text:
Calculating entropy change: reversible vs irreversible process
Since the change in internal energy and enthalpy, which are equal to the heats for a constant-volume and constant-pressure process, respectively, are state functions, the heats for a reversible v.s. irreversible process should be equal. Thus, the entropy change of the surroundings which equals the quotient of the heat and temperature are equal for reversible v.s. irreversible processes.
As Engel's Physical Chemistry states it:

Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly.

However, when the book proceeds to evaluating the change in entropy for the surroundings for a reversible vs irreersible process, the q used for evaluating the change in entropy of surroundings are not equal for the reversible v.s. irreversible process, which contradicts what the textbook previously stated.
The question:

One mole of an ideal gas at 300. K is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath thermal reservoir in the surroundings is very large, T remains essentially constant at 300. K during the process.

Calculation of change in entropy for reversible path:

Calculation for irreversible path:

It seems that the book used the actual heat for the irreversible process. However, I don't understand why the entropy's from the two calculations are not equal. Also, I fail to understand why the change in entropy for the surroundings are not equal for the two process for which the final and initial states are equal --> change in entropy is a state function and should be the same regardless of path.
References
Engel, Thomas, and Philip Reid. Physical Chemistry. San Francisco: Pearson Benjamin Cummings, 2006.

Update: I am starting to get the feeling that (somehow) the final state of the surroundings after an irreversible process is not the same as the final state of the surroundings after a reversible process. Something (for example, friction between the piston and walls of container) make the final states of surroundings different for irreversible vs reversible --> change in S is not the same. However, since the initial and final T and V are the same for the system in reversible and irreversible, the change in entropy is the same for reversible vs irreversible for system. Is this a valid argument?

Entropy change of your system will be the same for both the reversible and irreversible path. However, the entropy of the surroundings will not be the same , as you have seen. The system goes from the same state A to the same state B for both the reversible and irreversible paths, the surroundings are not in the same state after an irreversible process as they would be after a reversible one. @Ethiopius

@Tyberius What you mean is that the surroundings are not in the same state after an irreversible process  imposed on the system as they would be after a reversible process imposed on the system between its same two end states.

@ChetMiller That is a clearer description. The way I had phrased I was open to misinterpretation.

You are confused about how to calculate the entropy change for a system and surroundings that have experienced an irreversible process.  For the irreversible path, you correctly calculated the final state using the first law of thermodynamics.  From this point on, you must totally forget about the irreversible path and, instead, focus exclusively on the initial and final end states.  You must separate the system from the surroundings, and then devise an alternative  reversible path for each of them separately that takes each from its initial state to its final state.  You did that correctly for the system, and determined its correct entropy change (-7.62 J/K).  This is the entropy change for the system both for the reversible path you devised as well as for the actual irreversible path.
In the case of the surroundings, during the irreversible process, its internal energy increased by 3.741 kJ at the constant temperature of 300 K.  So, for its alternative  reversible path, to achieve this same internal energy change at 300 K, the amount of reversible heat it would have to absorb would be the same 3.741 kJ.  So, the change in entropy of the surroundings in the irreversible process was 12.47 J/K.
I know that this might all sound a little confusing.  So here is a more detailed explanation of the fundamentals as well as a cookbook (foolproof) recipe for determining the change in entropy of the system and surroundings for any arbitrary irreversible process on a closed system (including worked examples of the methodology):  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

I'll address part of your confusion here:
The following statement is not generally true:

Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly.

For instance, in a closed system $\Delta U = q + w$ so what is independent of path isn't $q$, it's the sum of  $q + w$.
However, there was no mistake or contradiction by Engel & Reid in making that statement, because in the sentence immediately prior they make it clear they are discussing changes done at constant V or constant p.  And, in those cases, their statement does apply.
Here is the entire quote:

If the surroundings are at constant V, $q_{surroundings} = \Delta U_{surroundings}$, and if the surroundings are at constant P, $q_{surroundings} = \Delta U_{surroundings}$. Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly.

I.e., the problem isn't with Engle & Reid's statement, the problem is that you quoted it while omitting a critical set of information.
There are errors in Engel & Reid, to be sure, but this isn't one of them.


text:
Effect of concentration on rate of reaction
I came across this problem:

The $\ce{2NO + Br2 -> 2NOBr}$ follows the mechanism below:
$$
\begin{align}
\ce{NO + Br2 &-> NOBr2} & &\text{fast} \tag{1} \\
\ce{NO + NOBr2 &-> 2NOBr} & &\text{slow} \tag{2}
\end{align}
$$
If the concentration of both $\ce{NO}$ and $\ce{Br2}$ are increased two times the rate of the reaction would become
A) 4 times
B) 6 times
C) 16 times
D) 8 times

Now I know that overall order of a reaction is equal to the molecularity of the slowest step, hence order is 2 and since the concentrations of both the reactants are being increased by the same factor, i.e. 2 , the rate should be $2^2$ times irrespective of what is the order with respect to any reactant.
But the answer given is 8. How is this possible?

The concentration of $\ce{NOBr2}$ is not the thing that is increased explicitly in the statement of the problem. You should express that in terms of reactant concentrations.

Actually, reading your reasoning again, you're just very confused about how to compute the rate of the reaction. For this kind of a simple system, the rate of the reaction is the product of concentrations of reactants raised to the power of their corresponding coefficients for the slow step. The overall molecularity for the step has nothing to do with the answer, except maybe in limited cases where the concentrations are all scaled by the same amount.

The fast reaction should be written as reversible reaction if $\ce{NOBr2}$ is a high energy intermediate:
$$\ce{NO + Br2 <=> NOBr2}$$
Then, the concentration of $\ce{NOBr2}$ can be estimated as:
$$[\ce{NOBr2}] = K_{eq} [\ce{Br2}] [\ce{NO}]$$
If the second reaction is an elementary step, the rate law follows the stoichiometry:
$$\text{rate} = k [\ce{NOBr2}] [\ce{NO}]$$
Substituting the concentration of the intermediate $\ce{NOBr2}$ into that rate law gives the overall rate law:
$$\text{rate} = k K_{eq} [\ce{Br2}] [\ce{NO}] [\ce{NO}]$$
and the dependency of the rate on doubling all reactants.

Now I know that overall order of a reaction is equal to the molecularity of the slowest step

This is only true if the slowest step is the first step. Otherwise, you have to consider all the reactants that make the intermediates necessary for the slow step.
What if the first reaction goes to completion?
In this case, the concentration of the intermediate would depend on the limiting reactant. Changing all reactant concentration by a factor of two, the concentration of intermediate would increase by a factor of two as well (not like the factor of 4 if the first step is a pre-equilibrium).
What else do we need to know to solve this problem?
The problem description does not tell us the physical state of the reactants, intermediates and products. If $\ce{Br2}$ is a pure liquid, we would not be able to change its concentration. I would guess the intention is to have all species in the gas phase, though.

For the reaction:
$$\ce{2NO + Br2 <=> 2NOBr}$$
the initial forward rate is given by
$$r_1 = k\ce{[NO]_1^2[Br]_1}$$
now if you double $\ce{[NO]_1}$ and $\ce{[Br]_1}$ in a second experiment, then
$$r_2 = k\ce{[NO]_2^2[Br]_2} = k\ce{(2[NO]_1)^2(2[Br]_1)} = 8k\ce{[NO]_1^2[Br]_1} = 8r_1$$

In (i), the concentration of both reactants is doubled so the rate of formation of $\ce{NOBr2}$  quadruples.
Since net rate is dependant on RDS which is (ii), the rate become 8 times as 4(As for $\ce{NOBr2}$) * 2 (due to the NO involved directly) = 8.
And I have done these calculations while assuming that all reactions are first order in each of the reactant.

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text:
Getting pH of a weak acid and conjugate base with a given ratio
A practice question reads

What is the $\ce{pH}$ of a solution of a weak acid and its conjugate base when the ratio of the concentrations of the conjugate base/weak acid is 1/3? (the weak acid $K_\mathrm{a} = 6.7\times 10^{-5}$)
a. 5.67
b. 4.65
c. 4.17
d. 3.69
e. 0.481

The answer given is d, I would appreciate it if someone would be wiling to walk me through the steps to solve this problem as my exam is tomorrow morning. Thank you in advance.

So what is the mathematical formula for the Ka and the various species?

Sorry, I figured it out. I used $pH = pKa + log([A-]/[HA])$ to get $pH = -log(Ka) + log(1/3)$.

No reason to be sorry. I was trying to get you to figure it out. That way you learned something.

Your comment about the formula for Ka is what reminded me to use the Henderson Hasselbalch equation so thank you. The given ratio was also confusing me even though you just simply plug it into the equation.

To solve this I used the Henderson-Hasselbalch equation:
$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\ce{A^-}]}{[\ce{HA}]}\right)$$
Because $\ce{pH} = -\log(\mathrm{p}K_\mathrm{a})$ we end up with:

$\ce{pH} = -\log(\mathrm{p}K_\mathrm{a})+ \log \left(\frac{1}{3}\right)=-\log \left(6.7 \times 10^{-5}\right) + \log \left(\frac{1}{3}\right)= 3.69$.

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text:
How do I reduce the smell of store-bought liquid soap?
I recently bought a brand of liquid soap I've never used before and it turns out they formulated it with a bit too much of the scent ingredients, to the point where I find it gets uncomfortably close to nauseating during use, and also it persists too long on my hands afterwards, compared to what I find normal.
How do I neutralize some of this scent, what can I try to mix in? All I can think of are baking soda or vinegar (but then vinegar also persists a lot and is unpleasant, plus it's acidic and I'm going to mix it with a probably-alkaline soap, and I don't want any crazy reactions and byproducts if possible).

Just dump it and buy something else. I dislike all of the scents that I have tried. :-(

This is like saying, I like to eat oranges, but I don't like the smell. Well, you can either not eat them or ignore the smell because your proposed solution is to remove orange smell from an orange.

Return it to the store? They don't have to take it back, but probably will if you tell them it stinks. Customer is king etc. ;-)

It may cost more than its worth to de-scent it. Better dump and buy a new as MaxW suggested.:-)

If the scent reaches out and touches you that much, it is likely volatile. Pour it in a pot, add a volume of water, heat on a hot plate outside your dwelling till the smell goes away. Evaporate to volume and check ingredients. You may want to add back a small amount of a desirable ingredient that boiled off. Raise the temperature slowly. A vigorous boil will make a real mess.

The OP is unlikely to have a fume hood. The smell would stink up the whole house. // You'd need a pot with a much larger volume, and stir, since this is going to bubble like crazy.

It is volatile at room temp, so should not have to heat too much to liberate the stench. I suggested slowly elevating the temp to avoid ever reaching a boil and do it outside. That said, I agree a fume hood would be a better option.

Thanks! It's not a "mix something in" solution like what I was going for, but it makes a lot of sense - I hadn't thought of the volatility aspect. So I should be able to pour this stuff into a flat, wide open container and let it off-gas and could expect some measure of results even at room temp, I'm thinking.


text:
How can a titration help me figure out the effect of rosmarinic acid on amylase activity?
So with my limited knowledge of rosmarinic acid, I am aware that it is an non-competitive inhibitor and can inhibit the enzyme, amylase. When starch is added to a solution of already mixed rosmarinic acid and amylase, it doesn’t break down due to the inhibition.
A titration was used to investigate the effect of rosmarinic acid on amylase. The solution containing the rosmarinic acid was put into a burette and was the titrant for this experiment whilst the potassium permanganate was the titrate.
I have a few questions about this experiment. Why was a titration used and how could it identify if the rosmarinic did have an effect on the amylase? Also, why was the solution containing rosmarinic acid the titrant instead of the titrate? What is the potassium permanganate’s role in this investigation?

Do you have a link to the experimental protocol? I have a few thoughts: 1. RA is easily oxidized, say by permanganate; 2. reducible sugar is also oxidized by permanganate; 3. RA interferes with conversion of starch to glucose; 4. pink/violet permanganate is reduced to colorless Mn2+ in acidic solution.


text:
Heat of hydrogenation
If benzene is more stable than cyclohexane then why is the heat of hydrogenation of benzene negative, ie heat is released during the formation of cyclohexane from benzene ?

What do you mean by "benzene is more stable than cyclohexane". Can you provide a source for that statement?

Related https://chemistry.stackexchange.com/q/108726/48509

If it is exothermic means cyclohexane must be more stable than benzene right? But it is not so. How is that possible?

"But it is not so". By what measure is it not so? What makes you say, "benzene is more stable than cyclohexane".

Because benzene is aromatic? That's what others told me but I'm not sure.

All hydrogenation reactions, not just for benzene, are exothermic. You can refer  Why are hydrogenation reactions exothermic .


text:
Aromatic enol is not possible in...?

How to solve this kind of problems. What determines their stability?

Welcome! Homework questions must display some effort. Please include your attempt/thoughts on the problem, to avoid the deletion of this question.

When a ketone undergoes tautomerism to form an enol, where is the new double bond?

This is a strange question - isn't it possible to form an aromatic enol in all 4 cases? #3 is the least obvious, but will still proceed:

Maybe I'm misunderstanding the question, vote me down if so!


text:
What are "O-hydrocarbons"? A pharma translator's question (Russian to English, "О-углеводороды")
A translator colleague is translating from Russian a plan for study of a number of leachable and extractable substances (pharmacology). This Russian text mentions one particular class of chemicals which theoretically could become leachables:

"Hydrocarbons (alkanes, alkenes, alkynes, O-hydrocarbons)".

She can't understand what this could mean.
The Russian term is О-углеводороды.

My guess is cyclic as opposed to linear HC ie cyclohexane.

I've never seen this term in Russian literature before. My first guess would be that it's any oxygen-substituted hydrocarbon ("кислородозамещенный углеводород"); it also might be leading zero (0) and then it might refer to [un]saturation degree, e.g. another name for alkanes ("насыщенный углеводород"); @NightWriter 's version also makes a lot of sense ("циклический углеводород").

Your first guess was correct, @andselisk

My fellow translator sent a letter to the authors, and they replied that they used this term to stand for "oxygen-substituted hydrocarbons", just like andselisk suggested.


text:
Why coefficient doesn't change amount of substance
$\ce{3Mg + 2As -> Mg3As2}$ ($10\ \mathrm g$ of $\ce{Mg}$ and $10\ \mathrm g$ of $\ce{As}$)
I don't understand why coefficient doesn't change the amount of substance, here's what I did:
$n_{\ce{Mg}}=3\cdot\frac{10\ \mathrm g}{24.31\ \mathrm{g/mol}}=1.233\ \mathrm{mol}$ of $\ce{Mg}$
But the correct solution is:
$n_{\ce{Mg}}=\frac{10\ \mathrm g}{24.31\ \mathrm{g/mol}}=0.411\ \mathrm{mol}$ of $\ce{Mg}$

The stochiometric coefficient only helps you realise the ratio of moles of reactant or products. 3Mg in the equation does not imply anything about the number of moles you have in your reaction vessel, but rather only tells you that you will need 3 moles of Mg to completely react with 2 moles of As.

related: https://chemistry.stackexchange.com/q/99809/23561

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

When you say number of moles, it is a number to express the number of constituent species (atoms, in this case) in the given substance. It does not have any correlation to where that substance will be used and what will be its 'coefficient' in that reaction.
When you say 10g Mg, you mean 0.411 moles of Mg (and vice versa) irrespective of whatever you are going to do with those 10 grams.
Even if you were to use 10g Mg in a reaction such as $$\ce{ 2Mg + O2 -> 2MgO} $$, you would still have 0.411 moles of Mg as it is still equivalent to 10g. Nothing changes on the basis of the 'coefficient'.
You can further try to look up more information to get a better understanding about 'The Mole Concept' because that is where I think your problem lies.

'The mole concept' appears to be such a loose invention of a particular brand of books. The concept/theory itself should be called stoichiometry. And I think calling it by the wrong name is where the problem actually starts. Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". And the base unit of the amount of substance is called mole.

Yes I agree, and I think I tried to make the same clear in my answer, but apparently it wasn't, so please feel free to suggest an edit if you want.


text:
Why is $1s-1s$ orbital overlapping stronger than $2p_x-2p_x$ orbital overlapping?
Assume the x-axis to be the internuclear axis. Hence, both $1s-1s$ and $2p_x-2p_x$ will form $\sigma$ bonds.
Also consider the fact that $p$ orbitals have more directional characteristics.
Given the above facts, shouldn't the $2p_x-2p_x$ overlapping be more effective than $1s-1s$ overlapping? Why is $1s-1s$ orbital overlapping stronger than $2p_x-2p_x$ orbital overlapping?

Who says this is true in the first place?

@IvanNeretin an exercise in my book, Resonance Inorganic Chemistry for IIT-JEE page #127

Your question is hinged on a false premise. On a deeper level, the premise makes no sense at all and can't even be classified as true or false. If a book says otherwise, then maybe it contains some very specific piece of context that would turn the situation upside down and make everything fall in place. Or maybe the book just says nonsense; such things happen, after all.
To begin with, orbital overlap depends on the internuclear distance more than slightly. Now, $1s$ orbitals are a good deal smaller than $2p$ (not like "you vs Dwayne Johnson", but more like "you vs an elephant"). Naturally, at a distance where $2p_x$ orbitals are in their best overlap, the $1s$ orbitals barely feel each other. Likewise, if the distance is good for $1s$, it is way too short for $2p_x$ and makes them feel pretty uncomfortable. So you see that considering one of these overlaps to be universally stronger than the other is plain untrue.
There is more to it, however. See, the $1s$ and $2p$ orbitals of the same atom are pretty far apart in energy. They never mix. In other words, they can never be important at the same time.

If both your atoms are $\rm H$, then their $1s$ orbitals overlap and make a good strong covalent bond, while the $2p$ orbitals are empty and hence effectively don't exist.
On the other hand, if both your atoms are (say) $\rm O$, then their $2p_x$ participate in bonding (together with other $2p$'s and $2s$), while the $1s$ orbitals are pretty much kept apart. What's more important, both $1s$'s are fully occupied, hence whatever little interaction they have will end up producing a bonding and an antibonding orbital which are also both occupied, thus negating the effect.

At this point you might be wondering what if one of your atoms is $\rm H$ and another $\rm O$, but that's a story for another day.
So it goes.


text:
Why acyl halides are not named as 1-substituted aldehydes?
For acyl halides, IUPAC uses preferably the "acyl halide" type naming (e.g. acetyl chloride). But they can be theoretically named as aldehydes, where a hydrogen atom in the $\mathrm{CHO}$ group (carbon 1) is substituted by a halogen atom (e.g. acetyl chloride $\mathrm{CH_3COCl}$ has the same structure as acetaldehyde (or ethanal) $\mathrm{CH_3COH}$, so it can be named 1-chloroethanal - the name is probably valid, some sources list it as a possible one).
Of course, functionally acyl chlorides and aldehydes are very different (since acyl chlorides are much closer to corresponding carbonic acids), but structurally they aren't (and the systematic name should probably express the structure of a compound at first). Consider for example the naming of alkyl halides (halogen-substituted alkanes), where IUPAC switched to the substitutive nomenclature (for example, from "ethyl chloride" to "chloroethane", despite that chemical properties of chloroethane and ethane are very different).
The question may be incorrectly stated, since I'm not a (professional) chemist. Thanks in advance!

It must have to do with traditional reasons, as it has always been easily synthesized from carboxylic acids rather than aldehydes. But again, that's speculation.

Principally because it is not at the aldehyde oxidation level, it is at the carboxylic acid level

Well, the rest of the substituted ethane in the chlorethane is still like an alkane, but the rest of to be 1-chlorethanal is not an alkanal in the first place. An alkanal must have the aldehydic group.
If there would be 1-chlorethanal, why not 1-hydroxyethanal ( aka known as the acetic acid  ), or 1-methylethanal ( aka acetone )  ?  :-)


text:
What is the ideal temperature and pressure to crack water?
Using sun mirrors and solar cells for electricity. What temperature, pressure and voltage ideal for cracking water.

Source: NPR

Source
Would it take more or less electricity to crack water under pressure or in a liquid or vapor state?

Check this link: https://en.wikipedia.org/wiki/Water_splitting#Solar-thermal, how will you pass electricity through water vapors?

@M.Farooq I want to say it could be done with laser filamentation.

@MaxW I don't think it so, water starts to spontaneously decay at $\approx 3000 ^\circ \rm C$. Plasma state starts to appear at $\approx 10000 ^\circ \rm C$. Mixed $\rm H$ and $\rm O$ plasma will be a neutral mix, there will be too hot for any chemical binding.

@MaxW Essentially, in a plasma state, all the elements are already long neutral chemically. Instead, new and interesting electromagnetic effects happen.

@M.Farooq It is a nice answer stub, in my opinion. I think it obviously closes out any gaseous phase. But it doesn't close out fluid water in elevated pressures. The question is, where is the efficiency optimum.

I think this question has potential, but you should organize your thoughts better so that the question is more clear.


text:
The van't Hoff rule for stereoisomers with planes of symmetry
We know that the maximum possible numbers of stereoisomers of a molecule is given by the van't Hoff rule, which says that:
$$Total \space stereoisomers = 2^{n}$$
Where $n$ is the number of stereocenters. However, this is just the theoretical maximum: in fact, if the molecule has a certain number of planes of symmetry, the number of isomers drastically decreases. Is there a formula that accounts for the number of planes of symmetry? The only contribution I could find was here:
How to derive these general formulae for number of stereoisomers of a compound with a possible plane of symmetry?
But this only accounts for one plane of symmetry. What if we have more? Things obviously get more difficult, as not only the number of planes, but also how they divide the molecule will influence how many stereoisomers we'll have. Do you think it would be possible to generalize the formula above anyway?


text:
Isoptope effect on electrophilic aromatic substitution
My book reads

Sulphonation will be the most effected electrophilic aromatic substitution reaction when benzene is replaced by hexadeutrobenzene.

I am not able to figure out how. All EAS reactions proceed by formation of an arenium ion and then the loss of an $H^+$ ion to regain aromaticity so all of them should be equally effected ( textbook does not mention how they are effected). I feel the only effect should be due to difference in bond enthalpies of $C-D$ bond and $ C-H$ bond.
I came across this reaction mechanism from chemlibretexts:

The hydrogen released is taken up by the $-SO_3H^-$
Can someone explain how and in what ways the replacement of hydrogen by deuterium effects the reaction ?

Noting edits is not necessary. Anyone can see the edit history.


text:
Strength of oxoacids of Chlorine
I have seen various different contradicting statements about oxoacids of chlorine regarding oxidising strength & reducing strength of these.
I understand the trend for acidity that HClO4 is strongest and HClO is weakest , but what about oxidisind and reducing character,
Some day that HClO4 is strong reducing as well as strongest oxidising agent as well,
I understand that it will perform reduction of itself and oxidise other element but what about reducing strength?

Some say that lion's bites heal wounds.

Yeah, saying that HClO4 is strong reducing agent makes as much sense as what Ivan mentions.

When you are considering redox reactions, you should get familiar with electrochemical series (Ref.1). Oxidizing and reducing characters of certain reagent also depends on the conditions  used as well. For example, according to Ref.1, $\ce{HClO}$ in acidic medium:
$$\ce{HClO + H+ + e- <=>  1/2Cl2 + H2O} ~~~~~~~~ \mathrm{E}^\circ = \pu{+1.594 V} $$
On the other hand, $\ce{ClO-}$ in basic or neutral medium:
$$\ce{ClO- + H2O + 2e- <=>  Cl- + 2OH-} ~~~~~~~~ \mathrm{E}^\circ = \pu{+0.890 V} $$
Consider the following half reaction:
$$\ce{Br2 + 2e- <=>  Br- } ~~~~~~~~ \mathrm{E}^\circ = \pu{+1.098 V} $$
Now, you can say that $\ce{HClO}$ in acidic medium spontaneously oxidizes $\ce{Br-}$ to $\ce{Br2}$, but $\ce{ClO-}$ in basic or neutral medium cannot spontaneously oxidize $\ce{Br-}$ to $\ce{Br2}$.
In similar way, you may also say that $\ce{HClO}$ in acidic medium is a better oxidizing agent than that of $\ce{ClO4-}$ in acidic medium, because:
$$\ce{ClO4- + 8H+ + 7e- <=>  1/2Cl2 + 4H2O} ~~~~~~~~ \mathrm{E}^\circ = \pu{+1.392 V} $$

Reference:

Steven G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at 298.15 K,” Journal of Physical and Chemical Reference Data 1989, 18(1), 1–21 (https://doi.org/10.1063/1.555839).


text:
Tollens' test with beta hydroxy ketones
My book mentions that 1,5-dihydroxypentan-3-one will not give silver mirror with diammine silver(+1):

This seems very counter intuitive to me as the alcoholic group can be easily oxidized to the aldehyde which will subsequently give the test.
What is the correct reasoning behind this?

The molecule as it stands is not an aldehyde.
If you turn it into an aldehyde before doing the test, you are not starting with the molecule that is stated to be negative in the test.
The chemistry of the Test will not convert your alcohol into an aldehyde and subsequently turn the test positive.


text:
Alkene production by dehydrohalogenation
What is the difference in the reaction mechanism when alkyl halides are reacted with aq.koh and alcoholic koh along with products

Why would you expect a difference?

The fundamental reason lies in the difference between the action of a base and a nucleophile.
There are two types of mechanism through which substitution proceeds:-

SN1 Reaction - Here, the substrate undergoes hetrolytic cleavage to produce a carbocation and an anion. The thing which solvents do is to stabilise the newly formed ions. A Polar solvent(One case-Water) stabilises the intermediate stage more than the reactant stage because the reactant state forms weak dipole-dipole bonds with the solvent where as the intermediate stage has full fledged proper charges on then to interact with the solvent and forms strong ionic bonds.
SN2 Reaction - Same thing but because of the changes in mechanism type of solvent changes but the essence is same.

This tells us that the solvent is crucial to generate a strong electric field which can help the reaction proceed further by ionising the substrate whereas a base is not so capable of handling this type of work
In elimination reaction, the solvent generates a strong electric field around the substrate which in turn helps in the cleavage of C-H bond so that it can stabilise the carbocation(in E1 mechanism) or help the transition state to get rid of the leaving group(In E2 mechanism).
Basic action is a matter of equilibrium which shows how readily it can accept protons and go forward whereas nucleophilicity is all about the rate how fast it can go which in turn shoes us how eager it is to form bond or get rid of its negative charge.

I will improve the answer over time, I feel that it is somewhat disconnected.


text:
How to keep a flask submerged in water?
I am trying to keep a 1000 ml Erlenmeyer flask submerged in a deep cooking pot filled with water. I tried tying the flask to a metal plate, but it was too light unfortunately.
Is there a purpose built piece of equipment for this or a commonly accepted method?

Why you need the flask submerged - what you are trying to accomplish will determine the answer. Is the submerged flask corked or is it filled with the water in which it is submerged.? Will you be boiling the water?

Try something like these: https://www.thomassci.com/scientific-supplies/Weight-Rings. Any major lab supply company should have something similar.

What is stopping you clamping the neck of the flask in a fixed position using a simple lab clamp and stand?

@matt_black: Is it possible to move the stand base out of the way so I can put my electric hotplate there instead? I don't have a stand at the moment.

@Raghib Clamps and stands are versatile and are designed to enable you to clamp things where they need to be clamped. You can arrange them whatever way you need to.

@matt_black: Does this look good? https://www.ebay.co.uk/itm/Chemistry-Laboratory-Retort-Base-Stand-Support-Lab-Flask-Clamp-3-Ring-Clamp-Z0HW/293004454736?hash=item44386d2b50:g:c~8AAOSwCPZb7meC

Yes. Those are the standard stands/clamps widely used in every lab to hold things where you want them to be.

The vinyl coated lead rings suggested in the comment are a great solution. They are, however, expensive and take time to order. There is also the cost of shipping.
A quick and less expensive route is to use a PVC flange and attach what ever heavy metal you have (steel, cast iron, lead fishing weights). Epoxy to flange or drill holes and tie. You may want to paint the metal to avoid rust or other reactions with the metal.
Home Depot has inexpensive 3" flange or 4" flange.

I would use a (coated) steel ring holder mounted to a heavy and secured rod holder.
Such a ring would be set to push and keep the Erlenmeyer flask down.
Such rings are commonly used in classing frame based settings to hold modular glass aparatures, e.g to support mesh plates with boiling flasks, or holding conical glass shapes.


text:
What happens to the pressure of gas when only one molecule of it is placed in a very large container?
We say that pressure due to gas is constant throughout the container. Pressure is created by molecules present in it. When there are many molecules I agree that they can apply the same pressure all over.
But when there are very few or just one molecule in the container, how can that single molecule create a uniform pressure in the container. I mean pressure will be created (say in a manometer) when  a molecule applies force on it (Hg). If we take a very large container and place just one molecule and also keep two manometers at the two extremes. Then how can that single molecule create the same pressure or apply the same force on both the manometers at the same time (container is very large)?

Well, not quite at the same time, but you are probably aware that molecules fly pretty fast.

And what if we kept the container as long as the circumference of the earth or the sun rather?

What you're saying is this fact of Uniform pressure is true for small containers and are not valid for larger ones

Well, that's pretty much so. Moreover, even for smaller containers and more molecules, pressure is measured only statistically.

Means average pressure?

Yes. BTW, the same applies to temperature.

You're right. One molecule can't exert uniform pressure. But in reality the pressure isn't from the molecule, but the nearly perfect vacuum. // The gas laws are based on overall statistical behavior. Having just few gas molecules doesn't give such statistical information.

Well thanks, so because the molecules move in straight line before collisions we can say that if a molecule has collided with the left extreme mercury it is only by chance that it will collide again with it after some time and we could also end up measuring vacuum even in the presence of one molecule

If you're using mercury manometers you'd never get to one molecule of gas. Mercury itself has a vapor pressure.

So it does have some errors ? How are those corrected ?

A vessel containing only one molecule of gas is way better than the most perfect vacuum we have ever been able to create. So there is really no need to worry about correcting for the situation. Even "empty" interstellar space contains between 1 and 1,000 atoms per cubic centimetre. Most concepts from equilibrium thermodynamics fall apart at such densities simply because there are too few molecular collisions to establish equilibrium.

https://chemistry.stackexchange.com/questions/97247/molecular-kinetic-interpretation-of-pressure

Because of quantum mechanics? This is probably more of a question for the physics SE site, but my intuition is that the particle would be effectively delocalized over the box. Even if my intutition is wrong, it would be interesting to know what the Heisenberg uncertainty principle has to say about this.....

What happens to the pressure of gas when only one molecule of it is placed in a very large container?

I think you have forgotten what pressure represents in terms of the kinetic theory of gases. When studying gas kinetics, you start by understanding the behavior of a single molecule/atom moving around a container and colliding with the surface and bouncing in random directions. Then you add a few more molecules/atoms so that now there are collisions and as long as the mean free path of the gas is less than the dimensions of your vessel you still have motion that acts in random diffusion-like behavior. Once sufficiently many atoms have been added that a given amount of area experiences a regular number of collisions per unit of time, then you have something that can be given as an average rather than a count of collisions. This is what pressure represents, an average force on a surface per unit area of molecular collisions on that surface. The instantaneous force can change for a given spot nanosecond to nanosecond, but averaging all spots over a span of microseconds will give an average force per area can be resolved which we call pressure.
The problem here that gives you a misunderstanding is that you have taken a behavior of an aggregation of many molecules and tried to apply it to a single molecule. This approach will disappoint. But to summarize and actually answer the question, pressure cannot describe a system of only one gas molecule, and is therefore undefined.

The gas pressure, the same as temperature, are statistical quantities. A single molecule has no temperature and no pressure.


text:
Derivation of Gibbs free energy formula
Okay, so I'm trying to derive an important formula which states that -
                             ∆G(r) = ∆G° + RTln( RQ )
where ∆G(r) is the instantaneous rate of change of Gibbs energy with respect to moles of reagent consumed (am I right on this?) I'm hight confused because I have 'almost' derived it but due to some conceptual difficulty I'm not able to get the exact formula!
NOTE:
(1) Please ignore any non expansion work done on the system and tell my mistake as simply as possible because I have just started thermodynamics.
(2) The line above G represent molar Gibbs energy of substance. The small letters on the side of the symbol of reactent are the stochiometric coefficients.
(3) Many times I see variations of this formula, instead of ∆G(r) they simply use ∆G, there is a difference between two. Is this formula valid if ∆G is used instead of ∆G(r) in left hand side of equation? Also don't see any 'sense' in using ∆G because if ∆G is there then we need the initial and final quantities of the substances in the reaction but in the reaction quotient we plug the instantaneous activities of reactents and products.

I'm getting the above equation but a stochiometric coefficient is also included in this ! Obviously it is wrong. Kindly tell me my mistake.

Comments are not for extended discussion; this conversation has been moved to chat.

∆G(r) is the instantaneous rate of change of Gibbs energy with respect to moles of reagent consumed (am I right on this?)

No, because then you would get different answers depending on which reactant you choose. Instead, it is with respect to change in amount divided by stoichiometric coefficient (sometimes called extent of reaction). This is similar to the way reaction rates are defined (change in concentration divided by stoichiometric coefficient).
If you update the definition of the Gibbs energy of reaction, you will get the expected equation in your answer.

Many times I see variations of this formula, instead of ∆G(r) they simply use ∆G, there is a difference between two. Is this formula valid if ∆G is used instead of ∆G(r) in left hand side of equation?

Textbooks and other sources are often fuzzy concerning the distinction of $\Delta G$ vs $\Delta_r G$. You are right in saying that the formula does not make sense if you are calculating the difference between two states with different concentration. The other thing that is fuzzy are the units, energy vs. energy per amount (e.g. kJ vs kJ/mol).

Exactly the answer I was looking for! Interpreting the wrong meaning of ∆G(r) was my mistake. Thanks!

I think you are using the first approach that I described.  So, here is the analysis based on the first approach.  Suppose you have the following numbers of moles of A, B, C, and D in the reaction mixture at some time: $n_A$, $n_B$, $n_C$, and $n_D$.
And the total pressure of the mixture (in bars) is held constant at P.  So the partial pressure of species A is $$p_A=\frac{n_A}{n_A+n_B+n_C+n_D}P$$And the partial molar free energy of species A is $$\bar{G}_A=G^0_A+RT\ln{\left(\frac{n_A}{n_A+n_B+n_C+n_D}P\right)}$$where $G^0_A$ is the free energy of pure species A at the reference pressure of 1 bar.  Similar equations can be written for the other species.  The total free energy of the mixture at any point in time during the reaction is:  $$G=n_A\bar{G}_A+n_B\bar{G}_B+n_C\bar{G}_C+n_D\bar{G}_D$$
Now, let the concentrations of the chemical species in the reaction mixture be expressed as follows:
$$n_A=n_{A0}-a\xi$$
$$n_B=n_{B0}-b\xi$$
$$n_C=n_{C0}+c\xi$$
$$n_A=n_{D0}+d\xi$$
where the subscript 0 refers to the initial state and $\xi$ is the conversion.  To find the equilibrium condition, you want to find the value of $ \xi $ that minimizes the free energy of the solution mixture.  To do that, you have a calculus problem to solve.

My understanding of the question is that the OP is trying to reconcile the different usages of $\Delta G$ and $\Delta_r G$, not trying to solve for the equilibrium concentrations.


text:
The role of temperature in Nernst equation
When measuring a non-standard voltaic cell’s electrode potential, regardless the standard conditions, as long as the concentration of product and reactant the same, a non-standard cell’s potential will be exactly equal to a standard cell’s potential.
$E= E^\circ -\frac{RT}{nF} \ln(Q)$
Q = 1 if the concentration of product equals to the concentration of reactant:
$E= E^\circ -\frac{RT}{nF} \ln(1)$
$E=E^\circ - 0$
In another word, no matter how temperature varies, as long as the concentration of both solution the same, temperature will have no effect on the voltage.
Is the understanding above valid? If it is, then my experimental result contradicted the Nernst equation. When I increase the temperature while having two half cell with the exact same concentration, the voltage increased as well. What is a possible cause for such result? (Assuming my voltemeter is funcotional one)

The standard cell potential is temperature-dependent itself, just like the Gibbs free energy. The easiest way to see this is to write the Gibbs free energy in terms of enthalpy, entropy and temperature:
$$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$
This shows you that $\Delta G^\circ$ is strongly temperature dependent (even if $ \Delta H^\circ$ and $\Delta S^\circ$ were independent of temperature, which they are not).
To relate this to the cell potential, remember that $\Delta G^\circ$ is a measure of the maximal work a system can do, and that the electrical work is electrical potential times charge moved across that potential.
$$\Delta G^\circ = w_\text{electrical} = -E^\circ n F$$
So $E^\circ$ is temperature dependent as well.

Similar answer to similar question: https://chemistry.stackexchange.com/a/84721


text:
Calculate pH at equivalence point

Calculate the pH at the equivalence point of a titration of 62 mL of 0.1 M $\ce{CH_3NH_2}$ with 0.20 M HCl. The $\ce{K_b}=4.4\cdot10^{-4}$.

At the equivalence point, the moles of CH3NH2 equals the moles of HCl. This is simple solution stoichiometry. Turns out, we require 62 mL or the CH3NH2 and 31 mL of the HCl for a total volume of 93 mL.
Now, construct an equilibrium equation for the weak base: $\ce{CH_3NH_2 + H_2O <=> HCH_3NH_2^+ + OH^-}\ \ \ \ \ \ $ or
$\ce{HCH_3NH_2^+ <=> H^+ + CH_3NH_2}$.
From here on out, I get really confused. You have to also make another equation that includes HCl to finally figure out equilibrium of $\ce{H^+}$, but I don't know how to get there.

I hope that these hydrogens before carbons are just "accident" and not some misconception - edit accordingly.

@Mithoron It helps me see it as a proton attached to a base, rather than an entirely new compound.

Edit: At the equivalence point, the solution contains the dissolved salt $\ce{CH3NH3Cl}$, dissociated to $\ce{CH3NH3+ + Cl-}$, as @MaxW noted.
You can calculate $pH$ of the conjugate acid $\ce{CH3NH3+}$ at its dissociation equilibrium.
$$\begin{align}
K_w &= K_a . K_b = 10^{-14}\\
pK_w &=  pK_a + pK_b = 14\\
\ce{ CH3NH3+ &<=> H+ + CH3NH2 }
\end{align}$$
With the simplifying assumptions
$$\begin{align}
 c_{\ce{H+}} &= c_{\ce{CH3NH2}} \\
c_{\ce{CH3NH2}} &<< c_{\ce{CH3NH3+}}
\end{align}$$
is the case rather simple:
$$\begin{align}
K_a &= \frac{ c_{\ce{H+}}.c_{\ce{CH3NH2}}}{ c_{\ce{CH3NH3+}}} = \frac{ c_{\ce{H+}}^2}{ c_{\ce{CH3NH3+}}} \\
pH &= 1/2.( pK_a - \log{c_{\ce{CH3NH3+}} } ) \\
pH &=7 -  1/2.( pK_b + \log{c_{\ce{CH3NH3+}} } )
\end{align}$$

Yes. The problem is essentially like you are dissolving $\ce{CH3NH3Cl}$


text:
Computational Efficiency of Periodic Boundary Conditions
From what I understand, in a molecular dynamics simulation the number of molecules included is limited by the computational power available.
In order to simulate macroscopic properties, one needs to ensure that the particles experience forces as if they were in the bulk, so the system is replicated in all directions through the application of a periodic boundary condition.
How is the creation of this larger, periodic system any more computationally efficient than just simulating a non-periodic system of similar size?

A non-periodic system will likely suffer from many particles being on the surface of the system. Consider 1000 particles in a cube: almost half of them will be on the surface. Their characteristics will be different from a particle in bulk.

The application of periodic boundary condition does not mean that the system is explicitly replicated. You only deal with the particles within the unit cell (the box that you replicate), but you let them interact with the replicated ones outside of this unit cell.
Consider a simple 1-dimensional case with three atoms in the unit cell: $\ce{A-B-C}$, and lets assume we only need to treat nearest-neighbor-interactions. So the "monomer" has only 2 interactions: AB and BC.
One option to simulate the macroscopic system would be to explicitly deal with $\ce{A-B-C-A-B-C-A-B-C-A-B-C}$ (or even longer), but this requires a lot of atoms with accordingly many interactions between them. And even then both ends of the chain would be missing one neighbor.
Using periodic boundary conditions, one explicitly treats $\ce{A-B-C}$ only. But this time we will allow $\ce{C}$ to interact with $\ce{A}$, since the periodic boundary conditions implicitly tell us, there is another $\ce{C}$ to the left. Similar, there is an implicit $\ce{A}$ to the right of $\ce{C}$. Compared to the monomer from above, we have again 3 atoms, but now there are 3 interactions to be considered: AB, BC and AC.

Just be aware that each entity will have multiple, in fact an infinity, of images and technically you should sum over the interaction with each and every one with the reference particle - but in practice for most force laws the potential decays sufficiently quickly that you can truncate that summation, most commonly just including the nearest image. The big exception is Coulomb interactions where special methods should be used - the magic word here is "Ewald"


text:
Is it possible to calculate electrostatic interactions using QM methods?
I'm trying to calculate an accurate electrostatic repulsion (and possibly attraction) energy term for various small molecules, using Quantum Mechanics (QM) methods. These may include molecules such as bifuran, bipyridine, etc... I'm mainly interested in, for example, the electrostatic interaction energy of the two oxygens in a bifuran.
I've already tried using Molecular Mechanics-based methods, which use point charges (ie. GAFF charges) but they seem to be unable to account for the nuance orbital interactions, especially when molecules are in close contact. I was wondering if there exists a robust QM method, which is able to give accurate electrostatic interaction energies for molecules?

With QM, be it HF, CC, MP2 etc, it will be hard to disentangle different contributions - You could compare fragments though. E.g. for 2,2'-Bipyridine: compare energy of conformer by dihedral angle as a proxy for N distance as well as comparison to energy of two seperate pyridine fragments.  But would also include dispersion and effects of aromaticity/conjugation in the molecule.

I agree that it is difficult to disentangle the different interactions. I was hoping that there would be a robust method (ie. NBO equivalent interaction energy for full orbitals), which may or may not be possible.

Although many localisation schemes exist (NBO, Mulliken population analysis, atoms in molecules) they each have various failing and I would be wary to call them robust for this purpose.  Though partitioning space/electron density does suggest ideas for DFT based schemes for calculating the interaction energy of two densities - but I'm not aware of the literature/previous work on that type of approximation.


text:
Perturbation theory for molecules, chromophore group, dipole approximation,
I am interesting in chromophore group and dipole approximation.
For example, i have a molecule (acetone or any other ketone/enol) which is belongs to some symmetry group. Because of the symmetry this molecule has degenerate energy levels. Now, i need to know the way how to destroy degeneracy.
The first and simple idea is to use perturbation theory and a dipole approximation, where perturbation caused by the light. This approximation let as to treat the interaction between light and molecule as dipole in  spatially-uniform electric field. Perturbation theory is common way to calculate absorption coefficients for systems that smaller than absorbed light (atom, diatomic molecule etc.) But the size of a molecule in example long compared with the size of a wavelength, so we  can't ignore the spatial variation of the electric field.
Solution of this situation is chromophore, part of a molecule where the energy difference between two  orbitals fits within the range of the UV or visible spectrum and responsible for the absorption of light. So now we can treat interaction between light and a molecule as an interaction between light and a part of that molecule which is responsible for the absorption. For example the C=O bond is a chromophore. So we treat absorption of light by a molecule as an absorption of light by a chromophore and save ourselves from complex calculations related to the size of the molecule.  The distance of a C=O bond is 1.22 Å while absorbance maximum at about 275 nm. Now the wavelength is long compared with the size of an chromophore, we  can ignore the spatial variation of the field.
My questions is:

Is this approximation correct? Can we approximate small perturbation by light applied to WHOLE molecule as a chromophore in spatially-uniform electric field? How rough is this approximation?
If this assumption is not right what else Way we can use/apply dipole approximation to a Whole molecule?
What else approximations is possible when we a dealing with a perturbation theory applied to a molecule?


text:
Determining the half-reaction equations for extraordinary redox reaction?
I have the following question stated in the science  teacher exam:

Write the half-reaction equations for  the following reaction  by using oxidation numbers:
  $$\ce{Na2SO3  +  Na2CrO4  -> Na2SO4 + Cr(OH)3 }$$

I am having some difficulty determining the half-reaction equations because the question doesn’t indicate either to the states of the species in the reaction nor to the medium of the reaction.
By determining oxidation numbers it is possible to see what is being oxidised and what is being reduced. I have indicated oxidation numbers below in parentheses:
$$
\ce{
\overset{\color{black}{(+1)}}{Na2}
\overset{\color{red}{(+4)}}{S}
\overset{\color{black}{(-2)}}{O_3}
+ \overset{\color{black}{(+1)}}{Na2}\overset{\color{green}{(+6)}}{Cr}\overset{\color{black}{(-2)}}{O_4}
->
\overset{\color{black}{(+1)}}{Na2}
\overset{\color{red}{(+6)}}{S}
\overset{\color{black}{(-2)}}{O4}   + \overset{\color{green}{(+3)}}{Cr}(\overset{\color{black}{(-2)}}{\ce{O}}\overset{\color{black}{(+1)}}{\ce{H}})3}
$$
By supposing the reaction takes place in an acidic medium, I failed to answer the question.
But, by supposing the reaction takes place in the basic medium, I conclude to following  two half-reactions:
Oxidation half-reaction:
$$\ce{2OH-  + Na2SO3  -> Na2SO4  + H2O + 2e-}$$
Reduction  half- reaction:
$$\ce{3e- + 4H2O  + Na2CrO4  -> Cr(OH)3 +2Na+  + 5OH-}$$
The  completely balanced equation:
$$\ce{5H2O  + 3Na2SO3  +2Na2CrO4 -> 3Na2SO4 + 2Cr(OH)3 + 4NaOH}$$
I am not sure about this solution. Are there, and where if there are, mistakes in the solution? What is the correct solution?

There's nothing "extraordinary" here and I don't see why you think there's something wrong. (Nice job with formatting though.)

Mithoron Thanks for the interest, what about the states of the species in the reaction? can a reaction occur in an acidic medium?

This reaction can't be in acidic medium because the chromium (III) hydroxide would dissolve.

@KarstenTheisHow the reaction proceeds?

You have rightfully assumed the reaction takes place in either basic or neutral medium, otherwise $\ce{Cr(OH)3}$ wouldn't be the product.
This automatically suggests $\ce{H+}$ are not used for balancing half reactions.
Your solution is fine for the provided reaction given that chromium(III) hydroxide stays precipitated.
Here I've basically formatted your solution in a form I'm more familiar with:
\begin{align}
\ce{\overset{+6}{Cr}O4^2- + 4 H2O + 3 e- &→ \overset{+3}{Cr}(OH)3 + 5 OH-} &|\cdot 2 \tag{red1}\\
\ce{\overset{+4}{S}O3^2- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} &|\cdot 3 \tag{ox1}\\
\hline
\ce{2 CrO4^2- + 3 SO3^2- + 5 H2O &→ 2 Cr(OH)3 + 3 SO4^2- + 4 OH-} \tag{redox1}
\end{align}
Resulting full redox equation:
$$\ce{2 Na2CrO4 + 3 Na2SO3 + 5 H2O → 2 Cr(OH)3 + 3 Na2SO4 + 4 NaOH} \tag{R1}$$
This implies that reaction begins in neutral medium; however there is remaining sodium hydroxide ($\ce{4 NaOH}$ on the right), so the pH will increase over time.
This will result in further dissolving of $\ce{Cr(OH)3}$ in $\ce{NaOH}$ yielding $\ce{Na3[Cr(OH)6]}$ complex, which is likely the final product.
Alternatively, addition of a base to this mix would also accelerate formation of chromium(III) hydroxocomplex.
With this in mind, I'd say the more appropriate way to write the reaction is
\begin{align}
\ce{\overset{+6}{Cr}O4^2- + 4 H2O + 3 e- &→ [\overset{+3}{Cr}(OH)6]^3- + 2 OH-} &|\cdot 2 \tag{red2}\\
\ce{\overset{+4}{S}O3^2- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} &|\cdot 3 \tag{ox2}\\
\hline
\ce{2 CrO4^2- + 3 SO3^2- + 2 OH- + 5 H2O &→ 2 [Cr(OH)6]^3- + 3 SO4^2-} \tag{redox2}
\end{align}
Resulting full redox equation:
$$\ce{2 Na2CrO4 + 3 Na2SO3 + 2 NaOH + 5 H2O → 2 Na3[Cr(OH)6] + 3Na2SO4} \tag{R2}$$


text:
Dissociation degree hell, Ostwald's law of dilution
Greetings dear chemists!
I got a nice exam problem on wich i am stuck for hours, well maybe the problem is with me. I know the problem can be solved. So here we go.
We got a weak monoprotic acid, we don't know its concentration, dissociation degree nor the dissociation constant. If we dilute it 3 times its volume, then the dissociated ions grow with an unknown constant, lets call it $y$. If we dilute the diluted solution again by 5 times its volume, then the dissociated ions grow 4 times. Whats the dissociation degree of the undiluted solution?
Yeah woah that looks tough. So lets give it my try.
Lets call the concentrations in order of dilution by $c_0, c_1, c_2$ and the dissociation degree by $\alpha_0,\alpha_1,\alpha_2$, and the dissociation constant of course $K_s$.
I managed to get three equations for this problem, but i feel i missed something or messed up my algebra someway i dont understand.
$$c_0=?,\alpha_0=?,K_s=?$$
$$\alpha_0=\frac{[A^-]}{c_0}$$
$$HA=H^++A^-$$
$$c_1=\frac{c_0}{3},\qquad c_2=\frac{c_1}{5} \to c_2=\frac{c_0}{15}$$
$$\alpha_1=y\alpha_0,\qquad \alpha_2=4\alpha_1 \to \alpha_2=4y\alpha_0$$
We call forth Ostwald;
$$K_s=\frac{c_0\alpha_0^2}{1-\alpha_0}\qquad \qquad \qquad \qquad (1)$$
$$K_s=\frac{c_1\alpha_1^2}{1-\alpha_1}=\frac{c_0y^2\alpha_0^2}{3(1-y\alpha_0)}\qquad \qquad(2)$$
$$K_s=\frac{c_2\alpha_2^2}{1-\alpha_2}=\frac{c_016y^2\alpha_0^2}{15(1-4y\alpha_0)}\qquad \qquad(3)$$
Equating the first and the second equation, i could get an expression for $y$. By setting $\beta=\frac{3}{1-\alpha_0}$ and ignoring the negative root
$$y^2+y\beta\alpha_0-\beta=0$$
$$y=\frac{-\beta\alpha_0+\sqrt{\beta^2\alpha^2+4\beta}}{2}$$
My problem starts when i equate the second and the third equation;
$$\frac{c_0y^2\alpha_0^2}{3(1-y\alpha_0)}=\frac{c_016y^2\alpha_0^2}{15(1-4y\alpha_0)}$$
$$5-20y\alpha_0=16-16y\alpha_0$$
$$11+4y\alpha_0=0$$
From wich i clearly see that im on the wrong track, since $\alpha_0$ and $y$ cant be negative, please help me on solving this nightmare. Maybe i missed something obvious, i dont really know that im on the right method to solve this either.

Is the original problem in English? Can we get it exactly as stated?

No sorry its in hungarian, these kind of problems pop out from my professors head, i wrote it from my memory, we talked about it a lot after the exam so im pretty sure its correct.

If you do ask the prof, and he clarifies the problem, I'd sure like to know what he proposed as the answer.

You are on the right track according to the directions given to you ( = the problem statement), but the directions themselves are wrong.
Say you had an equilibrium with $\ce{{[H^+][A^-]\over[HA]}=K}$. You dilute it five-fold and tell your particles to wait a bit with dissociation until you are done with the math. That makes all concentrations 5 times smaller, and consequently, the LHS is now $1\over5$ of what it used to be. Now you tell them to increase their dissociation by a factor of 4. In doing so, each of $\ce{[H+]}$ and $\ce{[A-]}$ increases four-fold, thus growing the LHS to $16\over5$ of its initial value, or more than 3 times greater than $K$, and they do that at the expense of the remaining $\ce{[HA]}$, decreasing the denominator and increasing the LHS further yet. Just how is it going to be the same $\ce{K}$ again?
I suppose it is some subtlety of wording that we are misinterpreting here. It might have been lost in translation, but can't we at least make a reasonable guess? Could it be that the 4-fold increase is compared to the initial state, that is, to $\alpha_0$ and not to $\alpha_1$? No, that would not suffice. Still, there might be other possibilities...

Thank you for your answer and time reading my post, your reasoning seems correct, obviously i misunderstood something or it might be but its more likely not that the problem itself i was doing is wrong. Feel free to delete my post, but i think your thoughts are a nice example for learning so i'll leave it.

@IvanNeretin - The degree of dissociation is relative not absolute. So for the diluted acid a greater fraction ionizes, but the concentrations are all smaller.

@MaxW They are all smaller in my answer. [H+] and [A-] are 4/5 of what they were before the dilution, and [HA] is less than 1/5 of its former value.

$K_a = \dfrac{\ce{[H^+][A^-]}}{\ce{[HA]}}$
Let's assume a weak acid so that we'll assume a negligible fraction ionizes. So for all concentrations we get
$\ce{[H^+] = [A^-]} = \sqrt{(K_a)\ce{([HA])}}$
and the dissociation degree (fraction) is:
$\dfrac{\ce{[A^-]}}{\ce{[HA]}} = \sqrt{\dfrac{K_a}{\ce{[HA]}}}$
So if the concentration of HA is reduced by a factor of 5, then the dissociation degree should only be $\sqrt{5} \approx 2.24$ greater, not 4 times greater.
Now to get a greater degree of ionization we could have a very dilute solution of acid, and get into a situation where the autodissociation of water is important.

Thank you for all your efforts, nice answer though i have tried it too with this shortcut but didnt gave me results i wanted, I'll look into it tomorrow, i didnt want to bother my prof with my stupid questions since he is the dean, but it looks like i have no other choice.

@AndrewKovács - As I said, the only way that I can think of to have a five-fold dilution get a four fold increase in ionization is to have the ionization of water come into play.


text:
Metal complexes and bonding of ligands
I am currently learning about transition metal complexes, however I keep seeing complexes like $\ce{[Al(H2O)6}]^{3+}$ but aluminium isn't a transition metal. Is this a case that any metal can form a metal complex or is it just transition metals and a few others, if so which ones and why?

Second part is duplicate of https://chemistry.stackexchange.com/questions/24873/why-do-copper-ii-complexes-contain-so-many-valence-electrons

Not only all "metals" but also most of "non metals" can form what can be called complexes, somewhat depending  on definition.

The nature of the bonding between metal ions and their ligands in complexes is probably of both the ionic type and the covalent type. In the complexes of the transition metals, it is clear that there is a large covalent contribution, in the form of dative covalent bonds. However, I believe that for ions such as $\ce {Mg^2+}$ or $\ce {Al^3+}$, the bonding may involve more of the ion-dipole interactions and less of electron donation via an orbital interaction.


text:
Textbook Claim: "... in all cases it is the electrostatic force acting between charged particles that is responsible for all the forms of bonding."
I am reading an introductory semiconductor physics textbook. The textbook states the following:

Several types of atomic bonding have been identified, including ionic, covalent, van der Waals, hydrogen, and metallic. Whatever the name given to the type of bonding, in all cases it is the electrostatic force acting between charged particles that is responsible for all the forms of bonding.

My concern is with the validity of the latter sentence; that is, that all cases of bonding are electrostatic forces acting between charged particles.
Wikipedia says the following:

A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bonds.

If electrostatic forces act between charged particles (ions), and atoms are, by definition, neutral particles, then, since atoms do bond, how can it make sense to say that all bonding is electrostatic forces acting between charged particles?
And my interpretation of the Wikipedia article is that the electrostatic force of attraction is responsible for bonding between oppositely charged ions, as in ionic bonding, whereas it is not responsible for bonding through the sharing of electrons, as in covalent bonding.
So is the textbook's claim that all types of bonding are electrostatic forces acting between charged particles false?
I would appreciate it if people could please take the time to clarify this.

Forget atoms. Look deeper. Then the claim is true.

@IvanNeretin Oh? How so?

Don't over think the scope here. The nucleus has a positive charge and the electrons have a negative charge. Hence the structure of atoms and molecules depend on electrostatic forces.

You ask how bonding between two neutral atoms can be electrostatic. In the case of covalent and van der waals the electrostatic basis is not so obvious as it is for ionic compounds. A covalent bond forms when two neutral atoms establish a shared molecular orbital which is much larger than the separate atomic orbitals that the bonding electrons contributed by each atom formerly occupied. Electrostatic repulsion favors electron pairs being in larger orbitals, because this increases average distance between the electrons. In the case of Van der Waals attraction, the neutral atoms are electrostatically attracted because when their electrons are not symmetrically distributed, there is a partial positive charge at one end of a molecule and a partial negative at the other, even though overall charge is zero. Electrons in adjoining molecules will tend to synchronize their movements so that the positive end of one molecule will be near the negative end of the adjoining molecule; this is called induced charge.


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How to Choose Orbitals for Hybridization
Recently, I was reviewing my textbook regarding orbital hybridization, and one of the problems was to determine the hybridization of each of the atoms in $\text{CO}_2$. I understand why the hybridization on the C atom is $\text{sp}$ and the hybridization on each of the O atoms is $\text{sp}^2$, but what I don't understand is how to determine which of the three p orbitals becomes hybridized.
This becomes especially relevant with the O atom, since in its normal state one of its p orbitals is full (2 electrons) while the other two p orbitals only have 1 electron each (meaning that the 3 p orbitals aren't interchangable). How do I know which of these 3 p orbitals become hybridized into $\text{sp}^2$ orbitals and which orbital stays as a regular p orbital?
Thanks for the help.

The O atoms are probably sp, not sp2. And the three p-orbitals are perfectly interchangeable; nobody said you have to put the paired electrons in one particular p-orbital, after all.

There is no choosing. You hybridize the orbitals, and then put the electrons there.

@IvanNeretin So do you fill the hybridized orbitals before the regular p orbitals, or vice versa?

Wait a bit more: you make bonding and antibonding molecular orbitals out of those hybridized atomic orbitals, and only then...

@orthocresol In this case, how would I determine the electron configuration for something like this oxygen, knowing that it is sp2 hybridized and that I have 6 electrons to fill everything with (also I think it is sp2 because according to the lewis structure there are 2 nonbonding electron pairs and each o atom is double bonded to the c atom)? Thanks


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Why is nucleus charge squared in Rydbergs formula?
According to coulumbs law, the energy changes linearly with charge. So why is the nucleus charge squared in this formula? Is there a way to derive it?
I was suspecting it might be because it pulls the electron closer too.

Your qualitative understanding is correct. As to the rigorous derivation, wait until you learn the internal structure of an atom, starting from the Bohr model (which then should be forgotten, but that's another story).

Derivation comes from the formula for electrostatic potential energy.
The electrostatic potential between the point charges of an electron, and a proton would be represented by the formula
$$U_E(r) = k_e \frac{q_eQ_p}{r}$$
However, in the derivation that we are looking at a difference of energy is from an emission or absorption, such that the energy concerned is
$$E_i - E_f$$
The subtraction appears in the Rydberg formula from the distance in the form
$$\left( \frac {1} {n_1^2} - \frac {1} {n_2^2} \right)$$
However, the terms other than the radius are multiplied together and come outside the parentheses.  Even though we are looking at only a single electron, the term for the charge in the nucleus is squared because of this. This can be seen in the Bohr derivation of the Rydberg formula.
$$E = h\nu = E_i - E_f = \frac {m_eq_e^2q_Z^2} {8h^2\epsilon_0} \left( \frac {1} {n_f^2} - \frac {1} {n_i^2} \right)$$
This shows the origin of the squared term for the nuclear charge which correlates with the Rydberg formula for a hydrogen-like element since all terms outside the parentheses other than charge are constant.
$$\frac {1} {\lambda_{vac}} = RZ^2\left( \frac {1} {n_1^2} - \frac {1} {n_2^2} \right)$$

Strictly speaking the nuclear charge $Z$ is not included in Rydberg's constant but multiplies this as $Z^2$ when used in the Balmer formula for a transition frequency $\displaystyle \nu = Z^2R_\infty\left( \frac{1}{n_i^2}-\frac{1}{n_f^2} \right)$ between levels $i$ and $f$. In a series of x-ray experiments Moseley, in 1914, was the first to confirm that the atomic number $Z$ was proportional to the square root of the transition frequency.
The $Z^2$ comes about in the Bohr model by considering the electrostatic force between the electron and nucleus, this is (in SI units) $\displaystyle \frac{Ze^2}{(4\pi\epsilon_o)}\frac{1}{r^2}$. This is the derivative of the electrostatic energy when charges are separated by a distance $r$. The nucleus has a charge $Ze$, the electron just $1e$.
This force is then equated with the centripetal acceleration giving  $\displaystyle \frac{Ze^2}{(4\pi\epsilon_o)}\frac{1}{r^2}=\frac{m_ev^2}{r}$ where $m_e$ is the electron mass and $v$ its speed. The total energy is the sum of kinetic and potential energy and so $\displaystyle E=\frac{1}{2}m_ev^2-\frac{Ze^2}{(4\pi\epsilon_o)}\frac{1}{r}$. Finally, Bohr assumed that only certain orbits were allowed and the equation for the energy leads to the transition frequency formula given above with $\displaystyle hcR_\infty=\left(\frac{e^2}{(4\pi\epsilon_o)}\right)^2\frac{m_e}{2\hbar^2}$ and $hc$ converts from energy to wavenumbers.
In the proper, quantum, derivation the same Coulomb electrostatic  energy  is used in the Schroedinger equation as was used by Bohr and is $\displaystyle -\frac{Ze^2}{(4\pi\epsilon_o)}\frac{1}{r}$


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Why is barium used instead of bismuth as a contrast in gastrointestinal xrays?
Why is barium used instead of bismuth as a contrast in gastrointestinal x-rays? Bismuth is a heavy metal that has a relatively low toxicity in comparison to barium. Is there a particular property of barium that makes it a better contrast? Is it less expensive?

I’d suggest looking at Wikipedia on ‘radiocontrast agents’. It has a very good reason given there.

Is it less expensive?

Yes, but not so much that it would be favored, i.e. you don't worry about saving \$5 when your doing a \$5000 procedure.

Is there a particular property of barium that makes it a better contrast?

No both absorb x-rays as a result of their high atomic numbers, in fact bismuth has more electrons per atom and thus is a better contrast agent..

Bismuth is a heavy metal that has a relatively low toxicity in comparison to barium.

Yes but this is for the soluble metal ion. Barium sulfate is inert under most conditions, importantly the acidic ones in the digestive tract. Because the sulfate binds the barium ions to the solid and thus barium cannot enter the blood stream to cause a physiological response. Contrast this to Bismuth (no pun intended) which is non-toxic but typically soluble in acid and if not soluble, the bismuth is at least leachable. Now typically bismuth ions are not an issue as they will harmlessly be excreted, but there is a potential it could interact with a patient or the drugs of a patient in a negative or uncontrolled way. Since doctors don't like to introduce variables when making diagnosis's nor do the like surprises, especially when it's caused by them, the fully inert barium sulfate is the favorite for radio contrast.


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The Action of Nitrogen Iodide on Methyl Ketones
I recently encountered two old publications on action of nitrogen iodide on various compounds (Ref.1 and Ref.2). Between them, Ref.1 described the iodoform formation by methyl ketones upon treatment of the conditions:

Ketones containing a methyl group react very readily with nitrogen iodide, iodoform, ammonia, an acid, and an amide being produced. In these reactions the methyl group appears first to be completely substituted by iodine, a tri-iodomethyl ketone being formed, which in presence of the ammonia simultaneously set free is hydrolysed to iodoform and an acid, a similar reaction between the substituted ketone and ammonia leading to the formation of iodoform and an amide.

The reactions were suggested as follows:
\begin{align}
\ce{RCOCH3 + NH3.NI3 &-> RCOCI3 + 2NH3}\\
\ce{RCOCI3 + H2O &-> RCO2H + CHI3}\\
\ce{RCOCI3 + NH3 &-> RCONH2 + CHI3}
\end{align}
This formation of iodoform is understandable. Yet, Ref.2 claimed that the authors have obtained iodoform from diethylketone (3-pentanone):

Like the other ketones, diethylketone when treated with nitrogen iodide, yields the iodoketones, having very irritable odor. On effecting the interaction in a dilute solution, iodoform is formed in good quantity. It has been observed that as we rise higher and higher in the homologous series of the ketones, the tendency of the formation of iodoform gradually diminishes and the yields also gradually become less.

The reported method the authors had been used: $\pu{2 mL}$ of diethylketone are taken in a beaker and diluted with $\pu{150 mL }$ of weak ammonia. The mixture is warmed on the water bath, and to it, iodide solution is added when at first nitrogen iodide is precipitated, which is immediately taken up. The addition of iodine has to be continued, and finally a copious yellow crystalline precipitate deposits. This is iodoform. On recrystallization from alcohol, this has been obtained quite pure, melting at $\pu{119 ^{\circ}C}$. The yield amounts to $\pu{0.8 g}$.
I’m very curious about their claim and like to hear from your theory of mechanism.

Cited References:

F. D. Chattaway, R. R. Baxter, “CCX.—The Action of Nitrogen Iodide on Methyl Ketones,” J. Chem. Soc., Trans. 1913, 103, 1986–1988 (DOI: 10.1039/CT9130301986).
R.L. Datta, N. Prosad, “Halogenation. XVI. Iodination by Means of Nitrogen Iodide or by Means of Iodine in the Presence of Ammonia,” J. Am. Chem. Soc. 1917, 39(3), 441–456 (DOI: 10.1021/ja02248a013).

In the second reaction i dont understand the significance of adding iodine continuously. If NI3 is already formed it has a property of dissociating again to I2. Then what is the reason for adding I2?

This question is hard to answer without more details from the two papers.

@Zhe hey i have a mechanism which gives us great yields of iodoform. The only problem is the details about dilution. I dont get why they dilute it.

@Som V. Tambe: I believe dilution is to prevent some unnecessary side reactions. I'm just speculating, though. :-)

@ Som V. Tambe: I believe the following statement in the publication may give what they meant by "the addition of iodine has to be continued": ...by means of a potassium iodide solution of iodine in conjunction with liquor ammonia, which represents nitrogen iodide in the nascent state... However, in the first paper they  had prepared nitrogen iodide by adding a solution of iodine monochloride in hydrochloric acid to strong ammonia where nitrogen iodide had formed as black precipitate.

@Zhe: Both papers did not indicate any mechanistic aspects. The first paper described the formation of iodoform from acetophenone and acetone with set of equations I have given in my question. The second paper described the iodination of phenol, nitrophenols, cresols, thymol, etc. In addition, they also described the idoform formation of ketones including methyl ketones and 3-pentanone. Preparation method only.

How do the authors in reference 2 know they have 3-pentanone instead of 2-pentanone using analytical techniques available in 1917? The boiling points are the same.

@KG3: By 1917, it they should be able to distinguish these two compounds by their physical and chemical properties. Also, they have checked the homologous compounds to 3-pentanone, such as 3-hexanone, etc.

@MathewMahindaratne Which physical and chemical properties are you referring to? The boiling point and refractive index are not distinguishing. There was no NMR, IR, or mass spec available to the authors in 1917. As the haloform reaction is a classical method for identification of methyl ketones it is most probable that the authors mistook 2-pentanone for 3-pentanone, which would be easy to do with the available analytical techniques of the time.

@ KG3: I cannot pinpoint how they distinguish those chemicals. Yet, chemistry is in prictice way before 1917. For example, how would you think Claisen recognized the rearrangement without NMR or IR or MS? It's not easy, but they have done that and results are very fruitful.

@MathewMahindaratne you are completely missing the point. The proposed reaction in reference #2 contradicts the entirety of the literature on the haloform reaction since 1870. Either we should (A) discard all subsequent literature examples that utilize the haloform test to identify methyl ketones as incorrect (B) presume Datta & Prosad made incorrect but completely understandable assumption. If you're going to make extraordinary claims you must supply extraordinary evidence. Claisen did. Datta and Prosad did not.

@KG3: Please read about my comment "...very curious about their claim.." part at the end. Doesn't it say something. I want to know if someone else come up with a reliable mechanism. I already knew there may be a mistake. But I cannot single handedly waive someone else's work. As a community, we all need to agree.

@KG3 and Mathew: My first thought about diethylketone giving CHI3 was that they had 2-pentanone. I am incredulous with the statement, "It has been observed that as we rise higher and higher in the homologous series of the ketones, the  tendency of the formation of iodoform gradually diminishes and the yields also gradually become less." It is conceivable that 3-pentanone is converted to the 2-iodide, hydrolyzed to the acyloin and oxidized to 2,3-pentanedione that gives iodoform. This procedure is improbable for higher symmetrical ketones. Of course, 2-pentanone is still a option.

@user55119: I appreciate your comment. It's make sense. I'll look at relevant results.


text:
Can the Lewis dot structure of SO2 be experimentally verified?
I have just watched 2 different videos on the Lewis dot structure of sulfur dioxide, $\ce{SO2}$. One says it is
$$\ce{O=S=O}$$
and the other says
$$\ce{^-O-S^+=O <-> O=S^+-O-}$$
The first paper acknowledges the second structure and says they are both right. Isn't there some way of experimentally determining the bond length, or the bond energies etc to tell which is the true structure?

Interesting question. In theory, yes; however, I'm not sure if there is actually an example of a true S=O double bond to compare the experimental bond length against. At least, one isn't occurring to me now. Most "S=O" bonds, e.g. in sulfoxides, sulfones, etc. are better represented by $\ce{^+S-O^-}$; and $\ce{SO3}$ has a S–O bond order of 1.33. The only recourse may be to computational chemistry.

Lewis dot structure of anything is too great a simplification to be experimentally verified.


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Should I include ions from the solvent in solubility product constant calculations?
I'm trying to determine the Solubility Constant $K_\mathrm{sp}$ of $\ce{Ba(NO3)2}$ dissolved in nitric acid solution $\ce{HNO3}$ from experimental data.
$\ce{HNO3}$ will be mostly dissociated so there are many $\ce{NO3-}$ ions already in the solvent before solid $\ce{Ba(NO3)2}$ is added.
When I calculate $K_\mathrm{sp} = [\ce{Ba^2+}][\ce{NO3-}]^2$, do I include in $[\ce{NO3-}]$ the ions already present in the solution before solid $\ce{Ba(NO3)2}$ is added?
For example, say I have dissolved ${0.1 mol}$ of barium nitrate in ${1 L}$ of ${1 M}$ nitric acid solution. There are now ${0.2 mol} + {1 mol} = {1.2 mol}$ of nitrate ion in the solution. Should my $K_\mathrm{sp}$ calculation be
$$K_\mathrm{sp} = 0.1\cdot 0.2^2,$$
or
$$K_\mathrm{sp} = 0.1\cdot 1.2^2?$$
I understand that the common ion effect in this case inhibits solubility somewhat, but it I think my professor told me to include ions from $\ce{HNO3}$ in the calculation, which seems wrong.

Yes, absolutely, all nitrate counts in solubility product regardless of source. The system cannot distinguish what a nitrate ion (or any dissociated species) was bonded to before it dissociated.

I get it now, thank you. My confusion was stemming from my misconception that the Ksp in HNO3 solution should be different from the Ksp in pure water.

@omzrs - At the risk of more adding confusion, you're not totally wrong. The Ksp actually depends on the chemical activity of the ions rather than their concentrations, and the activity is dependent on the ionic strength of the solution. Activity and ionic strength are concepts that you should study later. // I'll add that for a salt as soluble as barium nitrate the Ksp using concentration is really pretty useless.


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Find the number of electrons
I am not understanding,why do we have to subtract the charge from the atomic number while finding the number of electrons.

Could you edit your question and give an example so that we know exactly what you are asking?

In short, we don't.

Suppose an atom of element 'X'
Let's assume the charge on this atom is 0 - hence, the number of protons in the nucleus and the number of electrons revolving about the nucleus must be equal, which results in a net 0 charge on the atom.
Now, lets assume that another atom of 'X' has +1 charge over it.
Well, how could this happen?
There are two ways for this to happen.
1) Nucleus gains an addition proton (+1 charge)
2) Atom loses an electron(-(-1) charge)
Both of these cases result in the atom gaining +1 charge.
Case 1 is very difficult to achieve, hence its not a good idea to think about in any chemical reaction (in fact it is not one at all, see comments!), instead consider case 2.
Notice, losing an electron results in +1 charge over the atom - another way to think of this is that we have ruined the balance that existed between the number of electrons and number of protons.
So we could say the above sentence in your words, we subtract the total charge present on the atom from the atomic number (remember, originally when the atom had no net charge, atomic number = number of protons = number of electrons. This is no more the case now!) to get the number of electrons present in the atom.
This is a very basic form of conservation of charge.

RE: 1) Nucleus gains an addition proton (+1 charge) -- This is just not a chemical reaction. It is a nuclear reaction.


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H+ vs HCl in the gastric juice
Parietal cells in the stomach are known to expel $\ce{H+}$ ions into the gastric juice to keep the pH acidic. When $\ce{H+}$ $\ce{[H3O+]}$ can fully function as acid in the gastric juice, why do cells combine $\ce{Cl-}$ to form $\ce{HCl}$ in the stomach?

You can't expel ions of one sign only.

Well, Parietal cells use ATP to expel H+ generated from H2CO3 and chloride ions are transported into the gastric juice separately via an independent process. So in the juice they become HCl. The question is, why HCl is made when pumping H+ itself should be good enough to keep the stomach liquid acidic?

It is not an independent process. You can't just pump H+ and leave it at that. Or rather, you can, but this would have certain unintended consequences, like some negative ions creeping seemingly out of nowhere.

If you only pump ions of one charge, e.g. positive, into a system, you will find that there is attraction of those ions to negative ions elsewhere. Thus, the $\ce{H+}$ will "drag the $\ce{Cl-}$ ions along". Note that in the gastric fluid itself, I expect that the two kinds of ions will be fully dissociated due to the strength of hydrochloric acid and a sufficient concentration of water being present.


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What is the principle of “scrubbing” in liquid extraction or solvent extraction?
There are many steps in solvent extraction:

Extraction
Scrubbing
Back extraction

I understand the 1st and the 3rd.
What is the principle of scrubbing?
Is it just washing the organic phase?
How undesirable species get removed from organic phase?

"Scrubbing" is (in chemical technology) a method that removes a gaseous compound. I have never heard the term used in pure chemistry. Can you please give some context?

Dictionary.com confirms that (4)https://www.dictionary.com/browse/Scrubbing  I suppose it is used in transferred meaning, dealing with the emulsion of the other solvent and mainly with other co-extracted components.

As noted by Karl, scrubbing is usually used to describe a process in which a liquid is used to clean (scrub) a gas.

I think OP's question is about solvent extraction in hydrometallurgy. In that context, solvent extraction includes three steps: (1) Extraction; (2) Scrubbing; and (3) Stripping (back extraction). This solvent extraction is a unit operation for the purification and concentration of a wide variety of metals:

Extraction: The operation of transferring the metal of interest from the aqueous phase (SX Feed) to the organic phase. During the extraction, which contacts an organic
phase containing an extractant with an aqueous phase containing the metal of interest (with impurities). The extractant chemically reacts with the metal to form an organic-metal complex that is soluble in the organic phase. Impurities normally do not react with the extractant and remain in the aqueous phase (known as raffinate, which is either sent for further treatment or effluent).
Scrubbing: The selective removal of impurity metals from the loaded organic phase from extraction step, which contains the metal of interest (and impurity metals). Scrubbing is conducted by treatment with fresh scrub solution (aqueous phase), which contains chemicals to remove impurity metals from organic phase, selectively reversing the reaction. The spent scrub solution is normally combined with the SX Feed. The scrubbed organic containing the metal of interest is separated from aqueous phase, and is advanced to stripping.
Stripping: The process of removing the metal of value from the scrubbed organic phase by reversing the extraction chemical reaction. It is normally conducted under conditions in order to produce a strip liquor containing a high concentration of the metal value (aqueous phase).

(Source)

"Scrubbing" in this context is the treatment of the organic phase to remove impurity metals. This is as opposed to "back extraction" which is to remove the metal of interest from the organic phase.


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How can nitrogen be quadruply bonded in tris(8-hydroxyquinoloinato)aluminum?
Nitrogen typically has five outer valence electrons. So how does a triply-bonded nitrogen in 8-hydroxyquinoline end up quadruply bonded in tris(8-hydroxyquinolinato)aluminum?

Ever heard of a coordinate bond?

It may be easier to first understand a simpler system, particularly the pyridinium cation.

Come to think of it, why not start with ammonium  cation?


text:
How to extract pigment from juice
Let's say I have carrot juice, standard kitchen and access to grocery store. How do I extract pigment from Orange juice? Sugars need to be separated out.
What's the fastest option? I thought about filtering, sedimentation, heating.

Do you have access to organic solvents?

Like what? I don't know this class.

Solvents immiscible with water suc has ethyl acetate or dichloromethane

@Waylander Since they have standard kitchen equipment: what about using oil to create two phases and separate water-soluble (sugars)  content?

Could do, but then you have a solution of the pigment in oil and it will not be straightforward getting it away from the oil.

Is coconut oil Ok? How to do oil extraction?

If your coconut oil does not mix with water then the beta-carotene which is the main pigment will go into the oil phase when you thoroughly mix the carrot juice and water. The sugars are water-soluble.


text:
Can an element have variable oxidation states in a compound?
Take propanoic acid for example. If I assign oxidation states to the 3 carbons in it using the idea that the more electronegative element gets all the electrons in the bonds it makes, the 3 carbons get -3( the one in the CH3), -2( the one in the middle), and +3( the one in the acid group).
Is this conclusion correct? The reason I doubt is that all the examples we've ever been given in class, we've assumed all the atoms of the same element to have the same oxidation state.

Yes, the same element can occur in a compound in different oxidation states. Consider sulfur in thiosulfate  https://en.wikipedia.org/wiki/Thiosulfate  Your teachers have just been keeping it simple so that your classmates understand the basic principle.

@Waylander, thanks. Are my numbers of propanoic acid correct though?

Oxidation numbers are not a very useful concept when it comes to organic compounds but yes, your numbers are correct


text:
Combining Bond Enthalpies and Enthalpies of Formation
Can I mix formation enthalpies and bond enthalpies in the same calculation?
I think the answer is no, since they are relative measurements, and possibly relative to different things, but I haven't seen it stated explicitly anywhere.

You can combine this data. You just need to find a sensible reaction cycle, and you have to make sure the temperatures (usually 298 K) match.
To give a simple example, consider $\ce{CO2(g)}$.
What is the $\ce{CO}$ bond enthalpy $\mathrm{\epsilon_{C=O}}$in $\ce{CO2(g)}$?
It is one half of the enthalpy for the reaction
$$\ce{ CO2(g)->C(g) + 2O(g)}~~~~~\mathrm{\Delta  H^\circ = 2 \epsilon_{C=O}}$$
The enthalpy of formation of $\ce{CO2(g)}$ is equal to the enthalpy of combustion of graphite:
$$\ce{C(s) + O2(g) -> CO2(g)} ~~~~~~\mathrm{\Delta _f H^\circ = -393.51}\pu{kJ mol^{-1}}$$
The heat of formation of $\ce{C(g)}$ is:
$$\ce{C(s) -> C(g) }~~~~~~\mathrm{\Delta _f H^\circ = 716.68}\pu{kJ mol^{-1}}$$
We also need a value of the $\ce{OO}$ bond enthalpy in $\ce{O2}$:
$$\ce{O_2(g) -> 2O(g) }~~~~\mathrm{\Delta  H^\circ = \epsilon_{O=O} =  498.3}\pu{kJ mol^{-1}}$$
Adding things up we obtain $\mathrm{\epsilon_{C=O}}$:
$$\mathrm{\epsilon_{C=O}} = \frac12(393.51+716.68+498.3)\pu{kJ mol^{-1}= 804.2 kJ mol^{-1}}$$


text:
Average Molar Weight
We have a mixture of two gases, say nitrogen and hydrogen.
To calculate the average molar weight M , the formula says:
M (average) = (hydrogen pressure/100).M(H2) + ( 1 - nitrogen pressure/100).M(N2)
What I haven't understood is that why do we have pressure in the formula?
What I don't find logical as well is that the unit on the left side isn't the same as that on the right side of the equation.

I suppose the $100$ is not unitless, but implies to be $\pu{100 kPa}$.
Then both equation sides have the same dimension and average molar mass is a weighed average of gas molar masses.

Other option could be to use a molar or volume fraction.

Thank you :) I have read on wiki  (https://en.wikipedia.org/wiki/Mole_fraction) as well that : In a mixture of ideal gases, the mole fraction can be expressed as the ratio of partial pressure to total pressure of the mixture. But I still need to know why :)

It comes as consequence of the ideal gas laws and the ideal gas state equation $$pV=nRT$$

You are correct that the pressure units should cancel. However your formula
M (average) = (hydrogen pressure/100).M(H2) + ( 1 - nitrogen pressure/100).M(N2)
doesn't make any sense to me.
If we assume that the 100 is for 100 kPa, and that the total pressure of the gas is 100 kPa, then the formula should be:
M (average) = (hydrogen pressure/100).M(H2) + (nitrogen pressure/100).M(N2)


text:
Prince Rupert's Tears - with Alumina?
When molten glass is dropped into water you can get tear shaped objects which are both incredibly strong and which have massive internal stresses. Which means any scratch can cause them to explode.
Is this only a feature of glass, or would it apply also to molten Alumina (Aluminium Oxide)?

AFAIK, molten alumina does not tend to form amorphous phases upon freezing. Also, to melt it you have to go quite a bit higher than for melting glass. In short, alumina hardly has anything in common with glass.

@IvanNeretin So going to some 2000degC to <100degC in a couple of seconds still produces a crystalline solid?

Quite probably yes (though I'm not 100% sure).

@IvanNeretin I assume polychrystalline?

Yes, sure. $;$

YOu would have to cool in a couple of microsceconds to get amorphous phases in alumina


text:
How to calculate the equilibrium constant for nitrogen, hydrogen, and ammonia?

$\pu{2 mol}$ of $\ce{N2}$ is mixed with $\pu{6 mol}$ of $\ce{H2}$ in a closed vessel of $\pu{1 L}$ capacity. If $50\%$ of $\ce{N2}$ is converted into $\ce{NH3}$ at equilibrium, what is the value of $K_c$ (equilibrium constant) for the reaction
  $$\ce{N2 + 3 H2 <=> 2 NH3}?$$

This is what I tried to solve it:

The answer to this question is $\frac{4}{27}$ as given in the solution key.
It is only possible if the concentration of $\ce{NH3}$ is not raised to its stochiometric coefficient.
Am I doing it wrong or is the answer key wrong?

$[\ce{I2}]\mathrm{e}=(2-1)=\pu{1M}$,$[\ce{H2}]\mathrm{e}=(6-3)=\pu{3M}$,$[\ce{NH3}]\mathrm{e}=(2\times{1})=\pu{2M}$
$$k_\mathrm{C}=\frac{(2)^2}{(1)(3)^3 }=\frac{4}{27}$$

A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.

@Martin-マーチン Can you Please suggest a title for this question

I have already changed the title of the question; I am more concerned with the image, currently I am somewhat guessing what you are trying to convey.

@Martin-マーチン Should I add that "if only 50% of N2 is used"

No. If that really were important, I would have edited this in, while I was trying to improve the question. The title is okay as it is. This is really not the problem of the question. I really cannot understand everything in the image (I can't even read all of it), otherwise I would have already converted that.

Few notes about your solution:

"Both are limiting reactant"

You can neglect this idea in this problem, as the reaction is not a completion reaction, it is an equilibrium reaction.

Your solution indicates that the amount of $\ce{NH3}~ \text{at equilibrium}=n_\mathrm{e}(\ce{NH3})=\pu{4mol}$

It is not correct, it is equal $\pu{2mol}$, it can be calculated as the following:
\begin{align}
n_\mathrm{e}(\ce{NH3})
&=n_\text{ formed}(\ce{NH3})\\
&=2\times{n_\text{ reacted}(\ce{N2})}\\
&= 2\times \left(2\times\frac{50}{100}\right)\\
&= 2\times{(1)}\\
&= \pu{2mol}
\end{align}
Use the molar ratio to calculate the amount of $\ce{H2}$ reacted:
\begin{align}
n_\text{ reacted} (\ce{H2})
&= 3\times{n_\text{ reacted}(\ce{I2})}\\
&= 3\times{(1)}\\
&= \pu{3mol}
\end{align}
Calculate The amount of $\ce{H2}$ at equilibrium:
\begin{align}
n_\mathrm{e}(\ce{H2})
&=n_\mathrm{I}(\ce{H2})
  -n_\text{reacted}(\ce{H2})\\
&= 6-3\\
&=\pu{3mol}
\end{align}
Use the following table to determine the amount of each species at equilibrium
in order to  calculate $K_c$ :
\begin{array}{c | c c c c c}
         &\ce{N2}       & + &\ce{3H2}      &\ce{<=>}& \ce{2NH3}   \\\hline
\text{I} &(\pu{2 mol})  &   &(\pu{6 mol})  &        &(\pu{0 mol}) \\
\text{C} &(\pu{-1 mol}) &   &(\pu{-3 mol}) &        &(\pu{2 mol}) \\
\text{E} &(\pu{1 mol})  &   &(\pu{3 mol})  &        &(\pu{2 mol})
\end{array}
Use the expression $[i]_\mathrm{e}= \frac{n_\mathrm{e}(i)}{V}$
to calculate the concentration of each species at equilibrium:
\begin{align}
  [\ce{N2}]_\mathrm{e}  &= \frac{\pu{1 mol}}{\pu{1 L}}=\pu{1 M}\\
  [\ce{H2}]_\mathrm{e}  &= \frac{\pu{3 mol}}{\pu{1 L}}=\pu{3 M}\\
  [\ce{NH3}]_\mathrm{e} &= \frac{\pu{2 mol}}{\pu{1 L}}=\pu{2 M}\\
\end{align}
Substitute the concentration of each species at equilibrium in the following formula to calculate $K_c$:
\begin{align}
K_c &=\frac{[\ce{NH3}]^2_\mathrm{e}}{[\ce{N2}]_\mathrm{e}[\ce{H2}]^3_\mathrm{e}}\\
&=\frac{(\pu{2 M})^2}{(\pu{ 1M})(\pu{3 M})^3}\\
&=\left(\frac{4}{27}\right)\cdot \pu{L^2//mol^2}
\end{align}

The answer would be better, if it were defining the quantities and then using actual equations instead of 'word equations', like amount of $\ce{NH3}$ at equilibrium would be $n_\mathrm{e}(\ce{NH3})$. That would also get rid of MathJax not breaking properly/ flowing into the margins. ($\ce{I2}$ is probably a typo?)