text:
Why is Avagadro's number independent of substance?
In the book of Irey, Thermodynamics, in vol1, IA, page 4, it is given that
Molecular (atomic) weight $M$:

The ratio of the mass of a molecule (atom) of a substance to one twelfth of the mass of an atom of the most common isotope of carbon $^{12}\mathrm C$.

Mole:

A gram mole is the amount of a substance whose mass is equal to its molecular weight in grams. The mass of the one gram mole equals $M$ grams ($M$: molar mass).

Avogadro's number:

Since the mass of a mole is proportional to molecular weight, there are a definite number of molecules in any molar unit, called $N_\mathrm A$.

And in the remark, the author mentioned that $N_\mathrm A$ is independent of molar units, and we know that it is even independent of substance.
However, with these definitions, I cannot understand why $N_\mathrm A$ should be independent of substance.
Question:
With these definitions, why is the constant $N_\mathrm A$ independent of substance that is used to measure $N_\mathrm A$?

It's not related to your actual question, but your book is more than 40 years old. In particular, the concept of "gram mole" is obsolete. You might want to consider getting a new book.

In short, the constant $N_A$ is independent of substance precisely because it is defined to be independent of substance.

@Loong That is one of the reasons why I particularly have chosen this book, apart from that is a great book; I cannot understand newly written books, regardless of their content.

@IvanNeretin I don't see any such statement in the definition; note that you can define something in a way that would be independent of substance, in which case you still show that it is indeed is independent of substance, but what you are claiming is something like "let $N_A$ be a constant that is independent of substance" which is definitely not the case.

If you cannot see such a statement in the definition then you are looking at the wrong definition

What is the reason for the down votes ?

@Waylander How do you exactly justify your claim about that the given definition is wrong ?

Irey's definitions of mole and molar mass inevitably lead to $N_A$ being a constant. The definition is a bit cumbersome, because only a few weeks ago, $N_A$ has once and forever been defined to be one specific integer number.

@Karl  how  that lead N_a to be a constant is the main question, so an explanaion would be nice

Your "gram mole" Isn't mole! Mole is exactly Avogadro's number of anything! Loong comment is actually exactly the point of your miunderstanding...

@Mithoron Would you care to give an explanation why it is the case, or will you just state it and done with it ?

;/ There is one single specific proportion between atomic mass unit and gram and it's Avogadro's number.

@Mithoron How do you know that ? If you can explain that, you basically will be answering to my main question.

@Mithoron Plus, how does that explain the claim that "a gram mole is not a mole" ?

"Gram mole" is an archaic unit name. By Irey's definition, it is identical to the mole as defined in the SI system half a century ago.

It is the same reason that a dozen doesn't depend on whether you are counting grapes or elephants. That is how the mole is defined: it is a number, nothing else.
The confusion, I suspect, is because of how we measure that number (or, strictly, how we originally measured it as the definition changed recently). The intention of the unit was always to define a number of things independent of the thing being measured. But to measure that number chemists originally defined it as the count of atoms in a particular substance (originally on 16O and later 12C). Since November 2018 it is now defined as an exact number.
Presumably something similar happened as humanity learned to count beyond 10. To illustrate the concept of 12 they might have used 12 rocks or 12 wildebeest. But the idea of a dozen was independent of the things being counted: the things were merely a concrete illustration or example of the idea of a dozen. Same with the mole: the thing being counted is irrelevant even if you had to use a concrete example of counting the atoms in a real substance to illustrate the number.

So, If understand correctly, we experimentally found that this number was independent of the substance, and then defined it in a way which also described how to measure that number ? and this is the main confusion; there is nothing that forces that quantity be independent of substance, but it is an experimental fact that it is the case ?

@onurcanbektas No. The definition is not empirical. Chemists observed that things combine in fixed ratios long before they even had a concept of the atom. As the concept of the atom emerged, chemists wanted to count them. Then they developed the concept of the mole and needed to measure how many atoms were in it. Don't confuse the number (independent of what is being measured) with how it was first counted. A dozen is not an empirical concept depending on what thing we are counting: it is an abstract number.

"Don't confuse the number (independent of what is being measured)"; yes the definition does not specify any specific substance, but it might turn out that measuring that number for different substances result in different numbers. In that sense, the definition might not be even well defined (in the mathematical sense); that is why somehow, empirically or analytically we should show that that number is independent of the substance that is being measured.

For example, in your example of "dozen": we are explicitly defining dozen as a number of things, which is indeed independent of the substance since it is defined as a number; but in the $N_A$ case, the definition includes mass of the substance and its molecular weight, which are specific to the substance that is being measured.

@onurcanbektas You are missing the point. A mole has always been a number and is currently defined as a precise number. Initially chemists used specific substances to measure it but the definition didn't change because of the difficulty of measuring it. in fact, if a measurement of a different substance looked off, what chemists would change is the atomic mass not the count of moles.

@onurcanbektas a dozen is exactly the same sort of concept as the mole. The only issue is that counting a mole using your fingers is a lot harder. We don't need to check empirically that a dozen elephants had the same number of things in it as a dozen grapes. Replace dozen with moles and the grapes and elephants with different atoms and the sentence still works exactly the same way.

Matt, the number of people in a room is also a number, but it is dependent on which room are we talking about. You mentioned that in order to make it independent, they  change the atomic mass, but that change is still not arbitrary, like changing the atomic mass of He, but not H; a single definition works of atomic mass works for every atom, so there is question still remains why does (with the current definition of atomic mass of atoms) the avagadro's number defined as in the question independent of substance that is being measured.

@onurcanbektas The number of moles of a chemical in a test tube depends on the size of the test tube, as the number of people in a room depends on how big the room is. How we count the people does not nor does our definition of people. We measure atomic mass of a substance assuming we know how much (in moles) of that substance there is. The idea that substances are made of discrete units is an axiom of the atomic theory of matter not something we need to keep testing empirically. The mole is just how we count those units.

"the number of people in a room depends on how big the room is" really ? So you haven't seen an empty room whose size is nonzero ?

@onurcanbektas I assumed your point was that the room was as packed as possible with people. It is clearly possible that a room doesn't contain as many people as that. But it is people we are counting, not the floor area of the room, so my point stands.

@onurcanbektas You are veering off into persistent ignorance and rudeness. A mole is a certain number of things, there is probably not a mole of stars in the sky, there is a little bit more than a mole of water in a half full glass, there is exactly a mole of electrons in half a mole of hydrogen gas. Same as there are usually not a dozen pups to the dog, usually more than a dozen children in a regular sized 5th grade class and exactly a dozen in a 12 pack of eggs. a mole is a mole by definition, exactly 1 per atom in a dozen grammes of carbon atoms. https://what-if.xkcd.com/4/

Technically, "dozen" is just "two and ten".  It cones from a French word for "twelve" which, in turn, they derived from the Latin for "two and ten".

@StianYttervik have read the whole question by the way ? Note that my purpose is to learn, not to be rude; i think you guys get offended to quickly, which is not healthy in an academic discussion.

Let's look at the $\ce{^9_4Be}$ isotope and apply the definitions:
Relative atomic mass
The relative atomic mass of $\ce{^{12}C}$ is, by definition, 12. Looking at the periodic table, we find that the relative atomic mass of beryllium is 9.0121831(5). That makes sense because $\ce{^{12}C}$ has a mass number of 12 (6 protons and 6 neutrons) while beryllium has a mass number of only 9 (4 protons and 5 neutrons).
Mass of one mole of substance
According to the definition, one mole of beryllium has a mass of 9.0121831(5) g, just as one mole of $\ce{^{12}C}$ has a mass of 12 g (exactly, until May 20th 2019). These two samples have same number of particles because their mass ratios are equal to their relative atomic mass ratios (look back at the definition of the relative atomic mass, it relates one atom of beryllium to one atom of carbon-12).
Avogadro constant $N_\mathrm A$
A mole of anything contains the same number of particles as a mole of $\ce{^{12}C}$. This follows from combining the definitions of relative atomic mass and of the unit mole (pre May 20th, 2019). The Avogadro constant is the ratio of number of particles to amount of substance:
$$ N_\mathrm A = \frac{N_\text{particles}}{n_\text{sample}}$$
Before May 20th, 2019, the Avogadro constant was experimentally determined. You would have to count atoms in a sample of known chemical amount, typically indirectly. For example, if you know the mass, the radius, the atomic distances of a silicon sphere and the average molar mass of the silicon sample, you could figure out the Avogadro constant.
Since May 20th, 2019, the Avogadro constant is set to a constant value:
$$ N_\mathrm A = \pu{6.02214076e23 mol-1}$$
Your textbook is out of date, just like most textbooks were when you asked your question, which does not make sense anymore given the current definition of the Avogadro constant.

I understood why their mass ratios are equal to their relative atomic mass ratios; however, how does that imply that both samples have the same number of particles ?

The mass ratio of one atom Be to one atom C is 9 to 12. If I compare a million atoms of Be to a million atoms of C, the mass ratio will be 9 to 12. Conversely, if the mass ratio is 9 to 12, I will have the same number of atoms of Be and C, no matter what the absolute mass is.

You mean the same number of atoms of Be relative to C ?

Yes, in order to get the same number of atoms of Be relative to the number of atoms in a C sample, their masses must have the ratio of 9 to 12.


text:
Will (3R,4S,5R)-1,3,4,5,6-pentahydroxyhexan-2-one tautomerise in nitric acid?
The following question was sent to me .The question is " Which of the following is  A ?"
The given answer was a,b,c

Background
If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid.

source : https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/carbhyd.htm#carb2
My thoughts

Options a and c are direct reactions.
Option b gives 2 and  it can only rotated in the plane it is drawn (2D) by 180 degrees (http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-7.html) to give 1.
How ever option d in acidic medium could tautomerise to give 3d , that on oxidation again gives 1.

Therefore answer should be a,b,c,d.
My answer hinges on tautomerisation of ketone is acidic medium. Would this compound (3R,4S,5R)-1,3,4,5,6-pentahydroxyhexan-2-one tautomerise in nitric acid ?
Any help in this regard is very much appreciated. Thankyou.
reference :
1 http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-7.html

This link would suggest that direct oxidation of fructose leads to a trihydroxyglutaric acid. Thus, option d) appears to be eliminated. https://chemistry.stackexchange.com/questions/121328/oxidation-of-fructose-with-conc-hno3

I realize now that configuration at 3rd carbon will not change. While tautomerism can occur between 1st a d 2nd carbon. Good work guys, thankyou.

@Sir Arthur7 did a fine job of discrediting D-ketose (d) as a viable candidate for structure A. Belatedly, I will expand upon my comment above at the request of @Jan. [In my comment above, "glutaric" should read "glucaric".]
Assuming that $\ce{HNO3}$ is capable of tautomerizing the D-ketose via the enediol and accomplishing terminal oxidation, two epimeric aldaric acids could arise: D-gularic acid and D-idaric acid. Note that they only differ in the hydroxyl configuration at C2. Rotation of D-gularic by 180o about an axis perpendicular to the page results in a structure that is also recognized as L-glucaric acid. Thus, D-gularic and L-glucaric acid are identical. When the same rotation is performed on D-idaric acid, D-idaric acid is returned. Choices (a), (b) and (c) all produce D-glucaric acid (red box) while option (d) produces the enantiomer, L-glucaric acid, and the irrelevant D-idaric acid. Option (d) is not a viable structure for A.

Well I am just a high school student, so I may be wrong, and I beg pardon in advance, but I would like to share my notions about this topic.
@ChakravarthyKalyan I would like to ask you to review your last step in the conversion of 3d to 1. 3d on oxidation should yield this product (in the image)--which is by my knowledge not same as 1.

So even your theory does not yield (d) as the correct option, I assume. Also I doubt the double tautomerism of the compound in option (d) in acidic medium, though your mechanisms were convincing. This closely resembles Lobry-Debryn-van-Eckenstein rearrangement that occurs only in presence of alkaline medium.
As per my knowledge, ketoses in presence of strong oxidizing agents like conc. HNO3, undergo oxidative cleavage forming 2 carboxylic acids(acc. to Popoff's rule), or as @user55119 had pointed out to me, that alpha ketoacids can undergo oxidative decarboxylation and form CO2 and an acid(though its mechanism is quite unclear to me as well).

Oxidative decarboxylation is one consequence of nitric acid oxidation. If tautomerization occurs, D-idose (3c) and D-gulose (3d) would be formed. HNO3 oxidation of D-gulose would afford L-glutaric acid (aka, D-gularic acid) as 1, not D-glutaric acid as shown. Your structure as drawn id D-gularic acid. Rotated by 180o, It is L-glutaric acid.

This is essentially correct. The product 3d and 1 are enantiomers. This answer could expand on the fact more.

source: Stack Exchange

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text:
Do magnesium and tin have the same volume?
Having the same atomic radii of 145 pm, should one mole of tin and one mole of magnesium take up the same volume? Since tin is much heavier, that means tin is much denser than magnesium. Is this true, or is there an error in my thought process.

You have to specify conditions, and if we are speaking of crystalline matter, what polymorphs you are interested in. It's very unlikely that both are going to have the exact same molar volume.

I think this is a superb question! The measure of atomic sizes is done by many methods, all of which have value. So when you see a number like atomic radius = 145 pm, is that definitive? No, because the packing of atoms is not always the same, as noted below. But the contrast between densities and molar volumes is instructive. Tin is denser ("heavier") (7.31 g/cc) than Mg (1.731), but the molar volumes are fairly close (Sn: 16.29 cc) vs Mg (14.00 cc). So the numbers of atoms in a given volume are similar, but tin weighs 118.7, while Mg weighs a paltry 24.3.

I do not think this is a homework question. In fact it offers an opportunity to clarify some interesting issues.

No, that is not how it works.
Atomic radius means the radius of an atom. A single atom. It does not in and by itself mean that a solid chunk of a material has a particular density. A solid chunk of material has many atoms, and they are arranged in certain ways and sometimes more than one way of this arrangement is possible.
Tin and Magnesium are very different. Magnesium is arranged in a hexagonal close packed structure, basically just repeating units of skewed (rombohedric) prisms. This can be hard to imagine, but imagine just stacking perfect spheres so they rest on each other and likely you will end up at a close packing arrangement. (there are only two*, try to find them. Hint: one has three layers before it repeats the bottom, the other has only two layers then it repeats)
*: There are actually 4, but you'll need a lot of layers and a lot of spheres to find those.
Tin has two different ones, so already here your question has to be no'ed without further investigation, but lets anyway. The first one, grey tin (${\alpha}$) has a diamond like structure. I am finding it hard to describe them in a meaningful way, but they have examples over at the excellent Wikipedia. The other one, tin (${\beta}$) which is white has a body centered tetragonal - which is basically stack 4 - put 1 in the middle and then stack 4 on top of that again (yes, the spheres will roll off, but there are so many of them per layer that hardly any of them do.) The white one is amusing, if you bend it it screams like you are hurting it. Sounds absolutely disgusting, like nails on a chalkboard. This is related to it's structure.
Summary:
When you put a whole mole (ha... ha...) of atoms together, they organize in structures, and it is the number of atoms per volume that you have to multiply with the atomic weight to get the density of the chunk. Fortunately for most kinds of structures, you don't have to do this, someone already has done the work, by actually measuring the densities of crystals and made tables. Just look it up.

One should note that there are more than 2 close packed structures (see https://en.wikipedia.org/wiki/Periodic_table_(crystal_structure)). The main one, DHCP, found in several Lanthanides/Actinides, is ABACABAC. Samarium is particularly weird. In theory, any periodic arrangement of (A,B,C) would be close packing. In reality, the longer the period the less likely it will be found in nature.

@JonCuster Right you are. I was drawing from memory, didn't remember those weird ones. I am still leaning towards keeping it simple for the OP though. Should I edit what you suggest in or... ?

Up to you. A parenthetical comment added would be more than enough.

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text:
dipole–dipole interactions: OCR Sample Paper wrong?
I was doing OCR A Level Chemistry A, H432/01 Periodic table, elements and physical Sample Question Paper.
Question 6 was:
The boiling point of hydrogen bromide is –67 ºC.
The boiling point of hydrogen iodide is –34 ºC.
The different boiling points can be explained in terms of the strength of bonds or interactions.
Which bonds or interactions are responsible for the higher boiling point of hydrogen iodide?
I put answer C: permanent dipole–dipole interactions
?
The given answer was D: induced dipole–dipole interactions
[Revision post down-vote]
My learning to date is that:

dipole-dipole interactions could only be one or the other i.e. permanent or induced.
iodine is more electronegative than hydrogen, therefore the interactions are permanent.

Please would someone explain where how the induced interactions are caused?

Your understanding is wrong, the types are not mutually exclusive. Moreover, the induced dipoles are always there, no matter what molecules it is.

Related: https://chemistry.stackexchange.com/questions/43350/trend-in-the-boiling-point-of-the-hydrogen-halides

Ivan & Mathew - many thanks for the clarification. By way of explanation, I was asking on behalf of my daughter, interpreting what she had told me to the best of my understanding. Funnily enough I had noticed the 'trend in the boiling point of the halides' question and pointed it out to her, thinking it might be germane. She had dismissed that article but, in light of your information, I will redirect her to it - though, in my ignorance, I'm unsure if it's at a suitable level for an English A level student. Thanks both.

Possible duplicate of Why are the dispersion forces in CS2 stronger than the dipole-dipole forces in COS?

I think that questions like this are tricky, and somewhat unfair: both molecules have dipoles and somehow the student is supposed to know that they happen to be similar. The induced dipole-dipole  interaction depends on the polarisability which is proportional to an atom's volume and so  number of electrons. This is greater for HI than HBr but to know that it counteracts the difference in dipole is tricky as I have said. I cannot see that an A level  student could reasonably work this out as the question is written.

Permanent Dipoles occur when two atoms bonded to each other have significantly different electronegativities. When the atomic number of an element increases the number of electrons present in them also increases. These electrons are constantly moving around in the atom. At some point, there are more electrons on one side of the atom than the other. This creates an imbalance of charge, creating a small charge of $\delta ^+$ and $\delta ^-$. These small charges induce its neighbors to create additional charges of $\delta ^+$ and $\delta ^-$, thus creating a chain of induced dipoles. These induced dipoles are temporary as the electrons can move, and hence can be reverted. They are, therefore, known as induced dipole forces, or better, known as London Dispersion Forces. In reality, they are not really forces, but rather attraction between molecules.
Naturally, more the number of electrons in the atom, more the amount of London dispersion forces, and hence greater the attraction. The greater the attraction, the more energy you have to put to break the bonds apart. Between the considered hydrogen halides ($\ce{HBr}$ and $\ce{HI}$), iodine has more electrons than bromine, and hence $\ce{HI}$ has a greater attraction force between molecules than $\ce{HBr}$. As a consequence, $\ce{HI}$ has a greater boiling point between them. Appropriate values are listed in following table:
$$\begin{array}{cccccc}
\hline
\ce{H-X} & \text{BP}^\text{Ref.1} & \ce{X}\text{'s EN} & \mu ^\text{Ref.2} & \text{Polarizability }(\pu{10^{-24} cm^3})  & (r_\ce{H-X})^\text{Ref.3} \\ \hline
\ce{HCl} & \pu{-85.1 ^\circ C} & 3.16 & \pu{1.09 D} & 2.63 & \pu{1.27 Å} \\
\ce{HBr} & \pu{-67.1 ^\circ C} & 2.96 & \pu{0.82 D} & 3.61 & \pu{1.42 Å} \\
\ce{HI} & \pu{-35.1 ^\circ C} & 2.66 & \pu{0.45 D} & 5.44 & \pu{1.60 Å} \\
\hline
\end{array}$$
Moreover, a related question you can find here. Orthocresol has given very reliable answer, which included two equations to calculate the strength of permanent dipole interactions ($U_\mathrm{dipole}$) and the strength of dispersion interactions ($U_\mathrm{dispersion}$). The author has stated that:

If you plug all the numbers in for the hydrogen halides $\ce{HX}$, you will find that the magnitude of the dispersion forces is much larger than the magnitude of the dipole-dipole attractions, i.e. $\left|U_{\mathrm{dispersion}}\right| \gg \left|U_{\mathrm{dipole}}\right|$.

For example, for $\ce{HCl}$, dispersion forces contribute $86\%$ to the intermolecular attractions, and for $\ce{HI}$, they contribute $99\%$.

References:

J. F. Ogilvie, W. R. Rodwell, R. H. Tipping, “Dipole moment functions of the hydrogen halides,” J. Chem. Phys. 1980, 73(10), 5221 – 5229 (https://doi.org/10.1063/1.439950).
N. N. Greenwood, A. Earnshaw, “Chapter 17 – The Halogens: Fluorines, Chlorinee, Bromine, Iodine, and Astatine,” In Chemistry of the Elements; 2nd Edn.; Elsevier: Boston, MA, 1997, pp. 789–887.
K. Fajans, N. Bauer, “Electronic Structure and Stability of Hydrogen Halides and of Complex Ions $\ce{XO4}$,” J. Chem. Phys. 1942, 10(7), 410 – 415 (https://doi.org/10.1063/1.1723742).


text:
Mechanism of burning of hydrogen
So I was curious regarding a simple chemical equation :
$$\ce{2 H2 + O2 -> 2 H2O}$$
You add heat to the system (i.e fire) and you get water vapor.
I'm curious of understanding the mechanism, which I'm sure at this point has been well studied.
Suppose I keep track of the location of each one of these atoms, and the energy nearby - my anticipation is that the reaction occurs when the atoms get close enough and there is sufficient energy present to conduct the reaction. Is there any way to summarize this critical point of water creation mathematically in this framework (i.e the combination of atoms being close enough and sufficient energy will make this reaction take place) - or is there a superior way?
Does anyone have any insight/publications that can help me understand the steps of water synthesis via reaction mechanism?

Math is not about summarizing things (that's creative writing). Math is about more specific questions, which often start with "how many".

It seems to me your post is an example of what happens when asker doesn't know proper terminology. If you want reaction mechanism, or chemical reaction model or something elese, then make up your mind and edit the question.

@Mithoron reaction mechanism looks like what im interested in. I'll edit the question. Thanks - I'm from the mathematics community so not a very good chemist.

Martin - マーチン has given excellent answer here. I'll give here an supporting review article for your benefit, which also gives some mathematical overview:

Louis S. Kassel, “The Mechanism of the Combustion of Hydrogen,” Chem. Rev. 1937, 21(2), 331–345 (DOI: 10.1021/cr60069a010).

Overall, mechanism can be summerized as follows:
\begin{align}
\ce{H2 + \text{heat} &-> 2 H^.}\\
\ce{H^. + O2 &-> HO-O^.}\\
\ce{HO-O^. + H2 &-> HO-OH + H^.}\\
\ce{H^. + HO-OH &-> H2O + HO^.}\\
\ce{HO^. + H2 &-> H2O + H^.}
\end{align}
The propagation would continue with having $\ce{H^.}$ at the end and the presence of excess $\ce{O2}$.

It is somewhat interesting that the mechanism you outline is not quite that posted by Martin- any comments on which one is closer to the truth, or when differences might be seen?

@Night Writer: See attached review paper. There are a lot of hand waiving arguments but this come closed to the real one, I guess.

to me it seems unlikely that H atoms are formed in this way as the energy involved is huge (436 kJ/mole) More likely is that $\ce{H2 + O2 \to OH\cdot}$ is the first step 72 kJ/mole and H atoms produced by the propagation step. Also peroxide is not a major a component except possibly at the third explosion limit since  peroxide has a retarding effect on the reaction replacing reactive OH with less reactive $\ce{HO2}$ via $\ce{OH + H2O2 \to H2O + HO2}$

I have added the scheme to the question referred to in the first line of your answer.


text:
How can equilibrium lie to one side ? Surely it’s not at equilibrium anymore?
Surely they’re not at equilibrium any more?
I understand that a favourable condition can make a reaction in one direction go faster, but then surely either the forward or backward reaction is faster and the rates are no longer the same so there is no longer an equilibrium? I do gcse chemistry, I’m a bit confused :/

Reaction rate depend upon concentration.

This is more like a figure of speech. This is still an equilibrium like the one you can observe for an asymmetrically balanced seesaw

I think your confusion is coming from the idea that the rate of the forward direction isn't permanently increased for example. As an example consider $$\ce{N2 + 3H2 <=>2NH3}$$ as this is a reaction you may be familiar with. If I had the reaction in a sealed container meaning an equilibrium will be established then the moles and thus concentration of each component will remain constant.  If I then change a physical condition, say increase the pressure, the equilibrium position in the example will move towards the right - the rate of the forward reaction will increase. However just because the rate has increased the reaction will still reach a new equilibrium where the concentration of the species will remain constant. The equilibrium position will remain constant.
If you are interested maybe go ad read about equilibrium constants. for example the formation of ammonia will have an equilibrium constant $K_c$ which will remain constant regardless of any physical change made to the system, expect temperature. $$\ce{K_c}=\frac{[\ce{NH3}]^2}{\ce{[N2][H2]^3}}$$

So when people say the equilibrium has shifted to the right, does that just mean the concentration of the products has increased and not the rate? Thank you for your help by the way!! :)

Yeah exactly, the equilibrium position is the ratio of products to reactants so if the equilibrium position shifts right then the ratio of products to reactants increases. (we give the ratio in terms of concentrations). If this answered your question be sure to accept the answer too.

Ohhhhh! I get it now, thanks! And I just accepted it sorry, haven’t been on here in a while.

Thats great, glad you get it now

@H.Linkhorn In your answer you say that the equilibrium constant remains constant, but in your comment you say that the ratio of products to reactants increases. That is a contradiction. It is correct to say that when you change concentrations, there will be a net shift of the reaction to attain equilibrium again (with the same equilibrium constant, but a different set of equilibrium concentrations).

The concentrations remain constant when the system is at equilibrium. When you apply a change to the system the ratio will change to a new equilibrium position so the ratio will be different but still constant

What is equal at equilibrium?
The forward and reverse rates are equal at equilibrium.
What is constant at equilibrium?
The concentrations are constant, and the rates are constant.
What is typically not equal at equilibrium?
The concentrations of reactants and products are typically not equal at equilibrium. For a reaction with multiple species in solution or in the gas phase, there are multiple sets of conditions that are at equilibrium. With three or more species that can vary in concentration (i.e. not pure solid or pure liquid), concentrations don't have to be equal even if the equilibrium constant happens to equal one. So, equilibrium does not mean concentrations are equal.

[..] surely either the forward or backward reaction is faster and the rates are no longer the same so there is no longer an equilibrium?

A reaction can be at equilibrium when concentrations of reactants are different from concentrations of products because rate constants can be different too. For a simple (single-step) reaction $\ce{A(aq)<=>B(aq)}$, if the rate constant of the forward reaction is larger than that of the reverse reaction, then the product concentration has to be higher than the reactant concentration at equilibrium to make up for that.

Reaction rate depends on concentration of reactants.
Consider a simple reaction
$$\ce{A -> B} $$
In this reaction, rate in forward direction is given by rate law
$r_1 =k_1 \ce{[A] }$
And in backward direction
$r_2 =k_2\ce{[B]} $
Then eventually a time will come when $r_1=r_2$, since concentration of both A and B are either decreasing or increasing, if concentration of A increases then that of B will decrease or vice versa, (it is NOT necessary that $\ce{ [A] =[B]}$ at equibrium) at this time (when forward and backward rates are equal) the reaction is in equilibrium and no further products are formed, since both forward and backward rates are equal.
If you add catalyst to decrease activation energy of, say, forward reaction then forward rate $r_1$ will increase but concentration of $\ce{A}$ i.e., $\ce{[A]}$ is also decreasing and concentration of B is increasing so again equilibrium will be established when $r_1=r_2$ and we say that


text:
Microscopic and mechanical approach to boiling point
I am confused as to how I should visualize the boiling process throughout a liquid.
From my understanding, the boiling point of a liquid is when its vapor pressure reaches atmospheric pressure. This makes sense at the surface of the liquid where the pressure of the atmosphere no longer overcomes the pressure of the gas molecules escaping the surface of the liquid, so in some sense it can be thought of, on a conceptual level, as becoming a mixture of "air" molecules and "liquid" molecules.
BUT, how does this account for the formation of bubbles on the bottom of container holding the liquid? Wouldn't the gas that is trapped in these crevices need to overcome to pressure of the atmosphere AND the pressure of the liquid itself (looking at it from a physics perspective as well)? How would this phenomena be explained using vapor pressures?
Furthermore, when gas bubbles form within the liquid, should I think of it as the kinetic energy of those particular liquid molecules are so great that it's "pushing" the surrounding liquid molecules with less kinetic energy?
EDIT: Part of my confusion was that I wasn't sure what the composition of these gas bubbles would be exactly—and yes the assumption is that it is being heated from the bottom. Although now I'm curious what would happen if it was heated uniformly somehow.

Good question but I think there are two points you should clarify. 1. Are you heating the liquid from below? 2. Are your “air” bubbles air or vapour (e.g steam)?

https://chemistry.stackexchange.com/questions/38653/vapor-pressure-and-boiling-point-for-solutions-in-open-containers https://chemistry.stackexchange.com/questions/23566/will-the-water-at-the-bottom-of-the-tube-boil-at-the-same-temperature-as-the-wat

Let’s ignore the air, first of all, and put it back later.
When you heat the bottom of the saucepan, the first thing that happens to the water at the top is nothing. It stays cold, because water is not very conductive.
The next thing that happens is that the water at the bottom becomes so hot that it is less dense than the water above it. The first thing that happens then is that nothing happens; the second thing is that the situation is unstable and the tiniest irregularity in the layers makes a blob of less dense water float upwards. The colder denser water sinks to the bottom to replace it, and then is heated in its turn.
(The long thin tube mentioned in the link finds it far harder to convect in this way, which is why the boiling, when it happens, is so explosive).
The rising blob of hotter water can cool by mixing with the water it meets on the way up. Mixing is more or less efficient depending on various things (such as how big the difference of temperature is, and so how buoyant the blob is), so the blob may still be travelling upwards when it hits the surface. Look sideways at a saucepan near to boiling and you will see this.
The boiling point of water is higher at the bottom than at the top, so at a certain moment you will have a blob that is below boiling point when it leaves the bottom finding itself above boiling point when it gets higher up. So boiling can start above the bottom of the vessel in that case; and indeed once a saucepan starts boiling, you see it boiling at the surface and only then does the boiling spread throughout the liquid.
Alternatively the blob may boil on the way up, and the bubble of steam then float up into the cooler layers and condense suddenly. This is the noise a kettle makes when it is near to boiling, and why it goes suddenly quieter when it really starts to boil.
An extra detail: water heated a little above boiling point does not boil. It stays a superheated until some irregularity or impurity acts as a nucleus for the change of state. This is why if you get a saucepan at the late “hot convection” stage and throw salt into it, it fizzes violently. The salt nucleates the superheated water and a bubble of steam forms round each grain.
Now air. Air is less soluble in hot water than in cold, so as the water is heated the air comes out of solution. First, bubbles form nucleated by any roughness of the surface of the base of the pan (if your pan is too shiny, scratch it). Eventually they grow big enough to detach themselves and float to the surface. The air bubbles stir up the liquid they float up through, so that convection happens quicker and in smaller blobs. That is why when you boil already-boiled water the second time, it makes such explosive noises: you get much bigger blobs (because no air bubbles intervene) making bigger steam bubbles which then implode more violently.
Finally, not heating from the bottom. If you grill water, it doesn’t boil at all, it evaporates in a quiet and genteel way. If you microwave water then there is little or no convection and most of it is superheated. This when you take your mug out of the microwave, it is still and innocent and unboiling. When you put a tea-bag into it then its rough surface nucleates the boiling everywhere (like the thrown salt) and you get an eyeful of boiling water.


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Can Diatomaceous Earth be added to plaster of Paris to harden it?
If diatomaceous earth is 85% sodium silicate, could it be added to plaster of Paris, with water, to form a hard surface, with the benefit of the diatomaceous earth to act as an anti mold agent, for instance if making one's own coaster?

Yes. In fact, you can mix it with pure gypsum and water to create a hardened diatom surface. People make them into coasters all the time. I think the ration is 2:1 (de to gypsum) but I’d have to confirm that. I’ve been experimenting with making a custom drain board. I have a pretty absorbent hardened cement made from a slurry of Portland cement and Food grade de with bentonite clay. The silica is amorphous in the food grade vs calcined, but that combined with the already absorbent rapid cement (the ingredients are proprietary but I’m pretty sure they use DE in that mixture too) makes for a ver

From personal experience - the hardness of the plaster is markedly increased. Much more force required to drill a hole in it, or scratch it. No measurements, just using it around the house for general repairs.

Diatomaceous earth is 80-90% silica (SiO2), not sodium silicate Na2SiO3 (or polymers).
Diatomaceous earth could provide strength gain for portland cement concrete (if it doesn't increase the water content), but is an inert filler for plaster. It might improve the properties you want in the final product, but I am unaware of its use for improving specific characteristics.
There are better fillers for plaster and concrete.


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Preparation of sulfonyl chlorydes and mechanism of preparation of tosyl chloride
After searching about the preparation of sulfonyl chlorides, I couldn't find much. I found that for preparation of tosyl chloride, chlorosulfonic acid can be used, as in:

Can chlorosulfonic acid be used to prepare sulfonyl chlorides in general? What are some other ways to prepare sulfonyl chlorides? If there is any.
The other part of the question, I'd like to know how the mechanism of this reaction would work. Having the equilibrium of the reaction:
$$\ce{3 ClSO2(OH) <=> SO2Cl+ + 2 SO3Cl- + H3O+}$$
How exactly is the $\ce{SO2Cl}$ formed? I didn't quite understand how the $\ce{Cl}$ became a cation. Apparently the benzene attacks that cation in order to form the desired product.

As for other ways of making sulfonyl chlorides, that is too broad to cover in a single post (I suggest removing it from this question, in general we don't encourage having two questions in one), but https://www.organic-chemistry.org/synthesis/C1S/sulfonylchlorides.shtm has a good summary.

Related: What is the mechanism of chlorosulfonation of benzene?

Not so far mentioned are other ways to prepare sulfonyl chlorides. The most obvious of these is oxidation of thiols, hydrogen peroxide or Oxone are among the reagents that will do this, then chlorination with e.g SOCl2

The way to understand this is to think of the autoprotolysis of water:
$$\ce{H2O + H2O <=> H3O+ + OH-} \tag{1}$$
In general autoprotolysis can be represented in the following form (you can recover eq. (1) by setting $\ce{A} = \ce{OH}$):
$$\ce{HA + HA <=> H2A+ + A-} \tag{2}$$
Now set $\ce{HA}$ as $\ce{ClSO3H}$ (or $\ce{(HO)SO2Cl}$, for clarity) and you have
$$\ce{(HO)SO2Cl + (HO)SO2Cl <=> (H2O+)SO2Cl + ^-OSO2Cl} \tag{3}$$
The cation produced in eq. (3) fragments to give water and the reactive chlorosulfonyl cation:
$$\ce{(H2O+)SO2Cl -> ^+SO2Cl + H2O} \tag{4}$$
The water in eq. (4) reacts with a third molecule of acid in an ordinary acid-base reaction:
$$\ce{H2O + (HO)SO2Cl -> H3O+ + ^-OSO2Cl} \tag{5}$$
The combined equation you have is simply the sum of eqs. (3), (4), and (5).

What about the reaction with benzene? I think the $\ce{SO2Cl}$ reacts with the double ring. But how is the double ring formed again to get the desired product? Maybe the hydrogen of the carbon which got attached to the S is abstracted by the O of $\ce{OSO2Cl}$?

@DennerEvaristo, your image seems to be taken from Clayden, which is a good textbook. Please consider reading the chapter on electrophilic aromatic substitution, it will answer your question. Also, it's a double bond, not a double ring.


text:
Do maxima in Gibbs energy also correspond to equilibrium positions?
ΔrG=(∂G/∂ξ ) at constant p and constant T, where ξ is the extent of reaction.
By Second Law implication, ΔrG= 0= (∂G/∂ξ ) at equilibrium. If we consider a graph of G against ξ, this would mean  we have a turning point (where first derivative=0) when ξ is that extent of reaction when equilibrium attained:

(∂G/∂ξ )= 0 when ξ= extent of reaction until equilibrium
  attained

So, ΔrG=0 when equilibrium attained.
However, first derivative=0 does not mean minimum point only, it can mean maximum point too! Let's say you do work until you reach Gibbs energy maxima. My question: Does a Gibbs energy maxima correspond to equilibrium state or not?
(a) If yes, doesn't this violate the Second Law which implies that Gibbs energy should be minimized whenever possible? No matter which direction you move, your Gibbs energy will always decrease. Just simply move from the maximum point and you will at least be better than your current state:
(b) If no, then it will mean that we have to move from the maximum point, but in which direction? In case the graph is a symmetrical parabola, where same dξ results in same dG, what will you observe in real life? Shift to reactants or products?

G vs. $\xi$ does not have a maximum
If all reactants and products are pure liquids and solids, G vs. $\xi$ is linear. If some of the species are in mixtures, the entropy of mixing is responsible for the "sagging" shape of the curve. If the curve has an extreme value, it will be a minimum.
Analogy to mechanics
In mechanics, this situation would be called metastability. Giving a nudge in either direction will allow the system to reach lower potential energy.

So all G vs Xi graph do not have maximum point? I clicked the metastability hyperlink but saw there is a graph with local maximum point?

That's a transition state, I guess. Here, you are considering the ground states of reactants and products, and your x-axis is the extent of reaction. In the graph you are referencing, the x-axis is the reaction coordinate for a single reaction.

Does a Gibbs energy maxima correspond to equilibrium state or not?
(a) If yes, doesn't this violate the Second Law which implies that Gibbs energy should be minimized whenever possible? No matter which direction you move, your Gibbs energy will always decrease.

There are a number of central concepts in thermodynamics that are relevant here: equilibrium, spontaneity, reversibility, work, heat. The perfect example to illustrate how these concepts are related to each other is that of a voltaic (or galvanic) cell or battery. This example can also provide a good approximation to the situation you describe in your question.
When you charge a battery you are increasing its Gibbs free energy. If performed reversibly (usually this means sufficiently slowly), the electrical work done during charging equals the gain in Gibbs free energy.
Is this a violation of the second law? No, because you did work to charge the system. If you did so reversibly, work was efficiently converted into a gain in Gibbs free energy. Or, if you did not charge the battery reversibly, you did more work than was necessary and wasted some of the energy as heat.
Now what happens if you disconnect the charged battery, leaving the circuit open? Not much! The battery does not discharge$^\dagger$. Is the system at equilibrium at that point? Yes$^\dagger$, as long as the circuit remains open. You can think of a very large resistance keeping the system stable despite the higher free energy.
Finally, if you connect the leads again to close the cicruit, what will happen? The battery will discharge and return to its original state. This will occur spontaneously, driven by a decrease in the Gibbs free energy. And the battery will do work as electrons are transmitted from one to the other terminal.
$\dagger$ To a first approximation; batteries tend to self-discharge.


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Salt Bridge for Microbial Fuel Cell: Full or Partly Full?
I'm making a microbial fuel cell out of benthic mud and salt water in containers. My salt bridge will be made of agar and table salt solidified in PVC pipes. Does the salt bridge connecting the anode (benthic mud) and cathode (salt water) have to be full of the agar-salt solution, or can it be partly full?

Partly full bridges can be partly functional with being partly durable. :-)

The solution (gel) in the bridge has to make electrical contact with mud and salt solutions.

Does the salt bridge connecting the anode (benthic mud) and cathode
  (salt water) have to be full of the agar-salt solution, or can it be
  partly full

There should be no air gaps or cracks so that the circuit does not break anywhere. Take a narrow PVC pipe if your concern is using too much of agar-salt solution.


text:
Is it appropriate to approximate solute particles in a dilute solution as an ideal gas?
I was studying osmosis and I came across the formula for osmotic pressure. I decided to see the derivation for osmotic pressure in a dilute solution. The derivation involved an equation involving the chemical potential, fugacity and so on, which I am not familiar with. I have been introduced to very little thermodynamics. So I decided to come up with my own justification regarding the osmotic pressure formula. I want to know is there any thing wrong in my explanation.
Suppose we take a u-tube containing pure water. The level of water in both arms of the u-tube will be the same and hence also the pressure in both of the limbs. Now suppose I take a u-tube containing pure water and a dilute aqueous solution separated by a semi permeable membrane. The pure water from one limb will flow towards the other limb through the semi permeable membrane until equilibrium. In equilibrium the side of the u-tube containing the dilute solution will have a higher pressure (as seen due to greater rise in height). This extra pressure is due to the dissolved solute particles (if solute particles were not there both arms of the u-tube would have the same pressure). This extra pressure due to the solute particles is the osmotic pressure. Since the solution is very dilute, the solute-solute interactions can be neglected and we consider the solute particles as an ideal gas occupying the volume of the solution. Thus to find the pressure exerted by the solute particles we can use the ideal gas law. Hence this pressure is the osmotic pressure.

Be careful not to take "as an ideal gas" too literally. Obviously you have a solution, but the behavior of the particles in the solution is analogous to those of an ideal gas, such that when you compute its properties the equations are analogous. This is due to the analogous statistical properties in the two systems (dilute ideal solution and ideal gas)

A solution is by no means an ideal gas. An aqueous solution is nearly incompressible while ideal gases conform fairly well to Boyle's Law.


text:
What's the densest alkali metal?
This was asked in a school exam of mine, GCSE curriculum, exactly as it was worded on the title. I found this question a bit vague, and answered it francium, however of course francium is just an estimation. I couldn't be sure what the answer to the question would be, so can anyone give me some insight?

The question is more than a bit vague. Privately I consider Cs the right answer, but they most likely wanted you to say Fr, even though Fr is not really a thing.

According to https://de.wikipedia.org/wiki/Alkalimetalle#Trends Fr is even sligthly less dense than Cs. Also sodium is definitely denser than potassium. The upward trend just isn't perfectly clean. All in all, this is quite a stupid question to ask in an exam. Do they want you to learn these numbers by heart?

Otoh, all period 7 elements are considerably denser than their counterparts in period 6, and those than theirs in 5. Only periods 2-4 are not continguous. I think I knew this at the end of high school. ;-)

It's hard to tell what is the answer you were supposed to provide, but yes, generally speaking, you are correct.
According to CRC handbook of chemistry and physics [1, p. 4-14] (emphasis mine):

Francium, the heaviest known member of the alkali metal series, occurs as a result of an alpha disintegration of actinium. […]
Because all known isotopes of francium are highly unstable, knowledge of the chemical properties of this element comes from radiochemical techniques. No weighable quantity
of the element has been prepared or isolated.

As the result, the majority of physical properties of francium, including density, were obtained by extrapolation and not by direct measurements.
According to Bonchev and Kamenska [2], $ρ_\mathrm{calcd.}(\ce{Fr}) \approx \pu{2.9 g cm-3}$, which is in a good agreement with the trend for the measured densities across the group.
For comparison, the following data (NTP conditions) is collected from the table PHYSICAL CONSTANTS OF INORGANIC COMPOUNDS [1, p. 4-43]:
$$
\begin{array}{cl}
\hline
\text{Element} & ρ/\pu{g cm-3} \\
\hline
\ce{Li} & 0.534 \\
\ce{Na} & 0.97 \\
\ce{K} & 0.89 \\
\ce{Rb} & 1.53 \\
\ce{Cs} & 1.873 \\
\hline
\end{array}
$$
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; CRC Press, 2017; Vol. 97. ISBN 978-1-4987-5429-3.
Bonchev, D.; Kamenska, V. Predicting the Properties of the 113-120 Transactinide Elements. J. Phys. Chem. 1981, 85 (9), 1177–1186. https://doi.org/10.1021/j150609a021.

A linear extrapolation though Li-Cs runs almost exactly though the value of 2.9 at 223. Bit weird, that.


text:
Is activation energy time-dependent?
I was thinking lately about activation energy's dependency on time. I couldn't find any information on the internet. Only answer that came up to my mind is the fact that activation energy is constant, and doesn't usually change with time.
But what about auto-catalytic reactions? Maybe their activation energy changes with time after creation of small amount of product, at least theoretically. What do you think?

Can you provide an example of the process you are thinking about? Of course if the process changes in some way the activation energy can change. In particular, if you modify the transition path for a reaction. This is called catalysis.

Ican only think of iodination of acetone, which is some of the well known autocatalytic reactions.

Imagine that you repeatedly measure the height of a stone wall, and on one day you find it significantly lower than the day before. Do lengths and heights change with time on their own? Maybe they do. But maybe it was just a group of guys with hammers and other demolition equipment that did the trick. Same thing here.

As Ivan Neretin writes, the activation energy is constant and is caused by the electronic energy in the bonds. Time does enter into a reaction, however,  because only occasionally (say 1000 times/sec)  do the reactants, by random thermal motion, come together in the right orientation and with the right energy to react.

If you add catalyst you can certainly expect the activation barrier for those molecules reacting through the catalytically-activated process to be reduced. The fraction of those molecules will change over time in accordance to the catalyst concentration.

The activation energy of a single reaction does not change with time but there might be more than one reaction happening
An activation energy is always constant for a given single reaction. But that doesn't mean the reaction speed will be constant (it might depend on concentration or temperature).
But what about autocatalytic reactions? There are two possible sources of confusion here. A very simple autocatalytic reaction might involve a changing concentration of an intermediate. In this case the reaction might go faster because that concentration is changing as the intermediate builds up (but the activation energy stays the same). In more complex cases, the catalytic route to the product involves a different reaction. What you have is not a changing activation energy for a specific reaction but the addition of a different reaction pathway to give the same product. This reaction will have a different activation energy but the activation energy of the original pathway is not changed.


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Synthesis of Aspirin, made unknown fluid
When I tried synthesising aspirin, after quenching and cooling it I noticed a translucent, milky, white fluid floating in the solution.

The method used was:

Salicylic acid (2 ml) was added to acetic anhydride (2 ml) in a flask with 8 drops of concentrated sulfuric acid.
Warmed the solution in a water bath until all solid was dissolved.
Added 5 ml of cold water to the solution to quench.
Placed solution in water bath to dissolve the solids again. (I noticed that not all the solid dissolved)
Added 10 ml of water to solution and placed in an ice bath (This was when I noticed the fluid forming)

When the solution was filtered only that fluid remained as the residue. Another person also had the same problem but said they got rid of the fluid by scratching the side of the flask with a stirring rod after taking it out of the ice bath. Could someone please explain what it is?

If the problem was solved by scratching with a glass rod, probably it's a mixture of solvent and product. The ice bath (and scratching) are probably intended to allow the growth of large crystals, which can be easily isolated.
Did you perform any analysis on it?

Often, acetylizations are performed with acetic anhydride being deployed both as reagent, as well as solvent.  Consequently, acetic anhydride is used in excess when referring to the number of $\ce{-OH}$ to react.
The addition of water you describe in step 3) intends to hydrolyse this excess of acetic anhydride to form acetic acid.  Apart from acetic acid equally representing a good solvent for many organic products, it takes time to complete this hydrolysis of acetic anhydride. You may accelerate the hydrolysis of the acetic anhydride if the beaker is immersed into an ultrasound bath, or just continue stirring the mixture at room temperature as usual.
The second point is, albeit its high melting point (136 $\mathrm{^\circ{}C}$), the crystallization of aspirine equally takes some time, too. Crystallization of organic materials from water may demand some more attempts prior to success than from organic solvents, too.  Cooling the mixture often will lower the solubility of your product in the solvent, however, if cooled too quickly to too low temperatures you may obtain either just a glassy gel / syrup instead of a solid easy to filter off, or a solid that has to be recrystallized to purify it further.  Scratching the inner walls of the beaker with a glass rod intends to create fresh sites where crystallization may start.
For peracetylating glucose with acetic acid, for example, it is not too strange to stirr the quenched reaction mixture with an excess of water over night at room temperature to convert the initially "honey-like" product into a good solid.


text:
Relationship between electronegativity and atomic radius
Would someone please explain in detail the relationship between the two? I understand that atomic radius is related to ionisation energy but I can't see how atomic radius may be related to electronegativity.
And it appears that they are not inversely proportional to each other, but I couldn't find any explanation for this (or for the relationship between the two) online.
I'm a sixth form student (the equivalent of high school in the US I suppose).

Sorry Clément, but I can't even answer that question because it is too complicated to go into details. You may also see in the next years that nucleophilicity and basicity are closely related to that and equally difficult to grasp.

Welcome to Chemistry SE. They can both be explained by effective nuclear charge.  Is this a concept that was explained to you in class?


text:
On the stability of methyl mercury

Question: Given that $\ce{Zn, Hg}$ are in the same group, how to explain that $\ce{Hg(CH3)}$ is stable but $\ce{Zn(CH3)}$ is unstable and $\ce{Hg(CH3)Cl}$ is stable? $\ce{Hg(CH3)}$ seems to have an unpaired electron and hence is unstable.

Background and Thoughts
Some similarities of zinc and mercury: All of $\ce{M(CH3)Cl, M(CH3)2}$ and $\ce{MCl2}$ are stable, where $\ce{M=Hg, Zn}$.
For the causes of the differences of the stabilities of $\ce{M(CH3)}$, I can only think of the following reasons:
1. Mercury ion is soft and zinc ion is not very soft. $\ce{CH3}$ is very soft. According to the HSAB theory, methyl mercury is stabler than the other one. I'm not very sure of whether HSAB theory can be applied here.
2. Zinc is reactive while mercury is stable.
3. Perhaps it is related to $d-\sigma$ conjugation?

For why both $\ce {Hg(CH3)}$ and $\ce{Hg(CH3)+}$ are stable, I think maybe the reason is strongly related to the fact that $\ce{Hg-C}$ bond is covalent.

It's probably simpler than that and related to oxidation states (Hg +1 is much more prevalent than Zn +1, probably because of relativistic effects). That said, does Hg(CH3) actually exist as such? Is it some kind of dimer Hg2(CH3)2 or even a polymer?


text:
As you heat up a substance will it's bonds begin to dissipate?
As you heat something up does it slowly start to dissipate? Water, for instance, boils at 100 degrees Celsius, but if you are heating it up at 90 degrees will its bonds be weaker?

Remember the energy of the bonds never increase or decrease. They remain the same. When things are heated what actually happens is the molecules begin to move around (in solids they just oscillate while in liquids and gases the movement becomes much more prominent) and their kinetic energy increases and the molecules bump into each other harder and quite often which makes it increasingly difficult for the bonds to hold the molecules together (intermolecular forces). Once a certain threshold is reached the bonds become inefficient at their job and hence small lumps of molecules move apart enough for us to perceive the action as a phase change.


text:
Is this statement correct or not regarding SN1 reaction?
A primary alkyl halide may also undergo SN1 reaction in aqueous formic acid.
This statement is written in one of my organic textbooks. Is it true ?

What do you think?

I don't know. It was written in one of the reputed organic chemistry textbooks. I think It can be true. But How ?

SN1 and SN2 are selected according to the stability of halogenated hydrocarbons corresponding to carbon cations, and have little relationship with reagents.

If you're asking in a general sense, then no. Primary alkyl halides generally prefer SN2, over SN1 since it (generally - for primary halides) takes less activation energy to do the substitution in a concerted way, rather than a relatively unstable primary carbocation. But, if you (or your textbook) is more interested in some relatively rare example of some perfluorinated substrate using very poor nucleophiles that is almost fully solvated in aqueous formic acid, then maybe it's true. Please could you provide us with the reference of the reputed textbook so that we can see the context?

Quoting ORGANIC CHEMISTRY SECOND EDITION ,Jonathan Clayden ,Nick Greeves , Stuart Warren from page no.338 ,in 50% aqueous ethanol the values give you an idea of the relative reactivity towards substitution of the different classes of compound. These rates are mostly SN1, but there will be some SN2 reactivity with the primary compounds.

And in page 342 ,the rates of reaction of the following alkyl chlorides with KI in acetone at 50 °C broadly illustrate the patterns of SN2 reactivity are analysed. These are relative rates with respect to n-BuCl as a ‘typical primary halide’.

As you have not mentioned any alkyl halide,n-butyl chloride's relative rate of reaction is 0.07 in 50% aqueous ethanol.Its tendency to undergo SN2 is more then SNI.
Similarly a primary alkyl halide in acetone at 50 °C has greater tendency towards SN2.


text:
Which ring expansion in cyclobutyl(cyclopropyl)methanol is favourable?

Predict major product:

In this question first I've protonated the OH group and then water is removed to form a carbocation. Now I have to expand the ring, but I'm confused which ring to choose: 3-membered ring or 4-membered ring?

Note:  I can make an alkene (which would be the final product of the reaction) by removing an alpha hydrogen if it is known which ring to expand.

Perhaps there is more than one rearrangement?   Could both rings expand?

I think there will be two separate ring expansion products, but the major product will be the expansion of the cyclopropyl group. The reason ring expansions occur at all, is to relieve ring strain. The farther away you are from a hexagon, the more strain there will be.

It is clear to me that no matter which ring you open the first, you'd get the same product as A. Michael Lautman gave the correct product , first opening the cyclopropyl ring. Michael's mechanism is fair, but I'm reluctant to say it is the most preferable path because it gives you bicyclobutyl $2^\circ$ carbocation as an intermediate, which needed extra 1,2-hydride shift to get stabilized as $3^\circ$ carbocation (total of 5 steps). However, if you have opened the cyclobutyl ring first, you'd get the same final product A, through a relatively low energy cyclopropylcyclopentane carbocation (total of 5 steps) even though it is also needed a 1,2-hydride shift (in relatively more stable cyclopentyl $2^\circ$ carbocation when compared to $2^\circ$ carbocation in cyclobytyl ring in Michael's mechanism) to get stabilized as $3^\circ$ carbocation. Complete mechanism is depicted below:

This mechanism is supported by following reference:

G. K. Surya Prakash, V. Prakash Reddy, G. Rasul, J. Casanova, G. A. Olah, “The Search for Persistent Cyclobutylmethyl Cations in Superacidic Media and Observation of the Cyclobutyldicyclopropylmethyl Cation,” J. Am. Chem. Soc. 1998, 120(51), 13362–13365 (https://doi.org/10.1021/ja9828962).

How about this:

Ring expansion of the cyclopropyl group
1,2-hydride shift
Expansion of the cyclobutyl group
Expansion of the "new" cyclobutyl group
Elimination

The first two structures aren't really distinct.

@Zhe can you elaborate please?  The first structure is the initial carbocation formed in the reaction.  The second has undergone a ring expansion.  If you are going to downvote an answer, it would be nice if you provided some constructive feedback.

Cyclopropylcarbinyl, aka homoallyl, aka cyclobutyl cations are generally thought to be non-classical instead of distinct species.

The  $\ce{C4H7^+}$cation shown as the first entry in Scheme 1 is  described as a bridged structure that is isomeric with cyclopropylmethyl and cyclobutyl ions.1
Scheme 1

The product following this pathway is shown in the post by Michael Lautman.
Second pathway is shown here in Scheme 2 with 2 in ring expansion to give2a .Next is elemination to give 3 and 4 .
Scheme 2

2a could undergo a hydride shift to 2c ,that on elemination gives third product 5.

Referance
1 Advanced Organic Chemistry FIFTH EDITION
Part A: Structure and Mechanisms
FRANCIS A. CAREY and RICHARD J. SUNDBERG

Cyclopropyl carbinyl is generally thought to be a nonclassical cation and not two isomeric forms.

@Zhe i thought so too  ,but its a direct quote from advanced organic chemistry by francis carey and sundberg

Historically, there are opposing view points on this subject, which Carey/Sundberg does not capture. Cf. the excellent work of George Olah.

@Zhe, I understand that, but I would have appreciated some insight into what the concept of the non classical carbocation would change about the product.

@MichaelLautman For one thing, you don't ever have to consider ring expansion of the cyclopropyl carbinyl cation; the non-classical view would be that it exists in an intermediate form. On first principle, the unknown regarding this mechanism is which steps are under thermodynamic versus kinetic control?


text:
Why would water, and not saltwater, undergo electrolysis?
I am attending a cram school and there was a question that goes:

Electrolysis decomposes chemicals by passing an electric current through a liquid or solution containing ions. Which of the following can be decomposed by means of electrolysis?
  A. Water
  B. Cream
  C. Saltwater
  D. Carbonated drink

My answer was C since the first sentence explicitly stated that electrolysis decomposes chemicals by passing an electric current through a liquid or solution containing ions, but they said that the correct answer is A. Also, I was thinking that water in itself is not conductive and therefore will not facilitate electrolysis by itself. Is the anwer really Water? And if so, why water and not saltwater?

That's an odd question, are you sure it's written like that? Pure water isn't conductive (very high resistance) but it can be subject to electrolysis: It's ionic product is 10e-14, meaning there are ions present. They are converted to H2 and O2. Saltwater can also be subject to electrolysis forming H2 and Cl2, and leaving behind NaOH.

The question is written exactly like that and I was really baffled because I am very familiar with the chloralkali process which uses electricity and saltwater so I was certain that it should be saltwater. I, however, didn't know that pure water can also be subjected to electrolysis.

Scientifically speaking, that question does not make any sense, but it can be an approximation. What class level is it supposed to be?

It was meant to be a review for college entrance exams

It is a poor question written by an inexperienced person. Your logic and reasoning is perfectly fine. By that logic, carbonated drink can also be electrolyzed.

I am pretty sure that if one pushes enough some electrolysis will happen in cream too :)). Indeed a discouraging question (that one of the test).

@Jasper I don't believe the question is unclear. The test question may be predicated on incorrect logic, but the users question is precisely whether this question makes sense and he provides his thoughts on why he thinks it doesn't. I think a reasonable answer would be that that the test question (not the OP's) is faulty.

Thank you all so much for your insights!

What is the "correct" answer
You are asking why the answer key has water and not saltwater as an answer. Decomposition is defined as:

A reaction where a single compound breaks down into simpler compounds.

For salt water, the salt reacts at one electrode, and the water at the other, and you are making NaOH. For pure water, you are splitting water into the elements. Neither process matches the definition of decomposition that I quote exactly, but pure water is the better fit.
In your question, you write:

I was thinking that water in itself is not conductive and therefore will not facilitate electrolysis by itself.

The half reactions could be written as:
$$\ce{2H2O + 2e- -> H2 + 2OH-}$$ and
$$\ce{H2O -> 1/2 O2 + 2H+ + 2e-}$$
So there are protons produced at one electrode and hydroxides at the other (You could also have protons be reduced at one electrode, and hydroxide oxidized at the other - the net effect is the same). The protons are very mobile in water and will quickly diffuse and neutralize the hydroxide. This takes care of the charge transport through the solution.
Is this a good question?
No, but most multiple choice questions aren't.


text:
How often is the IUPAC Goldbook Updated?
I was wondering how often does the IUPAC update its Gold Book, ie, IUPAC Compendium of Chemical Terminology - the Gold Book.
It poped in my mind when I saw this question about jellium aromaticity, that the term should be there. However, it was not. Whereas, the term has been used several times in the linked references (1 2), which I like to consider credible.
So, I was wondering how up-to-date the book is and how much can one rely on it for latest information.
PS While I understand that this question is primarily and heavily opinion based, I think it is equally relevant to all of us as we delve further in the chemical world.

It looks like Gold Book, the online edition, is a rolling release with corrections and additions done when they are done.

As for your suggestion to introduce iupac tag, currently it looks like there is no need for it: Do we need the iupac tag?. If you disagree, feel free to discuss this matter on Meta. In fact, it's better to search or ask on Meta first whether tax $x$ or $y$ would be necessary beforehand.

And I don't think an iupac tag is needed. It wouldn't find much usage after a few questions, and there's no need to focus on a single organization as such. Anyway, more on this if it pops up on meta.

Oh, and, the Gold book is up-to-date, don't worry about that. It is the book which decides what's up-to-date.

@WilliamREBenzener, put it very well, it is IUPAC which decides what's up-to-date and not the other way round. So, does the question remains relevant, or should be closed, or deleted or left as it is?

There is a lot of terminology that probably never catches the attention of IUPAC because it is rarely used, or only in very special cases and by a small subset of chemists or physicists, especially if it is something new and in development. This is perhaps the case with jellium aromaticity. There is something called the jellium model, a model that's used in the solid-state materials field but might otherwise seem somewhat obscure to chemists. A search for this term should lead you to a world of hits.

see btw http://publications.iupac.org/pac-2007/1990/pdf/6201x0103.pdf

@user79161 Well it isn't a bad question - seeing that not many new users are familiar with stuff. If you have a good insight into the question already, you could answer your own question, or wait for a good answer and accept it.

IUPAC.org appears to have on line meeting minutes and activity reports of the various committees and sub-committees. It would also seem that work continues through the year. It would seem that updates occur as they come up and agreed upon, not a fixed times (as they used to do pre-web).

As various users have contributed to the arrival of this answer, it goes to the  community.
Firstly, the IUPAC is the authority of chemistry, which decides on and accepts terminology for international compatibility so that information can be unambiguously expressed by researchers, scholars, educators and students. So, it can be finally concluded that unless it is IUPAC approved, it is not official terminology.
Secondly, since the IUPAC is an enormous organisation, dealing at a global level, it can be understood that changes will be a little delayed as the information has to be checked for authenticity and a common end has to be reached. For example, it took quite some time to decide on the newcomers of the Periodic Table(Elements of Atomic Number 113-118).
Thirdly, a lot of terminology has a very limited scope and can be at the very least left out. For example, the iupac tag was dropped due to its limited scope.(See comments above).
Lastly, I thank the community for the comments which have fabricated the "bonds" of this answer.

I commend your efforts for this.


text:
Two different answers obtained by balancing this chemical equation!! Somehow
This is the concerned equation:
$\ce{ XeF4 +  H2O ->  Xe + XeO3 + HF + O2}$
This can be balanced as
$\ce{6 XeF4 + 12 H2O -> 4 Xe + 2 XeO3 + 24HF + 3 O2}$
or
$\ce{5 XeF4 + 10H2O ->3 Xe + 2 XeO3 + 20 HF + 2 O2}$
Former is what found in this book.  NCERT chemistry part one, the textbook for class 12th. Chapter 7 page number 205 (b) Xenon Oxygen compounds, at the top.
while the latter is the one I come up while I was messing the equation. Something is wrong here. But what?

Related: https://chemistry.stackexchange.com/questions/64202/more-than-one-way-of-balancing-a-chemical-equation/64233#64233

Even more related: https://chemistry.stackexchange.com/q/58069/

I like how the title sounds like "somehow duplicate"


text:
Retention time and dead time
Why does ts start just after tm ?
Does the analyte starts spending time in the stationary phase only after the mobile phase has passed and reached the detector ?
Wouldn't there be an overlap between the two ?

The column is filled with the carrier (liquid or gas) before the sample is injected. Thus if there is no interaction between the sample and the column, then the fastest that the sample can get to the detector is the dead time denoted by $t_M$ in the diagram. If the sample does interact with the column, then it is retarded by a time $t_S$  shown in the diagram. Thus the total retention time from injection to detection is $t_M + t_S$.

That seems more convincing. But then, what does ts mean ? The mentioned definition in the picture is of not much help.

Most of the separation techniques consist of a chromatographic column, stationary phase, and mobile phase, and use common terminology. Therefore, it is better learn some described in the text book for HPLC (Ref.1) as an overview:

The retention time $t_\mathrm{R}$ solute A can be defined as the time from the injection of the sample to the time of compound elution, taken at the maximum (apex) of the peak that belongs to the specific molecular species A (known or unknown). The retention time indicates how long it takes for a compound A to elute from the column (from the injector to the detector). The retention time of the last peak (if sample contains multiple compounds) in a chromatogram is used to estimate the necessary length of the chromatographic run. In general, for a molecular species A, the retention time can be indicated as $t_\mathrm{R}$(A) and time is usually measured in $\pu{min}$. However, the part "(A)" in the notation is sometimes omitted, yet $t_\mathrm{R}$ is always related to a specific molecular species.
Retention time depends not only on the structure of the specific molecule, but also on factors such as the nature of the mobile and stationary phases, the flow rate of the mobile phase, and dimensions of the chromatographic column. Retention time is usually characteristic for a specific compound in a given separation. For this reason, the retention time is critical in identifying analytes once their retention time is known (e.g., by using standards).
Of particular interest in a separation is the dead time $t_\mathrm{M}$, which is the time a nonretained molecular species needs to elute from the chromatographic column. The dead time is also known as void time or holdup time. The dead time $t_\mathrm{M}$ can be also interpreted as part of the retention time $t_\mathrm{R}$(A) for the analyte A, which the analyte spends in the mobile phase moving through the column (That is the reason for the subscrpt "M" meaning mobile). This parameter is not related to the retention process and depends on the flow rate and physical characteristics of the column (i.e. length, diameter, porosity of stationary phase). The difference between the retention time ($t_\mathrm{R}$) and the dead time ($t_\mathrm{M}$) represents the time the analyte A is retained on the stationary phase ($t_\mathrm{S}$). This difference is indicated as reduced retention time $t_\mathrm{S}$ (or $t'_\mathrm{R}$) and is expressed by the formula:
$$t_\mathrm{S}=t_\mathrm{R}-t_\mathrm{M}$$

The value for $t_\mathrm{M}$ is typically obtained as an approximation by using compounds that are very slightly retained, since it can be difficult to find a compound that is not retained at all on a chromatographic column. For example, during HPLC runs, the solvent used for injecting the sample (when different from the mobile phase) can be such a compound, and the retention time of this solvent peak can be taken as dead time.
Reference:

Serban C. Moldoveanu, Victor David, Essentials in Modern HPLC Separations; 1st Edn.; Elsevier, Inc.: Waltham, MA; 2013.

Still confused. ts gives the retardation in elution due to the stationary phase. How are we sure that ts is the time spent in stationary phase ? It could include the time spent in the column without contact with stationary phase (in case the column is not entirely filled with stationary phase). Or tr itself would be the time spent in stationary phase (in case the column is entirely filled with stationary phase).

@Zam: The column is not entirely filled with stationary phase. The solute's time spent in the column without contact with stationary phase is its time spent in the mobile phase because column is filled with stationary and mobile phases only (theoretically). No air bubbles. :-)

But how can time spent by analyte in mobile phase be equal to the time required for an unretained species to pass through (dead time, tm) ? Wouldn't the former be less than the latter ? Because former doesn't take, passage through stationary phase into consideration while the latter does.

@Zam: If you are taking this class, I assume analytical chemistry or instrumental analysis, you may able to think a little deeper. I can take the horse to water, but I can't make him drink. Sorry.

"The dead time tM can be also interpreted as part of the retention time tR(A) for the analyte A, which the analyte spends in the mobile phase moving through the column (That is the reason for the subscrpt "M" meaning mobile)." I have come across statements with this very same meaning in my class but not reasoning. That's what I am enquiring here.


text:
What is quasi-aromaticity?
During my study of aromaticity, I came through this topic of quasi-aromaticity. Can you please elaborate what Quasi aromatic compounds are and how are they designated?

Consider using google for such terminology. It has might show good results and is recommended before asking a question. If it is not, then the question is welcome.

This appears to be the original paper: Lloyd, D.; Marshall, D. R.  Quasi-aromatic compounds: a definition, Chem. Ind. (London) 1964, 1760. Also, there is a modern review by Krygowski et al. (PDF). Feel free to improve and narrow-down your question in the meantime.

@user79161 I completely disagree. We definitely welcome many questions that could "just be Googled." Many of these questions may bring users to less reputable sites like Yahoo answers. If we build up a repository of easily-Googled questions, we start showing up in Google which really grows our site. And it helps people get an authoritative answer.

@MelanieShebel This site "shows on Google" more then sci. papers, so that point is moot. While the fact that something can be googled isn't a reason to close a question, because like everything here can be, users should still provide some context and introduction into the topic in the body of question, otherwise downvotes are are proper reaction.

In general, we can describe a quasi aromatic compound as a compound, which is ionic in nature with a counter ion, and the $\pi$ electrons in such compounds follow Huckel's rule ($4n+2$).
In other words, quasi aromatic compounds are those in which the charges present on the molecule are a part of aromaticity of the compound. A few examples of such compounds are depicted in the diagram:

However, it has a deeper and broader meaning. For a more in-depth explanation, please read the given references.

References:

T. M. Krygowski, B. Bankiewicz, Z. Czarnocki, M. Palusiak, “Quasi-aromaticity—what does it mean?” Tetrahedron 2015, 71(30), 4895–4908 (https://doi.org/10.1016/j.tet.2015.05.074).
E. Kleinpeter, A. Koch, “Characterization and quantification of quasi-aromaticity by spatial magnetic properties (TSNMRS),” Tetrahedron 2015, 71(33), 5275–5284 (https://doi.org/10.1016/j.tet.2015.06.019).

The references do not seem to match the answer, as the former tend to emphasize secondary bonding as part of the quasi-aromatic structure and the species do not seem to be charged.  What am I missing?

@ Oscar Lanzi: I don't think you are missing anything. I think chemists are giving broader picture to existing simple definition. All are still a theory.

@Melanie Shebel: Thank you for careful (and ruthless! :-)) editing. I'll also take an advantage to congratulate you for your new moderator ship!

Even though the gold book calls it Hückel (4n + 2) rule, there are more components to aromaticity than a simple number. Yet it appears that all that ever is taught about aromatic compounds is that number, without any limitations of the generalisation and approximation of these rules. I think it is more important to understand that all compounds that follow Hückel's rules have aromatic character, but not all compounds where aromaticity is observed follow Hückel's rules.


text:
What actually happens when 'energy is released'?
Thermodynamics classes and equations always discuss a chemical reaction in the form of energy being released or absorbed. This is always hand-waved as just 'energy' from the surroundings.
What's actually happening at a molecular level? When a bond is broken and 'energy is absorbed' to accomplish this, is the kinetic motion of the resulting species lower than the total kinetic motion of the species before it broke? Does something else happen?
When 'energy is released' on a bond being formed, does the new species then have a greater kinetic motion than the combination of the two species alone? Does this energy somehow translate to a change in energy level of an electron and a release of a photon at the subsequent decay?
Thanks in advance.

Basically yes, of course! Particles have kinetic energy before a collision, and after, the difference must come from or go into a chemical bond or electromagnetic radiation.

Thermodynamics works perfectly well by assuming that matter does not need to be made of discrete particles, i.e atoms and molecule are not necessary. It was developed before the structure of atoms and molecules  were known so it cannot inform us about anything at a molecular level. It is concerned primarily about energy and entropy.
If you break a bond in a diatomic molecule then all that can happen is that the two atoms move away from one another with the excess energy used to break the bond  shared between the atoms in the form of kinetic energy. The 'excess energy' is the energy above the dissociation energy. The atoms will be 'hot' and if in a gas (or solvent) of inert molecules will collide with these and their energy will be spread out among all the molecules raising the temperature by just a little bit.
In a polyatomic molecule if the products are lower in energy than reactants the excess energy  will raise the temperature of the surrounding solvent or gas. To break a bond the activation energy has to be overcome, once this happens the product has more potential energy than the reactant, i.e. more vibrational and rotational energy levels are populated than in the reactant, see figure. This is because the potential well of the product is lower than that of the reactant. At the top of the activation barrier is the transition state.
Once in the product potential the extra energy is lost quickly, a few picoseconds, to any solvent surrounding the molecule. Energy is also lost to other vibrational /rotational energy levels in the product molecule. These are vibrations not involved in the bond breaking.  Eventually all this energy is spread out in the surrounding solvent (or gas) raising its temperature. The kinetic energy change in a reaction may not be that large if the energy is taken up by the vibrations and rotations. This is unlike the case when an atom is formed as a product.

The figure shows how the product has more vibrational energy than  the reactant after surmounting the barrier and transition state (blue dot) which is the energy and coordinates at which the molecule changes from reactant to product. The energy difference, leading to heating, is that between the two lowest energy levels in each potential.

yes, there was of course hypothesis, but no evidence for the structure of molecules until towards  the end of the 1800's.  But it is still true that thermodynamics works if matter were continuous not discrete.

I have a question about 'vibrational and rotational energy levels'. Are you speaking of electron energy levels (orbitals), or something else? I could use clarification on this - my understanding is only up to basic molecular orbital theory.

No not electronic. A molecule's bonds vibrate and the energy is quantised. (see harmonic oscillator). This quantisation is seen as lines in an infra-red absorption spectrum. Molecules also rotate, as do small groups such as methyl and this motion is also quantised but with smaller energy gaps that for vibration, generally, leading to microwave spectroscopy. Energy can be transferred between different vibrational and rotational levels within a molecule. A bond with lots of energy can lose this energy as it spreads around molecule. This 'hot' molecule will lose energy to solvent by collisions.


text:
Chirality with tautomers

This molecule has 6 peaks in its proton NMR. I think this is due to the following tautomerisation:

However, I am unsure as to whether this molecule would be chiral, as it can freely interconvert between these two forms, when each form is itself chiral.

The chirality is not to do with the phosphorus being a chiral centre, but rather the binaphthalene bit, see https://en.wikipedia.org/wiki/Axial_chirality Also, the naphthalene ring has six protons, so you see six peaks. The two naphthalene rings are equivalent by symmetry ($C_2$ rotation axis).

Thanks, I had forgotten about axial chirality. However when the -OH group is replaced with a -Cl, there are 12 peaks seen. Does this not have C2 symmetry?

Are you referring to a phosphoryl chloride $\ce{P(=O)(Cl)(OR)2}$? If so, then no, because the P=O becomes P–Cl when you rotate 180°. Right, here it is P=O and P–OH, which shouldn't be the same, but for all intents and purposes you might as well consider them to be the same thing, exactly because of the tautomerism that you pointed out.


text:
Determining the charge on an ion through mole calculation in electrolyis

During electrolysis $\pu{0.02 mol}$ of metal $\ce{X}$ is deposited on cathode when $\pu{0.12 mol}$ of electrons are passed through molten chloride of $\ce{X}.$ What is the formula of the chloride?

I have tried solving it and my answer is $\ce{XCl6}$ which is incorrect. The correct answer given is $\ce{XCl3}.$
In the explanation the book says that $\pu{0.06 mol}$ of electrons have flown to cathode and goes on to divide the amount of electrons at cathode by the amount of metal $\ce{X}$ $(0.06/0.02),$ so it's $3$ which means $1$ atom of $\ce{X}$ gains $3$ electrons, giving it $+3$ charge.
What I do not get is what about the other $\pu{0.06 mol}$ of electrons. Isn't metal $\ce{X}$ supposed to gain all of the $\pu{0.12 mole}$ of electrons? Shouldn't cathode have the total $\pu{0.12 mol}$ of electrons?

I think your interpretation is more reasonable than the apparent answer, tricky or not. Can you post the name of the book?

The suppose total and half reaction are:
$$\begin{align}
\ce{2 MCl_n &-> 2 M(s) + $n$ Cl2} \\
\ce{M^n+ + $n$ e- &-> M(s)} \\
\ce{2 Cl- &-> Cl2 + 2 e-}\\
\end{align}$$
The question is rather tricky.
Note that I am addressing the supposed question in hand, not the electrochemical evaluation.
In some sense, if we consider large scale motion, no electrons passed through the electrolyte, they just got on one side in and on the other side out.
In usual sense, there is 3:1 electron:halogenide molar ratio for $\ce{MCl3},$ therefore result would we $\pu{0.06 mol}.$
But in yet other sense, if we consider nanoscale motion, the $\pu{0.06 mol}$ of electrons passed through the electrolyte at the anode, and the same amount through the electrolyte at the cathode, therefore in total $\pu{0.12 mol}.$
By other words, $\pu{0.12 mol}$ is total molar amount of electrons involved in net electrochemical reaction, passing through the electrolyte between electroactive substances and electrodes.
I would say, the question is either intentionally tricky either badly formulated.

That's absurd, right? The net flux of electrons through the material is zero. The same number of electrons goes in through one electrode as comes out through the other, otherwise you end up with a charged substance. I can't see the justification for doubling the number of electrons.

You may not got my meaning. :-)

I did not say the 2 senses are only possible senses. And we cannot know what the author meant. Note also we do not speak about charge, but electron molar amount. No electrons passed. Some got in and some got out.

Ok let me think about it some more :-)

I suggest writing the half reactions that would be associated with this process.

Why not, but it will not help.:-)

I correct myself, I had found the question in notes compiled by a high school teacher. He made the question up himself so I too think he's probably adding the loss and gain of e- at both the electrodes (0.06+0.06) thus a total of 0.12 mole of e-'s. In this sense the answer is XCl3. Before the question he also wrote this concept that I do not get: Total no of e- passing through the electrolyte is twice the no. of e- removed from cathode or returned to anode. Author has used this concept.

It looks like it follows my tricky reasoning. :-)


text:
Why should AAS use element lamps?
AAS (Atomic Absorption Spectroscopy) is a quantitative analytical technique used to measure very small concentrations of ions in substances. The main idea is that the sample is atomised in a flame and then light is shot through the atomised sample and the absorbance is measured.

However I don't get why they use an atomic emission lamp (a lamp made of only 1 element). What's wrong with using a normal incandescent lamp or some lamp that produces all the wavelengths at once (it would save you having to change the lamp for every test)? I've heard that multi-element lamps make the machine less sensitive due to noise but I don't understand how. If we tune the monochromator to the particular wavelength we want, it should be the same as with using an atomic emission lamp?

Related: Why are hollow cathode lamps used for atomic absorption spectroscopy?

Absorption bands are very small: smaller than the band which a typical monochromator is capable of isolating.
If a continuum light source was employed, the absorbed band would be much smaller than the band which would be isolated by the monochromator and which would be read by the detector. This implies a low SNR (signal-to-noise ratio), and a very poor resolution.
Using a single element lamp (or lamps with combined elements but with distinct emission frequencies) overrides the problem: a monochromator is well capable of isolating the different frequencies, and absorption, which occurs in a narrow band of frequencies, occurs in the same narrow band of light from the source: this implies a much higher SNR, and a much better resolution.
To make an easy to conceive example, let's say that the effect of absorption by the sample is the effect of a star shining in the sky: using a continuum light source implies watching it through the midday sky. A HCL lamp would be a midnight sky, in the metaphor.
Note that, aside from what textbooks say, continuum souce AAS (HR-CS AAS) does exist, but it requires, as expected, a sophisticated monochromator (high resolution monochromator, with a resolution of a few pm!)

ah right, so in essence the monochromator isn't able to split the light well enough so we try to limit the wavelengths?

Exactly. We have a room full of each possible green fruit (a large frequency band), and we want to extimate by sight how many pears are being eaten over time. That's hard!
HCL gives you a room with some pears to analyze :)

One of my good acquaintances spent a considerable amount of time on continuum sources in atomic absorption spectrophotometry. It is very much possible, perhaps there is one commercial instrument as well by Jena (Germany). The source was a xenon arc lamp. In those cases, it is better to use a special monochromator, which is called an echelle monochromator. It is extremely high resolution one. This is one of the reasons why hollow cathode lamps are popular as stated in the answer by The_Vinz.
Now the alternative is atomic emission- you can detect all elements at once. High end atomic emission spectrometers use the same ultrahigh resolution echelle monochromators.
The punch line as stated above is that there is no simple/ "good enough" monochromator to isolate extremely narrow lines emitted or absorbed by the atoms. A hollow cathode lamp is an intelligent choice then.

"commercial instrument by Jena (Germany)" => would that be Analytik Jena? (Jena is the town where Analytik Jena has their headquarters, but so does e.g. Zeiss and a bunch of other optics companies)

You are right, it must be Analytik Jena.


text:
Hydrogen bonding of chlorine
Even though nitrogen is less electronegative than chlorine according to Pauling scale bit why don't it form hydrogen bonding as effective as of nitrogen ? ( Electronegativity valve of  N =3.0,Cl=3.2)

In addition to the answer given, in general small elements tend to interact more strongly with other small elements. The same goes for large-large interactions.

https://chemistry.stackexchange.com/questions/31591/why-is-hcl-not-considered-to-have-hydrogen-bonding https://chemistry.stackexchange.com/questions/21773/is-hydrogen-bonding-generally-defined-to-include-only-three-period-two-elements

There is more to hydrogen bonding than electronegativity.  To get the strongest hydrogen bonds you need a compact atom to concentrate the negative charge for electrostatic attraction and to overlap well with the hydrogen 1s orbital for the molecular orbital/delocalized covalent bond contribution (yes, the latter is a thing.)  Nitrogen has that, chlorine not so much.  So, you get stronger hydrogen bonds with nitrogen than with chlorine.
Did you catch the wording?  Though chlorine has disadvantages it can form (weaker) hydrogen bonds under some conditions.  See here; the abstract indicates that the interaction can exist in some chlorinated organic compounds.