text:
Calculating Gibbs free energy change when the state of ions is unknown
Problem

Calculate $\Delta G^\circ$ for the following reaction:
$$\ce{Zn(s) + Ag2O(s) + H2O(l) -> Zn^{+2}(aq) + 2Ag(s) + 2OH-(aq)}$$
Given $E^\circ_{\ce{Ag+/Ag}} = \pu{0.80 V}$, $E^\circ_{\ce{Zn^{+2}/Zn}} = \pu{-0.76 V}$, $K_\mathrm{sp}$ of $\ce{AgOH} = \pu{2\times 10^{-8} M^2}$
A. $\pu{-301 kJ/mol}$
B. $\pu{-213 kJ/mol}$

Answer

 B

Question
This reaction can be split into two redox reactions:
$\ce{Zn -> Zn^{+2} + 2e-}$
$\ce{Ag+ + e- -> Ag}$
Accordingly, my initial approach was to take the $E^\circ$ of the entire cell, which comes out to be $\pu{1.56 V}$. Then using the formula $\Delta G = -nFE$, We get option A as the answer.
What am I missing?

Please note that this question is currently self-answered. I found the correct numerical value for the answer, but I'm still unsure if I have solved it correctly (sometimes wrong methods give the right end-answers too). Hence I've posted it here so that other users can verify if my method is correct or wrong. If wrong, the problem still stands, and of course others' insights and answers are welcomed!

Very good way of using this site @TRC.

@Mathew For future reference, you can make that even shorter (if you like): \pu{2\times 10^{-8} M^2} -> \pu{2E-8 M2} $\pu{2E-8 M2}$

The key thing to note here is that $E_\mathrm{cell} = E^\circ$ only when the ions involved are all in their standard states, i.e. concentrations of all ions is $\pu{1 M}$. But here we need to find $\Delta G^\circ$ for the reaction:
$$\ce{Zn(s) + Ag2O(s) + H2O(l) -> Zn^{+2}(aq) + 2Ag(s) + 2OH-(aq)},$$
whereas the redox reactions involved in the cell would be:
\begin{align}
\ce{Zn &-> Zn^{+2} + 2e-},\\
\ce{Ag+ + e- &-> Ag},
\end{align}
which requires $\ce{Ag+}$ to be present in its standard state. However, the first reaction given only tells us that $\ce{Ag2O}$ and $\ce{H2O}$ on the side of reactants are in their standard states. We hence don't know anything yet about the concentration of $\ce{Ag+}$, required in the cell.
Using the expression of $K_\mathrm{sp}$, $\ce{[Ag+][OH-]} = \pu{2E-8 M2}$:
From the products side of the first reaction, we know that $\ce{OH-}$ is in its standard state, so  $\ce{[OH-] = 1 M}$, which gives $\ce{[Ag+]} = \pu{2E-8 M}$.
Now we use Nernst equation for this cell:
\begin{align}
  \Delta G^\circ &= -nFE_\mathrm{cell}\\
  \Delta G^\circ &= -nF\left(E^\circ_\mathrm{cell}
    - \frac{0.059}{n}\log\frac{\ce{Zn^{+2}}}{\ce{Ag+}}\right)
\end{align}
$\ce{[Zn^{+2}]} = \pu{1 M}$ as it is in its standard state. Using $\ce{[Ag+]} = \pu{2E-8 M}$, we get:
$$ \Delta G^\circ = -2 \times 96.5 \times (1.56 - 0.059(8-\log2)) = \pu{-213 kJ/mol}$$

standard state oh- is 1?

@Buraian From what I've learnt standard state of an ion is defined to be a concentration 1 M at STP. If you feel there's a mistake in that, please let me know.

@TRC. The concentration of $\ce{Zn^{2+}}$ cannot be $1$ M, as you write. In the presence of $\ce{OH^-}$ ions, zinc makes a rather insoluble precipitate of $\ce{Zn(OH)2}$. The solubility product of $\ce{Zn(OH)2}$ is equal to $1.24 ·10^{-17}$. So the solubility of $\ce{Zn(OH)2}$ cannot be higher than $\ce{10^{-6}}$ M.

@Maurice You could run it in a cell with acidic pH in in zinc half cell and basic pH in the silver half cell.

@Karsten Theis. You are right. I have not thought of this possibility.


text:
How to draw a skeletal formula with thick lines and high resolution?
Consider the caffeine molecule skeletal formula as depicted in the Wikimedia Commons .svg file. Suppose that I want to replicate this skeletal formula with a high resolution and thicker lines.
I have found several online tools to draw molecules, but none of them allows the user to adjust line thickness (and, moreover, the resolution of the final output is generally not very high). I have also thought of using Tikz in LaTex, but going this route is really difficult and time-consuming, particularly if one wishes to replicate several skeletal formulae.
How would you draw a skeletal formula with thick lines and high resolution?
P.S: I’m sorry if this is not the correct place to ask such a question. If this is the case, feel free to tell me which SE site might be adequate for my question.
Thank you all very much for your time.

Software list for drawing from scratch with desired graphical features. As for this caffeine SVG, you can tweak line thickness in any vector editor such as Inkscape.

@andselisk  I disagree with closing this question because its answer can be found within a list of source materials, since that's true of many questions answered on Chem SE.   For instance, if you're going to close this question because its answer can be found within the materials listed at https://chemistry.stackexchange.com/questions/112760/is-1-ppb-equal-to-1-μg-kg/112761#112761, shoudn't you also close https://chemistry.stackexchange.com/questions/112760/is-1-ppb-equal-to-1-μg-kg/112761#112761? Its answer can also be found within one of the sources in the former link (IUPAC Green Book).

@Hector Here's how you can make use of the resources listed in the link andselisk supplied: (1) go to the software list he linked. (2) Identify the subset of those that do chemical drawing. (3) Google: '[software name] adjust line widths in structural formulas'. If you do this with, e.g., MarvinSketch, you will find this link, docs.chemaxon.com/display/docs/drawing-settings.md , which indicates these can be adjusted in MarvinSketch using "Wireframe bond thickness", which "Sets the width of all bonds and lines."

@theorist I see your point. Let me reframe my closing decision as motivation for OP to narrow down their request. Shall they decide to ask how to tweak the detailed appearance of a bond within a specific piece of software, I'd immediately reconsider my close-vote. Currently I see the question broad enough to be a duplicate without much of a stretch. Anyway, please feel free to cast a reopening vote if you still disagree.

Thank you all for your comments. I ended up using a vector editor to change the undesired features of the original svg files I found in Wiki Commons.

source: Stack Exchange

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

sort: votes


text:
Eigenenergies and states of Jahn–Teller molecule
I'm generally interested in numerical chemistry. In our lecture, we have discussed the $E \otimes e$ Jahn–Teller molecule with the following Hamiltonian
\begin{align}
 H = \epsilon n_d + \omega ( a^{\dagger}_x a_x + a^{\dagger}_y a_y) + \lambda (Q_x \sigma_x + Q_y \sigma_y )n_d.
\end{align}
So we have two degenerate electronic levels and two degenerate vibrational modes. The operators $Q_{x/y}$ are the position operators and $n_d$ is one when the molecule is in the charged state.
I'm interested in both the eigenenergies and eigenstates of this Hamiltonian, I initially thought, that this is not complicated, as we have just the operators given in second quantuzation.
I found in some publications that it is convenient to write down this Hamiltonian in the polar basis of the harmonic oscillator. So I transformed the Hamiltonian in this basis. There the basis states look like $|n,m\rangle$, with $n$ the radial excitation and $m$ the angular momentum. So, as a first try, I expressed the Hamiltonian for the uncharged molecule, in order to be sure that I understood the Hamiltonian in a correct way and I receive the correct energies according to $H|n,m\rangle = \omega n |n,m\rangle$.
Now I'm trying to express the charged states in this basis. Again, $m$ is the angular momentum of the state in the 2D isotropic harmonic oscillator and ranges in $n-1,n-3,\ldots,-n+1$. Now in the lecture, we claimed that the total angular momentum is conserved, i.e. I also need to consider the spin of the electron. However, I'm not sure how to adopt this in the basis. For sure, I need to introduce an additional parameter in my basis states, namely the spin.
My only try was to introduce states $|n,j\rangle$ with the total angular momentum $j=m\pm \sigma$, since I really have no idea to do so. However, the $j$ would be a half-integer, what is impossible for the 2D harmonic oscillator states. So there is something that I'm not getting correct.
Has someone a hint that could bring me back to the track? Do you have any references that discuss the eigenstates of this Hamiltonian?

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


text:
Coordination number of metals in complexes
The co-ordination number of a metal is fixed in which way?
My chemistry textbook says, " The secondary valence is equal to the coordination number and is fixed for a metal."
Obviously, the coordination number for a metal is not constant in all its compounds. So does it mean to say that it is constant in just the isomers of the compound? Or,does it mean that it is constant in all compounds with the same central atom/ion, anions and ligands as seen for Co in CoCl3.6NH3 , CoCl3.5NH3 and CoCl3.4Nh3(=6)? So all complex compounds with Co, NH3 and Cl only(in various proportions)will have coordination number of Co equal to 6?

It means the first.

It's usually constant but there's always an compound that shows exception. What I have been taught is that the coordination number is in most cases is the number of ligands directly bonded to the CMA. So for something like [PtCl6]-2 the coordination will be 6, as 6 ligands are directly attached to the CMA, also keep note the coordination number for a particular metal atom won't be more than it's valance electrons.
Similarly [PtCl4] will have coordination number of 4.
To summarize, it is not constant, but due to most CMA showing same CN over and over it is generally said to be so.


text:
Determining ion charge using rate constants and ionic strength
I am given the following data:

The rate constant $k$ at $\pu{25 ^\circ C}$ in a reaction of persulfate ions and iodide ions in a water solution is assumed to vary with the total ionic strength of the solution which follows:
$$
\begin{array}{l|cccccccc}
\hline
I/\pu{mol dm^-3} & 2.45 & 3.65 & 4.45 & 6.45 & 8.45 & 12.45 &\\
\hline
k/\pu{dm^3 mol^-1 s^-1} & 1.05 & 1.12 & 1.16 & 1.18 & 1.26 & 1.39 \\
\hline
\end{array}
$$
Determine the value of the product of the ion charges $(z_Az_B)$ for the two ions in the reaction.

From previous problems I thought this would be a straight forward problem solved by using the equation:
$$\log k = \log k_0 + 1.02z_Az_B(I)^\frac{1}{2}$$
So I plotted $\log k$ against $I^\frac{1}{2}$ and got that the slope is $0.0584$. This is not the correct answer, the correct slope should be around $2$ but how this slope is obtained is not given to us. I was wondering if anyone here sees what I am doing wrong? All help is appreciated!

This extrapolation of the Debye–Hückel equation is too far stretch. There is already significant deviation for c=0.1 M.

It seems there is an error in the concentration. OP please check if the conc. is right.

@M.Farooq I have written the problem exactly as given to me


text:
Can NaH open the epoxide ring to form alcohol? If so, how?
I was looking at the organic chemistry 8th Ed textbook by Paula Yurknais Brucie. On page 509, I found this:

I think about it for a while and I am doubting that it is wrong. I believe the hydride ion itself cannot act as a nucleophile for SN2. (not really sure but I guess it is too small and too basic to work as a nucleophile, it should be much easier to attack hydrogen rather than carbon for hydride ion)
Is this reaction is possible with $\ce{NaH}$ in solvents other than water and alcohol? If so, how does the hydride ion open the epoxide ring? (hoping for an answer with mechanism)
And I am specifically talking about hydride ion (or hydrogen anion). Not sodium borohydride $\ce{NaBH4}$, lithium aluminum hydride $\ce{LiAlH4}$, or other metal hydrides.

Well, there's no proton acidic enough, so it's gonna open the ring.

Going to have to be in aprotic solvent. NaH reacts vigourously with alcohols and water

I have been looking for this reaction and I find no examples. Nor in my limited experience of 40+ years can I ever recall seeing sodium hydride used as a source of nucleophilic hydride without some modifying agent (such as this example https://www.sciencedirect.com/science/article/abs/pii/S0040403900986313). Perhaps someone who has better database access than I can find it.

@Mithoron I don't think so. It is going to sit there and do nothing until it is worked up.

This is a duplicate of this. It has a satisfactory answer.

@MathewMahindaratne  the answer you refer to concerns LiAlH4 opening of epoxides. It specifically excludes any reference to NaH even though the OP asked about NaH.

Further to my comments above, I am going to offer my opinion that the book is wrong and the epoxide is not opened by sodium hydride as depicted. I have failed to find any literature examples of such a reaction and, in 40+ years as a synthetic chemist, I can recall no reactions in which $\ce{NaH}$, unmodified by other reagents, acts as a source of nucleophilic $\ce{H-}$.
Can anyone produce a literature example?

@user55119 do you know of any examples?

I was as nonplussed as you when I saw this reaction of "naked" NaH. In my 40++ years I have not seen such a transformation. My Chem. Absts. search turned up only the reference in your Comment (DOI:10.1016/S0040-4039(00)98631-3). Textbooks are invariably riddled with errors and omissions.

Apropos of my final remark above, I once used a textbook that interpreted in great detail the fragmentation pattern of the EI mass spectrum of methyl propyl ether. Only one problem. The spectrum was that of diethyl ether!

@Waylander, maybe the textbook is incorrect and just makes up this reaction out of nowhere. I also cannot found any example in the literature. Thanks for the answer.

Kicked it around this evening at the pub with a synthetic chemistry colleague of similar experience - he also thinks this reaction will not occur.


text:
Total orbital angular momentum in a filled p subshell
From "Physical Chemistry", P. Atkins, J. De Paula, 9th edition, page 358

"A closed shell has zero orbital angular momentum because all the
individual orbital angular momenta sum to zero. Therefore, when
working out term symbols, we need consider only the electrons of the
unfilled shell."

So, a configuration like $[\text{He}]1s^2$ or $[\text{He}]1s^2 2p^6 $ should have $L = 0$, right? The first presents $L = 0$ for sure, since $l = 0$ for $s$ orbitals and applying the Clebsch-Gordan series we obtain $0$. Doing the same for the second configuration, on the other hand, I obtain values other than $0$.
Could someone explain to me better?

One way of looking at this is to consider that if $L = 1$ (for example), then there must be a state with $M_L = -1$ (because $M_L$ ranges from $-L$ to $L$ in integer steps).
However, $M_L$ is not obtained by a Clebsch–Gordan series: instead, it's just a simple addition of the individual $m_l$'s of each electron. That is:
$$M_L = m_{l,1} + m_{l,2} + m_{l,3} + m_{l,4} + m_{l,5} + m_{l,6}$$
and for a p subshell, the allowed values of $m_l$ are $-1$, $0$, $+1$ (there are two of each). So $M_L$ for a full p subshell must be $0$, and it's then a contradiction to have $L \neq 0$ as that would imply that $M_L$ can have a nonzero value.
This is basically a manifestation of the Pauli exclusion principle, because it is precisely that which forces the sum of $m_l$'s to be zero. If you read around in Atkins, it will mention how certain term symbols are forbidden due to the Pauli exclusion principle. This is a similar case.


text:
when copper oxide is dissolved in vinegar does it release copper oxide gas
i was cleaning my pennies in a bowl of vinegar and table salt and was wondering when the copper oxide on the pennies dissolves does it release copper oxide gas or a gas i should be worried about. I was sitting outside with a fan on above me but i was worried that when i leaned over the bowl i might have inhaled somthing harmful. I tried to look up what happens but i could not find anything that states what happens to the copper oxide when its "dissolved".

Nothing to worry about: there is no “copper oxide gas”.

You're healthy enough to have posted about it here. Carry on...

I wish this degree of encouragement is also given to other new posters by moderators :-)

Thank you all i was just a little paraniod there was some sort of harmful gas

The best defence against the fear of unknown is obtaining knowledge about it. If you studied basics of copper chemistry, you would realize that copper oxide is not a gas and that there is no easy way to obtain a volatile copper compound.
The only potentially dangerous gas there is vapor of acetic acid, natural and essential part of vinegar, with low enough concentration to be safe(*).
Insoluble solid copper oxide (possibly together with basic copper carbonates) gets transformed to blue solution of non volatile copper acetate.
$$\ce{CuO(s) + 2 CH3COOH(aq) -> (CH3COO)2Cu(aq) + H2O(l)}$$
$$\ce{CuCO3.Cu(OH)2(s) + 4 CH3COOH(aq) -> 2 (CH3COO)2Cu(aq) + 3 H2O(l) + CO2(g)}$$

(*) If you ever sniffed vapor over pure acetic acid, the perception would be very unpleasant and your nasal mucous membrane would be for some time very unhappy.


text:
What does the following diagram represent?
Suppose, the following is a diagram of a protein's polypeptide chain:

What does this diagram represent?

What are the letters A, E, M, W, L, N, S, etc. represent?  (I suppose these are amino acid codes. Right?)
Which one is the c-alpha atom here?
What are the sequence numbers i-2, i-1, i, i+1, i+2, etc. represent? Why there are negative and positive increments from i?

The following is a related diagram:

Note the schematic is from https://chemistry.stackexchange.com/questions/153935/what-is-the-meaning-of-this-data-from-a-protein-data-bank-pdb-file

What is the origin of the diagram?

@BuckThorn, some book. i don't know the exact source.

Yes, those are 1-letter amino acid codes
The image just shows each amino acid as a single point, so it doesn't the individual atoms, including the C−α.
This may require knowing where this figure came from to be more specific, but they seem to be interested in the distance between amino acids before and after some particular point in the chain.

i don't know the exact source. all I know is that it's from a book.


text:
Why does chlorine show a mesomeric effect in haloarenes and an inductive effect in haloalkanes?
Chlorine shows the +R effect in haloarenes and the -I effect in haloalkanes. Does chlorine do the same for all cyclic compounds or only for aromatic rings? Is this just another experimentally seen fact about chlorine or is there a higher mechanism involved?

Chlorine is always +R and -I, regardless of the compound.

In aromatic rings like benzene it shows less pulling of electron towards itself when compared to alkanes why is this so? Sorry if I caused any confusion, it'd be a great help if you could elaborate on your answer!

Because in the aromatic case it other than pulling the electrons also gets a chance to donate electrons whereas in the aliphatic case it can only pull.

Quoting myself, "Theoretical reason why ...."

Why, because it has a lone pair of electrons which are almost in plane with the aromatic system. Chlorine lp can also resonate.

Oh man ;) I'm the one who added that into the question but sure :P @DhruvDesai

The thing is to show +R , you need to have something into which the lone pairs can resonate into , maybe you should compare arenes with other conjugated systems

@Buraian yes all credits go to them to correct my broken chemistry english.

There is a book named David R Klein's Organic Chemistry, you should check out the conjugated dienes chapter. It starts a bit heavy with MO theory but it is understandable with HS knowledge only. I think that would give you some more insight into how conjugation works.

So, if I'm not wrong its because of the partial double bond nature of benzene?

Not just partial, it's in the way that the electrons are pushed that a meaningful delocalization of electrons take place

Translation: +/- R effect = +/- M effect -> Mesomeric effect, see Wikipedia https://en.wikipedia.org/wiki/Mesomeric_effect


text:
What is the theoretical reason for the dependence of the surface tension on the alcohol concentration in water?

The qualitative sketches I, II and III given
below show the variation of surface tension
with molar concentration of three different
aqueous solutions of$ \ce{ KCl, CH_3 (OH) }$and
$ \ce{CH3(CH2)_{11}  O SO_3^{-} Na^+}$
at room temperature

What is the correct assignment(s) of sketches?

It is understandable that $ \ce{CH3(CH2)_{11}  O SO_3^{-} Na^+}$ will have a sharp reduction surface tension with concentration (reference) and $\ce{KCl}$ increases surface tension with concentration since it is an inorganic salt. Through option elimination, we conclude that I must be methanol but would there be an independent way to conclude the same (without option elimination)?
I found this  but the answers are not really authoritative.

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

Curve I could be that of a detergent, but sodium dodecyl sufate decreases the tension faster than methanol and displays a very distinct change in behavior with onset of micellization at the critical micelle concentration. Above that concentration the amount of free solute - the concentration of which is accountable for the change in surface tension - remains approximately constant.
Methanol on the other hand is not a detergent and does not micellize. It does however reduce the surface tension of water, presumably because it disrupts the hydrogen bonding network in water. The boiling point of water is similarly decreased by addition of methanol.
Note that a similar decrease is observed for ethanol solutions, but that in some regards the ethanol-water system is more complicated (forming for instance an azeotrope).
Beyond this it is not possible to generalize, as hydrogen bonding renders water a very complicated liquid. There are something called structure forming (kosmotropic) and structure destroying (chaotropic) solutes. The ions dissociating when $\ce{KCl}$ dissolves jointly appear to have a net kosmotropic effect, but you'd have to distinguish this from the boiling point elevation expected on statistical grounds (Raoult's law). For any non-associating (eg associating by H-bonding) liquid adding a solute is expected to raise the boiling point. But in the case of water the effect of adding a solute might be that you enhance H-bonding. So it can be an effect beyond the statistical one.

If it disrupts and hence decreases, then can't we argue the same for KCL since the ions interrupt the hydrogen bonding?

Well, there is no simple answer. There are something called structure forming (kosmotropic) and structure destroying (chaotropic) solutes. The ions dissociating when KCl dissolves jointly have a net kosmotropic effect, and you'd have to distinguish this from the boiling point elevation expected on statistical grounds (Raoult's law).

Where does raoult  law come in here?)

Because for any non-associating (eg H-bonding) liquid adding a solute is expected to raise the boiling point. But in this case the effect might be that you enhance H-bonding. So it is an effect beyond the statistical one.

Oh , so an negative deviating azeotrope. I see, this question is more complex than I had thought initially.


text:
Reactivity of Benzaldehyde vs Acetaldehyde and Benzoic acid vs Acetic acid
I was trying to compare the reactivity between acetaldehyde and benzaldehyde. After some googling, I found out that benzaldehyde is less reactive than acetaldehyde because the benzene ring can stabilize the partially positive Carbon through resonance. However, if I apply the same logic to benzoic acid vs acetic acid, I find that benzoic acid should be less acidic than acetic acid, but that's not true. What am I missing?

Does this answer your question? Why is benzoic acid a stronger acid than acetic acid?

@Mithoron I saw that question before. The answer did not satisfy me. In the comments, A.K said "The aromatic ring is so thermodynamically stable that the phenyl ring donating electrons to the carboxylate is not a major resonace structure. Though it is enough to dictate meta directed electrophilic aromatic substitution", by this argument, can't it be said that the phenyl ring should not stabilize the carbonyl carbon in benzaldehyde through resonance?

The reactivity of an aldehyde group (or a ketone group for that matter) depends on the strength of the partial positive charge on the relevant carbonyl carbon. Thus, when comparing the reactivity of benzaldehyde and acetaldehyde, it is a valid argument that benzaldehyde is less reactive than acetaldehyde because the adjacent phenyl ring can stabilize the partially positive charge on carbonyl carbon through resonance (decreasing the strength of positive charge). That is possible because carbonyl group and phenyl ring is directly conjugated.
However, that same argument is not valid when you compare the reactivity of benzoic acid and acetic acid. That's because the reactive center of acidity is not the carbonyl carbon anymore as it was for aldehydes and ketones. The reactivity of carboxylic acid is depend on the strength of the $\ce{O-H}$ bond on carboxylic group, a bond away from carbonyl carbon. Since it is not directly conjugated to the other group attached to the carbonyl carbon, the resonance is not playing the role as that is in aldehyde and ketones. The inductive effect is the one playing strong role in this case. The more the electron withdrawing ability of the group attached to $\ce{COOH}$ the better the acidity. Thus, when comparing the electron withdrawing ability of phenyl and methyl groups, $\ce{Ph \gt CH3}$. Therefore, benzoic acid is more acidic than acetic acid by that reason alone.

"The reactivity of carboxylic acid is depend on the strength of the O−H bond on carboxylic group, a bond away from carbonyl carbon" but O-H bonds in carboxylic acids are weak because of the partial charge developed in the carbonyl carbon. So if resonance can stabilise the carbonyl carbon, then won't the O-H bond be stronger?

I'm confused as to why direct conjugation is so important, because in my point of view, the acidity of the group mainly depends on the stability of the carbonyl carbon. O-H group is present in alcohol too, but they aren't very acidic.

@Tahsin Choudhury: Remember, the acetic hydrogen is equally distributed between two oxygen in $\ce{COOH}$ group. You may need more reading on this subject.

Sorry, what do you mean by that? Could you share some resources to know more about it?


text:
Do we prefer the main functional group or the longest chain when identifying the main carbon chain IUPAC nomenclature?
Sorry if this question was answered before, but I could not find a satisfactory answer from related questions.

Anyways, I have trouble identifying the main carbon chain in above structure. As you can see, there is a carboxylic substituent connected to the longest carbon chain which has 7 carbon atoms. So when numbering the chain the active functional group should receive the smallest number according to IUPAC rules. So, if we begin from the carbon atom in the carboxylic substituent , there will be 6 atoms in the carbon chain but, the main functional group will receive the smallest number. Or, should I choose the longest chain possible despite above facts.
Also , I am quite unclear which number does the carboxylic group get if we choose the longest chain?
Could anyone answer the above 2 questions please...

Does this answer your question? When we name an organic compound ,if there is double bond and a functional group which is now a substituent who has priority?

@NisargBhavsar no, actually my question is ,do we consider giving the carbon atom of  main functional group the lowest number possible.....despite it is not the longest chain?

Did you read the answer to the linked question fully? The first rule clearly states that the primary functional group is given priority over chain length.

I am adding an answer to make my point clearer.

The details in this anser gives the recent IUPAC recomondations of how to choose the principle chain.

The following are the rules given by Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013[1] for selecting the primary/parent carbon chain/skeleton:

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)

It is clear from the first two rules itself that the primary functional group is given priority over the length of the parent chain and thus in your example the carboxylic acid should be given priority over the possible longer parent chain.
Thus the preferred IUPAC name of your example will be 2-ethylhexanoicacid.

Reference:
(1)   Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013; Royal Society of Chemistry: Cambridge, 2013.


text:
Can I dissolve IBA from talc powder?
My container of in total 25 grams of rooting powder contains 2.5 mg IBA (Indole-3-butyric acid), which is mixed with talc powder and a tiny bit of chromium(III)oxide. I know talc is not soluble in water or ethanol. But is it possible to add ethanol to dissolve only the IBA? The talc will then remain in my solution. Is this correct?

Since IBA is not completely soluble in water, it is typically dissolved in 75% or purer alcohol for use in plant rooting, making a solution of between 10,000 and 50,000 ppm. This alcohol solution is then diluted with distilled water to the desired concentration. IBA is also available as a salt, which is soluble in water.  https://en.wikipedia.org/wiki/Indole-3-butyric_acid

The question remains,how much ethanol is needed for the near complete extraction, as it is not the same as dissolving pure compound. Probably the best way would by slow column continuous extraction,similar as elution of ions captured in ion exchaging resin.

chromium(III)oxide in rooting powder? That is not possible.

Rooting powder can have fungicides, it is often just not talc, it is advisable not to play around with it.

Thanks all for the answers. I checked the SDS, according to that it contains:

TALC >50%
Chromium(III)oxide <1%
IBA <1%
I think the chromium oxide is there to give it a bit of color. Its a nice green powder. I added to 0,8 grams of powder 2ml pure ethanol, shaked it and added it up to 10 ml with destilled water. Let's see if my in-vitro cultivation works...

It is very strange to find Chromium (III) oxide in rooting powder. I cannot find any role of chromium. Be careful, chromium (III) is toxic and known to cause cancer.

@M.Farooq Interesting. It is well known about Cr(VI) compounds, but I have never heard it about Cr2O3. It is widely used as a pigment.

Be aware the talc matrix may play an important role in the kinetics of IBA release, similarly as for medicine pills. In both cases, the way how it is released can be as important as the presence of the active component itself.


text:
Do decomposition potentials change during electrolysis?
I was just wondering if decomposition potentials actually change during electrolysis. For example, if we reversed the Daniell cell reaction so that copper is oxidized, the half cell reactions would be:
$$\ce{Cu <=> Cu^2+ + 2e-} \qquad \pu{+0.34 V}$$
$$\ce{Zn^2+ + 2e- <=> Zn} \qquad \pu{-0.76 V}$$
if we assume both of the solutions are at $\pu{1 M}$ at $\pu{298.15 K}$, the decomposition voltage should be $\pu{1.1 V}$.
If we applied the voltage until half of the zinc cation solution reacted, we would have $\pu{1.5 M}$ $\ce{Cu^2+}$-solution and $\pu{0.5 M}$ $\ce{Zn^2+}$-solution, so that the resulting decomposition voltage should be higher:
$$E_\ce{Cu} = 0.34 + \pu{0.0295 V}\times \log (1.5) = \pu{0.345 V},$$
$$E_\ce{Zn} = -0.76 + \pu{0.0295 V}\times \log (0.5) = \pu{-0.769 V}$$
This should result in a decomposition voltage of $\pu{1.115 V}$ approx. Am I correct on that matter? I haven't found a single article that talks about the actual potential changes during electrolysis, even in my exam I only had to calculate the initial voltage, not what happens afterwards.

You can simply use Nernst equation for the cell reaction, $\ce{Zn + Cu^2+ <=> Zn^2+ + Cu}$ with $n = 2$.

Hah! I was just having the same question a few days ago. Your conclusion is correct.


text:
How is electron richness defined?
What does it mean for an atom/molecule to be electron rich? For instance, why is the methyl carbanion considered electron rich (having excess electrons) even though it has a complete octet, no more, no less? I know that if the octet is incomplete, an atom/molecule is considered electron deficient, but how exactly is electron richness defined? I was under the impression that electron richness referred to atoms/molecules which had more than 8 electrons on the central atom.

These are just wording used by organic chemists. It has nothing to do with reality. What is experimentally realizable is the electron density by X-ray measurements. The rest is all hand-waving spam.

It is electron rich as it still has electrons to spare for bond formation.

Like temperature, without established reference, there is no comparison possible.  A reasonable comparison would be like «There are more $\pi$ electrons per non-H atom in pyrrole than in benzene; thus the $\pi$ electron density in pyrrole is higher than the one in benzene and considered electron rich.»  If not reformulated, the question will attract opinions instead of answers backed by evidence.


text:
How many 4th nearest neighbours are there in an hcp lattice?
My teacher told us that there are 18 4th nearest neighbors in an hcp lattice, at a distance of √3a . I have only been able to identify 12.
Can you help me find all the 4th nearest neighbors?

With HCP, things get hairy pretty quickly.
This is the first coordination sphere, with 12 atoms at the distance $a$ from the origin:

These are the second nearest neighbors, i.e. the 6 atoms at $\sqrt{2}a$ from the origin, each touching 4 atoms of the first sphere:

These are the third nearest neighbors, which are the 2 atoms at $\sqrt{8\over3}a$ from the origin, each touching 3 atoms of the first sphere:

And finally, here are your 18 atoms which are the $4^{th}$ nearest neighbors of the atom at the origin. Each touches 2 atoms of the first sphere:

I found one of the early Olex2 advertising cards mentioned.  A small scale photo is here in the chatroom.

I'm sorry but I did not understand the figures for the 2nd and 4th nearest neighbors. Has the atom at the center been taken as the origin? The atoms linked by the black lines represent the required neighbor?

Yes and yes. Also, note that the currently added atoms on each stage are shaded, while the previously added atoms are empty.

In the figure for second nearest atom, there are 18 atoms linked by the black lines. However, there are only 6 second nearest neighbors. Similarly, in the the figure for 4th nearest neighbors, there are more than 18 atoms linked by the black lines. In the figure for 1st and 3rd nearest neighbors, I can make out the required atoms. Could you please clarify the required neighbor in the 2nd and 4th one.

When a new atom is linked to an old atom, they both are linked by a black line. There is no going around it. Just compare any picture to the previous one to see which atoms are new.


text:
Can an alloy catalyst change during a reaction?
I am studying the reaction pathway of oxygen evolution reaction (OER) on a PtNi alloy catalyst using density functional theory (DFT). What I found is that, for different intermediates, the most stable configuration of the PtNi(111) surface can be different. Let's say I have a $2\times2$ cell, and the most stable surface composition for the HO* intermediate is Pt$_4$, whereas for the O* intermediate, the most stable surface composition is Pt$_3$Ni (O* attached to Ni). My question is as follows:
Is the alloy catalyst allowed to change during the reaction? If we consider the reaction step HO* → O*, the easiest way is to assume the catalyst doesn't change during the reaction. However, if the energy barrier of the reaction is lowered so much when O* bonds Ni that it compensates the energy barrier of swapping a surface Pt atom with a subsurface Ni atom, does it suddenly become favorable? I think it comes down to comparing one large energy barrier vs. two small energy barriers, but I can't find any theories regarding this. Any suggestion is appreciated.

I am not very familiar with the catalyst system you mention, but if the catalyst is changed by one step of the reaction, then won't the catalyst be deactivated eventually? Change of surface structure of catalysts is quite common, but if that deactivates the catalyst, there has to be a way to return to the original configuration.


text:
Ring Flip of glucose
I have recently read about mutarotation of glucose specifically
I am familiar with how cyclic structures like substituted cyclo-hexanes exist in an equilibrium of both their chair conformers.
The thing is, for mutarotation, the interconversion takes place via an open-chain form, which is I think supported by enough experimental evidence like the specific rotation of equilibrium mixture.

But, can ring flip take place the old-school way, that is without opening the ring in a solvent like CCl4 or Benzene
Or is the ring too strained to cross the energy barrier?

Cyclohexane does not require change of chiral centers to change conformation. R stays R and S stays S. For mutatotation, you need a change in a chiral center, which requires a bond to break.

The short answer is no.

Experiment with a model kit (example) and recognize changing conformers only is about wiggling/rotating bonds which may be gradual in many tiny steps.

The transition from one isomer to an other (e.g., $\alpha{} \rightarrow \beta{}$-glucose) is binary like using a light switch; instead of on/off, it is either (R), or (S) – or, in case of double bonds, either (E), or (Z).  This requires breaking existing/creating new bonds.  Solvent, temperature, radiation, pressure, reagents, catalysts may affect the energetic barrier to achieve this, but normally this change requires surpassing a much higher energetic barrier than the typical rotation around a $\ce{C-C}$ single bond.

While simplified,* this description often is sufficient for everyday like chemistry, including your example about the inversion/mutarotation of glucose dissolved in water.  In benzene or $\ce{CCl4}$, this reaction however probably will proceed at a much lower rate of conversion than in water because the assumed mechanism includes steps of protonation/deprotonation and these two solvents are not as good "proton shuttles" (as a Bronsted acid/base) as is water.
From a practical perspective, either form of glucose will not dissolve well in benzene, or $\ce{CCl4}$.  Mutarotation however is observed for many sugars e.g., permethylated glucose in benzene and surfactants (e.g., described by Fendler et al.), or pyridone (e.g., Rony) and still of interest today (e.g., Dujardin et al.).
*) My answer here points to some exceptions from this simplification.
References
Fendler, E. J.; Fendler, J. H.; Medary, R. T.; Woods, V. A. Catalysis by reversed micelles in non-polar solvents : mutarotation of 2,3,4,6-tetramethyl-α-D-glucose in benzene. J. Chem. Soc. D 1971, 1497–1497; doi 10.1039/C29710001497.
Rony, P. R. Polyfunctional catalysis. I. Activation parameters for the mutarotation of tetramethyl-D-glucose in benzene. J. Am. Chem. Soc. 1968, 90, 2824–2831; doi 10.1021/ja01013a018.
Dujardin, N.; Dudognon, E.; Willart, J.-F.;  Hédoux, A. H; Guinet, Y.; Paccou, L.; Descamps, M. Solid State Mutarotation of Glucose. J. Phys. Chem. B 2011, 115, 1698–1705; doi 10.1021/jp109382j.


text:
Water density in the calculation of the ionic product of water
In 2019 the International Association for the Properties of Water and Steam released this document for the semi-empirical calculation of the ionization constant of water: http://www.iapws.org/relguide/Ionization.pdf
My idea was to insert this formula in my spreadsheet for calculating the pH of an acid base mixture, where until now $K_w$ I entered it manually.
On the other hand, this formula, in addition to requiring the inlet temperature (between $\pu{0°C}$ and $\pu{800°C}$), also requires the density of the water (between $0$ and $\pu{1.25 g/cm3}$).
Since the density of water in turn depends on temperature and pressure, I wondered if in the literature a sufficiently compact formulation could be found that would give me the density of the water at the input temperature and pressure.

Before asking here of course I searched far and wide:

in MapleSoft here they show their procedure, which in turn refers to this other routine which is apparently secret;

in WolfraAlpha I found this widget which in turn refers to a long line of books in which I have not found anything that can do for me;

this is the only page where without many turns of words they show a simple algorithm, too bad it is only valid for temperatures between 0°C and 40°C, if it had been valid at least up to 100°C it would have been enough.

This is why I was wondering if anyone was aware of some more general formulation of the latter.

The problem with equations for a wide range is that they are either not  accurate enough either not simple enough.  There is always possible to involve numerical interpolation of tabelized values that are available, e.g. by a cubic spline.

Here is the scanned PDF on nvlpubs.nist.gov with the formula from 1975 for 0-150 deg C range.


text:
How do buffers work? and why only solutions of a weak acid and its salt with a strong base or vice versa work as a buffer
Please note that the definition of a buffer in my textbook is given as

The solutions which resist change in pH on dilution or with the addition of small amounts of acid and alkali are called Buffer Solutions

Is this the right explanation of a buffer?
If not what is the right explanation?

In contrary to some other Q/A or forum sites, answers on CH SE site are figuratively paid by the user's own effort. When you ask, it is expected you have already searched for and thoroughly thought about the topic, providing explicit compact summary of partial answers or at least ideas or thoughts you have got until then. Effort not shown may be considered as effort not done and such a question may be closed.  How do I ask a good question.

pH basically means "how much of $\ce{H+}$ there is in the solution". And the job of the buffer is to keep this concentration more or less constant: if someone "consumed" $\ce{H+}$ our buffer has to generate new $\ce{H+}$; and if someone produced $\ce{H+}$, our buffer needs to "consume" the extra amount.
Suppose you have a solution with $\mathrm{pH=4}$ and you add a weak acid (let's call it $\ce{HA}$) with approximately the same $\mathrm{pK_a=4}$. Because $\mathrm{pH=pK_a}$ half of that $\ce{HA}$ substance will be dissociated into anion ($\ce{A-}$) and $\ce{H+}$ and the other half will stay intact as $\ce{HA}$. But if something changed the $\mathrm{pH}$ of the solution, then the proportions of Dissociated:Intact versions of the molecule would shift.
So if we then try adding another acid which is stronger ($\mathrm{pK_a=2}$):

We'll get an excess of $\ce{H+}$ from that new acid
And those extra $\ce{H+}$ will be consumed by the original $\ce{A-}$ turning it back to $\ce{HA}$. Therefore $\mathrm{pH}$ of the solution will stay approximately the same.

If we add a base ($\mathrm{pK_a=6}$):

Those $\ce{H+}$ that left the original $\ce{HA}$ will be consumed by that base
Which means the concentration of $\ce{H+}$ drops, but this in turn leads to our original $\ce{HA}$ dissociating further producing more $\ce{A-}$ and $\ce{H+}$

Therefore $\mathrm{pH}$ of the solution again stays more or less the same. Here are some video explanations of Acid-Base equilibrium and Buffer Solutions.


text:
stereoselectivity in aldol reactions
in "Advanced Organic Chemistry, part B" fifth edition book, page:103, the authors compare between aldol reactions where 1-substituted nucleophile is used. They stated that when the 1-substituent contains benzyloxy group stereoselectivity is less than that of siloxy group, the problem in the following drawing is that reaction with the chelated TS shows that 2,2'-syn-2,3-syn is preferred than 2,2`-anti-2,3-syn. But I wonder how is that possible with  steric interaction between axial-like hydrogen in the aldehyde and (R-) group?
also I would like a better source for studying stereochemistry because this book is a little ambiguous in explaining it.

Erik Carrera has a great book called Classics In Stereoselective Synthesis. Dave Evans Harvard lecture notes are also widely accessible

Here is the source if you have access. https://www.sciencedirect.com/science/article/pii/S0040402011008805 DOI:10.1016/j.tet.2011.06.019

What is the reference given for this reaction?

@MathewMahindaratne ,S. Figueras, R. Martin, P. Romea, F. Urpi, and J. Vilarrasa, Tetrahedron Lett., 38, 1637 (1997).

I do not have the recent or the fifth edition of Advanced Organic Chemistry: Part B but have the forth edition (Ref.1) to answer this question. First, this book is not for for studying stereochemistry, but for studying reactions used in synthesis. In that category, this book can't be a bad book because al most all universities in US use this book for advanced post-graduate teaching.
Said that, let's look at the question in hand. I suppose the authors expect the interesting students to look at the relevant reference (Ref.2) for further learning.  The given transition states and relevant diastereo-selectivity $(de)$ are given as below:

The OP's question as follows:

when the 1-substituent contains benzyloxy group stereoselectivity is less than that of siloxy group, the problem in the following drawing is that reaction with the chelated TS shows that 2,2'-syn-2,3-syn is preferred than 2,2`-anti-2,3-syn but how is that possible with steric interaction between axial-like hydrogen in the aldehyde and (R-) group?

I highlighted the main group in the question for convenience, which is the $\ce{R'}$ group in all three transition states. The so-called chelated transition states are TS-II and TS-III (Ref.3). If you look at carefully, you'd see TS-II would have substantial steric hindrance due to $\ce{R}$ group and $\ce{i-Pr}$ group. The free rotation of $\ce{CHR-OR'}$ group is restricted due to the $\ce{OR'---Ti}$ chelation. As a consequence, $\ce{R}$ group is situated just above the bulky $\ce{i-Pr}$ group. On the other hand, under the same situation with chelating of $\ce{OR'---Ti}$ in TS-III, the steric hindrance for $\ce{R}$ group is minimal because there is no group below it.  The same situation dictates the non-chelated-transition state, TS-I.
When $\ce{R'}$ group is $\ce{TBDMS}$ group, the 5-membered chelate is likely not so abundant in the equilibrium (Ref.4 and references therein). For instance, it has been stated in Ref.4 that:

Alkyl substitution on oxygen is known to permit chelation with common Lewis acids, e.g., $\ce{MgBr2, SnC14}$ or $\ce{TiC14}$, whereas substitution with trialkylsilyl apparently attenuates or even eliminates the ability of an oxygen to chelate.

Thus, it is safe to assume that When $\ce{R'}$ is $\ce{TBDMS}$ group, non-chelated transition states are likely dominate the stereochemical outcome with higher syn-syn isomer.
References:

Francis A. Carey and Richard J Sundberg, In Advanced Organic Chemistry, Part B: Reactions and Synthesis; Fourth Edition, Kluwer Academic Publishers: New York, NY, 2002 (ISBN: 0-306-46244-3).
Sergi Figueras, Ricardo Martín, Pedro Romea, Fèlix Urpí, Jaume Vilarrasa, "Highly stereoselective aldol reactions of titanium enolates from ethyl α-silyloxyalkyl ketones," Tetrahedron Lett. 1997, 38(9), 1637-1640 (DOI: https://doi.org/10.1016/S0040-4039(97)00108-1).
Note that according to Ref.2, the meaning of the non-chelated- and chelated-transition states mentioned here is the five-membered chelation within enolate $\ce{O}$, $\ce{OR'}$, and the metal ion. In all three transition states, the six-membered transition chelation exists.
S. D. Kahn, G. E. Keck, W. J. Hehre, "The Effect of Protecting Groups on Chelation Control," Tetrahedron Lett. 1987, 28(3), 279-280 (DOI: https://doi.org/10.1016/S0040-4039(00)95706-X).

I understand the chelation extent of TBDMS what i can't understand is that if TS III (anti) is more stable than TS II (syn), then why is the ratio of syn : anti (in the data table) still high for TBDMS

@mohamed: When R' = Bz, all three TSs have involved to give final products. The out come depends on the size of R as well. When R' = TBDMS, non-chelate group is favored al most exclusively.  That is the major focus of the paper.


text:
Justifying assumptions about method to find equilibrium pressure for co-existence of graphite and diamond

The standard state Gibbs free energies of formation of $\ce{C(graphite)}$ and $\ce{C(diamond)}$ at $T = \pu{298 K}$ are $\pu{0 kJ mol-1}$ and $\pu{2.9 kJ mol-1}$, respectively.
The conversion of graphite $\ce{C(graphite)}$ to diamond $\ce{C(diamond)}$ reduces its volume by $\pu{2e-6 m3 mol-1}$.
If $\ce{C(graphite)}$ is converted to $\ce{C(diamond)}$ isothermally at $T = \pu{298 K}$, the pressure at which $\ce{C(graphite)}$ is in equilibrium with $\ce{C(diamond)}$ is:
(A) $\pu{14501 bar}$
(B) $\pu{58001 bar}$
(C) $\pu{1450 bar}$
(D) $\pu{29001 bar}$
JEE Adv, 2017, related 1, 2

In the solution stated in this site, there are two assumptions made about this process.

$\Delta S_r = 0$ i.e. the total entropy change of the process is zero.
The internal energy change is zero due to it being an isothermal process.
$\Delta H_r= P \Delta V$, that is we assume pressure is constant for the process and solve for $P$. The total pressure in the final state is apparently the initial pressure plus the pressure calculate from the ratio $\frac{\Delta H_r}{\Delta V}$.

How do we justify these two assumptions?
Firstly, how can we justify the entropy change being zero? And, the second point I don't get how we can claim $\Delta U = 0$, just because it is isothermal. I know it is true for ideal gases but how does that apply here? About the last assumption, I can not understand at all what the logic is behind finding the final pressure as initial plus the calculated from the ratio.

P.S: I know the given links answer the question completely but I want to figure out how to reason these assumptions myself.

Why is a diamond crystal "much more ordered" than a hexagonal crystal? Both are crystals, with atoms sitting on specific lattice sites (ignoring point defect energetics).

Hmmm good point I guess

This is definitely not the approach that I would have used to solve this problem.  I would have used the condition that, at the equilibrium pressure $G_g=G_d$ or $\Delta G=0$.  The free energy of graphite at the equilibrium condition would be $$G_g=G_g^0+v_g\Delta P$$where $G_g^0$ is the molar free energy of graphite in the standard state (0 kJ/mol), $v_g$ is the molar volume of graphite, and $\Delta P$ is the increase in pressure relative to the standard state.  Similarly, for diamond, at the equilibrium condition, $$G_d=G_d^0+(v_g-\Delta v)\Delta P$$where $G_d^0$ is the molar free energy of graphite in the standard state (2.9 kJ/mol), and $\Delta v=2\times 10^{-6}\ m^3/mol$.  So setting $\Delta G =0$ gives $$0=(G_d^0-G_g^0)-(\Delta v)(\Delta P)$$

hmm but what about justifying the assumptions? Nice approach though. Didn't think the moles would be the slope the slope of G with varying pressure

The only assumption I can think of is that the change in molar volume between graphite and diamond over the range of pressures between the standard pressure and the equilibrium pressure is constant.  Is that what you were referring to?

Yes, I understand, you took the first order variation of G with respect to moles. In the quoted answer, they made some assumptions, I do understand that the way shown would not be how you would solve it but could you see if you can give explanations as to why they made the assumptions that they did?

I didn't take the variation of G with respect to moles.  There is no change in the number of moles.  I took the variation of G with respect to pressure at constant T, which is given by dG=vdP.

I see, so it is actually that variation of G with pressure is the moles

I don't know what this means.  All I did was find the pressure at 298 K at which the molar G of graphite is equal to the molar G of diamond.

I am just saying that $\frac{\partial G}{\partial P} = v_g$, therefore the rate of change of G with pressure is moles

v is not moles; it is volume per mole.


text:
Need good aproximation to the atomic composition of late 1980's expanded polystyrene foam
I'm not a chemist and what little I know about expanded polystyrene (EPS) foam has been found by looking at articles on the internet.  At work I need an approximation to the atomic composition (at %) of foam that dates to the late 1980's up to early 1990's.  Beyond the fact that it's 'EPS,' I have a density of 0.05g/cc, and that is it.  It is important to get a reasonable estimate of higher Z elements for this foam.
I've read that EPS is 95-98% air.  Therefore, when I look on the Pharosproject.net at the common material EPS Insulation, I am confused in that air components, $N_2$, $O_2$, and $Ar$ seem not to be in the list.  I'll stop at the top three components by weight percent in that list and suitably approximated: (polystyrene=$CH_2$, 96%), (flame retardant containing bromide, 1.0%), (3 blowing agents, 2%) This totals 99%, and for the sake of the ensuing calculation, let the extra 1% fall in the category of blowing agents.  The lack of the atmospheric components seems suggestive that this is some kind of pellet that has yet to undergo a chemical process.  I don't know a chemical formula for the bromide-containing flame retardant, but let's take HBCD (hexabromocyclododecane) with formula $C_{12}H_{18}Br_{6}$.  I'm assuming a closed-cell foam.  I'm guessing that some of the blowing agent and all of the flame retardant ends up in the cells, along with the air. Someone with expertise might be able say how much blowing agent ends up in the cells.
Now in the 1980's through the early 1990's, I've read that hydrochlorofluorcarbons were the blowing agents.  Therefore assume 3% $CF_3 CH_2 F$ (refrigerant R134a) to replace the blowing agents of the previous paragraph.  If this is a bad assumption, please correct it.
Assume air is, in vol %,  ($N_2$,78.08%), ($O_2$,20.95%), (Ar, 0.93%), ($CO_2$, 0.04%) where the $CO_2$ percentage has been made to take up the slack in trace elements so the total for air is 100% by volume.
I want a table in atomic percent for each element for a block of EPS.  The elements are C, H, N, O, Ar, Br, F.  I especially want feedback if the assumptions are way out-of-line.
ADDED:  Here is a quote that suggest the blowing agent, or at least part of it, stays in the foam.  As stated, the foam is meant to be 'old' as in late 1980's to early 1990's.  'A wide range of concentrations of CFC-12 (dichlorodifluorocarbon) was detected in styrofoam ware.  It fluctuated between a low of 32 ppm and a high of 35471 ppm in two different samples of styrofoam plates.' In abstract of Environ Monit Asses. 1991 Oct; (1-3): 413-22 'The determination of chlorofluorocarbons and extractives in the expanded and disposable polystyrene ware' by C.T. Sam & T.H. Chua.
It should be mentioned that contemporary blowing agents, like that found on Pharosproject, seem to be $CH_2$ based, like pentane.  Since the vast majority of atoms of the styrene foam will be C and H, these kind of blowing agents are not of a concern to me, and they may well completely diffuse out.

The blowing agent has long diffused out of the foam.The only heavy element in there is bromium, and the air inside has the same composition as outside. ... What do you want to do with this?

@Karl Thanks for response. I have a potential identification process that depends on <Z^2>.  Therefore I want an estimate of common impurities in EPS or 'styrofoam' of the late 80's - early 90's.  I have added one reference that suggests that chlorine might be present.  When you say Br remains, does it stay in its unit of $C_{12}H_{18}Br_6,$ or is the C&H driven off in the foam-making process?

$\ce{C_12H_18Br_6}$ is a moleculear compound. And I should probably have said could have diffused out. How much still resides in there depends on a lot of things. Blowing agents e.g. do not diffuse through a metallic layer. Size of the EPS item, environmental conditions in the past 40 yrs, ... Even the flame retardant could be leached out, potentially, although unlikely. Why don't you just run the analysis? Then you'll know.

@Karl  I'm doing a simulation and no product is available.  I think you are right about too many unknowns.  I guess I should figure the composition with no assumed Cl, F, or Br; then assume 1 part/thousand (taken out of the C &H)  for each and run a sim; then 10/thousand for each and do a sim, and see if makes any difference.  (The sims are expensive however.) Thanks for your comments.

What kind of a simulation is that?

@karl While I would expect some diffusion of blowing agents, this might depend strongly on how the foam has been treated since it was made. So there might still be some of the agent in closed cells in much of the foam. The key problem is the amounts are likely to be very variable depending on the history and size of the sample.

Btw. EPS afaik was always made with pentane as blowing agent. Fluorinated hydrocarbons were used for XPS. @matt_black You cannot keep pentane in a PS container with micron-thick walls. No, no way. Diffusion rate is proportional to the partial pressure difference, and these hydrocarbons mix very well.

You do not explain the context of your analytical problem. One cannot go to an analytical chemist/analysis service and state that tell me all the compounds which are present in this foam. This can take a year. So you have to simplify the problem.
(i) If you are solely interested in atomic compositions, all you have to do is to request elemental combustion analysis for
Carbon, H, Nitrogen, Oxygen, Sulfur and Halogens. This is a standard process. They will burn a small amount of your sample and tell you the percentages of these elements.
(ii) If you interested in other elements of the periodic tabke, you will have to request inductively coupled plasma with mass spectrometric analysis. You can find what elements are there up to Z=90 (or perhaps even higher).
(iii) The third problem is the most difficult one. What compounds are present? like the halogenated compounds mentioned in your post. This may require specialized gas chromatography-mass spectrometry analysis with targeted analysis. One has to specify what the analyst should look at.
(i)-(iii) will cost a lot.
From the comments, you mentioned simulations without having a real sample?? What is the basis of those simulations. You do need some starting real data and real data can only come from the actual analysis.

Thanks for your answer; I've accepted it.  I agree with your points.  Unfortunately, I don't have a time machine to go back and get a sample.  As mentioned in a correspondence with Karl, I think I'm forced to take a reasonable guess, and then 10 times the amount to see if it matters in my simulation.

I am still unsure of what the actual problem was with the foam from the 80s. Were you concerned due to health issues? What were you modelling or simulating?


text:
Does hydrogen bonding affect the structure of a molecule?
We know an individual molecule of $\ce{H2O}$ has a bent structure, but if we consider a volume of water, then there are many molecules having hydrogen bonding with each other. Due to this, does the structure of a single molecule, on average, deviate from bent? How severe are the deviations?
I took water as a simple example, but I mean my question can be considered more generally for other compounds as well.

This question is mainly inspired by the following question from the (2014) JEE advanced paper:

The correct statement(s) for orthoboric acid
is/are (Multi correct)
(a) It behaves as a weak acid in water due to self
ionisation
(b) Acidity of its aqueous solution increases upon
addition of ethylene glycol
(c) It has a three-dimensional structure due to hydrogen
bonding
(d) It is a weak electrolyte in water

Option 'c' in particular is what made me start thinking about this.

It looks as if the five answers are correct. And the deviations of the angle H-O-H must be rather weak.

Compare water bond angles in vapour and ice.

When a molecule hydrogen bonds, intramolecular bonds tend to be weakened, and such bonds become longer. Hydrogen bonds can also result in distortion of internal bond angles. However, H-bonds are typically weaker than covalent bonds by an order of magnitude or so, so H-bonds will perturb but hardly alter the conformation of rigid molecules, exceptions including rotational (conformational) potentials with low energy states differing in energy by an amount comparable to the H-bond energy.
In some macromolecules such as proteins, H-bonds occur extensively and often cooperatively (synergistically), both internally and with the solvent, altering (possibly even "flattening") the potential energy landscape, and resulting in 3D structures that would not be stable in their absence. H-bonding is one contributor to the acquisition of specific secondary to quaternary protein structures (another complementary one being hydrophobic interactions).
Note that molecules are in general inherently dynamic. You can explain this using quantum mechanical (QM) arguments, but you might simply think of molecules somewhat roughly as mechanical arrangements of atoms bonded by spring-like bonds. The vibrational state of a rigid molecule usually occupies a single energy minimum at room temperature. The QM wavefunction describes a distribution of possible observable instances (conformations) of the molecule, from which we may compute an average and other statistics such as a most likely state. Hydrogen bonding perturbs this distribution by altering the underlying potential. The perturbation due to a single H-bond, however, is relatively small.


text:
Selectivity in tosylation of sulphur and oxygen
Problem

Answer

 C

Question
The first step involves a nucleophilic substitution on the sulphur atom of the tosyl group. I reasoned that in the cyclohexane substrate, the sulphur atom (thiol group) will be more nucleophilic than the oxygen atom (alcohol group) due its larger size, higher polarizability and less solvation. Accordingly I proceeded by the following mechanism -

which gives D as the answer. Where am I going wrong?

There are not many examples of alkyl thiols reacting with tosyl chloride. There's this http://orgsyn.org/demo.aspx?prep=cv6p1016 but it is not very similar. It seems that tosyl chloride is selctive for alcohols over thiols.

I was thinking the same @Waylander - I've seen plenty of questions where OH is converted to OTs, almost always for making it a leaving group, but it was the first time there was this possibility of sulphur acting as nucleophile.

I cannot recall using or seeing -S-Ts used as a leaving group

If -S-Ts is formed, it may act as tosylating functionality toward -OH but poorer than TsCl..


text:
Can electrons change principal quantum number upon excitation?
I wanted to know when can excitation of electron take place i.e excitation of electron only takes place between subshells having same principal quantum number or is it seen by difference in energy level?
For example look at configuration of Pt- [Xe] 6s2 4f14 5d8. It has vacant 5f, 6d, 6p. I know Pt is a noble  metal and generally excitation may not take place, but for a moment let us forget it and assume excitation can take place. Can electrons of 6s excite to 5f? Or only 6d and 6p. Can electrons of 4f excite to 6d or 6p?

All these excitations are possible. Some may need more energy than others to take place. And some are not really probable, because they may violate selection rules. Do you know the selection rules ?

The selection rules govern how an electron is observed to transition (excite upwards or relax downwards) from one orbital to another.
Formally, they are written as:
ΔS=0
ΔL=0,±1
L+S=J
∴ΔJ=0,±1
where
ΔS
is the change in intrinsic angular momentum of the electron (spin multiplicity is 2S+1),
ΔL
is the change in orbital angular momentum, and
ΔJ
is the change in the total angular momentum.

are you talking about these ?

A transition is allowed if it respects the selection rules that you have mentioned. If they don't, the transition is forbidden. it does not mean that they do't take place. It only means that their probability is small, or that the absorption coefficient is much lower than for allowed transitions.

Search Grotrian diagrams.

"I know Pt is a noble metal and generally excitation may not take place," - if you're shining light / EM radiation on the atom, then excitation of an electron is perfectly possible as long as photons of the correct wavelength are supplied (and as long as the selection rules are obeyed). It doesn't matter how noble the element is. "Noble" is merely a crude description of their chemical properties; it has no bearing on the underlying, more fundamental, physics.


text:
Lithium Oxide Colour
According to Chemistry LibreText:

Lithium is the lightest of all metals and is named from the Greek work for stone (lithos). It is the first member of the Alkali Metal family. It is less dense than water (with which it reacts) and forms a black oxide in contact with air.

However, if I'm not wrong, all alkali-metals react with non-metals to form white oxides.  $\text{Li}_2 \text{O}$ is a white chemical compound, isn't it? What's with the site stating the opposite -- anything I'm possibly missing out on?

It is white, as you say. The black stuff, as per wikipedia, is several substances.

Aldrich sell Lithium Oxide as a "white crystalline solid" https://www.sigmaaldrich.com/GB/en/product/aldrich/374725

One has to remember lithium reacts with nitrogen as well.

From Handbook of Inorganic Chemicals by Pradyot Patnaik:

LITHIUM OXIDE
White cubic crystals; refractive index $\pu{1.644}$; density $\pu{2.013 g/cm3}$; melts at $\pu{1,570 °C}$; dissolves and decomposes in water ($\pu{6.67 g/100g}$ at $\pu{0 °C}$ and $\pu{10.02 g/100g}$ at $\pu{100 °C}$).

What the quote has written is not correct. Lithium on contact with air forms both lithium oxide and lithium nitride which looks black during the reaction perhaps due to constant oxidation (actually pure lithium nitride exist as red hexagonal crystals). The reaction is exothermic.
$$\ce{6Li + N2 -> \underset{black}{2Li3N}}$$


text:
Looking for volume
The question in my assignment is as follows. You have 5.5 M $\ce{NaOH}$. The volume of $\ce{NaOH}$ is 0.5 L. How many mL of the 5.5 M $\ce{NaOH}$ solution do you need if you would like to make 300 mL of 1.2 M $\ce{NaOH}$? I thought you could use the $C_1V_1= C_2V_2$ formula and rearrange it in there to get the answer, 720 ml. But that doesn't quite feel right. Does anyone have hints on what I'm doing wrong?

You may take 65 ml of the solution and top it up with 235 ml of water.

It is very simple algebra of junior high school, just darkened by almighty magic, because it is about chemistry.

@Shub No. That's not how it is done.

Take care ! The volumes are not additive! A $5.5$ M solution of $\ce{NaOH}$ has a density is $1.205$ and contains $18.1$% $\ce{NaOH}$ by weight. So let's start from the beginning!
A volume $\pu{V}$ of this concentrated solution has to be taken for the dilution. It contains $\pu{5.5·V}$ moles of $\ce{NaOH}$. The final solution must contain $\pu{1.2 mol L^{-1}*~ 0.3 L~ =~ 0.36~ mol}$ $\ce{NaOH}$. As these two number of moles are equal, the volume $\pu{V}$ of the concentrated solution to be taken for the final dilution  is given by : $\pu{5.5 mol L^{-1}*V~ =~ 0.36 mol}$, and $\pu{~ V = \frac {0.36~ mol}{5.5~ mol~ L^{-1}} =~ 0.0655 L~ =~ 65.5 mL}$
To obtain the final solution you should not mix it with $\pu{300 mL - 65.5 mL = 234.5 mL}$ water, as the volumes are not additives. $1 $ liter of the concentrated solution plus $1$ liter water does not make $2$ liters diluted solution. In this case choose a flask containing exactly $\pu{300 mL}$. Transfer $\pu{65.5 mL}$ of the concentrated solution, and then add some water and mix. Then add enough water and mix to get to the exact final volume $\pu{300 mL}$.

Maurice has already explained it nicely!
You have the right dilution formula, but most students even get this wrong anyway. I substitute the indexes and it helps students remember. You should be able to derive this relationship otherwise it will keep haunting as a mystery formula.
$$C_iV_i= C_fV_f$$
where
$C_i$ is the initial concentration of the stock solution
$V_i$ is the initial volume you need from the stock solution
$C_f$ is the final desired concentration of the diluted solution
$V_f$ is the final desired volume of the diluted solution
If you now plug in the numbers you should the answer right, 65.45 mL or 65.5 mL.


text:
Why London Dispersion force is attractive not repulsive
The London Dispersion force is caused by the formation of temporary dipoles in the electron cloud of neutral atoms that attract each other.

Why is this only attractive? Why don't the like poles repel, thereby nullifying the net effect?

I fail to understand the answer (the only one and unaccepted) in this post.

Do not try to understand an answer, try to create your own with the reasoning. This simple picture is self-explanatory, based on basics of electrostatics. Hint: Replace the left temporary dipole by permanent one. What happens?

To add to Poutnik. If you have a partial positive charge pa A, it can only induce a negative one on B. Tough the treatment of dispersion forces is hard, the above is quite a valid and general principle.

The underlying quantum mechanics of electron clouds can be repulsive. That is why solids and liquids have specific volumes: the electron clouds in different molecules can't get too close. The same underlying quantum interactions also create weak London dispersion forces. A complete quantum treatment would show both repulsion and attraction. But we separate the components of those interactions to make them easier to understand as the full QM analysis is far too complicated.

Dispersion forces are caused by induced dipoles. They are basically electric dipoles. Think of two magnets- why when you bring them close in any manner/alignment, they still attract?

A simple demonstration in electrostatics can convince that London dispersion forces are attractive. Has your physics teacher shown "charging by induction"?
Charging by induction video
Now replace the rods by molecules. Note the rods always attract. If one molecule develops a dipole or negative charge (on one side), it will induce an oppositive dipole in its neighbour.
Second demo, which you should yourself. Open tap in such a way that only a thin stream of water is flowing, but not dripping. Charge a comb with your hair and bring it close to the water stream. You will see water is attracted to the comb. No matter what you use for charging, water always gets attracted. There is no repulsion, because the permanent dipole of water always orients itself in such a way that its end is opposite in sign to whatever is the charge on the comb.
Water attracted to a charged comb
Now van der Waals attraction is not due to permanent dipoles but transient dipoles.
BTW, atoms and molecules do repel each other if you try to bring them very close. Can you compress water easily? Why and why not?
P.S. Please don't consider the answers on the web as answers from the Heavens. Online answers are written by students, teachers, researchers, former chemists etc. and all of us can be right or wrong.

Please add a description of the video if possible. Such links may become defunct and not everyone wants to wander off to youtube.

Is the attraction of pieces of paper by a charged comb also due to van der Wals force?

No, electrostatic. van der Waals and electrostatic forces of attractive have a different distance relationship.

I would add two points to MFarooqs' answer.
First, I would emphasize that the basis for attractive van der Waals interactions is the polarizability of molecules. Polarizability is the property of having flexible charge distributions which may be distorted by interacting with charges outside of a molecule, leading to a more stable electronic state. Polarizability is responsible for macroscopic phenomena such as the response of an electrically neutral material to an applied electric field.
The second point is that repulsive interactions are unsustainable, as they lead to separation of the interacting particles, whereas attractive interactions are self-sustaining or even re-enforcing as they bring the particles closer together.
When two neutral atoms (noble gas atoms are a good example) approach each other, they are, at long distances, electrostatically "blind" to each other. If sufficiently close though, their outer electron clouds will begin to significantly repel each other and their nuclei will do the same, and simultaneously opposite charges will attract. This does not seem to bode well for the prospects of forming a net attractive bond. However, since the electron clouds can be distorted (polarized), if a spontaneous asymmetry exists or arises in the charge distribution (easier to visualize if one atom is larger than the other, for instance He approaching Ar), then a net attractive force can develop and become reinforced. This is an induced (as opposed to permanent) type of polarization responsible for attractive van der Waals interactions.


text:
Why aren't both alpha-carbons halogenated when acetone reacts with chlorine under UV?
My book shows the following reaction:

Why is only one alpha carbon halogenated? I googled, but couldn’t find anything on it.

UV light does not necessarily mean that the reaction is totally governed by a radical chain mechanism. The formation of HCl allows for acid-catalyzed chlorination. By either mechanism, the first chlorinated methyl reacts faster than the "naked" methyl.. With  this in mind, look at the stoichiometry of the reaction. Not enough Cl2 to form hexachloroacetone. Acetone can be perchlorinated given sufficient chlorine.

No, there is no stopping the hydrogen to get replaced with chlorine. Given enough chlorine, eventually all the hydrogen can get replaced to form hexachloroacetone. The formation of HCl helps and using catalyst can be good in accelerating the reaction like pyridine or triphenylphosphine. Some literature for you: https://cdnsciencepub.com/doi/pdf/10.1139/v86-208 https://www.researchgate.net/publication/243873792_Chlorination_of_Acetone_in_Gaseous_Phase_to_Mono-Chloroacetone_-_Process_Design_and_Simulation


text:
Isn't the change in Gibbs free energy for a reversible process zero?
In problem 6, part 6.5 of the 2020 IChO (PDF), a thermodynamic cycle is given for “one mole of monoatomic perfect gas”:

$\mathrm{A \to B}$; isothermal reversible expansion receiving $\pu{250 J}$ by heat transfer $(q_H)$ at a temperature of $\pu{1000 K}$ $(T_H)$ from a hot source.
$\mathrm{B \to D}$; reversible adiabatic expansion.
$\mathrm{D \to C}$; isothermal reversible compression at a temperature of $\pu{300 K}$ $(T_C)$ releasing some amount of heat $(q_C)$ to a cold sink.
$\mathrm{C \to A}$; reversible adiabatic compression.

6.5 Calculate the entropy change $(\Delta S)$ for $\mathrm{A \to B}$ and $\mathrm{D \to C}$ processes in the heat engine in terms of $\pu{J K^-1}.$

To my understanding, the answer should be $\pu{0 J}$ for both steps since $\Delta G = 0$ for any reversible process, which $\mathrm{A \to B}$ and $\mathrm{D \to C}$ are. But the solution below gives non-zero values of $\Delta G.$ $ΔS$ for both steps was calculated in previous parts, $\pu{+0.25 J}$ for $\mathrm{A \to B},$ $\pu{-0.25 J}$ for $\mathrm{D \to C}.$ What is wrong with my reasoning here?
$$\Delta G = \Delta H - T\,\Delta S,\tag{1}$$
but $\Delta H = 0$ for an isothermal process, so
$$\Delta G = - T\,\Delta S.\tag{2}$$
$$
\begin{align}
\Delta G_\mathrm{A \to B} &= \pu{-0.25 J K^-1}\times\pu{1000 K} = \pu{-250 J}\tag{3}\\
\Delta G_\mathrm{D \to C} &= \pu{0.25 J K^-1}\times\pu{300 K} = \pu{75 J}\tag{4}
\end{align}
$$

From what I've learnt in high school, the formula for entropy change is $\int\frac{dQ}{T}$. Here T is constant in the two processes asked for, so it simply becomes $\frac{\Delta Q}{T}$. I've never done entropy calculations with Gibbs free energy so far... that's probably just my inexperience, but then IChO is meant for high school students

$\Delta G=0$ for process at equilibrium NOT for reversible process.

@Jay The working fluid experiences a closed cycle, so of course $\Delta G=0$, since G is a function of state.

The reason the calculation is wrong is that the heat removed at the cold reservoir is not 250 J.  It is something less than this.

The cycle shown is a classic reversible Carnot cycle. An intuitive reason for the change in Gibbs free energy not being equal to zero for such a step is that the pressure is not constant: the internal and external pressure are throughout in  equilibrium, but since $p_\mathrm{ext} = p = nRT/V = \mathrm{constant}/V$, $p$ is not constant. A condition for applying $\Delta G$ as a criterion of spontaneity (reversibility) is that the pressure is constant.

In fact you can show (by extension of the calculations show below) that the free energy change is equal to the expansion or compression work done in that step: $$\Delta G_\mathrm{isothermal} = w$$ as follows.
Computing the entropy changes for the isothermal steps from the heats is straightforward using the first law of thermodynamics, since

for an ideal gas undergoing an isothermal expansion or compression $\Delta U=0$ so that $w=-q$
for a reversible isothermal process $q=T \Delta S$

such that we can compute

for the first step, $\Delta S = \frac{\pu{250 J}}{\pu{1000 K}} = \pu{0.250 J K-1}$

For the second isothermal step ($D\rightarrow C$) we use the fact that the total entropy change for the system, given that it is a cyclic process, is $\Delta S = 0$, therefore

for the second isothermal step, $\Delta S_\mathrm{DC} = - \Delta S_\mathrm{AB} = \pu{-0.250 J K-1}$

Regarding now the free energy computation, you would use (sure enough)
$$\Delta G_\mathrm{isothermal} = \Delta H_\mathrm{isothermal} - T \Delta S_\mathrm{isothermal} = - T \Delta S_\mathrm{isothermal}=-q=w$$
since $\Delta H_\mathrm{isothermal}=0$ for such a process.

It maybe of interest to note that, in this case, the efficiency of the engine is given as $\frac{\Delta G}{q_{in} }$ that is ratio of the gibbs free energy by the heat put into the sytem


text:
Why is it that not all metals produce hydroxides when reacting with water?
I was taught, as a rule, that:

metals + water $\rightarrow$ metal hydroxide + hydrogen gas

$$\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2$$

However:
$$3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2$$
Why is this so? Why isn't $\ce{FeOH}$ produced? Is it because $\ce{FeOH}$ is unstable and it thermally decomposes the moment its formed?

There is no compound called FeOH.

Many rules taught at lower education levels are simplified, often oversimplified and have ( often many ) exceptions. They definitely cannot be taken as laws, because exceptions are behind near every corner.

Anyway the stability of $\ce{Fe3O4}$ is something like a challenge, because it contains two atoms $\ce{Fe}$ with oxidation number $+3$ and one $\ce{Fe}$ with oxidation number $+2$. This is hard to justify.

Not even talking about many metals ( like gold and platinum ) which do not care about water. IMHO, the author of the above rule should reassess his/her teaching abilities, unless it is quoted out of proper context. As metals reacting by the mentined way are rather exceptions than rule.

This question is well asked, unfortunately the duplicate is not. But that is the one that covers your question. You still get a +1 from me, but I also will close the question.

Most likely, the initial deprotonation of water gives the hydroxide. But if the reaction requires high temperature and the hydroxide is not strongly thermally stable, it could go over to the oxide. The boundary is actually not sharp; see this answer.


text:
Why is the thermal stability of calcium carbonate higher than that of magnesium carbonate even though lattice energy decreases down a group?
The difference in thermal stability of $\ce{CaCO3}$ and $\ce{MgCO3}$ can be explained using Fajans' rules, but why not using the lattice energy method?
Why is the thermal stability of $\ce{CaCO3}$ higher even though the lattice energy decreases down a group (as the size of cations increases)?

Smaller magnesium ion destabilises the carbonate anion more. Electrostatic and chemical stabilities go against each other here, with the latter having the upper hand.

The so called "lattice energy" method doesn't work for all compounds, same is the case with Fajan's rule. These are theories which have been developed to explain the characteristics and make it somewhat believable to the general audience but we cannot compare all compounds using one theory or the other because each theory has its own drawbacks.
Lattice energy is directly proportional to the mod of product of charges on individual ions, which is same in this case, further it is inversely proportional to the radius of the resulting compound, and you're right that $\ce{CaCO_3}$ has lower lattice energy implying that it has lower ionic character, which further implies that it should have lesser thermal stability but here's the drawback. Lattice energy isn't the dominating factor when it comes to ionic character.
Diamond has very high lattice energy, you can't say it is ionic, can you?
Fajan's rule along with size compatibility and polarization decides the ionic character of a compound. By Fajan's rule we have: covalent character directly proportional to the extent of polarization, implying that the smaller cation will cause  greater distortion of electron cloud as compared to the larger cation. This means that $\ce{CaCO_3}$ would have higher extent of polarization, i.e. higher covalent character. This would cause slight thermal instability, and $\ce{MgCO_3}$ would be relatively more thermally stable.

Welcome to Chemistry StackExchange and thanks for posting an answer! You might want to use \ce{} to format chemical formulae.

The rules properly predict the relative stabilizes we observe for calcium and magnesium salts, such as the carbonates and hydroxides, but they actually got lucky in that case. If you try to bring in the beryllium compounds, which should be less stable than the magnesium compounds according to these rules, you run into problems. The issue is that kinetics, which the rules do not really account for, in reality enters the picture.
Let's look at the decomposition of  beryllium and magnesium hydroxides.
According to the respective Wikipedia articles, beryllium hydroxide decomposes at 400°C whereas magnesium hydroxide does so at only 350°C even though magnesium hydroxide "should,be" more stable.
If you look up the thermochemistry data for $\ce{BeO,Be(OH)2,H2O}$ you find that the standard free energy change for the decomposition
$\ce{Be(OH)2\to BeO + H2O(g)}$
would reach zero at about 125°C, so the oxide and hydroxide would be in equilibrium under one atmosphere water vapor pressure at this temperature. With magnesium in place of beryllium the corresponding equilibrium temperature is higher, about 207°C with the data I used. These temperatures are lower than the observed hydroxide decomposition temperatures but in the "right order" according to Fajan's Rules.
What happens is both decompositions are kinetically hindered, as we would expect with solid phase reactions, and the kinetic hindrance is greater with beryllium than with magnesium. The net result is that beryllium hydroxide is more stable kinetically whereas magnesium hydroxide would have been more stable based on thermodynamics. Because the lattice energy and Fajan rules are generally thermodynamically based, they miss on the kinetically dominated cases given here.


text:
What is a spiroketal compound?
I read a medicinal chemistry book about SGLT2 inhibitors. One of them is tofogliflozin. In that book, they refer to it as spiroketal. Is the spiroketal functional group the red part in this picture? What is special about this group (to deserve a unique name)?

yes the red part  is a spiroketal

So, what would you like to know about it? Yes or no questions are not OK here. I guess what a spiro (not spiral!) compound is, for starters?

This ain't a name. "Spiro" is for two cycles connected in such fashion, and "ketal" is for ketal.

Quoting the entry on spiroketal from ChEBI:

A cyclic ketal in which the ketal carbon is the only common atom of
two rings.


text:
Upon the addition of metallic sodium to cold water, is an explosion possible?
In a Cambridge IGCSE question paper, the following question was asked:

Describe the reaction of sodium with cold water. Include the observations that can be made.

I was thinking of writing that "sodium explodes" when it comes into contact with water, but the marking scheme says that point is unacceptable:

Common incorrect answers include "sodium bursts into flames or explodes"

Doesn't sodium react explosively in cold water (at STP/RTP) or is that reaction limited to water at warmer temperatures?

Note that sodium explodes even in inert atmosphere, as it is primarily a coulombic explosion.

Other cases: https://chemistry.stackexchange.com/questions/108332/would-sodium-explode-in-salt-water ... https://chemistry.stackexchange.com/questions/72234/will-sodium-metal-stored-in-mineral-oil-explode-on-contact-with-water

From having tested with ca 5 pounds of sodium into a fjord of the atlantic ocean in autumn, I can say that it makes a horrendous noise, sputters and hisses, and eventually catches flame, carried upwards by the column of hydrogen that emerges from it. It spreads outwards quickly, setting aflame your boat and your oars. At this point there is a general sense of urgency as to the "getting the f out of dodge with flaming oars" going between you and your mates in your pathetic little on-fire skiff.

Edits made to the question are not notified to the users who answered and so the edit request you made might just turn futile.

Depends on the form of sodium, a single chunk - no, sodium sand probably will

The question how sodium reacts with cold water does not necessarily imply it reacts (qualitatively) differently in hot water. BTW, it is question of time when cold water around sodium becomes hot water.

I suspect that wording is part of the game. Chemically and pedantic sodium doesn't explode even in hot water but the reaction is explosive speaking, similarly sodium doesn't catch fire but hydrogen does. I copy the answers by DrMoishe Pippik below. When things like this occurs , if possible try to add notes explaining your line of thinking. Quiz like exercises have this limitations...

I suspect the question/answer rejects the "it explodes" option because it wants a proper description of what chemical reactions are happening (in detail) not because an explosion is not one possible outcome. Even if there is an explosion, why does it happen, what chemical reactions cause it?

The question is vague. Does the sodium burn, or does the hydrogen, liberated by the sodium, burn? If you were to completely immerse sodium in water, with no air above (or with an inert gas, e.g., argon), would even the evolved hydrogen burn?
In your answer, try to distinguish amongst the various reactions, including the amount of energy from $\ce{2Na + H2O -> 2NaOH + H2}$ and from $\ce{2H2 + O2 -> 2H2O}$.

Would enough energy be liberated to vaporize the sodium in that first reaction?
Is the yellow glow from vaporized/ionized sodium

from the heat of that first reaction, or
from sodium reacting with air above the water, or
from the metal spattered by exploding hydrogen...

My complaint with the question is that it does not specify the conditions, e.g., temperature, amount of sodium, atmosphere... but then, perhaps, it's meant for the essayist to state.

From personal experience doing a demonstration I can tell you that for similar amount of sodium (approx. 0.5 cm3) if you use cold water you would not see an explosion.
Using small volumes of water at room temperature (around 20 °C) or hot water the explosion is almost guaranteed.
I think that is all due to kinetics of the reaction and thermal diffusion, thought it is a complex process and there is plenty of nuances.
I recommend you to access the article Exploding some myths publish by Declan Flemig in Education in Chemistry  https://edu.rsc.org/feature/exploding-some-myths/2000082.article.


text:
Bond Angle of NH3, NCl3 and NBr3
I have read the other answer on chemistrystackexchange regarding the bond angles of $\ce{NH3}$ and $\ce{NCl3}$. The bond angle of $\ce{NCl3}$ is more than $\ce{NH3}$ due to back bonding between p-orbitals and $\ce{N}$ and d-orbitals of $\ce{Cl}$. But recently I came across a question asking for the relation between bond angles of $\ce{NH3}$ and $\ce{NBr3}$. I thought that due to the same reason of back bonding described earlier the bond angle of $\ce{NBr3}$ would be greater than $\ce{NH3}$. But the answer key gave the correct answer as bond angle of $\ce{NH3} > \ce{NBr3}$. Why is it so that the back bonding reason does not work here? Is there any other influencing factor?
And also why is the bond angle of $\ce{PBr3} > \ce{PH3}$ ... which is opposite to the trends shown by corresponding compounds of Nitrogen?

You should link these answers

https://chemistry.stackexchange.com/questions/92323/bond-angles-in-nh3-and-ncl3#:~:text=The%20bond%20angle%20in%20a,H3%3ENF3.

The second part of question is already answered here :https://chemistry.stackexchange.com/a/125871/80950

https://chemistry.stackexchange.com/questions/65285/why-does-nf3-have-a-smaller-bond-angle-than-nh3#:~:text=NH3%3A%20Here%20N,into%20the%20N%E2%88%92F%20bonds. and this is the first part

A quick answer one could give to this is due to the large differences in the size of orbitals. We will only get meaningful back bond bonding if the size of orbital involved is close by (i.e: $n$ is nearby). See this answer as a reference.

Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

Hey...try to search the questions first... then ask...btw good qsn.

Does this answer your question? Why does NF3 have a smaller bond angle than NH3?

no the above links do not completely answer my question. I have also asked for the reason for bond angle of NBr3< NH3


text:
What causes the voltage potential in a chemistry battery
Consider a simple chemistry battery: Zn and Cu with H2SO4 as the electrolyte. The electron will travel from Zn to Cu. And at the Cu side, H ion will get electron and turns to H2. If the circuit is open, Zn itself will react with the acid, and H2 will be generated at the Zn side.
Why there is voltage potential when the circuit is closed? The voltage force the electron to travel around the circuit.

Your hypothetical cell is just zinc and copper in a beaker with sulfuric acid and no salt bridge. So the zinc reacts with the sulfuric acid. Now look up the Daniell cell, which is discussed in many places, including here. The idea is to not just waste the reaction’s free energy as heat. Instead, electrical energy is desired.

You can benefit greatly by getting this and working through it: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95 (10) (2018) 1801-1810. The Zn and Cu Daniell cell is addressed in great detail. TL; DR Cohesive energy differences are the major factor in explaining the behavior of this famous galvanic cell.

Electrodes can be considered as chemically powered electron pumps, due different tendency of metals performing reaction $\ce{M(s) <=> n e-(s) + M^\mathrm{n+}(aq)}$

Ed V, Thank you for the info. I just have a quick look at the Daniel cell. In that cell, it seems there is no reaction when the circuit is open, right?

Open circuit, the Daniell cell (note spelling of Daniell) just slowly dies the diffusion death: some copper ions get into the zinc electrode’s solution reservoir and oxidize the zinc electrode. If the cell was always open circuited, the solutions in the electrode reserviors and salt bridge would all become the same, in the limit, except that copper ions would oxidize zinc metal until one was exhausted. So all the free energy would be wasted as slow heat release.

By the way, the scenario you describe is basically the lemon ‘battery’, with dilute sulfuric acid in a beaker instead of citric acid in a squishy (to rupture membranes) lemon. So it would work, like the lemon cell, but not well: it is internally partially shorted out.

In your setup, the metal zinc is transformed into $\ce{Zn^{2+}}$ ions plus electrons. The electrons can and will react with $\ce{H^+}$ ions in contact with the zinc plate, and this takes place when the circuit is open and when it is closed. It produces bubbles of $\ce{H2}$ gas. But this reaction is slow, because the reaction produces a positive charge ($\ce{Zn^{2+}}$) which repels the ions $\ce{H+}$, and prevent them from touching the zinc plate.
On the contrary, if the circuit is closed, the $\ce{H^+}$ ions are in contact of the copper plate without being repelled, and the reaction can proceed with a higher rate, as nothing prevents the electrons from reacting with the $\ce{H+}$ ions present in the solution. So Hydrogen is produced with a higher rate on the copper plate, even if it is far away from the zinc atoms which have produced them.


text:
Plotting second order reaction with different reactants

In a laser flash photolysis experiment, $\ce{C6H5NH2}$ $(C_0 = \pu{60 \mu M})$ is oxidize to its corresponding radical cations. There are three possible reaction pathways for the formation of the cations:

Dimerization of the radical cations;
Reaction between the cations and the initial substance; and
Reaction between the cations and the solvent.

Determine with the help of the experimental data which is the most
likely pathway and the rate constant of the reaction.
$$
\begin{array}{l|cccccccc}
\hline
t \text{ after the pulse (in µs)} & 0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 \\
\hline
\text{[Radical cation] (in µM)} & 20 & 11.12 & 6.86 & 4.47 & 3 & 2.06 &1.43 &1 \\
\hline
\end{array}
$$

I understand how to plot for (1) and for (3), however how do I plot for the second pathway? I know that I will have the reaction type:
$$\ce{ A + B -> P}$$
and then I should plot $\ln\left(\frac{a-x}{b-x}\right)$ against $t$ however, how do I actually calculate $(a-x)$ and $(b-x)$?

Does it make sense to consider that two radical cations may dimerize ? As they are charged, they repel one another, which prevents a dimerization ...

@Maurice my question is more, how do I get (a-x) and (b-x) I don't understand how they are calculated. I tried to say that $a = 20$ µM and $b = 60$µM then for $b$ I thought to get (b-x) I should take (b- the concentrations of the radical ion) while (a-x) is just the concentration of the radical ion (which I know is not correct but it is the closest I get to the correct answer). But what is the correct way of calculating (a-x) and (b-x)?

@Katara form the stoichiometry of the reaction I guess that $$x  = [\ce{C6H5NH2+}]_0 - [\ce{C6H5NH2+}]_t$$

@Katara, are you sure that you got the right units for the concentration of the radical cation? I would have expected that the concentration of the radical cation generated in the excitation step was small compared to the concentration of aniline. In that case, $\ce{[C6H5NH2]} \gg \ce{[C6H5NH2+]}$ and you would observe pseudo-first order kinetics  A semi-log plot of the data will show a linear behaviour.