text:
Activity of a chemical species and compound
While reading a book on Instrumental analysis by Skoog, in appendix part they explain activity of a species.

The relationship between activity of a species and it's molar concentration $[X]$ is given as $a_x=\gamma_x[X]$ where $\gamma_x$ is a dimensionless quantity called the activity coefficient of $X$.

Basically $\gamma_x$ tells about measure of effectiveness with which that species influence the equilibrium in which it is a participant.They say that in the expression of equilibrium constant, activity of the compound is used instead of its concentration.
In high school, we use the concentration of the compound and the activity of a compound is not introduced at that time. But it is also the fact that the concentration of compounds given at that time is very low for which $\gamma_x \to 1$. So $a_x \approx [X]$.
Basically, the question is that what is the need of the concept of activity for using in the expression of equilibrium constant? Without using this concept when the concentration of compounds is high (in this case $\gamma_x<1$ and $a_x<[X]$), what contradicting result we would get?
Also please tell me some reference  where I get the derivation of mean activity coefficient ($\gamma_{+-}$) of an electrolyte $A_mB_n$, given as $\gamma_{+-}=(\gamma^m_A \gamma^n_B)^\frac{1}{m+n}$.

A physical chemistry textbook like Atkins introduces how mean activity coefficients arise naturally when dealing with the thermodynamics of electrolytes.

Regarding your question on activity, the concept is introduced to deal with non-ideal behavior that might be encountered in non-ideal solutions such as non-dilute or electrolyte solutions. You can approximate activity with concentration in ideal dilute solutions.

Basically, the value of equilibrium constant for a weak electrolyte can be found by calculating the degree of dissociation which is derivable using Kohlrausch law($\alpha=$(Conductivity at concentration C)/(Conductivity at infinite dilution)) and then we can calculate K. But the value which we get does it is in accordance to activity of the electrolyte not the concentration (when high concentration of electrolyte is taken because then $a_x\neq [X]$), so the concept of activity is introduced?

This paragraph might have confused some people: "Basically, the question is that what is the need of the concept of activity for using in the expression of equilibrium constant, without using this concept when high concentration of compounds are taken (in this case γx<1 and ax<[X]), what contradicting result we would get?" I think I understood it after rereading, but you may want to edit it to make it clearer (perhaps by breaking it up into two sentences?)

You may want to try a small change: "Basically, the question is that what is the need of the concept of activity for using in the expression of equilibrium constant? Without using this concept when the concentration of compounds is high (in this case γx<1 and ax<[X]), what contradicting result we would get?"

OK. I have made the edit. Please answer the question.


text:
SMILES string convertors
I'm new to SMILES strings, and have noticed that they're not standardised.
Firstly, what is the reason for different structures of SMILES strings to represent the same compound?  Why not have a standardised format?
Secondly, are there any tools to convert between different SMILES strings for a given molecule?

https://chemistry.stackexchange.com/questions/34563/pubchem-inchi-smiles-and-uniqueness

SMILES doesn't define a particular order in which the atoms show up in the string. Hence the position in the molecule where software starts printing will determine the string. E.g. for CO2:

Ketcher prints C(=O)=O (starts with C and considers one of the O's a branch)
MarvinJS prints O=C=O (starts iteration with O, hence no branching)

Both of these representations are valid, they simply start with different elements. Each tool would recognize SMILES string generated by another tool, so there's no need to convert between strings created by different software.
In order to make different tools generate the same string a so called Canonical SMILES was developed. Not every tool can generate it, but if they do - they'll spit out the same string (in theory).

Thanks Stanislav, that's really helpful. Useful to understand the different structures, and that Canonical SMILES exists.
I'm interested in a converter for a different reason (expanding a machine learning dataset, where it's valid to have more than one SMILES representation for a specific molecule), but I'm guessing as different software recognises strings generated by another tool, there's no demand for such a convertor and one therefore doesn't exist?

@GEOGrey, yeah, there's no need for such converter. For the dataset you better clean it first before feeding it to the algorithm. So all the SMILES that you get you should convert to the same form (regardless if it's Canonical). If you use some specific library (CDK, OpenEye) they'll probably generate the same SMILES string for the same molecule.

Rdkit, which is widely used in cheminformatics, can generate a list of valid smiles strings for a given molecule. That sounds like exactly what you are looking for. More info found here:
https://iwatobipen.wordpress.com/2018/11/15/generate-possible-list-of-smlies-with-rdkit-rdkit/

source: Stack Exchange

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

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text:
How do I analyze 17 different gases from a sample?
I am designing a test setup in which I heat a battery until it releases its gases, and I need to analyze the composition of the fumes. Below is the list of gases that are believed to be present in the battery's fumes. I was planning on just using a bunch of gas detection tubes, but they don't seem to be available for all these gases. I was thinking about moving on to buying a gas chromatograph instead since, so far, it seems cheaper than getting a bunch of tests, but I have never used a GC, so I don't know if this would even work for so many gases.
If I were to go the GC route, what would I need to learn and buy in order to test for these gases? Is there a handbook that demonstrates the proper setups for different gases since I think you need to change some physical components for proper testing of different gases (I didn't find any)? Or is there a better method to test for this?

Well, this should be easy for GC-MS, but unless you'd use the hardware for many years, you should "outsource" the tests.

Yeah, that was my thought too. My boss wanted me to figure out if it's possible for us to do the GC-MS ourselves so he can sell this service himself. I work at an aerospace flammability testing lab, so we burn lots of stuff, and we could get more business if we also offer to analyze the gases off other samples we burn up.

GC without MS is sufficient to recognize the presence of the mentioned gases. GC-MS is better. But GC alone is sufficient, if you have samples of the pure gases to determine their retention times.

I think I would have highly pure gases since I am planning on conducting this test in a vacuum chamber. Would GC be enough to tell me the composition of the gas, ie 20% of one thing, 10% of another, etc? That is the purpose of the spike heights in the data, right?

I read a bit more about the differences between GC and GCMS. I have a lot of molecules that only vary by an atom; would this make it too difficult for the GC by itself to analyze the gas composition? Or since the molecules are simple and the variation in weight is a large percentage of the total weight, the GC doesn't struggle with finding the difference between CO and CO2 or C3H6O3 and C4H6O3?

GC can sometimes distinguish deuterated vs. non. deuterated compound. Your analysis must be a very simple exercise for GC people. The key question is choosing the right column.

Sounds good. Where can I find sources on choosing the right column and sampling techniques and all other things that would be important with using the equipment properly?

This is, I´m sure, a solution you can buy, with all equipment and a manual. Don´t reinvent a wheel unless you already know very well what you are doing.

@AlbertGarcia, Search the literature. Use Google Scholar. Also contact Supelco and other GC column manufacturers. They will tell you the right column.

You could use a permanent gas analyzer that has an FID channel for hydrocarbons and CO2, with a TCD detector for H2 and HF and a third MS channel if needed. Basically, this would mean you split your sample into three different columns in the oven to analyze them for different things. I agree with Karl and don't reinvent the wheel. I've done the hydrocarbons (C1-12) and CO2 and permanent gases using a Thermo Trace 1310 with an FID and TCD detector and would recommend that for ease of use. Then I would probably add an MS detector to it in case you need it.
How are you hoping to get the battery gases into the GC? I think you could fit a battery into a headspace vial and heat it up to release the gases then directly inject the headspace into the GC.

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


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On drug-like criteria and QM9 dataset
According to the book Machine Learning Meets Quantum Physics

QM9 is built up from $\sim 134 \mathrm{k}$ drug-like molecules with up
to nine heavy atoms (C, N, O, or F).

I'm assuming that drug-like molecule is referring to Lipinski's rule of five. I used RDKit to plot 2D histogram of octanol-water partition coefficients (LogP) based on molecular weight (MolWt) and Polar Surface Area (PSA) based on molecular weight (MolWt)

These plots show that all molecules in the QM9 are below the suggested molecular weight interval of 180 to 480 and half the molecules have LogP below the -0.4 to 5.1 interval suggested by Ro5. I was wondering why they are called drug-like molecules? I'm using the following code to compute the properties:
from rdkit.Chem import Descriptors, Crippen

Descriptors.TPSA(mol)
Descriptors.MolWt(mol)
Crippen.MolLogP(mol)

Both the criteria you listed seem to under the variants of rule. The initial rules just say below 500 dalton and below 5 partition coefficient.

You are right. Seems my mind was biased by the numerous references that gave a lower bound of 200 for MW and looked over the earlier section. Now it makes sense why every paper that I read gave a slightly different definition.


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Identifying unknown compound using mass spectrum
I am currently studying the textbook Mass Spectrometry - A Textbook, third edition, by Jürgen Gross. In chapter  1.4.5 Statistical Nature of Mass Spectra, there's a following example:

It is important to be aware that a single molecule can only yield one ion of one $m/z$ value. This ion may either reflect the intact molecule or any of its fragment ions. Statistics on thousands of ion formations and dissociations are needed to generate a useful mass spectrum exhibiting signals at different $m/z$ where each of them can be assigned a relevant relative intensity. To understand this, simply imagine a single methane molecule that gets ionized and detected as a molecular ion: this would lead to a spectrum showing a single line at $m/z$ $16$, the intensity of which would have no meaning beyond ‘yes’ and ‘no’ (Fig. 1.7a). Alternatively, the molecular ion might fragment to yield a single $\ce{CH3+}$ ion at $m/z$ $15$; again, the spectrum would show only one line. In fact, such a spectrum could not tell us whether the peak is caused by a molecular or a fragment ion. Eight ions might lead to a spectrum where each ion corresponds to $33.3 \%$ relative intensity (as in Fig. 1.7c) even though other distributions would be possible. Eleven ions could scale the intensity in $20\%$ steps, while 23 ions could lead to a spectrum with $10\%$ steps. It is evident that a spectrum with intensity levels as accurate as $0.1\%$ has to be based on thousands of ions (Fig. 1.7f).

Reading this, I do not understand how the author used this mass spectrum to infer that the molecule in question is $\ce{CH3+}$. The author seems to imply that, given this mass spectrum ($m/z$ values of $16$, $15$, etc.), we are able to infer that the molecule is $\ce{CH3+}$, but he doesn't really explain how this is done.
I would greatly appreciate it if people would please take the time to explain how the author identified $\ce{CH3+}$.

So what atom is the mathematical difference between 44 and 28? Then identify 28.

It seems you have trouble to fit our Q&A model. Usually 2 questions is one too many. In this case, if you don't get the spectrum of methane, there's IMO like no chance to solve this monster you mentioned earlier.

@user55119 I've cut it out, for more focus.

@Mithoron Yes, I assume that if I understand the methane case, then I'll be able to understand the other case.

@user55119 Well, we can see that the difference is 16, but what can we infer from this? That there is an oxygen molecule present (since it has atomic mass of 16)?

If it is oxygen, then what is 28?

@user55119 I'm not sure. Maybe we next consider that 28 - 22 = 6, but then what is 6? This just seems like guesswork. Surely analyzing mass spectra is not done in this way? It seems very "hand-wavy".

Focus on the 4 strongest peaks, 12, 16, 28 and 44. Peak 22 (m/2z) is a doubly charged 44 ( m/z). Look here: https://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Mask=200#Mass-Spec

@user55119 Ahh, I understand now! The lowest peaks are 12 (carbon) and 16 (oxygen), so 28 = 12 + 16; and 28 + 16 (oxygen) = 44, so we have $\ce{CO2}$.

Yup! The 28 peak is CO. The peak 29 is C(13)O. Peak 45 is O=C(13)=O and peak 46 is O=C=O(18). Natural abundance of C13 = 1.1%; O18 = 0.2%.

@user55119 Ok, thanks for taking the time to clarify this.


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How does copper rich "weld thru" primer protect iron / steel?
In the automotive industry, to protect steel which cannot be painted (e.g. the gap or seam between two sheet panels), a product known as weld thru primer is sometimes used.
This is a coating of zinc-rich or copper-rich material which is meant to prevent rusting of the steel, especially at the weld points.
From the reactivity table below, zinc is more reactive than iron, so I can understand how this would protect the iron via sacrificial corrosion. Similar to galvanized steel.
However, copper is less reactive than steel. When salt water get's in between the sheet metal seams, would this not corrode the steel? How would applying a layer of copper to the steel protect it?

Well there are also many different kinds of steels with different galvanic characteristics are well. For example, all of these metals have a higher reduction potential than copper:
1. Red Brass
2. Stainless steel 347 (active)
3. Molybdenum, Commercial pure
4. Copper-nickel 715
5. Admiralty brass
6. Stainless steel 202 (active)
7. Bronze, Phosphor 534 (B-1)
8. Monel 400
9. Stainless steel 201 (active)
10. Carpenter 20 (active)
11. Stainless steel 321 (active)
12. Stainless steel 316 (active)
13. Stainless steel 309 (active)
14. Stainless steel 17-7PH (passive)
15. Silicone Bronze 655
16. Stainless steel 304 (passive)
17. Stainless steel 301 (passive)
18. Stainless steel 321 (passive)
19. Stainless steel 201 (passive)
20. Stainless steel 286 (passive)
21. Stainless steel 316L (passive)
22. AM355 (active)
23. Stainless steel 202 (passive)
24. Carpenter 20 (passive)
25. AM355 (passive)
26. A286 (passive)
27. Titanium 5A1, 2.5 Sn
28. Titanium 13V, 11Cr, 3Al (annealed)
29. Titanium 6Al, 4V (solution treated and aged)
30. Titanium 6Al, 4V (anneal)
31. Titanium 8Mn
32. Titanium 13V, 11Cr 3Al (solution heat treated and aged)
33. Titanium 75A
34. AM350 (passive)
35. Silver
36. Gold

(source: Corrosion Doctors)
So, as long as you're welding any of those under copper's boiling point of 2,562°C (or 4,644°F, at which point it could just float away from your intended target), then it should be fine for your application.


text:
Why is the water solubility of CO worse than that of CO2?
I tried my best to ask SciFinder, Google, and Google Scholar the wrong questions ...
I found many paper on the fitting of the solubility to some equations to be able to calculate the amount of $\ce{CO}$ in water, but I could hardly find any explanation on the low solubility. Only one paper [1] dealt with it.
Why is $\ce{CO}$ way less soluble in water than $\ce{CO2}$, despite, e.g., the first having a dipole moment and the latter doesn't? What am I missing?

[1] H. Sato, N. Matubayasi, M. Nakahara, F. Hirata, "Which carbon oxide is more soluble? Ab initio study on carbon monoxide and dioxide in aqueous solution," Chemical Physics Letters 2000, 323(3-4), 257–262 (https://doi.org/10.1016/S0009-2614(00)00508-X).

Is this gonna be both Q&A?

related https://chemistry.stackexchange.com/questions/35603/why-is-co-practically-nonpolar https://chemistry.stackexchange.com/questions/35603/why-is-co-practically-nonpolar

Well, CO2 is a great deal more polar.

Because if it's serious then your comparing it with wrong compound - comparison between CO and N2 would be interesting, but CO2 reacts with water, which changes everything. BTW @IvanNeretin don't even start that again...

@Mithoron I will if I have to.

I was going to comment on the polarity discussion here, but there's clearly some cross-talk I'm missing... :/

Polar does not mean anything. It is among the most ambiguous words used in chemistry.

@M.Farooq IMHO, "stable" is much worse. It is not used here, though.

@M.Farooq Do not confuse lack of meaning with multiple meanings :-) It's also a relative location on planet earth or an adjective, or the name of various commercial brands.

Please note that OP is correct as referring to the dipole moments and not to a more ambiguous (for some) polarity. The linked article find its justification in dealing with ab initio methods rather than common sense. When we speak about CO2 solubility I do not think we intend "free" CO2. Perhaps the article does but it is not what common intuition means. That is the answer is probably that by @Oscar Lanzi integrated by Mithoron comment.

@Alchimista, I was referring to comments. OP is using correct wordings.

@Mithoron unfortunately it is not Q&A

But it could be - discuss reaction with water and relevant equilibria, and it would be done.

Most CO2 dissolved in water is actually CO2, not reacted to form carbonic acid. So a reaction-based explanation misses.  For an alternative hypothesis see the answers.

The fact that $\ce{CO2}$ does not have a dipole moment does not really mean it's nonpolar.  What $\ce{CO2}$ has is a quadrupole, two opposing dipoles that occupy different portions of space (opposite sides of the carbon atom).
Ordinarily we expect an electrostatic interaction with a quadrupole to be relatively weak, but in $\ce{CO2}$ the component dipoles are strong and at close range the water molecules can "home in" on one end of the molecule and interact selectively with the corresponding component dipole.  Thereby the electrostatic interaction between water and the $\ce{CO2}$ quadrupole is stronger than the interaction between water and the weak $\ce{CO}$ dipole.

Related: https://chemistry.stackexchange.com/questions/54115/quadrupole-moment-of-a-molecule

Yes carbon dioxide has a weak microwave rotational spectrum as well.


text:
Is it possible to separate Lead(II) sulfate from sulfuric acid?
To prepare lead(II) sulfate $\ce{PbSO4}$ from lead(II) carbonate $\ce{PbCO3}$, can I add excess sulfuric acid to lead(II) carbonate so that the products formed will be excess acid with lead(II) sulfate salts? Is it possible to separate lead(II) sulfate from the acid to obtain the pure salts?

https://en.wikipedia.org/wiki/Lead(II)_sulfate tells all you need.

This might help: https://www.quora.com/Why-is-sulfuric-acid-not-used-to-prepare-lead-sulphate

The reaction will lead to formation of lead sulfate but what happens is that the lead sulfate formed gets deposited over the lead carbonate and thus it gets disconnected form sulfuric acid and the reaction stops. It happens regardless of the concentration and amount of sulfuric acid used. So, what is done is that lead carbonate is converted to lead nitrate using nitric acid and then it is converted to lead sulfate using sulfuric acid.

Lead sulfate is not soluble in moderately concentrate or dilute sulfuric acid. So a simple filtration will allow the separation lead sulfate - sulfuric acid.


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What is the name of this compound?
This doesn't look like an amide. What is its IUPAC name? Is this compound even possible?

Without hydrogen at nitrogen, as you draw, uncharged – likely impossible.

Is it a homework? Just asking.

No .I asked it to one of my teachers, he said it is impossible to solve! I made up this question.

Just assume that hydrogen to be there.

You likely mean $\ce{CH3-CO-NH-CO-CH2CH3}$.
$\ce{R-CO-NH-CO-R}$ compounds are called imides (in contrast to amides $\ce{R-CO-NH2}$, like in inorganic chemistry, $\ce{M^INH2}$ vs $\ce{M_{2}^{I}NH}$). They can also be „intramolecular“, i.e. cyclic (look at succinimide or phthalimide).
A similar symmetric compound $\ce{(CH3CO)2NH}$ „diacetamide“ exists and can be bought (melting point 79 °C). Its current preferred name is N-acetylacetamide.
Your mixed imide of acetic and propionic acid is currently preferably named N-acetylpropanamide.

Just asking , what about something like (CH3CO)2N.

I mean (CH3CO)3N.

@compenthusiast Regarding the name, you should be able to solve it, the is analogous, and, according to the current IUPAC recommendations "asymmetric" as well (unfortunately, IMHO).

http://www.molbase.com/cas/641-06-5.html


text:
Stability of cyclohexane type species
I came across a question where we have to find out the stability of species that are in chair conformations. The species are as follows:

Comparing the 'A' values tells me that compound I, III and VI should be most stable, but the answer given is: I,IV and V.
Can anybody tell me where did I go wrong?

You can't use A-values directly except for the first row. The others have oxygen atoms in the ring. The oxygen atoms lack flaghole hydrogens, so the steric repulsion is less. Off the top of my head, I don't remember ever seeing numbers for this, so I don't have intuition on how much it changes the A-value. However, for V/VI, I still think VI. And the repulsion between the t-Bu and the lone pairs in IV might still strongly prefer III, but I'm just guessing.

Why is V according to you is correct?

Sorry, I meant to write VI!

Wow, there were a lot of typos in that original comment... :(

Isn't the case same for III and IV? Because tert butyl is in equatorial position in III, so less steric repulsion will be in this case. Isn't that?

Yeah, I meant to type III... :(

@Zhe: I think H-bonding with ring O makes V more stable than VI.

But H-bonding is also possible in VI and it will even be more favourable energetically due to equatorial substituents. Isn't it @Mathew?

@Sanu_012: No, VI can't make H-bonding. I'm talking about 1,3-O atoms in the ring. It is like 1,3--axial-H interactions. Here it is lone pair of O instead. Imagine the picture. Of course, these are solvent dependent.

And what about III and IV? Which one do you think is favourable energetically? @Mathew

And sorry @Mathew, this tiny head of mine couln't visualize that picture you are talking about. It would be great if you write it as an answer with essential diagrams.

@Sanu_012: My opinion, III is more stable than IV. I'd answer the question, but you have to wait for a while because, here, now 2:00 am, so I gotto have a sleep. :-)

Okay sure. Good night :) But I also think III is more stable.

Maybe you can consider anomeric effect in V-VI.(Consider the dipole moment reason)

@B.Anshuman Can you please elaborate on this?

Anomeric effect Acts on such structures like VI (but anomeric effect is explained only if hydroxy group is adjacent to any of the ethers). Then the axial position of -OH group becomes more favourable. This is explained in L G Wade due to two factors: Hyperconjugation of oxygen in the ring to the ABMO of hydroxy group, and the net dipole moment of both -OH and -O- being in the same direction. As the compound looked similar, I thought to mention this effect.

@mathewmahindaratne Good call on hydrogen bonding. I missed that.

According to the Ref.1, 5-hydroxy-1,3-dioxane (the compound of structures V and VI) both in the gaseous state and in dilute $\ce{CCl4}$ solution exists as a chair conformer, the hydroxy group in an axial position with an intramolecular hydrogen bond of the $\ce{O—H⋯O}$ type:

Accordingly, the axial conformer (V) is more stable by $\Delta G^\circ = \pu{-1.2 kcal mol−1}$). Microwave spectroscopy and the $^3J_\ce{H(5eq),OH}$ coupling constant suggest that the $\ce{OH}$ group lies in the plane of symmetry $(\ce{C_{(2)}-C_{(5)}-O}\text{-plane})$ and is a part of a bifurcated hydrogen bond to the two ring oxygen atoms (Ref.2). Therefore, it is safe to say the conformer V is more stable than conformer VI between two possible conformers of 5-hydroxy-1,3-dioxane (marked with $\color{green}{\text{green}}$ oval).
It is well known that the most stable conformer of tert-butylcyclohexane is tert-butyl in equatorial position. In a 1,4-substituted version of 1-tert-butyl-4-methylcyclohexane, since methyl group is so small compared to tert-butyl group that I is almost exclusive in equilibrium (marked with $\color{green}{\text{green}}$ oval).
However, I wasn't so sure about stability between III and IV. There is clearly no H-bonding opportunities except theoretical chemist may argue hyper conjugation with ring oxygen by MO calculations. I leave it open for any computational chemist to prove that possibility, yet my best guess is III more stable between them (marked with $\color{orange}{\text{orange}}$ oval with question mark).
References:

J. C. Jochims, Y. Kobayashi, “Bifurcated hydrogen bonds in z-2-phenyl-1,3-dioxan-5-ols,” Tetrahedron Letters 1976, 17(24), 2065-2068 (https://doi.org/10.1016/S0040-4039(00)93819-X).
Jose L. Alonso, E. Bright Wilson, “Study of an intramolecular, bifurcated hydrogen bond in 1,3-dioxan-5-ol by microwave spectroscopy,” J. Am. Chem. Soc. 1980, 102(4), 1248–1251 (https://doi.org/10.1021/ja00524a005).

Okay, this is understandable. But in para 2, I think you meant to write VI instead of IV. Isn't it?

@Sanu_012: Good catch! Yes, I meant VI. I corrected it.:-)

That means, according to you the answer should be I, III and V and I can assume that the answer I, IV and V provided in the the answer key is wrong? Isn't it?

@Sanu_012: To my knowledge, yes. But, they might have a reason. Let's wait and see how others have responded.

Okay. Let's wait some days :)


text:
Theoretical plates in Plate theory of chromatography
In "Chromatography theory" by Scott and Cazes,
The plate volume is given as ($v_m+Kv_s$) where $v_m$ and $v_s$ are the volume of stationary and mobile phase in the column and $K$ is the distribution coefficient (basically the equilibrium constant of the solute concentration in stationary and mobile phase).
They give the definition of plate volume as

The plate volume is defined as that volume of mobile phase that can contain all the solute in the plate at the equilibrium concentration of the solute in the mobile phase.

The plate concept is introduced to explain the equilibrium attained by the solute in distributing between stationary and mobile phase. We can't assume this equilibrium in the whole column as mobile phase continuously flows during the process and disturbs the equilibrium, so column is considered to be made of number of theoretical plates in which equilibrium is attained for some finite time.
Basically, I am not able to understand the definition of plate volume. Because as the name suggests one can interpret as the column to be made of number of plates which is made of the volume of mobile phase and stationary phase in which the solute is distributed between the two phases. So, I initially misinterpreted the formula of plate volume as $v_m+v_s$.
Please explain the definition of the plate volume. It seems somewhat confusing.

I shall take the example of the mixture ethanol-water, first without chromatography. If I start distilling 250 mL of a mixture 10% ethanol and 90% water, the solution boils at 96°C, and the vapor contains 55% water + 45%  ethanol. This vapor is condensed by passing through the cold condenser. So the first fraction of maybe 10 mL condensed vapor contains 45% ethanol. Suppose 250 mL of this first mixture (45% ethanol) is heated in another distilling apparatus. At maybe 85°C, the liquid boils and the first vapor will contain 70% ethanol and 25% water. If this vapor is condensed, the liquid so obtained contains 70% ethanol. These distillation - condensation operations can be repeated, with the following results.
1st distillation. Original liquid : 10% ethanol. Distilled liquid : 45% ethanol
2nd distillation. Original liquid : 45% ethanol. Distilled liquid : 70% ethanol
3rd distillation. Original liquid : 70% ethanol. Distilled liquid : 83% ethanol
4th distillation. Original liquid : 83% ethanol. Distilled Liquid : 90% ethanol.
etc.
nth distillation. Original liquid : 96% ethanol. Distilled liquid : 96% ethanol.
Now suppose all these distillations are carried out in the same apparatus. Instead of having the vapor go through a cold tube and be recovered in a flask out of the condenser, imagine that the vapor is sent into a long vertical tubes situated just above the first container half filled with the mixture 10% ethanol. Suppose this vertical tube contains a lot of successive floors or plates, punched with one or two small holes. In the beginning the vapor made of 45% ethanol will pass through the holes of the first floor or plate, and will be condensed just above this level and may partly go down to the bottom of the setup. But a fraction of this liquid will now boil again and produce a vapor made of 70% ethanol. This vapor passes through the second floor and be condensed just above the second floor. On this second floor the liquid (containing 70% ethanol) will later on be evaporated and produce a vapor containing 83% ethanol. And so on. After n floors the vapor and the liquid will contain 96% ethanol. As a result, the separation ethanol-water improves if the number of floors or plates increases. And you may imagine that the height of the room between two floors is critical. If it is too small, it will be immediately filled. If it is too big, some secondary condensation will happen on the wall of each room.
Now ethanol is not the best example for my purpose because the successive distillations cannot get to pure ethanol. Suppose for the rest of my description that after n distillations, a pure amount of a solvant X can be obtained. Suppose that to get a better that 99% separation you need 8 floors, which are all 5 mm high.
Suppose now that instead of having used a distillation apparatus, you are using of chromatographic column. In order to get a better than 99% separation of pure compounds you need a column measuring 60 cm. The plate volume is : 60 cm / 8 plates = 7.5 cm. Of course, if you use a longer column, the separation might be 99.9% or better.
Of course this explanation is an over-simplification of the reality. But it may help you understand what is a plate volume.
Did you follow me ?

Thanks for the explanation. Basically you use the analogy of  fractional distillation. This helps a lot. But one thing which I am not able to understand is that if we consider 1 theoretical plate of the chromatographic column then it has stationary and mobile phase with solute distributed between the two phases, so shouldn't the plate volume be $v_m+v_s$ instead of $v_m+Kv_s$. In derivation of the plate theory, there comes an expression of $v_m+Kv_s$ and they defined it to be plate volume. But I am not able to link this expression to plate volume?

Sorry ! I don't know this theory.


text:
Why there is no position of H and He in Lothar Meyer's curve?
I see Alkali metals occupy the peak points. $\ce{H}$ is not an Alkali metal. There's no $\ce{H}$ in the graph. So no point before $\ce{Li}$.
Inert gases occupy place on the ascending portion of graph.
As there is no graphical line before $\ce{Li}$, where can $\ce{He}$ be put? There's no place for $\ce{He}$.
But why there's no plot for $\ce{H}$?

There are two options. Maybe the plot was drawn by the hand of a genius many years ago, and can't be reproduced. But then again, maybe the graph represents actual physical quantities that may be looked up for all elements. Well... which will it be?

The question seems a bit ill defined. Meyer was working during the time that the modern periodic table was created. Chemists were still fumbling around trying to find the overall pattern for the elements. Lothar Meyer seem to have published several curves of atomic volume versus atomic mass. Are you referring to one of his such graphs in particular, or a like graph that can be constructed with modern data?

A German chemist, Julius Lothar Meyer was contemporary and competitor of Russian chemist Dmitri Mendeleev to draw up the first periodic table of chemical elements. In 1864, he published the first version of his periodic table through his book Die modernen Theorien der Chemie. This abbreviated version of a periodic table consisted of 28 elements, about half of the known elements, arranged from the left to right classifying by their properties (mainly their valences) as shown below:
$$
\bf{\text{Lothar Meyer's 1864 table}}\\
\begin{array}{|c|c|c|c|}
\hline
\text{Valance}  & 4 & 3 & 2 & 1 & 1 & 2  \\
\hline
\text{Element} & - & - & - & - & \ce{Li} = 7.03 & \ce{Be} = 9.3   \\
\hline
\text{ Element} & \ce{C} = 12.0  & \ce{N} = 14.04  & \ce{O} = 16.00  & \ce{F} = 19.0  & \ce{Na} = 23.05 & \ce{Mg} = 24.0   \\
\hline
\text{ Element} & \ce{Si} = 28.5  & \ce{P} = 31.0  & \ce{S} = 32.07  & \ce{Cl} = 35.46  & \ce{K} = 39.13 & \ce{Ca} = 40.0   \\
\hline
\text{ Element} & -  & \ce{As} = 75.0  & \ce{Se} = 78.8  & \ce{Br} = 79.97  & \ce{Ru} = 85.4 & \ce{Sr} = 87.6   \\
\hline
\text{ Element} & \ce{Sn} = 117.6  & \ce{Sb} = 120.6  & \ce{Te} = 128.3  & \ce{I} = 126.8  & \ce{Cs} = 133.0 & \ce{Ba} = 137.1   \\
\hline
\text{ Element} & \ce{Pb} = 207.0  & \ce{Bi} = 208.0  & -  & -  & \ce{Ti} = 204.0 & -   \\
\hline
\end{array}
$$
It is evident from this periodic table that Meyer had arranged the elements not only with their increasing atomic weights, but also arranged them according to their valencies. For example, the first column contain the elements $\ce{C, Si, Sn}$, and $\ce{Pb}$, all of which form tetrahydrides ($\ce{CH4, SiH4, SnH4}$, and $\ce{PbH4}$) when combined with hydrogen. Thus, their valency is 4. Similarly, the elements of the second column, $\ce{N, O, F, Na}$, and $\ce{Mg}$, since their valency is 3, would combine with hydrogen to form corresponding trihydrides such as $\ce{NH3}$ and $\ce{PH3}$, and so on. Meyer's grouping of elements based on their valency, and this type of grouping is based on the chemical properties of the elements.
Later, Meyer has plotted the graph of the atomic volume (molar volume) vs atomic weight of elements, which is called Lothar Meyer's  Curve:

He has not plotted hydrogen on the curve probably because it is considered to be a specific element, which is able to react with most of elements to form hydrides (see above). On the other hand, in the case of helium, it wasn't discovered at the time of Meyer's (technically speaking). The existence of helium was first discovered in 1864, when a French astronomer, Jules Janssen has detected a bright yellow line with a wavelength of $\pu{587.49 nm}$ in the spectrum of the chromosphere of the Sun during the solar eclipse. Around 1870, when Meyer plotted his curve, $\ce{He}$ wasn't still discovered on the Earth (the year of that discovery was not until 1895). Also, the discoveries of all noble gasses have been done after Meyer's time and none of them were in his curve.
However, the supposed places of two elements in question, $\ce{H}$  and $\ce{He}$, have been plotted by me in the original Lothar Meyer's curve for convenience (see scatted yellow lines). As evident, there is no maximum (with $\ce{H}$) or minimum as a series of maxima and minima shown by other elements (alkali metals at the peaks). The absence of all noble gasses in the curve is justified by Mendeleev's Periodic Table, first published in 1869:


text:
Moles of chlorine gas needed to make up an equivalent amount of saturated salt solution
I’m trying to solve a question about a saturated salt solution but am unsure of how to answer one of the parts or what it’s asking for.
The Problem
The maximum solubility of table salt (NaCl) in water at 25 degrees Celsius is approximately 359 g/L. Assume the volume of the solution does not change with the addition of salt.
a) What is the molar concentration of the mixture in mol/L?
b) What is the molal concentration of the mixture in mol/kg? Assume that the density of water is 980 kg/m3.
c) How many moles of Cl2 gas would need to be present in order to make up an equivalent amount of saturated salt solution?
d) What is the density of the mixture?
My Attempt
a) molar concentration
359 g NaCl * 1 mol NaCl / 58.44 g NaCl = 6.14 mol => 6.14 mol/L
b) molal concentration
(1 L H2O * 1 m3 H2O * 980 kg) / (1000 L H2O * 1 m3) = 6.14 mol / 0.980 kg = 6.26 mol/kg
c) moles of Cl2 gas
I thought it was asking how many moles of chlorine gas is equivalent to the amount of salt in the 1 L solution (359 g NaCl). There are 70.9 g in 1 mole of Cl2, so that gives me an answer of 5.06 mol Cl2 gas (359 g NaCl / 70.9 g Cl2).
But the answer key states the correct solution as 3.07 mol Cl2 gas. How should I go about this part?
d) density of the mixture
density of water = 980 kg/m3
density of salt = 359 g/L => 359 kg/m3
density of mixture = (980 + 359) kg/m3 = 1339 kg/m3

a) molar concentration
There are indeed 6.14 moles of $\ce{NaCl}$ in 359 grams of $\ce{NaCl}$.
I'd prefer to write the molar mass as "$\pu{58.44 g mol^-1 }\ \ce{NaCl}$" or "$\pu{58.44 g/mol}\ \ce{NaCl}$" to keep it all together.
b) molal concentration
You got the right value, but you did it the wrong way. For the solvent:
$$\dfrac{\pu{980 kg/m^3}}{\pu{1000 L/m^3}} = \pu{0.980 kg/L}$$
c) moles of $\ce{Cl2}$ gas
The book is correct. There are 6.14 moles of $\ce{NaCl}$. Each molecule of chlorine gas contains two atoms of chlorine, so you need half as many moles to get the same number of atoms of Cl. So 6.14/2 = 3.07
This part of the question has poor phrasing. You can't dissolve 3.07 moles of $\ce{Cl2}$ in water and get a solution of $\ce{NaCl}$.
d) density of the mixture
Your answer is correct.


text:
By what mechanism is ammonia liberated in the Kjeldahl Method of protein determination?
I was looking at the Kjeldahl digestion of proteins, and I very much do not understand the mechanism by which it works. The reaction scheme is as follows:
$$\ce{Protein-N + H_2SO_4 ->[K_2SO_4, CuSO_4][380^0C] (NH_4)_2SO_4 + CO_2 + H_2O}$$
I read that the sulphate salts are used in order to raise the boiling point of the sulphuric acid (I'm not particularly concerned with how this works). What confuses me is that the reagents used are anhydrous, so to me this takes acid-catalysed hydrolysis off of the table.
Even if the amide linkage was cleaved, the only way I could think of by which ammonia could be released is an acid-promoted E1 elimination. The amine at a secondary carbon centre would be protonated so is a possible leaving group under the harsh conditions, and the absence of a nucleophile (since water is driven off by evaporation) but presence of sulphate ions lead me to theorise an E1 style elimination of ammonia. This would then be protonated in the acidic conditions to form the salt. The intense heating also lead me to believe elimination would be favoured:

This looks like a rather unconventional mechanism but I assumed the high temperature would allow for something like this to occur. I suspect this mechanism is going in the complete wrong direction considering carbon dioxide is not produced in it, but I'm unable to think of or find anything else.

Well I doubt it's really known, but it sure doesn't look like that.

Kjeldahl digestion includes heating to 400 celsium.  Oxidation of peptide isn't really all that out place at those conditions.


text:
Dependence of Heat Capacity on Temperature
I was studying about the graph of dependence of heat capacity on temperature.

As you can see in this picture, heat capacity with constant volume increases near 0K.
As the temperature keeps rising, it converges to 3R.
I was finding out why these changes happen. In the textbook, the author explained like this.
For the increasing part,

This corresponds to an increased ability of the lattice waves to enhance their average energy with increasing temperature.

And, for the converging part,

The quantity of energy required to produce a 1-degree temperature change is constant.

However, I could not understand about the reason of converging part.
I think the energy required to produce a 1-degree temperature change can be constant, but how this can be only applied only for converging part?

That's the definition of converging: it approaches and then reaches (matches) a determined value. To understand this you need to study statistical mechanics, which provides theories that explain the changes in $C_V$ with temperature.

The heat capacity is the slope of internal energy with temperature ($C_V=\partial U/\partial T|_V$ ) so as more energy is added by increasing temperature the rate at which this increases eventually becomes constant  (as energy levels now v numerous) and hence the $C_V$ is constant. Similarly at v low $T$ v few energy levels can be populated and $U$ is almost constant, as they become populated $U$ increases and so does $C_V$.


text:
Effect of steric effect on acidic and basic strength
I found a sentence given in my textbook:

More is steric effect more is acidic strength of carboxylic acids and
more is steric effect less is basic strength of aniline and other
amines.

Could someone please explain with an example?
I guess it might seem I am asking about ortho effect but I am not. I have gone through the post suggested but I am basically asking about subcases of ortho effect as in if we are asked to compare between ortho tertbutyl benzoic acid and ortho methyl benzoic acid then how would we go about it?  Or suppose the questions asks comparing the basic strength between orthotert butyl amine and ortho methyl amine how would we do it ?

Please check out I have edited my post

This kind of sentence is also mentioned in Finar, though not for carboxylic acids(as I have seen), reason was mentioned as steric acceleration as somehow the loss of H+ relieves the molecule of steric strain(I don't know the exact reason that's why I am commenting) and as amines are Lewis bases, steric retardation acts(groups are hindered during their approach).

Please could you provide a complete answer . The reason I am not able to get properly

Yes I looked up IL finar but the book doesn't really explain this matter. I am still confused

Ortho tert butyl benzoic acid will be more acidic than the other one .because former is more bulkier so it will rotate the COOH group out of the benzene ring plane to a more extent .
For NH2 group Ortho to CH3 or C(CH3)3 .Ortho effect never applies NH2 can never go out of the plane .rather this time CH3 and C(CH3)3 goes out of plane .so hyperconjugation effect is lost only inductive effect is present .still tert one will be comparatively less basic because of STERIC INHIBITION OF PROTONATION (SIP) . Due to hindrance nitrogen can't donate it's electron pair easily to a proton


text:
Can a charge smaller than that of an electron exist? If so how?
In school I've always been taught that the smallest charge possible is that of an electron( or proton) however I recently solved a question regarding a dipole and the charge came smaller than that of a proton, after  the solution given in the book however, there was a little statement written saying this can also exist in real life as well which seems off to me.. like how can you get less charge then the smallest unit of charge? If the answer is quarks well don't they exist in a combined state and wouldnt they combine to form protons and stuff (don't know too much about this topic tho.. I know a little thru YouTube videos but that's it) and hence change the charge to a multiple of that of a proton? Any insight is welcome
*Please note that this is not a homework question, it's just out of my curiosity.. thx for understanding!
Also not rly sure if I should ask this on chem stack exchange or phy..

Forget quarks, it is much simpler than that. See, if I work for some company part-time, say 1 day a week, they can claim that they hired 0.2 of a person...

It's pretty weird that a quark should have a fractional charge, once you get used to the idea of an electron having unit charge. But if we'd discovered quarks first, then the charge of an electron might be regarded as -3 (I leave it to the physicists to correct me).

As for the dipole, those aren't "real" partial charges (fractions of electrons or protons, say), it's a symbolic representation, as Ivan suggests not translatable into tangible objects.

@Dp03, Don't forget to do a dimensional analysis of dipole moment and charge. What was the exact problem?

@M. Farooq, it was a basic problem and I was able to do the calculation easily but what intrigued me was the last line of the solution which I mentioned above.. I'd also like to thank Ivan ,Buck and M. Farooq for ur comments, now getting back the question was-

HCL molecule which had a dipole moment of 3.4 * 10^-30 Cm, distance between them was given as 1.0 * 10^-10 m, my answer came to be 3.4 * 10^-20 C ( it was given that we can assume the charges to be equal and opposite).

I think Buckthorn has answered it well. I think it is historical that electrons were discovered first.

Dipole moments have a distance times an implicit charge, but you could give the moment as unit charge separation times a distance which is not the bond length. This would eliminate fractional charges and suggest some distance, which suggests that an electron and the nucleus are point charges separated by something less than the bond length. This is difficult to visualize. The electron is not really a point charge. Dipole moment is not separable into fractional charges or tiny distances, it measures a distortion of charge.
And when you think about it, an undistorted charge distribution (like electron density in benzene) is still "distorted" in that electron density is between atoms and above/below the atomic plane.
Other fractional charges might be imagined in compounds like magnetite: $\ce{Fe3O4}$. Each iron has a charge of $\pu{2.666}$. But if you describe magnetite as $\ce{Fe2O3.FeO}$, then two thirds of the iron atoms have a charge of +3, and one third have a charge of +2. More satisfying.
And then there is cyclopentadienyl anion:

where each carbon has an extra 0.2 extra electronic charge, and there are other ions like cycloheptatrienyl cation, cyclopropenium cation, etc. that have a surplus or deficiency of electrons.
But there is no fractional electric charge, just fractional molecules, or combined valences. Although, I still wonder how quarks can have a partial charge!


text:
Hydrolysis of (chloromethyl)cyclopropane: Is but-3-en-1-ol a possible product?
This is from one of the practice papers for the JEE , asks for possible hydrolysis products of (chloromethyl)cyclopropane .

Product A happens to be due to stability of cyclopropyl methyl carbocation.
Product B happens to be due to relaxation of ring strain of cyclopropane ring (though A is major even this is possible)
Pathway for C doesn't seem to be anywhere in my books , I tried googling it but found nothing.
So, is but-3-en-1-ol a possible product of hydrolysis of (chloromethyl)cyclopropane?

Bad news. Cyclopropylmethyl cation doesn't exist in a single form. It's nonclassical. It's equivalent to the cyclobutyl cation and equivalent to homoallyl cation (which gives product C).

I didn't know this , hence I didn't know the answer , thanks for your help

I'd say, yes it is possible. The pioneering work on interconversion reactions of cyclobutyl, cyclopropylcarbinyl, and allylcarbinyl derivatives has been done by John D. Roberts and coworkers (Ref.1). In these works, they have checked reactions under both $\mathrm{S_N1}$ and $\mathrm{S_N2}$ conditions. They have observed the reactions expected to involve carbonium ion intermediates have shown interconversion, which have been correlated to factors such as carbonium ion stability, ease of interconversion, and stability of possible products. One such interconversion have been shown in the reaction of cyclopropylcarbinyl-,  cyclobutyl-, and allylcarbinylamines with nitrous acid, which gave relevant carbonium ions. Similar approach was recently taken by Arca, et al. 2014 using solid catalysts (Ref.2). They have proposed a bicyclobutonium cation transition state to explain the product formation:

The coorelation to liquid phase reaction is well explanied in their abstract:

Rearrangement of cyclopropylcarbinyl bromide over proton and ammonium-exchanged mordenite ($\ce{H}$-MOR and $\ce{NH4}$-MOR) was studied at different temperatures. The product distribution analysis revealed a similar profile to what is found in solution, with prevalence of cyclobutyl over allylcarbinyl bromide. The activation parameters showed that compared with 80% aqueous ethanol solution, the zeolites present lower enthalpy of activation and higher entropy of activation, but yielding a significantly lower $\Delta G^‡_{298}$. The results may be explained in terms of the ability of zeolites to serve as solid solvents, providing a polar nanoenvironment for ionic reactions to occur.

Therefore, we can assume the reaction in hand have also unedergone with similar mechanism, giving same three products, $\bf{1}$, $\bf{2}$, and $\bf{3}$ (probably in different ratios). The formation of  $\bf{3}$ can also explained by the possible resonance contribution as depicted in above $\color{red}{\text{Red Box}}$. Note that two open resonance structures are essentially the same, but shown because it would make a double contribution due to the symmetry.
It's noteworthy that the open-chain derivatives have been achieved from the reaction of cyclopropylcarbinyl halides with amines (Ref.3).
References:

John D. Roberts, Robert H. Mazur, “Small-Ring Compounds. IV. Interconversion Reactions of Cyclobutyl, Cyclopropylcarbinyl and Allylcarbinyl Derivatives,” J. Am. Chem. Soc. 1951, 73(6), 2509–2520 (https://doi.org/10.1021/ja01150a029).
Hugo A. Arca, Gabriel C. C. Gomesab, Claudio J. A. Mota, “Solid solvents: activation parameters for the rearrangement of cyclopropylcarbinyl bromide on mordenite zeolite,” New Journal of Chemistry 2014, 38(7), 2760-2762 (https://doi.org/10.1039/C4NJ00287C).
Michael B. Smith, Robert T. Hrubiec, Charles A. Zezza, “Reaction of amines with cyclopropylcarbinyl halides: $\mathrm{S_N2'}$ or solvolysis?,” J. Org. Chem. 1985, 50(24), 4815–4821 (https://doi.org/10.1021/jo00224a033).


text:
Is it possible to produce an epoxide from 2-bromoacetyl bromide using methyl magnesium bromide
This is also one of the questions from practice papers for the JEE , asks for 2 step reaction for epoxidation of ketene

For product A , π complex is generated on c-c bond since carbon labelled 1 is electron deficient(inductive effect of sp carbon C2 and oxygen)

Is it possible to produce an epoxide from 2-bromoacetyl bromide using methyl magnesium bromide

Suppose you generated the epoxide. Notice, very importantly, that the question's author bothered to stipulate 2 equivalents of the Grignard reagent. Why do you think that is?

Probably 1st equivalent to substitute methyl at acyl position and 2nd equivalent to make addition reaction on  carbonyl ---- but this gives me bromohydrin which is wrong

See first comment. "Suppose you generated the epoxide."

How stable is the bromohydrin alkoxide? Can it react further?

Since no-one has been back to wrap this up here is how is goes: The addition of 2eq of Me Grignard generates the Mg alkoxide of the bromohydrin which undergoes an internal cyclisation by loss of Br- and Mg++ to generate the epoxide.

So one one bromine is left behind  ?


text:
When is a buffer solution considered efficient?
By this I mean what is the rigorous definition of a buffer solution?
I know that a buffer is a solution able to resist changes of $\mathrm{pH}$, but how much resistance to $\mathrm{pH}$ does a solution need to be considered a buffer?
For example I was told that one cannot make an acetic acid buffer $\left(K_\mathrm{a} = 1.8 \times 10^{-5}\right)$ at a desired $\mathrm{pH}$ of 6.5. Is this true?

https://en.wikipedia.org/wiki/Buffer_solution#Buffer_capacity

@Blade: See how I manipulate $K_\mathrm{a}$ using $K_\mathrm{a}$ formatting. Rate constants and equilibrium constants should be itallics such as other variables (e.g., $V, T, R, P, x, y, z,$ etc.).

Who "told you", instead of teaching you how buffers work?

Buffer solution Is considered efficient, if it meets your criteria for pH stabilization. Note the buffers are defined qualitatively, not quantitatively. 1 M acetic acid buffer at pH 6.5 is roughly comparable in buffer capacity to 0.01 M acetic acid buffer at ideal pH 4.75. So it is inefficient buffer, but still a buffer.

Well, I´m not blaming you, but he seems to have left out a few important points in his teaching. A buffer is obviously most efficient when it is not too dilute and has not already used up part of its buffer capacity in one direction, i.e. it´s fresh. A fresh buffer solution is an equimolar mixture of an acid and its conjugate base. That means you have no choice in the pH. It´s a property of the compound pair.

Efficiency of a buffer is not an absolute parameter, but it is conditional and case dependent.
If buffer differencial capacity ( $\frac{\mathrm{d}n}{\mathrm{dpH}}\cdot \frac{1}{V}$, where $n$ is molar amount of an added base/acid and $V$ is buffer volume)
or integral capacity ($\frac{\Delta n}{\Delta \mathrm{pH}}\cdot \frac{1}{V}$ per $\Delta pH = \pm 1$) meet criteria required by the solution of interest, the buffer is efficient and vice versa. The same buffer may be efficient for one application, while not for another.
As $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[A-]}}{c-\ce{[A-]}}=\mathrm{p}K_\mathrm{a} + \log {\ce{[A-]}} - \log{(c-\ce{[A-]})}$
then relative differential capacity wrt the maximal capacity at $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$ is $\frac{4}{ \frac{c}{\ce{[A-]}} + \frac{c}{c-\ce{[A-]}}} $, where $c$ is the total molar concentration of the buffer conjugate acid + base pair.


text:
How can a minor constituent of an alloy, like carbon in steel have such a profound effect?
In steel that is 0.65% carbon by weight, barely one in 34 atoms is carbon (assuming no other additives/impurities). How can such an outnumbered constituent make such a big difference? Surely many of the iron atoms are not touching any carbon.

Hint: crystal boundaries; edge vs. bulk properties

https://en.m.wikipedia.org/wiki/Steel Discusses it. A few precipitates have a large effect.

Think about the big strength difference 0.1 % C makes in "micro  alloyed "steels.


text:
Does Cl- Have more -I effect or +M effect ( Resonance effect)?

In this question my answer was (A) because I thought $\ce{NH2}$ group would give more electrons than $\ce{Cl-}$. But the correct answer is (D) and my teacher's reason is that chlorine has more negative Inductive effect than positive Resonance effect. Is that true?

Yes as you said the inductive effect of chlorine exceeds the strength of it's +R effect.
This effect becomes clear when you study aromaticity, that chlorine is a deactivating group (i.e it destabilizes the carbocation formed) but is still Ortho - para directing, because it still causes stabilization by the +R effect in those positions.
However unlike other +R groups which are also activating, chlorine is deactivating, because its inductive effect is slightly stronger than it's +R effect
( Note that all comparisons are based on which effect causes more stability)

Chloro group is considered a deactivating group according to Hammett plots of each possible reaction. The $\sigma_\mathrm{para}$ of $\ce{Cl}$-substituent is listed as $+0.227$ while $\sigma_\mathrm{meta}$ of $\ce{Cl}$ is listed as $+0.373$. It is known fact that $\sigma_\mathrm{meta}$ is an indicative of how much inductive effect contribute to the reaction, while $\sigma_\mathrm{para}$ is a combination of the strength of both inductive and mesomeric effects (more resonance effect than inductive because the substitution is one carbon away compared to meta-position). For comparison, when the substitution is $\ce{H}$, both effects are zero while when the substitution is $\ce{NO2}$ group, $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ are $+0.710$ and $+0.778$, respectively (resonance effect predominates). Based on Hammett's original definition, when $\sigma$ of any group has positive sign,it is electron withdrawing group (EWG) and the numerical value is an indication of hoe big electron withdrawing ability compared to that of $\ce{H}$ (value of which is always zero). Note that Hammett has assigned these sign and value according to his research on the acidity of substituted aromatic carboxylic acids (EWG substitution makes acid stronger).
Contrary to carboxylic acids, EWG substitution at aromatic nucleus makes amino bases weaker because EWG dilutes the electron density on active $\ce{N}$ atom by inductive and/or resonance effects. Therefore, largere the positive number of $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ of EWG, weaker the base. Also keep in mind that $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ of $\ce{NH2}$ group are $-0.161$ and $-0.66$, respectively. Hence, basic strength of given benzamidine are decreased in order of their 4-substitution of $\ce{NH2} \gt \ce{H} \gt \ce{Cl} \gt \ce{NO2}$ meaning your correct answer is $\bf{(D)}$. It is noteworthy that this order remains the same even if these substitutions are at 3-position of the aromatic nucleus (based on $\sigma_\mathrm{meta}$ values).
I can't really find the $\mathrm{p}K_\mathrm{a}$ of relevant compounds. However, I found $\mathrm{p}K_\mathrm{a}$ values of some guanidines, $\ce{X-C6H4-N=N(N(CH3)2)N(N(CH3)2}$ (parent compound: 2-phenyl-1,1,3,3-tetramethylguanidine, $\ce{C6H5-N=N(N(CH3)2)N(N(CH3)2}$) which would prove my point (Ref.1):
$$
\begin{array}{c|ccc}
\hline
\ce{X} & \sigma_{(m,p)}^a & \mathrm{p}K_\mathrm{a} \ \text{of 2-Ar-guanidine} & \mathrm{p}K_\mathrm{a} \ \text{of aniline}^b  \\
\hline
\ce{H} & 0 & 12.18 & 4.62 \\
\text{3-}\ce{CH3} & -0.06 & 12.25 & 4.70 \\
\text{4-}\ce{CH3} & -0.14 & 12.37 & 5.11 \\
\text{3-}\ce{OCH3} & 0.11 & 11.96 & 4.23 \\
\text{4-}\ce{OCH3} & -0.28 & 12.57 & 5.34 \\
\text{3-}\ce{Cl} & 0.37 & 11.47 & 3.52 \\
\text{4-}\ce{Cl} & 0.24 & 11.70 & 3.98 \\
\text{4-}\ce{NO2} & 1.26 & 9.78 & 0.99 \\
\hline
\end{array}\\
^a: \text{Ref.2 and } ^b: \text{Ref.3.}
$$

References:

Przemyslaw Pruszynski, "Synthesis and properties of phenyl substituted derivatives of 2-phenyl-1,1,3,3-tetramethylguanidine," Canadian Journal of Chemistry 1987, 65(3), 626-629 (https://doi.org/10.1139/v87-107).
D. D. Perrin, Boyd Dempsey, E. P. Serjeant, In $\mathrm{p}K_\mathrm{a}$ Prediction for organic acids and bases; Chapman and Hall: London, England, 1981 (ISBN 978-94-009-5885-2).
D. D. Perrin, In Dissociation constants of organic bases in aqueous solution, Volume 1; International Union of Pure and Applied Chemistry. Commission on Electroanalytical Chemistry, Butterworths: London, England, 1965 (Suppliment 1972)(ISBN-13: 978-0080208275).


text:
2 problems with elimination-addition of halobenzene using NH2-
When converting chlorobenzene to aniline, I drew the following mechanism and the book says it is correct, but there are 2 questions that I do not understand:

In the 2nd step, I am thinking that it is a nuclophilic attack, since $\ce{NH2^-}$ is a good nucleophile. However, I do not understand what makes the $\ce{C#C}$ bond in benzyne electrophilic. Doesn't a triple bond have high electron density and render it more nucleophilic instead? How can the reaction happen?

In the 3rd step, I don't know why proton transfer happens here. Isn't forming $\ce{NH2-}$ from $\ce{NH3}$ not thermodynamically viable as $\ce{NH2-}$ is a strong base?

My thought on (2): forming $\ce{NH2-}$ ion from $\ce{NH3}$ is more stable than leaving the carbanion (the aniline with an extra electron) there, but I am not sure about whether this is a correct explanation.

Consider the stain on the benzyne triple bond, does this make it unstable?

1. Electrophilicity of the triple bond
Hard-soft acid-base theory ties in with this question. The general trends in nucleophile-electrophile reactivity show that hard (highly charge dense) nucleophiles have a tendency to react well with hard electrophiles, and soft (low charge density) nucleophiles with soft electrophiles. This relationship is due to two competing factors; electrostatics and overlap of frontier orbitals. Hard-hard interactions lend their stability to electrostatic attraction, while soft-soft interactions rely on a good HOMO-LUMO overlap.
A π-bond is generally considered a soft species that doesn't reply on electrostatics very strongly. Good soft nucleophiles have relatively high HOMO's (highest occupied molecular orbitals) and good soft electrophiles have relatively low LUMO's (lowest unoccupied molecular orbitals). The third bond in benzyne is abnormal in that it doesn't involve overlap of 2 unhybridised p-orbitals, but 2 hybridised $sp^2$ orbitals that aren't completely parallel. From Clayden Organic Chemistry:

This will, as expected, result in a very weak orbital overlap. As a result, the split in energy between the resulting bonding and antibonding orbital is very small, giving a very low LUMO which can easily accept a nucleophile's electrons. From here:

This is the source of the triple bond's electrophilicity.
2. The final protonation step
(Image from Clayden Organic Chemistry)

Your assumption is correct that the anion is very unstable. According to this source, , benzene has a $\mathrm{p}K_\mathrm{a}$ of 43. This is due to the negative charge being situated in an $\pu{sp^2}$ orbital, which is highly charge dense, with no delocalisation into the ring (since it is not a part of the π-system). Ammonia has a $\mathrm{p}K_\mathrm{a}$ of 38, so its conjugate base, $\ce{NH2^-}$, is an extremely good base. This value is still lower than benzene's, so the benzene anion is still less stable. By a quick calculation, this gives the reaction above an equilibrium constant of roughly $\pu{10^5}$, heavily in favour of ammonia deprotonation.


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Determination of spinel structure by CFSE

$\ce{CuFe2O4}$ has inverse spinel structure, but $\ce{ZnFe2O4}$ has normal spinel structure. Explain.

I know that normal spinels have a structure like $\ce{(A^{II})^{tet}(B^{III})^{oct}_2O_4}$-like structure, while inverse spinels have a $\ce{(B^{III})^{tet}(A^{II}B^{III})^{oct}O4}$-like structure.
Also, I know that if $\ce{B^{III}}$ ion has greater CFSE gain in octahedral site than that of $\ce{A^{II}}$ ion, a normal spinel structure is expected. If $\ce{A^{II}}$ ion has greater CFSE gain in octahedral geometry than the trivalent $\ce{B^{III}}$ ion, then we get an inverse spinel structure.
For the given question, $\ce{Fe}$ has a $\mathrm{3d^6}$ configuration having a structure $\mathrm{t_{2g}^4 e_{g}^2},$ giving a CFSE
$$2(0.6 \Delta_\mathrm{o}) + 4(-0.4 \Delta_\mathrm{o}) = -0.4 \Delta_\mathrm{o},$$
while $\ce{Zn}$ has a $\mathrm{3d^{10}}$ structure having 0 CFSE. So, CFSE of $\ce{Zn}$ is greater than CFSE of $\ce{Fe}.$ So, according to the rule stated before, it gives me an inverse spinel structure.
Similarly for $\ce{Cu},$ it should have a $\mathrm{3d^9}$ $(\mathrm{t_{2g}^6 e_g^3})$ structure which gives a CFSE of
$$3(0.6 \Delta_\mathrm{o}) + 6(-0.4 \Delta_\mathrm{o}) = -0.6 \Delta_\mathrm{o} < -0.4 \Delta_\mathrm{o},$$
which gives a normal spinel structure.
What is wrong with this calculation so that I'm getting opposite structures to those predicted in the question?

Your calculations are wrong because you should be looking at $Fe^{3+}$ which is $d^5$ and you are using $d^6$

I'm not really getting your numbers here.
In your situation the iron is Fe(III), so a $d^5$ system. For $d^5$ it does not matter if it's in an octahedron or a tetrahedron since the energies will just cancel out. So in both situations it will be determined by the other ion. For Zn(II) it also doesn't really matter. So I guess here the regular spinel dominates as Zn(II) just prefers more tetrahedral coordinations in general, especially in solid states while Fe(III) is happy with an octahedron.
For the Cu(II) however you've got a $d^9$ system. So the octahedron will give you something like $(6*2/5)-(3*3/5) = 3/5$ (ignoring things like spin pairing energ here).
But in the tetrahedral case you also have to consider that the ligand field splitting for a tetrahedron is only $4/9$. So it will be something like $4/9*[(4*3/5)-(5*2/5)] = 8/45$ and therefore lower.
So the Cu(II) will prefer the octahedral position here and hence form an inverse spinel. Feel free to correct me here I was just drawing the diagrams as I read the question, perhaps I did a mistake somewhere.

Also discussed here:https://chemistry.stackexchange.com/questions/71935/why-is-znfe2o4-a-normal-spinel


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Why does FeCl3 react with carboxylic acid, but not with sulphonic acid in this question?
I was solving a question where I was asked to identify the reagents given in the question which could react with a compound provided.
The first compound was methanoic acid. One of the reagents given was $\ce{FeCl_3}$. According to the answer key, these 2 could react. On googling the reaction, I found that methanoic acid loses its acidic proton and is attacked by $\ce{FeCl_3}$ and the proton released then removes chlorine to form the complex. It all made sense because $\ce{CH_3COOH}$ was an acid.
But the next compound was 3-carboxy-4-nitro-benzene sulfonic acid. And I wanted to check its reactivity with $\ce{FeCl3}$. So, I felt that the compound being acidic due to the sulfonic group would work similarly as the first compound and thus react with $\ce{FeCl_3}$. However on verifying with the answer key, I found that it would not react with $\ce{FeCl_3}$. But I couldn't think of a reason why.

Looks like carboxylates make stronger complex with Fe than sulfonates.

Methanoic (=derived from methane ) acid is formic acid $\ce{HCOOH}$, while $\ce{CH3COOH}$ is ethanoic ( = derived from ethane ) acid aka acetic acid.

Acetic Acid reacts with $\ce{FeCl3}$ to form Iron (III) acetate complex. I think the sulfonate with it's large benzene ring won't be able to form a similar complex.

@AniruddhaDeb What could have prevented it?

@AniruddhaDeb iron does form Sulfonate complex. See this: https://www.nrcresearchpress.com/doi/pdfplus/10.1139/v81-098.

"Please help i still couldn't get the answer" Just adding this statement does not improve your question. A question is edited when one needs to add extra informations. So, if you have any information to add or if you went through any sources or if you have done any research, then edit the question accordingly. You need to be patient in this process of answer-seeking. I am sure someone will answer your question. BTW, did you went through the linked paper in my previous  comment?

@NilayGhosh okay I have removed that statement. And yes I went through that paper. So it definitely has something to do with the benzene ring, as sulphonates themselves form iron complexes.

It all made sense because $\ce{CH_3COOH}$ was an acid

$\ce{CH_3COOH}$ being an acid facilitates the complex formation as the chlorido ligands can be easily removed as $\ce{HCl}$. But, that's not the only reason why the red colored complex $\ce{[Fe(OAc)_3]}$ was formed.
You have to understand that the stability of the complexes is also determined by how well the ligand interacts with the central metal ion.Here, the ligand forming the coordination complex is $\ce{OAc^-}$ and this is associated with enough negative charge density to stabilize the complex formed.

So, I felt that the compound being acidic due to the sulfonic group would work similarly as the first compound and thus react with $\ce{FeCl_3}$. However on verifying with the answer key, I found that it would not react with $\ce{FeCl_3}$

This is because although as you have said has a sulfonic acid group, it is not the acidic strength of this ( or acidic strength of $\ce{CH_3COOH}$ in the first question) that drives the complex formation.It mainly depends upon the strength of the ligand and effective nuclear charge of the central metal ion.The ligand in this case, is associated with lot of electron withdrawing groups like $\ce{-NO_2}$ and that makes this a poor ligand.Also, the steric hindrance due to the bulky groups also has a role to play for the complex formation not to take place.
NOTE:
All my arguments made regarding stability of complexes is as per the Crystal Field Theory which assumes the interaction between metal ion and ligand to be purely ionic.That's why, I was comparing the charge density of the ligands in the 2 cases.

@VamsiKrishna Where did you find this question btw? And I hope you understood the reason :)

Amazing I am finally relieved! It was in a worksheet that my sir had provided, but he left shortly so I couldn't get the solution. Thanks a lot fpr your answer!

Happy to know that my answer helped you! :)


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Is this a correct way to write a chain reaction?
I don't know much about chain reactions and how I understand them, they are just a series of chemical reactions where the products of the preceding reaction act as the reactants for the succeeding reactions.
Now, let's assume that $\ce{A}$ and $\ce{B}$ are two compounds that chemically react under certain conditions to give $\ce{X}$ and $\ce{Y}$. This chemical reaction can be written as:
$$\ce{A + B -> X + Y}$$
Now, let's also assume that $\ce{X}$ and $\ce{Y}$ react under the very same conditions to give $\ce{M}$ and $\ce{N}$. This chemical reaction can be written as:
$$\ce{X + Y -> M + N}$$
Now, can the series of these reactions be written as follows :
$$\ce{A + B -> X + Y -> M + N}$$

We tend not to combine two reactions on the same line.

So, writing them in two separate lines would be more appropriate. Thank You!

Your mechanism is not a chain reaction. An example of a chain reaction may be the radioactive decay chain, like U-238 -> Th-234 -> Pa-234 -> U-234 -> etc.

Is chain reaction a term specific to nuclear reactions? I assumed  that any reaction whose products react to form another set of products without any external conditions would be called a chain reaction.

No and no. The term is indeed used in chemistry, but it has a different meaning. https://en.wikipedia.org/wiki/Chain_reaction

Your scheme is one of sequential reactions. The chain reaction has feedback so that one to two reactions occur repeatedly until the reactants are all consumed.
The general scheme is shown in the diagram where $R\cdot$ represents any radical species ( H or Cl in Maurice's answer) and $\alpha$ has to be at least 1 for species $R\cdot$ to continue the chain reaction.

A good example of chain rection is the photochloration of $\ce{H_2}$ by $\ce{Cl_2}$. The first reaction is due to the first photon striking a chlorine molecule, producing two chlorine atoms Cl. Then each Cl atom reacts with a $\ce{H_2}$ molecule producing one HCl and one H atom. This H atom reacts with another $\ce{Cl_2}$ producing one HCl and a new Cl atom that repeats the first reaction with $\ce{H_2}$ $$\ce{Cl_2 ->  2 Cl}$$$$\ce{Cl + H_2 -> HCl + H}$$$$\ce{H + Cl_2 -> HCl + Cl}$$ And the wole process repeats in a so-called chain reaction, up to the destruction of all reagents. The terminal reaction is $$\ce{H + Cl + M -> HCl + M}$$ and M may be any molecule or substance. The chain reaction is maintained by the production and destruction of H and Cl free atoms.


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Why is some ionic substance insoluble, such as BaSO4?
If Ba is less electronegative than Mg, then why is BaSO4 insoluble in water and MgSO4 is soluble? I thought the greater the EN difference the more soluble the compound is because of how unequally shared the electron is.

The electronegativity is not the main factor. The entropy factor is much more important. When a salt like $\ce{BaSO_4}$  tries to get dissolved, its ions are attracting water molecules that get fixed or attached around each ion. This phenomena creates order, and decreases the entropy of the solution. The same thing happens around sulfate ions $\ce{SO_4^{2-}}$. If there is too much water attached around each ion, the substance does not dissolve. This is the case for $\ce{BaSO_4}$, because the ion $\ce{Ba^{2+}}$ is bigger than $\ce{Mg^{2+}}$, and so creates more order in attracting more water molecules around it. The entropy decreases too much if $\ce{BaSO_4}$ gets dissolved. This salt is practically not soluble. This is not the case for $\ce{MgSO_4}$.

Do you have numbers for their relative enthalpies and entropies of solution?

but why does the same phenomenon not explain the relative solubility of $\ce{Ba(OH)_2}$ and $\ce{Mg(OH)_2}$, in which the latter is less soluble than the former?

@theorist and WHOOP. I don't have numerical values to back my answer. Sorry ! And I must admit that I am unable to explain why $\ce{Mg(OH)2$ is nearly insoluble in water.


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Red P+HI reduction
My teacher told us that Red P and HI can reduce any functional group. Is it true? Or are there any groups it can't reduce?(I know this exact question has been asked before but I couldn't find the exact thing I wanted)
I need a list of functional groups which cannot be reduced but red P HI

One group that P/HI won't reduce is -COOH. Kiliani reduced glucose to hexanoic acid with this reagent.

The linked question answers your question to the point. What could you not find there?

@AniruddhaDeb the ans referred is scattered. I couldn't find many results from that

To OP:See this. To @user55119: carboxylic acid can be reduced (for ex-benzoic acid reduced to toluene and another source saying lauric acid can be reduced)

@Nilay Ghosh: Aromatic acids may be an exception. As to lauric acid, need it carefully. The commentary is shear speculation. Not a reference in sight.

@user55119 forget to edit the previous comment. Granted that 2nd source was full of skeptical inforrmation but the 1st source also mentioned stearic acid with benzoic acid. Maybe, I need to find some authentic sources.

Imines survive, that's for sure. Some 20 years ago junkies developed a way of turning a relatively innocent drug, which was then legal (and which has since ceased to be legal in many jurisdictions for this very reason) into a much more potent substance. I'd rather not identify it precisely; let's just say it is prominently featured in a TV series with name that is an anagram of "Baking Bread".
In short, the first substance contained the $\ce{-OH}$ and $\ce{-NHCH3}$ groups, and the reduction took away the former, but not the latter. The synthesis could be done with surprisingly crude equipment, so the junkies did it every time they needed the product. They bought red phosphorus and crystalline iodine by kilograms, and carried them around in a form of pre-prepared mixture ready to go into the reaction. They used to call this mixture "Stendhal", because of "The Red and the Black".
So it... went for a while.


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Role of potassium hydroxide in alkaline battery?
Is the role of potassium hydroxide ($\ce{KOH}$) in an alkaline battery to provide hydroxide for the reaction with zinc? Would the battery cease to work if $\ce{KOH}$ was removed and only $\ce{OH-}$ from autoionization ($\pu{10^{-7} M}$) remained?

The role of $\ce{KOH}$ is to provide enough ions for redox reactions. Remember that redox reactions come in pairs, for every reduction there must be oxidation. If $\ce{Zn}$ reacts with $\ce{OH-}$ to form a hydroxide, what would be a counter-reaction?
A battery also needs to have certain ionic conductivities to be functional. Could you ever get enough $\ce{OH-}$ ions through autoionization?

Thanks. Yes I'm aware of the conductivity. The cathode reaction is MnO2 + H2O + 2e-.

I think the reason I asked is because potassium hydroxide is labeled as the electrolyte, whereas it is actually the anode reaction, its the oxidation of hydroxide that releases electrons. The anode reaction, the O that reacts with Zn comes from 2 OH- --> O + H2O + 2 e-.

The oxidation state of the zinc does not change. ZnO has 38 electrons, O has 8 and Zn(s) has 30, and the lewis structure of ZnO is analogous to Zn2+ (28 electrons) and O-2 (10 electrons), still 38 in total.

Or, the oxidation state of zinc does change. But the total number of electrons in it (ZnO) remains the same, zinc is actually releasing its electrons into the oxygen it bonds with. The source of the electrons released from the anode is the hydroxide ions.


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How can I simulate chromatograms and spectra?
Is there a tool which allows to generate simulated chromatogramms and spectra and export it into a mzML file? There is e.g. psims which allows to write mzML on a low level. However I was not able to find a high level tool which allows to define chromatogram shapes via a e.g. a GUI.

What is the purpose to generate fake chromatograms and spectra?

It's about testing if the library I'm using is not manipulating the data by mistake. I'm using pyopenms which uses OpenMS under the hood. I could not find tests for the lib yet.

OK, I found the tests: https://github.com/OpenMS/OpenMS/tree/develop/src/tests I'm trusting the devs that they are doing their job properly :)

Sorry, I do not know that programming stuff but you can "make" chromatograms yourself in MATLAB by summing up several Gaussians as a function of time. Also add some random noise. Note that the peak width increases linearly with time. You can also explore PeakFit as GUI for chromaatographic peak analysis.

Instead of using the word "fake" use the word simulated. Fake has a negative connotation and there are tons fake papers out there.

The words "fake"/"mock" are the correct ones in the software development domain. However I understand your concerns w.r.t. "fake" in the chemistry/research domain :)


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Why do most character tables include an $(x^2 + y^2)$ term?
Why do most character tables (e.g. $C_{3h}$ but not $C_{2h}$) include an $(x^2 + y^2)$ term?
Is it an abbreviated form for $d_{z^2}$, applying only where $3d$ (and higher) orbitals might be involved, or just a generic binomial term with no specific relevance?
The only clue I can find is on this page where it lists the full expression for the angular wave function for $3d_{z^2}$ as
$$Y_{3d_{z^2}} = √(5/4){2z^2-(x^2 + y^2)}/r^2 × (1/4\pi)^{1/2} $$
where it also comments that "the $3d_{2z^2-x^2-y^2}$ orbital is abbreviated to $3d_{z^2}$ for simplicity".
Thoughts appreciated.

Thanks to initial editor of the question, for showing me how to improve super- and sub-scripting

The functions on the right hand side of the table are the 'basis functions' for the irreducible representation, which are the rows of character, say $A_g$ or $B_g$ in C$_{2h}$. The basis functions have the same symmetry properties as the atomic orbitals with the same name (but other functions in $x, x^2, x^3$ etc and similarly for  $y,z$, could be used).
To determine where  an $xy$ function goes, look at a $d_{xy}$ orbital shape and subject it to each of the symmetry operations in turn,$E, C_{2}, i$ etc depending on point group.  (Assume z is the principal axis). Count +1 if the orbital (or function) is unchanged and -1 if not. So in C$_{2h}$, a $C_2$ operation leaves $xy$ unchanged as does $\sigma_h$ and $i$ so $xy$ is a 'basis' for A$_g$. The $xz$ transforms as B$_g$; a $C_2$ operation gives -1, as does $\sigma_h$, but $i$ produces +1.
(note in cases such as $z^2 -(x^2+y^2)/r^2$ the $x^2+y^2$ part is totally symmetric in the $xy$ plane and does not change anything so can be ignored; think of the donut around d$_{z^2}$.)


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Is entropy of surroundings a state function?
It is a pretty common well known fact that the entropy of a system is a state function ie. it doesn't depend upon the path taken by the process as long as the initial and final states are the same.
I have a doubt whether we can say the same for surroundings or even the universe.
My first thought is that the entropy of the universe would be continuously increasing or at best it would be constant. It can never decrease. Thus the universe can never be at a state where it was before. The talk of state function seems strange here.
Still I am not very sure and I have no idea whether entropy of surroundings is a state function or not.
Any help would be appreciated.
Thanks

In reality, when we talk about the surroundings, what we are referring to isn't really the entire universe, outside what we call "the system."  What we are really referring to is a subregion of the entire universe, wherein it is possible to carry out changes reversibly without encompassing the whole rest of the universe.  So, when we talk about the entropy change of the surroundings, we are not including spontaneous processes occurring light years away in other galaxies.  And, when we say that the entropy change for system plus the surroundings is zero, we are really only including the selected subregion of surroundings chosen to represent our sub-universe.

A proper appellation would be perhaps a light cone.

I don't feel this answers the question. Can you give a mathematical argument that the entropy of surroundings is a state function? Because I'm under the impression the entropy change of the surroundings has a different value for reversible and irreversible processes, making it path-dependent.

@electropusher Can you please give a specific example to illustrate your contention?


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Finding heat of formation
The heats of combustion of $\ce{C_xH_y}$, $\ce{C}$, and $\ce{H2}$ are $a, b,\text{and } c$ cal respectively. The heat of formation of $\ce{C_xH_y}$ will be:
The answer is
$-\left(x b+\frac{y c}{2}-a\right)$ cal
But I am getting negative of it by adding equations:
$$\ce{C_xH_y} + (x+\frac{y}{4})\ce{O2->}x\ce{CO2}+\frac{y}{2}\ce{H2O} \text{ , }\Delta H=a \tag{1}\label{1}$$
$$\ce{C} + \ce{O2->}\ce{CO2}\text{ , }\Delta H=b \tag{2}\label{2}$$
$$\ce{H_2} + \frac{1}{2}\ce{O2->}\ce{H2O} \text{ , }\Delta H=c \tag{3}\label{3}$$
$\Delta H_f=-1\times\eqref{1}+x\times\eqref{2}+\frac{y}{2}\times\eqref{3}=-a+bx+\frac{cy}{2}$

What are your x and y?

The same x and y which is in question

Using photos of text instead of text itself is strongly discouraged for multiple reasons. 3 main reasons are  1/  Impossible to search for it 2/ Impossible to be reused in answers 3/ Laziness of writing puts extra burden on shoulders of responders while reading and evaluating it.

Be aware combustion and formation heat have switched products and reactants when the value is evaluated from component contributions.

All should be in units like cal/mol or cal/kg, not just cal.

You didn't do anything wrong excep for not applying correct sign. All values of $a,b,c$ should be negative values because for each equation, heat gives away. It seems, the values given in question is just the numerical value without a sign. Asker expects you to apply correct sign when you do your calculations. Fault is on who wrote the question.

it seems to me your assessment is correct, and the given answer is incorrect.

The question: The heats of combustion of $\ce{C_xH_y}$, $\ce{C}$, and $\ce{H2}$ are $a, b,$ and $c$ $\pu{cal mol-1}$, respectively. What will be the heat of formation of $\ce{C_xH_y}$?

We have three combustion reactions to consider:
$$\ce{C + O2 -> CO2} \quad \Delta H_{\mathrm{comp}(\ce{CO2})} = b \tag{1}$$
$$\ce{H2 + 1/2 O2 -> H2O}  \quad \Delta H_{\mathrm{comp}(\ce{H2})} = c \tag{2}$$
$$\ce{C_xH_y + ($x$ +y/4) O2 -> x CO2 + y/2 H2O}  \quad \Delta H_{\mathrm{comp}(\ce{C_xH_y})} = a \tag{3}$$
Assuming the given answer is correct, I have to tell this straight: The wrong doing in this question is giving all three heat of combustion values as their numerical values and not mentioning it in the question (thus, solver has to assume it). That, I called being tricky at this level of education. For example, if the question is a multiple choice, then it may include both $-\left(x b+\frac{y c}{2}-a\right) \ \pu{cal mol-1}$ and $+\left(x b+\frac{y c}{2}-a\right) \ \pu{cal mol-1}$ as chices to pick as an answer. Too bad for students.
Let's assume $\Delta H_{\mathrm{comp}(\ce{CO2})}$ is $-b$ in the equation $(1)$, and so on. Now we can rewrite above equations as:
$$\ce{C + O2 -> CO2} \quad \Delta H_{\mathrm{comp}(\ce{CO2})} = -b \tag{4}$$
$$\ce{H2 + 1/2 O2 -> H2O}  \quad \Delta H_{\mathrm{comp}(\ce{H2})} = -c \tag{5}$$
$$\ce{C_xH_y + ($x$ +y/4) O2 -> x CO2 + y/2 H2O}  \quad \Delta H_{\mathrm{comp}(\ce{C_xH_y})} = -a \tag{6}$$
From the equation $(4)$:
$$\Delta H_{\mathrm{comp}(\ce{CO2})} = \Delta H_{f(\ce{CO2})} - \left(\Delta H_{f(\ce{C})}  + \Delta H_{f(\ce{O2})}  \right) = \Delta H_{f(\ce{CO2})} - \left(0  + 0  \right) = -b\\ \therefore \ \Delta H_{f(\ce{CO2})}  = -b \tag7$$
Simillarly, from the equation $(5)$:
$$\ \Delta H_{f(\ce{H2O})}  = -c, \tag8$$
and from the equation $(6)$ :
$$\Delta H_{\mathrm{comp}(\ce{C_xH_y})} = \left(x\Delta H_{f(\ce{CO2})}  + \frac{y}{2}\Delta H_{f(\ce{H2O})}  \right) - \Delta H_{f(\ce{C_xH_y})} = \left(-xb - \frac{yc}{2}\right) - \Delta H_{f(\ce{C_xH_y})} = -a$$
$$\ \Delta H_{f(\ce{C_xH_y})}  = +a - \left(xb + \frac{yc}{2}\right) \tag9$$
The equation $(9)$ is the correct answer.


text:
How many gauche butane interactions are present in cis-1,2-dimethylcyclohexane?
We were asked to find out the number of Gauche-Butane interactions for this compound:

I proceeded by drawing the Newman Projection for this chair conformation as follows and marked the Gauche-Butane interactions (as shown by red curves):

They told me that I was almost correct and asked me to try once again but I can't find out where I probably went wrong.
Can someone guide me out in this regard?

#3 looks dubious. Consider that the gauche interaction stretches over four carbons: C–C–C–C.

@orthocresol, I thought that the methyl group involved in the step 3 suffers steric repulsion from that carbon like in step 1. Isn't it so? If not, what should it be?

Which 4 carbons should I choose? Should I start from the carbon in step 3 to all the way left ?

Ahem, well, I don't really want to give you the answer straight up (and indeed, I don't think you need it). The equatorial methyl group is only involved in one gauche interaction, which is your #2. The other carbons are anti to it, so that's not an issue. Your #1 and #2 are quite correct. Your #3 isn't. Have a think about it. By the way, are you supposed to ignore gauche interactions that are within the ring itself? Even without the methyl groups, unsubstituted cyclohexane has some gauche interactions, which you haven't included.

Is this from an entrance exam preparation manual/worksheet?

@orthocresol, I don't know if I am supposed to ignore gauche interactions within the ring. But, after giving a thought about #3, I still couldn't figure out the total nos. of interactions :(

While cyclohexane does have gauche interactions, they are irrelevant because chair cyclohexane is taken as the reference point. The axial methyl group in cis isomer 1 has two gauche interactions with the ring as illustrated in red in 1a and 1b. A third gauche interaction is between the two methyl groups in 2. By comparison the trans isomer 3 only has gauche interactions between the methyl groups.

The difference in the number of gauche interactions - namely, two -  between these epimers is approximated by the difference in their heats of combustion[1] which is $\pu{1.54 kcal/mol}$. [1,2 cis: $\Delta H^\circ_\mathrm{comb} = \pu{-1248.31 kcal/mol}$; 1,2 trans: $\Delta H^\circ_\mathrm{comb} = \pu{-1246.77 kcal/mol}$.] A typical value for a gauche interaction is $\pu{0.9 kcal/mol/interaction}$.
References:

Walter H. Johnson, Edward Prosen, and Frederick D. Rossini, Heats of Combustion and Isomerization of the Eight $\ce{C8H16}$ Alkylcyclohexanes. Research Paper RP1S12, Volume 39, July 1947. PDF via NIST

Should I consider 1a and 1b as equal?

They are mirror images but additive. Axial methylcyclohexane is less stable than the equatorial conformation by 1.8 (2 x 0.9) kcal/mol.

So you're saying that there are 4 gauche-butane interactions in total?

Not at all! Structures 1a, 1b and 2 are all the same cis isomer showing the three gauche interactions. Compound 3 is the trans isomer with one gauche. The difference between the two is a total of 2 gauche interactions. This difference is seen in the difference in the heat of combustion of the cis and trans isomers. See the 1,2-cis and trans isomers here: http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#Conformation

So, you are saying that there are only two gauche-butane interaction?

No! cis-1,2 has 3 gauche interactions as illustrated. Trans-1,2-has one. If you want to figure out the difference in their heats of combustion, the methyl/methyl interactions cancel each other leaving a net of 2 gauche interactions. The number 2 only applies to the net difference between the two isomers as far as the heats of combustion and heats of formation, for that matter., are concerned. You need to do some reading. Google the issue.


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Are electrostatic effects temperature dependent?
A charged particle like an ion, does its repulsive or attractive effect on other particles change with temperature? Are electrostatic effects temperature dependent?

What "electrostatic effects"?

There is nothing in the Coulomb's law that is temperature dependent.

Well ...... there is a relativistic effect that means the force between two charged particles does depend upon their velocity (see e.g. https://newt.phys.unsw.edu.au/einsteinlight/jw/module2_FEB.htm), and higher temperature does mean a wider velocity distribution ... But for any chemical system under reasonable conditions this can be ignored.

The strength of the electric charge does not change with temperature (unless approaching that of the Big Bang). However, effects of the charges do change with temperature. For example, at low temperatures (near 0 K), ions can form arrays in space, though their charges repel. The history of Bose-Einstein condensates begins with trapped ions, extending to neutral gases.

What I'm looking at is similar to ions being able to form ordered structures at low temperatures, but, that their charges repel at higher temperatures. Is that supported by theory? If yes, what is the reason for that?

No. The interaction is temperature independent. But higher temperature means higher thermal energy, which may be able to overcome the electrostatic interactions.

@Moishe I object. Your first link tells about arrays of neutral atoms (sure, they are in a highly unusual state, but still).

@IvanNeretin, yes, but a Rydberg atom is almost an ion, so please don't get excited ;-)

An ion with an electron around it! Well, you may have the same in any piece of any metal. There is no need to look for exotic states of matter.


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Which way is major in this interesting intramolecular SN2 reaction to form different five memeber ring?
Which direction of the intramolecular reactions is major and why? Could someone could depict the mechanism step by step and it is better to draw out the transitional state configurations

The two lone pairs are obviously not equivalent. Any stabilization would lead to decreased nucleophilicity. Can you identify any factors that might stabilize two lone pairs?

@Zhe Can you elaborate a little more? I am not exactly sure about the factor you're referring to, is it the conjugation between both the oxygens?

I want to know what the difference between these two lone pairs was, and could anybody sketch out the molecular orbitals of these two lone pairs located in?


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What limits supersaturation in the absence of nucleation sites?
Imagine a scenario where we have a chamber with air that is supersaturated with water vapour. In this ideal scenario there are no nucleation sites (e.g no suspended particles or nucleation on the walls of the chamber).
What is the physical/chemical mechanism that limits further supersaturation?

The wall of the chamber will serve as a nucleation site. If you magically make that impossible, then I guess you may get insanely high supersaturation, before molecular fluctuations in the air will become new nucleation sites.

There is still ionization due cosmic rays and natural ( mostly radon ) radioactivity, which serves as nucleation centre source. It was used earlier as detector of energetic charged particles.

Okay, since it's a hypothetical scenario, we forbid also those. Basically the question then is if there exists a mechanism that enforces an upper limit or not (e.g. as pointed out by @Ivan Neretin molecular fluctuations)?

There are many such mechanisms; you turn one off, then another steps in. It is pretty much like death.

@IvanNeretin - now, you just have to prevent the molecules from interacting with each other. Sounds easy to me...

@JonCuster Why, of course, that would mean turning water into an ideal gas, so no more phase change whatsoever. It is pretty much like heaven or hell.

@IvanNeretin Do you have a source for the "insanely high" supersaturation that one could reach? It is rather simple to make a perfectly clean experiment, dust-free  and deionised, and natural radioctivity is easily excluded, because you can see it when it occurs. https://en.wikipedia.org/wiki/Cloud_chamber

@Karl - things like drop towers and fast pulse-laser melting can easily reach 100's of K of supercooling of liquids below the melting temperature.

@JonCuster That I know, the records for supercooled water are falling faster than one can keep up with them. But supercooling gets increasingly (sic) easier, as everything slows down, supersaturation does not advantage itself like that.


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Understanding formulas for PZT
In the slides i'm studying (about the production of PZT), is given the following formula for lead zirconate titanate :
$$\ce{PbZr(Ti)O_3}$$
with the explanation:

It is a solid solution of two oxides: lead zirconate $\ce{PbZrO_3}$, and lead titanate $\ce{PbTiO_3}$

Later on, it's used an other formula (acceptor doping: $\ce{K^{1+}}$ replacing $\ce{Pb^{2+}}$ site)
$$\ce{(Pb_{1-x}K_x)(Ti,Zr)(O_{3-x/2}V_{O\,x/2})}$$
where $\ce{V_O}$ indicates oxigen vacancies
even later (acceptor doping: $\ce{Fe^{3+}}$ replacing $\ce{Ti^{4+}}$/$\ce{Zr^{4+}}$ site)
$$\ce{Pb\{(Ti,Zr)_{1-y}Fe_y\}(O_{3-3y/2}V_{O\,3y/2})}$$

Now, what's the meaning of parenthesis in this context?

Is this the standard notation to indicate solutions? In this case, which is the general rule? Because it seems that quite multiple ways were used here..

In chemistry, much like in math, parentheses can mean a lot of things. The multiple notations for solid solutions are used more or less interchangeably. As for the oxygen vacancies in the formula, they are needed because oxygen vacancies exist in the real compound. Also, charge balance.

So you say there is not a standard way? The problem arises because for example from the first i would understand that $AB(C)D$ means: '$ABD$ in solution with $ACD$', and from the second one that $A(B,C)D$ means: '$ABD$ in solution with $ACD$', but if now i use this understandings in the third then $A((B,C)D)(EF)$ would mean: '$A$ in solution with $BD$, $CD$ and $EF$', which is certainly not the case!

Lol what's the meaning of -1? Please, if the question is found not clear just propose some improvements

Your parens/braces don't match for the Fe3+ case. But, in general, the point is to group what is on various distinctly different crystal sites. So, you have (Ti,Zr) meaning they occupy that site. But {(Ti,Zr),Fe} you now have a different charge state ion (Fe) that somehow fits onto the (Ti or Zr) site, leading to changes in the oxygen site occupancy. It is pretty clear once you know what has to go where, and can't go over there.

Ah ok now i start to understand, thanks! So the parenthesis used for ${(Ti,Zr) Fe}$ have also the meaning of substitution and they should have a comma probably? i.e. ${(Ti,Zr), Fe}$

But why the first formula is not just written as $Pb,(Zr,Ti),O_3$?


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Cram's rule vs Felkin Anh's model
I am getting two different products based on the Cram and Felkin-Anh models as shown in the diagram. Any explanation?

If they always gave the same answer, we wouldn't need two models...

@Zhe it’s actually Anh :-)

@orthocresol [face palm]

@orthocresol Of all the references in the Wikipedia article on asymmetric induction to "Anh," one of them was misspelled and for whatever reason, that's the one that I found on my search. Obviously, the correct thing to do was not to fix it here but to fix it in Wikipedia...

@Zhe, I had to check to be sure (when I was making my first comment), and I found plenty of websites with Ahn, so it seems pretty common...

The Newman projection you have used for the Felkin-Anh model is -the one giving minor diastereomer. Attack on the less hindered side on the appropriate Newman projection would give the same product as the cram model.


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work and chemical bonds
If breaking bonds is endothermic (takes energy), and work is force over a distance-
Would breaking bonds require "work"? Or do we only need heat energy? I'm trying to understand how to relate work to chemistry.

Macroscopic heat means microscopic work, that's really one and the same, only perspective is different.

You start with a very narrow definition of work.

When modeling a Diesel engine you might think of the compression step as being adiabatic, in which case you are using work to heat the gas which ultimately results in combustion. As matt_black suggests, you have to consider who is doing the work.