text:
NMR definition and meaning
honestly, I have seen a lot of problems with "NMR" spectrums,
I looked for this title in Charles Mortimer's general chemistry book,
but I didn't see any sort of data, can anyone unfold what does NMR spectrum consist of?
thank you very much

When a radiation like the light is sent on some colored substance, part of the light goes through, and the other radiation (other colors) are absorbed. The energy of the absorbed radiation is used to push orbital electrons to a more excited orbital, may be at a greater distance from the nucleus.
The visible light is characterized by a wave length situated between $400$ and $700$ nanometers. Radiations having shorter wavelength exist but they are not visible with our eye : it is ultra-violet light. Radiations having longer wavelength exist too, but they are also invisible : they are called infra-red. Unfortunately they don't have enough energy to excite electrons. Infra-red rays are just sufficient to make the chemical bonds vibrate more strongly. Their wavelength are about $10 - 100$ micrometers. At bigger wavelength ($0.1 - 10$ cm), the energy of the radiations is too weak to excite vibrations. They are called micro-wave and they are just sufficient to promote rotations.
At still longer wavelengths (about $0.3$ to $10$ meters), the energy is too weak to promote vibrations or rotations. The radiations are called radio waves and all they can do is to change the spin of the protons in the center of the Hydrogen atom (and of some other atoms). But as half of the H atoms have spin up and half spin down, they have first to be all (or nearly all) alined in the same spin state for the radiation to produce a change of this spin orientation. This alinement is done by a strong magnetic field. The substance is then placed between the North and the South pole of magnet. And a radiation is sent on the sample, whose energy (or wavelength) can be changed at will, as the color is chosen with the visible light. The intensity of the absorbed radiation is measured versus the wavelength or the energy the incoming radiation. This gives a spectrum, called nuclear magnetic resonance, because the absorption is due to the nucleus, and that it is due to the interaction of a magnetic field.
Sorry for the specialist reader : the previous explanation has been over-simplified. May be too much ! Sorry in advance !

There are atom nuclei with a non-zero magnetic spin, for example $\ce{^1H}$ (the most frequently seen isotope of hydrogen), but not $\ce{^2H}$ (deuterium).  You may think about these atoms like small magnets.  The orientation of these little magnets in space does not matter until these are brought into a strong magnetic field of the NMR spectrometer; here, they either orient parallel or antiparallel in respect to the magnetic field applied externally to the sample.  Thus the «N» in «NMR».
The «MR» for magnetic resonance is derived from the possibility excite these little magnets by radio waves; these «atomic magnets» then may start to wiggle, and eventually, a detectable portion of these magnets' orientation flips from «up» ($\uparrow$) to «down» ($\downarrow$) in respect to the outer magnetic field at will.  Thus, you record the number of these atomic nuclei (magnets) participating in these transitions by their decay signal, when relaxing to the initial state; again, this is in the frequency range of radio waves.  (NMR is not about radioactivity, neither bonds are broken / newly formed, nor new elements generated.)
For one, the energy required to yield these magnetic transitions depend on the experimental setup; thus -- by convention -- the abscissa traces a dimensionless number (ppm) accounting e.g., for the outer magnetic field strength the spectrometer may apply to the sample.  (But note, the scale has direction, this real number may be either positive, or negative.)  For two, and thus so useful to elucidating structures (both organic, as well as inorganic samples), the transition energies depends both on the nature of the nuclei examined (e.g., $\ce{^1H}$, $\ce{^{13}C}$, $\ce{^{19}F}$, $\ce{^{31}P}$), and their chemical environment.

These little magnets experience interaction with nuclei of the same sort (e.g., coupling between $\ce{^1H}$ and $\ce{^1H}$ in proximity), which may yield typical coupling pattern of diagnostic value.

Atom nuclei are surrounded by an electron cloud, which shields the atom nuclei from the spectrometer's magnetic field.  Atoms nearby the one examined may increase or decrease this electron density, relatively to a neuter, unbound and isolated atom.  (Possibly you heard about electron donating / electron withdrawing groups, or how difference in electronegativity yields polarized molecular bonds).  This equally influences the energy required to yield such a flip, thus shifts the signal of said atom in the spectrum, too.  Thus you get to know, e.g., protons bound to a benzene are typically recorded around $\pu{7.26 ppm}$, while those bound, e.g., to an alkine around $\pu{1.7 ppm}$.

Beside the «typical suspects» in Wikipedia (English, simple English), the ChemistryLibreTexts offer a textbook like approach to the large topic.


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Isn't the change in entropy zero for a reversible process?
I'm calculating $\Delta G$, $\Delta H$ and $\Delta S$ when an ideal gas expands isothermally and reversible at $20$°C from  $5$ MPa to $1$ MPa. I used the relation that:
$$\left(\frac{\partial G}{\partial P}\right)_T=V$$
This relation can be shown from the defintion $G=H-TS$. Then I integrated G so that:
$$\Delta G=\int_{P_1}^{P_2}VdP$$
We have an ideal gas $=>V=\frac{RT}{P}$. Which gives us that:
$$\Delta G=\int_{P_1}^{P_2}\frac{RT}{P}dP=RT\ln\frac{P_2}{P_1}$$
If you calculate this you'll get that $\Delta G\approx3.9$ kJ/mol. This was correct.
However I then tried to calculate $\Delta H$. Since $dH=dQ+VdP$ and $dQ=TdS$:
$$dH=TdS+VdP$$
After this we know that we have a reversible process which according to me means that $dS=0$. Which means that $dH=VdP=dG=>\Delta H=\Delta G$. However the answer was that $\Delta S= 13.38$ J/(Kmol) and $\Delta H=0$. How is this possible? Isn't $dS=0$ for a reversible process?

Delta S is zero for the combination of system plus surroundings, not for each individually.

As mentioned in the comments, there is an entropy change that occurs in the system, however, when combined with the entropy change of the surroundings, is zero.
When solving for the change of entropy in the system, you can use the equation below,
$ \Delta S = R \ln(\frac{P_1}{P_2})$
converting MPa to atm and plugging in, we get
$ \Delta S = 8.314 \ln(\frac{49.3}{9.87})$
which yields an entropy change that is equal to the answer you provide.
The above equation can be rearranged to have different variables instead of pressure. Here is a link for a ChemLibretexts article on entropy and reversible processes if you're interested.

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Entropy of a van der Waals' gas
I would like to ask how I could derive the entropy change $dS$ for a van der Waals' gas.
I think I would need to use the total differential of entropy and then use the Maxwell relation shown below.
Firstly, I would write down the total differential for $S$.
$$dS = \left(\dfrac{\partial S}{\partial V}\right)_T dV + \left(\dfrac{\partial S}{\partial T}\right)_V dT$$
Then I would substitute $\left(\dfrac{\partial S}{\partial V}\right)_T$ by $\left(\dfrac{\partial P}{\partial T}\right)_V$ and $\left(\dfrac{\partial S}{\partial T}\right)_V$ by $\left(\dfrac{C_V\ \partial T}{T\ \partial T}\right)$.
But then I am stuck. May anyone gives me some hints on how to derive $dS$ for a van der Waals' gas?

Oh yes, I am sorry. Let me change it back. Thanks!

Derive the partial derivation dp/dT  atV=const for the vdW equation. Be aware it will not be probably explicitly resolved.

But then I do not have $C_V$... That's where I am stuck.

You can derive it from the experimental Poisson coefficient for real gasses $\gamma=\frac{CV+R}{CV}$. Or use tabelized heat capacities. But all the task may  easily lead to numerical solutions only.

BTW you should always include all relevant circumstances into the question, to avoid guessing, wrong assimptions and clarifications.

You need to know Cv which, for a Van Der Waals gas is independent of specific volume.  So it would be the same as the ideal gas Cv.

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Why trans-1-chloro-4-methylcyclohexane have a higher relative energy than trans-1-chloro-2-methylcyclohexane?
Both can be A,a or e,e, in 1,4 the groups are further apart -> less steric strain, why is 1,4 less stable according to experimental values?
1,2 high energy conformer's relative energy (kj/mol) is 4.05 and
1,4 high energy conformer's relative energy (kj/mol) is 4.50

It is not clear what you are comparing, what are the sources of your numbers and why you think that the 1,4-isomer is higher in energy. The e,e-isomers should be compared. The 1,4-e,e-isomer is 0 kJ/mol while the 1,2-e,e-isomer has a value of whatever a gauche CH3/Cl is worth. Both isomers have the same a,a energy which is 9,7 kJ/mol. If you mean why is the difference in energy between the e,e and a,a conformations for both isomers greater for the 1,4-isomer than the 1,2, then the difference is the gauche interaction in the 1,2-isomer. I don't think that you asked that question.

trans-1-Chloro-4-methylcyclohexane (1) has two chair conformations, 1-ee and 1-aa. Conformation 1-ee has two non-interacting functional groups, chlorine and methyl, both of which have anti-relationships with the $\ce{C2-C3 and C5-C6}$ bonds of the ring. Accordingly, the energy of this conformation is set at 0 kJ/mol. There is no need to try to set some absolute value to the energy because the only issue is how 1-ee compares with conformation 1-aa. Conformation 1-aa has both substituents axial in gauche-relationships with the aforementioned C-C bonds. This conformation behaves as a composite of axial methylcyclohexane and chlorocyclohexane. Its energy is merely the sum of their respective A-values, 9.7 kJ/mol.
Diaxial conformation 3-aa of trans-1-chloro-2-methylcyclohexane has the same energy value of 9.7 kJ/mol because both groups are axial and non-interacting with each other. However, the diequatorial conformation 3-ee has a gauche interaction between the chlorine and the methyl group. The magnitude of this interaction may be estimated.
The value of the $\ce{CH3/CH3}$ interaction in conformation 4-ee of trans-1,2-dimethylcyclohexane may be estimated by dividing the energy of 4-aa by 4 or by halving the A-value for a methyl group. The value of 3.8 kJ/mol is in accord with the difference in the heat of combustion of dimethylcyclohexanes 2 and 4. Indeed, the heat of combustion of 4 is greater than 2 by ~5 kJ/mol.  [https://nvlpubs.nist.gov/nistpubs/jres/39/jresv39n1p49_A1b.pdf
]. Similarly, division of the energy of 3-aa by 4 gives an estimated value for the $\ce{CH3/Cl}$ gauche interaction of 2.4 kJ/mol. The heat of combustion of 1,2-isomer 3 should be slightly greater than that of 1,4-isomer 1 by ~2.4 kJ/mol. In the event, trans-1-chloro-4-methylcyclohexane is more stable than trans-1-chloro-2-methylcyclohexane by ~2.4 kJ/mol. By more stable is meant, more negative heat of formation and smaller heat of combustion.


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Why is Tyrian purple called 6,6′-dibromoindigo?
If the two bromines are attached at opposite ends of the molecule, why are they both labelled with a 6?
Also, why is 6,6′-dibromoindigo not called 1,1′-dibromoindigo if carbons in a ring are numbered according to their substituents, and bromine is the substituent, therefore the carbon it is attached to should be called 1, not 6?

Because of Indole numbering: N is 1 and the other atoms (except the C that bridges both rings) are numbered in ascending order, so the Br substituent is at position 6 in both rings.

Br atoms are not both numbered 6. The other one is 6', where the apostrophe means the same position but on the other indol structure.

Beside the above comments, rings numbering is always, although "fixed" and coded, quite a "mess".


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Why is Bonding between H and Cl not considered as H bonding?
Cl is an electronegative element and hydrogen is electropositive. Why are intermolecular interactions involving H and Cl not considered as H bonding? I read it in a book but there was no reason given for it.

H bonds have their bond energies in the order of ~10-20 kJ per mol. The bond energy of H-Cl is ~427 kJ per mol, which is too strong to be a H-bond

Why does H bonding depend on bond energy?

I am asking this because there are the molecules needed to form H bond.Then why does the bond energy I.e the energy required to break their bond matters here?

Also Cl has less charge density.How is the bond so strong?

Please see the definition of Hydrogen Bond here (the first paragraph).

I did read it .

Cl is more electronegative than N and less than O.

You can't have hydrogen bonds as the basis for forming the molecule itself, there has to be at least one covalent or ionic bond inside the molecule. The hydrogen bond then features as an electrostatic interaction between a bonded hydrogen and another sufficiently electronegative group, which will act as the H-bond acceptor

What do you mean by inside the molecule?

Cl and H are forming polar covalent bond.

A hydrogen bond by definition needs three centers to be formed: the donor atom(Dn), the hydrogen itself (H) and an acceptor atom (Ac) to form a setup like Dn–H···Ac .A bond between H and Cl has only two centres: the hydrogen(H) and chlorine(Cl). Without a third center to act as the donor or acceptor (depending on the convention you have followed) you can't have a "hydrogen bond" per se, it will just be a regular covalent or ionic bond

H20 has bonding because there is one Ac(O) and one Dn(H) with H.Does it mean there have to be 3 atoms for H bonding to form.

Yes, in each hydrogen bond of H2O, you can trace out three centres : O-H···O, where the O from one water molecule acts as the donor and also the first center, the hydrogen attached with donor oxygen forms the second center, while the oxygen from another water molecule as an acceptor and the third center

Ok thank you very much.Just one question , there are no 2 shared pair of electrons in Cl to share with HCL whereas in H20 O could bond with 2 H atoms.IS THIS RIGHT TO SAY?

@YusufHasan, you basically have an entire answer there. Can I suggest converting it to an answer?

@Mithoron, surprised you didn't choose one of the better questions linked on that one. :-)

@orthocresol I am afraid that the meaning of the question is slightly different than what I originally assumed after looking at the edit. I believed the OP was asking why is the intramolecular covalent bond b/w H and Cl not called an H-bond, but after the edit it seems that the question was about the intermolecular interactions between multiple HCl molecules. Hence I feel my answer above was not what the OP wanted

@YusufHasan, hmm... that isn't OP's edit so arguably it doesn't tell us anything about what OP intended. In any case, I see what you mean.

@orthocresol Maybe closing as unclear would be better idea...

As per my teacher, the reason we say that Chlorine does not form hydrogen bonds, even though its electronegativity is almost same as N (both have electronegativity values around 3.0 as per the Pauling Scale) is because :
Chlorine being a third period element,
has a relatively larger size as compared with Nitrogen and Oxygen, so it has a quite large and more diffused electron cloud.
Therefore the dipole-dipole interactions in HCl are not as strong as the ones encountered in HF and H₂O, and so it wouldnt be quite accurate for them to be called as Hydrogen Bonds, as the ones in HF and H₂O are much stronger.
Hope this answered your question.

This is not the inherent reason why the bonding between H and Cl is not considered hydrogen bonding, as on the Pauling scale , the electronegativity comes quite close to N, which is often involved in H bonds. The reason in this case has more to do with the definition of what we call a hydrogen bond, check my comments above for more details on the same. I believe that you have answered something else, while the OP has a misconception on a different aspect

As to why HCl might not show H-bonding in an extended network at low concentrations,yes, your reasoning could work as explained here in a related question. Although do note that clusters of HCl in a concentrated form may be obtained as shown here.As I said, I believe you have answered something else, and the OP was confused on a different point.Please modify accordingly

I do believe this was the actual question which the OP intended to ask...! I myself have had this doubt before.

I think the question is ambiguous enough to admit both interpretations. I wouldn't downvote just based on that. But also, this question has been addressed elsewhere on SE.


text:
In the DNA backbone, which bonds exactly are considered ester bonds?
Following are the three different answers to this question that I reached by googling. Which of them is right?

The covalent bond between the 3' C and the O, and the covalent bond between the 5' C and the O:

2. The covalent bond between the P and the O that is bonded to the 5' C, and the covalent bond between the P and the O that is bonded to the 3' C (deduced from the following image):

3. The -O- (two covalent bonds with an O) that link between the P and the 5' C, and the -O- that link between the P and the 3' C (deduced from the following image):

Maybe my confusion stems from a more general misunderstanding:
When two alcohols are linked through an ester linkage (i.e., both OH groups are replaced with an -O-), we get R1-O-R2. In this case, do we say that each of the covalent bonds with the O is an ester bond? Or is the whole -O- considered a single ester bond?

(Note that I am not asking what exactly a phosphodiester bond is, which is (if I understand correctly) everything that connects the 5' C and the 3' C of adjacent nucleotides.)

There's no such thing as "ester bond", there's ester group.

@Mithoron That's very helpful. Thank you! Could you provide a reference? The internet seems to be full of people that think that "ester bond" is a thing...

@Mithoron:  By the group, you mean O atom, right ? Like in $\ce{CH3-O-SO2-O-CH3}$.
IMHO, $\ce{COO}$ is a group, $-\ce{COO}-$ is a bond,  $\ce{O}$ in $\ce{R-O-X}$ is a group, $-\ce{O}-$ is a bond.

@Poutnik Biochemists may call that "a bond", but from chemical point of view it's not correct.

@Mithoron Who says a bond has only one meaning ? In fact, it has multiple meanings. We can be both true, speaking about different terms of the same name, that are implicitly defined by their context.

I.e. the scope of bond meanings goes from as small as gluons to as big as galaxy clusters. I would limit it to the interatomic bond only if we speak about interatomic bonds. In this case, it is the bond of molecular structures.

All of them are right, regarding ester bonds themselves. But only one of them is right, regarding DNA context. AMP/ADP/ATP/GMP are based on ribose ( see RNA ), not on deoxyribose ( DNA ).
Phosphoric acid has 2 ester bonds in DNA.
You have somehow confused an ester and an ether bond.
$\ce{R_1-O-R_2}$ is an ether bond,
$\ce{R-O-X}$ is an ester bond, where $\ce{X}$ is a carbonyl ( from carboxylic acids or carbonic acid ) or some nonmetal like $\ce{S}$ or $\ce{P}$ for esters of inorganic acids like $\ce{(CH3O)2SO2}$.
If we are to consider it as a bond or a group depends on the point of view. It is both.
It is a group, if we focus on the $\ce{CO-O}$ atom group of esters of organic acids or $\ce{O}$ for mineral acid esters.
It is a bond ( or a bridge ), if we focus on what the group connects.

Thank you. What about the general misunderstanding I mentioned in the end? Would you say in this case we have a single ester bond or two ester bonds? Or are both acceptable?

"R−O−X  is an ester bond" - so are you saying that the ester bond is the -O- that links between the R and the X? (by the way, a comment to my question says there is no such thing as "ester bond"..)

@OrenMilman The answer has been updated.

I will first talk about peptide bonds and ether linkages before circling back to the question. All of these (and the ester group) can be synthesized by a condensation reaction, and the compounds containing these can be split in two by hydrolysis.
Peptide bonds
Peptide bonds are a clearly defined term in biochemistry. The peptide bond is the bond between the carbonyl carbon of one amino acid and the main chain nitrogen of the next amino acid in the protein sequence. When you hydrolyze the peptide bond, you end up with a carboxylic acid group (or carboxylate, depending on solvent and pH) on the one hand, and an amino group (or ammonium depending on solvent and pH) on the other hand.
Ether linkages
In contrast to the amide functional group in proteins, which are hydrolyzed into two different functional groups, ethers are symmetric in the sense that hydrolysis yields two alcohols. Which bond breaks (or where the bridging oxygen ends up) depends on the specifics of the ether and the reaction conditions. Therefor, it is hard to call one or the other bond of the bridging oxygen the "ether bond", it could be either (pun intended).
The situation is different for glycosidic bonds (often incorrectly described as ethers). If you hydrolyze them, you get a hemiacetal and a hydroxyl group (for e.g. a 1,6 linkage in cellulose or starch). In some cases, like sucrose, both sugars use the hemiacetal group in the condensation reaction, and you could describe the linkage as containing two glycosidic bonds.
"Ester bonds"
If you want to be exact in describing the functional group called ester, you should give the name of the acid that is obtained after hydrolysis. For example, there are carboxylic acid esters (like in triacyl glycerides) and phosphoric acid esters (like in nucleic acids). When ester hydrolize, both products carry an oxygen, so like with esters, you have to figure out where the bridging oxygen goes. This depends, again, on mechanism influenced by reaction conditions and the nature of the ester, but mostly it stays with the alcohol product.
So even though ester bond is not a technical term, you will find it used (even on this site), and for a phosphate ester, it would typically be used to talk about the pond between phosphate and the bridging oxygen atom. By that logic, a phosphodiester would have two ester bonds. For the three possibilities the OP offered, number 2 would be the best one because it describes which atoms end up with which hydrolysis product (for the typical hydrolysis mechanism).


text:
Selection of parent carbon chain for IUPAC naming

How to select the parent carbon chain for IUPAC naming of following compound?

I think the red numbering is correct, but in the answer key, they have used the blue numbering.
Which is correct?
My question is regarding selection of parent chain in an alcohol. While selecting the parent carbon chain (base name) to write the IUPAC name of an alcohol, do we have to give preference to the one which gives the lowest locant to the $\ce{–OH}$ group (viz. primary functional group here), or the one which has the maximum number of carbon atoms?

MarvinSketch program gives 2-bromo-3,6-diethylnonan-5-ol

When assigning numbers to carbon atoms in an aliphatic skeleton, your first priority must be the longest carbon chain. The chain with red numbers is one carbon atom smaller than the one with blue numbers. Blue - 9, red - 8.
It's also helpful to know that when you have two carbon chains of the same length, the one that's the most branched must be assigned the highest priority. Here we have exactly this case, where two chains of the same length can be found. One of them has 4 branches (2 ethyl radicals, 1 alcohol and 1 bromide) and the other one has 3 branches (1 hydroxyl, 1 ethyl and 1-(1-bromoethyl)).


text:
How to balance the reaction between ethanol and chlorine using oxidation number method?

Balance the following chemical equation using oxidation number method:
$$\ce{C2H5OH + Cl2 -> CHCl3 + HCO2- + Cl- + H2O}~\text{(In basic medium)}$$

Chlorine in $\ce{Cl2}$ is in oxidation state $0$ and it is changing from $0$ to $-1$ in both $\ce{CHCl3}$ and $\ce{Cl-}.$
$\ce{C2H5OH}$ has two carbons and both of them have different oxidation states: one has $-3,$ the other has $-1,$ and they are changing to $+2$ in both the products $\ce{CHCl3}$ and $\ce{HCO2-}.$
So, it is definitely a redox reaction and thus I want to use oxidation number method to balance this equation.
I am finding it hard to balance because of multiple products of $\ce{C}$ and $\ce{Cl}$ in the same equation.

This problem is rather difficult to solve, because some hypotheses have to be made, as will be explained in the following text.
Let's first admit that, in the molecule $\ce{C2H5OH}$, the two carbon atoms are, on average, at oxidation state $-2$. It helps solving the problem. By chance, these carbon atoms are going to the same oxidation state $+2$ at the end, both in chloroform $\ce{CHCl3}$ and in the formate ion $\ce{HCOO^-}$. So each carbon atom has a leap of $+4$ in their oxidation number. In total it makes a leap of $+8$. And this may be carried out by $\ce{Cl2}$ if $4~\ce{Cl2}$ do the job,  introducing $3$ $\ce{Cl}$ atoms in the $\ce{CHCl3}$ so that the $5$ remaining chlorine atoms are transformed into chloride ions. We also have to admit here that the first carbon atom will always produce $\ce{CHCl3}$ and that the second carbon atom will always producing the $\ce{HCOO-}$  ion. As a consequence the relative amount of $\ce{CHCl3}$ and $\ce{HCOO-}$ are equal. At this step the equation is $$\ce{C2H5OH + 4 Cl2 + ... -> CHCl3 + HCOO- + 5 Cl- + ...}$$ The problem of the oxidation is solved, but some oxygen, hydrogen, and charges are still missing. Six OH- ions have to be added at left to satisfy the stoichiometry. It gives the final equation $$\ce{C2H5OH + 4 Cl2 + 6 OH- -> CHCl3 + HCOO- + 5 Cl- + 5 H2O}$$ I would be interested to know if this problem can be solved without the present hypotheses.

Not that it is substantially different from what you did, I wrote a half reaction where ethanol is converted to 2 hydrated forms of formic acid, CH(OH)3. After balancing O and H, there is a loss of 8 electrons that requires 4 Cl2. Now substitute 3 Cl- for 3 OH's in one hydrate, lose water from the other and add OH- to both sides of the equation to allow deprotonation of formic acid and increase the amount of water on the right by one equivalent. I arrive at your answer.

@User55119. The author of the problem did not want half reactions. He wanted to solve the equations with oxidation states.

Using whole molecules as oxidizing or reducing agents is my favorite method for dealing with redox reactions where atomic oxidation numbers are not straightforward.  Here we use ethanol as a reducing reagent.  With that let us design our half reactions:
Oxidation of ethanol to formate ion
We suppose ethanol is oxidized to give formate ion according t
$\ce{C2H5OH\to 2 HCO2^-}$
where we have balanced the carbon atoms.  Introducing the water, hydroxide ions (in basic solution), and electrons we would have the fully balanced equation
$\ce{C2H5OH + (a)OH^-\to 2 HCO2^- + (b)H2O + (c)e^-}$
We expect the hydroxide ions to be on the left to balance all the negative charges on the right.  To balance the hydrogen atoms we need $a+6=2b+2$, for oxygen we need $a+1=b+4$. Thus $b=7,a=10$ and when we put these results in the charge balance will give $(-10)=(-2)-c$ so $c=8$.  Then
$\ce{C2H5OH +10 OH^-\to 2 HCO2^- +  7 H2O + 8 e^-}$
Reduction of chlorine
This is straightforward sincevinly chlorine atoms are reduced:
$\ce{Cl2 + 2e^-\to 2 Cl^-}$
Combine this with the oxidation half-reaction in the usual way and you have a balanced equation ... without the chloroform.
What happened?
We have a combination of two reactions, one oxidizing the alcohol to formate ion and the other oxidizing it to chloroform.  We balanced just the oxidation to formate ion.  Let's now look at the component where the alcohol is oxidized to chloroform.
Oxidation of ethanol to chloroform
Again we use ethanol as axwhole-molecule reducing agent.  We must combine it with some chloride ions to balance carbon and chlorine, leading to an equation with the form
$\ce{C2H5OH + 6 Cl^- + (a)OH^-\to 2 CHCl3 + (b)H2O + (c)e^-}$
Balancing hydrogen requires $a+6=2b+2$ (again), and the oxygens give $a+1=b$.  So $b=3,a=2$ and the charge balance then gives $(-6)-2=-c$ or $c=8$:
$\ce{C2H5OH + 6 Cl^- + 2 OH^-\to 2 CHCl3 +  3 H2O +  8 e^-}$
Combine this with the chlorine reduction given earlier and you have the balanced equation for the oxidation to chloroform.  Note that in both component reactions each ethanol molecule gives off the same number of electrons and therefore should react with the same amount of chlorine.  Thus the ratio of chlorine to ethanol is independent of how you combine these component reactions.  (The actual value should emerge when you combine each of the oxidation half-reaction with the reduction.)
How to combine the two balanced equations
So you have two balanced equations, one with formate and one with chloroform.  How do you combine them? Since any combination satisfies all the laws of physics, the answer is you can't with just pencil and paper.  You have to do an experiment to determine the relative amounts of formate and chloroform you get, and match your combination of the two equations with those relative amounts.  Don't forget to include your reaction conditions, as changing conditions could easily change the formate/chloroform ratio.

The key  is to remembering that  $n \cdot m = m \cdot n$.
It seems obvious, but you would not believe how much struggles people suffer when trying to apply this trivial equation to chemical reactions.
In this special case, The total change of oxidation numbers is zero, otherwise, it would mean we have lost some electrons somewhere.
$$\sum{n_i \cdot \Delta ON_i } = 0$$.
The stoichiometric coefficients are in the reciprocal ratio to ( absolute value of ) the oxidation number change ratio.
So the sum of the oxidation number change of carbons in 1 ethanol molecule is $( (+2 - -3) + ( +2 - -1) ) = +5 + +3 = +8$
The sum of the oxidation number change of chlorine in 1 chlorine molecule is $2 \cdot (-1 - 0) = -2$
We have then:
$$n_{\ce{C2H5OH}} \cdot 8 + n_{\ce{Cl2}} \cdot -2 = 0$$
$$n_{\ce{Cl2}} = 4 \cdot  n_{\ce{C2H5OH}} $$
Note that you are missing negative ions of the left side for charge balance, also overall mass balance is missing.

Enumeration of stoichiometric coefficients have also a general, analytical mathematical solution, but personally, I never used it, as it would take much longer time than just enumerating it on the fly by the chemist's heart:

You are solving a set of linear algebraic equations for multiple variables.
Each variable is a stoichiometric coefficient for a reagent or a product.
Each equation represents some kind of balance.
1 equation for the atom count balance for each of the elements
1 equation for the charge balance, if applies.
In most cases there will remain 1 degree of freedom, so 1 stoichiometric coefficient must be chosen and some others will depend on the choice.
E.g. there can be $\ce{n H2 + n F2 -> 2n HF}$, so we choose $\ce{n=1}$ for $\ce{H2}$  and rest will follow.

But then according to the oxidation number method, we should multiply 4 on the product side(CHCl3 and Cl-) also, and thus we won't get a balanced chemical equation. If we follow the steps.

I have intentionally not completed the task, as we try to help, we do not solve homework-like tasks on behalf of the question author. What do you think is the reason we should multiply stoichiometric coeefficients of both products by 4 ? It applies just for the ethanol:chlorine ratio.


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Dissolving Polystyrene with D-Limonene
I have a product, which according to the MSDS is mostly D-Limonene. I don't have the means to chemically test it, but I did try dissolving polystyrene with it. Before I come check it again, should I expect how polystyrene "dissolves" in acetone? As a goop at the bottom, even as I add much excess acetone? Or, should I see it truly dissolved, a translucent, thick liquid, as with using MEK, or xylene?

Limonene and acetone do dissolve polystyrene.

Time to get it dissolved should depends on the batch as well as the form (foam, grains, pellets...).

Foot note: After squeezing lemon into tea in a Styrofoam cup, one's hands stick to the cup as the limonene, among other "essential oils", dissolves the polystyrene.


text:
Does the octet rule only apply to the second and third shell or does it apply further?
In my textbook, examples for the octet rule are only given for up to the third shell.
What about the fourth and fifth shell?
I still get confused over the fact that the $n=3$ shell has 18 electrons yet for chemical purposes atoms seek out 8 electrons. My understanding is that the energy difference between the 3p and 3d orbitals are just so large that it creates a certain stability when the 3p orbital is full. If that is true then things should get messier for the fourth and fifth shells due to more overlapping.
The first shell contains the 1s orbitals. The second shell the 2s and 2p. The third shell the 3s and 3p. What orbitals do the fourth and fifth shells include and how many electrons fill these shells?
Another way of viewing my issue. Take the electronic configuration of Calcium which is 2,8,8,2. There are 2 electrons in the fourth shell. What is the maximum number that can go in that fourth shell before a fifth shell is introduced?

The shell concept is useful rather for 1-electron hydrogen-like atoms, where energies of ns, np, nd, nf orbitals are all the same ( well, not fully true due the very tine QED effect of interaction of electrons with quantum vacuum ).
For multielectron atoms, energy wise, $\mathrm{nd}$ orbitals from originally nth shell belong now rather to the (n+1)th shell. Similarly, $\mathrm{nf}$ orbitals from originally nth shell belong now rather to the (n+2)th shell.
So the octet rule for the 4th and higher periods have limited applicability for the $\mathrm{s}$ and $\mathrm{p}$ orbitals only for groups 1-2 and 13-18. It is not applicable for transition metals involving partially occupied $\mathrm{d}$ or $\mathrm{f}$ orbitals.


text:
How is temperature constant in enthalpy change formula?
I have learnt that temperature sort of indicates the internal energy of a system, or at least this is my intuition. So if temperature is constant, then internal energy is constant as well.
But in the derivation of the following formula
$$\begin{align}{
{\Delta H = \Delta U + \Delta n_g RT}}
  \end{align} $$
T doesn't have a delta, which means it's treated like a constant value, but then wouldn't this make $\Delta U = 0$? In fact most questions in my textbook have the statement "at constant pressure and 298/300 K", but $\Delta U$ isn't 0 despite the temperature being constant.
Am I missing/neglecting something here?

As long as the number of moles of all species is constant, U is a function of T and P only.  But, if chemical reaction occurs, you make and break chemical bonds, and this changes the number of moles of each species.  This causes U of the mixture to change  at constant T and P.

@ChetMiller We have also to exclude spontaneous processes like phase changes ( 1 kg of ice at 1 atm and 0 deg C has dfiferent U than liquid water ) or compound redistribution between phases ( diffenent chemical potentials in phases before equilibrium is reached .

@Poutnik Yes, I agree with this.  But sometimes people treat liquid water as a different chemical species than ice, such that a "reaction" took place.  This is an OK approach too.

Is there a reason why U is a function of T and P only when the number of moles of all species is constant?

The internal energy of a system is in general not only a function of temperature. This means that holding the temperature constant does not guarantee that the energy will be constant. The statement that the energy can be expressed as a function of only the temperature applies to the particular case of ideal gases.
Assuming that gases are ideal simplifies expressions for the change in internal energy and enthalpy. For ideal gases at constant temperature and composition ($\Delta T=0$ and $\Delta n=0$):
$$\Delta U = 0\\\Delta H = 0 \tag{1}$$
These expressions are not general. They apply to ideal gases at constant temperature and composition.
In general,
$$\Delta H = \Delta U + \Delta (pV) \tag{2}$$
For an ideal gas at constant temperature,  $ \Delta (pV)= \Delta (nRT)= RT\Delta n$
so that
$$\Delta H = \Delta U + RT\Delta n \tag{3}$$
It is also worth remembering that at constant pressure $w=-p \Delta V$ so that  equation (2) becomes $\Delta H = q_p$ (in general for any substance, provided p is constant and only pV work is done). But what is $\Delta U$?
We can show from the ideal gas law that it is only a function of T for a constant amount of a pure ideal gas, and, from the kinetic theory of gases, one can also show that for one mole,
$$U_{m,i} = f_iRT$$
where $f_i$ is a constant that depends only on the type of gas (not on thermodynamic variables). Then
$$\Delta U = \sum_i U_{m,i}\Delta n_i = RT \sum_i f_i \Delta n_i $$
for a mixture of ideal gases.


text:
Why do certain compounds dissolve in oil? Why do some types of compounds dissolve in oil, yet not water?
Sorry if this is an obvious answer. I tried rewording this question every way I could, but search engines came up with nothing- they all skirted around this exact question.
Why do certain compounds, such as fatty acid or Vitamin A, dissolve in oil?
Polar substances can dissolve in water because water molecules are polar and literally pick apart the [polar] molecules in the solute by attracting to them. However, oil is nonpolar and has no such ability.
How, then, does it make certain compounds dissolve? Is there some other bond it can form with certain compounds?
On a second question, I read something about London dispersion and Van der Waals forces being the main force of attraction between oil solvents and solutes that can dissolve in oil. If this is the case, that means the electrons in orbit around the solute's molecules shift around and form temporary dipoles with each other.
But what makes it so that some solutes can only dissolve in oil-- and not water [molecules], which are polar, and could also interact with these temporary dipoles?
edit: A comment said that this not energetically preferable. So I think the second question can be considered answered.
Thanks!

Water molecules create mutual relatively strong hydrogen bonds what releases bond energy. Weak bonding with a nonpolar molecule is not energetically preferred.

Indeed is the way things are presented. If I get it, you see an active mechanism for the solubilzation of salt and polar molecules in water / polar solvents but the counterpart for apolar stuff seems to be missing. Indeed, there are other forces at play - still electrostatic in nature although transient. Also entropy favours dissolution. Remember the "rule" like dissolve like.

@Poutnik Thanks. I think that answers the second question. I'm not sure if the question should be deleted now?

No, keep it. Questions and answers are aimed to have permanent value, if others have similar questions in future.

@Alchimista Would the main cause of a solute's dissolution in oils be the Van der Waals forces, then? Or do the other forces you mention play more of a role? Edit: Yes, the counterpart for the apolar stuff is missing because I know very little about how apolar solvents and solutes interact.

@Anise yes let us call them VdW. Their treatment in details is not in the current scope. Look at the problem in light of Maurice answer (ie if something stay togheter there are forces keeping it as an ensemble. While it is easy to see them in ionic crystals or polar compounds, is more subtle to analyse them in apolar molecular lattices. Still they are el.magn. in nature. Also as Poutnik said, there is alway an energetic balance.

I propose a logical approach, to explain the rule "the like dissolve like". Why don't we admit that all liquids should be able to mix, exactly like gases may be mixed in all proportion ? This is the case for non polar molecules, like oils. And if two liquids don't mix, this is due to the existence of attracting forces between molecules A (hydrogen bonds for exemple), and that mixing them with another stuff B may destroy too many of these bonds. If the mixing operation A + B creates about as many new bonds as those broken during the mixing, the liquids A and B may be mixed. This is the case in the mixture water - ethanol. If the mixing operation A+B does not create enough new bonds between molecules A and B, like water and oil, Nature refuses to break the bonds between A molecules without any counterpart, and prevents the A molecules (water) to be separated from one another in order to accept B molecules, and A and B are not miscible.

Thank you. Your answer puts a whole new perspective on my current understanding of how bonds are made. Out of curiosity... is it possible for the mixing operation A + B to create more bonds than those broken? If so, what will happen?

@Anise. If the mixing operation creates more bonds than those broken, this means that a new chemical species has been obtained containing A and B.


text:
Double Bond Geometry
The double bond at position '4' appears to be showing a 'E' geometry because the priority 1's are on opposite sides of the double bond ie top-left of bond4 is '1' (because Oxygen > Carbon)); bottom-left of bond4 is '2'; top-right of bond4 is '2' and bottom-right of bond4 is '1'(because it has more carbons than the other side).
I am unable to see why 'Z' is the correct geometry.
Appreciate any suggestions.
Thank you

Welcome to ChemSE!  The tetrasubstituted double bond is E (CH2OH>CH3CH2 and CH=CHCH3>CH3CH2). The disubstituted double bond is Z (C4>H and CH3>H). BTW: Your numbering is incorrect. The actual name is (2E,4Z)-2,3-diethylhexa-2,4-dien-1-ol. The cmpd. is an alcohol.

Thank you for the welcome and response. Appreciated.


text:
Calculating change in state function following different paths
in thermodynamics, we learn that change in state function is independent of the path/process, and thus we can conveniently pick any arbitrary path that allows easier calculation. I'm TA'ing a class and found that it would be great to actually show students an example where that a simpler path can indeed be constructed to calculate the change of state function. However, I feel that there must be some blind spot that I myself am struggling with. And would like to get everyone's feedback.
Example and detail
We are asked to calculate $\Delta H$ for an isochoric process where a known amount of energy is transferred into the system (of a monatomic ideal gas) as heat. Assuming only P-V work is allowed in the system.
I can first calculate change of internal energy ($\Delta U$) using the first law, and subsequently $\Delta T$ with the known heat capacity (over constant volume) $c_V = \frac{3R}{2}$. Then to calcuate $\Delta H$, I'll follow an isobaric process instead (with $c_P = \frac{5R}{2}$), but it will give the same result. Here's the detail:

$w = -\int_i^f P_\text{ext} dV = 0$ (no work is done at constant volume)

$\Delta U = w + q = 0 + q = q$

$\Delta U (= q_V) = n c_V \Delta T \implies \Delta T = \frac{\Delta U}{n c_V} = \frac{q}{n c_V}$

$\Delta H = q_P = n c_P \Delta T = n c_P \frac{q}{n c_V} = \frac{c_P}{c_V} q$

Question
My question - under the context of this example - is how to make up a path connecting the same endpoints (the initial and final states) that is under constant pressure?
Without loss of generality, assume that $q > 0$, and thus the temperature at the end increases as the internal energy increases (no work is done due to the constant volume constraint). This suggests that the pressure also increases. But I just said that I'd like to construct an isobaric path that connects the initial and final state..., which now seems to be contradicting the fact.
My thought is - that instead of a single isobaric path (which connects the initial state and potentially an intermediate state), maybe I need another segment that connects the intermediate state and the final state? Such that overall the volume stays the same? If this is the approach, what would the strategy be for picking the intermediate state and the two segments making up the simpler path?
Would the following work?

1st segment: isobaric expansion to an intermediate state whose temperature is the same as the final state. The relevant heat is $q_P$ which will contribute to the overall change in enthalpy with its contribution $\Delta H_1 = q_P$.

2nd segment: isothermal compression to the final state (i.e., need to volume to go back to its initial/final state volume). For this part - there is going to be additional heat and work involved, but internal energy and enthalpy should remain unchanged (as same as the intermediate state), resulting in $\Delta H_2 = 0$.

Overall, $\Delta H = \Delta H_1 + \Delta H_2 = \Delta H_1 = q_P$.

I'm not very sure about this as it invokes that enthalpy change is zero for an isothermal process -- which I don't know if it will hold for non-ideal gas? And if not - this suggests my proposal doesn't work universally...
[Edits for fixing typo]

You wrote, in one of your comments, "My biggest question is how to demonstrate the proper use of $\Delta H = n C_p \Delta T$irrespective of the path...".
So:
In a closed system (no flow of matter across the boundaries) of constant composition (no chemical reactions),
$$dH = \left(\frac{\partial H}{\partial T}\right)_ p dT +  \left(\frac{\partial H}{\partial p}\right)_ T dp$$
$$= C_ p dT +  \left(\frac{\partial H}{\partial p}\right)_ T dp$$
$$= C_ p dT +  \left(V - \alpha T V \right) dp$$
$$= C_ p dT +  \left(V - T \left(\frac{\partial V}{\partial T}\right)_ p  \right) dp$$
Thus, at constant $p$,  $dp =0$, and
$$dH_p = C_ p dT \textit{, for all substances.*}$$
It is worth noting, as Chet Miller mentioned in the comments, that $C_p \equiv C_p(T,p)$ (i.e., $C_p \text{ is a function of both } T \text{and } p$ for real substances).
If there are no inter-particle interactions (e.g., an ideal gas), then $V = T \left(\frac{\partial V}{\partial T}\right)_ p $, and thus
$$\left(\frac{\partial H}{\partial p}\right)_ T = 0$$
You can confirm this yourself for an ideal gas by substituting $V = \frac{n R T}{p}$ for the second occurrence of $V$ in: $\left(V - T \left(\frac{\partial V}{\partial T}\right)_ p  \right)$.
Hence, for an ideal gas,
$$dH = C_ p dT \text{, always,} \textit{ even if the pressure isn't constant.}$$
In summary, if the substance is an ideal gas, $dH = C_ p dT$ regardless of whether there is a change in pressure.  If the substance is real, $dH = C_p dT$ only if the pressure is constant.

*Note that this relationship holds for all types of work.  If we add the constraint that the only work is $pV\text{-work}$, then we have the additional nice relationship that:
$$dH_p = C_ p dT = q_p \textit{, for all substances.}$$
If we keep the constant-$p$, closed system, and $pV\text{-work}$-only constraints, but remove the constant-composition constraint (allowing chemical reactions), then it will no longer be the case that $dH_p = C_ p dT$, but it will be the case that:
$$dH_p = q_p \textit{, for all substances.}$$

I think it is worth mentioning that, for real gases, Cp is a function not only of temperature, but also of pressure.  Only in the ideal gas region, in the limit of  low pressures, is Cp a function only of temperature.  So, at higher pressures, the ideal gas heat capacity will not be an accurate representation of Cp.  Also, usually, heat capacity data is only available in the ideal gas limit.  So the equation you developed for the effect of pressure on enthalpy needs to be used to calculate and apply Cp at higher pressures.

@ChetMiller  Good point.  I added that as an explicit note to my answer.

You have teen talking about two different processes with two different final states, two different amounts of heat, and two different amounts of work.  The only thing that's the same is the temperature change, which, for an ideal gas makes $\Delta U$ and $\Delta H$ the same.  If it were not an ideal gas, these two would not even be the same either.
Your mistake is associating the amount of heat q with the heat capacity C.  In thermodynamics, q is dependent on path, and can vary between two different paths between the same two end states, while C is a state function, defined in terms of the path-independent state functions U and H:  $$C_v=\frac{1}{n}\left(\frac{\partial U}{\partial T}\right)_V$$and$$C_p=\frac{1}{n}\left(\frac{\partial H}{\partial T}\right)_P$$
For an ideal gas, U and H are functions only of T, but this is not generally true.
In the specific process you analyzed, once you changed the temperature at constant volume, there was no isobaric path between these same two end states.  The equation of state requires that P=P(V,T), and, once you said that P is constant along your alternate path, for the same temperature change, V would have to change.

Thanks for your reply! I'm aware of work/heat being path-dependent. My biggest question is how to demonstrate the proper use of $\Delta H = n c_P \Delta T$ irrespective of the path, other than proofing the equation is valid under the assumption (only P-V work is allowed)? This led to my thought of wanting to construct an additional path consist of one or more segments with the net heat equal to $n c_P \Delta T$.

It is correct only for an ideal gas.  Is that what you are trying to prove?

$H = U + PV$ so $(dH)_P = (dU)_P + (d(PV))_P = -PdV + \cancel{d}q_P + PdV  = \cancel{d}q_P$. Does this somehow suggest $\Delta H = q_P$ is valid only for an ideal gas? I don't see it, could you point out during which step did I inexplicitly assume ideal gas? (PS - I don't know how to properly make distinction between exact and inexact differential on this platform and thus the weird \cancel command... sorry)

The derivation starts with $dH=TdS+VdP$ and ends with $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P$$ The right hand side of this equation is equal to zero for an ideal gas.  This derivation is in every thermodynamics book.

I'll need to think a little bit more about what you mentioned. But are you saying that $\Delta H = q_P$ is only valid for ideal gases? This sounds a bit confusing to me - because in practice we conduct most experiments under constant pressure and report $\Delta _r H$ for various chemical reactions. It would not be as useful if $\Delta H = q_P$ is indeed true only for ideal gases...?

No.  That's not what I'm saying.  For  both ideal gases and any other fluids, this equation is correct.  This derivation focuses on the other side of the equation:  how $\Delta H$ relates to the change in temperature and pressure.

Bot for a real gas, this equation is true only for a constant pressure path.  Here's the problem:  for a real gas, Cp is a function  of both temperature and pressure, but you only know Cp in the limit of low pressures where ideal gas behavior prevails.  So what do you do if you have an isobaric temperature change at a high pressure and you need to calculate $\Delta H$ and qp?

@ChetMiller  Particularly since he's a TA, I wouldn't recommend referring to heat capacities as state functions, since that is non-standard usage that will confuse his/her students.  Rather, they are inherent properties of the material that can be expressed in terms of a combination of state functions (U and H) and state variables (T, V, p, and n).  They are not themselves state functions.

@theorist Maybe my terminology is not in perfect agreement with your, but I think you know what I meant, and I'm hoping the OP, who is a TA, also got the message.  I was actually shocked that a TA for a thermodynamics course was asking such fundamental questions.

@ChetMiller I taught thermo at a top-10 US chemistry dept., and I've gotten a mix of TA's, ranging from those who knew the material well to those who didn't know the subject at all, but where the Dept. needed to put them in my class because they were the only ones left available. At the same time, I wouldn't be so hard on this questioner, because he's at least showing a deeper interest in the material (unlike some TA's who get put in thermo, don't really understand the subject, and don't want to make the effort to remedy this).

And regarding "state functions", I did know what you meant, but the issue there isn't that your terminology wasn't in concert with mine, but rather that isn't in concert with what is standard in the field (and in pchem textbooks, at least all those I've seen), and thus could lead to confusion, which is a particular issue when teaching new students.


text:
Catalyze dehydration of formic acid
I know that formic acid will dehydrate upon heating to 100°C (under normal pressure) or adding sulfuric acid. What other catalysts can be used to induce this reaction?

Activated alumina, silica, thoriated silica gel, phosphorus pentoxide-silica gel, aluminum phosphate, titanium dioxide
Catalytic Dehydration of Formic Acid, E. G. Graeber and D. S. Cryder
Industrial & Engineering Chemistry 1935 27 (7), 828-831, DOI: 10.1021/ie50307a019
Mechanism of dehydration decomposition of formic acid over alumina and silica. Adsorption measurements during the catalytic reaction, Kenzo Fukuda, Yuko Noto, Takaharu Onishi and  Kenzi Tamaru, Trans. Faraday Soc., 1967,63, 3072-3080, DOI: 10.1039/TF9676303072

Mixed titanium-vanadium oxide
Sadovskaya, Ekaterina & Chesalov, Yuriy & Goncharov, V. & Sobolev, Vladimir & Andrushkevich, Tamara. (2017). Formic acid decomposition over V-Ti oxide catalyst: Mechanism and kinetics. Molecular Catalysis. 430. 52-64. 10.1016/j.molcata.2016.12.010

Various homogenous transition metal catalysts
Homogeneous Catalytic Dehydrogenation of Formic Acid: Progress Towards a Hydrogen-Based Economy, Gábor Laurenczy, Paul J. Dyson, J. Braz. Chem. Soc. vol.25 no.12 São Paulo Dec. 2014, DOI: 10.5935/0103-5053.20140235

Monoclinic bismuth chromate hydroxide proto-catalyst, $\ce{m-Bi(OH)CrO4}$
Two-step catalytic dehydrogenation of formic acid to CO2 via formaldehyde, Viasus et. al., International Journal of Hydrogen Energy, Volume 44, Issue 3, January 2019, Pages 1534-1543, DOI: 10.1016/j.ijhydene.2018.11.178


text:
Cp - Cv for a real gas
I am trying found a relation between cp - cv for a real gas. I know how to calculate for an ideal gas, but when I try to do the same for a real gas I stopped at some point and I don't know how to continue.
The EOS used is Van der Waals equation.
If someone could help me, please, I am glad for it.

You need to show that $$\left(\frac{\partial U}{\partial V}\right)_T=-P+T\left(\frac{\partial P}{\partial T}\right)_V$$


text:
Partial molar volume of binary mixture
A 2% (weight %) solution of a salt with Mr = 190.6 g mol-1 was prepared at 25 0C. The density of the prepared solution was 1.00926 g cm-3. The density of water at 25 0C is 0.99707 g cm-3, and the density of the crystalline salt was 3.00845 g cm-3.
a) ) What is the molar volume of the crystalline salt?
molar volume of salt=(190.6g/mol)/(3.00845g/cm^3)= 63.4cm^3 /mol
b) what's the partial molar volume of the salt at infinite dilution
assuming the mass of solution =100g, w%:(mass of salt)/(mass of solution) x 100
w%:  (mass of salt)/100g*100= 2%
mass of salt= 2g
mass of water=100-2=98g
no.of mole salt= 2/190.6=0.010493mol
no.of mole of water=  98/18.02=5.4384mol
x_salt=  0.010493/(5.4384+0.010493) = 0.001925
x_water=1-0.001925=0.99881
these are my workings for part a and b. My understanding of part b is to find partial molar volume via dV/dn, but I can't seem to find the difference in volumes, let alone infinite dilutions...I'm not even sure if I'm on the right track. Do I have any misconceptions?
-struggling physical chem student

Please, always use as the first step a general algebraic solution, for 4 reasons.: 1/better grasp of principles for OP  2/ better orientation and spotting eventual mistakes 3/solution reusability and permanent value
4/ The site focus is on helping understanding principles and solution procedures, not on solving particular tasks. Use eventually MathJax for mathematical/chemical expression/formula formatting.

Hint: What would be the solution volume without the salt ?

I see. Sorry about that


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Sodium Sulfate or Sodium Carbonate for electrolysis in water
I am not a chemist but I have a question regarding electrolytes in electrolysis.
I have done a lot of reading but cannot find the answer I'm looking for regarding Sodium Sulfate (or Sulfide)
I intend to clean a film off a Stainless Steel plate, so that will be my cathode.
For my Anode I will be using a Platinum coated Titanium plate.
At a small scale I have been using Sodium Carbonate (Na2CO3) and it has worked fine, however I intend to scale it up to ~1m^2 plates for the Anode and Cathode, as well as 150A @ 15v DC for ~3min in Deionised water with many repeats, so temperature rise of the water should also be accounted for.
I would like to make this as safe/clean as possible and am considering switching out the electrolyte at this stage to something more stable. I have read that Sodium Carbonate can precipitate directly onto the electrodes at high currents. I believe this is not an issue with Sodium Hydroxide but want to avoid that for obvious reasons. I have stumbled upon Sodium Sulfate as an alternative but very little information on its breakdown and potential pollutants. Is there any other that I am missing/suggestions?
I will be using a fume hood, however Sodium Chloride is definitely not an option.
Thanks for your time,
Tea
Edit:
So I have done some reading and have mostly worked out the half equations of Sodium Sulfate (Na2SO4):
Cathode:
(2)H+ + (2)e- -> H2(g)
Anode:
(2)O2- -> O2(g) + (4)e-
Remaining in aqueous solution:
NaOH (aq) + S (?)
As an update then, can I ask what, if anything, happens to the NaOH and Sulfur?
Also, what form will the Sulfur remain in? I assume it will try to re-bond with oxygen, which will be abundant in the system? Will either be outgassed?
Assuming the NaOH is constantly building up, at what point does it become overwhelming in the system and need to be removed?
Thanks so much for reading this far.
tea

If you electrolyze a solution of sodium carbonate, it will be fine at the beginning. Later on, some problem may happen around the anode where $\ce{O2}$ is released. Indeed some acid ($\ce{H+}$) is produced in solution according to the half-equation : $$\ce{2H2O -> 4 H+ + O2 + 4 e-}$$ and these acidic ions $\ce{H+}$  react with the carbonate ions $\ce{CO3^{2-}}$according to $$\ce{CO3^{2-} + H+ -> HCO3^- + H2O}$$ In presence of the ion sodium $\ce{Na+}$, the ion hydrogenocarbonate $\ce{HCO3^-}$ reacts to produce sodium hydrogenocarbonate $\ce{NaHCO3}$ which is not very soluble in water, and may make a white deposit around the anode $$\ce{Na+ + HCO3^- -> NaHCO3(s)}$$ Furthermore if the electrolysis is continued a long time, a supplement of acid ions will react with these $\ce{HCO3^-}$ ions producing some $\ce{CO2}$ by the reaction  $$\ce{H+ + HCO3- -> CO2 + H2O}$$
$\ce{CO2}$ is a gas, and it is relatively soluble in water. In the beginning the $\ce{CO2}$ produced will stay dissolved in the liquid. But on the long run, $\ce{CO2}$ will get out of the solution and get mixed with the oxygen produced at this electrode. So instead of getting pure $\ce{O2}$ at the anode, you will obtain a mixture $\ce{O2 + CO2}$.
So you better use a solution of sodium sulfate $\ce{Na2SO4}$ instead of sodium carbonate. The above mentioned drawbacks do not occur with sodium sulfate.
In a $\ce{Na2SO4}$ solution, the ions $\ce{Na+}$ are moving towards the negative electrode (cathode), and the ions $\ce{SO4^{2-}}$ are moving toward the positive electrode (anode). They move but they don't touch their electrodes and they are not discharged. Instead that is water that either give or absorb electrons.
At the negative electrode (cathode) where $\ce{Na+}$ ions are approaching, the following reaction occurs :  $$\ce{2 H2O + 2 e- -> H2 + 2 OH-}$$ and the $\ce{OH-}$ ions neutralize the approaching $\ce{Na+}$ ions. Of course some $\ce{H2}$ bubbles are appearing on the cathode.
At the positive electrode (anode), the following reaction happens $$\ce{2 H2O ->  O2 + 4 H+ + 4 e-}$$ and the $\ce{H+}$ ions neutralize the approaching $\ce{SO4^{2-}}$ ions. Of course some $\ce{O2}$ bubbles are produced on the anode. Half as much as $\ce{H2}$, because $4$ e- produce $1$ $\ce{O2}$ and $2 \ce{H2}$
To summarize, electrolyzing a $\ce{Na2SO4}$ aqueous solution will produce $\ce{H2}$ and a solution of $\ce{NaOH}$ at the cathode. And it will produce $\ce{O2}$ plus a solution of $\ce{H2SO4}$ at the cathode.
Now if the anodic solution is stirred and mixed with the cathodic solution, the following reaction happens $$\ce{H2SO4 + 2 NaOH -> Na2SO4 + 2 H2O}$$ As a consequence, $\ce{Na2SO4}$ is regenerated, and the only chemical that is consumed in the whole process is water. The only product getting out of the whole process is the formation of $\ce{H2}$ at the cathode and $\ce{O2}$ at the anode.

Thank you Maurice, for the detailed breakdown.
If you have time and for curiosities sake, could you break down what happens with Sodium Sulfate under these conditions?
Appreciate your time, tea
Edit: Also, would you suggest a ratio of electrolyte to De-Ionised water? I have been running Sodium Carbonate at aprox 5%.

I have updated my question with an edit after doing some more homework. I want to thank you again for your time, and if you have any more please could you give the edit a quick read/comment.

@Tea1023. The electrolysis of a $\ce(Na2SO4}$ solution does not produce any sulfur. And this solution contains neither $\ce{O^{2-}}$ nor $\ce{H^+}$ ions. So all your chemical concepts are pure fantasy. The electolyzing a sodium sulfate solution, the ions $\ce{Na+ + SO4^{2-}}$ are not modified. They jus approaches their respective electrodes without being chemically modified. I continue this comment as an edit to my previous answer.


text:
Why are square planar molecules drawn from the side plane as opposed to the top plane?
Simply put, why can't you draw square planar molecules top down (so you don't have to use depth bonds)?

You can do that. For example, a Google Image search for "cisplatin structure" suggests that although the hashed + wedged depiction is more common, the four plain bonds depiction is also in regular use.
For more detail on what's "acceptable" or not, refer to: Graphical representation of stereochemical configuration (IUPAC Recommendations 2006). Pure Appl. Chem. 2006, 78 (10), 1897–1970, DOI: 10.1351/pac200678101897. They recommend the hashed + wedged depiction, but do not prohibit four plain bonds (section ST-2.6, pp 1942–3). There is some rationale for this, which may go some way towards explaining why the hashed/wedged depiction is more common: it explicitly differentiates the square planar structure from other tetrahedral centres in organic structures.

The first recommended depiction should be drawn with two solid wedged bonds and two hashed wedged bonds. The two solid wedged bonds should be positioned below the central atom, while the two hashed wedged bonds should be positioned above. [...] It is also acceptable to draw similar structures with plain bonds. There are some cases where this drawing style would even be preferred, particularly when all four bonds are required to be in the plane of the paper by virtue of other aspects of the structural diagram [...] The similar depiction with the bonds themselves positioned along the axes is acceptable but should be avoided as it is extremely commonly used to represent tetrahedral centers, either if the tetrahedral configuration is nonstereogenic (carbon tetrachloride) or in Fischer projections.


text:
What do γ (gamma) and λ (lambda) mean in organic chemistry?
From the Wikipedia entry on Cyhalothrin:

Pyrethroids such as cyhalothrin are often preferred as an active
ingredient in agricultural insecticides because they are more
cost-effective and longer acting than natural pyrethrins. λ-and
γ-cyhalothrin are now used to control insects and spider mites in
crops including cotton, cereals, potatoes and vegetables.

Image: γ-Cyhalothrin, λ-Cyhalothrin

Image: λ-cyhalothrin (racemic)

Gamma-cyhalothrin and lambda-cyhalothrin are the active ingredients in
the current commercial products based on cyhalothrin. Both are
cyanohydrin esters of
cis-3-[(Z)-2-chloro-3,3,3-trifluoropropenyl]-2,2-dimethylcyclopropanecarboxylic
acid. All of the insecticidal activity is due to the proportion of
absolute stereochemistry (1R) in the mixture.[7] The active isomer of
deltamethrin,
(1R)-cis-3-(2,2-dibromovinyl)-2,2-dimethylcyclopropanecarboxylic acid,
has the same stereochemistry.
γ-cyhalothrin (a single chiral isomer) is indeed twice as active as
λ-cyhalothrin on a weight-for-weight basis. The latter is racemic and
contains the (1R) and inactive (1S) isomers in equal amounts.

Image: Cyhalothrin and deltamethrin acids
I have tried and tried on my own to find out what lowercase gamma and lowercase delta could mean here but I can't find an answer.
Also what does "AND Enantiomer" mean? Or is it "A.N.D. Enantiomer"?

Can you please indicate which parts are quoted? You can do this by placing a right angle bracket > in front of the paragraph. Right now, it's rather difficult to parse your question. I don't know which parts are you talking and which parts are you telling me about the context.

Appears to be purely marketing, chemistry-unrelated in this particular case.

The greek letters usually denote on which carbon atom on a chain some functional group is located, often counting from a carboxy group. The next - alpha, the one after the next - beta, the third - gamma, the very last on a long chain - omega. I don't understand the naming here, honestly.

I don't have the full answer but I guess it might be to differentiate the stereoisomer by using Greek letters. One example I can think of is Lindane which is normally the "gamma" isomer of hexachlorocyclohexane and is written "γ-HCH". It has 2 isomers: α-HCH and β-HCH.


text:
How to draw a reaction coordinate diagram for SN1 mechanism?
How would you draw a reaction coordinate diagram for an SN1 mechanism with the following conditions?

each step in the mechanism (there are 2 steps) is favorable
the second step of the mechanism is the rate-limiting step

I believe the RC diagram should look some thing like this, but I am not completely sure.

Yes, that diagram would be correct for the generic reaction. However in the specific case of an SN1 reaction the intermediate is almost always higher in energy than the reactants or the products, as they usually involve a carbocation. It is also typical of SN1 reactions for the first step to be rate-limiting, as carbocation formation from a neutral reactant is thermodynamically unfavorable unless specific reaction conditions are present. The three general rules for reaction coordinate diagrams are as follows:

The number of “peaks” should be equal to the number of steps in the reaction.

The height of each peak relative to the valley to its left is representative of the activation energy of that step. Larger vertical distance between peaks = greater activation energy = slower step. The rate-limiting step of any reaction will have the greatest distance between its peak and the valley preceding it.

The height difference between your reactant(s) and product(s) is equal to your total change in enthalpy (delta H) for the reaction. If the products are below the reactants then enthalpy change will be negative, and the reaction is exothermic. The inverse is true of endothermic reactions.

It may be worth clarifying that this diagram is supposed to represent an SN1 mechanism and the second step is rate-limiting. The diagram of these reactions should look like this:

Hopefully this is helpful!


text:
Splitting of ethyl ester proton NMR
I synthesised a dihydropyridine molecule today and recieved my NMR. I am wondering where the complex splitting for the 7 and 22 protons is coming from.
My spectrum:

1.26ppm, triplet, 6H
2.33ppm, singlet, 6H
4.16ppm, multiplet, 4H
5.20ppm, singlet, 1H
5.65ppm, singlet, 1H
5.94ppm, doublet of doublets, 1H
6.21ppm, doublet of triplets, 1H
7.21ppm, singlet, 1H

Here is the multiplet form 4.16ppm:

The molecule:

Thanks very much!

The protons at 7 and 22 are diastereotopic because of the stereogenic center at 9. The hydrogens at the -CH2- are non-equivalent, have different chemical shifts and couple with each other in each -CH2-. Look here: https://chemistry.stackexchange.com/questions/95027/magnetic-non-equivalence-of-acetal-protons/95044#95044

On closer inspection C9 is non-stereogenic but that doesn't change the argument. Each methylene has -CH(a)H(b).


text:
Mechanism of acid-catalyzed Robinson annulation?
My professor posed my organic chemistry class a challenge problem for bonus points and I've been racking my brain over the last three days trying to work it out and I want to know if my solution is plausible. The problem asks us to draw the best curved arrow mechanism for the reaction in the bottom left corner of the photo included in this post (a substituted cyclobutane molecule to a substituted cycloalkene). At first, this seemed like a simple intramolecular Robinson annulation, however it appears to be catalyzed using the hydronium ion, rather than the hydroxide ion. No other reactants are stated, including light, heat, nor a double equivalence of the starting compound. Working through every mechanism that I could think of, I eventually ended up with the solution in the picture below. I assumed that reaction rates and percent yields were unimportant, as long as the mechanism is plausible. Does this solution break any rules of organic chemistry? Is there a better solution?

Overall, a good solution. You don't need the enol in the first line. The allylic anion in the fourth structure is too basic. Protonate as you did in the first step but push electrons to fragment the ring. Enols do not sit around for several steps. They tautomerize to carbonyls. Formation of the cyclohexenone is fine except I would have opted for the enol of the ketone in the 3rd structure of the penultimate line to assist in the loss of water. This would lead directly to 3-methylcyclohexenone.

As I mentioned in my Comment the only serious issue was the formation of an allylic anion in acid medium. You seem to know how to form enols from ketones and how to form ketones from enols. I will ignore those steps in the diagram. Protonation of the carbonyl in 1 can cause ring fragmentation to form transient enol 2 that rapidly tautomerizes to diketone 3. Cyclization via enol 4 leads to β-hydroxyketone 5. Although loss of water from 5 as you have described is viable, I prefer formation of the enol 6 because loss of water is facilitated over a tertiary cation given the allylic nature of the species 7. Loss of a proton from carbocation 7 leads to 3-methylcyclohex-2-en-1-one (8).
Addendum: The process is not a Robinson annulation, which involves an initial Michael addition followed by an aldol condensation. The latter reaction includes the transformation 3 $\rightarrow$ 8. The transformation 1 $\rightarrow$ 3 may described as a retro-aldol reaction.
You may want to see the formation of mesitylene (1,3,5-trimethylbenzene) from acetone under acid conditions.


text:
Cause of photoluminescence in CaV2O6
I had prepared the sample of Cav2O6 using solid-state reaction route. I had done XRD, SEM-EDS which all matched with the previous results. There was something strange when I did the PL analysis. I saw that most of the vanadates gave the yellow-orange spectra but my sample was giving dark blue. I was not able to find anything related to that in previous researches.
So, I thought that CaV2O6 must also have the properties of CaO as (CaV2O6=CaO-V2O5). I searched the PL studies of CaO and the results were matching closely. So, can I conclude that at a particular excitation wavelength, the properties of CaO component of my sample are more prominent.

Did you took an excitation spectrum? Literature?

Yes, I took the excitation spectrum. For CaV2O6, only one previous research has taken Photoluminescence and excitation spectrum but with CaV2O6 in the form of nanorods. Other papers include the excitation and emission of CaV2O7.

So you likely have a two (or more) phase sample, so are happily seeing PL from the CaO bits.

But the XRD results does not indicate any additional sample. The variations between the undoped and doped XRD are extremely minor (marginal peak shifts and increase in intensity).


text:
Acid boric preparation
Is there a simple reaction to yield boric acid ($\ce{H3BO3}$) from boron trifluoride ($\ce{BF3}$)?

"Quick questions" without elaboration effort are not very welcome, and may be closed.

Yes, hydrolysis of boron trifluoride.
$$\ce{4 BF3 + 3 H2O → 3 HBF4 + B(OH)3}$$
The reaction commences with the formation of the aquo adduct, $\ce{H2O−BF3}$, which then loses $\ce{HF}$ leading to formation of the products.
The heavier trihalides do not undergo analogous reactions, possibly due to the low stability of the tetrahedral ions $\ce{BCl4-}$ and $\ce{BBr4-}$.

I can imagine a less weird way to prepare  boric acid. :-)

Wikipedia gives several methods to get boric acid.  The first one uses a relatively common natural source of boron.  Curiously compared with the other answer, boron trifluoride is excluded among the halides.

Boric acid may be prepared by reacting borax (sodium tetraborate decahydrate) with a mineral acid, such as hydrochloric acid:

$\ce{Na2B4O7·10H2O + 2 HCl → 4 B(OH)3 [or H3BO3] + 2 NaCl + 5 H2O}$

It is also formed as a by product of hydrolysis of boron trihalides and diborane:[1]

$\ce{B2H6 + 6 H2O → 2 B(OH)3 + 6 H2}$
$\ce{BX3 + 3 H2O → B(OH)3 + 3 HX (X = Cl, Br, I)}$
Upon further review, boron trifluoride can be used to give boric acid as well, as shown in the other answer, but at reduced yield as most of the boron forms $\ce{HBF4}$ instead.  Using a heavier halogen favors all the boron going to boric acid.
Cited reference:
1.
Housecroft, C. E.; Sharpe, A. G. (2008). "Chapter 13: The Group 13 Elements". Inorganic Chemistry (3rd ed.). Pearson. p. 340. ISBN 978-0-13-175553-6.


text:
A problem regarding solubility and concentration

A solution is $\pu{0.10 M}$ $\ce{Ba(NO3)2}$ and $\pu{0.10 M}$ $\ce{Sr(NO3)2}$. If solid $\ce{Na2CrO4}$ is added to the solution, what is the $\ce{[Ba+]}$ when $\ce{SrCrO4}$ begins to precipitate?  $\mathrm{K_{sp}}=\pu{1.2 \times 10^{-10}} \text{ for } \ce{BaCrO4}$
$\mathrm{K_{sp}}=\pu{3.5 \times 10^{-5}} \text{ for } \ce{SrCrO4}$

The solution given in my textbook is:

$\mathrm{K_{sp1}}=\ce{[Sr^2+][CrO4^2-]}$ $$\ce{[CrO4^2-]}=\frac{3.5\times 10^{-5}}{0.1}=3.5\times 10^{-4} \pu{M}\tag{1}$$ $\mathrm{K_{sp2}}=\ce{[Ba^2+][CrO4^2-]}$ $\ce{[CrO4^2-]_{total}}\approx\ce{[CrO4^2-]}\text{ from } \ce{SrCrO4}$ $$\ce{[Ba^2+}]=\frac{1.2\times 10^{-10}}{3.5\times 10^{-4}}=3.4\times 10^{-7}\pu{M}$$

First of all, in equation $(1)$ why did they put $\ce{[Sr^2+]}$ same as the initial concentration (before the addition of $\ce{Na2CrO4}$)? When $\ce{Na2CrO4}$ is added,  $\ce{[Sr^2+]}$ will react with $\ce{CrO4^2-}$ (obtained from $\ce{Na2CrO4}$) to form precipitate. The concentration of $\ce{[Sr^2+]}$ will undoubtedly change until the equilibrium is established.
This given solution is totally confusing, I would appreciate if you could tell me the proper way of solving this problem. I tried this on my own, but after calculating the concentration of $\ce{CrO4^2-}$, I couldn't make sense out of this question.

It is supposed addition of solid $\ce{Na2CrO4}$ would cause negligible volume change, so we can afford to consider the same $[\ce{Sr^2+}]$ as the initial one.
The task is rather theoretical, supposing the time point when $\ce{[Sr^2+][CrO4^2-]}$ just reached $K_\mathrm{sp,\ce{SrCrO4}}$, while $K_\mathrm{sp,\ce{SrCrO4}}$ and $\ce{[Sr^2+]}$ are given.
The proper computation procedure is already presented in you solved task. What  else is unclear ?

I am going to assume that you already understand the theory behind this equilibrium problem. I wanted to first say that you are absolutely correct in that the [Sr2+] will "undoubtedly change until the equilibrium is established".
The short answer to your question is that the solution in the textbook used an assumption that we often use in equilibrium calculations for simplification.
Different textbooks have different rules, but the general idea is that, for an equilibrium constant K, such that K = [x][I + x], we can approximate K ~ [x][I] given that K << [I]. I have often seen that the condition for K << [I] holds for when [I] is at least 10^2 times greater than K.
So, basically, your textbook just simplified the work. Instead of writing [Sr2+] = 0.10 + x, they wrote the initial conc (0.10) because Ksp = 3.5*10^-5 << 0.10 to avoid solving an annoying quadratic.
Hope that helps.

Before we begin, there are two implicit assumptions:

That there is not a significant change in the volume of the solution by either adding $\ce{Na2CrO4}$ nor the precipitation of the $\ce{BaCrO4}$
That concentrations can be used instead of activities.

First of all, in equation $(1)$ why did they put $\ce{[Sr^2+]}$ same as the initial concentration (before the addition of $\ce{Na2CrO4}$)?

The problem statement asks for the $\ce{[Ba^{2+}]}$ when $\ce{SrCrO4}$ "begins to precipitate." Given that "some" $\ce{SrCrO4}$ must precipitate to reach the point where $\ce{SrCrO4}$ has already begun to precipitate, let's define some variables:
$\ce{[Sr^{2+}]_{init}}$ -- The initial concentration of $\ce{Sr^{2+}}$, 0.10 molar.
$\ce{[Sr^{2+}]_{ppt}}$ -- the concentration of $\ce{Sr^{2+}}$ removed by ppt.
$\ce{[Sr^{2+}]_{final} = [Sr^{2+}]_{init} - [Sr^{2+}]_{ppt}}$ -- the concentration of $\ce{Sr^{2+}}$ after some $\ce{SrCrO4}$ has precipitated.
Now here comes the rub. Chemistry isn't like math. The value of $\pi$ has been calculated to millions of digits. $\ce{[Sr^{2+}]_{init}}$ was given as 0.10 molar. So there are only two significant figures in this value. If 0.01% of the $\ce{Sr^{2+}}$ is used to form the initial precipitate then $\ce{[Sr^{2+}]_{final} \approx ([Sr^{2+}]_{init}} = \pu{0.10 M})$
Note that the problem could have been written differently. The problem could have asked for $\ce{[Ba^{2+}]}$ when $\ce{SrCrO4}$ reaches its $\mathrm{K_{sp}}$ (or some variation thereof). This would be right at the equilibrium point, before any $\ce{SrCrO4}$ has formed.

When $\ce{Na2CrO4}$ is added,  $\ce{[Sr^2+]}$ will react with $\ce{CrO4^2-}$ (obtained from $\ce{Na2CrO4}$) to form precipitate. The concentration of $\ce{[Sr^2+]}$ will undoubtedly change until the equilibrium is established.

Obviously when $\ce{SrCrO4}$ nucleates there is some finite size of the $\ce{SrCrO4}$ particle which will be stable in an aqueous solution. So if we use a nanoliter of the cation solution then the given answer will be wrong. Thus there is a implied assumption  that a macro volume of the solution is being used. "Macro" in this case meaning that creating a detectable amount of the $\ce{SrCrO4}$ does not change the $\ce{[Sr^2+]}$ from its initial nominal value of 0.10 molar.

I would appreciate if you could tell me the proper way of solving this problem.

The book gives the proper answer albeit without a full explanation of the additional assumptions.
The notion is that $\ce{BaCrO4}$ is more insoluble than $\ce{SrCrO4}$, so most of the $\ce{Ba^{2+}}$ will be removed before any $\ce{Sr^{2+}}$ starts to precipitate.
I'll add that this assumption isn't really true. As the $\ce{BaCrO4}$ precipitates, it will incorporate some of the $\ce{Sr^{2+}}$ into the $\ce{BaCrO4}$ precipitate. Again, not an appreciable amount compare to what is left in solution since initially $\ce{[Sr^{2+}] = [Ba^{2+}]}$. However if initially $\ce{[Sr^{2+}] = 0.01\cdot [Ba^{2+}]}$ then there would be a significant loss of $\ce{Sr^{2+}}$ in the solution.


text:
Can a metal which contains impurity of another metal that is more reactive than itself be electrorefined?
So I know that cathode is made of pure metal and anode of impure. For example take Cu. Cu will be oxidised to Cu+2 at anode and will deposit at cathode.
But now if it contains impurity of , let's say, Zn (even if that doesn't occur in nature just consider hypothetically) then won't Zn be oxidised at anode instead of Cu? Will electrorefining work in such a case?

Yes it will work. Everything will dissolve, then Cu will reduce.

"Electrorefining of copper was first demonstrated experimentally by Maximilian, Duke of Leuchtenberg in 1847." (en:wp)

The key point to remember here is that any metal that is oxidized more readily than copper will be reduced less readily. So yes, it will be oxidized before the copper in the impure sample, but it will not be reduced and deposited on the pure metal electrode. It will simply stay in solution as an ion as long as the process is stopped before all of the copper has deposited.


text:
Best way of studying protonation in non-polar solutions with ~100ppm of water present using computational methods?
I am currently looking at the mechanism of the formation of storage deposits in jet fuel. Interestingly, the concentration of acids in the fuel correlates strongly with the total insolubles formed over testing periods. As such, several papers propose acid catalysed deposition processes occurring. However, since jet fuel is a non-polar solvent, with very little water present, this would be an extremely slow process. Alternative, acids could be taking part in an alternate mechanism leading to deposits.
Currently I am employing DFT methods in my work to explore the deposition process from a mechanistic perspective. I represent the bulk jet fuel solvent with an n-dodecane PCM solvent model. N-dodecane was chosen as it is the average chain length of jet fuel bulk alkanes. To explore the protonation in these solvents with small amounts of dissolved water, I have been explicitly adding water molecules around the conjugate bases/acids and looking at how successive water molecules affect the ΔfG (Gibbs energy of reaction). However, I am unsatisfied with this method. At current I have been placing the water molecules at what seems like the most stable configurations. In fact there could be several stable configurations, and putting on calculations one-by-one in various configurations seems cumbersome and unsystematic.
Is there a more statistical approach to this problem? Or, would stepping outside of computational chemistry and performing a study in the lab be the only way to tackle this?
Thanks.

I think that traces of water in apolar fluids can have non standard behaviour. Surely the activity a of protons must be very high. While a barrel can store water indefinitely, non anhydrous hydrocarbons or alogenated solvents can literally drill a hole through it. But I don't know the details. Just know that because of this a quick dessication is needed before storing alogenated wastes in big steel barrels.

Interesting, I will try and find any papers looking into this.


text:
How can we prove the inequality TdS > dU + pdV?
I know that for reversible process
$$T\,\mathrm dS = \mathrm dU + p\,\mathrm dV,$$
but since all the terms in the given equation are state variables so the above equation should also be true for irreversible processes but in my textbook for irreversible process the inequality is used for irreversible process.

Lowercase letter $s$ is used to denote specific entropy. Are you absolutely sure this is what you are asking about? Just in case I corrected the symbol for the entropy to $S.$

The equality is always true.  The equation just describes the relationship between dS, dU and dV that must exist between two closely neighboring (i.e., differentially separated) thermodynamic equilibrium states of a closed system.  It has nothing to do with any process, no matter how complicated or tortuous, that took the system from the first state to its neighbor.

For a closed system with a single phase of a pure substance, when only expansion work is possible
$$\mathrm dU= -\mathrm p\,\mathrm dV + T\,\mathrm dS, \label{eqn:1}\tag{1}$$
where $p$ (the state variable) is the internal pressure and $T$ (the state variable) is the internal temperature of the system.
The first law, however, says more generally that
$$\mathrm dU = \mathrm dw + \mathrm dq. \label{eqn:2}\tag{2}$$
Equation \eqref{eqn:2} was derived by applying equation \eqref{eqn:1} to a reversible process.
For the reversible process, thermal and mechanical equilibrium between internal and external pressure means that $p_\mathrm{ext} = p$ and $T_\mathrm{ext} = T,$ and, combined with the second law, that
$$\mathrm dw = -p_\mathrm{ext}\,\mathrm dV = -p\,\mathrm dV \tag{3}$$
and
$$\mathrm dq = -T_\mathrm{ext}\,\mathrm dS_\mathrm{ext} = T\,\mathrm dS. \tag{4}$$
More generally, the second law says that
$$\mathrm dS_\mathrm{tot} = \mathrm dS_\mathrm{sys} + \mathrm dS_{surr} \ge 0 \tag{5}$$
the equality holding for a reversible process. Dropping the label for the system
$$\mathrm dS \ge -\mathrm dS_\mathrm{surr} \quad\leftrightarrow\quad \mathrm dS \ge \frac{\mathrm dq}{T_\mathrm{surr}}. \tag{6}$$
Therefore for an isothermal process (thermal equilibrium condition applies)
$$\mathrm dS \ge \frac{\mathrm dq}{T}. \tag{7}$$
The inequality applies for instance to an isothermal free expansion.

In a reversible process, the pressure $p$ of $p\,\mathrm dV$ is the internal pressure of the gas $p = nRT/V.$ In an irreversible process, the pressure $p$ of $p\,\mathrm dV$ is the value of the outer pressure which is usually much bigger than the internal pressure.
The piston is quickly pressed and the gas is quickly compressed. This irreversible compression produces more heat than the reversible compression, where the internal and the external pressure are continuously adjusted to be equal (or nearly equal) to each other.


text:
Does potential energy of molecules increase or decrease when covalent bonds are formed or broken?
My sir told me that when bonds are formed or broken , average separation between particles comes closer or decreases when they are broken.I see there is a lot of change in K.E but how is there change in P.E is not what I am able to understand.My sir also told that in case of liquids , P.E would be less as compared to gases. Pls help me understand how does that happen.
Online it says that since the gaseous atoms are attracted to each other , they have P.E.Now P.E =mgh right?So if we say bonds come closer , We know that M is mass of body , g is gravity , H is distance from the earth.As you go up , P.E increases and down,P?e decreases.Just saying that they are attracted to each other (in all cases liquids , gases and solids), they posses P.E.But how does the distance of force of attraction that the molecules have matter?.That i what I am not able to understand.I hope you have got it.

The one on the fridge doors have higher P.E.Energy is released while attaching or detaching the magnet yes.

I don’t understand how can this be compared to my question.

Please tell what have I not elaborated ?

I’d be happy to improve it.

Thank you.When someone tells what I have to change,then only I understand what to do.

I have written about what I got online in the question itself.

Ok sir then . .

Just don't overthink that potential/kinetic energy similie. The force fields are not gravitational, and have totally different formula. E=mgh is useless. You are still quite a few lecture hours away from calculating anything, give it time.

Means mgh is not used in cases of attraction forces.

See e.g. https://courses.lumenlearning.com/boundless-chemistry/chapter/bond-energy-and-enthalpy/ Skip the text, if too difficult, and see the chart of bond energy versus distance. See also https://en.wikipedia.org/wiki/Potential_energy

Note that $E_p= mgh$ is simplification of $$Ep=\kappa Mm \left(\frac{1}{R}-\frac{1}{R+h}\right)$$
where M and R are the Earth mass and radius, $\kappa$ is gravitational constant, m and h are the object mass and relative height wrt Earth surface

Your main mistake is to believe that the potential energy is only due to gravity $(E_\mathrm{pot} = mgh).$ But there exist other forms of potential energy. Potential energy may have another origin. It may have an electric origin.
When you take a H atom (containing one proton and one electron) in order to separate the proton and the electron, you must give some energy to the H atom (as light for example). When the proton and the electron are separated, far from one another, they usually do not have any kinetic energy $(E_\mathrm{kin} = mv^2/2),$ they do not have more gravific energy $(E_\mathrm{pot} = mgh).$
But they do have electric potential energy. Potential energy means that the sample has "something" that is invisible, but could be delivered to the surrounding if it has an opportunity to do it. A bead on a table has a gravific potential energy, because if you give it the possibility of going to the ground, it will deliver energy which is first in the form of kinetic energy $(E_\mathrm{kin} = mv^2/2),$ then this energy is transformed into heat when touching the ground (if it does not rebound).
That is about the same for the proton and the electron. When infinitely separated and not in movement, the only energy they have got is the electric potential energy. If you give them the opportunity of approaching one another, they will be attracted, loose their electric potential energy which is transformed into kinetic energy or into light during this approach.
When the H atom is created and stabilized, the atom may still have some kinetic energy, but does not have electric potential energy any more. The gravific energy $(E_\mathrm{pot} = mgh)$ of the proton and electron may have changed a bit during the process, but this change is not significant.

Why is it that the gravific P.E does not change much ? In your last sentence. @Maurice

The variation of the gravific potential energy is $mg\Delta h$, with $m$ is the mass of the electron ($\ce{9·10^{-31}} kg$, and $ \Delta h$ is not far from the dimension of the atom, about $10^{-10} m$. So the energy change is  about $\ce{10^{-41} }J$. This is so small that it can not be measured.


text:
MO diagram when three orbitals of the same symmetry is allowed to mix
How exactly should we combine three orbitals when we have three orbitals that are of correct symmetry to interact?
Take $\ce{H2O}$ for example. $\mathrm{2s},$ $\mathrm{2p}_z$ and the $A_1$ group $(\ce{H2})$ orbitals can interact. It would seem trivial that the lowest bonding orbital is $\mathrm{2s}$ in phase with $A_1,$ but what about the anti binding orbital? Most resources suggest $\mathrm{2p}_z-A_1$ out-of-phase interaction, but is $\mathrm{2s}$ with $A_1$ out of phase not correct?
Could this be due to we try to construct the anti bonding orbital with least overlap possible?

Three AOs produce three MOs.

Yea, so there will be a non-bonding orbital in the middle right, but what about the anti bonding orbital ?

It depends on the geometry. You might have a non-bonding orbital in the middle and an antibonding one above, or you may have two roughly equal no-so-antibonding orbitals.

What I am asking is how will they mix? If you have two orbitals their linear combination is trivial, one combination is in phase while the other is not. What if there are three?

It will be something like $(+++),;(+0-),\text{ and }(+-+)$.

You might look at http://mcgrady.chem.ox.ac.uk/lectures/Handout%202019.pdf, especially sections 14 and 15

One factor that muddies this situation is that MOs of the same symmetry can mix, which happens in H2O.

You can estimate what happens by using a Huckel MO approach and, even though it is approximate, it shows the trends. With three similar orbitals on three atoms, such as 3 p orbitals these can be combined in two ways depending on whether the molecule is triangular or linear in shape. The MO's produced always sum to the same energy as that of the three p orbitals.
Without going into details, if we place them in a triangular arrangement each of the p orbitals can interact with each of the other two and the MO orbitals produced have relative energies -2 ,+1, +1 so when three electrons are added there is an overall lowering of energy with two in the lowest energy orbital and one in either of the two degenerate orbitals.
If linear, the three p orbitals will only interact significantly with the one next to it, 1 to 2, 2 to 3 for example. The MO orbitals all have different energy with one lowered a lot and two raised less so, so that the total energy is the same as the initial p orbitals. None are degenerate.
The conclusion is that the energy of the MO orbitals produced depend on the interactions between the p orbitals and this often depends on the geometry with closest p orbitals interacting most.
The Huckel method is described in many places, Wikipedia and most phys. chem. textbooks, if you want to try the calculation for yourself.

Another answer addressed the situation of three p orbitals, each on separate atoms. However, the example of $\ce{H2O}$ given in the question is a bit more complicated, since two of the orbitals are on the same atom (s and p on O). The short answer is that we cannot determine qualitatively what the orbitals look like, specifically the nonbonding one that is intermediate in energy.
If we first consider the s and p orbital independently, we have four possible combinations:

s + A1 (where " + " means same phase) [bonding]
s - A1 [antibonding]
p + A1 [bonding]
p - A1 [antibonding]

But we know that only three orbitals actually result, since three went in. SO already we have a problem that we don't know which three to use. It gets even more complicated when we realize that the s and p orbitals can both contribute to the same MO. (This is often described as "s-p mixing", though it formally is a mixing of the MOs, not the AOs.)  If we don't worry about the exact value of the coefficients of each orbital and just focus on sign, we have four distinct results:

s + p + A1 [strongly bonding]
s - p + A1 [essentially nonbonding, since the portions of s and p that are oriented towards A1 are out of phase with each other and at least partially cancel out so there is very little overlap with A1]
s + p - A1 [strongly antibonding] here the portions of s and p oriented towards A1 are additive to make a large lobe that is out of phase with A1
s - p - A1 [essentially nonbonding, since the portions of s and p that are oriented towards A1 are out of phase with each other and partially cancel out as in (2)]

Of these, (1) and (3) are good representations of the known bonding and antibonding orbitals in $\ce{H2O}$ that you will find in pictures of MOs of $\ce{H2O}$, although the contribution of p to (1) is small enough that it closely resembles a simple s + A1 orbital.
The challenge is the third orbital. Determining whether it will more closely resemble (2) or (4) is not something that can be done with simple qualitative analysis. We must instead do a more quantitative analysis, which tells us that (4) is more accurate; the density on O has a small lobe pointed towards A1 that is in phase with A1 but overlaps very little and a large lobe that points away from A1 that is out of phase with A1. This quantitative analysis is covered in advance MO theory texts.


text:
Molecular orbital diagram for carbene-metal and carbodiphosphorane-metal interaction
I am interested in understanding how carbenes and carbodiphosphoranes interact differently with a transition metals. I would like to use qualitative molecular orbital theory arguments (with an emphasis on qualitative) in order to highlight key differences.
A carbene is a di-valent carbon, with a long pair (in some structures it may be a diradical). It is a very strong $\sigma$-donor, in addition to being a $\pi$-acceptor: It can form a nice bond with an empty $d_{z^2}$ orbital from a metal, and it can accept two pi electrons into its empty p orbital ($p_x$ or $p_y$ depending on our coordinate system).
The geometry of the carbene, including the lone pair, is trigonal planar.

A carbodiphosphorane, on the other hand, is zero-valent, and has two lone pairs. These may be used in various different bonding schemes (see for example introduction of this paper).
The geometry here, including the lone pairs, is tetrahedral.

Assuming it binds to a metal in the following manner:

then I expect that the $d_{z^2}$ orbial is not involved in the bonding, but rather that for example $d_{xz}$ and $d_yz$ interact with the two lone pairs from the carbon.
But I find it difficult to summarize the bonding for a carbene and carbodiphosphorane into (over)simplistic MO diagrams.

"The geometry here, including the lone pairs, is tetrahedral." - I doubt that. Carbon may well be sp2.

It is a bit far from my expertise but I thought this papers may be of help: Stable Carbenes and Persistent Triplet Carbenes

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