text:
Could polydiazene exist?
I am a bit new to the world of chemistry, and after messing around on ChemSpider, I had a thought. Could a polymer like this exist:

Due to VSEPR, it would be shaped pretty weirdly and N-N single bonds are known to be atrociously weak, but is there a reason as to why this polymer has not been observed at all? It seems like it should be possible (though very unstable), so I'm not sure why it hasn't been seen at all.

Polymerization from $\ce{H-N=N-H}$ would be endothermic reaction and the monomer itself is unstable, forming mostly nitrogen as hydrazine.

This question touches upon a concept of catenation - an ability of atoms of the same element to make bonds between them that could produce series constituting a chain or ring structure.
Something like that is quite common for nonmetal chemical elements like silicon (polysilanes), phosphorus (red allotrope consisting of interconnected P4 units) or sulfur (S8). In case of chemical elements from 2nd row of periodic table thing are a bit different - we have carbon, which is famous for its catenation abilities, while other elements - nitrogen and oxygen are very limited in this regard. This is often explained by repulsion of lone electron pairs that are present on oxygen and nitrogen atoms, which would destabilize the molecule if present in a series of neighboring atoms (in case of alkanes there are no lone electron pairs on carbon atoms - there are C-H bonds instead). For elements in lower parts of periodic table atom sizes are larger and bonds are longer and thus repulsion of lone electron pairs seem not to be such issue anymore. (Interestingly, boron also makes structures with B-B bonds, but there are no lone pairs on boron atoms and these structures are not linear.)
Going back to nitrogen - there have been attempts on making compounds with a series of nitrogen atoms one after another, i.e. making a chain:
https://chemistry-europe.onlinelibrary.wiley.com/doi/abs/10.1002/chem.202004830
Also, there have been some theoretical studies on the matter:
https://pubs.acs.org/doi/10.1021/acs.jpcb.3c00620

A magnesium nitride compound with the empirical formula $\ce{MgN4}$, having polymerized nitrogen chains,  is known under high pressure.
Under about 50 GPa pressure magnesium can form a pair of nitrides other tgan the common $\ce{Mg3N2}$[1]. $\ce{[Mg^{2+}]2[N4]^{4-}}$ contains discrete  $\ce{N4^{4-}}$ ions, which are trapezoidal in shape with the two end nitrogens cis relative to the central nitrogen-nitrogen bond. This geometry, which would not be expected based on steric considerations, may be stabilized by homoaromatic coupling as the ion has six pi-symmetry electrons. In the other compound, $\ce{[Mg^{2+}][N4]^{2-}}$, the nitrogen forms an infinite chain of overlapping  $\ce{N4}$ trapezoids sharing their lateral edges. Including shared electrons along the shared edges, each trapezoid again effectively has access to six pi-symmetry electrons and may be homoaromatic.
Reference

Laniel, D., Winkler, B., Koemets, E. et al. Synthesis of magnesium-nitrogen salts of polynitrogen anions. Nat Commun 10, 4515 (2019). https://doi.org/10.1038/s41467-019-12530-w


text:
Calculating the Barrier Between Rotamers
There are two stable conformations of a medium-sized organic molecule roughly 5 kcal/mol apart in ground state energies. One is probably the global minimum conformer identified through a conformer search— the other is the reactive conformer that would eventually lead to product formation. Both have been optimized with DFT. Say I want to calculate the barrier to accessing the higher energy, reactive conformer, how would I do that? I suspect that although the higher energy conformation is thermodynamically feasible, perhaps it is not kinetically accessible. I know from basic conformational analysis that these barriers can be quantified from the energetic cost of rotating about dihedral angles, however I can’t think of any systematic way to do this for a larger system. Any ideas? Or am I thinking about this incorrectly?

Depending on which program you are using, there will be a way to calculate the potential energy surface (PES) by scanning along a particular coordinate (bond length, angle or dihedral). For instance, if you wanted to look at the PES of the stretch between hydrogen and oxygen in water, you can tell a program to calculate a single point energy (SPE) when the O–H bond length is 0.7 Å, then calculate a further 10 SPEs after increasing the bond length by 0.01 Å per scan step (for example). This is called a 'rigid scan'.
While you can make sure that the only change in geometry is the coordinate you are scanning, programs will generally also allow you to alter the coordinate and, at the same time, optimise the geometry of the rest of the molecule to compensate for the change in scanned coordinate. This is called a 'relaxed scan'.
If you go to the documentation of the program you are using and search for 'scan' you should be able to find it. (Relaxed scans are generally run through the geometry optimisation procedure, so you might have to look there if you want to run a relaxed scan.) All you need to do is specify the coordinate you want to scan across, tell the program how many scan points you want, and work out your start and end points.
Good luck!

source: Stack Exchange

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text:
Aliphatic alcohol analysis in capillary gas chromatography

4.5. Standard solutions of trimethylsilyl ethers of aliphatic alcohols from C20 to C28.
They may be prepared from mixtures of pure alcohols at the time they are required for
use.

Reference solution for thin-layer chromatography: C20-C28 alcohols
0,5 % in chloroform, or a fraction of alcohols obtained as indicated in
point 5.2 from the unsaponifiable matter of an olive-pomace oil.

And

reference sample: pure monoglycerites or mixture of monoglycerites which the quantity in the solution is known,

something which I can translate like this. But all in the olive oil analysis (for this one, for analysing % glycerile monopalmitate).
I searched a lot but I could not find what these are exactly. We are searching for the olive oil analysis. Anyone who knows what it means?

It could be more clear what exactly is the question.

The first two quotes refer to solutions of aliphatic alcohols; here "C20 to C28" indicates a range of the number of carbon atoms in each of the molecules, which can be 20, 21, 22, 23, 24, 25, 26, 27, or 28.  If it is a sample to eventually calibrate the GC, the supplier will document its composition, i.e. the percentage of each fraction. To reduce the risk of decomposition of the trimethylsilyl ethers, they are prepared just before their intended use.
To illustrate the concept, Agilent is one source of such standards for GC currently compiled here.  One of them is Part Number:5080-8716 to simulate a distillation with 19 analytes in reference to ASTM D2887 and ASTM D7213:
| analyte           | concentration | CAS number |
|-------------------+---------------+------------|
| n-Decane          | 9.7 % (w/w)   |   124-18-5 |
| n-Dodecane        | 19.9 % (w/w)  |   112-40-3 |
| n-Dotriacontane   | 0.9 % (w/w)   |   544-85-4 |
| n-Eicosane        | 1.3 % (w/w)   |   112-95-8 |
| n-Heptadecane     | 5.2 % (w/w)   |   629-78-7 |
| n-Heptane         | 4.6 % (w/w)   |   142-82-5 |
| n-Hexadecane      | 10.3 % (w/w)  |   544-76-3 |
| n-Hexane          | 4.4 % (w/w)   |   110-54-3 |
| n-Hexatriacontane | 0.9 % (w/w)   |   630-06-8 |
| n-Nonane          | 4.8 % (w/w)   |   111-84-2 |
| n-Octacosane      | 0.9 % (w/w)   |   630-02-4 |
| n-Octadecane      | 2.2 % (w/w)   |   593-45-3 |
| n-Octane          | 4.7 % (w/w)   |   111-65-9 |
| n-Pentadecane     | 5.1 % (w/w)   |   629-62-9 |
| n-Pentane         | 8.2 % (w/w)   |   109-66-0 |
| n-Tetracontane    | 0.9 % (w/w)   |  4181-95-7 |
| n-Tetracosane     | 0.9 % (w/w)   |   646-31-1 |
| n-Tetradecane     | 10.2 % (w/w)  |   629-59-4 |
| n-Undecane        | 4.9 % (w/w)   |  1120-21-4 |

A standard (as in protocol) usually describes the necessary equipment (as in chemicals, machinery) and procedure how to yield a specific result. ASTM, ISO, FDA are a few of larger organizations to convene and publish standards relevant to chemistry.

The third describes either the solution contains only one monoglycerite, or multiple monoglycerites.  But in either case their concentration is known; for instance 3% by weight is monoglycerite A, 36% by weight refers to monoglycerite B.

Based on the snippets you shared, it seems possible their source is European Regulation 2019/1604 of 27 September 2019 amending Regulation (EEC) No 2568/91 on the characteristics of olive oil and olive-residue oil and on the relevant methods of analysis.  In ANNEX VII, there is a section describing the required reagents for to perform the next analyses including

4.8. Silylation reagent, consisting of a 9:3:1 (V/V/V) mixture of pyridine/hexamethyl disilazane/trimethylchlorosilane.

It is followed by a description how to prepare the ethers in question and setup of the analysis (temperature of injector and detector, temperature ramp, carrier gas composition, flow, split ratio, volume of injection).  Section 6.4. Peak identification, Table I listing relative retention times, and Figure 2 of an annotated chromatogram for instance name some the sterol alcohols analyzed, e.g., cholesterol ($\ce{C27H46O}$), brassicasterol ($\ce{C28H46O}$), ergosterol ($\ce{C28H44O}$); in the above definition, one C27, and two C28 alcohols.
If this European norm is the protocol you have to follow, check it again for the chemicals mentioned.  (A larger public database to recognize chemicals than Wikipedia is e.g., chemspider -- if there is a public record there, it will report e.g., the Hill formula, and a structure formula.)  Else, get in touch with your supervisor or/and client to identify the relevant standard to use.

Buttonwood, thank you for the answer. But as a fresh beginner in laboratory, I do not have a whole understanding of what you wrote. Could you please write me these specifying how to find them (maybe chemical identity) or how to mix what kind of solvents etc.? Thank you very much.

@wise The answer was amended; likely, the next work ahead of you is more evident now.  And in general (not only for chemistry.stackexchange.com): if you quote, name/link the source of the information.

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


text:
Is the compound optically active?
My work -
The compound seems to have a plane of symmetry(POS)(The plane of the compound itself seems to be the POS)
Because- all carbon atoms in the benzene ring, the - COOH, the C=C, are all SP2 which means they are all planar.
The CH3's may be oriented so that the 2 H is in the plane, 1 H above, and 1 below - meaning, they are symmetric about the plane.
So I concluded that it is optically inactive but the answer states that it is indeed optically active.
Where am I wrong?
Thank you in advance

This compound sure would "like" to be flat, but is it able to be?

Won't the Br being large  try to avoid the CH3 and so become out of plane along with the rest of the group.


text:
Calculating the enthalpy of formation for ethane in Gaussian
I am attempting to run a benchmarking calculation in Gaussian software to obtain the enthalpy of formation for a series of molecules. I am using the example calculation found  here for ethane (C2H6). The author utilizes non-optimized structures at the HF/STO-3G level of theory. I have attempted the same calculation, however, at the B3LYP/6-31G* level of theory. I have calculated individual electronic energies for both C and H using this method as well as for ethane, whose structure has been optimized. To find the enthalpy of formation at $0$ K, the author uses this equation:
$$\Delta_\mathrm{f}H^\circ = 2\Delta_\mathrm{f,exp}H_{\ce{C}}^\circ + 6\Delta_\mathrm{f,exp}H_{\ce{H}}^\circ - 627.5095\times(2\mathcal{E}_{0,\ce{C}} + 6\mathcal{E}_{0,\ce{H}} -(\mathcal{E}_{0,\text{eth}} + \mathcal{E}_{\text{ZPE,eth}}))$$
The first terms in the expression are obtained experimental values for the heat of formation regarding the individual atoms of C and H. The term 627.5095 is used to convert from Ha to kcal/mol. The next terms pertain to the total electronic energy for the individual elements, not including the ZPE which is $0$ for these. Finally, we have the electronic energy and the ZPE for the ethane molecule, obtained from a frequency calculation. Following this work, I have obtained for ethane, at $0$ K, a heat of formation of -223 kJ/mol, which seems wildly different than the experimental value of -68 kJ/mol.
Does anyone have any tips on how to accurately calculate thermochemical values using Gaussian? I am using the example given by the company, but following the work there appears to be yielding inaccurate values. Any help would be much appreciated!

Much too tired to think about it properly but have you mixed up your units? You talk about kcal earlier in and kJ at the end, and there is very roughly a factor of 4 difference between what you get and what you expect.

Are you obtaining the same values in Ha for the individual terms shown in Table 1 ?

I cannot check the whitepaper right now, but if the author comes close with the HF level of theory, which is more like educated guessing, you should come close with B3LYP, too. Can you post the values you have used, please.

@IanBush Oh I so wish that were the case! I have double checked these pretty thoroughly however.

@BuckThorn So, using the same level of theory as the author, I was able to obtain the same total electronic energy for the hydrogen atom, I did not observe the same value for the carbon atom, nor did I get the same value for the ethane molecule.

@Martin-マーチン Sure thing! I have elected to do this at the B3LYP/6-31G* level of theory now since I am following a paper who is doing similar calculations at this level. The total electronic energy for the H atom was -0.0500273 Ha, for the C atom -37.776008 Ha, and the sum of electronic energy and ZPE for ethane was -79.679935 Ha. I believe that contains all the necessary ingredients to calculate the heat of formation at 0K. Experimental values for the elements are found in Table 3 if you are looking to do the calculation yourself.

I cannot do the calculations unfortunately, but others might. I recommend adding all values to your post. You should also include a sample input file. And have you tried another DFA? Maybe B3LYP really does perform crappy on this task. It wouldn't be too surprising.

@Martin-マーチンI am also noticing that the data the author uses in as an example output does not match the data they are using to do these calculations, which seems quite strange.


text:
Which of these monomers can polymerize by chain addition reaction?
Problem:
Which of these monomers can polymerize by chain addition reaction?

Solution:
The monomers with red crosses underneath them can polymerize by chain addition reaction.
Question:
So, how I am thinking going about solving this problem is by using the definition for an addition reaction, which is when a molecule is added to another molecule on f.ex. its double bond and turning it into a single bond. I can see how that is possible for (i) and (ii), but how come the same doesn't apply to (iii) and (iv) where double bonds are also present? The method I'm thinking of obviously needs some tweaks, but maybe it should be scrapped entirely because it's not the intended method to use for these kind of problems?

Consider that those active sites are

Carbon-carbon double-bonds
Terminal
Not in a ring, stabilized by resonance

Is a good rule of thumb then that the monomer should be unsaturated for it to polymerize by addition reaction?

A start, but not sufficient.


text:
synthesis of lead hydroxide
what is the best and cheapest way to do it? I tried solving lead in HCl but only a little of lead reacts with HCl so I used electrolysis to speed up the reaction but still a little lead reacted. is it because of soloubility of PbCl2? if I use nitric acid will it work better( soloubility of lead nitrate is more than others)?

Yes, poor solubility of PbCl2 is quite probably the issue.

The insoluble salt increases the oxidation potential but can cause mechanical problems if the salt is adherent; the reaction takes longer. For a small amount electrolysis in HCl and patience should work, otherwise nitric acid. Just curious why make lead hydroxide?

can I directly make Pb(OH)2 by electrolysing water? for example use CuSO4 as electrolit?

by adding H2O2 to Pb(OH)2, red lead is made. I want for making red lead.


text:
Why is hydrogen bonding stronger than dipole-dipole interaction of carbonyls?
My understanding was that hydrogen bonding is much stronger than other dipole-dipole interactions since the dipole between hydrogen and O/F/N is much greater than other dipoles. However, the dipole between C and O in carbonyl is apparently greater than that of a hydroxy group in, say, an alcohol. Hydrogen bonding is still said to be stronger. Why is that?

If it was just a  dipole-dipole interaction there would be no reason to set it apart.

https://chemistry.stackexchange.com/questions/143419/why-hydrogen-bonds-are-stronger-than-van-der-waals-forces

https://chemistry.stackexchange.com/questions/5503/what-exactly-is-hydrogen-bonding-and-does-it-really-need-fluorine-oxygen-or-n


text:
Who codified/how was it codified that oxidation is electron loss and reduction is electron gain (OILRIG)?
As far as I can tell from some reading on the subject, the terms oxidation and reduction were first used by Antoine Lavoisier and contemporaries to discuss combustion in the late 1700s. The term electron wasn't coined for another 100 years, so I'm wondering who came up with/proved the above concept, commonly codified as "OILRIG".

https://www.britannica.com/science/oxidation-reduction-reaction/Historical-origins-of-the-redox-concept will be of interest

As you state, in early chemistry, oxidation meant combining with oxygen, e.g., burning, and reduction meant removing oxygen from an ore, e.g., reducing iron oxides to metal.

Oxidation/reduction may be the lose/gain of electrons, but need not to. More general is increase/decrease of the oxidation state/number, what is the formal charge after heterolytic bond breaking according electronegativities of atoms. //  Note that electrolysis as the link between charge and oxidation/reduction is known about 100 years longer than an electron too. // The question is more suited for History of Science and Mathematics SE. I have never heard about "OILRIG", even if it is clear what it means.

i agree its more complicated than this rule of thumb poutnik. i found the mnemonic here though it is well known https://en.wikipedia.org/wiki/Redox#/media/File:Redox_example.svg


text:
Stereochemistry in bicyclo compound
In the below bicyclo compound , there are two chiral carbon atoms. So, total combinations of configurations are four : RR, RS, SR, SS. However according to answer key , only two isomers of this compound exist. I am confused that how that can be ? Are some of the above configurations give identical compounds ? I cannot figure any simple symmetry which makes so.
Thanks very much.

The two chiral centres are not independent. The stereochemistry of one affects the other.

@An_Elephant Can you show how does it affects ?

It is trivial. You must be able to see it.

@An_Elephant It is not clear to me. Can you please write a detailed explanation ?

@DoraCake One approach: 1) assign configuration of the stereogenic centers in the structure shown. 2) imagine how the four substitutents around each of them should be attached to interconvert a centre presently (R) would be (S), and vice versa a centre (S) as one  (R).  How would this look like joining the pieces together?  Best thing for many is to build a model (no affiliation, but as an example this one).

@DoraCake, to see that chiral centers in this compound are not independent try to draw a molecule with just one chiral carbon with opposite configuration. This way you will see that orientation of bond closing one of the rings from bicyclic system would be such that it will be impossible to connect it with the other atom that it is supposed to be connected with.

I think you could use a better textbook. What kind of lame question was that? There's certainly no stereoisomers at all, except on paper, but you could have some trouble even with drawing any.

This site is instructive.


text:
Redox half-reactions involving gold at three distinct oxidation states
I am preparing for my exam, and for some reason really struggle with half-reactions in the context of Galvanic cells. I do manage to solve the more straight forward questions, but really struggle with some.
Question asks to "write the half-reactions, balanced formula and Galvanic cell structure".
Given reaction is
$$
\ce{Au+ (aq) ->  Au (s) + Au^3+ (aq) }
$$
I managed to successfully identify that the cathode is: $\ce{Au+ (aq)  + e- -> Au (s) }$
But I got the anode wrong, I thought it should be $ \ce{Au+ (aa) ->  Au^3+ (aq) + 2e- }$, while book says it should  be $ \ce{Au (s) ->  Au^3+ (aq) + 3e- } $
What am I missing?

The book is wrong. Those half reactions can't be combined to give the total reaction.

Both half-reactions involving Au(III) are correct. However, one is easier to find in a table of standard potentials:

Source: http://ch301.cm.utexas.edu/data/standard-potentials.pdf
If you combine your set or their set of half-reactions into a balanced chemical equation, you get the same answer. This is because you can combine two of the half reactions to get the third (they are linearly dependent):
$$\ce{Au(s) -> Au+ (aq) + e-}\tag{1}$$
$$\ce{Au+(s)  -> Au^3+ (aq) + 2e-}\tag{2}$$
$$\ce{Au(s)  -> Au^3+ (aq) + 3e-}\tag{1+2}$$

[OP] What am I missing?

You are missing the practical part. It is more straightforward to build a cell where the electrode is one of the reacting species. For the Au(I) to Au(III) reaction, you would need an inert electrode. Also, if you wanted to calculate the cell potential, you'd have extra work figuring out the missing electrode potential of half-reaction (2).

The reaction is a disproportionation Au+ gives Au and Au+3
Oxidation: Au+ = Au+3 + 2e-; Reduction:  Au+ + e- = Au. Multiply reduction by 2 to balance charge and add equations.
3Au+ = 2Au + Au+3 These are the correct equations and I do not think the book is correct about the anode reaction.
Lets try it their way!
Oxidation Au = Au+3 + 3e- ; Reduction Au+ + e- = Au  Multiply reduction by 3 and add
Au + 3Au+ = Au+3 + 3Au Subtract a gold from each and... There is an extra oxidation of Au metal. The initial oxidation to Au+ has the highest potential about so should be inhibiting.
The problem is that to develop a cell the oxidant and reductant must be separated in space and that is not possible here since they are the same. The book's solution is to supply a different reductant, Au metal, for the anode. The only way I see to do this is a concentration cell where the cathode reaction is reduction of aurous ion 2Au+ +2e- = 2Au, Rate = Kr[Au+]^2; the anode reaction is oxidation of aurous ion Au+ = Au+++ + 2e-, rate = Ko[Au+]. When the concentrations are equal there is no cell potential and the disproportionation proceeds in both cathode and anode compartments. Increasing the concentration of Au+ in one of the compartments Increases the rate of the reduction reaction in that compartment removing electrons from the other compartment increasing the rate of the oxidation reaction in the other compartment. The compartment with the higher concentration loses Au+ ions thru reduction twice as fast as the other compartment loses Au+ from oxidation. The cell voltage decreases until the concentrations of Au+ are the same in each compartment and the cell voltage becomes zero. the disproportionation is continuing behind the scenes.

The anodic reaction cannot be : $\ce{Au -> Au^{3+} + 3 e-}$, because there is no neutral gold atom $\ce{Au}$ in the original mixture to start the oxidation reaction of $\ce{Au}$. So the book is wrong, and the anodic reaction is $\ce{Au^+ -> Au^{3+} + 2 e-}$.
But the reaction $\ce{Au -> Au^{3+} + 3 e-}$ is necesary to know, just to get the potential of the anodic reaction $\ce{Au^+ -> Au^{3+} + 2 e^-}$. As Karsten states, the potential of $\ce{Au^+ -> Au^{3+} + 2e^-}$ is not given in the tables. But it can be deduced by combining the two potentials of the reaction ($1$) and ($1+2$) from Karsten's answer. As the potential of ($1$) is $+1.69$ V, and the potential of ($1$) + ($2$) = $1.42$V, the potential of ($2$) (which is also the potential of $\ce{Au^+ -> Au^{3+} + 2 e-}$), is $\frac{3·1.42 - 1,68}{2} = 1.285 $ V.
The necessity of this last calculation may have been at the origin of the wrong statement, contained in the textbook.

A more complete table is needed Lange's  Handbook gives "ca. 1.29v"  suspiciously close to your calculated value. Please read my answer above; it was difficult to figure out how to separate something from itself and reunite it.


text:
Reaction between lead acetate and carbon dioxide
As we know, $\ce{CO2}$ will not react with $\ce{Pb(CH3COO)2}$ but if I start electrolyzing lead acetate solution and during the process add $\ce{CO2}$ to the solution will $\ce{CO3^2-}$ reacts with $\ce{Pb^2+}$? Can it be a method to synthesise of lead carbonate?

Why would it? Electrolysis adds nothing to the picture.

Why do you want to make two different operations simultaneously, namely electrolysis AND adding CO2 ? Why ?

by using electrolysis I want to replace Pb2+ ions that react with (CO3)2- sth like a circulated situation because I don't want to use lead acetate more and more. is it possible?


text:
How long does it take for calcium carbonate to calcium oxide if constant heat of 850 degree Celsius is given?
I am using a muffle furnace and the temperature is set for 850 degree Celsius .
I have placed 100g of Calcium carbonate ( in the form of seas shells) in the furnace .
How long will it take to completely convert calcium carbonate into calcium oxide ?

Check periodically the mass and evaluate the time series. It cannot be determined from simple principles and it depends on many variables. Even if the estimation is based on empirical knowledge like in chem. engineering, different scenarios can lead to different results.

Be aware that sea shells are natural composite material, containing organic components protecting them against fragility. there is possibility the result could be blackened by not oxidized carbon or organic origin.

At this temperature it is likely to take forever, since You asked for complete decomposition. However, partial decomposition into impure product strongly depends on the seashell thickness and other substances present in the biological composite.

Since the calcination is a heterogenous process, the reaction rate depends on the surface area of the shell material exposed to the heat. Therefore you have to consider not only the size of calcium carbonate crystals, but also their macroscopic arrangement, size of the reactor etc. Secondly, the reaction is reversible. The faster $\ce{CO2}$ leaves the system the better (for example, a nitrogen flow might help).
Because of these considerations - as Poutnik has already mentioned in his commentary - it would be hard to tell when your reaction would be complete.
I would say that the shells are unlikely to decompose within minutes - rather, you should count on several hours. If it is safe, leave your setup overnight. Otherwise, run it from morning till evening.
When the decomposition is complete the sea shells will be cracked into greyish mat pieces (not glossy anymore). You could of course test the action of an acid onto your product (to be sure that the starting carbonate has gone).
Lastly, you could consider the calcination of the chicken egg shell. As a rule, it is thinner than the sea shells and should decompose faster.


text:
Are these stereoisomers considered geometrical isomers?
I came across this compound in the Black Book of Organic Chemistry by Ashish Mishra. The question asked whether this compound could show geometrical isomers.

Now, I know the compound has the following two stereoisomers:

However, these are optical isomers, they look like cis/trans compounds, but the distances between any two atoms are identical for both structures, unlike in normal cis/trans isomers. So, according to me the answer should be no, this compound cannot show geometrical isomerism, but according to the answer key, it can show geometrical isomerism.
So, can someone tell me why I am wrong, or is it the book that's wrong?

They are stereoisomers that are enantiomers. Each one is optically active. The geometrical E-double bond doesn't exist. See this.

First of all you cant say wether cyclohexane derivatives has geometrical isomers or not just by drawing the structure given in question. That structure is wrong it exists in chair form its most stable form. As there is no pure enantiomer given in question the upper methyl can be at axial or equatorial position. Hence in these cases the distance will be different. Try drawing its conformer

The question is about a cyclohexene derivative


text:
How to use the thermodynamic identity in a reversible process
I have trouble wrapping my head around the thermodynamic identity.
From Schroeder's book "An Introduction to Thermal Physics," I understand that entropy is defined as
$$\mathrm dS=\frac{δQ}T\tag1$$
Later in the book, however, the author warns that $Q=T\,\mathrm dS$ only holds when the change in volume of the process takes place quasi-statically because we then know that $W=-p\,\mathrm dV$. So, from the thermodynamic entity
$$\mathrm dU=T\,\mathrm dS−p\,\mathrm dV\tag2$$
we can associate the heat and work with the first and second terms, respectively.
These two affirmations seem contradictory to me, and I can't seem to figure it out.
My Questions:

Under which assumptions is equation $(1)$ true?

When this equation isn't true, how can we find the change in entropy of a system?

More specifically,  $\mathrm{d}S=\dfrac{\delta Q_\mathrm{rev}}{T}$ or $\mathrm{d}S \gt \dfrac{\delta Q_\mathrm{irrev}}{T}$

The two equations you presented are valid only for reversible processes (i.e., processes in which the system passes through a continuous sequence of thermodynamic equilibrium states).
For a system experiencing an irreversible process, your two equations are not valid.  For an irreversible process, the change in entropy of the system is determined by doing the following:

Use the first law of thermodynamics to solve for the final thermodynamic equilibrium state of your system (starting from the initial state).

Completely forget about the irreversible process in step 1. because it is no longer of interest, except for the initial and final states.

Devise (dream up or conceive of) a reversible thermodynamic process between the same two end states.

Calculate the integral of dQ/T for the reversible process you devised in step 2.  This is the change in entropy of the system.

Thank you for the answer! So you are saying that once I know of the initial and final state of my system, I can then approximate it by a reversible process, right? Just to make sure, wouldn't that cause some issues with non-state variables that depend on the process? Or am I missing something?

You are not approximating the entropy change of the irreversible process.  You are determining it exactly.  The entropy change is a physical property of the substances comprising the system, and the change is independent of any process.  I might also mention that there are an infinite number of reversible processes going between the two end states, and the all give the same value for the integral of dQ/T.

The entropy of a system is only defined (strictly for equilibrium thermodynamics) only for reversible processes as:
$$\text{d} S=\frac{\delta Q}{T}$$
Due to Clausius inequality, for any other process entropy is actually defined by the statement:
$$\text{d}S\geq \frac{\delta Q}{T}$$
Where the equality is true for perfectly reversible processes (those that you call Quasi-Static).
For any other process that is not reversible, you might however still obtain a value for Entropy by considering an analogous process that is reversible and has the same starting and ending point in the space of thermodynamics variables.

Thank you, I'm going to edit that.

Thanks for the answer, I didn't understand the full meaning of Clausius' inequality, now it's clearer!


text:
Define a rate-determining step and an energy barrier for a multi-step reaction profile
I am currently making reaction profile by means of DFT calculations. I am able to correctly draw a whole reaction profile.
However, I am struggling to find what exactly is the energy barrier of a reaction. It is pretty difficult to find this kind of information on Internet or in literature because most described systems are often one-step simple cases. When multi-step reactions are described, I noticed researchers seem to follow different rules ?
To give an example, here is a reaction pathway with the energies attributed to the intermediates and TSs. The pre/post-reactive complexes were determined by IRC followed by optimization.

We see that the separate reactants are defined as the reference (0 kJ/mol) for the reaction pathway. The pre-reactive complex has an energy of 10 kJ/mol and follows a TS of 30 kJ/mol. We then arrive to an intermediate of 20 kJ/mol. The same intermediate in a different configuration following the pre-reaction has an energy of 25 kJ/mol. Finally, the last TS has an energy of 40 kJ/mol. The post-reactive complex of this second step has an energy of -5 kJ/mol. The products have an energy of -2 kJ/mol. This is clear.
Now here is where I am questioning:

What is the energy barrier of the forward reaction ? Is it 20 kJ/mol corresponding to the energy barrier of the first elemental step ? Is it 30 kJ/mol it is the energy of the first elemental step vs reactants ? Is it 40 kJ/mol as this energy is the highest of the pathway ? This is kind if misleading as the second step has a smaller elemental energy barrier of 15 kJ/mol.

What is the energy barrier of the reverse reaction ? Is it 40.0 + 5.0 = 45.0 kJ/mol or 42.0 kJ/mol (compared to the products ?

Overall, I would say I am pretty lost if I have to select the separate reactants / products as references for calculating barriers and how to select the rate-determining step when the second TS has a larger value of absolute energy but is a small elemental step.
Researchers in the literature have tendency not to report energy barriers when they compare pathways, or just mention these of elemental steps.
Others take account of the energy barrier of an elemental step (example here: 20 kJ/mol) to define the global energy barrier but others take the highest value (40 kJ/mol) here.
If researchers do different things, isn't there a rule ? Is it misleading for most people ? So many people doing DFT nowadays to support mechanisms without having ground rules or proper tutorials for organic chemists is clearly missing to my view.
I already thank you for your explanations.

Not sure exactly how one would find the rate-determining step with that data. For example, the single biggest barrier is from pre-reactive complex 1 to TS1, 20 kj,/mol, but that does not describe how quickly that reaction occurs. It would be a rate limiting step, in that until 20 kj/mol is available the reaction cannot proceed.


text:
Why are there 2 spots in my TLC of furan?
I'm working on a monoalkylation of furan, and I'm trying to separate my product from byproducts. When comparing the TLC (10% EtOAc in hexane) of my mixture with that of furan (starting material), I observe 2 spots, one at the bottom line and one around rf~0.8.
Any clue of what this can be?
Thanks!

Try to test each of these two spots with appropriate reagents. Are they alcohols, ketones, aldehydes, etc, ?

Does the lower spot move off the baseline if you increase the polarity of the eluent?

I suspect your product is the high running spot. What you may have is a small amount of polyfuran that has a very strong UV chromophore and is sitting on the baseline

When comparing the TLC of your mixture with that of furan as you say – where is the sport of furan?


text:
Hydrolysis of $A_3B$ type weak acid-weak base salt
I know the formulae for weak acid-weak base salt of AB type. A peculiar question made me ask this. Do the formula for derived for AB type also hold for A3B type sal.
For example:
This is the question, I got from my teacher
My attempt at solution:
Equation-1:$$\ce{3NH^+_4 +PO_4^{-3} + 3H_2O\rightleftharpoons 3NH_4OH + H_3PO_4}$$
$$K_h=\ce{\cfrac{[NH_4OH]^3[H_3PO_4]}{[NH_4^+]^3[PO_4^{-3}]}}=\cfrac{[3ch]^3×[ch]}{[3c(1-h)]^3[c(1-h)]}$$
$$\implies K_h=\frac{27c^4h^4}{27c^4(1-h)^4}\approx h^4$$
Equation-2:
$$\ce{NH_4OH\rightleftharpoons NH_4^+ + OH^-}$$
$$K_b=\ce{\cfrac{[NH_4^+][OH^-]}{[NH_4OH]}}$$
Equation-3:
$$\ce{H_3PO_4\rightleftharpoons 3H^+ + PO_4^{-3}}$$
$$K_a=\ce{\cfrac{[H^+]^3[PO_4^{-3}]}{[H_3PO_4]}}$$
Equation-4:
$$K_w=[H^+][OH^-]$$
From the above equations,
$$K_h×K_a×K_b^3=K_w^3$$$$\implies h^4=\cfrac{K^3_w}{K^3_b×K_a}$$
$$[H^+]^3=\cfrac{K_ach}{c-ch}\approx K_ah\implies [H^+]=\left(\cfrac{K_wK_a}{K_b}\right)^{(1/4)}$$
$$pH=\frac{1}{4}\left(pK_w+pK_a-pK_b\right)$$
This gives half of the correct answer.
Teacher's Solution
The formula given here is applicable to all scenarios so use
$$pH=\frac{1}{2}\left(pK_w+pK_a-pK_b\right).$$  This gives $7.24$ which is the correct answer.
My question is :
Is the formula for $AB$ type weak acid-weak base also applicable to  $A_3B$ type weak acid-weak base salt? What is the correct formula for $A_3B$ salts? How to derive pH of $ A_xB_y$ type weak acid-weak base salt??

There's no such thing as "NH4OH" and acid-base equilibria here should not be called hydrolysis.

@Mithoron but isn't that ammonium hydroxide . But ammonium cyanide, ammonium acetate and ammonium carbonate undergoing the same reaction is hydrolysis? My book explains weak acid-weak base salt hydrolysis by ammonium acetate and ammonium cyanide $$NH_4^+ + CN^- + H_2O\rightleftharpoons NH_4OH+HCN$$

If it is not a question of hydrolysis, how do i solve it ?  In link it the formula for $AB$ salt hydrolysis is derived similar to my method?

There's no "ammonium hydroxide" and calling all the equilibria hydrolysis is a misnomer. These are rather objections to your nomenclature than actual issue, though.

The question itself is unclear and I recommend using a newer textbook of higher quality.

$\ce{NH4OH}$ assumes $\ce{N}$ is able to form 5 bonds. It is not.
The $\mathrm{p}K_\mathrm{a} = 5.23$ for phosphoric acid is nonsense; the task is made up. Additionally, the above provided formulat is not applicable on salts of multiprotic acids like $\ce{(NH4)3PO4}$
The respective $\mathrm{p}K_\mathrm{a}$ for each of $\ce{H3PO4}$ hydrogen are   $\mathrm{p}K_\mathrm{a1}=2.14$,  $\mathrm{p}K_\mathrm{a2}=7.20$ and  $\mathrm{p}K_\mathrm{a3}=12.37$.
It is possible to have a combined $\mathrm{p}K_\text{a,tot}$  directly for $\ce{H3PO4 \to PO4^3-}$, but it would be their sum, 21.73.
The task can be simplified to ( informed guess assuming $1 \lt x \lt 2$):
$$\ce{3 NH4+ + PO4^3- -> (3-x) NH4+ + x NH3 + (2-x) HPO4^2- + (x-1)H2PO4-}$$
and
$$\mathrm{pH} \approx \mathrm{p}K_\ce{a,NH4+} + \log {\left(\frac{x}{3-x}  \right)} \approx \mathrm{p}K_\ce{a2,H3PO4} + \log {\left(\frac{2-x}{x-1}\right)} $$
$$\mathrm{p}K_\ce{a,NH4+} - \mathrm{p}K_\ce{a2,H3PO4} =     \log {\left(\frac{2-x}{x-1}\right)} - \log {\left(\frac{x}{3-x}  \right)}=\log{\left(\frac {(2-x)(3-x)}{(x-1)x}\right)}$$
Let substitute $A = \mathrm{p}K_\ce{a,NH4+} - \mathrm{p}K_\ce{a2,H3PO4}= +2.05$
$$10^{A} \cdot x(x-1) = 6-5x+x^2$$
$$(10^{A}-1)x^2 + (5 -10^A)x  - 6 = 0$$
$$(10^{2.05}-1)x^2 + (5 -10^{2.05})x  - 6 = 0$$
$$111.2 \cdot x^2 - 107.2x  - 6 \approx 0$$
$$x \approx 1.01708$$
$$\mathrm{pH}  \approx \mathrm{p}K_\ce{a,NH4+} + \log {\left(\frac{1.01708}{3-1.01708}  \right)} \approx 9.25 - 0.30 = 8.95$$
$$\mathrm{pH}  \approx \mathrm{p}K_\ce{a2,H3PO4} + \log {\left(\frac{2-1.01708}{1.01708 - 1}  \right)} \approx 7.20 + 1.76 = 8.96$$

Ignoring the mistakes in the question or textbook data , Is the formula I derived $$pH=\frac{1}{4}\left(pK_w+pK_a-pK_b\right)$$ correct if I write $\ce{[NH_3]}$ in place of $\ce{[NH_4OH]}$ ie replace $NH_4OH$ by $NH_3+H_2O$ and ignore $ [H_2O] $concentration

No, it is wrong.


text:
How is it even possible that vapour pressure of liquid and vapour of solid are equal at freezing point?
My text book states The freezing point is defined as "the temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase" Also the vapour pressure is the pressure at phase equilibrium by definition.
By looking at a typical phase diagram
Its clear that theres no single temp or even different temp corresponding to same pressure at equilibrium for solid with gas and liquid with gas (except the triple point) . The how can the vapour pressure be assumed equal at freezing points. Many sources explain this in term of chemical potential but doesn’t the phase diagram contradict the statement .Please correct if anything i stated is incorrect. Thanks

https://chemistry.stackexchange.com/questions/65325/at-the-freezing-point-why-is-the-vapor-pressure-of-the-liquid-phase-equal-to-th

https://chemistry.stackexchange.com/questions/55819/how-can-water-exist-in-three-states-in-freezing-point

@Mithoron thanks the  question “ https://chemistry.stackexchange.com/questions/55819/how-can-water-exist-in-three-states-in-freezing-point “  is very similar to one i asked but my question is concerned with the contradiction between phase equilibrium graph and definition of freezing point (definition according to some sources )

I know. There is no contradiction. I've seen other similar issue, and you're looking at wrong chart.

How does not come into it. It is an experimental fact that at equilibrium,  and only at equilibrium, will the vpr pressures be the same.

The liquid-solid line in the phase diagram consists of the two phases with no gas phase present so the question of having the same vapor pressure is moot. If the physical constraint is removed, the common example is the freezing point of water in the open air, there are the additional equilibria involving water vapor. The vapor pressures of the liquid and solid must be equal. In essence a new triple point is defined for every inert gas pressure. This is why the freezing point of water for the initial Celsius scale had to be specified at a specific pressure [this pressure was the sum of the inert atmosphere and the vapor pressure]. Water in an inert atmosphere at a temperature has a definite vapor pressure. When both water and ice are present at a given T and P a triple point is established. Each inert gas pressure establishes a unique triple point, a distinct vapor pressure; solid and liquid must have the same VP at equilibrium.

Which physical constraint you are referring to ? Also will you  please explain at undergraduate level. Thanks

This is at grade school level. Please explain if it needs more in depth discussion or further simplification. The physical constraint is the piston compressing the liquid and solid. If this piston is removed and an inert atmosphere [usually air] applied then liquid and solid can both vaporize. More detail is in several PChem books, GET them Read about equilibrium and get totally confused ponder until the light bulb lights. then do it all again.

Below the melting point, the vapor pressure over the solid phase is lower than over the respective liquid phase. And vice versa above the melting points. At the melting point, they are equal.
Look at it from the different angle. At the freezing point, the solid and liquid phase are at equilibrium. How could they be at equilibrium if each had different vapor pressure? The one with higher pressure would have tendency to transform to the one with the lower pressure. Similarly as if jars with water and saturated salt solution were place in the common, closed container.

To understand what you are asking, you need to remember two things:

Each point in a phase diagram is not related to a certain state function, but it represents the phase state (solid, liquid, vapor) that is thermodynamically stable at a fixed pair of values $(p, T)$. For a single component (i.e., a simple pure substance), this means that if the point lies in a certain region of the phase diagram (not specifically on any single curve or the triple point), then it is in that physical state (single phase).
Each point in the phase diagram lying on a curve represents an equilibrium between two different phases, such that if you artificially maintain $(p, T)$ constant, the phase transition between these two states will be perpetually ongoing. To be clear, for a freezing (melting) transition, you would observe that in the bulk material certain volume elements melt while others freeze until you let the system evolve to a new pair $(p', T')$, thus allowing for the phase transition to end.

Now, if you understand that each point on the separating curve between the two regions associated with compressible liquid and solid phases is equal to a pair $(p_{\text{melt.}}, T_{\text{melt.}})$ of parameters at which solid and liquid phases coexist, and that for the coexistence of these two in an isolated system, you need them to be in equilibrium, this means that their chemical potential is actually the same. So, at different fixed freezing temperatures, there exist a set of freezing (melting) vapor pressures that define the separation curve between the phases, meaning that for each point on the curve $p_{\text{solid}} = p_{\text{liquid}} = p_{\text{on the curve}}$. The key point is to note that in the phase diagram, $p$ is not associated with any of the phases but it is a generic parameter (so on the curve it must be equal for the two coexisting phases at a specific melting temperature $T$).
EDIT:
I got the topic wrong.
Sorry, but I was checking out my notes to try and understand if I was totally on the wrong track and as @jimchmst correctly said that there is just one point at which 3 phases do coexist for a pure substance. However, considering the atmosphere, you should look at a different graph, namely the one depicting a two phase mixture of an inert atmosphere and the pure substance. Honestly I tried to find one that depicts it correctly but I did not succeed. However, consider that the phase diagram you depict considers just the pure substance.
There is however also the possibility that your text is just using an old convenction: to consider the vapor pressure as the pressure of the pure component in an inert atmosphere in equilibrium with the pure substance (for example at the melting point for your substance in an atmosphere the vapour pressure of the liquid and the vapour pressure of the solid must be equal to allow for the transition to happen).

Im sorry , Inspite of your answer by  using chemical potential, Doesn’t the definition of freezing point  still contradicts  the graph.

The point is that freezing point (also boiling point) is defined at a certain pressure. At atmospheric pressure you get a certain freezing point, however this means that at a different pressure (say one half of the atmospheric one), the temperature will be different. You might want to have a look at Clausius-Clapeyron equation to better understand this concept. There is no such thing as a single freezing point for a substance.

Im sorry for any confusion I didn’t said theres a single freezing point. I’m very well aware that every point on the line between liquid and solid represent freezing points at different pressure. The question is vapour pressure of solid and vapour of liquid can never equal at any temp which is the definition of freezing. Hopefully this clears the confusion. Thanks a lot  for trying.

With only one component there is only one triple point among vapor, liquid water and ice one. addition of a second component, the atmosphere, introduces an additional degree of freedom. The solid-liquid curve does not define vapor pressures.

Oh, right, I should read again my pchem notes, I've got it mixed up, thanks. I'm deleting the other comments to avoid confusion in new readers.

I made an addendum to the answer to address the errors and now it should be clearer, I wil check out for some references. Thank you @jimchmst


text:
Name of cyclic tetraatomic carbon molecule
I came across this article on stack exchange which mentioned a cyclic compound of four carbon atoms each connected with a double bond to the two adjacent carbon atoms. However, this made me think what would be the name given to such a compound. Also, could this similar compound exist? If so, what would be its name?

This abomination ist called C4, and it is very unstable. Joking aside though, his is just one possible structure of $\ce{C4}$, and they all are unstable. The image given here likely is not a good approximation to what a "quadratic" arrangement of 4 carbons would be like.

One could call such structure cyclobutadiyne. Except it's organic nomenclature name.

https://chemistry.stackexchange.com/questions/19475/hydrocarbons-with-only-4-carbon-atoms


text:
Why can't elements and atoms be separated by physical and chemical means?

Our textbook tells us ''that elements and atoms cannot be separated by chemical and physical methods. He says only atoms can be separated by nuclear methods.''

First of all, I would like to start my question as follows. We call the smallest unit of an element an atom. We call a collection of single types of atoms with the same number of protons an element. At this point, let's consider, for example, the molecular element oxygen. Isn't separating this element into its molecules or separating the molecules into their atoms decomposing this element?
My other question is: isn't removing an electron from an atom a decomposition? So, when we say decomposition in chemistry, do we mean extracting different types of substances from one substance?
Edit:
https://en.wikipedia.org/wiki/Chemical_element

Kimyasal element, kimyasal reaksiyonlarla başka maddelere parçalanamayan kimyasal bir maddedir. Kimyasal elementi oluşturan temel parçacık atomdur.

Edit:2

We all know that atoms can never be created or destroyed by any physical or chemical method. Any type of atom can only be converted from one form to another.
Elements are substances that contain only one type of atom. As we have previously stated that atoms cannot be destroyed in a physical or chemical reaction, elements cannot be broken down into simpler substances. Breaking down an element gives us atoms of the same element, so no new substance is obtained.
For example, water actually consists of two main elements: hydrogen and oxygen. Therefore, water is a compound that can be broken down into simpler substances such as oxygen and hydrogen. However, we cannot separate the elements hydrogen and oxygen into simpler substances.
Therefore, we cannot break down elements into simpler substances.

https://www.vedantu.com/question-answer/can-elements-be-broken-down-class-11-chemistry-cbse-60235daff865ac1e018d5242

Oxygen can be an element, as it is the collection of atoms having 8 protons. The same name is usually used for describing the molecules $\ce{O2}$ made of two atoms oxygen. But oxygen cannot be called a molecular element, as you are stating. Oxygen is either an element, or a molecule. Not both simultaneously. And removing an electron from an atom is not a decomposition. It is an ionization. the word decomposition is used when separating one or several atoms from a molecule.

@Maurice Well, can't we separate an element with a single atomic structure, such as iron, into its atoms? Wouldn't this be segregation?

Either your textbook by "separation" means breaking up atoms, or you translated it badly. No matter what you do with iron, it's still iron, be it liquid, gas or plasma. Unless you manage to cause a nuclear reaction.

@Mithoron Isn't separating an atomic element into atoms a decomposition in chemistry? The expressions in the book cannot be decomposed into simpler units. How can the atom be separated into simpler units with nuclear energy? Should a substance with different properties emerge as a result of this separation? For example, if we divide the molecular element oxygen in nature into its atoms, wouldn't this be decomposed into simpler units?

@Maurice Isn't oxygen O2 an element with a molecular structure? Let's also consider an element with an atomic structure: iron. Isn't separating the element iron into its atoms a decomposition? Or do we not take this as separation because atoms show the properties of the element? Finally, if we consider 1 iron atom, is this 1 atom also an iron element? In other words, aren't the concepts of 1 atom and element the same concept?

Asking on CH SE site should not be the first choice to find answers. It should be rather the last choice, after thorough consulting available offline and online resources before asking. Include compact summary of this research as the starting point.  If you found nothing, provide what/where you were searching for.  // This way you will learn the independent studying and saving resources of people telling you, what you could find on your own. Additionally, such a question would be a bigger contribution to the community.

This is a question about English, not about chemistry.

It was not an answer, but a comment. Btw metals do not have atomic structure. If they had had one, their melting and boiling points would have been very low, lower than for many organic compounds. They are interconnected by metallic bonds.

If things connected by metallic bonds are not atoms, what are they? What I mean by atomic structure is a structure made up of atoms. Also, please keep your comments that constantly correct me to yourself. I may make mistakes. My English is not as good as yours. Please keep this in mind when commenting.

What book is this? Could you provide a full citation?

@BuckThorn It is the national education book in this country. While explaining the properties of elements and atoms, he says that they cannot be separated into simpler units by physical and chemical means. This information is also available on wikipedia and on English websites.

Well, that's nonsense, depending on how you define "separate", "units", "simpler", "chemica methodsl" and "physical methods". If you use standard definitions, it's nonsense.There is something called "sub-atomic structure". It's basic knowledge that atoms consist of nuclei and electrons, that nuclei consist of neutrons and protons (which in turn are also divisible), that a process called radioactivity allows conversion between elements, etc. Atoms might be defined as the simplest or smallest chemical structures. There might be better definitions.

@BuckThorn The book says, Atoms cannot be broken down physically or chemically. However, with nuclear energy, atoms are broken down into simpler units. However, the book states that elements cannot be separated by chemical and physical methods. The point I'm asking is, can't we separate an element into its atoms? In the source I have given below, it says that since elements are composed of the same type of atoms, separating elements into atoms will not create a new substance, chemistry does not call this decomposition.

@BuckThorn https://www.vedantu.com/question-answer/can-elements-be-broken-down-class-11-chemistry-cbse-60235daff865ac1e018d5242

"Separate" and "decompose" have different meanings (and even individually can be used to mean different things). I understand decompose to mean "to break down". Separate means to bring apart. A single atom cannot be decomposed chemically. Two atoms can be separated from each other, however. The term "element" requires some care (see the answer you received). If you have matter in elemental form, all atoms in that matter are of the same type and cannot be decomposed chemically into other types of atom.

Be aware that the term element has several meanings in chemistry, depending on context. It is mandatory and essential to clarify in any discussion, which of meanings you have in mind.

An element(1) is an atom, a ion, or a nucleus itself, having the particular number of protons in its nucleus.
An element(2) is a molecule or molecular ion consisting of atoms of the same element(1). Such a molecule can be huge, if you consider graphite or diamond.
An element(3) is a substance, consisting of element(1) or element(2), with all atoms of the same proton number. Various forms of element(3) are called allotropes of element(1).
An element (3a) is any of element(1,2,3)
An element(4) is an element(1) integrated in molecules of compounds or substances.
An element(5) is a basic building block of a more complex system or structure.

Element(4) can be chemically isolated as element(3).
Element(3) can be chemically(*) or physically separated to atoms/ions of element(1), in many cases via element(2).
An element(1) cannot be further divided by chemical means. As a molecular entity is the smallest particle of a compound, still having it's properties, an atom/ion/nucleus of element(1) is the smallest particle of the element(3a), having still its properties and can be still recombined to element(2,3).
$\ce{Hg(l)->Hg(g)}$ is decomposing of element(3) to a set of atoms=element(1).
$\ce{C(s)->C2(g)}$ is decomposing of element(3) to element(2).
$\ce{O2(g)->2 O(g)}$ is decomposing of element(2) to element(1).
Further dividing is possible, involving physical methods of breaking or transforming the nucleus, forming one or more nuclei of different elements(1).
It does not mean one element(1) consists of other elements(1). It means the building blocks of elements(1), subatomic particles protons, neutrons and electrons, form different elements(1) by different ways in different numbers they get together via strong and electromagnetic force.
(*) - Chemistry is a part of physics, focusing on mutual interactions of atomic  electrons and their implication for matter behaviour on micro and macros scale.

As far as I understand, there is a misconception here. Now let's think of an element with a 2-atom molecular structure. Here, this element can be decomposed into atoms, or even into simpler substances through a nuclear reaction after the atoms. However, since the atoms of an element show the same chemical properties as the element, chemistry does not consider this as a dissociation.

But if we separate a compound chemically, we can reveal 2 or more elements. And the chemical and physical properties of these elements can be very different from the properties of the compound. So chemistry takes this as an example of dissociation. At least that's how textbooks get it?

Where exactly did you get that O2 -> 2 O is not dissociation? HCl is a compound and can be dissociated to atomic or molecular chlorine and oxygen. O2 is not compound, it is one of allotropes of oxygen, that can be dissociated to atomic oxygen.

See my update. By definition written on Wikipedia and all other textbooks and chemistry sites. Elements cannot be broken down into simpler substances by chemical or physical means.

Avoid using screenshots instead of copy pasted text or easy retyping. // Elements(1) cannot be. Elements(2,3) can be broken to elements(1,2). Context rules. E.g. O3, O2 and O,  or S8, S2 and S are different substances, being different allotropes of respective elements. // You consistently avoid distinguishing of meanings of "element" what leads to vague statements.

Could you please check the new update?

It is not correct very much much.

One way to discern a separation by physical means and a separation by chemical means is to monitor if (valence)* electrons are redistributed, or not.
By this criterion, passing for instance a mixture of water and sand through filter paper is purely physical: sand remains sand, and water wasn't changed and still is water.  On the other hand, the electrolysis of molten sodium chloride (i.e. mobile ions of $\ce{Na^+}$ and $\ce{Cl^-}$) to yield elemental sodium in one part, and elemental chlorine in an other part of the Downs cell is a chemical process; here, the sodium cations obtain an electron by chloride anions; you converted the compound sodium chloride (a salt) into new compounds sodium (a metal) and chlorine (a gas).
The separation of elemental** nitrogen gas and elemental oxygen gas by fractional distillation, for instance while running a Linde process equally is a physical process: there is no transfer of electrons between the atoms, no old/already existing chemical bonds between the atoms are broken, and no new chemical bonds between atoms are formed.  This invalidates the statement elements couldn't be separated by physical means alone.
* Except nuclear chemistry, most chemistry involves the most accessible electrons in highest atomic/molecular orbitals.  For instance, the reaction of sodium and chlorine to yield sodium chloride is about the transfer of sodium's electron in the $3s$ orbital into a $3p$ orbital of chlorine with preservation of the $1s^2 2s^2 2p^6$ configuration of neon.  Hence some tables abbreviate the report of the electron configuration e.g., for chlorine with $\ce{Ne}\ 3s^2 3p^5$.
** "element" as in case (3) of the answer by @Poutnik

I understood how to separate a compound into its elements by chemical means. But can't we separate an element into its atoms? Can't an atom be a simpler substance than an element? Or are we saying that since atoms represent that element chemically and physically, separating an element into its atoms does not mean decomposing it into a simpler unit? Again, what do we mean by decomposing an atom into its simpler units? When we split the atom nuclearly, do we create 2 different types of atoms?

Some elements can occur as isolated atoms (metals, helium); other only as polyatomic molecules (e.g., nitrogen). The nuclear split of atoms is nuclear fission as e.g., in an nuclear power plant. Energies involved are much greater than typical bench chemistry. But today chemists know than an atom itself consists of smaller (subatomic) particles, elementary particles.  These (at present state of research) are recognized as the smallest units of matter.


text:
What is the cause of different metal densities?
What causes differing density?
Is one metal more dense than another because the atoms are larger (more mass per atom), or because the atoms are closer together?   Or both?
Example:
Aluminum
Density: 2.7 g/cm³
Atomic mass: 26.981539 u
Protons: 13
Neutrons: 14
Copper
Density: 8.96 g/cm³
Atomic mass: 63.546 u
Protons: 29
Neutrons: 35
So the atomic mass of copper is higher than aluminum because it has more protons and neutrons.
I have also read that the atomic spacing in copper is approximately 2.56 angstroms (Å), while the atomic spacing in aluminum is approximately 2.86 Å.
So it seems the atoms in solid copper are closer together than in aluminum.
Can you confirm this all sounds right and add any comments to help us understand why different metals have different densities?
Thanks!

Both.$\mathstrut$

Your explanation is right. I must add that most metals also contain differences in their chemical bonding, which affect their bond lengths. Being a metal does not exclude also having covalent bonds. Covalent bonds are shorter than purely metallic bonds.


text:
-I effect vs +M effect of Chlorine on acidity of benzoic acid
The -I power of Halogens like Chlorine is stronger than their +M power. Therefore, we consider -I effect for comparing acidity of benzoic acid [Ka-> ortho > meta > para]. But shouldn't -I effect be weaker in para-chloro benzoic acid, it will pull electrons, from the ring, but not so much directly from -COOH. On the other hand, the +M power will be same as it is range independent.

Using the above example to emphasize on the point, the -I power on -COOH will be extremely weaker than the +M power. Chlorine would pull electrons from the attached ring, but not so much from -COOH group and its +M effect should make it less acidic.
So why do we always take halogens like chlorines to be always electron withdrawing group?
Which structure is more acidic? The left one will have +M effect but stronger -I, right one will have no +M effect but sligthly worse -I effect


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Metallic nanostructures exhibiting charge separation
Mostly ions consist of a few atoms. It is easy to see, why the solids they form must have simple stoichiometries. It is true even for polymeric ions, such as some complicated organic molecular ions as long as there are only covalent bonds inside these.
It is definitely possible to have two different metals with a miscibility gap even in their liquid state. Let's imagine one of these has higher chemical potential for electrons. Therefore opposite charges build up on their interface. Opposite charges obviously change their components' chemical potentials. In the liquid state, there is also a tendency of increasing the area of such a charged interface. However, their surface tensions may oppose that tendency.
The question is: Does this process always end with some kind of intermetallic compound with a single (possibly defected) lattice structure in some cases and complete phase segregation in those cases in which electron chemical potentials in both phases are very similar to begin with?
Any counterexample would be nice to see.
Silver subfluoride has both similarities and differences from what I'm looking for: Similarities are non-integer oxidation state and layers of silver thicker than one atom, differences are fluoride ions' non-bonding to themselves, a simple fraction stoichiometry of 2:1, and the lack of incommensurability of the lattices they form.

@PaulKolk OK, then I did not get your question. What are you asking about, really?

@IvanNeretin The possibility of something that resembles two-phase metal alloy, but forms a few-atom thick layers of incommensurable lattices spontaneously due to opposite charges on these phases.

So, you mean something like Ag2F but fully metallic and with some charge separation between phases? No one would guess it from the title... So, better title, paragraphs, highlighting important stuff, maybe cutting out quite a bit in the beginning.

I edited title and tags based on your last comment. Check it out and edit further. This may become a useful question, but still needs work.

Would the impact of gold on the adsorption of hydrogen on magnesia be in scope? See https://pubs.acs.org/doi/10.1021/jp4037588.

@OscarLanzi It is relevant, even though I'm asking about macroscopic phenomenon (or a theoretical reason for the lack of such phenomenon). One may imagine these protons behave differently if a solid is built by adding more gold and magnesia-hydrogen layers on top of each other.

The type of charge separation described in the question has been identified between hydrogen and gold, the latter a relatively electronegative metal, by Chen et al[1].
Hydrogen adsorbed on neat magnesia breaks up into atoms which are separately bonded to magnesium or oxygen; these hydrogen atoms are respectively polarized negatively and positively. Putting the magnesia as a thin film on silver does not change this but using a gold support instead of silver does: the gold takes up electrons from the magnesium-hydride bond, so the hydrogen remains adsorbed on the oxygen only and thus becomes entirely protic. The gold takes up the negative charge that would otherwise have been on the hydridic hydrogens. In effect a magnesia-hydrogen auride sandwich is created.

Reference

Hsin-Yi Tiffany Chen, Livia Giordano, and Gianfranco Pacchioni. "From Heterolytic to Homolytic H2 Dissociation on Nanostructured MgO(001) Films As a Function of the Metal Support." J. Phys. Chem. C 117, 20, 10623–10629 (2013). https://doi.org/10.1021/jp4037588


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Why are alcohols considered slightly basic (based on pKa values)?
I understand why it makes sense for an alcohol to be slightly basic given the hydroxyl group and the electron-donating properties of the alkyl chain. But I'm a bit confused as to why it's the case based on the pKa of ROH and of ROH2+. pKa of ROH is approx. 16, meaning half of the alcohol would be deprotonated to become RO- at pH 16, while the pKa of ROH2+ is approx. -3, meaning half of the alcohol would accept an extra proton to become ROH2+ at pH -3. A pH of 16 is much less extreme than a pH of -3, so doesn't alcohol more readily give up a proton than accept an extra one, if one were to look only at the pKa values?

Alcohols are not slightly basic. They are about as basic as water (which means: not much at all), only less. Also, they are about as acidic as water, only less. Taken together, this reads: neutral.

A  pKa of 16 means equal concentrations of alkoxide and alkonium ions of about 10^-8. Please explain the pKa of ROH2+.

@jimchmst at pH = pKa for ROH, [RO-]/[ROH] = 1, and at pH = pKa for ROH2+, [ROH]/[ROH2+] = 1 (I'm a bit confused to as to why you're saying RO- and ROH2+ would be equal)

Alcohols especially methanol and ethanol self-protonate as does water. There are kinetic differences, superdry alcohols have a sharp OH NMR peak that couples to the adjacent carbon H's; the coupling vanishes with a trace of water. The equation is similar to water: 2R-OH = R-OH2+ + RO-. Alcohols are Bronsted acids and bases in equal amounts.
Alcohols do not furnish OH- ions so they are not Arrhenius bases. [The hydroxyl group is not the hydroxide ion.]  In many organic reactions alcohols react as Lewis bases but not as Lewis acids so maybe this is the reason for the idea. Potentiometric titrations in ethanol-water are very similar to just water and my limited experience with the pH of ethanol-water mixes is that they are effectively neutral to slightly acidic from CO2 absorption.
Here is one article to confuse the issue: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC8201227/ This indicates that at low concentrations in water alcohols are acidic but pure water extrapolates to an acidic pH so CO2 may be involved. At high concentrations the solutions are basic. This could possibly be caused by the lowering of the water chemical activity and not the basicity of the alcohols.


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Doubt in Zeeman effect for sodium spectrum
I am reading an atomic molecular physics book (for Indian universities). You may not familiar with it. But you can download it from the link if you wish.
So in the book it is mentioned that for normal Zeeman effect, transition between any two spin singlet states $ (S=0) $ results in a 3 component splitting, one $  \pi $ and two $ \sigma $ depending upon the selection rules.
They also tell us about the polarisation state of emitted radiation.
The $\pi$ component transition refers to electric field parallel to applied magnetic field and $\sigma$ component transition refers to electric field perpendicular to applied magnetic field.
Same kind of components appear in case of Sodium atom for anomalous zeeman effect, as shown in fig. below.

But their polarisation conditions are not explicitly mentioned. So can I assume that in this case of Sodium atom the $\pi$ and $\sigma$ components have the usual polarisation meaning i.e. parallel and perpendicular to magnetic field, as in normal Zeeman effect.

+1 clearly stated question

May I direct you to the article? The life of π and σ—A tutorial review of the ubiquitous use of these symbols in Zeeman and magnetic-resonance spectroscopy, https://onlinelibrary.wiley.com/doi/epdf/10.1002/cmr.a.20105. It addresses your query in detail.

@AChem Thanks. Really helpful


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Why doesn't β-elimination occur in Cr(CMe3)4?
By $\beta$-elimination, I am refering to the reaction where a M-C-C-H structure reduces to ($\eta^2$-alkene)-M-H.
My lecture notes claim that $\ce{Cr(CMe3)4}$ is stable and does not incur such elimination reactions due to "geometric constraints", but I am having trouble understanding why that is.
Whether the molecule is tetrahedral or square planar, it feels like there should be enough room for hydrogen to bond to the metal.
As the molecule is unsaturated, the metal should be able to provide an empty orbital to accept the hydrogen's electron pair.
It certainly doesn't help that I can't search for this molecule, either as "tetra-t-butyl chromium", "tetra-tert-butyl chromium", "tetrakis-t-butyl chromium" or other variations of these.

There should not be enough room. That's the point. Too little space for it to be planar, too. It's tetrahedral and Cr surrounded by ligands almost completely. It's not really an issue about this complex, you were already told the reason, but some misunderstanding how bulky ligands can be.

@Mithoron Thanks a lot! You are right; I am just beginning to study organic chemistry and have little intuition on the bulkiness of ligands.

A paper on obtaining the compound can be found at https://doi.org/10.1016/S0022-328X(00)81814-8 . It claims half-life of 4.5 min at 70 C in heptane. It offers minimal discussion on the fact, but the intro contains some links that might be useful

https://www.sciencedirect.com/science/article/abs/pii/004060909290719R


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Potassium Ethoxide Synthesis: Does the Reaction Involve Radical Processes?
Can Potassium Ethoxide be obtained by reacting the alcohol with metallic potassium?  Considering that potassium, as a group one metal, has only one valence electron, would this reaction therfore involve a radical process?

It is a common practice to let alkali metals react with alcohols as weak acids to get rid of remnants,  as the reaction is much slower than with water.  $\ce{2 M + ROH -> 2 ROM + H2}$. The reaction involves free solvated electrons, similarly as reaction with water or liquid ammonia. Electrons then reduce alcohol to hydrogen and alkoxide.

The reaction can be very dramatic. Sodium metal reacts smoothly with dry ethanol generating hydrogen gas. Potassium reacts violently bouncing around on the surface. I recommend reacting potassium in small amounts under a N2 atmosphere.
Reaction of most active metals with water or acid most likely goes thru one electron transfers to give hydrogen atoms that then react further usually to form H2 molecules.
I worked for a company that made pressure transducers that were filled with silicone oil with a thin flexible stainless steel diaphragm that was in contact with the process. They had an interesting failure mode. They filled with H2 gas at eventually high pressure. This was first noticed in refineries in hydrogenation processes and was dismissed as H2 migration thru the steel from the high pressure in the process and was partially solved by using lower nickel stainless and a competitor actually gold plated theirs. Then failures happened in sugar beet plants and breweries in processes that had no H2 gas. The reaction was fastest with the corrosion resistant high nickel steels. What was happening was that the transducers were in electrical contact with steel tanks the tanks were corroding and the hydrogen was generated on the SS diaphragms and migrating thru the metal as H atoms and forming H2 inside with no means of returning. This was reproduced in the lab using iron and zinc in acid in contact with the transducers. Problem solved by using the lowest nickel SS suitable and carefully insulating the transducers from the metal tank. The evidence for a one electron reaction even for the divalent metals is that hydrogen transferred thru the metal extremely rapidly at very low hydrogen pressures and even at high inside pressures it did not transfer back.


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What is the reason for the difference in melting point between oleic acid (9-octadecenoic acid) and isooleic acid (10-octadecenoic acid)
Isooleic acid only differs from oleic acid in the position of the double bond. I don't understand why this might lead to an increase in the melting point. The degree of saturation hasn't changed and nor has the length of the carbon chain.

Ask yourself the different way - why should the change of both chains at the double bond have no influence? 1-butene has m.p. -185.3 Deg C,  and 2-butene -138.9 deg C, both the same chain length and double bond count.

What is this difference, then? I found so far only m.p. for trans isomer of your isooleic acid.

Another issue is that it could have polymorphs with different m.p.

https://scripts.iucr.org/cgi-bin/paper?S0108270191012921

Oleic acid is not just 9-octadecneoic acid it is (Z)-9-Octadecenoic acid (cis at the double bond). isooleic acid is usually (E)-10-Octadecenoic acid (trans at the double bond). are those the two compounds you are comparing? If so, do you see an obvious major difference?


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Reversible and irreversible reactions
I searched that an reversible reaction can be made irreversible by removing its product.
I am wondering if catalysts can help produce the same effect. And I'd also like to know, if actually an irreversible reaction can be made reversible.
Please give me some examples if possible, thanks!

The equilibrium of a specific chemical reaction cannot be in general 'changed from the outside', unless the temperature is changed accordingly. However, an equilibrium reaction may seem to be irreversible if we remove the products selectively, although I don't like the wording that you use. We are not altering the equilbrium, rather allowing the forward rate of reaction to keep marching on.
One simple application of this is a membrane catalytic reactor for dehydrogenation of alkanes. These reactions are limited by equilibrium. In the following image we can read the equilibrium conversion $X_\mathrm{eq}$ of several alkanes as a function of temperature $T$:

Even though the temperature is very high, more than $\pu{50 \%}$ of the reactant will just sit in the reactor if it is long enough. But, in this particular reaction we may eliminate the hydrogen gas generated through a permeable membrane, which we illustrate for ethane:
$$ \ce{C2H6(g) <=> C2H4(g) + H2(g) v} \tag{1} $$
A scheme of the procedure is the following:

The idea is very simple, we remove hydrogen so the reaction can continue. If we keep doing this, the reaction may seem to be irreversible as we will find a negligible amount of reactant at the end of the reactor. We will walk at the outlet of the reactor and find $\approx \pu{0 mol dm^{-3}}$ of ethane, but the reaction in inherently reversible.
As a remainder, this is a very delicate process, because it would be a huge problem if ethene or any other alkene entered the gas stream along with hydrogen. Since the wanted product is the alkene, we would further need a separation process of it an hydrogen, which would be a waste of money. The key element is the solid membrane, whose structure must not intefere with the reaction and only serve the purpose for the small hydrogen molecules to pass through them via a mass transport process.

References
The $X_\mathrm{eq}$ vs $T$ graph was taken from:

This link.

The image of the membrane catalytic reactor is found in this work:

Shelepova, E. & Vedyagin, Aleksey. (2021). Theoretical Prediction of the Efficiency of Hydrogen Production via Alkane Dehydrogenation in Catalytic Membrane Reactor. Hydrogen. 2. 362-376. 10.3390/hydrogen2030019.

Keeping a reaction away of its equilibrium by removing products does not make it irreversible, just "effectively irreversible". Catalysts do not change it, as they affects reaction rates, but not the equilibrium position. Removal of reaction products can be faster, as they would be formed faster.
Note that truly irreversible reactions are rare. They are those that do not have an equilibrium. Typical (nuclear, not chemical) irreversible reactions are radioactive decays.
It is usually like the equilibrium of a reversible reaction is shifted so far to right it is effectively irreversible.

An equilibrium reaction is by definition reversible. At equilibrium forward and reverse reactions have equal rates.

@jimchmst Sure, this is widely known.

Then revise you last sentence to show that.

Or, revise your interpretation of it.

Radioactive decay is technically reversible too, and has an equilibrium. It's just that these equilibria tend to be very hard to reach outside the core of a star or a supernova explosion. The reason why nuclear decay is practically irreversible is basically the same as for e.g. the decomposition of nitroglycerine or other similar explosives: in either case, both entropy and energy very much favor the products.

@IlmariKaronen I am afraid matter quantization can contradict this idea. A hypothetical equilibrium shifted so far right that nonzero count of reactant nuclei or molecules means nonzero net rate until there is none. It can be hardly considered reversible reaction with equilibrium.


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Naphthalene absorbed by HDPE?
I placed some empty HDPE plastic vitamin bottles in a sealed closet
with moth balls, (naphthalene, d-diclorobenzne.) Years later I found
the mothball smell had somehow gotten into the bottles, and no amount
of airing them out would get rid of it. I assume the
naphthalene etc. has actually been absorbed by the HDPE itself and is there
to stay, no?
Come to think about it, it all makes sense: when deploying mothballs one can simply toss their plastic packages into the closet, not bothering to tear open their wrappers. The smell gets out of the bag and into the closet just about the same as if one tore them open.

As it had years to come in the HDPE, it would have years to come out. There is also the question why to store vitamines for years, unless just being empty bottles.

Yes, and now the bottles are useless for storing vitamins, spices, etc. Even for storing mothballs themselves, as we now see while water cannot pass through their walls, naphthalene can!

PE is famous for absorbing smells. In communist era, when original bottles for bicycles were not easy to get, my colleage and me often used some empty PE bottles. (PETS were not common yet at all).  After months, they got all various smells one cannot get rid of, so then had to be replaced.

@Poutnik. Please explain your experiment with plastic bottle in the communist regime. For exemple, what is  a "bottle for bicycles" ?  Were they in or not in PE ? Why did you replace them by PE bottles ? Were PE bottles easier to find ? Were they used for doing chemistry experiments ? Or for keeping beverages for a long time ? What sort of beverages ? Did you find differences in behavior if you compared PE bottles and the other bottles ?

@Maurice It is rare one simple comment brings so many questions.. :-)  // It was not an experiement, rather practical necessity due availability. I assume bottles placed in bicycle holders are, at least mostly, PE/PP, the replacements may be just of different qualities. I also assume all these bottles capture smells in long term, especially if not cleaned properly, being nonpolar and absorbing nonpolar/little polar molecules of smalls. PET as more polar plastic is far less vulnerable to catch smells.

BTW, polyethylene also absorbs some natural pigments well. Store tomato products in a polythene container and lycopene will permanently permeate into and stain the plastic. Some plant pigments are also quite stable: https://www.sciencedirect.com/science/article/abs/pii/0031018288900612

@DrMoishePippik Or carotens from carrots.


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How to produce AgI nanoplates?
I'm trying to understand the procedure used to produce AgI nanoplates described in the linked paper. Specifically, I have a few questions about the following text:

When silver acetate is in deficiency, all of silver acetate reacts with KI and only forms pure AgI nanoplates. ... However, when KI is in deficiency, i.e., silver acetate is in excess, the redundant Ag ions may be reduced by the component of EG to produce Ag NPs on the surface of AgI nanoplates (i.e., formation of Ag/AgI nanoplates).

...

The AgI and heterostructured Ag/AgI nanoplates were synthesized by using simple aqueous solution, as illustrated with the schematic images in Figure S1 (Supporting Information). For the synthesis of AgI nanoplates, 9 mL of EG was taken with the addition of 0.5 × 10−3 M of silver acetate and 160 mg of PVP (K29-32) in a 20 mL vial. The solution was stirred with a magnetic stirring bar at a rate of 400 rpm for 4 min. In the second stage 1 M of KI was dissolved in deionized water, and 150 µL was added and stirred at room temperature for 10 min until the formation of a homogeneous mixture. In the next step, the vial was put into an oil bath keeping at 140 °C for 90 min with magnetic bar at a rate of 400 rpm.  ... For the synthesis of heterostructured Ag/AgI nanoplates, the concentration of silver acetate was changed from 0.5 × 10−3 m to 2 × 10−3 m. The procedure and other conditions were kept same with the synthesis of AgI nanoplates.

My questions are:

If I'm understanding correctly, they are reacting around 5(20) micromoles of silver acetate with 150 micromoles of Potassium Iodide. But why is 150 micromoles of KI considered a deficit for 20 micromoles of silver acetate? It seems like equals moles of both should be balanced (CH3CO2Ag + KI -> AgI + CH3COOK). How can I compute the "balanced" amount if I am wrong?
Is the choice of solvent (Ethylene Glycol) and silver precursor (Silver Acetate) important for the production of AgI here? I understand that Ethylene Glycol can be a reducing agent, but is that only relevant for the production of AgI? Does the ethylene glycol do something else? I guess it provides the ability to heat it to 140C (vs water). Same on the silver precursor? Is there some significance to silver acetate (vs. for example, silver nitrate?)

Just in general, if you have any information of the production of silver halide nanoplates on the scale of 100 nm- several ums, let me know!
Thanks in advance!
Paper in question: M. U. Khan, H. J. You, X. T. Liu, L. L. Zhang, J. X. Fang, Small 2018, 14, 1702948. https://doi.org/10.1002/smll.201702948

Thanks for the reply. My confusion is that the article appears to produce AgI nanoplates with a 5/150 ratio of silver to iodide, but produces AgI/Ag composites with 20/150.
To me, both of these seem like an excess of iodide, and I would expect both to produce AgI nanoparticles. I am unsure of why 20 micromoles of silver acetate is seemingly considered an "excess" for 150 micromoles of KI.

Oops... my misreading. Do your calculations take into account the four-fold increase of Ag? "For the synthesis of heterostructured Ag/AgI nanoplates, the concentration of silver acetate was changed from 0.5 × 10−3 m to 2 × 10−3 m."

Yeah, my math is that (Volume solvent)(M)=9ml2 × 10−3 m ~ 20 micromoles for the silver excess case (vs 5 micromoles for the silver deficient case). Perhaps my understanding is off on that.


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Is Fajans Rule valid for polyatomic anions?
In my textbook its mentioned that BeCO3 (beryllium carbonate) is an ionic compound, but going according to fajans rules, small cation with high charge density and large anion should give a majorly covalent compound.

Relatively small, electronegative oxygen atoms in large anions are less polarizable than big single-atom anions like iodide.

Beryllium carbonate is unstable. Mismatch is size is too large, and carbonate have a general tendency to break down to CO2 and oxides.

For delocalized electrons, it does provide some insight. However, strictly speaking, Fajan rule is only applicable to atoms and ions. Hard-Soft Acid-Base theory is more widely applicable for polyatomic ions.


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The formula of ammonium phosphomolybdate
It is written in my qualitative chemistry textbook that phosphate could be tested by introducing phosphate, nitric acid and ammonium molybdate which gives (NH4)3H4[P(Mo2O7)6].
I found out in the qualitative chemistry book of professor Nordman Joseph that the yellow precipitate would be widely accepted as (NH4)3[P(Mo3O10)4].
*Question:

Is there any strong evidence for the existence of the precipitate whose formula is given in my textbook?

The yellow precipitate could be further reduced to blue complex when added reducing agent like SnCl2 or FeCl2. I cannot seem to find any chemical equation which could depict this statement.

Review also search results for site:stackexchange.com OR site:libretexts.org OR site:wikipedia.org phosphomolybdate formula

BTW, the structure shown at https://www.chemicalbook.com/ChemicalProductProperty_EN_CB6162083.htm is pretty fancy for an inorganic compound!

The simplest way of writing the formula of ammonia phosphomolybdate is the Brutto formula, $\ce{(NH4)3[PMo12O40]}$. The phosphomolybdate ion is a polyoxometalate but we tend to stick with $\ce{[PMo12O40]^3-}$ for simplicity. Sometimes hydrate/water of crystallization is appended to the formula or the individual components are written separately like below:

$\ce{(NH4)3PO4.12MoO3.6H2O}$
$\ce{(NH4)3[PO4(MoO3)12]}$

Older versions had wackier formula probably because they didn't realize the "complex" factor:

$\ce{12MoO3.(NH4)3PO4.2NHO3 +H2O}$
$\ce{(NH4)3PO4.10MoO3.3/2H2O}$
$\ce{(OH)2PO-O-[MoO2-O]11-MoO2(OH)}$

Regarding the reduction of ammonia phosphomolybdate with tin(II) chloride, unfortunately I couldn't find any single step reaction, possible because the product molybdenum blue is very loosely defined and thus there is no concrete formula. It can be presented with a formula $\ce{[(MoO3)154(H2O)70]^{y-3}}$.
References

Pharmaceutical Chemical Analysis: Methods for Identification and Limit by
Ole Pedersen, CRC Press, 2006
Ionic Equilibria in Analytical Chemistry by Jean-Louis Burgot
Springer Science & Business Media, 2012
A Dictionary of Chemical Solubilities: Inorganic by Arthur Messinger Comey, Macmillan and Company, 1896
Analytical Chemistry by Nikolaĭ Menshutkin, Macmillan and Company, 1895


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Is there a plane of symmetry in [Ma3b2c]?
We have been taught to look for plane of symmetries in inorganic complexes. If there is a plane of symmetry then the complex is optically inactive, otherwise it is optically active. In this particular complex, I am unable to find the plane of symmetry. I have tried creating mirror images in my head and they were superimposable.
This was actually a test question and the answer was that this is optically inactive
Can someone help me find the plane of symmetry here? Or is this method not applicable to all compounds?

Near duplicate: https://chemistry.stackexchange.com/questions/146929/why-are-complexes-of-type-ma3b3-not-optical-active. You don't have the threefold symmetry described there, but one of the symmetry planes still works.

There is indeed a plane of symmetry. The symmetry axis passes through the top-most 'a', 'M', and 'c' (as you correctly surmised it must). Have a look at this diagram to see where it goes:

For octahedral complexes, it sometimes helps to visualise the structure from a different angle. If you 'look' at the complex with 'b', 'c'. and 'b' facing towards you, all three 'a' atoms will face away from you, and the symmetry axis might be easier to see:

You can see how to re-orientate the molecule by following the atom numbering.
Finally, the best way to visualise sterics is in 3D. This isn't much help for an exam, but to convince yourself of the symmetry this can be a great tool. This visualisation is in Avogadro, which you can download and play with for free. This first image is in the same orientation as your question:

And here it is re-orientated:

Yes, there is an oblique plane of symmetry passing from between the two b ligands. Ergo, the compound is optically inactive.


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What materials can be used to store supercritical water
I am wondering which materials might be suitable to contain SCW (supercritical water). Here, we are assuming that the pressure is almost equal on the outside of the container than the inside. I have already considered plastics and some metals but these seem likely to corrode due to how reactive SCW is. Also, it needs to remain sufficiently strong at temperatures around 400°C but not much higher. I've tried to Google it but there doesn't seem to be much literature on the subject

What specific metals and plastics have you considered? How strong is strong enough?

Unlikely any organic compound could be used as a container. Even polytetrafluoroethylene (PTFE) has melting point of 327 °C. Polyimides can be extruded at 400 °C, so would be too soft.
Some materials for supercritical water include Inconel and Hastelloy, nickel-chromium steels.
BTW, if you need an observation port or a non-conductive feed-through, silica-based glasses won't do, since they dissolve in supercritical water. Perhaps sapphire ($\ce{Al2O3}$, corundum) might survive for a time, or not -- supercritical water can etch it under certain conditions.
You might research the Benson boiler.