text:
Calculating numbers of moles at equilibrium given the volume of mixture and initial concentration of reactants

The equilibrium between three substances, A, B and C Is shown below.
$$\ce{A(g) + B(g) <=> C(g)}$$
Initially there were $\pu{0.1 mol}$ of A and $\pu{0.2 mol}$ of B in the reaction mixture. A and B reacted together to produce an equilibrium mixture containing $\pu{0.04 mol}$ of C. The total volume of the mixture was $\pu{2.00 dm3}$. Calculate the number of moles of A and B at equilibrium.

I honestly don't know how the answers are $\pu{0.08 mol}$ and $\pu{0.18 mol}$. I got the answers to be $\pu{0.06 mol}$ and $\pu{0.16 mol}$ by subtracting $\pu{0.04 mol}$ from each of $\pu{0.1 mol}$ and $\pu{0.2 mol}$.
Can someone please help?

Always consider the option there is an error in the task description or the key results.

Does that mean that 0.06 and 0.16 are correct?

Do you really need to ask for verification of trivial subtractions ? If anybody tells you 0.1-0.04=0.08, you must know they are wrong. They have probably erroneously omitted coefficient 2 in the equation.

OK I thought the method I used was wrong. Thank you for clarifying.

If given answers are correct, as Poutnik pointed out, your equation should be : $$\ce{A (g) + B (g) <=> 2C (g) }$$.

Could you explain how you arrived at that conclusion ?

I upvoted the question because it is surely a question that can only be answered (as it has been, in the comments) by someone who has a good grasp of the meaning and the wording of the question and the given answer, and how the answer or the question can be misleading or just wrong.

Following a discussion that took place in the comments:

If given answers are correct, as Poutnik pointed out, your equation should be :
$$\require{cancel}\ce{A(g) + B(g)<=>2 C(g)}$$ - Mathew Mahindratne
Could you explain how you arrived at that conclusion? - Ollie

This answer solves this question for a general case:
$$\ce{a A(g) + b B(g) <=> c C(g)}$$
Let's assume that initial concentrations of A, B, C are $[A_0]$, $[B_0]$ and $[C_0]$ and the final concentration be $[A]$, $[B]$ and $[C]$.
Now, the given variables are $[A_0]$, $[B_0]$, $[C_0]$ and $[C]$, assuming similar data to the question. Let $\mu$ be change for every mole of C
\begin{align}
\therefore\ &\text{Change}=[C]-[C_0] \tag{1}\\
\implies &\text{Change/mol of C} = \frac{[C]-[C_0]}{c}\tag{2}
\end{align}
Now, building a RICE table we get:
\begin{array}{lc} \hline
 &\ce{a A &+ &b B &<=> &c C} \\ \hline
I & [A_0] && [B_0] && [C_0] \\
C & -a\cdot\frac{[C]-[C_0]}{c} && -b\cdot\frac{[C]-[C_0]}{c} && +c\cdot\frac{[C]-[C_0]}{c} \\
E & [A_0] - a\cdot\frac{[C]-[C_0]}{c} && [B_0] - b\cdot\frac{[C]-[C_0]}{c} && [C_0] + [C]-[C_0] \\ \hline
\end{array}
Therefore, the final concentrations of A and B would be:
\begin{align}
[A] &= [A_0] - a\cdot\frac{[C]-[C_0]}{c} \\
[B] &= [B_0] - b\cdot\frac{[C]-[C_0]}{c}
\end{align}

Now, onto the specific solution, how to figure out $a$,$b$ and $c$ given the above convoluted generalisation?
Point 1: The change in $[A]$ and $[B]$ are both equal. Looking at change in the above RICE table, we deduce:
\begin{align}
a\cdot\cancel{\frac{[C]-[C_0]}{c}}&=b\cdot\cancel{\frac{[C]-[C_0]}{c}} \\
\implies a&=b
\end{align}
Point 2: We notice that for the change of $\pu{0.04 mol}$ of C, we observe a change of $\pu{0.02 mol}$ in A (or B, both work). Similar to point one, we equate the equations of change of A and C.
\begin{align}
2a\cdot\cancel{\frac{[C]-[C_0]}{c}}&=c\cdot\cancel{\frac{[C]-[C_0]}{c}} \\
\implies 2a&=c
\end{align}
Now, we substitute the values of $a$,$b$ and $c$ in terms of $a$, we get:
$$\ce{\cancel{a} A + \cancel{a} B <=> 2\cancel{a} C}$$
which implies:
$$\ce{A + B <=> 2C}$$
Therefore, your textbook is clearly in the wrong, since it states $\ce{A + B<=>C}$.

It seems to me to be overly complicated. It could be described in just few lines.

@Poutnik I know.. You can just drop the entire change in C part and talk about ratios instead of actual values. I have no idea why I made it so complicated. I was sleepy at the time.. Also, I want to make sure that there would be no more doubts on OPs side..


text:
What explains the case hardening color effect?
Year 11 chemistry student here.
I'm curious about how the case hardening effect works on metals.
I know that when the electrons gain an energy 'level' or lose an energy 'level' they emit light at differing frequencies.
So why does the outside of metal, after being heated, stay those pretty rainbow colors after they cool down?

I have never heard about "case hardening", but rainbow colours are interference patterns of reflected light on thin oxide layers.Nothing to do with the actual metal, except that it is below the oxide and reflecting.

Steel tempering, rainbow colors, and electron transitions are very different things, almost unrelated to each other. That is, tempered steel does show some colors, but it doesn't tell us much of its internal structure and the mechanism of hardening. Also, thinking of colors in terms of electronic transitions doesn't help us much in this case.

Case hardening (carburizing or induction or flame) does not color steel. You are seeing “temper” colors which are unrelated to case hardening. It is caused by different thickness of very thin layers of iron oxide. These colors were developed during tempering heat-treatment after case hardening. Starting with yellow, then blue and purple in a continuous transition. Faint yellow starts about 300 °F, longer times give more color development of color.
The same colors also develop on stainless steels and titanium at different temperatures and times. A stainless motorcycle exhaust pipe will show the range of colors as temperature changes along its length. If a titanium weld shows yellow color it has been exposed to air at elevated temperature (and is likely defective because of absorbed oxygen and nitrogen).

blacksmith37's answer is quite correct. An article from 2015 (Ref 1) shows the colors developed on stainless steel over a period of 10 minutes at 650ºC in laboratory air.

Ref 1. https://www.tandfonline.com/doi/abs/10.1179/0960340914Z.00000000083   This page lists several options, one of which is to download several articles; the one from which the color chart was obtained is the first, by Higginson.

source: Stack Exchange

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

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text:
Strength of nucleophiles among SN1 and SN2 reactions
As per the theory of chemistry, we say that unstable carbocations undergo SN2 reactions as it involves strong nucleophile. As a consequence, it does not lead to the formation of reaction intermediate, i.e. a carbocation.
Similarly, in stable carbocations undergo SN1 reaction by substitution of a weak nucleophile.
Why necessarily weak nucleophiles are involved? In case of SN2, since the carbocation is unstable, a strong nucleophile which prevents the formation of carbocation. But in SN1, since stable carbocation is involved, why shouldn't both strong and weak nucleophiles be involved?

Don't use Capital letters in titles.

I find your question confusing. Could you clarify it?

Why necessarily weak nucleophiles are involved - not always: you can have an SN1 reaction with strong nucleophiles as well eg cyanide.

@AniruddhaDeb Doesn't cyanide in Sn1 conditions form isonitrile compound? Since it is ambient I mean.

@AdityaRoychowdhury you mean ambivalent? yes, it can form an isonitrile, but if a covalent cyanide eg silver cyanide is used. With KCN/NaCN it forms a nitrile.

This seems to be true only because SN1 reactions prefer polar protic solvents which stabilize both the leaving group and the carbocation. But such solvents also surround strong nucleophile and hinder it from the carbocation.
Weak nucleophiles on the other hand don't interact with protic solvents as much and thus stand a better chance of getting close to carbocation.

This feels like a comment more than an answer.. Some references attached to it would make it better. Not downvoting since it seems to answer the question though..

@Safdar, not sure what to reference for generic chemistry question. I added links to wikipedia, hopefully this is enough.

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text:
Turn Alpha Helix into Beta Pleat
Beta sheets have the peptide backbone parallel whilst the alpha helix has a curled backbone.
But the thing I can't get around my head is why would the same protein always fold into one of these shapes?
It looks to me like I could grab an alpha helix and just straighten it out to form a beta pleat. I can't see anything that wouldn't work - so if I could turn alpha into beta, then what ensures that these proteins will fold the same way each time they're made? I think there must be something that prevents me from turning alpha into beta but I'm not sure what.

This paper might have more details on the conversion of alpha helices to beta sheets.

You ought to read a bit further in your textbook. No more than one and a half pages, I guess.

Look at https://www.ncbi.nlm.nih.gov/books/NBK22342/figure/A355/?report=objectonly in this chapter: https://www.ncbi.nlm.nih.gov/books/NBK22342/

In the alpha helix structure as well as the beta pleated structure, the atoms are arranged in such a way that the number of hydrogen bonds is maximum. H bonding is a stabilizing factor, and consequently, any molecule (or polymer) would like to maximize their number. This answers your question as to why the protein would fold into one of these shapes.
Coming to why alpha and beta structures are not easily interconvertible, it is once again because of hydrogen bonding. The H bonds between H and electronegative atoms like oxygen and nitrogen are difficult to break, and hence we cannot simply smash an alpha helix into a beta pleated structure, since in doing so, we would inevitably have to break some hydrogen bonds. Thus, whether a protein exists in alpha or beta form depends on how it was formed.

You provide a kinetic argument but not a thermodynamic one. Why "choose" one secondary structure over another?

Backbone h bonds are just one of many factors determining protein structures


text:
Atoms visualisation
In 7th class I was been taught that in an atom, electrons revolve around shells and the electrons with higher energy level are placed likewise in the shell above of that electron:

In 11th class I have been taught that electrons in atom are found in orbitals within shells.
But I am finding it difficult of how to visualise this electrons within shells with different energy level.
Can you provide me a simple way of visualising this atomic model?
As I know the atom seen through microscope is too complicated.

If you search for 'afm molecule images' you will see lots of atom's images by atomic force spectroscopy. These are experimental observations as opposed to the image you give which is not representing anything real other than in a very very approximate way.

Duplicate of https://chemistry.stackexchange.com/questions/33971/how-do-i-visualize-an-atom

Does this answer your question? How do I visualize an atom?

I have been taught that electrons in atom are found in orbitals within
shells

This sentence is perhaps the source of all problems. Atomic structure has a history of 100 years. Best minds of the world spent their life in trying to understand atoms. The problem is that in the sub-atomic world, good teachers will teach you not to make a literal picture of anything. Unfortunately, this is not the case in schools, colleges and early university classes. Often a literal picture is presented.
If we were by some magical or miracle slice see the atoms and the electrons, you may not see a particle going around a nucleus or the electrons arranged like onion rings around the nucleus. Forget about all these imaginations.
Contrary to the common opinion, orbital was not coined by Schrodinger. Robert Mulliken (another Nobel laureate) proposed this term.
An orbital is not a physical object or structure inside an atom, no matter how many useless general chemistry books or websites show them as balloons and other drawings. Even leading journals fell prey to this non-sense saying that orbitals have been observed or imaged, only to be embarrassed later. Orbital is a short form name of "one-electron orbital wave function". It is a mathematical construct which exists on paper and in the mind of a physicist.
Roughly you can imagine orbital as a pattern of electron density around a nucleus. But again, there is no actual " cloud" of an electron surrounding the nucleus (if we were able to see it). Do not take the small world of particles in a literal sense. Much is known and much more is not known.
The only real thing which one can perhaps observe is electron density using X-ray techniques. Electrons are indistinguishable particles. You cannot say this is a K-shell electron or P-shell electron. What is available experimentally is that not all electrons have the same energies when they are bound to the nucleus.
See Moseley's experiments on Youtube and if you are more interested see X-ray photoelecton spectroscopy.

An atom cannot be seen with a microscope, using ordinary light. Grains of light, also called photons, are simply too big to give an image of atoms. As it is difficult to understand for beginners, I could use a simple image.
Imagine you are living in a cavern like our ancestors, the prehistoric men. To draw a hand on the wall of the cavern, they filled their mouth with wet charcoal powder, and they spit this black mixture on the hand against the wall. They obtain a black-and-white image of the hand. If they had used pingpong balls or basket balls, instead of charcoal powder, they would never have obtained an image. These balls are simply too big to pass between the fingers.
It is exactly the same for us. Photons from usual light are too big to allow us take an image of the atoms.  Much smaller photons are needed. For example, photons from the X-rays are small enough to allow a photographe of the atoms, But these X-rays are not part of the usual spectrum violet-blue-green-yellow-orange-red. So, unfortunately we will never develop a microscope which could show us atoms with ordinary light.
Now, to go back to your question, the first scientist to give a description of the electron in an atom is Niels Bohr. He proposed that electrons are turning around the nucleus like the planets around the Sun. He made calculations about the mass, the charge, the speed of the electron turning around the Hydrogen nucleus, and the distance electron-nucleus. These calculations give good results with the Hydrogen atom when compared to energies and distances as measured by X-ray or similar techniques.
So this theory was soon extended to other atoms giving electrons rotating on circular orbits as you know. But the calculations made with other atoms gave results that are not so good. Qualitatively they are good enough to make drawings like the page you mentioned, where electrons are like planets. It allows to show nice description of molecules, using Lewis method. You probably know these drawings, with dots and bonds.
Today we think that electrons are not turning. They "exist" on sorts of clouds, called orbitals, centered around the nucleus, and whose average center of mass is near the nucleus for the first electron, so that the elctrons are more and more repelled away from the nucleus, when an atom has a lot of electrons. Alas, this description is not easy to visualize. That is why high school chemistry is still often taught with the old Niels Bohr theory, although it is not perfect. Showing the shape of these orbitals in 3 D is not easy to do at a high school level.

Thank you but I was asking how to visualise an atom in a simple way possible

This is not the point of this question.

x-rays are not used to photograph atoms but atomic positions are determined by diffraction from crystals. As scattering is proportional to electron density different types of atoms can be distinguished from one another.

@porphyrin The picture is even worse: the atomic positions modeled represent an average, both across the space (all the atoms of the accessed unit cells of the sample) and time because (even if using the N2 cooling nozzle cools the sample) atoms vibrate around equilibira positions, which is seen in the non-spherical ellipsoids draw, the ADP.

Hydrogen atom and X-rays- they don't go well together.

@Buttonwood Quite so. How can one put it without being rude or gratuitously critical;  most users have no idea how the software works and there does need to be a concerted  effort to examine and then update algorithms to incorporate new ideas.

All you say is perfectly right. I was expecting your critical reactions. But try to be one minute in front of a class of non-scientific students, who are not really interested by science and chemistry ! You have to teach them the basic concepts of the atomic structure. You simply have to convince them that the atom cannot be seen. What do you do ? Of course you may to teach them the usual theory. They will not been convinced about the value of your development. But if you mention the prehistoric painting, they accept the message, even though it is not perfect. See next comment !

Continuation. Of course you may try to teach the usual theory to the average student. They will not been convinced about the value of your development. But if you mention the prehistoric painting, they accept the message, even though it is not perfect. I know that Hydrogen atom and X-rays do not get well together. I also know Bragg's law, and the law of X-ray diffraction by crystals. I teach it to advanced students. But a good teacher must find a way to be on a level with his students. That is for the average or poor student.

@porphyrin It is a general problem. To seldom a paper addressing «what can be done wrong» (https://doi.org/10.1080/0889311X.2019.1569643), or an audience already sensible to think and act accordingly.  Because (not only, but in part) because it takes time and effort to digest critical thinking and then to act accordingly while today's distractions are literally on your fingertips 24/7 (e.g., cell phone).  The attention span among the younger tends to shorten, too.  If you can't get them, be happy if they just knit silently a scarf with little chitchat with their neighbours during the lectures.


text:
Kinetics question regarding Rate Law

The acid catalysed hydrolysis of an compound $\ce{A}$ at $\pu{303K}$:
$$\ce {A ->[$k$] B}$$ has a half-life of $\pu{100 min}$ when carried out in a buffer solution of $\mathrm{pH=5}$ and $\pu{10 min}$ when carried out at $\mathrm{pH=4}$. Both the times the half-life are independent of the initial concentration of $\ce{A}$. Find its Rate Law.

I am planning something like this:
Let $$\text{Rate} = k [\ce{A}][\ce{H+}]^x\tag1$$
Now let:
$$k'=k[\ce{H+}]^x\tag{constant}$$
And then:
$$k' t_\frac 12 = \ln2$$

Is it conceptually correct?

Note:

I have found tons of solutions over the internet all with different answers.

The answer given in my book is x=1 (x is same as is used in eq(1)) and I found the same myself but some sites give answer of the same question as x=2.

Umm, you have a rate law in first eq. so...

could you share the different answers given? just a link to contradictory answers would be enough..

@Safdar Check the note section in my question.

@ Tony Stark. As the rate is independent of the concentration of A, it means that the rate is only dependent on the concentration of $\ce{H+}$ and first order.

We can do this question in two steps, finding the rate law with respect to both A and $\ce{H+}$ individually.
$$\text{Rate}=[A]^y[\ce{H+}]^x$$

The first part is simple, they've given that the half life is independent of $[A]$, therefore it is first order with respect to $\ce{A}$.
Therefore as you've found,
$$\text{Rate}=[A][\ce{H+}]^x$$

Now, onto finding the order with respect to $\ce{H+}$.
From what I've seen on the contradictory answers on the net that OP has mentioned, they state that since the rate is inversely proportional to concentration of $\ce{H+}$, it is second order with respect to $\ce{H+}$ by the relation:
$$t_{1/2}\, \alpha\, [A]^{1-n}$$
where $n$ is the order with respect to A in a reaction.
Is this true? Yes. Is it correct in this context? No. This is because this is true only when the reactant in question actually decreases in the mixture, i.e.
$$-\frac{\mathrm d[A]}{\mathrm dt}=k[A]^n$$
Here, it is specifically given that it is in a buffer of $\mathrm{pH} = 4$. This means that there is no change in the actual concentration of $\ce{H+}$.
Therefore, what you've done is conceptually sound and is the correct method.
$$k'=k[\ce{H+}]^x = \text{constant}$$
\begin{align}
\frac{t_{1/2}(\mathrm{pH}=5)}{t_{1/2}(\mathrm{pH}=4)} &= \frac{k'(\mathrm{pH}=4)}{k'(\mathrm{pH}=5)} = \frac{k[\ce{H+}]^x(\mathrm{pH}=4)}{k[\ce{H+}]^x(\mathrm{pH}=5)} \\
\implies \frac{100}{10}&=\left(\frac{10^{-4}}{10^{-5}}\right)^x = 10^x \\
\implies x &=1
\end{align}
Therefore, the rate would be:
$$\text{Rate}=k[A][\ce{H+}]$$

There is a similar question -- Rate Law for inversion of cane sugar -- https://chemistry.stackexchange.com/questions/104543/how-to-derive-rate-law-expression-for-inversion-of-sucrose-with-changing-ph#comment196359_104543  . Applying the same method does not give me the answer. Please look into it.

@TonyStark It is the exact same question, poorly formatted.. except A and B are given (conditons seem similar as well). Did you read the comments? this is same.. Also, I seem to get the correct answer there as well.

Check this : https://www.toppr.com/ask/question/the-inversion-of-cane-sugar-proceeds-with-the-halflife-of/ . This is the same question as https://chemistry.stackexchange.com/questions/104543/how-to-derive-rate-law-expression-for-inversion-of-sucrose-with-changing-ph#comment196359_104543 but well formatted.

@TonyStark What part of my above statement did you not understand? "Is this true? Yes. Is it correct in this context? No. This is because this is true only when the reactant in question actually decreases in the mixture," the other answers are wrong.. the internet isn't always right.. one group adds a solution, others plagiarise. nobody cares if they are correct since they aren't community vetted..

Actually I was looking for this statement clearly stated by some experienced person - " the other answers are wrong.. the internet isn't always right.. one group adds a solution, others plagiarise."

@TonyStark you didn't need to wring that out of me.. Did you? ;D

Let us continue this discussion in chat.


text:
How to use NMR readings to determine the molecular structure of a compound?
A question in my exam came as follows:

How is NMR used in the determination of structure of organic compounds?

I have studied NMR spectroscopy and I know how to determine values using an NMR spectroscope. However, none of my textbooks talk about how to actually use these readings to determine the actual molecular structure of the compound that is under study. My professor has not taught this either. I was unable to find any external sources that talk about this as well.
Could someone explain the steps on how to determine the molecular structure of a compound after knowing the spectrum?

Then you should already know all you need about that.

I’m closing this question because Stack Exchange isn't an essay-writing service.

Let us continue this discussion in chat.

In addition to @M.Farooq 's advice, my recommendation is to flip the problem on its head and use your intuition. If you know what parameters can be measured by NMR and their origin, then think about how those parameters might be used to determine (intuit) structure. For instance, shifts depend on and therefore give you information about the local chemical environment, including bonding, vicinity of substituents, etc. J-couplings (including peak multiplicity) depend on and therefore give you information about connectivity (bonding networks).

Browsing this site and looking up NMR questions and answers might help give you an idea of how nmr is used. Your question as it stands is a little broad. There isn't just "the spectrum", there is a slew of possible experiments that can be performed depending on the complexity of the molecule.


text:
Finding the chemical formula for an ionic compound made using iridium and nitrite ion

What is the most likely chemical formula for an ionic compound involving the elements iridium and nitrite?

The charge on the polyatomic ion $\ce{NO2}$ is $-1$, whereas $\ce{Ir}$ is a transitional metal that has more than one oxidation state, and in this case it is $+1$, I believe.
Is $\ce{Ir(NO2)}$ the correct answer?

No it isn't, and this question doesn't make sense.

I think it is unlikely that any of such compound would exist.

Nitrite is not an element, and the question is indeed weird at best.

Iridium forms many oxidation states, +3 and +4 are the most common. https://en.wikipedia.org/wiki/Iridium

There is possibility of a redox reaction between both.

The exact composition of iridium nitrite hasn't been determined because the compound has been poorly studied, although it is predicted to be $\ce{Ir(NO2)3}$ or $\ce{Ir(NO2)4}$ as +3 and +4 are the most common oxidation state. It is prepared by reacting iridium oxide with sodium nitrite at 333 K(1) or digesting "chloro compound of iridium" with hot sodium nitrite at pH 1.5(2). The compound is mostly studied in the form of double salts(3) and complexes(4).
I did find another compound called Hydrogen Iridi-nitrite which serves as a starting material for the formation of metal-iridium nitrite double salts:

Hydrogen Iridi-nitrite, $\ce{H3Ir(NO2)6}$, is stated by Gibbs to crystallise in yellow prisms which are readily soluble in water. The hydrogen may be replaced by metals yielding a series of salts known as iridi-nitrites, of general formula $\ce{M3Ir(NO2)6}$, and analogous to the cobalti- nitrites and rhodi-nitrites already considered.

References

ELECTRODEPOSITION OF IRIDIUM, G SHEELA, MALATHY PUSHI'AVANAM AND S PUSHPAVANAM,
Bulletin of Electrochemistry 15 (5-6) May-June 1999, pp 208-210 (PDF)
Separation Procedures in Inorganic Analysis: A Practical Handbook, Roland Stansfield Young,Wiley, 1980

Double nitrite of iridium and rhodium: Inorganic analysis, Analyst, 1901,26, 107-110, DOI: 10.1039/AN9012600107
Ammonium iridi-nitrite: Journal of Research of the National Bureau of Standards, Volume 10, U.S. Government Printing Office, 1933

The dissolution and recovery of iridium in a nitrate system, G.L.Silver, Journal of the Less Common Metals, Volume 40, Issue 2, April 1975, Pages 265-267, DOI: 10.1016/0022-5088(75)90069-7


text:
Finding concentration of hydronium ions in water after adding hydrogen chloride gas

A sample of hydrogen chloride gas, $\ce{HCl}$, occupies $\pu{0.932 l}$ at a pressure $\pu{1.44 bar}$ and at a temperature of $\pu{50^\circ C}$. The sample is dissolved in $\pu{1 l}$ of water.
What is the resulting hydronium ion ($\ce{H3O+}$) concentration?
Ammonia gas also dissolved quantitatively in water.
If it is measured at $\pu{0.720 bar}$ and $\pu{50^\circ C}$, what volume of $\ce{NH3}$ gas is required to neutralize the solution prepared in the above question?
For present purposes, assume that the neutralization  reaction occurs quantitatively.

My attempt:
$\pu{0.05 mol}$ of $\ce{HCl}$ is dissolved in the $\pu{55.55 mol}$ of water. Balancing equation is
$$\ce{HCl(g) + H2O(l) <=> H3O+ + Cl-(aq)}$$
Is there a mistake in this assumption? I ask this because I am unable to proceed further.

Did you forget to complete the attempt, or is this it?

@Safdar, I didn't forget to complete the attempt. I just don't have any idea to proceed further. I know molarity formula, but i am doubtful about its usefulness here.

I've removed the second part of your question since it is basically the same thing but using $\ce{OH-}$ instead.. Also, HCl is a strong electrolyte. Does this help?

@Safdar, I think the second part is required because the author is asking what volume of ammonia gas $\ce{NH3}$ is necessary to neutralize the solution created in first part.

Is there a reason to suppose you get negatively charged gas consisting of chloride ions, while solution would be charged positively ? BTW, I replaced the character you have used by the regular minus to be displayed properly.

For the task purpose, you need not to work with water molar mass nor molar amount of 1 L of water.

You correctly found that amount of $\ce{HCl}$ gas in the container is $\pu{0.05 mol}$. When it dissolves in $\pu{1.0 L}$ of water you get $\pu{0.05 M}$ $\ce{HCl}$ solution. Does it matter for this calculations? No, but I did it anyway. However, it gives us very important information. $\ce{HCl}$ is a strong electrolyte and it dissolves in water as high as $\pu{12 M}$. When dissolve in water, it completely dissociates:
$$\ce{HCl (g) + H2O (l) -> H3O+ (aq) + Cl- (aq)} \tag1$$
For the second part of the question, you need to know what is the reaction between $\ce{HCl}$ and $\ce{HH3}$:
$$\ce{HCl (aq) + NH3 (aq) -> NH4Cl (aq) -> NH4+ (aq) + Cl- (aq)} \tag2$$
Thus, mol% of the reaction is $1:1$. Therefore, the amount of $\ce{HH3 (g)}$ you need to complete neutralization is the amount of $\ce{HCl}$ in the solution, which is $\pu{0.05 mol}$. Now, you can use ideal gal law to calculate the $V$ under given conditions.

Note: Ammonia in water is in equilibrium with:
$$\ce{NH3 (g) + H2O (l) <=> NH4+ (aq) + OH- (aq)} \tag3$$
The complete nutralization is the reactions of $(1)$ and $(3)$ where net ionic reaction is:
$$\ce{H3O+ (aq) + OH- (aq) -> 2 H2O (l)} \tag4$$
However, since $\ce{NH3}$ is a weak base, its conjgate acid $\ce{NH4+}$ is acidic. Therefore, this titration would not pass $\mathrm{pH = 7}$ mark at the end point (thus need an appropriate indicator).

So, the concentration of $\ce{H3O+}$ in the solution of HCL created in first part is $\pu{0.05 mol l^{-1}}$. Isn't it?

That is the key ($\ce{H+ + H2O -> H3O+}$).

Sorry, I am beginner in chemistry study. So i don't know what is $\ce{HH3}$ and $\ce{HH4+}$? Would you tell me?

See equation (3) under notes (similar to $\ce{H2O}$ and $\ce{H3O+}$).

So, the volume of ammonia gas $\ce{NH3}$ required to neutralize the solution prepared in the first part at 0.720 bar and $50^\circ C$ is 1.866 liters. Isn't it?

That is correct. Best way to do word problem is better understanding of the problem and try what you can do with the date with your knowledge of chemistry. Good luck!


text:
What's the usual agent added to discourage drinking of hazardous substances
I've once read somewhere that in materials that are hazardous to swallow, a chemical is used in very small quantities to assure any accidental drinker to retch. This also helps to prevent suicide by drinking. Am I correct about this info or just myth? And what is the name of the substance?

Maybe you're looking for a class of substances called emetics, not a specific compound per se. Or you're thinking of alcohol denaturing.

Denatonium benzoate (the most bitter compound) is added to toxic alcohols like methanol or ethylene glycol in order to avoid its consumption.

Ipecac syrup is an effective medicine.

Many compounds are added to dangerous substances that might be accidentally or deliberately ingested
There isn't just one, universal, answer to this question.
The problem is that there are a variety of different hazards and specific countermeasures may be required for each. And ingestion of some dangerous materials occurs deliberately not just accidentally and providers need to act to minimise those risks as well as the accidental ones.
Take the classic example of the weedkiller paraquat. It took a while after it was first introduced for people to notice its toxicity, possibly after accidental ingestion of the solution which was often stored by domestic users in old bottles previously used for soft drinks. It is fairly deadly if ingested and allowed to get into the bloodstream.
Initial solutions were to add material such as Bitrex

which is so bitter it should provide adequate warning that the solution should not be drunk. But, once the deadly nature became public knowledge, many people deliberately took paraquat as a method of suicide. This prompted the addition of other ingredients designed to minimise the hazard even when taken deliberately.
The first key step was to add an emetic (designed to encourage the user to vomit, reducing the volume of the solution that can be absorbed in the stomach.) FAO agricultural standards actually specify the nature an emetic requires for paraquat though only one compound met the standard:

The compound known as PP-796 or 2-Amino-5-methyl-7-propyl-7H-1,3,3a,7-tetraazainden-6-one is a powerful, fast acting emetic.
Even this, though, is not fully effective and the manufacturers have explored a variety of other strategies including adding a natural alginate to the formulation that gels in the acid environment of the stomach slowing the absorption into the body.
While all these strategies have been applied to the particularly hazardous compound paraquat, they are representative of the other steps manufacturers will take to minimize the risk of accidental or deliberate use of their products.

It is denatonium benzoate. This comes from wikipedia page of denatonium:

Denatonium, usually available as denatonium benzoate (under trade
names such as Denatrol, BITTERANT-b, BITTER+PLUS, Bitrex or Aversion)
and as denatonium saccharide (BITTERANT-s), is the most bitter
chemical compound known, with bitterness thresholds of 0.05 ppm for
the benzoate and 0.01 ppm for the saccharide. It was discovered in
1958 during research on local anesthetics by MacFarlan Smith of
Edinburgh, Scotland, and registered under the trademark Bitrex.

With respect to discouraging drinking of hazardous substances it again comes from wiki:

The bitterness of the compound guides most applications of denatonium.
Denatonium benzoate is used to denature ethanol so that it is not
treated as an alcoholic beverage with respect to taxation and sales
restrictions.


text:
Fischer indole synthesis: significance of choice of acid catalyst
Lately, as a matter of interest, I have been studying sigmatropic rearrangements. I came across the Fischer indole synthesis, whose Wikipedia article states:

The Fischer indole synthesis is a chemical reaction that produces the aromatic heterocycle indole from a (substituted) phenylhydrazine and an aldehyde or ketone under acidic conditions. ... The choice of acid catalyst is very important. Brønsted acids such as $\ce{HCl}$, $\ce{H2SO4}$, polyphosphoric acid and p-toluenesulfonic acid have been used successfully. Lewis acids such as boron trifluoride, zinc chloride, iron chloride, and aluminium chloride are also useful catalysts for this reaction.

The significance of the choice of acid catalyst to be employed in the reaction is beyond my scope of knowledge. Browsing available articles, it would seem that the [3,3]-sigmatropic rearrangement is stereoselective. In addition, if I am not mistaken, regioselectivity of the Fischer indole synthesis could be increased depending on which acid catalyst was used, such as methanesulfonic acid. However, I was still unable to understand the overall importance of using an appropriate acid catalyst.

I'd like to take this opportunity to give a nutshell review of the Fischer indole synthesis for benifit of future readers. This highly useful synthetic tool, which was named after the famous chemist, Emil Fischer who discovered it in 1883 (Ref.1 & 2)), is considered to be one of the best methods for preparing indoles. Hence, needless to say, it is the most widely used and versatile method for indole synthesis. The method simply converts converts ayrlhydrazones ($\bf{1}$) of an aldehyde or a ketone into corresponding indole ($\bf{8}$) in the presence of an acid catalyst (Ref.3 & 4). The overall reaction is depicted in the top box in following image:

The proposed mechanism of The Fischer indole synthesis is also included for the convenience. As indicated, ayrlhydrazones ($\bf{1}$) are formed from a condensation reaction between an arylhydrazine and an aldehyde or ketone (reference to 2,4-dinitrophenylhydrazine test for aldehydes and ketones). In the presence of acid (Brønsted or Lewis), imine $\bf{1}$ is in equilibrium with $\bf{2}$ (protonated imine), $\bf{3}$ (enamine form), and $\bf{4}$ (protonated inamine). Once formed, $\bf{4}$ undergoes [3,3] sigmatropic rearrangement (ene-reaction; could visualize as azo-Cope rearrangement), forming a new $\ce{C-C}$ bond to give $\bf{5}$ (this is a key step of the mechanism). Imine $\bf{5}$ gains back aromaticity to form $\bf{6}$, which is the most stable isomer between two. The anilinic $\ce{NH2}$ of $\bf{6}$ attacks electron deficient imine group in the side chain ($\ce{C}$ of $\ce{CNH2+}$ group) intramolecularly to give the corresponding dihydroindole derivative, $\bf{7}$ (which is also another key step), which eventually release $\ce{NH4+}$, $\ce{NH3 + H+ <=> NH4+}$, to give stable indole derivative,  $\bf{8}$. Reference 3 has given excellent review on the mechanism and hence, for benefit of interested reader, the PDF has been attached.
The key feature to produce corresponding indole is need of a $\ce{CH2}$ group as noted (as indicated by a green circle; $\alpha$ to the original ketone or aldehyde). If this $\alpha$-$\ce{C}$ is $2^\circ$, then $\bf{7}$ would still release $\ce{NH3}$ to give corresponding cyclic imine derivative (Ref.5). For example, compound $\bf{10}$ was formed ( Ref.5) when $2^\circ$ $\alpha$-$\ce{C}$ is attacked selectively at $\bf{3}$ to $\bf{4}$ conversion (thermodynamically stable enamine):

However, except for the situation to make $\bf{10}$, and the intention is making indole derivative of interest, the stereo selectivity at sigma-tropic rearrangement to give $\bf{5}$ and thereafter would not matter since at the end, all of stereocenters involved would be diminished (see the scheme above). Therefore, the choice of acid to get selectivity in general synthesis of indoles is not as important as it was told elsewhere. In reality, a variety of catalysts (Table I in Ref.3) have been used to effect the indolization of arylhydrazones. According to the Ref.3 (an excellent review of the Fischer indole synthesis):

Generally speaking, it may be said that acids and many compounds capable of complex formation with organic ligands may act as catalysts.

In addition to the variety of catalysts listed in Table I of Ref.3, there are wide variety of other acid catalysts including organic acids such as tartaric acid (e.g., Ref.5) and p-toluenesulfonic acid (e.g., Ref.6). The bottom line is the indole formation by this method is independent of acid catalysis (e.g., Ref.7), although it is markedly accelerated by it.
It is also worth noting that this reaction has proven very useful for the pharmaceutical synthesis of antimigrane drugs (e.g., Ref.8 & 9).

References:

Emil Fischer, Friedrich Jourdan, "Ueber die Hydrazine der Brenztraubensäure," Berichte der deutschen chemischen Gesellschaft 1883, 16(2), 2241-2245 (https://doi.org/10.1002/cber.188301602141).
Emil Fischer,  Otto Hess, "Synthese von Indolderivaten," Berichte der deutschen chemischen Gesellschaft 1884, 17(1), 559-568 (https://doi.org/10.1002/cber.188401701155).
B. Robinson, “The Fischer Indole Synthesis,” Chem. Rev. 1963, 63(4), 373–401 (https://doi.org/10.1021/cr60224a003)(PDF).
B. Robinson, “Studies on the Fischer indole synthesis,” Chem. Rev. 1969, 69(2), 227–250 (https://doi.org/10.1021/cr60258a004).
Sangram Gore, Sundarababu Baskaran, Burkhard König, “Fischer Indole Synthesis in Low Melting Mixtures,” Org. Lett. 2012, 14(17), 4568–4571 (https://doi.org/10.1021/ol302034r).
Seble Wagaw, Bryant H. Yang, Stephen L. Buchwald, "A Palladium-Catalyzed Strategy for the Preparation of Indoles:  A Novel Entry into the Fischer Indole Synthesis," J. Am. Chem. Soc. 1998, 120(26), 6621-6622 (https://doi.org/10.1021/ja981045r).
Okiko Miyata, Yasuo Kimura, Kanami Muroya, Hajime Hiramatsu, Takeaki Naito, "Thermal cyclization of N-trifluoroacetyl enehydrazines under mild conditions: A novel entry into the fischer indole synthesis," Tetrahedron Lett. 1999, 40(18), 3601-3604 (https://doi.org/10.1016/S0040-4039(99)00583-3).
Paul R. Brodfuehrer, Bang-Chi Chen, Thomas R. Sattelberg, Patrick R. Smith, Jayachandra P. Reddy, Derron R. Stark, Sandra L. Quinlan, J. Gregory Reid, John K. Thottathil, Shaopeng Wang, “An Efficient Fischer Indole Synthesis of Avitriptan, a Potent 5-HT$_\mathrm{1D}$ Receptor Agonist,” J. Org. Chem. 1997, 62(26), 9192–9202 (https://doi.org/10.1021/jo971368q).
Majid M. Heravi, Sahar Rohani, Vahideh Zadsirjan, Nazli Zahedi, “Fischer indole synthesis applied to the total synthesis of natural products,” RSC Adv. 2017, 7, 52852-52887 (DOI: 10.1039/C7RA10716A).


text:
Why do we use molality on describing colligative property?
I'm satisfied with the explanation about Raoult's law that the chance of vaporizing will depend on the surface exposed to the atmosphere and solute will decrease such chance by taking portion of the surface at rate of $X_{solute}$.
However, incase of elevation and depression of boiling point, I can't find any explanation why it should depends on the molality. There's a molarity, molality, mole fraction, etc.. why molality among them?
I know that since it's microscopic, we can't be sure about anything since we cannot observe. But still I'm looking for some satisfactory explanation.
And below is the example of boiling-point-elevation equation that I made by looking at the graph of tripoint and considering the fact that boiling point elevation occurs due to the vapor pressure change. Why can't we use such method? it seems accurate to me though..
Let $$P(t):T \to P, when$$ $$t_{tri}<t, t \in T (temperature), t_{tri} \in T (temperature), p\in P(pressure)$$,
$$P^{-1}(p): P \to T $$ $$p_{tri}<p, t \in T (temperature), p_{tri} \in P (pressure), p\in P(pressure)$$
$p_{tri}, t_{tri}$ are pressure and temprature at tripoint $P, P^{-1}$ are function for solvent before dissolve.
According to the Raoult's law, we can calculate vapor pressure of solution's solvent $P_{2}(t)$ as such
$$P_{2}(t)=P(t)*X_{solvent}, X_{solvent}=mol_{solvent}/(mol_{solvent}+mol_{solute})$$
Let's say that we want to know the elevation of boiling point at pressure $P(t_0)$ when $t_0$ is former boiling point. According to the Raoult's law stated above, we can say that boiling point elevation $\Delta t=P^{-1}(\frac {P(t_{0})}{X_{solvent}})-t_0$.
(I'd tried to write down the specific process. But I couldn't do it well on words. Still you can see it by looking at the graph of any tri point.)

Molality is a useful unit in colligative properties because it is independent of temperature, it is not affected by volume changes upon mixing and it will allow you to calculate mole fractions. Keep in mind that colligative properties only depend on the number of particles. Fortunately, mass is additive but volume is not.
Molarity is way more complex, where 2+2 is not equal to 4. For example mix 50 mL water+50 mL ethanol. The volume is quite less than 100 mL but if you mix 50 g of water with 50 g of ethanol, it will always be 100 g.
I am not sure why set theory and mapping notation was needed here?

It is much easier to work with molality than molarity when measuring freezing point depression. Imagine. You weigh $10$ g, or $20$ g ice (or any other amount); you add $1$ g salt, $2$ g salt, or any other amount of any sort of solute. You drop a thermometer in the mixture, and you measure the fusion temperature. It is done quickly, and the scientist does not mind about the molarity of the obtained solution.


text:
Do we need to check for all kinds of symmetry in a compound to declare it achiral?
I know that a compound is achiral if it contains any one of the following kind of symmetry.
1.Plane of symmetry
2.Centre of symmetry
3.Normal axis of symmetry
4.Alternate axis of symmetry
It is difficult to write, for many compounds, its mirror image and then check using visualisation that if the image and compound are same or not, so we check for these 4 symmetries in molecule. My question is that for any compound do we need to check for all the symmetry elements one by one till we get a symmetry possessed by the molecule to declare it achiral, for example, does there exist any compound which has only alternate axis of symmetry and not any other kind of symmetry, so that any one will have to check all the symmetries one by one, till he reaches alternate axis of symmetry, to declare it achiral?
I know compound which have centre of symmetry, but not plane of symmetry, but  I don't know about a compound which has only alternate axis of symmetry and not any other kind of symmetry. Most of the achiral compounds that I have come across have plane of symmetry or centre of symmetry. So, do we need to check for all axis of symmetry or only centre and plane of symmetry, or any other shorter combination?

https://chemistry.stackexchange.com/questions/104409/use-of-axis-of-symmetry?noredirect=1&lq=1
You need to check for actual symmetry elements..

We need to check all 4 symmetries to determine whether a compound is optically active or not.
Consider the following compound. It has only Alternating Axis of Symmetry

Note that the rings are in perpendicular planes. The following diagrams illustrate the top view of the molecule.The stereochemistry of the methyl groups can be:

$S_4$ axis of symmetry is seen only in the following diastereomer where the methyl groups are anti to each other in each ring and the view looks like a Nazi cross from the top.

Umm.. So? do we need to check or not.. you might have left that out.. also references please regarding the chirality of since it may may depend on the diastereomer taken..

Your compound has undefined chiral centers; in which direction are the methyl pointing?

I have edited the answer listing all possible cases where only one of the diastereomers has the S4 axis of symmetry

But that compound also has two fold axis of symmetry.

Each ring coincides with itself when compound is rotated 180 degree about vertical axis.

Yes, that is a $C_2$ axis. But C axes are not taken under consideration while determining optical activity/chirality

@SoumyadwipChanda A better choice of term would be 'swastik' rather than you know what :)

卐 Yeah... But the world knows it better as Nazi-Cross


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In electrophoresis, are positive and negative molecules obviously separated?
I have just studied electrophoresis in biochemistry. The charged particles move to the oppsite charged electrodes whose speed is depended on their particle size. If that so, are negative and positive particles obviously separated into two distinct boundaries? As if we will find only negative particles near positive electrode and only positve particles near negative electrode.
Some searching I have done:

We cannot separate any two positive and negative ions because of their huge Coulombic force.
Exerted from electric field, a charged particle will move to the electrode and carry   counterion cloud together.

If (2) is true, why doesn't counterion move to its electrode but, instead, move along with charged particle?
If (1) is true, how is it possible that charged particles move to the electrode without separating from its counterion?

The process is similar to electrolytic cell. Did you think about how negatively charged ions go to positive electrode while positively charged ions go to negative electrode? Isn't it there possibility of ion separation? But it happens, How?


text:
How does Al-Hg react with carbonyl in presence of concentrated HCl?
In Clemmensen reduction, $\ce{Zn-Hg}$/ $\ce{ conc. HCl}$ reduces the carbonyl $\ce{>C=O}$ to $\ce{>CH2}$, by the following mechanism(?):

What happens to the carbonyl group if we replace Zinc with any other metal such as Al or Na?

Will this reaction take place if X is $\ce{Al-Hg / conc. HCl}$ ?

A more interesting reaction to explore is first react a sheet of Al in HCl. Extract the Al sheet which now holds (for a limited time) H chemisorbed hydrogen. The latter .H surface radical acts per the reaction: H+ + e- = .H, as working with an e- followed by H+. For example, with PbS + 2 .H -> Pb + H2S. That is, Pb2+ + 2 e- = Pb and S(2-) + 2H+ = H2S. Depending on the organic exposed to the Al sheet, expect a variation of the inorganic chemistry which may (or may not) display particular selectivity to produce one or more products.  The example I cited comes from application in Hydrometallurgy.

Alluded to source reference: 'Hydrometallurgy 2008: Proceedings of the Sixth International Symposium', p. 818, a commercial reductive leaching equation, to quote:
" PbS + 2 •H = Pb + H2S (5) "  (see https://books.google.com/books?id=1etfSdk55SYC&pg=PA818&lpg=PA818&dq=what+metals+react+with+hcl+to+form+nascent+hydrogen&source=bl&ots=yEBZaIFL_T&sig=9383ZW7NCXAoSNmmny2sdrxXIF0&hl=en&sa=X&ved=0ahUKEwjEjpnWjejaAhVP4VMKHe1gBZE4FBDoAQg6MAU#v=onepage&q=what%20metals%20react%20with%20hcl%20to%20form%20nascent%20hydrogen&f=false ).

In comparision to Clemmensen reduction, both $\ce{Na-Hg}$ and $\ce{Al-Hg}$ cannot be used as direct substitutes for $\ce{Zn-Hg}$.
Reaction with $\ce{Na-Hg}$ amalgam:
$\ce{Na-Hg}$ amalgam is also a good reducing agent and is capable of reducing ketones/aldehydes to alcohols1. However, Do not use conc. $\ce{HCl}$ with $\ce{Na-Hg}$ amalgam, as there would be a vigorous reaction between the acid and the sodium in the amalgam. Instead, use an organic solvent such as absolute alcohol.
The mechanism for this reduction is a bit different than standard Clemmensen reduction:

\begin{align}
\ce{2(C6H5)2CO +2Na &->2(C6H5)2C-ONa} \\
\ce{2(C6H5)2C-ONa&<=>(C6H5)2C(ONa)-C(C6H5)2-ONa\tag{1}} \\
\ce{2(C6H5)2C-ONa + 2C2H5OH &-> 2(C6H5)2C-OH + 2C2H5ONa} \tag{2a} \\
\ce{2(C6H5)2CO &-> (C6H5)2CO +(C6H5)2CHOH \tag{2b}}
\end{align}

Reaction with $\ce{Al-Hg}$ amalgam:
Aluminium amalgam, like Sodium amalgam, vigorously reacts with water, and you cannot use concentrated $\ce{HCl}$ with it. With ethanol as a solvent, it is a mild reducing agent, can reduce cycloalkanones and aldehydes to alcohols. Acyclic ketones remain almost inert2

In some cases, it behaves similarily to $\ce{Mg-Hg}$ amalgam and promotes dimerization of ketones to form pinacols. Thus, your reaction would not take place, and instead pinacol would be the major product.

This question and this one cover some other reduction uses of $\ce{Al-Hg}$, such as reduction of imines to amines.
References:

Bachmann, W. E. “The Mechanism of Reduction by Sodium Amalgam and Alcohol. I. The Reduction of Aromatic Ketones to Hydrols.” Journal of the American Chemical Society, vol. 55, no. 2, Feb. 1933, pp. 770–74. doi:10.1021/ja01329a051.
Troyansky, Emmanuil I., and Meghan Baker. “Aluminum Amalgam.” Encyclopedia of Reagents for Organic Synthesis, John Wiley & Sons, Ltd, 2016, pp. 1–6. doi:10.1002/047084289X.ra076.pub2.


text:
Analysis of cations using sodium thiosulphate
A question in an exam was as follows:[1]

How many of the following salts are first soluble in $\ce{Na2S2O3}$ solution, followed by the formation of a white precipitate and finally the precipitate turns black on standing
$\ce{AgNO3}$, $\ce{Pb(CH3COO)2}$, $\ce{Hg(NO3)2}$, $\ce{BiCl3}$, $\ce{CuSO4}$

The answer given is:

 $5$, which means that all of them give this test.

I understand that this is a test for cations since the reagent is sodium thiosulphate.
A pre-emptive guess on what the reactions would be that the first reaction forms the thiosulphate of the cation followed by the conversion of thiosulphates into sulphides (from the color change given).
However, I was unable to find any text referencing this test and so haven't been able to verify the actual reactions that take place.[1]
What cations does this test give a positive result for? What are the reactions that take place?

[1]: The previous version of the question asked for a test using sodium dithionite which was a typo (due to reasons provided in the answer below), confirmed after M.Farooq's suggestion to ask the professor who set the exam.

I do not think the question had sodium dithionite. This is a mistake. It must be sodium thiosulfate, because sodium dithionite is such a strong reducing agent that it will not solubilize these metals. The properties mentioned in the answer match with thiosulfate.
Sodium dithionite is such a strong reducing agent that it will reduce copper (II) to copper.
With this correction...
Interesting question but requires some knowledge of colors of sulfides. Sometimes you would have to use intuition in such exam questions. The hint is "...finally turn black on standing". Recall that silver tarnish is brown black, so are lead, copper and bismuth sulfides. Mercury also forms black sulfide. So in a state of exams, a safe bet would be all cations.
Thiosulfate has complex properties (a) it is a strong reducing agent (b) it forms complexes and (c) it decomposes under various conditions to pure sulfur and sulfite ion in acidic medium.
From Jander and Blasius, Textbook of Analytical and Preparative Inorganic Chemistry (in German)

Von den Salzen sind nur $\ce{BaS2O3}$, $\ce{Ag2S2O3}$ und $\ce{PbS2O3}$ schwer löslich. Sie werden jedoch durch Sodalösung quantitativ in das lösliche Natriumsalz überführt. Mit vielen Schwermetallionen (z.B. $\ce{Ag(I)}$, $\ce{Fe(II)}$ und $\ce{Cu (I)}$) bildet Thiosulfat im Überschuss lösliche Komplexe.

An english translation of this would be

Of the salts only $\ce{BaS2O3}$, $\ce{Ag2S2O3}$ and $\ce{PbS2O3}$ are poorly soluble. However, they are quantitatively converted into the soluble sodium salt by soda solution. With many heavy metal ions (e.g. $\ce{Ag(I)}$, $\ce{Fe(II)}$ and $\ce{Cu(I)}$) an excess of thiosulfate forms soluble complexes.

In the book the equations of conversion to sulfide are given as. Ask your teacher if this a typo in the exam.

4. Nachweis als $\ce{AgS2O3/Ag2S}$
\begin{align}
\ce{S2O3^2- +2Ag+ &-> Ag2S2O3\downarrow} \\
\ce{Ag2S2O3 + 3S2O3^2- &-> 2[Ag(S2O3)2]^3-} \\
\ce{Ag2S2O3 + H2O &-> Ag2S + H2SO4} \\
\end{align}

That is all you have to remember that thiosulfate will precipitate $\ce{Ba}$, $\ce{Ag}$ and $\ce{Pb}$ initially but excess would dissolve it.
I would be interested to know otherwise.

The question is messy because it should clarify that thiosulfate is added in excess. I do not know enough of thiosulfate chemistry. As per the book and with rudimentary knowledge of German, it is mentioned that thiosulfate complexes with many heavy metals under the heading of silver. It says similar equations can be written for other heavy metals. So assume, thiosulfate is added in excess, this will make answer to 5.

Not always. When an ion or molecule can complex, as well as it can precipitate then you can see this phenomenon. Recall ammonia soln being added to copper sulfate solution.


text:
Help with iodine potassium-iodide / hydrogen peroxide reaction lab?
The setup: As part of my chemistry course, we're required to complete an in-home lab where we measure the reaction rate of an iodine potassium-iodide (IKI) solution with a hydrogen peroxide solution. We daisy-chain 2 Erlenmeyer flasks together with a system of rigid & flexible tubing, with a tube extending to an open graduated cylinder at the end. The chemical mixture is in the first flask, 200 mL of water is in the second flask, and about 10 mL of water is in the graduated cylinder. Here's how it all looks:

What's SUPPOSED to happen: once we mix the IKI/hydrogen peroxide solutions in Flask #1, we swirl this flask on and off for something like 20 minutes to start the reaction. The reaction should build up pressure (I think?), causing the water to be forced up a rigid tube in Flask #2 and over to the open cylinder. Note that the longer of the 2 rigid tubes in Flask #2 extends below the surface of the water. The water will then start to drip into the graduated cylinder and we measure the displacement, which will then give us a table of data points to determine the rate constant, rate order, and many cheerful facts (about the square of the hypotenuse...)
What's ACTUALLY happening: nothing. I've been swirling Flask #1 for over an hour, and not a single drop of water has been displaced into the graduated cylinder.
Question: any idea what might be going on? Would there be something wrong with the setup? All the flexible tubing fits pretty tightly over the rigid tubing, and the rubber stoppers on the flasks also fit pretty tightly. I don't know how to troubleshoot this, so any tips would be helpful!

What are the concentrations of the hydrogen peroxide, of the KI/I2 solution ? Which volume of these solutions have you used ?  As far as I can judge from here, I guess the concentrations have been too low. I have done this experience often. The reaction is rapid, and the volume of Oxygen is important.

For the first trial, we use 10 mL of undiluted (1.0-2.0%) IKI and 5 mL 3% hydrogen peroxide

If you tape "Catalytic Decomposition of Hydrogen Peroxide" in Google, you will find an article saying that using 20 mL H2O2 30%, then adding 5 mL KI 2M, the decomposition is nearly immediate. It lasts less than one minute, and it produces a huge amount of Oxygen gas. There is also a second reference by R. Conkin, J. Chem. Educ. 73, 9, 838 (1996), showing that the reaction is rapid with H2O2 3% mixed with an equal volume of a saturated solution of KI. In both cases, the volume of O2 is important. Why don't you have a look in the last publication ?


text:
How to use Clausius-Mossotti equation to determine the molecular structure of a compound?
A question in my exam came as follows:

Explain the Clausius-Mossotti equation. How is it used in establishing the molecular structure of a compound?

I have studied the Clausius-Mossotti equation and I know how to derive its expression. I know that it deals with the Dielectric Constant, Molecular weight of the given molecule, the Density of the dielectric or medium between the two parallel plates of condenser, Avogadrov Number, and Molecular polarizability of medium or dielectric. However, none of my textbooks talk about how to actually use this equation to determine the actual molecular structure of any compound that is under study. My professor has not taught this either. I was unable to find any external sources that talk about this as well.
Could someone please explain the steps on (or concept about) how to determine (or establish) the molecular structure of a compound after knowing the required values?

The question asks 'how it is used in establishing the structure', not in determining it, so you would use it to estimate the molecular weight.

@porphyrin  Ohh Thank you for clarifying the difference. If you don't mind, an elaborated answer will be so helpful.

I’m voting to close this question because this question has become a question more on the mathematics of the Clausius-Mossoti relation and not the concepts behind it considering OP has an understanding on the different parameters required.

@Safdar  After reading your comment, I posted this question on Mathematics Stackexchange. They are saying that it is about establishing 'Molecular Structure' and not about deriving the equation itself. So, they closed my question and asked me to post this question on Chemistry Stackexchange. As you had already closed my question here too, I would request you to kindly guide me that in which forum should I post this question?
Thanks in advance!

I meant that once poryphyrin told you that it is used to estimate the molecular weight and as you said you knew about all the conditions and how to obtain them, I thought all you had to do was plug in values. However I do not know about this and was going from the statement in your question. I am sorry about the inconvenience for having to ask in Math SE..


text:
Why does there need to be a planar intermediate in 2,2'-dibromobiphenyl when one of the phenyl groups rotates?
I was in class and my professor brought up the example of 2,2'-dibromobiphenyl:

In her example, one of the bromines was actually pointing into the page and the two phenyl groups were not actually on the same plane of the page. Let's call the phenyl with the bromine that points into the page group 1 and the one that is on the page group 2. She then said that in order for the group 1 phenyl to point out from the page (a conformer) there will be a planar intermediate where the bromines are very close to each other during the rotation (just like in the picture), and this will require a large amount of energy (20 kcal/mol) which will again drop off once the group 1 rotates till it's pointing out of the plane. My question is, why doesn't it just rotate the other way where the bromines will not come into close contact and there will not be a planar intermediate? Am I overthinking this?

It will rotate both ways. If it rotated the other way then the H atom (not shown in your picture)  becomes next to the Br, this has a smaller interaction than the Br but it is still important because bromine is a large atom. This happens on both sides of the molecule.

I think that your professor was just mentioning a rotational angle vs relative energy plot. It is really important to understand relative energy difference between different points in such plots. The situation when two bromines are in the same plane and close to each other will represent the highest energy spot, this actually means that the probability of the molecule to be in such spot is extremely low, but technically not zero, that is why some reaction that require such spot may happen. In the picture below you can see the plot for biphenyl, in case of the dibromobiphenyl one of the peaks will be much higher then the others. But you are actually right, molecule itself would prefer to rotate the other way.

Ah ok yeah I was wondering if my prof just wanted to demonstrate that scenario, but it makes more sense in the context of an energy vs. rotation graph. Thank you!


text:
Why do the lanthanides and actinides have a 5d and 6d orbital, respectively?
I was taught in school that the rare earth metals were the $\mathrm{f}$-orbital group. Additionally, the Aufbau principle states that the order of orbitals based on energy levels is $\mathrm{6s4f5d}$ and $\mathrm{7s5f6d}$ for the lanthanides and actinides, respectively. However, upon inspection, both periods have 15 elements whereas the $\mathrm{f}$-orbitals can only hold 14 electrons. I was able to resolve this by assigning one electron to the $\mathrm{5d/6d}$-orbital, which is correct according to chemistry libretexts. Why does this occur for the rare earth metals?
Furthermore, according to libretexts, only some of the rare earth metals have a $\mathrm{5d/6d}$-orbital, whereas others have the $\mathrm{6s/7s}$-orbital and proceed to only fill their outermost $\mathrm{f}$-orbital. Why is this the case, and is there a way to determine which elements are like this short of simple memorization?

There are 14 elements in the lathanide/actinide series, excluding Lanthanum and Actinium. Also, the aufbau principle pretty much breaks down after period 4, don't use it for anything other than main group elements.

https://chemistry.stackexchange.com/q/47542/16683 https://chemistry.stackexchange.com/questions/74880/going-from-la-to-ce-why-is-the-extra-electron-added-into-the-4f-orbital

Yep, all of these help out a lot. Thanks!


text:
What is the name origin of ester?
Ester is quite a random nomenclature for a compound derived from a parent acid and parent alcohol.
Is there any reasoning behind using the word 'ester' to name such compound (for memorising purpose)?
One explanation for such nomenclature I found is quoted below from this website

The word "ester" was coined in 1848 by German chemist Leopold Gmelin,
probably as a contraction of the German Essigäther, meaning acetic
ether.

But even then, how does 'acetic ether' has to do with anything about ester being a nomenclature for a compound derived from a parent acid and parent alcohol?

The thing you have to remember is that most of the 'simple' compounds and functional groups were discovered before the structures were figured out. The names, therefore, would have been descriptive based on some quirk of the chemical. As musicians and coders know, naming things can be hard. You look at the etymology of most early elements and compounds, the names tend to work out to 'smells bad' or 'green flame' or 'oil from Tolu'

The term "Essigäther" is actually the German name for ethyl acetate i.e Essig = "vinegar" + Äther = "ether". "vinegar" becomes acetic, hence ethyl acetate becomes "acetic ether". (etymoline.com)

But, ethyl acetate is an ester. So, why is it named "acetic ether"?

According to Leopold Gmelin, ester was called "oxy-acids ether" or "ether of third type". The original German text is:

b. Ester oder sauerstoffsäure Aetherarten. Ethers du troisième genre.
Viele mineralische und organische Sauerstoffsäuren treten mit einer
Alkohol-Art unter Ausscheidung von Wasser zu neutralen flüchtigen
ätherischen Verbindungen zusammen, welche man als gepaarte
Verbindungen von Alkohol und Säuren-Wasser oder, nach der
Radicaltheorie, als Salze betrachten kann, in welchen eine Säure mit
einem Aether verbunden ist.

which translates to:

b. Ester or oxy-acid ethers. Ethers of the third type.
Many mineral and organic acids containing oxygen combine with an
alcohol upon elimination of water to [form] neutral, volatile ether
compounds, which one can view as coupled compounds of alcohol and
acid-water, or, according to the theory of radicals, as salts in which
an acid is bonded with an ether. (italics mine)

You see the name "ether" was used to define an ester but this usage has become obsolete to avoid confusion.
Other compound classes originally named ethers
Gmelin reports in his 1848 Handbook of Chemistry that French chemists use the terms ether of the first, second and third type to refer to ethers, thio- and halo-alkanes, and esters, respectively. While the modern term ether refers to compounds with a specific functional group, the term ether was historically used to refer to a variety of volatile substances.

The names ether and ester are currently used, while the name "afer" for the thio- and halo-alkanes has not caught on.
Source: https://www.google.com/books/edition/Hand_book_of_Chemistry/U4sMAAAAYAAJ?hl=en&gbpv=1&bsq=Ethers%20du%20troisi%C3%A8me%20genre
Reference

Leopold Gmelin, Handbuch der Chemie, vol. 4: Handbuch der organischen Chemie (vol. 1) (Heidelberg, Baden (Germany): Karl Winter, 1848)

Esters are volatile, ethereal, just like ethers.

Thanks for the answer. The translated text of the german quote you provided said 'Ester [is the] Ethers of the third type," would you know what is the second type of ether (if there is one)?

@BobbyLeung these are very archaic terminologies. Only Gmelin knows what is he referring to ;-)

@NilayGhosh Second type of ether are haloalkanes like chloroform.

Just to add another resource besides the excellent find by Nilay, in Organic Chemistry, The Name Game, it mentions

ester from German Essigäther (acetic ether) an early name for ethyl
acetate

The unabridged Oxford English Dictionary (by subscription only) also mentions that Gmelin called these compounds as napthas but later he changed it to ester.

H. Watts tr. L. Gmelin Hand-bk. Chem. VII. 190   Ethers du troisième genre. I formerly distinguished these compounds by the name of Naphthas produced by oxygen-acids (Naphthen durch Sauers[t]offsäuren erzeugt); but I now propose for them the term Ester.]

Naphthas referred to

naphtha was used in Latin and Greek for a flammable, volatile liquid derived from the
distillation of coal tar.

So it sort of supports what Karsten is saying that the volatility of esters must have impressed Gmelin. Ethers is very volatile!

I updated the other answer to include an excerpt of the English translation of Gmelin's handbook where he talks about ethers of the first, second and third kind.


text:
Why do you count both bonding electrons towards an atom's octet, but you only count one of them towards the atom's charge?
For example, oxygen is neutral when it has just 6 electrons. But the pink O in the molecule pictured has 3 bonds giving it 8 electrons, a full-octet. It's more electronegative than all the atoms it's bonded to also. It would make sense if an O with a full-octet had a -2 charge since it has an extra 2 electrons.
But because it "owns" only 1 electron from each bond, the O has a positive charge. This doesn't make sense to me: if we don't count the extra electrons in its charge, why do we count them towards its octet?

Where do the bonding electrons come from? This question might help.

Umm, such issue winding up in sugar hydrolysis of all things is rather surprising...

https://chemistry.stackexchange.com/questions/14478/the-definition-of-formal-charge

That charge labelled on oxygen atom is formal charge, right?

@Tge99sLearner No it is a real charge, which can lead to the confusion.

Your nick fits well with the colourful sugar :)

I learned that formal charge is actual charge... do both electrons in a bond count towards an atom's charge, or only the one it owns?

@histrionics no it's not. A formal charge is of course formal. The oxygen in your sketch has a real charge. Or the ensemble. But there is a charge somewhere. And it is more around that oxygen.

Alright thank you. How do I reopen this question? The linked question does not address my question about octets (which are used to tell whether the last shell is full). Both formal charge and oxidation state are contradicted by how we count an atom's electrons in an octet. That is what my question is about.

@Mithoron I vote to reopen the question. The edited question asks about the difference in counting electrons when looking for octets as opposed to assigning formal charges. The linked question compares counting electrons for assigning oxidation states or formal charges. Ideally, there would be a question comparing all three ways of counting to direct all further questions there.


text:
describing in words a primitive heated water vapor lift

The primitive lift illustrated in the left hand side of the following figure operates using heated water vapor ("gas") that pushes on a sealed piston. During its operation, the vapor undergoes the rectangular cyclic process shown in the right hand side of the figure.
I'm trying to describe in words how one might realize each of the four steps.
My idea is : We assume the system is isolated entirely from its surroundings such that it obeys the ideal gas law. First, heat is added through an external source to the gas while the piston is held fixed such that Temperature increases while the volume remains constant, therefore the pressure increases from $p_{1} \to p_{2}$ (A). Then, the piston is released and the gas is no longer heated. The volume will increase from $V_{1} \to V_{2}$ and the pressure will remain constant. When the piston reaches the top it is stopped and held still.
Here I'm blocked, I don't know how to explain a situation in which pressure decreases while volume is held constant .
Any help would be appreciated.

It would help, if the system was not closed, but there was a source of the external gas of the pressure p2, and a ventil keeping pressure p1.

To carry out the B transformation, you have to heat a lot. If you don't heat, the volume will of course increase, but the pressure will decrease. B is a process difficult to carry out, because it is not easy to  heat enough a gas who is inflated to as to maintain the inner pressure. Then, at the end of B, the temperature is very high. You block the position of the piston, and you cool the system : the pressure will decrease from P2 to P1, during the process C. The last process D is also difficult to carry out. Because in D you must compress the gaz so as to decrease its volume, but the pressure must not increase. It means that the system should be cooled during D. At the end of D, the temperature will be extremely low.


text:
How can you calculate the tendency of one compound to become another?
Excuse my inexperience with chemistry here. I've been poring over papers with terms like "Standard Gibbs free energy of formation", "Hess's Law", "standard enthalpy of formation", "entropy", and several others, and I haven't been able to find an answer to one simple question:
Given I have a system with compounds A, B, and C, and there is an alternative set of compounds X, Y, and Z involving the same amounts of the same elements, how do I calculate how much the ABC system wants* to become XYZ?
As an example, if I have a system with propane and oxygen, with an excess of oxygen, this system wants to combust to become carbon dioxide and water vapor. A quick Wikipedia check reveals the balanced equation for this reaction:
$$\ce{C3H8 + 5O2 \rightleftharpoons 3CO2 + 4H2O}$$
Anyone who's played around with a camp stove knows that propane and oxygen like to burn to become carbon dioxide and water vapor, and that carbon dioxide and water vapor really don't like to "unburn" to become propane and oxygen. How would I measure this tendency for the propane-oxygen system to become the carbon-dioxide-water-vapor system? Similarly, how would I measure the distate for the carbon-dioxide-water-vapor system to become the propane-oxygen system?
If I remember correctly, the tendency of a reaction to happen and the speed of the reaction are entirely unrelated, one being a thermodynamic property and the other a kinetic property, but if you wouldn't mind throwing in just enough info about reaction rate in the above case that I can look up and learn myself on that topic as well, I'd really appreciate it.
*I'm falling victim to the oh-so-elegant abstraction of molecular anthropomorphism here. Yes, I understand that chemicals don't want anything, but given I'm looking for a unit most easily expressed as the degree of desire for chemicals A, B, and C to become X, Y, and Z, these human terms play out quite well. Please do let me know if there's better terminology I could be using

Well, with something like that: https://chemistry.stackexchange.com/questions/39479/finding-equilibrium-constant-from-standard-free-energy

I find it interesting no one mentioned the equilibrium constant, which measures the extent to which the reactants "want" to become the products, or reaction kinetics, which measures how quickly the reactants become the products. I don't think an answer to this question is complete without address these constructions.

Depending on the constraints you impose on the system, there are different criteria for spontaneity, however, the commonly running theme is that it is related to minimizing some state function. For example, as a simple example from classical mechanics, the time evolution of the system is such that the potential energy of the mechanical bodies is minimized.
When we do a chemical reaction, we usually do it at constant temperature and pressure. In this case, the evolution of the system is determined by a potential function known as Gibb's free energy. If there is a state in which the system which reduces Gibb's free energy, then the system will evolve to reach that state. When I say evolve I mean to have processes such as chemical reactions.
I had actually written a very detailed answer about this on physics stackexchange, check it out for more information (here)

Refer these notes by MIT for the criteria for spontaneity for different constraints :
MIT ocw notes
Discussion about constant P and T chemical reactions:
Here

Well, essentially, the Gibb's Free Energy Change of a reaction is the property that determines how much a reaction wants to progress in any particular direction.
It's a property denoted by ΔG.
Basically, the more negative the value of the Gibb's Free Energy change, the more the reaction "wants" to happen. Essentially, the reaction tends to achieve a state of minimum Gibb's Energy
You can calculate ΔG using the enthalpy change, entropy change and temperature by the relation:
ΔG = ΔH - TΔS
ΔH is the enthalpy change
T is the temperature
ΔS is the entropy change
Hope this helps!


text:
Mechanism of decomposition of alkali metal halides in flames
A student recently asked an interesting question What happens when KCl is introduced in the Bunsen flame?. This apparently simply question led me to an interesting debate. We all know that emission of characteristic colors of alkali metal salts especially the halides and carbonates in ordinary Bunsen flames comes from the atomic emission not their oxy compounds (like the alkaline earth salts) or their ions.
The point of debate is point number 3:

KCl solution has to atomize and become KCl (solid).
We would get KCl in gas phase from the heat of the flame.
How does KCl (gas) convert to K atoms in a flame? What is the path? Note Thermal decomposition of salts without a flame is another story. Please do not mix, decomposition in a flame versus isolated vapor phase studies of salts.

Flame chemistry is quite complex and my understanding is that flames are electrically conducting from a very small amount of free electrons and ions. Certainly, we are not talking about cool flames like candles. Flame conductivity is a well known phenomenon and it is utilized in gas chromatography's "flame ionization detector" as well.
Herein, I am talking about spectroscopic flames used in atomic emission spectroscopy.  It is known for more than 100 years that when we introduce alkali metal salts into the Bunsen flames, the conductivity of flame increases as a function of salt concentration. For alkali halides, it is a square root function of concentration. This clearly shows that there are free ions and electrons in a flame in which alkali metal salts are introduced. The amount must be really small but significant enough to alter flame properties Electrical properties of flames.
There must be certainly thermal decomposition but it is not a complete story at all. The key question is the path of KCl(s) to K atoms!
(a) KCl goes to potassium ions in the flame, by thermal decomposition, and it becomes neutral potassium atom in flame after gaining an electron (from flame components)?
(b) KCl is thermally decomposed directly to potassium atoms and there are no intermediate potassium ions?

The Saha ionization equation may be of interest.

Dear Ed, This is just one step before the Saha equation. How does the KCl (particle) turn into a free K atom in a flame? What is the mechanism?

If you look at my KF answer here, it is energetically favorable for the ions to become neutral atoms. Same for KCl. So once a KCl particle melts and vaporizes, you get the neutral atoms, some electrons, some ions and some polyatomic species (neutral and ionic). That is my crude picture of it. ;)

So what would be source of "reduction" of potassium in a flame, for a KCl particle? I mean thermodynamics is predicting the energetically favorable final state. It is not telling us how we reached there. It is impossible to even ask or to distinguish, but does the electron comes from the chloride ion or it comes from the flame? This would settle the issue.

It is complicated. Assuming local thermodynamic equilibrium (LTE) and reasonable flame temperature (without LTE, temperature is not even well defined), the chlorine species is predominantly in the form of neutral chlorine atoms. That sort of thing is what all those Boltzmann plots are about. My simplistic assumption is that any KCl moiety in the flame will be collisionally dissociated predominantly into K and Cl atoms, variously excited, and their ions will be present in lesser number densities. The electrons are just bit players, so to speak.

The spectroscopy of some of these halides has been studied in detail and the potential energy profiles determined for the ground state and 1 or 2 electronically excited states. In the case of NaI the ground state has a lot of ionic character  but given enough energy Na +I atoms are produced and only at higher energy Na$^+$ + I$^-$ ions. Time resolved experiments show how dissociation occurs and for which A. Zewail received the Nobel Prize, see Faraday Disc. Chem. Soc. 1991, v91, p201 for a review. A few energetic electrons in the flame could easily produce atoms, in a similar way to photons.


text:
How can I solubilize fragrance oil into polyvinyl acetate/clear glue?
I'm a high school student who makes slime for fun, using polyvinyl acetate (PVA) glue and borax as my main ingredients.
Earlier, I had the idea of adding fragrance oils to clear glue. However, because they aren't soluble, it makes the whole thing cloudy. I tried using a fragrance oil solubilizer, polysorbate 20, however I soon found out that it wasn't soluble in it the clear glue either!
This hit me as odd, since polysorbate 20 is perfectly soluble in water. I'm trying to find out what exactly makes up clear glue, but all I know is that it contains PVA. I read somewhere that since PVA isn't super soluble in water, it's the reason why white glue is white. In addition, I also read that clear glue might be solvent based altogether. I'm not sure if any of this is actually true, but if anyone has insight on this it would be really helpful!
Should I be trying a different solubilizer, or am I missing a crucial step when I mix the two ingredients? Or is the issue not even related to solubility altogether?
Here is a short list of what I have tried and what happened:

Experiment
Observation

Mixing the oil directly with the glue
Glue mixture ends up cloudy

Mixing the oil into the polysorbate 20, and then adding water
Water mixture ends up clear

Mixing the above water mixture into clear glue
Glue mixture ends up cloudy

Mixing polysorbate 20 directly into clear glue
Glue mixture ends up cloudy

Mixing water directly into clear glue
Glue mixture ends up clear

Some other information you might find helpful:

I use Darice clear school glue.
All of the ingredients on their own are already clear.
I have run each of those above trials multiple times and have gotten the same result each time.


text:
Textbook problem with molar mass of carbon dioxide

$\ce{K2CO3}$ and $\ce{HCl}$ react to produce $\ce{CO2}$. In an empty flask ( m = $\pu{85.431 g}$ ) we put in the produced gas and the mass of the flask is now $\pu{85.510 g}$. After that we fill the flask with water. Volume of the flask filled with water is $\pu{122 ml}$. Pressure is $\pu{101325 Pa}$, temperature of air is $\pu{300 K}$. Density of water is $\pu{0.99893 g cm-3}$, and the molar mass of air is $\pu{28.8 g mol-1}$. Find the molar mass of carbon dioxide.

I have calculated the mass of $\ce{CO2}$, using the mass of the flask filled with gas and subtracting the mass of the empty flask. I have figured out that somehow through $pV = nRT$ I am supposed to get to $n$ and take into consideration that $n$ is not only the number of moles of $\ce{CO2}$ but also of $\ce{H2O}$ and air.
What I don't understand is the volume of water used to fill the flask. In that volume do we only have $\ce{H2O}$ or is that volume of 122 mL also the volume of air + $\ce{CO2}$ + $\ce{H2O}$?
Any hint is appreciated.

Hint: Find the density of carbon dioxide. You know mass of the CO2 added and the volume of the flask. Using pV=nRT find out the molar volume of carbon dioxide. What is the relation between density and molar volume. Can we get molar mass of CO2 from that?

Molar volume of CO2 implies that n=1 in PV=nRT.

Ok. Thanks. And what am I supposed to do with the volume of water added? That's what's bugging me.

Please refrain from using the homework tag, it was removed a long time ago.

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

Having done the experiment, the water is used to determine the volume of the flask.

Ok. I understand that part. But what about molar mass of air and density of water? How do I include those values in my calculation?

The pyknometric experiments usually work with the mass and density of water.If the vine is known, density is not needed.

https://chemistry.meta.stackexchange.com/questions/566/houston-we-have-a-problem-in-question-titles

The problem is that, when the flask is empty, it weighs $85.431\ \mathrm g$. But this weight is the weight of the empty flask plus the air included. So the first thing to do is to calculate the weight of the air, and the corresponding number of moles of air in the flask before adding $\ce{CO2}$. The volume of the air is the same as the volume of water when it is filled with water, namely $122\ \mathrm{mL}$.
The amount of air is $$n = pV/RT = \frac{122\times10^{-6}\ \mathrm{m^3}\times101325\ \mathrm{Pa}}{8.316\times300\ \mathrm{J/mol}} = 4.956\times10^{-3}\ \ \mathrm{mol}$$ The mass of this air is : $$m = 4.956\times10^{-3}\ \mathrm{mol}\times28.8\ \mathrm{g/mol} = 0.1427\ \mathrm g$$ So the mass $m_0$ of the empty flask without air is : $$m_0 = 85.431\ \mathrm g - 0.1427\ \mathrm g = 85.2883\ \mathrm g$$
A s consequence, the mass of the $\ce{CO2}$ in the flask is : $$m(\ce{CO2}) = 85.510\ \mathrm g - 85.2883\ \mathrm g = 0.2227\ \mathrm g$$
The molar mass of this $\ce{CO2}$ is: $$M = \frac{m}{n} = \frac{0.2227\ \mathrm g}{4.956\times10^{-3}\ \mathrm{mol}}= 44.8\ \mathrm{g/mol}$$
This is nearly the expected value for the molar mass of $\ce{CO2}$. If you take into account the fact that the volume $122\ \mathrm{mL}$ is known with a precision $\pm1\ \mathrm{mL}$, or $\pm1\ \%$, the final result will be defined with an uncertainty of $\pm1\ \mathrm{g/mol}$. It is: $$M(\ce{CO2}) = 44.8\ \mathrm{g/mol} \pm 1\ \mathrm{g/mol}$$

I would rather say 122 mL means  +- 0.5 mL, as the more different value would be estimated rather as  121 mL or 123 mL.

The question should have explicitly stated the meaning of "Empty" flask. Empty as  in vacuum or empty in a non-scientific way.

@M. Farooq  Empty as in a lab way, filled by air at lab conditions. :-)

@Poutnik, vacuum techniques? And how could they ensure carbon dioxide filled it completely?

@M. Farooq sure, vacuum techniques are possible, but not necessery. If filled from the bottom, flushed by a multiple of the volume, it will be.

Right Poutnik, I was just saying that the exam question writer should make it explicit. For example, was carbon dioxide bone dry when it came to the flask and so on? Good exam questions do not require assumptions. They should write all this.


text:
Comparing stability of 2-propenyl cation and 1-propyl cation

Compare the stability of 2-propenyl cation and 1-propyl cation:
$$\underset{\text{2-propenyl cation}}{\ce{CH2=CH-CH2+}}\qquad \underset{\text{1-propyl cation}}{\ce{CH3-CH2-CH2+}}$$

In 2-propenyl cation we get resonance. So, 2-propenyl cation seems to be more stable, but due to the double bond (or, more precisely, the $\mathrm{sp}^2$ carbon) in 2-propenyl cation, an electronegative carbon atom which makes cation less stable.
In case of 1-propyl cation due to inductive effect cation is getting stabilized, but we know resonance is more stabilising than inductive effect.
I would answer that 1-propyl cation is more stable than 2-propenyl cation because we get $\mathrm{sp}^2$ or a more electronegative C in 2-propenyl cation making it less stable.
I'm really confused which factor is more effective: stability of resonance or destabilisation due to the $\mathrm{sp}^2$-hybridized carbon in 2-propenyl cation.
What would be a valid zero-order approximation for this?

Stability order -  $2$-propenyl cation>$1$-Propyl cation
Resonance has way more effect on the cation than that tiny-whiny little weak $\mathrm {sp}^2$ hybridised carbon

Electronic configurations are to be written upright, See here for a list of what should be upright and what shouldn't..

@OscarLanzi Dear Sir, i'm at very initial stage of learning organic Chem. We haven't covered whole IUPAC in our school but i saw a similar question on chegg(link provided below) according to them it's name is 2-propenyl cation. If i'm wrong please correct me it will help me alot in future. Thanks,

@OscarLanzi link of question -   https://www.chegg.com/homework-help/2-propenyl-cation-appears-primary-carbocation-yet-considerab-chapter-6-problem-49p-solution-9781133952848-exc


text:
Geometry of a free radical at the bridgehead carbon
With reference to J. Am. Chem. Soc. 1968, 90 (19), 5266–5267, they state:

The most favorable geometry of aliphatic free radicals remains an unsolved problem, since existing data are consistent with a geometry which is either pyramidal and rapidly inverting, or planar but easily deformed.2-4 One approach to the problem is generation of the radical
at a bridgehead of a bridged bicyclic system. [...] From their aldehyde decarbonylation study, however, Applequist and Kaplan concluded that 1 -adamantyl and 1 -bicyclo[2.2.2]octyl were appreciably more stable than t-butyl. 3
3. J. Am. Chem. Soc. 1965, 87 (10), 2194–2200

which is later concluded as:

[...] The extent of geometrical destabilization is considerable, and we conclude that alkyl radicals tend toward planar geometry. The lesser extent of destabilization, relative to that for the carbonium ions, might mean that radicals seek a flattened pyramidal state. [...] This issue can evidently not be settled without spectroscopic information about alkyl radicals.

Has there been any further studies on the geometry of free radicals at the bridgehead carbon?

If you have access to Science Citation Index, you may locate more recent work.

@user55119 unfortunately I do not.. Could you share a reference if possible?

I found 115. I'll see what I can do.

These 2 are dated but perhaps of interest. STABILITY OF ADAMANT-1-YL RADICAL
JOURNAL OF THE CHEMICAL SOCIETY D-CHEMICAL COMMUNICATIONS  Issue:  5   Pages:  216-&   Published:  1969
.
BRIDGEHEAD FREE RADICAL STABILITY . DECOMPOSITION OF T-BUTYLPEROXYESTERS OF 1-ADAMANTYL 1-BICYCLO[2.2.]OCTYL AND 1-NORBORNYL SYSTEMS
CANADIAN JOURNAL OF CHEMISTRY  Volume:  46   Issue:  19   Pages:  3099-&   Published:   1968

@user55119 Thanks.. I was also able to find a comprehensive paper regarding bridgehead radicals Chem. Soc. Rev. 1992, 21 (2), 105 by John C. Walton

@SafdarFaisal: So is the answer to the question "why can free radicals form at bridge head carbon whereas carbocations cannot...?" is because a carbocation is $\ce{sp^2}$ hybridized whereas a free radical would not be $\ce{sp^2}$, it's not $\ce{sp^3}$ as well, but something in between. And $\ce{sp^2}$ hybridization is not allowed as per Bredt's rule. Am I right?

@user55119 Sir please see my above comment, is my explanation for stable existence of free radical at bridge head positions correct? If not, why can a free radical exist at bridge head position then?


text:
Comparing melting points of substituted benzaldehydes

The correct decreasing order of melting point for the given organic compounds is:

3 has maximum melting point because of intermolecular hydrogen bonding.
I am unable to predict rest of the order correctly.
4 should have higher melting point as it shows greater intermolecular hydrogen bonding than the rest, but this contradicts the experimental melting point (mp) data [1]:
$$
\begin{array}{llr}
\hline
\text{#} & \text{Compound} &\text{mp}/\pu{°C} \\
\hline
\textbf{1} & \textit{o}\text{-hydroxybenzaldehyde} & −7(2) \\
\textbf{2} & \textit{o}\text{-methoxybenzaldehyde} & 37.5(0.5) \\
\textbf{3} & \textit{p}\text{-hydroxybenzaldehyde} & 116.0(0.2) \\
\textbf{4} & \textit{p}\text{-methoxybenzaldehyde} & 0 \\
\hline
\end{array}
$$
Reference

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.

melting point depends on how well the molecule packs in a lattice, not neccessarily on strength of H-bond


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What characterises the smell of alkane hydrocarbons?
I have heard that aromatic hydrocarbons smell fruity and like petrol. However, I understand that there is a class of hydrocarbons called alkane, as for example propane or butane. I wondered if I might identify them by smell or sight in the same way as aromatic hydrocarbons. If so, what would characterise them?

Methane is odourless. Higher terms are easy to smell and I would define them fuel like, but "less round - more dry" and not very intense. Think of lighter fuel. Note that alkane become non volatile already at n=7. I also suspect that the smell of oil gases is due at least in part to sulphur and other stuff.

Depends on the terroir... speaking crudely, some may be high sulfur (of course, that's only an impurity).

Related: https://chemistry.stackexchange.com/questions/23842/why-does-heptane-have-an-odour... https://chemistry.stackexchange.com/questions/9278/are-all-pure-alkanes-alkenes-and-alkynes-tasteless-odourless-and-colourles

Usually, some other compounds like thiols are added to natural gas like butane and propane in order to ensure that they have a smell in the event of a gas leakage for safety reasons. Alkanes usually do not have a smell, at least not that I know of.

Cigarette lighters are typically fueled with butane. I don’t believe there are any odors added, so you can get an idea of the smell from one of those

Let us put it this way. Petroleum ether certainly smells even at analytical grade. It can be called fuel like. Smelling compounds can be present from the natural crude product and lowered down, or conversely added for safety. I must admit that I don't know if pure alkanes intrinsically smell, or it is always due to the presence of traces, but they at least smell fuel for real. If they don't smell it is ok, they are paraffines at the end. And it means they have low affinity to many chemicals, and perhaps to smell receptors, too :)


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How does the calorimeter affect heat released by the system (q)?
Here are the 2 questions I am pondering about

How would the calculated quantity of heat (q) released by a system differ if the calorimeter's heat capacity was taken into account?

My thought process is that the calculated quantity of heat (q) released by a system would increase if the calorimeter's heat capacity was taken into account, b/c the calorimeter would aid in absorbing the heat along with the surroundings, so change in temperature would increase ($q=m c \Delta T$)

How would quantity of heat (q) released by a system differ if the calorimeter was not a good insulator?

Here, I am not sure how to word the answer, but if the calorimeter wasnt a good insulator, heat released by a system would be lost to the environment, and thus less heat released would be calculated via $q=m c \Delta T$ due to a smaller change in $\Delta T$.

Let's take a numerical example. $0.2432$ $g$ magnesium metal ($0.01$ mol) is introduced into a calorimeter containing $50~ g$ HCl $1$ M. The chemical reaction will be $$\ce{Mg + 2 H^+ ->  Mg^{2+} + H2}$$ The composition of the solution is changing between the beginning and the end of the reaction. There is an excess of HCl.  But let's admit in first approximation that the specific heat of the solution does not change and remains as in water ($C_w = 4.185~ J g^{-1} K^{-1}$). The tabulated enthalpy change of the reaction is - $462~ kJ/mol$. The total amount of heat $Q$ produced by the reaction of $0.01 mol$ Mg is : $$Q\ce{= 0.01 mol·462 kJ/mol = 4.620 kJ = 4620 J}$$
If the calorimeter was ideal, without any weight, and with the already mentioned approximation, the change of temperature should be : $$\Delta T = \frac{Q}{C_w·m} =  \frac{4620~ J}{4.185~J g^{-1} K^{-1}·50~ g} = \frac{4620~ J}{209.25~ J K^{-1}} = 22.07 K$$
Now suppose that the calorimeter is made of copper and weighs $m_{Cu} = 20~ g$. The heat capacity of copper is $C_{Cu} = 0.385~ J g^{-1} K^{-1}$. The change of temperature will be reduced and becomes $$\Delta T = \frac{Q}{C_w·m + C_{Cu}·m_{Cu}} =  \frac{4620~ J}{(209.25 ~J K^{-1})~ + (0.385 ~J g^{-1} K^{-1}·20~ g)} = 21.3 K$$
And now if the calorimeter is not well insulated, the final temperature change will be smaller than $21.3 K$.
Sorry for my poor management of the italicized letters. Any help would be appreciated.

The calorimeter heat capacity is always taken into account.

In real measurements, there are plotted temperature trends before and after action,so it can be eliminated.


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Is Cu stable as Cu(H2O)2+ for determination on MP-AES?
I have an unknown sample of mineral which I need to dissolve in nitric acid. I need to analyze $\ce{Cu}$ from it on MP-AES. Do I need to add $\pu{1ml}$ of $\ce{HCl}$ to stabilize $\ce{Cu}$, or it is enough just $\ce{HNO3}$?

This procedure from an MP-AES manufacturer features copper analysis (among other elements) without the use if hydrochloric acid.  Hydrogen peroxide is used in this procedure but would not affect the extraction of copper.  A copper standard is also available in "5% nitric acid" solution from the same company.
So the people who are selling this technology do not believe hydrochloric acid is needed for copper analysis, you can try nitric acid alone if this gets your sample to dissolve.


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Separation techniques
I need to identify separation processes that begin with each letter of the alphabet. I am missing an example for B, J, K, O, Q, X, Y, and Z. Any help will be greatly appreciated. Thank you.

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


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Regarding comparison of ionization energies

Why is the ionization energy of $\ce{Mg^2+}$ greater than the ionization energy of $\ce{Ne}$ (neon)?

My teacher said the answer was $\ce{Mg^2+}$ but I have no idea why as my general knowledge tells me that $\ce{Ne}$ has the greatest ionization energy since ionization energy increases across the group? Would someone explain why $\ce{Mg^2+}$ has a greater ionization energy?

How many electrons have to be removed, and what will be the net charge when done?

When you talk about the ionization energy of $\ce{Mg^2+}$, it is the third ionization of $\ce{Mg}$, which is equal to $\pu{7734 kJ/mol}$. When you talk about the ionization energy of $\ce{Ne}$, it is the first ionization of $\ce{Ne}$, which is $\pu{2081 kJ/mol}$. Thus, actually, $IE_\ce{Mg^2+} \gt IE_\ce{Ne}$.
Keep in mind that the first and second ionization energies of $\ce{Mg}$ are equal to $\pu{738 kJ/mol}$ and $\pu{1451 kJ/mol}$, respectively, both of which are smaller than the first ionization of $\ce{Ne}$ ($\pu{2081 kJ/mol}$).
Simplified reason: Both $\ce{Mg^2+}$ ion and $\ce{Ne}$ atom have the electronic configuration ($\mathrm{1s^22s^22p^6}$). Yet, $\ce{Mg^2+}$ has extra two protons in its nucleus than that of $\ce{Ne}$, thus attraction forces on electrons in $\ce{Mg^2+}$ are higher than that of $\ce{Ne}$ (e.g., $F_{q_+q_-} = \dfrac{1}{4 \pi \epsilon_\circ} \cdot \dfrac{q_+q_-}{d^2}$; also keep in mind that $r_\ce{Mg^2+} \lt r_\ce{Ne}$). By that reason alone, one can assume that the ionization energy of $\ce{Mg^2+}$ is higher than that of $\ce{Ne}$.
Note that your statement of "$\ce{Ne}$ has the greatest ionization energy" is incorrect. In that concept, the first ionization of $\ce{He}$ ($\pu{2372 kJ/mol}$) is greater than that of $\ce{Ne}$ ($\pu{2081 kJ/mol}$).