step is for a complex reaction unless you also know all the other species present and rate constants.

@porphyrin I've been searching Wikipedia bearing your comment in mind, and I think I understand. So I shouldn't have assumed that ${k_{\ce{RDS}}=k_{\ce{overall}}}$... I tried again, and I was able to derive ${K}$ successfully as I originally intended. Thank you.

I will first try to answer your second question.
There's absolutely no mistake you made! As you assumed first step is RDS(rate determining step), You would expect $K=K_{f1}$.
Continuing from where you left, $$K=K_{r2}\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}$$
$$=\frac{\left[\ce{B}\right]^{b}\left[\ce{E}\right]^{e}}{\left[\ce{D}\right]^{d}} \times \frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}}$$
$$ = \frac{\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}{\left[\ce{A}\right]^{a}} =K_{f1}$$ as expected.
You seem to assume that the the slowest step is the RDS always, and that it only governs the rate of the reaction. This isn't always true. Only if the slowest step(RDS) is MUCH slower that all the steps this (approximation) that rate of reaction = rate of RDS is valid.(hopefully this answers your first question).
More generally(ALWAYS TRUE, no approximation), for your second question, (assuming no RDS anywhere)
$K_{f1}=\frac{\left[\ce{C}\right]^{c}\left[\ce{E}\right]^{e}}{\left[\ce{A}\right]^{a}}$
$K_{f2}=\frac{\left[\ce{D}\right]^{d}}{\left[\ce{B}\right]^{b}\left[\ce{E}\right]^{e}}$
$K_{f1}K_{f2} =\frac{\left[\ce{C}\right]^{c}\left[\ce{D}\right]^{d}}{\left[\ce{A}\right]^{a}\left[\ce{B}\right]^{b}} = K$

Thank you for the detailed explanation! Actually, I've found my mistake on the second question. I'm still confused about the first question, though. Could you elaborate, please?

The rate of the reaction depends only on the concentration. Energy doesn't affect it directly. You mentioned that "the 2nd step is the rate-determining step for the forward reaction, while the 1st step is the rate-determining step for the reverse reaction." which isn't necessarily true(we can't conclude this from energy profile diagram). The difference in energy is what we call the "Activation energy" and difference between activation energy of reactants and products is called "enthalpy of the reaction". This is  constant for a given reaction.

There are two equations: Arrhenius equation and vant Hoff equations. The former relates rate constant with activation energy, temperature and a constant called Arrhenius factor or frequency factor. The vant Hoff equation relates equilibrium constant with temperature and enthalpy for the reaction. You would learn more about this in chemical kinetics chapter.

Your diagram is of a first order reaction which takes place in two steps. It's been long since I revised chemical kinetics so extremely sorry if I mentioned anything wrong.

No need to be sorry and thank you! I guess I'll have to study harder... Just for clarification, so it is possible for the RDS of the reverse reaction to be different from the RDS of the forward reaction?

Yes and usually it is (because if for a reaction step, if the forward reaction is fast, the reverse is slow)

Hmmm... Then the "Kinetic Derivation of the Equilibrium Constant" I quoted above would be extremely flawed... I guess the method you stated at the bottom of your post (${K}$=${K_{1}K_{2}K_{3}×⋯×K_{s}}$) is the most accurate for deriving the equilibrium constant. Thanks.

There is no reason to invoke a rate-determining step in making the connection between kinetic constants and equilibrium. In fact, you might get the wrong answer.
The easiest way to make the connection is to separate the reaction into elementary steps. The principle of micro-reversibility states that any elementary reaction will also have a elementary reverse reaction. At equilibrium, the rate of each step is equal to the rate of the reverse step (all intermediates at constant concentration).
Once this is set up, you can combine all the kinetic equations to arrive at the equilibrium constant equation. The equilibrium constant itself will be a ratio of the product of the forward rate constants divided by the product of the reverse rate constants. Note that even rate constants of steps after the "rate-determining step" will be part of that expression.
The math is shown and illustrated for a simple example in this answer. Alternatively, you could first derive the equilibrium constant expression for each elementary step (as the ratio of forward and reverse rate), and then combine these to arrive at the equilibrium constant expression of the net reaction (see bottom of answer by @thewitness).
Some remarks about the rate-determining step
In a multi-step reaction, the overall rate law will change when concentrations change. For example, if under some condition the rate is zero order in reactant X, if I lower the concentration of X (or leave it out completely), the reaction with X as reactant will become rate-determining step. However, the equilibrium constant expression and its value are independent of the rate-determining step. Therefor, deriving the equilibrium constant expression by relying on a rate-determining step is logically flawed.

I have carefully read "Kinetic Derivation of the Equilibrium Constant" I quoted in the question once again after reading your answer. What I've realized is that even though the whole process is logically flawed, it is algebraically accurate because it basically shares the same method with the examples you provided in your answer. Thanks for sharing your explanation.


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Naming of a dispiroalkane compound
Please help me in naming a dispiro compound:

I propose bispiro[3.0.4.1]decane. Is this correct?

There are instances where a photo might seem an alternative to depict a chemical formula here than a dedicated editor (often this is not the case, and have a look at https://en.wikipedia.org/wiki/Molecule_editor and http://www.gunda.hu/dprogs/ to familiarize yourself with a few of them).  But if you do consider a photo for upload on ChemSE, consider to scale and crop the depiction to what is about the question you address here.  More general, however, post your question altogether with your first attempts to solve the question.

All the special rules you need for naming of such dispiro cycloalkane systems can be found in one chapter of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-24.2.2 Linear polyspiro alicyclic ring systems
Polyspiro parent hydrides consisting of unbranched assemblies of three or more saturated cycloalkane rings are named using the nondetachable prefixes ‘dispiro’, ‘trispiro’, etc., according to the number of spiro atoms present, cited in front of the name of the acyclic hydrocarbon that has the same number of skeletal atoms. The von Baeyer spiro descriptor indicates the number of carbon atoms linking the spiro atoms by arabic numbers that are cited in order, starting at the smaller terminal ring if one is smaller, and proceeding consecutively, always by the shorter path, to the other terminal ring through each spiro atom and then back to the first spiro atom; the numbers are separated by full stops and enclosed in square brackets. The compound is numbered in the order in which the numbers of the von Baeyer spiro descriptor are cited, including spiro atoms when encountered for the first time. Each time a spiro atom is reached for the second time its locant, which has already been assigned, is cited as a superscript number to the number of the preceding linking atoms.
(…)

Following these rules step by step, you will find that the name for the compound that is given in the question is dispiro[3.0.45.14]decane.

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Why is t-BuOK necessary for this synthesis?
1-chloro-2-methylpropane --> 2-chloro-2-methylpropane.
My understanding is that an E2 reaction produces two types of alkenes, the major product for non-sterically hindered bases is the more substituted product (Zaitsev) while the major product for sterically hindered bases is the least substituted product (Hofmann). In this instance, there is only one beta carbon and thus, one product. Wouldn't sodium ethoxide be a better choice as a reagent then tert-butoxide as tert-butoxide will run into steric hindrance from the orbitals of the surrounding methyl groups?
Reading from Klein, Organic Chemistry as a Second Language, First Semester Topics, 4e, p.242, 11.80

Ethoxide is not as strong a base as t-BuOK and is a decent nucleophile which will give significant amounts of the substitution product.

I see it now! Thank you for the help Waylander.

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Why are wet things harder to ignite than dry things?
Specifically wood and paper and other hydrocarbons, but not necessarily excluding other things that burn. We all know wet wood is very hard to ignite. Wetness here refers to water.
But why is that the case? At the chemical level, what is it about water that makes it so effective at inhibiting the reaction of hydrocarbons with oxygen? I guess this also ties in to questions like "Why can't you strike an ordinary match underwater?"

Many hydrocarbons are less dense than water, float on water's surface as seen so often and still may burn (e.g., after oil spills).  But the evaporation of water is a significant heat sink, disfavouring continuation of combustion. And a «blanket» of finely dispersed water droplets may shield the flame from oxygen often required to continue combustion, too.

In general, you can make many things burn all right under conditions which intuitively seem impossible, you just have to find a better oxidizer or a way to improve kinetics. Even water itself burns all right in a flow of fluorine gas: $$\ce{2 H2O + 2 F2 -> 4 HF + O2}$$

To have combustion, you need three things: fuel, oxidizer, and heat. To ignite a piece of wood, you need to raise its temperature to the combustion temperature, where a significant amount of the solid hydrocarbons can vaporize, and then ignite. However, if this wood is wet, the energy input will distribute among the water and solid hydrocarbon. Water is a great heat sink (large heat capacity), so it can absorb a large amount of energy. Thereby, the wet piece of wood is inhibited from achieving the combustion temperature until a large amount of the water has been evaporated off.
One analogous example is heating a full and open, aluminum can of soda (sugar water) with a butane torch. The aluminum won't melt, initially, since the water in the can absorbs the energy. However, if you boil off the water, then the aluminum can will melt after a certain amount of heating.

Burning (=oxidizing with oxygen from the air) organic material generates heat, because the reaction is exothermic, CO2 and H2O being energetically more stable than the organic material before combustion. Of course, despite of not being in thermodynamic equilibrium, we all don't ignite spontaneously, as there is an energetic barrier preventing this. In order to overcome this energetic barrier, we would have to put in some energy first, usually in the form of heat.
In the end, if we want to burn something, we want to create a chain reaction. Create enough energy (heat) to overcome the energy barrier for the oxidation of organic matter with oxygen from the air. Then, if more matter oxidizes, more heat is created, facilitating the oxidation of even more matter, and so on. Water on the other hand serves as a very efficient heat sink, as it is one of the most (energetically) expensive materials to warm up. And evaporating water needs a lot of energy, it really likes it liquid state because of the hydrogen bonds between its molecules.
So if you hold your lighter on a piece of wet wood, most of the energy will be taken up by heating or evaporating water, and almost nothing will go into facilitating the oxidation of the wood.

Perhaps your A is getting diwnvoted because your emphasis on activation energy though correct neglects the pyrolysis /evaporation of organic matter (no one knows what exactly :)) . I did up vote because I find much downvotes not fair. Perhaps it was your style. The answer could have been better but this is rather general.


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When to use chromatographic separation vs vacuum-assisted rectification distillation?
I'm trying to understand a bit more about industrial scale separation processes. One particular element I'm interested in is the question as to when chromatography (HPLC or Simulated Moving Bed) may be preferable to vacuum-assisted rectification distillation.
My first thought would be that chromatography sees its greatest industrial use when processing & separating easily damaged compounds, or to select an individual compound for separation through ion exchange chromatography. This, by comparison to vacuum-assisted rectification distillation, where differing vapor pressures allow for high-purity separation -- but requires the addition of heat to material that may destroy easily-damaged compounds. Would love some help on this! Thank you!

It is always the main question: What are we trying to separate, otherwise the question is too broad. I am afraid the comparison of HPLC  with distillation is like comparing apples and oranges. Even if we talk about preparative HPLC, and simulated moving bed, the scale of quantities separated cannot be compared. I can tell you where distillation will fail...separation of enantiomers. Secondly, most compounds analyzed on HPLCs cannot be distilled. Unfortunately GC cannot be operated at preparative scale. The comparison of the separation techniques is not fair at all.

The decision to use a particular separation technique industrially is based on multiple factors. To speak generally, this comes down to ease of separation and process economics. The simplest and cheapest techniques would be ones like biphasic separation or filtration.
The distillation you refer to bears more relation to bulk chemical manufacture such as oil refining.  Process HPLC and simulated moving bed (SMB) are more involved techniques,  requiring more sophisticated technology and hence, is a more expensive technique. This puts it into the realms of higher value products which would tend to be pharmaceuticals or ultra high purity display chemicals.
So, for process economics, if we think in terms of Input-Process-Output we have cost of input material,  the cost of processing and the cost of output material.  If Input + Processing are greater than the ultimate commercial cost that the product can bear, then you don't have a product or it would need a key differentiator (which could be purity), to justify the extra cost.
If the costs justify it, then using higher vacuum can offset degradation risk involved in a fractional distillation. Physicochemical factors come into play to select this technique (under the ease of separation banner). Industrial distillation would tend to be used for more complex mixtures of desired volatile and semivolatile components.  A good Quora discussion on the topic is here:
https://www.quora.com/How-are-rectification-and-distillation-related
The technique, obviously relies on boiling point and exploitation of potential synergies of compound (usually liquid/oil) mixtures.
Chromatographic separation relies on adsorption/desorption by judicious selection of stationary phase and solvent (mobile phase). For commercial processes, this needs to be repeated over and over in a completely reproducible way with minimal variation. If the stationary phase is cheap, then less material has to go through the material before being changed. If the stationary phase is expensive (decent spherical functionalise silica can easily be over £2000/kg, chiral phases over 10-20k per kg. Irregular silica can be £500/kg), then you want a faster,  cleaner separation with minimal fouling of the support. Over 200 injections onto the same stationary phase is not unheard of.
These injections would also be maximised on production scale vs the standard analytical technique.
For SMB,  typically four HPLC columns are used and a more sophisticated algorithm is required to inject the right amount of material onto the correct column at the right time.
As commented on, isomeric separation,  particularly if the materials are solids,  then  favours process HPLC rather than high vacuum distillation. But again, the process economics would help decide this.
Other factors are internal expertise in the relevant techniques and/or capability of outsourcing partners (contract manufacturing organisations) in more niche techniques. SMB would probably fall into this category i.e. most manufacturing companies would look to a specialist 3rd party unless they were already engaged with this technology.


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Gelifying hydrogen peroxide
What substances could be used to gelify 60% hydrogen peroxide? I imagine anything organic like agar would be broken down by the radicals. I tried looking up what dentists use and the only thing that came up was glycerine, which I know from experience doesn't make gels with water, so I doubt it would work with $\ce{H2O2}.$

Dentists use what they call carbamide peroxide to avoid premature decomposition of the gel itself.

Since 60% H2O2 decomposition is highly exothermic and produces reactive oxygen rather violently, I'd question adding anything that might catalyze that reaction.


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What is the order of stability of the cyclobutyl and cyclopropyl carbocations?
I know  about the exceptional stability of Cyclopropylmethyl carbocation.
In some books it is given that cyclopropyl and cyclobutyl carbocations are highly unstable due to their high strain.
That is true and quite acceptable, but just like cyclopropylmethyl carbocation they can also have resonance structures
and the cyclobutyl carbocation can rearrange itself to the cyclopropylmethyl carbocation by ring contraction.
My question is, what will the stability order between the
cyclopropyl, cyclobutyl, cyclopentyl carbocations?
Also if anyone please can help me to understand why resonance structures in cyclopropyl carbocation will not be possible?

I have tried to fix up your question as good as I could. Please try to use proper spelling and interpunctuation, otherwise it is really hard to understand what you are trying to ask. Also not that the homework tag should no longer be used. Please note that 'resonating structures' is a wrong term. It is resonance structures, as these are structures that take part in resonance. Additionally, please check what you mean with 'cyclopropanium carbocation'; to my knowledge no such thing exists. It appears that this is a completely new question, please ask it separately.

@Martin-マーチン I don't know why I am not able to upload the image. It is repeatedly showing error.


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Making aspirin without acetic anhydride
I want to make aspirin. Normal method is to react salicylic acid with acetic anhydride, but it is extremely hard to find acetic anhydride. How may I synthesize aspirin with acetic acid and salicylic acid only?
I have watched a video where someone makes it with sulphuric acid, salicylic acid and acetic acid. Can I use phosphoric acid insted of sulphuric acid as a catalyst?

This seems like the kind of question that could only be answered by searching in the literature. Without trying it out, it's pretty hard to say. However, it does seem plausible and worth investigating or giving it a try.

@Raphaël Is it theoratically possibble that I can use phosphoric acid insted of sulphuric acid. I will try it If it is theoratically plausible. Check this out https://www.youtube.com/watch?v=M5dQMvPsTEk

Is it not possible to use salicylic acid and acetyl chloride ?

I just looked in the literature, and it has been done even without catalyst in a "reactive distillation column" at high temperatures (patent US201113194873). If you can remove the water formed, you can technically drive the reaction forward, even if it is slow. A catalyst will just make that process easier, requiring less heat and time.
Phosphoric acid is a much weaker acid than sulfuric acid, and it does not have the same dehydrating properties. However, if you have a proper setup, it should do the trick. A big problem with phosphoric acid is that the most concentrated you can normally get is 85%, which means you still have to remove 15% of water somehow. If you don't, you'll likely hydrolyse any aspirin formed and get your starting material back.

Polyphopshoric acid would be a better option

@Raphaël Can I use 99 percent sulphuric acid + salicylic acid + glacial acetic acid because I dont think I can find reactive distillation column

@cccccccaaaaccc Yes, that would be ideal. That patent was mostly to make very large quantities.

@Raphaël OK thank you very much. Another question do I need to heat the mixture of sulphuyric acid, salicylic acid and acetic acid maybe to boil the water off.

The water will co-distill with acetic acid, you may need to add more anhydrous acetic acid as the reaction proceeds

@cccccccaaaaccc The concentrated sulfuric acid can act as water trap for small amounts (that is, what is produced in the reaction). You might have to try it and see if your reagents are water-free enough. If you have anhydrous magnesium sulfate or some other dehydrating agent, you could add that instead.

@Raphaël Would Calcium chloride work. Do I add it with sulphuric acid or how am I going to add it?

@Waylander I didnt understand what you meant. I am not distilling???

You said you were boiling off acetic acid. The water will boil off with it. It would be deeply inadvisable to simply boil it off as that fills your working area with AcOH fumes. Thus you need to distill it off via a simple condenser

@Waylander I dont think my process involves boiling acetic acid? I asked If I needed to heat the solution to boil off water. Is heating necessary in the first place?

you cannot remove water from acetic acid by boiling it off. The acetic acid comes too.


text:
Why polar aprotic solvents don't surround negatively charged nucleophiles?
Protic solvents can form a tight shell around a nucleophile. But what's stopping any polar solvent from doing so? Shouldn't a polar solvent surround negatively charged $\ce{F-}$ with its partially positive side?

Found a good explanation in Oxford University presentations.
It appears that partially positive side of aprotic solvents is hindered in the middle of the molecule. Only negative charge is on the surface. I've looked at DMSO, DMF, acetone, acetonitrile - all of them share this feature. So in case of a negatively charged ion - such solvents are not good at surrounding it.
It looks like elements that end up partially positive (except for $\ce{H}$) are those that can have 3+ bonds ($\ce{C}$, $\ce{S}$), so they are mostly surrounded by other atoms. It's only $\ce{H}$ who has 1 bond and therefore ends up at the surface of the molecule and can interact with negatively charged nucleophiles.


text:
Isothermal vs adiabatic compression of an ideal gas
Could someone please explain these specific questions physically (especially the first question!)? I understand the corresponding mathematical proofs, but not the physical reasoning? I have looked online and can only find answers to what I believe correspond with my second question.

Why is isothermal work larger than adiabatic work for the compression of an ideal gas for the same change in pressure?
Why is isothermal work less than adiabatic work for the compression of
an ideal gas for the same change in volume?

There are a number of ways to look at this. One is to look at the equations. Another is to plot the pressure p as a function of volume V. The area under the pressure curve on the plot is the total work done on the system.

It is very easy to answer your second question by inspection of this figure, but perhaps not the first question without actually evaluating integrals (areas).
Yet another way to answer this, but perhaps hand-wavy, is to think of what happens as you compress the ideal gas either adiabatically or isothermally. In the adiabatic compression no energy is lost to the surroundings as heat. All of the energy transferred to the gas as compression work is retained by the gas. But the only way to increase the internal energy of an ideal gas is to raise its temperature. Therefore the effect of the adiabatic work is to increase the temperature of the gas. The increase in temperature is accompanied by an increase in pressure above that due to Boyle's law. This follows from
$$p=\frac{nRT}{V}$$
In the adiabatic process p increases with the higher temperature above the level it would have if the heat had been lost to keep the temperature constant, as it would in an isothermal process. In that way the adiabatic compression is a more efficient way of increasing the pressure. This means less work is required to arrive at the same final p. By the same logic, if you want to achieve a determined final volume, you will have to work harder in an adiabatic setting because the pressure you are working against increases more rapidly as the gas gets hotter.


text:
Is it compulsory to have the coefficients of a balanced chemical equation in the simplest ratio?
The thermal decomposition of siderite equation: $\ce{FeCO3 -> FeO + CO2}$ looks correctly balanced to me, you can notice $\ce{1 Fe , 1 C , 3 O}$ on the Left side and the same number of molecules on the other side.
However my text book writes it as $\ce{2FeCO3 -> 2FeO + 2CO2}$
and I wanted to know why mine is not correct.

Textbook's version is wrong, as written, because you have to write the smallest possible coefficients, therefore 4A -> 4B+4C is also incorrect for something which should be  A -> B+C.

@M.Farooq So  FeCO3→FeO+CO2 is the correct answer ?

If the products are carbon dioxide and FeO, then your version is correct. There is not need of an additional 2. It must be a typo in the book.

@M.Farooq thanks for your time , that helped me a lot.

Please do not use MathJax formatting on the title.

Well, just like any other reactions, it depends upon the the heating temperature and the reaction environment. Contrary to what the reaction tells, the decomposition products of the reaction is actually iron(II,III) oxide, $\ce{Fe2O3}$ if it is heated in oxygen environment$\ce{^{[1]}}$ and $\ce{Fe3O4}$ if it is heated in $\ce{CO2}$ environment$\ce{^{[2]}}$.  $\ce{FeO}$ also formed during the reaction but for a small period of time. Hence it is an intermediate product.

Figure 1 shows the thermal decomposition of iron carbonate at
  temperature at 600°C. It is shown from one DTA curve in Fig. 1 one
  exothermic peak at 495°C accompanied with a weight loss of about 32%
  (theoretical value 31.1%). This step is due to the decomposition of
  iron carbonate to $\ce{Fe2O3}$.
X-ray diffraction patterns of the decomposition products of iron
  carbonate were determined for the solid calcined at 550 and 800°C
  Figure 3 (a) shows X-ray diffractogram for the sample dried at room
  temperature corresponding to iron carbonate. The diffractogram of the
  sample calcined at 550°C consists of very poorly crystallized α-Fe2O3
  phase. .Heating of ferrous carbonateat 800°C led to the formation of
  well crystalline $\ce{α-Fe2O3}$

In air, the X-ray pattern of $\ce{FeCO3}$ was replaced by that of
  $\ce{Fe3O4}$ as the sample was heated. The change was essentially
  complete at 500°C. No further changes were noted. If, however, the
  furnace atmosphere was changed from air to $\ce{CO2}$ or helium above
  1000°C. the sample transformed slowly to $\ce{Fe3O4}$. Restoring the
  air atmosphere again resulted in the pattern of $\ce{Fe2O3}$. When
  $\ce{FeCO3}$ was decomposed in a $\ce{CO2}$ atmosphere, the first
  decomposition product observed was $\ce{Fe3O4}$.[...]the brief
  appearance of $\ce{FeO}$ was suspected in the decomposition. By
  oscillating the Geiger counter through a very small arc encompassing
  the strongest line of FeO, the presence of this oxide was indicated.
  The line appeared for only a very short time and was at no time very
  strong. In the inert atmosphere, the products of decomposition were
  $\ce{FeO}$ and $\ce{Fe3O4}$. The $\ce{Fe3O4}$ in the sample could not
  be converted to $\ce{FeO}$ even by heating to the highest temperature
  available with the furnace (about 1150°C. in helium).

So, the reaction indeed forms $\ce{FeO}$ but it gets oxidized depending on the environment the reaction is proceeded.
$$\ce{FeCO3 -> FeO + CO2 }$$
$$\ce{4FeO + O2 -> 2Fe2O3}$$
$$\ce{3FeO + CO2 -> Fe3O4 + CO}$$
References

A.A. El-Bellihi, "Kinetics of Thermal Decomposition of Iron
Carbonate", Egypt. J. Chem. 53, No. 6, pp.871- 884 (2010) DOI: https://ejchem.journals.ekb.eg/article_1268_82097e00ce1c4e475f17cfec61edb6d2.pdf
H. E. KISSINGER, H. F. McMURDIE, and B. E. SIMPSON,Thermal Decomposition of Manganous and Ferrous Carbonates,DOI: https://ceramics.onlinelibrary.wiley.com/doi/pdf/10.1111/j.1151-2916.1956.tb15639.x

The thermal decomposition of siderite is interesting aspect. First, the answer to your question is your textbook has mistakenly put a common factor (of 2) to balance the chemical equation. It is not wrong but traditionally we do not put a common factor on equations, unless we intended to add or subtract two of more equations. Said that, I have a doubt that this equation is correct at all, based on literature. Although this must have been too much for the OP, I'd like to make my point.
One such publication reveals the following (Ref.1):

Abstract: In order to provide a better theoretical foundation for utilisation of Xinjiang siderite resources in China, its thermal decomposition behaviour was studied in neutral and oxidising atmospheres by employing thermodynamics analysis, chemical titration, thermogravimetric, and X-ray diffraction means. Isothermal experiments were conducted to investigate the thermal decomposition kinetics of siderite lump in a weakly oxidising atmosphere at $\pu{500–850 ^\circ C}$. The results reveal that siderite has self-magnetisation characteristics under controlled conditions, and the phase evolution process and final products of decomposition depend temperature and atmosphere. The phase transformation process in weak oxidising atmosphere follows the steps as: $\ce{FeCO3 -> Fe3O4 -> \gamma-Fe2O3}$ at $\pu{550 ^\circ C}$, and $\ce{FeCO3 -> FeO + Fe3O4 -> Fe3O4 -> \gamma-Fe2O3 -> \alpha-Fe2O3}$ at $\pu{800 ^\circ C}$. In inert atmosphere, the decomposition pathway is $\ce{FeCO3 -> Fe3O4}$ below $\pu{733 ^\circ C}$ and $\ce{FeCO3 -> FeO + Fe3O4}$ above $\pu{733 ^\circ C}$. The molar ratio of $\ce{FeO/Fe3O4}$ increases with temperature. The decomposition kinetics of siderite lump in oxygen-deficient atmosphere is consistent with chemical reaction control in the temperature range $\pu{500–700 ^\circ C}$ and nucleation and growth mechanism in the $\pu{750–850 ^\circ C}$. The corresponding activation energies are $53.73$ and $\pu{38.15 KJ mol−1}$, respectively.

It is clear from this reference that decomposition products of siderite ($\ce{FeCO3}$) are dependent on the temperature. Yet, the authors have not obtained iron(II) oxide ($\ce{FeO}$) as a sole product. Another recent publication (Ref.2) claimed $\ce{Fe2O3}$ as the final product when $\ce{FeCO3}$ is heated above $\pu{400 ^\circ C}$, based on its thermal decomposition studies in air by DTA-TGA, XRD, SEM, and Mossbauer measurements. The authors have observed a weight loss of about $32 \%$ at the single exothermic peak at $\pu{495 ^\circ C}$. They claimed the theoretical weight loss is $31.1 \%$, thus decomposition should be consistent with the equation:
$$\ce{2FeCO3 -> Fe2O3 + CO + CO2}$$
This finding is also consistent with an old Science publication claiming $\ce{FeCO3}$ decomposed to magnetite ($\ce{Fe3O4}$) and graphite when heating in equilibration with graphite at $\pu{455–465 ^\circ C}$ (Ref.3).
All these evidence directed to doubt the simple equation given in OP's textbook.
References:

Y. H. Luo, D. Q. Zhu, J. Pan, X. L. Zhou, “Thermal decomposition behaviour and kinetics of Xinjiang siderite ore,” Mineral Processing and Extractive Metallurgy - Transactions of the Institutions of Mining and Metallurgy: Section C 2016, 125(1), 17-25 (https://doi.org/10.1080/03719553.2015.1118213).
A. A. El-Bellihi, “Kinetics of Thermal Decomposition of Iron Carbonate,” Egypt. J. Chem. 2010, 53(6), 871-884 (DOI: 10.21608/EJCHEM.2010.1268).
Bevan M. French1, P. E. Rosenberg, “Siderite ($\ce{FeCO3}$): Thermal Decomposition in Equilibrium with Graphite,” Science 1965, 147(3663), 1283-1284 (DOI: 10.1126/science.147.3663.1283).

Both equations are balanced, both equations are correct. Using other coefficients such as $0.5, \sqrt{2}$ or ${e^3}$ would also be correct. Only if the ratio of coefficients were different (i.e. not $1:1:1$) it would be wrong.
However, it is commonly preferable to write the smallest possible integers, so $1,1,1$ is preferred over $2,2,2$.


text:
If LPG gas burners can reach temperatures above 1700 °C, then how do HCA and PAH not develop in extreme amounts during cooking?
According to What is the temperature of heat generated from LPG gas?, the temperature of (1) LPG gas is above 1700 degree Celsius.
Many of us are familiar with pan-frying meat (steak, pan-frying the chicken breast etc).
Next, consider this: when you're (2) cooking meat at high temperatures ("above 200 °C) then it'll start forming heterocyclic amines (HCAs) and polycyclic aromatic hydrocarbons (PAHs), according to healthline.com and precisionnutrition.com.
Note: The way they're formed is when fat drips onto the flame or the utensil, which is at very high temperature, and then reacts with the high heat to form HCAs and PAHs.
Lastly, there is research that (3) corelates HCAs and PAHs with causing cancer in our bodies.
Sources: They're are all mentioned in the two articles that I linked above.
Question: If you combine these 3 facts, then we must be cooking a huge amount of HCAs + PAHs. It's a wonder that we aren't dead yet! How is it that we haven't developed cancer by eating meat cooked over a pan on an LPG gas (which I believe is quite common)? Or are we all consuming HCAs / PAHs slowly over time and will develop cancer? Or are one of the 3 facts above wrong?

It is the temperature of the pan that matters not the flame
The flame temperature is irrelevant if you are cooking in a vessel. the only temperature you need to worry about is the temperature of the surface of the pan (or–even more importantly–the temperature of the meat). The surface of the pan will rarely get above around 220 °C if you are monitoring it. An unattended pan can get hotter, but, if you let it get too hot, your food will rapidly burn and will not be edible. Conversely, if the food is palatable, you probably haven't heated it enough to get lots of nasty HCAs and PAHs.
But the research that caused your worry is also frequently overstated. Yes, large quantities of PAHs or HCAs may be nasty, but the amounts in food–even food cooked on an open flame barbecue–are very small and there have been no convincing studies showing a notable effect on health. There are some studies showing  a very small effect of meat on health (diseases such as bowel cancer have been linked to some meats but the studies are statistically weak and the effects are very small. Moreover these studies link to meat not cooking products but would probably have spotted any effects based on nasty cooking by-products). It is worth remembering that people have been cooking on open flames since we invented fire. If that were really dangerous, there would be strong evidence of harm and/or primitive man would have developed good defences to avoid the harm (as milk-drinkers evolved lactose tolerance in adults because their diets consisted of a lot of dairy products).
There have been many food scares based on observations of known nasties in cooked food. Acrolein and Nitites, for example. None of these have been shown to have any notable measurable effect on people in the concentrations present in the diet.

Note that teflon begins to break down into rather nasty substances at just over 250°C, so it's reasonable to assume that your pan is below that when cooking.

@R..  Thanks! That's very helpful to know.

Nitrites can be converted to nitrosamine in the stomach and those may be carcinogenic. It would help if you could add some references to support some of your claims.

@JJJ For a good summary of the evidence see this question on skeptics.SE. The bottom line is that epidemiology does not see the effects a simplistic chemical hypothesis suggests might exist.

Nothing with more thermal mass than a needle will reach the temperature of the flame. Note that if the stove or utensils would be this hot they would be glowing blinding white, while IRL they won't even glow a dim red. Granted if you dripped fat on a hot stove it would turn into a black tar containing PAH's, and a microscropic fraction of this could end up in your food. We are exposed to carcinogens and radiation all the time, theres no reason to be alarmed.

Okay this was helpful. You make a valid point. The utensils are at a much lower temperature. I'm not sure how much but I'm assuming that crossing 200 degree Celsius is possible. What do you think? Also, the black tar is what contains the PAHs is it? I've seen that forming every time I cooked chicken on the pans. But what makes you say that a microscopic fraction of it would end up in our food? If you see the article I linked (source1) it says that any fat that drops on a hot pan is going to create PAHs which will end up in your food. It doesn't talk about quantities at all though. What say?

@Mugen If you let the pan get that hot, and then drip the fat on the pan, sure. But why (and how) would you ever do that? Francis (and your reference) talks about dripping fat on the stove/flame, not the pan. If the pan was at 1500 °C, it would glow orange hot, surely you don't normally cook that way :) Finally, don't forget that the utensil reaching 200 °C isn't enough - you need it hot enough to heat the oil fast enough to reach (and maintain) 200 °C. Unless you have a very tiny droplet of oil, this isn't going to happen. Heat takes time to transfer, and you've added cold mass.


text:
Spectrometer vs Spectrometry vs Spectroscopy
How do you differentiate between the three?
I read that "Spectrometry deals with the measurement of a specific spectrum. There are four primary types of spectrometers:

Mass spectrometry  2. Neutron triple axis spectrometry 3. Ion-mobility spectrometry 4. Rutherford backscattering spectrometry"

How about Raman Spectrometers and Infrared Spectrometers. Don't they belong to spectrometry?
Some believe that the main difference is that spectroscopy typically deals with light, mass spectrometry does not.
So do you called it Raman spectroscopy or it is not right to refer it as Raman spectrometry?
It's confusing.

'Spectrometry' means measuring spectra of any sort and more specifically  'spectroscopy' measuring spectra as a result of interaction of light or radiation in general. A  'spectrum' means measurement of some quantity on a scale  between two limits.

Long time ago, I wrote mass spectroscopy, by mistake, in an undergraduate exam and the instructor told me "You were the only one in my class who wrote mass spectroscopy." The explanation was that MS is mass spectrometry because it does not deal with light or electromagnetic radiation and that's what the textbook said. At that time internet was not that common. Now I searched the term "mass spectroscopy" in papers using Google Scholar and there are 1,100,000 results with this exact phrase. There are 2,330,000 results for mass spectrometry.
It seems, given the usage, and the difference is not that strict. Consulting the controller of all  chemical terminologies, IUPAC says
https://goldbook.iupac.org/terms/view/S05848

The study of physical systems by the electromagnetic radiation with
  which they interact or that thay [spelling mistake] produce. Spectrometry is the
  measurement of such radiations as a means of obtaining information
  about the systems and their components. In certain types of optical
  spectroscopy, the radiation originates from an external source and is
  modified by the system, whereas in other types, the radiation
  originates within the system itself.

You will see famous books, especially by Griffiths with the title "Fourier Transform Infrared Spectrometry" and nobody raised an eyebrow.

In the Export Strategic dual list items controlled listing, there is this passage:
"3A999 Specific processing equipment, n.e.s.,
as follows (see List of Items Controlled).
f. Chromatography and spectrometry analytical
instruments."
Do you think the spectrometry above also cover Raman spectrometers? Or only mass spectrometers or alike those not related to light?

@Jtl likely yes. While chromatography equipment may serve for actual preparation of strategic compounds, mass spect. does not. And the text specifies "analytical", so all spectro(photo)meters could be included.

@Alchimista Thanks. Please look at this HTS (Harmonized Tariff Schedule). https://imagizer.imageshack.com/img921/9388/N8HbhY.png Do you consider Spectrometer as "Electrical" or "Other"?  I know spectrometer is electronic. But it doesn't run on electricity even when there is a CCD sensor. Or is it electrical since the adapter or USB can be powered by electricity? Or is there other distinction I missed?

@Jtl I have no idea what they mean. Neglet power source as for nothing is run by a crank, it is obscure to me. If you have a raman I would call it spectrometer or spectrophotometer (by means of which you can do spectrometry/spectroscopy, in  be in line with the answer above).

An spectrophotometer is defined as "an apparatus for measuring the intensity of light in a part of the spectrum, especially as transmitted or emitted by particular substances.". Raman spectrometers don' do that. so why do you think a raman spectrometer is also called a raman spectrophotometer?

While I'm not sure about English usage, I think there may have been a shift in the usage of the corresponding terms in German (Spektrometer, Spektroskop, Spektrometrie, Spektroskopie). I first learned spectrometer/spectrometry => implies quantitation (of intensity), which was not (necessarily) the case with spectroscopy and certainly not with a spectroscope (an instrument for visual comparison of e.g. atomic emission lines against reference; skop IIRC is greek meaning "look at"). Later I've also met "if it's about electromagnetic waves, it's spectroscopy" instead.


text:
Finding original concentration w%/w%
A question from a grade 9 admission exam:
When 1 kg of salt is added to a solution of salt
and water, the solution becomes 33 1/3% salt by
mass. When 1 kg of water is added to the new solution, the resulting solution is 30% by mass. The percentage of salt in the original solution is...
I tried:
(Salt orig +1)/(Water orig + salt orig + 1) = 33 1/3. (S + 1)/(W+S+2) = 30. Which means 33 1/3*(W+S+1)=30*(W+S+2); 33 1/3 W+33 1/3 S +33 1/3 = 30W + 30S + 60; 3 1/3 W + 3 1/3 S = 26 1/3; S+W = 79. Putting that into the 1st eq gives S+1 / 80 = 33 1/3 meaning S= 26 2/3. Solving for W = 79-26 2/3 = 52 1/3. So the original concentration = 26 2/3 / (52 1/3 + 26 2/3) = 33 3/4%. But that's not one of the multiple choice answers (I did get further this time!).

We have a policy which states that you should show your thoughts, effort and attempts to answer your question yourself. It'll make us certain that we aren't doing your homework for you, and that the Q/A is beneficial for broad audience.   As "homework-like questions" are considered literal homeworks, self-study questions, puzzles, worked examples etc.  Please [edit] in your full reasoning or thoughts on this. See Homework.  Otherwise, the question may get closed as "low effort homework-like question class."

I tried using algebra to solve for original concentration - using the formula salt orig +1/ salt orig + water orig + 1 = 33 1/3. Then S(o) + 1/ s(o) +1+w(o)+1 = 30. I went down a rabbit hole that told me w(o) = .77w(o) -1 which is clearly wrong! We haven't done anything like this in school and my science teacher says he has no idea. This question is from a practice exam for a science/math high school admission. Help!

I should also say I'm in grade 8. Thanks.

Put your failed solution attempts to the question, and you can ask for the reopening the question. Note that the needed algebra is very basic.

You should either calculate with mass fractions instead of percentages, either multiply the mass fractions by 100.

$$\frac{S +1}{W + S + 1} = 1/3$$
$$\frac{ S + 1}{W+S+2}= 3/10$$

$$S +1=\frac {W + S + 1}3$$
$$S + 1=(W+S+2) \cdot 3/10$$

$$3S +3=W + S + 1$$
$$10S + 10=(W+S+2) \cdot 3$$

$$W=2S +2$$
$$10S + 10=(2S+2+S+2) \cdot 3$$
$$10S + 10=9S+12$$
$$S=2$$
$$W=6$$

The initial concentration is 25 % (w/w).

Thanks!!! Like Sherlock Holmes once you explain it, it's so easy! Using 1/3 instead of 33.3% is brilliant. And the question isnt really chem, it's math! The original equation for w/w percent is chemistry though.

You then can mark the answer as accepted.


text:
How to weight picric acid?
Picric acid is stored under a layer of water to prevent an explosion.
What is the right way to measure the amount of picric acid that taken out? It is wet and containing water.

Only about 1–2% of picric acid would be actually dissolved in water at room temperature, so without consulting any literature sources I guess it would be enough to just distribute the dispersed substance on a surface and let it dry for a while.

To let it dry is simple, but SDS says not to dry as it is very unstable and may cause explosion ...

Well, you have to take certain precautions (maintain purity, avoid rapid heating/sparks/shocks etc.) and don't accumulate substantial amounts over months.

You may have to dry it in an inert and safe place and then re-wet it/make a known solution in a controlled manner so you can weigh it safely. I worked with another shock/friction sensitive explosive and that's how the lab techs and researchers worked with it. You have to have very good safeguards and lab practices for the drying portion. You may want to consult with your lab/university safety person, they may be able to find more resources on how to best handle the material.

Find some nice instruction:
http://nzetc.victoria.ac.nz/tm/scholarly/tei-Bio06Tuat03-t1-body-d2.html
As Picric Acid stored under layer of water, we could know it's concentration in the water. It should be enough for most of the applications.
Temperature ° C.    grams picric acid/100 grams solution

0    | 0.67
10   | 0.80
20   | 1.10
30   | 1.38

If accuracy is not essential, weigh a small amount drained for a fixed time but still wet, and then weigh again dried, to find the ratio of wet-to-dry weight (carefully disposing the dried, sensitized, explosive without destroying the scale or personnel).
Though this could give a rough idea of the actual dried weight, it would vary from batch to batch and with temperature.


text:
Can someone walk me through this gas mixture question?

Argon $(\ce{Ar})$ and helium $(\ce{He})$ are initially in separate compartments of a container at $\pu{25 °C}.$ The $\ce{Ar}$ in compartment A which has a volume $V_\ce{A}$ of $\pu{9.00 L}$ and a pressure of $\pu{2.00 bar}.$ The $\ce{He}$ in compartment B of unknown volume $V_\ce{B}$ has a pressure of $\pu{6.00 bar}.$ When the two compartments are connected and the gases allowed to mix, the total pressure of gas is $\pu{3.60 bar}.$ Assume both gases behave ideally.
$$
\begin{array}{|l|l|}
\hline
\text{Chamber A} & \text{Chamber B} \\
V_\ce{A} = \pu{9.00 L} & V_\ce{B} = \pu{??? L} \\
p_\ce{A} = \pu{2.00 bar} & p_\ce{B} = \pu{6.00 bar} \\
\hline
\end{array}
$$
[4 marks] Determine the volume of compartment B.

I've been stuck on this problem for a while.  Could someone walk me through the beginning steps? I should be good to go with some guidance.  Would I use Boyle's law in attempt to find the final volume once the partition is removed?

Have you watched this Youtube mini 8 min lecture, it addresses exactly the same problem on board... https://www.youtube.com/watch?v=uFwFxgWDm5U.

@M.Farooq  I watched the lecture, but didn't know how to approach without any values for n.  I can determine the number of moles in chamber A, but I'm not sure how I would do that for chamber B or the final mixture.

You can determine the moles of Ar from its pressure temperature and volume using PV=nRT.

@M.Farooq Using the equation Pfinal = RTfinal[(n1+n2)/(V1+V2)], I still have 2 unknowns, n2 and V2. Is there a way to find n2 with my current information?

You need one more equation to set-up. Two unknowns require two equations. What if you set up PV=nRT for helium? P is given, V is unknown, n is unknown and RT are known. See if that solves the problem.

When I used those equations in a system of equations, I got negative values.  Ignoring the negative sign, it still didn't provide the correct final pressure when input into the ideal gas law for the final values.

Getting a negative sign is a red flag that some arithmetic or algebra is wrong. I don't see any other approach besides solving two equations with two unknowns. Did you convert temperature to Kelvin and the units of R are correct?

Solve for the amount of gas in A: $n_A=p_AV_A/RT=\pu{0.727 mol}$. You'll need to perform necessary unit conversions or use R in units that match the given T, p and V units.
Write the volume of chamber B in terms of knowns: $V_B=n_BRT/p_B=n_B \times\pu{4.13\times 10^3 m^3/mol}$.
Write the total volume when mixed: $V_{tot}=n_{tot}RT/p_{tot}=n_{tot} \times\pu{6.89\times 10^3 m^3/mol}$.
Solve this last expression for $n_B$ after inserting the equalities $V_{tot}=V_A+V_B$, $n_{tot}=n_A+n_B$  and the result of step 2:
$$\begin{align} V_A+V_B &=(n_A+n_B)\times \pu{6.89\times 10^{-3} m^3/mol}\\ &=V_A + n_B \times\pu{4.13\times 10^{-3} m^3/mol}\\\rightarrow n_B &= \frac{V_A - n_A\times \pu{6.89\times 10^{-3} m^3/mol}}{\pu{2.75\times 10^{-3} m^3/mol}} = \pu{1.45 mol}\end{align} $$
Compute the volume in chamber B: $V_B=\frac{n_{B}RT}{p_B}=\pu{6.00\times 10^{-3} m^3}$

It's usually a good idea to perform a check to see if we did things right. Compute back the final pressure: $p_{tot}=\frac{n_{tot}RT}{V_{tot}}=\pu{3.60 bar}$.
Answer: $V_B = \pu{6.00 L}$


text:
Liquid crystals for academic outreach?
I am interested in making a Schlieren texture of liquid crystals to show kids at an outreach event (see image below). I have the polarizers and camera for the imaging part, but I don't know of a place to find liquid crystals.
I would prefer something cheap and chemically inert because I don't want to bring anything sensitive or hazardous. I would imagine that it should be really easy to get solutions or powders of liquid crystal materials, but I am having a hard time finding a source.
Any suggestions? I don't have access to a proper chemistry lab, but I can mix basic things in my kitchen.

Could you disassemble a cheap calculator so that you have access to the LCD display? It comes with polarizers already attached, but if you can remove the top one, you can use your polarizer. It doesn't have Schlieren, though.

Did you try right the title in Google? Teachers have wrote about. Don't know if setups are simply enough for you, though.

I don't understand how this question is too broad, no one has given a single example of a purchasable liquid crystal solution. If it was too broad I would expect too many options if anything.

Try this site. Merck are a major manufacturer of LCD materials: https://www.merckgroup.com/en/expertise/displays/solutions/liquid-crystals/lcd-technologies.html

Assuming you already have the optical setup built around, aiming for the least amount of expensive chemicals, you may try semicrystalline extruded polymers, e.g. polyethylene foils, or PET bottles.
The «trick» is that you do not melt them, but that you reach just the glass transition temperature ($T_g$) which is below melting. Here, crystalline parts in the polymer break up, rendering the material isotropic and in transmittent observation with crossed polarizer / analyzer black.  Depending on material and degree of polymerization, this solid-solid transition already may happen in a range of 60 to 80 Celsius.  Put the material on a glass slide which is heated gently from beneath with an low-powered heating plate.  With a tooth-pick you may test that the material becomes softer.
Upon cooling you will see how spherulites re-appear, e.g.

(source)
Size and shape of them will depend on the rate of cooling, for example if let cool at air (or on the table of your microscope), or if you cool the glass slide by putting it on an ice bag.
If not overheated this may be repeated over and over easily for a dozen of times.  Apart from the heating plate, there is little potential danger.
A low-cost set up including a heating stage, where you could continue the observation of the sample's recrystallization was described in the February issue of the Journal of Chemical Education (J. Chem. Educ., 2019, 96, 823-826, doi 10.1021/acs.jchemed.8b00879).

Thanks, but isn't this an example of a crystal, not a liquid crystal? That is, it is actually on the solid state at the end of the day. In that case, I think salt crystal formation from water drying would be the typically demonstration they use for outreach. What I'm struggling on is the liquid crystal (e.g. nematic, smectic) phase.


text:
Is there any publicly available software that can do deterministic CCSDTQP other than MRCC?
MRCC can do arbitrary order coupled cluster but much of the code is computer-generated (not hand-coded by a human) so it can be slow. There are some stochastic codes for high-order coupled cluster, but as my question states, I am looking for a deterministic code that can be an alternative to MRCC.

I had performed a similar search and found nothing. Many programs capable to compute CCSDTQP do this by calling MRCC (for CCSDTQ is different). If you look for extreme precision maybe you can success by following other approaches, like selected CI or similar.


text:
How does one electroplate Lithium onto a conductive metal?
In this case the anode for electroplating would be the Lithium and the cathode would be any conductive metal.
Unfortunately Lithium reacts violently with water, which is the basis of most electrolyte solutions that are used for electroplating.
So is there an anhydrous electrolyte solution or an alternative method for electroplating Lithium onto a conductive metal?

Molten salt maybe?

Lithium hexafluorophosphate dissolved in dialkylcarbonate. That is bases of LiIon/LiPol cells. But it is highly flammable.

I'd rather use the ambient temperature molten salt ethylammonium nitrate. Easy to synthesized and no need for high temperatures. Thoughts?

Nitrate is oxidizing, maybe?

Ethylammonium is a weak acid. Lithium would not survive there. I would expect reactions: $$\ce{2 EtNH3+ + 2 Li -> 2 EtNH2 + 2 Li+  + H2}$$   $$\ce{2 EtNH2 + 2 Li -> 2 EtNHLi  + H2}$$ and reduction of nitrate. Additionally, there would be preferred $$\ce{2 EtNH3+ + 2e- -> 2 EtNH2 + H2}$$

I would expect lithium to react with ethylammonium nitrate more violently than with water. Additionally, an inert atmosphere is needed, as lithium reacts even with nitrogen.

Why alternative ? Note that even dialkylcarbonate very slowly react with lithium. But considering your estimated chemistry knowledge, I rather discourage you from such experiments.


text:
Is the correlation energy of a system always negative?
The correlation energy of a system is defined as the difference between the exactly energy and the energy in the Hartree-Fock method: $E_\mathrm{cor} = E - E_\mathrm{HF}$. In the case of an atom or a molecule, is it possible say that $E_\mathrm{cor} < 0$, since the interaction between the electrons is positive?

Being nit-picky, there are systems in which HF provides the exact answer and thus $E_{\text{cor}}$ is zero and thus not negative.

The answer to this question rests on the fact that Hartree-Fock is a variational method.
A variational method is one where you supply a guess wavefunction ($|\Psi\rangle$) with some parameter(s) ($\alpha$) and then use those parameter(s) to minimize the energy. Doing this, you are guaranteed that your resulting energy is an upper bound to the exact ground state energy.
$$
E_{\text{exact}} \leq \frac{\langle\Psi_{\alpha}|\hat{H}|\Psi_{\alpha}\rangle}{\langle\Psi_{\alpha}|\Psi_{\alpha}\rangle}
$$
So, if we are talking about the exact correlation energy (such as one given by full CI) then the answer is that the correlation energy must always be negative. This is  because Hartree-Fock always provides an upper bound to the exact ground state energy. However, if we use a non-variational method to approximate the correlation energy (such as coupled cluster) then we could find cases where we calculate a positive correlation energy, but this is distinct from the exact correlation energy.
(For an authoritative description of this in the context of quantum chemistry with examples and practice problems, I recommend the first chapter of Szabo and Ostlund's Modern Quantum Chemistry)


text:
Can anybody help me out with this chain conversion?
I can understand that grignard reagent will attack the oxygen breaking the double bond of the oxygen forming a diol but that is unstable so it should form a Ketone but after that I can't really understand.

It is a bit of a trick question. Write a balanced equation for the reaction between methyl magnesium chloride and benzoic acid, then refocus your attention!


text:
Effect of chloromethyl substituent vs methyl substituent on yields of nitration
The effects of $\ce {-CH3}$ and $\ce {-CCl3}$ substituents on the product distribution of an electrophilic aromatic substitution reaction are unambiguous. The $\ce {-CH3}$ group exhibits a clear electron-donating effect and directs electrophiles to the 2- and 4- positions. Similarly, the $\ce {-CCl3}$ group exhibits a clear electron-withdrawing effect and directs to the 3- position. Additionally, due to the steric bulk of the three chlorine atoms, there is significant steric hindrance that electrophiles would face when they attack the 2- position. Hence, the proportion of ortho nitration products would be very low.
Some data on the production distributions for nitrations of particular substituted benzenes, taken from Carey & Sundberg (2007), p. 786, is shown below. Observe that the ratios given are consistent with what we have discussed above.
\begin{array}{|c|c|c|c|} \hline
\text{Substituent } & ortho & meta & para \\ \hline
\ce{$\ce {-CH3}$} & 56-63 & 2-4 & 34-41 \\ \hline
\ce{$\ce {-CCl3}$} & 7 & 64 & 29\\ \hline
\end{array}
However, we are interested in the $\ce {-CH2Cl}$ substituent, not the $\ce {-CCl3}$ substituent. We would, of course, expect the steric effect to be lesser than that of the $\ce {-CCl3}$ substituent and we would also expect different electronic effects exerted on the ring. If we were to look at more data from p. 786, as shown below, we would observe that for a substituent like $\ce {-CHCl2}$, it still remains as an electron-donating group as the proportion of meta nitration is still lower than that of ortho and para nitration. It seems that even with one $\ce {C-H}$ bond, there is still sufficient hyperconjugation with the benzene ring to outweigh the inductive pull of the two electronegative chlorine atoms.

I have no data for the $\ce {-CH2Cl}$ substituent but I believe that based on the above analysis, we predict that $\ce {-CH2Cl}$ would also be an electron-donating group that is more electron-donating than $\ce {-CHCl2}$ but less so than $\ce {-CH3}$. Additionally, we would expect to also display directing effects to the 2- and 4- positions However, I have great uncertainty regarding the steric effect of this substituent. How would the ratio of ortho to para nitration be like?
Reference
Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.

The following is the abstract of K.M. Zhang "Selective Nitration Of Benzyl Chloride And Its Application In The Synthesis Of Nitrobenzaldehyde"(link). The full pdf is behind paywall.

The selective nitration of benzyl chloride and its application in the synthesis of 4-nitrobenzaldehyde were studied. Nitration of benzyl chloride with $\ce{HNO3}$ in the presence of $\ce{Nb2O5}$(6% weight percent) at 0-5℃ for 1.5 hours gave mononitro products in 94% yield with 77-80% para selectivty. $\ce{Nb2O5}$ was recovered by filtration and reused in the nitration. The yield of mononitro products and para selectivity were above 90% and 76% respectively for 5 times repeated use of $\ce{Nb2O5}$. Nitration of benzyl chloride with urea nitrate(UN) in $\ce{H2SO4}$ was also studied. The nitration was carried out at 25℃ for 24 hours and gave mononitro products in 98% yield with 90% para selectivity. Base catalyzed hydrolysis of 4-nitrobenzyl chloride at 75-80℃ for 22 hours gave 4-nitrobenzyl alcohol in 77% yield with 98% purity. Oxidation of 4-nitrobenzyl alcohol in dilute nitric acid at 55℃ for 3 hours gave 86% 4-nitrobenzaldehyde, compared with industrial oxidation of 4-nitrobenzyl alcohol,whichwas carried out at 70℃for 4 hours and gave 70% 4-nitrobenzaldehyde. 4-nitrobenzaldehyde was also obtained in 86% yield with 92% purity via ionic liquid immobilized $\ce{TEMPO/CuCl}$ catalyzed oxidation of 4-nitrobenzyl alcohol with molecular oxygen in ionic liquid $\ce{[bmim]PF6}$. The reaction condition is mild, the selectivity is high, and no 4-nitrobenzoic acid is detected in any case.

Ok thanks for the provision of data. May I clarify the inferences to be made from such data: Firstly, we can assert that, based on the high para selectivity, the chloromethyl group still provides significant steric hindrance to the ortho positions. Secondly, the chloromethyl group is indeed still rather electron-donating and thus reacts under mild conditions of 25 deg C for 24 hours to give the nitration products?

that seems a reasonable conclusion


text:
Is "density per unit volume" something?
I was reading articles on the preparation of shellac-based composites when I encountered the following statement (in ref. 1):

Wood flour and fibres are interesting because of their low cost, good specific  strength, low  density  per  unit  volume,  along  with  their  renewable  and degradable  features. The  drawbacks of  natural  organic  fillers include poor  adhesion between the filler and the polymer matrix, and poor thermal and water resistance (Saba et al. 2014).

Note the same phrase is used in the reference cited in the passage quoted above (Saba et al. 2014) but then the trail goes dead.
Is "density  per  unit  volume" a typo or other error that's been propagated during citation or a real concept of concern for material scientists? If it is, what is the meaning of this concept?
References

Obradovic et al. (2017). “Cellulose-shellac composites,”BioResources12(1), 1943-1959.

I suppose it is a common error, when a writer thinks about 2 related things or sentence variants, ending up somehow with their hybrid, that does not make sense. I do such errors quite often.:-)

Unless it is a measure of density variability across the bulk object volume.

@Poutnik I don't know either way. I make such mistakes myself often enough, but this article is very clearly written otherwise, so it makes me wonder...

If you can, you should call the vendor and ask them to connect you with a technical support person who can answer this question. I assume the vendor literature is correct and ask them to clarify. Vendor literature is not always written to be easily understood by people outside their market.

@J.Ari Thanks? The article I reference is academic, but I suppose I could contact one of the authors.

I'm sorry, I read "manufacture" as "manufacturer" in your first sentence. Hopefully you are successful in contacting one of the authors. Maybe my mistake is also happened with the authors x)

Note that a quantity defined as A/B is called ‘quotient of A by B’ or ‘A per B’, but not ‘A per unit B’.

@Loong Are you sure that usage is so strict? (Or are you saying that's how it should be used? And is this directly relevant to the question or another example of possibly confusing subtleties of word usage?)  Basically I've concluded that the text I quote meant to say "density, in this case amount per unit volume, ..." in an abbreviated manner. It turned out to be somewhat weird phrasing, if my interpretation is right, but less interesting than I thought, not a mistake or weird property, just potentially confusing phrasing (to me at least, obviously not to those accustomed with it).

It's a direct quote from ISO 80000. Anyway, I don't think that your actual question is about this detail of the wording.

Here, "density per unit volume" implicitly indicates that the densities of the substances being compared have been normalized.
It is redundant but safe - it rules out the possibility of having compared the density of one pound of something with that of a kilogram of another, for example.
The authors probably should have used the phrase "specific density" instead, although, as seen below, "specific density" is equated with "density."
From the Grundfos company website:

The specific density of a material or liquid is the term used to describe the unit mass of a material or liquid.
The technical term for specific density, also referred to as density, is $\rho$ and is measured in $\mathrm{kg\cdot m^{-3}}$.
The specific heat capacity is dependent on the temperature of the medium. The specific density of water in a heating system at a temperature of between 20–90$^\circ$C is 1,000 $\mathrm{kg\cdot m^{-3}}$.

In the case above, specific density appears to always reference the mass (in kilograms) of one cubic meter of a substance (here, water in a certain temperature range).
In the literature, the phrase "density per unit volume" appears to be used when describing bulk properties of materials which consist of a wide range of particle sizes (as in your case, wood flour and fiber).
I note that the widely-used phrase "energy density per unit volume" and other variants such as "energy density of states per unit volume" are not included in my explanation here, but do apply to the field of materials science and many others when describing systems of interest.
See here, in reference to different particle sizes of wood contributing to "density per unit volume" when looking at bulk properties of adsorbates; and here for dry-matter "density per unit volume" of forest biomass.

Have you seen usage of this specific term "density per unit volume" anywhere else? Any references? Your explanation is reasonable otherwise.

It is a kind of average density, is this that you mean? Still I don't see why it should be called something else that simply density. I am not convinced, but it is just an opinion. I see density per unit volume as redundant while in some cases, I see the use proposed by @Buck Thorn as common, though mass density require a very special contest to be meaningfull, eg the amount of mass ablate off a thin layer on a surface.

Some years later.... I believe a common alternative term is "bulk density": https://en.wikipedia.org/wiki/Bulk_density You correctly refer to bulk properties in your post.

I'm unconvinced. If "density per unit volume" has some special meaning, that is different from "density" than what is it? Density is density; it definition is mass per unit volume. And what units you measure it in is irrelevant. The density of a kilo of lead is the same as the density of a pound of lead if measured in the same units.


text:
Cold process to make glass
Since glass is a mix of silicates of sodium and calcium produced by heating or fusing together $\ce{Na2O}$ (Lewis base) with sand or $\ce{SiO2}$ (Lewis acid). Since $\ce{Na2O/NaOH(aq)}$ is a strong base and is used to create water glass(es)$(\ce{Na2SiO3},$ $\ce{Na_xSiO_y}$ etc.).
Is there some procedure in place to create glass by mixing (a base rich) water glass at room temperature with calculated amounts of $\ce{CaO}$ in mould to prepare float glass of definite shape without heating or mild heating?

Bulk materials are not always easily created from the components. For example, is it easy to create bulk iron from iron filings with heating?

Not glass per se, but silica is often produced chemically by silicon alkoxide gelation to create aerogels.
The problem with creating glass by some cold process would be that the reaction side-products, such as water, would have to be removed, leaving holes in a froth of glass. This is fine for creating an aerogel, but is not helpful in making solid objects, such as test tubes or lenses. Of course, you could sinter the glass frit or froth, but that still requires heating.


text:
Bond energy and enthalpy of formation
Is the following equation true?
$$\ce{∆_fH} = \ce{∆_{BE_{reactants}}H} - \ce{∆_{BE_{products}}H}$$
where $BE$ stands for Bond energy and $f$ stands for formation.
I read that,
$$\ce{∆_fH} = \ce{∆_{products}H} - \ce{∆_{reactants}H} = \ce{-∆_{BE}H}$$
But how is the first equation true?

Is it because of this $\ce{∆_f_{reactants}H} = \ce{-∆_{BE}_{reactants}H}$ ? When we substitute this in the second equation we get the first. Is it?

This equation is valid. Recall that the enthalpy of formation can be written as: $\Delta H_{break} - \Delta H_{form}$. When you translate this into bond energies, bonds breaking corresponds to the reactants and bond forming corresponds to the products. Thus,
$$\Delta_f H = \Delta_{BE, react} H - \Delta_{BE, prod} H$$
as you have. I don't think your second equation is correct however. If you're looking at enthalpies of products and reactants, the correct equation should be:
$$\Delta_f H = \Delta_{f, prod} H - \Delta_{f, react} H$$

second equation i actually meant to mention what you mentioned. Thanks!


text:
Why do strong polyprotic acids not dissociate completely, whereas strong bases with multiple (OH)- groups do?
$\ce{H2SO4}$ is a strong acid, which follows the reaction
$$\ce{H2SO4 -> HSO4- + H+}$$
Not
$$\ce{H2SO4 -> SO4^2- + 2 H+}$$
However,
$$\ce{Ba(OH)2 -> Ba^2+ + 2OH-}$$
Is there a reason why these "polyprotic analogs" in bases completely dissociate but polyprotic acids do not?
Note: This is in context to titrations, where $\pu{0.10M}$ of $.\ce{Ba(OH)2}$ is $\pu{0.20M}$ of $\ce{OH-}$, and leads to different values of titrant needed.

Ba(OH)2 has pKb1 = 0.15, pKb2 = 0.64, So There is no complete dissiciation. https://en.wikipedia.org/wiki/Barium_hydroxide For Ca(OH)2 it is 1.37 and 2.43 https://en.wikipedia.org/wiki/Calcium_hydroxide pKa2 for H2SO4 is 1.99.

And you don't get 0.10 M Ba(OH)2 either at ambient temperature.  Saturation is closer to 0.05 M.

In simplistic terms, Arrhenius acids are molecules while Arrhenius bases are ionic compounds. Much like $\ce{BaCl2}$ – another ionic compound – it is a good first rule of thumb to assume it completely dissociates into its respective ions: one $\ce{Ba^2+}$ and two single-charge anions. You could say that barium were never really connected to either hydroxide; ionic compounds are often shown as collections of balls simply in a certain arrangement (crystal structure) but easily separable.
Arrhenius acids, which are all molecules, are different. There is no inherent separation point; dissiociation in water actually requires the transfer of a proton, i.e. the breaking of an $\ce{X-H}$ covalent bond. Two such bonds don’t get broken simultaneously, instead in a first reaction one proton is removed – but then you end up with an anion. Removing another $\ce{H+}$ has to go against electrostatic attraction and that is a harder job to do so it happens less and the $\mathrm pK_\mathrm a$ values differ.
In actual fact, the picture I have drawn for Arrhenius bases is simplistic. Ionic bonds also involve orbital interactions; no bond is purely ionic or purely covalent. Cations and anions obviously also electrostatically attract each other. As barium is twice positively charged, it does tend to bind the hydroxide fragments rather tightly. Overall, upon dissolution it is much more likely to end up with what effectively is a coordination compound of the $\ce{[Ba(OH)(H2O)_n]+ + OH-}$ type.


text:
What is mass-independent isotope separation/fractionation?
How can any separation/fractionation of isotopes be mass-independent, if the only difference between two isotopes is mass?
The only clue(s) on Wikipedia mention molecular symmetry, 'forbidden' spin-flip transitions/reactions and photochemical dissociation....
But how is any of that affected by an extra neutron?

It can be direct versus indirect dependence. E.g. centrifuging of UF6 to get 235U, versus electrolysis of water to get heavy water.

The answer is "quantum mechanics, quantum mechanics, quantum mechanics".  A good paper: https://www.pnas.org/content/110/44/17708


text:
Are chlorinated solvents generally avoided in reducing reactions?
Are chlorinated solvents incompatible with reducing reagents? Is reductive dechlorination the reason?
There also seems to be exemptions, in J. Org. Chem. 1996, 61, 3849-3862, 1,2-dichloroethane was used as a solvent for a reductive amination reaction. NaBH(OAc)3 was used as the reducing agent in this paper.

Dibal-H is commercially available as a solution in dichloromethane   https://www.sigmaaldrich.com/catalog/product/aldrich/214973?lang=en&region=GB


text:
Electrolytes in electrochemical half cells
In most of the examples in my book I have come across, the electrolyte for an electrochemical half cell has always been the aqueous version of the metal used. For example, a silver electrode is immersed in a silver solution and so on, but is why is this necessary? Surely any positive metal ion will do?

For example, a silver electrode is immersed in a silver solution and so on, but is why is this necessary? Surely any positive metal ion will do?

There is a fundamental reason, which is the Nernst equation. It requires that the metal be in contact with its own ions. What is the functional meaning of the electrode potential? It is the tendency of an ion to reduce itself. Yes one can make salt water battery by dipping an iron nail (or copper) in one half cell and a piece of aluminum in another half cell but you cannot predict the cell potential beforehand.


text:
Coffee Degassing
I roast coffee, and currently package it in plastic bags with one way 'degassing' valves.
As soon as coffee is roasted, it starts giving off $\ce{CO2},$ a process which continues for days, and sometimes weeks. The degassing valve allows the $\ce{CO2}$ to escape the bag, while it prevents oxygen, which is the staling enemy of coffee beans, from entering the bag.
I fold the top of each bag over with three folds and then tape it with a piece of tape.
Due to composting concerns (the valves are not compostable) I would like to stop using the valves and package the coffee in plain bags with the three folds and piece of tape.  My theory is that the pressure of the off-gassed $\ce{CO2}$ will allow it to escape out of the three folds at the top of the bag, and that the oxygen not being under as much pressure from the outside, will not find it's way into the bag for a long time (months, hopefully), thereby maintaining a $\ce{CO2}$ flush of the beans in the bag, which prevents them from staling.
Does this seem like a plausible, likely scenario to keep the beans fresh, by keeping the oxygen away from the beans, allowing the $\ce{CO2}$ to escape under the initial pressure, and then maintaining the $\ce{CO2}$ environment in the bag?
Your informed answers are appreciated.

I think the easiest way to answer this would be via an experiment. What you're hoping for doesn't seem implausible. Even if oxygen does seep in, maybe it's not enough to matter, etc...

Hmm,but if your concern are non compostable valves, then it looks like the bags are compostable. If the bags are compostable, they must be made from paper or organic material. If they are from paper or similar, my concern would be oxygen diffusion through bags.....

Indeed, where the valves really necessary? Likely nothing would change, just the valve did nothing in the past (if the bag was already compostable). I agree with the Poutnik comment as well as with the suggestion to experiment.

Another thought: Maybe the buildup of CO2 (higher pressure) will inhibit O2 from reaching the beans?

@Tunk But gas diffusion follows partial pressure gradient, not total pressure gradient. Oxygen would diffuse through bag in spite of higher pressure inside.

@poutnik Ah, my mistake.

If you fold the top, open end of the bag over itself 3 times and then tape that resulting fold of bag, you are causing a restriction to flow. Depending on how strong the tape is, you can build pressure until the tape fails or it will retain pressure that will "puff" out the first time the bag is open. Perhaps I don't understand your folding scheme? The off gassing valve you had before may have just been a check valve that lifted at a certain pressure, or did it also have a membrane behind it to prevent oxygen permeation?

If we consider this from a diffusion perspective, your concept might not be as effective as you might think. Graham's Law of diffusion and effusion of gases states that the rate of diffusion for a gas is inversely proportional to its molar mass ($M$):
$$rate_\text{gas A} \propto \frac{1}{\sqrt{M_\text{gas A}}}$$
So, if we want to compare the rate of diffusion for $\ce{O2}$ ($rate_\ce{O2}$) against that of $\ce{CO2}$ ($rate_\ce{CO2}$), it would become:
$$\frac{rate_\ce{O2}}{rate_\ce{CO2}} = \frac{\sqrt{M_\ce{CO2}}}{\sqrt{M_\ce{O2}}}$$
Since $\ce{CO2}$ has molar mass $\pu{12 g/mol}$ more than $\ce{O2}$, we can see that $rate_\ce{O2}$ is going to be greater than $rate_\ce{CO2}$, and indeed, when we plug in the molar mass values ($\pu{32 g/mol}$ for $\ce{O2}$ and $\pu{44 g/mol}$ for $\ce{CO2}$), this ratio $\frac{rate_\ce{O2}}{rate_\ce{CO2}} = 1.173$ (In this simplified understanding of diffusion and gas interactions). This means $\ce{O2}$ diffuses $17\%$ faster than $\ce{CO2}$.
You might be able to use this information to determine if the amount of $\ce{CO2}$ being produced is significant enough to overcome the inward diffusion of $\ce{O2}$ into the coffee bags without the valves.
For reference: https://openstax.org/books/chemistry-2e/pages/9-4-effusion-and-diffusion-of-gases
As a possible suggestion, you could easily set up an off-gassing box with pure $\ce{CO2}$ by have a supply of dry ice (solid $\ce{CO2}$) in the bottom of the chamber and a vent at the top. As the dry ice sublimates its colder and denser vapor will displace the oxygen from the local environment of the box and allow the coffee bags to off-gas without oxygen present. Dry ice can be available at certain markets.


text:
Why does cyclobutadieneiron tricarbonyl behave aromatically?
It is said that $\ce{(\eta^4-C4H4)Fe(CO)3}$ can undergo electrophilic substitution reactions. Therefore, it displays aromaticity.
For the iron atom, it has $8$ electrons in its outer shell initially and it receives $6$ electrons from three carbonyls. It achieves the stable state of $18e$ after bonding with cyclobutadiene. Therefore, the oxidation state of iron should be zero, and the $\pi$ electrons of cyclobutadiene in the complex is still $4$.
But this result contradicts with Huckel's rule, which states that the $\pi$ electrons should be $4n+2$. Where did I go wrong?

You imply that during bonding the C4H4 gives its 4 electrons to iron for nothing in return. Now there is a catch: bonding is a game for two.

@IvanNeretin One of my professors once said ‘you notice that that, what doesn’t work with humans, namely socialism, works beautifully with electrons.’

@Jan True, or so it seems. Then again, I never had a chance to discuss that with an electron.

@IvanNeretin Exactly, this was what I thought. I have never thought about the delta bond. Now I understand, with the help of the paper, the beautiful game for these two fragments and the exactly same symmetry of MOs.

In the paper "(Cyclobutadiene)iron Tricarbonyls - A Case of Theory before Experiment", Organometallics 2003, 22, 2-20 https://pubs.acs.org/doi/pdf/10.1021/om020946c, at the page 12 nice explanation of bonding in this compound is provided:

The electronic structure of (cyclobutadiene)iron tricarbonyl has been the subject of many papers. He I and He II low-energy photoelectron spectra provided useful information.77,78 The eight observed bands in the lowenergy PE spectrum of  (cyclobutadiene)iron tricarbonyl in the range 7.65-20.31 eV all were assigned.78 The
  calculations (ab initio SCF MO) showed that there is a net negative charge on the cyclobutadiene ligand that results from π back-bonding from the iron atom into an
  antibonding MO of the ligand (δ bond). Better agreement between the experimental assignments and theoretical calculations was obtained by Chinn and Hall using generalized MO calculations with configuration interaction.79 A simple textbook approach to the bonding is shown in Figure 8.80 It is assumed that it is triplet state cyclobutadiene that is involved with unpaired electrons in the two degenerate ψ2 and ψ3 molecular
  orbitals. These interact with two singly occupied iron orbitals, generating two covalent bonds.


text:
Why does reducing the pressure in an equilibrium reaction favour the side with more moles of gas?
I was trying to explain it in terms of reaction rates and such; more than just 'Le Chatelier's principle states ...', but I can't figure out the underlying reason as to why. I can explain why for increasing the pressure, temperature changes and such, but not for this specific one. If there is more space for the particles to move without collided, this reduces the frequency of successful collisions, reducing the reaction rate of both forward and reverse reactions. But, in $$\ce{A + B<=>3C}$$
there is initially more moles of C than A and B combined. So, isn't it reasonable to think C will collide more frequently than A and B, meaning the reaction rate of the reverse is quicker, rather than the reaction rate of the forward being quicker?

By your reasoning, the equilibrium is never possible at all, because there are still 3C on the right, hence more collisions. But no, it doesn't work quite like that.

For your reaction, $\Delta n_g = 3-(1+1)=1$ (assuming all A,B,C are gases). $K_p = \frac{[C]^3}{[A][B]} (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$, where $\Sigma n$ is the total number of moles of gases in the reaction mixture(even of those species which are not taking part in the reaction). Here, it is a constant.
When pressure is decreased($P_{total}$), $ (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$ decreases as $\Delta n_g >0$.
In order to keep $K_p$ constant (it is always constant at a given temperature), $[C]$ has to increase or $[A],[B]$ has to decrease. Therefore the reaction proceeds/shifts more in the forward direction, which has more number of gaseous moles.
Similarly, increase in pressure favours the side with lesser number of gaseous moles.

You can relate an equilibrium constant $K_p$ written in terms of partial pressures to one written in terms of mole fractions $K_\chi$, as follows:
$$K_\chi=K_p (p/p_0)^{-\Delta \nu}\tag{1}$$
where $\Delta \nu$ is the difference in the stoichiometric coefficients for a molar amount of reaction. $K_p$ is independent of the pressure $p$ of the system but depends on the temperature.
Now take the logarithm on both sides of Equation (1) followed by the derivative with respect to pressure, keeping temperature constant. Remembering that $K_p$ is not a function of pressure at constant temperature, you can show that
$$\left(\frac{\partial \ln K_\chi}{\partial p}\right)_T=-\frac{\Delta \nu}{p}$$
This equation can be rewritten as
$$d(\ln K_\chi)=-\Delta \nu \times d (\ln p) \tag{constant T}$$
This expression shows that if $\Delta \nu>0$ as in the example you provide, then increasing p (or equivalently ln(p)) will cause $K_\chi$ to decrease, implying that the mole fraction of product will diminish.
This is of course a mathematical explanation and not entirely satisfying, but there are various ways to justify it heuristically, the simplest being that (in your example) by reverting to reagents the number of molecules is reduced, decreasing the total volume, and thereby reducing the amount of work and heat exchanged during the compression.


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Reduction of carbamate with LAH
The products of the reduction of esters with $\ce {LiAlH4}$ and the products of the reduction of amides with $\ce {LiAlH4}$ are vastly different. The former reduction cleaves the ester and produces two alcohols while the latter reduction produces an amine with the carbonyl group of the original amide replaced with $\ce {CH2}$. A carbamate seems to display both chemical behaviour of esters and amides. I am curious to know what would be the mechanism by which reduction of carbamate with $\ce {LiAlH4}$ takes place and what would be the products of such a reduction.

Carbamate reduction with LiAlH4 gives N-methylation,  see https://www.nrcresearchpress.com/doi/pdfplus/10.1139/v66-043  and Tet. Letts vol 26 (1985) 5367

Carbamates are usually reduced to N-methyl groups. There are numerous examples:

J. Am. Chem. Soc. 2012, 134 (16), 6936–6939

Org. Lett. 2012, 14 (18), 4834–4837
But it is not always a given. In this next example, the nitrogen is part of a three-membered ring (aziridine). These nitrogens are better leaving groups than usual, cf. Ketone/aldehyde synthesis from N-acylazetidines or aziridines where the same kind of reactivity is observed:

Angew. Chem. Int. Ed. 2002, 41 (24), 4683–4685

What if the "ester part" of the carbamate is also linked to another part of the  molecule such that it doesn't simply go away after reduction? What would happen to the "ester part" of the carbamate upon reduction?

You get the alcohol. e.g. N-CO2Et -> EtOH There aren't really other ways to link a carbamate to some other part of the molecule; the central carbon (which becomes the NMe carbon) has to have four bonds to either N or O. The N bit becomes NMe and the O bit becomes the alcohol.

@orthocresol: H. C. Brown, et al., argue that it is the inability of the aziridine to form the iminium structure that represses reductive alkylation in favor of aziridine formation. J.A.C.S, 1961, 83, 4549.

@user55119 thank you for the reference (I just noticed this). That is consistent with the rationale provided in the question I linked, which is good!