text:
Enthalpy in the van't Hoff equation
In the van't Hoff equation, why do we say that the change in enthalpy at standard pressure is constant? My book derives the equation for a mixture of ideal gases reacting with each other. The proof itself is good, until the authors declare that the equation says that the graph of $\ln K$ with $\frac{1}{T}$, where $K$ is the equilibrium constant and $T$ is the temperature of the mixture, will be a straight line. This would only be true if the change in enthalpy is constant (since the slope of the curve is proportional to this change). But isn't the enthalpy of an ideal gas proportional to temperature? So why is the change in enthalpy constant in $T$, then?
I've attached an example of this effect, as presented by the book. See the image below.

It is a simplifying assumption. Could you provide more details about the gases involved, the pressure, and the temperature window?

@BuckThorn I don't have any particular gases or temperature windows in mind (simply because that wasn't mentioned in the book), but they do provide an example of this effect. Do you think it's a good idea to include that in my question?

Good question. Enthalpy of reaction dependence on temperature is given by Kirchoff's equation (you can check it) and it is true that enthalpy of reaction changes with temperature. For most reactions if temperature change isn't too big, enthalpy of reaction won't change much so we can regard it as constant in that temperature span on which we are plotting lnK vs 1/T for sake of simplicity.

Ah, okay, so it's just an approximation. But shouldn't the span of the temperature in the example I gave (see the edit) be big enough for the approximation to go wrong?

Well, it is pretty big temperature span. However, for this reaction difference in cp between products and reactants is probably small since you have only dissociation of hydrogen to atoms. Since products and reactants are similiar, I would say their cp isn't much different because of which enthalpy of reaction doesn't change much with temperature. How much enthalpy of reaction changes with temperature is determined by stochiometric  difference in cp between products and reactants (Kirchoff's equation).

Thank you for that. Yeah, that makes sense. Also, thank you for the suggestion to look up Kirchhoff's equation.

The $\Delta C_p\Delta T$ is quite small compared the $\Delta H$  even though the temperature range is large. Typically $C_p$ is in tens of J/mol/K where as $\Delta H $ is in hundreds of kilo J/mol.


text:
Cationic vacancies created when non-stoichiometric iron oxide is doped with silicon(IV) ions
Problem

It is believed that non-stoichiometric compound $\ce{F_{0.93}O}$ forms by doping of $\ce{Fe^3+}$ ions in $\ce{FeO}$ crystal by replacement of $\ce{Fe^2+}$. Calculate the number of cationic vacancies if all of the $\ce{Fe^2+}$ ions are replaced by $\ce{Si^4+}$ ions in $\pu{0.1 mol}$ of $\ce{Fe_{0.93}O}$. Express your answer as a multiple of Avogadro's Number.

Answer stated by the book

 0.0465

My Solution
As the compound is electrically neutral I equated the net charges due to the cations and anions like this, assuming $x$ to be the fraction of $\ce{Fe^2+}$ in the compound.
$$[2x + 3(0.93-x)] - 2 = 0$$
This gave $x = 0.79$. Now, in $\pu{0.1 mol}$ of the compound there would be $\pu{0.079 mol}$ of $\ce{Fe^2+}$. As each silicon ion would replace two ions of $\ce{Fe^2+}$, one vacancy would be created for every replacement.
The number of replacements would be $\left(\frac{0.079}{2}\right)\times N_\mathrm{A} = 0.0395 N_\mathrm{A}$, and we would have as many vacancies. ($N_\mathrm{A}$ is Avogadro's Number)
Can anyone point out the error in my solution?

You have not considered the vacancies formed when ferrate ions were doped with ferric ions.

@NisargBhavsar Thank you for improving so many posts. Please have a look at these guides for even better edits. And again, I am really, really not fond of this spoiler thing.

@ NisargBhavsar Thank you, that solves my problem.

Your approach is correct, however, you forgot to account for the vacancies created when Fe3+ replaces Fe2+. First, calculate the cationic vacancies formed due to the replacement of Fe2+ by Fe3+.
Assuming  to be the fraction of Fe2+ in the compound, the fraction of Fe3+ in the compound is (0.93 - x).The total charge on x Fe2+ and (0.93 – x) Fe3+ should be equal to the charge due to O2- ions.
Hence, 2x + 3(0.93 – x) = 2
=> 2x + 2.79 – 3x = 2.00 or x = 0.79. So, fraction of Fe3+ ions in the compound is 0.14
Now, 2 Fe3+ ions replace 3 Fe2+ ions, so there's one vacancy created for every 2 Fe3+ ions. That means for 0.1 mole of F0.93O, the number of vacancies will be 0.007A due to Fe3+ replacing Fe2+ in FeO. This is the part you missed.
Now, as you mentioned, each Si4+ ion will replace two Fe2+ ions in the compound, so again, one vacancy per replacement. And, as you already solved, The number of replacements would be (0.079/2)A = 0.0395A. So, the total number of vacancies are 0.0395NA + 0.007NA = 0.0465NA.
Hope this helps!

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text:
Which has higher priority between unsaturated ring and unsaturated chain in the nomenclature of organic compounds?
According to the IUPAC rules of nomenclature, alkenes have higher priority over rings but what if the ring also contains a double or a triple bond?
For eg:

Is this compound going to be named as 1-(2-chlorocycloprop-2-enoxy)propene or 1-chloro-3-propenoxycyclopropene?

"According to the IUPAC rules of nomenclature, alkenes have higher priority over rings" – where did you find this rule? It's not in the current IUPAC recommendations.

@Loong Check this nomenclature for eg https://images.app.goo.gl/C4EEzNMs5zoy2BmV6

or this one https://images.app.goo.gl/FhmRDtSUG8wZ72LY7 they clearly exhibit that rings have lesser priority than alkenes (in my question I asked about the unsaturated rings)

Both examples "1,2-dicyclopropylethene" and "2-cyclopropylpropene" are not preferred IUPAC names.

I'm unaware then but in all my books the nomenclatures are done in a similar way.

for eg:- in the book Modern's ABC of chemistry Pg-1195 it is clearly mentioned that : if the side chain contains multiple bond or a functional group the alicyclic ring is treated as the substituent irrespective of the size of the ring

The name given by ChemDraw is: (E)-1-chloro-3-(prop-1-en-1-yloxy)cycloprop-1-ene.
The ring is determined to be the parent structure.  This is rationalised in the 2013 edition of the IUPAC Blue Book on organic nomenclature by the following rules (pg 451):

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent.  Rings and chains are chosen regardless of their degree of hydrogenation.  As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated a cyclic structure may be recognised, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

So either the chain or ring may be used as the parent for the name, but the ring will define the preferred IUPAC name.
The preference of ring over chain is further clarified in the following rule (pg 596):

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for tha same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a prefered IUPAC name. In general nomenclature, a ring or a chain can be a parent hydride (see P-44.1.2.2).

All emphasis added.

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Why is change in volume of the solution zero in case of an ideal solution?
In my textbook it is written that $ΔV_{mix} = 0$ for ideal solutions i.e. $V_{solution} = V_{A} + V_{B}$. But why is this true? There is literally no explanation in my textbook as to why this is true. I am so confused. Please can someone explain it rigorously. Please help.

In simple words because the average intermolecular interaction is the same in X, in Y, and in X+Y. It is the analogue of the ideal gases case.

If you have two component mixture of A and B, most important reason for volume of mixing being zero is that when solution is ideal heterogeneous interactions (A - B) are the same as homogeneous (A - A and B - B). If interactions are the same than volume doesn't change when you mix components. Other reasons which contribute to non - ideality of mixing is size and shape of the molecules. If you mix water with some liquid polymer, you can expect that  due to big difference in size and shape of water and polymer molecules, volumes will not be additive because water molecules being much smaller than polymer molecules can easily fit between them in free volume of polymer liquid. Such packing/fitting of water molecule is quite different than if you just look at water molecules in pure water or if you look at pure polymer since bulky polymer molecules can't just fit easily in free volume of other polymer molecules like water can.


text:
Determining dissolution in solvents
In recent chemistry class, I learnt that like dissolves like I was told following points:

Ionic solutes are highly soluble in polar solvents like water, ammonia etc.

Reason: The ion-dipole interactions between solvent molecules and solute's ions rip away the electrostatic forces between them and further keeps them apart, as a result the solute dissolves

Covalent compounds are soluble in non-polar solvents. (no reason given for this)

So as I was told that covalent compounds are soluble in non-polar solvents, I thought of $\ce{HCl}$ soluble in water and this left me in confusion.
I think the correct statement should be polar covalent compounds dissolve in polar solvents while non-polar covalent compounds dissolve in non-polar solvents. What's correct?
Also is it possible for a polar covalent compound to dissolve in non-polar solvent(eg- $\ce{HCl}$ in $\ce{CCl4}$)?
Are the reasons written correct?

Pure HCl is a covalent substance, and it is a gas, that could be dissolved in covalent solvants, like toluene or carbon tetrachloride. In the presence of water, $\ce{HCl}$ immediately reacts to produce $\ce{H3O^+}$ and $\ce{Cl-}$ ions, which are soluble into water.
You should do the following reasoning. In principle, liquids should be like gases : Gases are all entirely miscible, and in any proportion. Liquids should have the same properties, and they have with non-polar molecules. But if the liquid contains hydrogen bonds between its molecule, the mixing operation requires to break these bonds. If the solute is able to recreate new bonds to replace the destroyed hydrogen bonds, Nature accepts to make a solution. This is the case with ionic compounds and polar molecules in water. If the solute is not able to make new bonds, Nature refuses to accept these foreign molecules as solute in the liquid. The mixture becomes a two-phases system.

1,4-dioxane and diethyl ether.

@Buttonwood What do you want to show with these two substances ?

It was just an addition about HCl in solution of toluene or tetra (wait, your lab is still allowed to use it?) in these more polar organic solvents.  C'est tout.


text:
Entropy Changes in Electrolytic/Galvanic Cell?
One of the most fundamental equations in chemical thermodynamics states: $$ \Delta_rH_m^⦵ = \Delta_rG_m^⦵ + T\Delta_rS_m^⦵ $$
If we look at this equation in context of net chemical reaction in electrolytic or galvanic cell, it is usually interpreted as follows: Enthalpy of reaction denotes total amount of energy at constant temperature and pressure which needs to be supplied (electrolytic cell) or which is released (galvanic cell) during a reaction, standard Gibbs energy of reaction denotes what amount needs to be supplied or which is released in form of electrical energy (electric potential difference between electrodes), the last term including standard entropy of reaction denotes what amount of heat is exchanged with surroundings during a process.
In electrolytic cell, as entropy of reaction is mostly positive, last term is usually interpreted as heat which comes from surroundings and as such it increases entropy of the system. As heat comes to the system, it helps us during electrolysis since we don't need to put in the whole enthalpy of reaction in form of electrical energy, but only Gibbs energy. In galvanic cell it is the other way around.
What I don't understand is the interpretation of that last term ,which includes entropy of reaction, which I found on hyperphysics page. According to hyperphysics, this term denotes heat exchanged with surroundings and they say that entropy change in the system is due that heat exchanged.
I would say that entropy change is not due to heat exchanged, but due to the fact that during chemical reactions entropy changes because products and reactants have different entropies since they are different compounds with different structure and aggregate state. Entropy of reaction denotes such entropic changes and not entropy changes due to heat exchanged. What are your thoughts?

This wall of text is hard to read: please use paragraphs.

I have separated your question into paragraphs, I hope that's okay. If not feel free to edit it again. But please pay attention to the formatting in future, no one would bother to read a wall of text.

Thanks, man. I've completely forgotten about it.

$\Delta G^0$ is the amount of electrical work that must be done to bring about the dissociation and $T\Delta S^0$ is the amount of heat that must be added to the system to hold it at constant temperature T.  $\Delta H^0$ is the enthalpy of the products of the dissociation minus the enthalpy of the reactants.

As I asked in my post what does entropy change of reaction have to do with heat exchanged with surroundings? Change in entropy due to reaction doesn't happen because of heat transfer, it has to do with the fact that products and reactants have different entropies because of their structure and aggregate state regardless of heat transfer.

It is also equal to the heat transferred from the surroundings to the system divided by the temperature for a process that takes the system reversibly from its initial state to its final state.

TdeltaS does often denote reversible heat transfered at constant T, but in this context entropy change has a different origin (in my opinion) and I don't see connection between the two.

I guess we will have to agree to disagree.

That is also okay.

Let us continue this discussion in chat.

Classical thermodynamics does not need molecules to exist, just substances, i.e. gas, liquid, crystal. The entropy is defined as heat divided by temperature  with no further explanation because molecules do not exist. After kinetic theory and Quantum Mechanics (and heaps of experiments) molecule do exist and have discrete energy levels and we can now interpret entropy as a measure of the distribution in space and among energy levels in molecules.  The thermodynamics calculation takes all of this into account, and it is quite remarkable that it does. This I think is what Chet was hinting at.

Standard entropy of reaction can't be treated that way since entropy change is exactly due to molecules, their structure and their aggregate state. Entropy change of reaction isn't due to heat tranfer.

[OP] One of the most fundamental equations in chemical thermodynamics states: $$ \Delta_rH_m^⦵ = \Delta_rG_m^⦵ + T\Delta_rS_m^⦵ $$

This is just the definition of the Gibbs energy. By itself, it does not give you any insight into chemical reactions.

[OP] If we look at this equation in context of net chemical reaction in electrolytic or galvanic cell, it is usually interpreted as follows: Enthalpy of reaction denotes total amount of energy at constant temperature and pressure which needs to be supplied (electrolytic cell) or which is released (galvanic cell) during a reaction, standard Gibbs energy of reaction denotes what amount needs to be supplied or which is released in form of electrical energy (electric potential difference between electrodes), the last term including standard entropy of reaction denotes what amount of heat is exchanged with surroundings during a process.

This is not the usual interpretation. Instead $ \Delta_rG_m^⦵ $ is the maximal work you can get out of the reaction (or, if the reaction moves away from equilibrium as written, the minimal work you have to put in to run it in that direction). $ \Delta_rH_m^⦵ $ is the sum of heat and non-PV work exchanged with the system. It could be negative or positive irrespective of whether this is an electrolytic or galvanic cell. Finally, the entropy of reaction is the same as for any reaction - roughly how much the dispersion of energy and particles changes within the system when the reaction proceeds.

[OP] In electrolytic cell, as entropy of reaction is mostly positive, last term is usually interpreted as heat which comes from surroundings and as such it increases entropy of the system. As heat comes to the system, it helps us during electrolysis since we don't need to put in the whole enthalpy of reaction in form of electrical energy, but only Gibbs energy. In galvanic cell it is the other way around.

You can't make any generalizations about the heat transfer when there is also work exchanged between system and surrounding. In many thermodynamic arguments, you talk about processes that happen in the absence of non-PV work, and in a reversible manner (i.e. the entropy of the universe is near-constant, which is the same as saying it is a near-equilibrium process).

[OP] What I don't understand is the interpretation of that last term ,which includes entropy of reaction, which I found on hyperphysics page. According to hyperphysics, this term denotes heat exchanged with surroundings and they say that entropy change in the system is due that heat exchanged.

Without a specific link, it is hard to understand what statement on the hyperphysics site you are referring to, and it what context it was made. In general, the entropy change of the system reflects changes in the system.
EDIT: The remainder was written after OP provided the link.

[Hyperphysics] Since the electrolysis process results in an increase in entropy, the environment "helps" the process by contributing the amount TΔS. The utility of the Gibbs free energy is that it tells you what amount of energy in other forms must be supplied to get the process to proceed.

If there is no work and the reaction is run near equilibrium, the environment shows a change in entropy of $-\frac{\Delta H}{T}$. TΔS is not directly related to the change of entropy in the surrounding. The reason that $\Delta H$ is related to the entropy change of the surrounding is that for a boring surrounding (i.e. one that only shows changes in temperature, and very little at that) the change in entropy is proportional to the heat transferred.

I would say that entropy change is not due to heat exchanged, but due to the fact that during chemical reactions entropy changes because products and reactants have different entropies since they are different compounds with different structure and aggregate state. Entropy of reaction denotes such entropic changes and not entropy changes due to heat exchanged. What are your thoughts?

That is correct. Entropy is a state function, so if the entropy of the system changes, it is because the system changes, no matter what happens in the surrounding (you can't break the second law, though). The hyperphysics quote does not say that the entropy change of the system is associated with heat exchange. Their (implicit) argument is the second law:
$$T \Delta S_\mathrm{system} + T \Delta S_\mathrm{surrounding} > 0 $$
So if we have a positive change in entropy of the system like in the example, we can get away with cooling down the surrounding a bit (negative change in entropy of the surrounding). Typically, though, we don't want a super-slow reaction, so you would apply a higher voltage than necessary, and the heat transfer would go in the other direction (as you can tell by the heat typically dissipated from a fast battery charger).
Applying this to the first law, some of the energy input for the reaction can be in the form of heat transfered from the surrounding, it does not have to be all work in this case. (In other cases you have to put in more work, not for the first law energy balance but to satisfy the second law.) For a graphic illustration of the different cases, see https://chemistry.stackexchange.com/a/112958/72973.

Yes, thanks for your notes. My question remains what does entropy change of reaction have to do with heat exchanged with surroundings? Change in entropy due to reaction in a system doesn't happen because of heat transfer, it has to do with the fact that products and reactants have different entropies because of their structure and aggregate state regardless of heat transfer (which you agreed with me). If so, why is TdeltaS(reaction) interepreted as heat transfered since entropy change here has a different origin which doesn't come from heat transfer.

You asked for a link to hyperphysics page: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html    Last two paragraphs in section electrolysis talk about what I mentioned/quoted.

No non-PV work? They are talking about Gibbs energy of reaction being a electrical work which is a non-PV work.

I am not sure I get what you are saying. I thought you said they didn't take into account electrical work.

I updated my answer now that I think I understand what they are saying.

Yes, thank you for your update. I actually thought they said that as well. Since entropy of isolated system must increase we can take some heat from surroundings (decrease its entropy) to help our process since entropy does increase in electrolytic cell. Problem here is that what stops from just contiously taking heat from surroundings? If we take heat from surr than entropy of the system increases even more since entropy is increased due to both reaction and heat transfer. So, if we just continue taking heat from surroundings, entropy change of isolated system would always be positive.

@DarioMirić If there is no reaction (to simplify things), you can't transfer heat from cold to hot place because the entropy gain in the hot place is smaller than the entropy loss in the cold place (because T is in the denominator of $\Delta S = \frac{q_\mathrm{rev}}{T}$). So what stops you is the second law.

I agree with that. If so, than in our example temperature of surroundings should be bigger than of system and we take that during heat exchange temperature of surroundings is constant since its heat capacity is uncomparably bigger than of system. Problem here is if we bring heat to the system its temperature will change, but in this analysis temperature of the system should remain constant since standard enthalpy, entropy and Gibbs energy change of reaction are applicable at constant temperature.

@DarioMirić Isothermal is an idealization. The temperature of the surroundings does not have to be higher than the system for heat transfer to happen. It will just happen with an entropy cost. An example is a cold pack at room temperature. The reaction cools it to lower than room temperature without breaking the second law.

Let us continue this discussion in chat.

The link between entropy and heat follows from the balance between entropy change of system and surroundings during a reversible process. During such a process  the total change in entropy is zero, and the change in the entropy of the surroundings is given by the second law of thermodynamics as $\Delta S_\mathrm{surr}=-q_\mathrm{rev}/T$ (assuming constant T), from which it follows necessarily that $$\Delta S_\mathrm{system}=\frac{q_\mathrm{rev}}{T}$$
When there is non-expansion work under constant T and p, the enthalpy change can be equated with the heat plus the non-pV work. When in addition the process is reversible, the equation you posted follows by writing $\Delta G = w_\mathrm{rev,non-pV}$.


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The reaction of Alkyl Halides with sodium nitrite in DMF

How does the reaction of Alkyl Halides with sodium nitrite in DMF proceed?

I was aware of the reaction of alkyl halides with aqueous sodium nitrite which results in the formation of an alkyl nitrite but I came across the following reaction and was confused of how that would work?
$$\ce{R-X + NaNO2 ->[DMF] R-NO2}$$
DMF being the solvent here, I'm pretty sure it's safe to assume that it will be following $\mathrm{S_N2}$ path here but since I have never seen this reaction before, I'd appreciate if someone could enlighten me.
Also, looking at the reaction, I thought maybe DMF makes the sodium nitrate covalent somehow? (I know this is baseless but that's all I could think) which looks quite confusing.

I came across a post very similar to this but I think that didn't answer what's being asked. So I was hoping if someone could explain the mechanism of this (including if its $\mathrm{S_N1}$ or $\mathrm{S_N2}$)?

Well it would depend on the nature of R, if it is tertiary, it's probably SN1, otherwise SN2.

It is a bit more complicated than that as nitrite is an ambident nucleophile so you can get R-NO2 or R-ONO depending on substrate and reaction conditions.

It is super complicated. This paper goes deeply into the theories and, as far as I can tell, their conclusion is that they do not have a satisfactory explanation  https://www.researchgate.net/publication/7740284_Ambident_Reactivity_of_the_Nitrite_Ion_Revisited

The $\mathrm{S_N1}$ or $\mathrm{S_N2}$ reaction of alkyl halide with nitrite ion is complicated one since nitrite is an ambident nucleophile. For instance, it was reported that silver nitrite reacts with alkyl halides to give nitrites while sodium nitrite gives more nitroalkane than alkyl nitrite (based on Hard Acid/Hard Base concept; Ref.1):
$$\ce{AgNO2 + R-Br -> R-ONO + AgBr \tag1}$$
$$\ce{NaNO2 + R-Br -> R-NO2 + NaBr \tag2}$$
However, these results were contradicted by work of Kornblum who compared the reactions of iodoalkanes with $\ce{NaNO2}$ and $\ce{AgNO2}$ (Ref.2):
$$\ce{C7H15CH2-I + AgNO2/NaNO3 -> C7H15CH2-NO2 + C7H15CH2-ONO + AgI \tag3}$$
It was shown by the work of Kornblum and coworkers that both primary alkyl bromides and iodides have given excellent yields of pure nitroalkanes (73-83% as major product) when $\ce{AgNO2}$ is used as the nucleophile (Ref.3). In contrast, primary chlorides are completely inert under the conditions $(\pu{0\!-\! 25 ^\circ C}/\ce{Et2O})$:
$$
\begin{array}{l|r|r}
\hline
\text{Heloalkane} & \% \ \ce{R-NO2} & \% \ \ce{R-ONO}  \\
\hline
\text{$n$-Butyl bromide} & 73 & 13 \\
\text{$n$-butyl iodide} & 74 & 12 \\
\text{$n$-Hexyl chloride} & 0 & 0 \\
\text{$n$-Hexyl bromide} & 76 & 10 \\
\text{$n$-Hexyl iodide} & 78 & 13 \\
\text{$n$-Heptyl bromide} & 79 & 11 \\
\text{$n$-Heptyl iodide} & 82 & 10 \\
\text{$n$-Octyl chloride} & 0 & 0 \\
\text{$n$-Octyl bromide} & 80 & 14 \\
\text{$n$-Octyl iodide} & 83 & 11 \\
\hline
\end{array}
$$
Based on the results obtained, the authors had concluded that this is a good method to prepare primary-nitroalkanes (Note that nitroalkanes, $\ce{R-NO2}$, are thermodynamically more stable than alkyl nitrites, $\ce{R-ONO}$; Ref.1). Some of the further work by Kornblum et al. have noted that use of tertiary alkyl halides to prepare corresponding nitro compounds is useless (0-5% yield) while those reactions with secondary alkyl halides have also given only less than 15% yield of expected nitroalkanes (Ref.4). After extensive  work on synthesis of nitroalkanes and alkyl nitrites, declared that:

In the synthesis of saturated primary nitro compounds, silver nitrite gives nitroparaffins in about 80% yield as against about 60% yields obtained with sodium nitrite. While silver nitrite is, therefore, the reagent of choice here, the lower cost and ready availability of sodium nitrite, the not very large disparity in yield, and the shorter reaction time all combine to make sodium nitrite an excellent second choice*" (p.114, Ref.2; and Ref.5).

In their mechanistic studies, Kornblum et al. (Ref.6) have found that yield of $\ce{R-NO2}$ isomer falls progressively when substrate changes from primary to secondary to tertiary halide. In contrast, the yield of $\ce{R-ONO}$ counterpart increases with the same order of primary to secondary to tertiary. Although there is a steric effect in play, the authors conclude that the reaction proceeds via a transition state, which has both $\mathrm{S_N1}$ and $\mathrm{S_N2}$ character in proportions that vary gradually with the structure of the halide. The greater the $\mathrm{S_N1}$ character of the transition state, the greater is the preferences for covalency formation with the atom with higher electronegativity (e.g., $\ce{R-ONO}$). Conversely, the greater the $\mathrm{S_N2}$ contribution to the transition state, the greater the preferences for covalency formation to the atom with lower electronegativity (e.g., $\ce{R-NO2}$).
In this regard, I'd say the reaction, $\ce{R-X + NaNO2 ->[DMF] R-NO2 + NaX}$ would proceed differently under different conditions. Since the solvent is (assuming) anhydrous DMF, the condition is set for $\mathrm{S_N2}$ reactions. Thus, for instance:

If $\ce{X = Cl}$ and $\ce{R-X}$ is a primary alkyl chloride, the reaction would not proceed.
If $\ce{X = Br or I}$ and $\ce{R-X}$ is a primary alkyl halide, the reaction would
proceed rapidly to give nitroalkane, $\ce{R-NO2}$, as the major product but in lower yield $(\approx60\%)$. There should also be some alkyl nitrites, $\ce{R-ONO}$ as well (Ref.3).
If $\ce{X = Br or I}$ and $\ce{R-X}$ is a secondary alkyl halide, the reaction would
proceed to give both nitroalkane, $\ce{R-NO2}$, and alkyl nitrites, $\ce{R-ONO}$, as the product. The yield would be lower $(\approx20\%)$ (Ref.4).

I think that would be enough of discussion since very little information is given in the question. For example, we cannot decide whether reaction goes in ether $\mathrm{S_N1}$ or $\mathrm{S_N2}$ without knowing the nature of $\ce{R}$ group. Knowing the structure of $\ce{R}$ group is extremely important since even some specific primary halides such as neo-pentyl halide do not react under same conditions where other primary halides proceed rapidly.
Finally, it is noteworthy to mention following reaction, which play a significant role in the product ratio of $\ce{R-NO2}$ to $\ce{R-ONO}$:

Reactions of the blue benzhydrylium salts, $\bf{1}$-$\ce{BF4}$, with $\ce{n-Bu4N+NO2-}$ in anhydrous acetonitrile gives colorless adducts, $\bf{1}$-$\ce{NO2}$ (Ref.5):

The monitoring method in Ref.5 does not suggest the product is direct formation of $\bf{1}$-$\ce{NO2}$ or through the kinetic product, $\bf{1}$-$\ce{ONO}$. It was suggest that when use more reactive benzhydrylium salts for the reaction, the monitored reaction process is the formation of $\bf{1}$-$\ce{ONO}$, which successively rearrange to the observed nitro compounds $\bf{1}$-$\ce{NO2}$. To prove this theory, the authors prepared pure $\bf{1}$-$\ce{ONO}$ in different method (Reaction $\bf{A}$), and monitored its rearrangement to thermodynamically stable $\bf{1}$-$\ce{NO2}$ (Reaction $\bf{B}$):

These experiments clearly suggest that preparation of nitroalkane is not simple as the reaction suggests, but rather complicated process!

Edit: In reaction $\bf{A}$, the aromatic ring of starting material $\bf{1i}$ and $\bf{1j}$ have two different para-substitution $\ce{R}$s: $\ce{R} = \ce{OMe}$ for $\bf{1i}$ and $\ce{R} = \ce{Me}$ for $\bf{1j}$ (Ref.5).

References:

Ian Fleming, In Frontier Orbitals and Organic Chemical Reactions-Student Edition; John Wiley & Sons, Ltd.: Chichester, West Sussex, United Kingdom, 2009, p. 122–123 (ISBN 978-0-470-74660-8).
Nathan Kornblum, “The Synthesis of Aliphatic and Alicyclic Nitro Compounds,” Organic Reactions 1962, Vol 12, 101 – 156 (DOI: https://doi.org/10.1002/0471264180.or012.03).
Nathan Kornblum, Bernard Taub, and Herbert E. Ungnade, “The Reaction of Silver Nitrite with Primary Alkyl Halides,” J. Am. Chem. Soc. 1954, 76(12), 3209–3211 (DOI: https://doi.org/10.1021/ja01641a029).
Nathan Kornblum, Robert A. Smiley, Herbert E. Ungnade, Alan M. White, Bernard Taub, and Stephen A. Herbert Jr., “The Reaction of Silver Nitrite with Secondary and Tertiary Alkyl Halides,” J. Am. Chem. Soc. 1955, 77(21), 5528–5533 (DOI: https://doi.org/10.1021/ja01626a028).
Alexander A. Tishkov, Uli Schmidhammer, Stefan Roth, Eberhard Riedle, and Herbert Mayr, "Ambident Reactivity of the Nitrite Ion Revisited," Angew. Chem. Int. Ed. 2005, 44(29), 4623 –4626 (DOI: https://doi.org/10.1002/anie.200501274).
Nathan Kornblum, Robert A. Smiley, Robert K. Blackwood, and Don C. Iffland, “The Mechanism of the Reaction of Silver Nitrite with Alkyl Halides. The Contrasting Reactions of Silver and Alkali Metal Salts with Alkyl Halides. The Alkylation of Ambident Anions,” J. Am. Chem. Soc. 1955, 77(23), 6269–6280 (DOI: https://doi.org/10.1021/ja01628a064).


text:
Why lowering of vapour pressure is not a colligative property?
In my textbook, it is written that lowering of vapour pressure is not a colligative property but relative lowering of vapour pressure is a colligative property. But why is that? There is literally no explanation given in my textbook as to why this is true. Please someone explain. I am so confused. Please help.
(By lowering of vapour pressure I mean $P_{0} - P_{S}$ where $P_{0}$ is the vapour pressure of pure solvent and $P_{S}$ is the vapour pressure of the solvent containing non-volative solute. Relative Lowering in Vapour Pressure means $\frac{P_{0} - P_{S}}{P_{0}}$).

Could you clarify with respect to the difference between 'relative lowering' and 'lowering'?

Consider reporting the whole sentences

Yeah, like @Alchimista said, please give us the complete quote;  plus please include a full citation to the textbook (author, name, publication year, publisher, edition, etc.).

Both 'lowering of vapour pressure' and 'relative lowering of vapour pressure' are colligative properties.

In my textbook of Maharashtra state board it is mentioned that both lowering in vapor pressure and rative lowering in vap pressure are colligative properties.

A colligative property is a property which depends on number of solute particles and not on nature of solute particles. Relative lowering in vapour pressure is equal to the mole fraction of the solute . This can be derived using $P_s=P_0\cdot x_\text{solvent}$ where $x$ is the mole fraction and Po is the vapour pressure of pure solvent. Lowering of vapour pressure is also a colligative property as it is equal to $\frac{x_\text{solute}}{P_0}$. $P_0$ is a property of the solvent, not the solute.


text:
Reactivity order of DIBAL-H and diborane with esters and acids
I have been taught in class that

DIBAL-H is more reactive towards esters than carboxylic acids while opposite is true for diborane.

Also DIBAL-H mentioned is taken under low temperatures (−78 °C).
I was given no reason for the difference in preference.
Now I would suppose that ester would be slightly more nucleophilic and bulky than acids. From these conditions it is hard to conclude the above mentioned order.
According to me opposite order could be true because as DIBAL-H is bulky so it would be easier for it to approach a less bulky molecule.
Who do we rationalise the mentioned order? Besides is the difference between preference very large or is there just slight more preference for one than the other?

You may read long list of references here to find a viable answer.

The mechanisms of the two reagents are different. Compare https://www.ch.imperial.ac.uk/rzepa/blog/?p=5114   with https://www.masterorganicchemistry.com/2011/08/26/reagent-friday-di-isobutyl-aluminum-hydride-dibal/


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Increase in volume of solution in neutralization reaction and its effect on weak acid
If 100ml of 1M acetic acid and 20ml 0.01M NaOH is taken in a reaction medium,they will neutralize each other. I don't know for sure if this reaction is completely neutralized or not just took as an example. Now the water molecules produced in this reaction would increase slight amount of volume though they were present in the medium as ions but let us consider that the formation of molecule made those more spacious. Weak acids have a very low amount of dissoiation constant which depends on the concentration of the acid. How significant will be the effect of the slight change in volume of the solution to the dissociation of acid? I mean since the volume is increasing,concentration of the acid (ions) is supposed to decrease. Again due to the dissociation constant being too low, how significant will the change in volume of the solution be in the dissociation of acid?

Short answer: Not very much.

«Not very much», as stated by @Nisarg Bhavasar's comment below the OP's question intrigued me to provide an estimate about how much, or better, how little the relative amount of «new water», obtained by the neutralization is.  The estimation assumes the absence of water's autoprotolysis, and indeed, the amount is tiny.

With the neutalization of
$$ \ce{HOAc + NaOH -> H2O + NaOAc} $$
and the assumption of $\ce{HOAc}$ ($\pu{1 M}, \pu{100 mL}$) and $\ce{NaOH}$ ($\pu{0.01 M}, \pu{20 mL}$) it is evident that the molar amount of lye is the limiting factor in this reaction:
$$ c = \frac{n}{V}  \leftrightarrow n = c \cdot V$$
$$ n(\ce{NaOH}) = \pu{10 \times 10^{-3} mol/L\,  \cdot\, 20 \times 10^{-3} L} = \pu{200 \times 10^{-6} mol}$$
This equally is the amount of «new water»,

Assuming a molecular weight of water of $\pu{18.0 g/mol}$ (reference) and for simplicity a density of $\approx \pu{1.0 g/cm³}$ (loc. cit.), the concentration of water in water equates to
$$c = \frac{\pu{1000 g}} {\pu{18.0 g/mol}} = \pu{55.6 mol/L}$$

With a molar mass of $\pu{60.1 g/mol}$ ($\ce{HOAc}$) and $\pu{40.0 g/mol}$ $\ce{NaOH}$ and the molar concentrations given, an estimation still may assume for the density of these solutions $\approx \pu{1.0 g/mL}$ (reference). Thus, as they total for $\pu{120 mL}$, this is equivalent to $\approx \pu{120 g}$ «old water»,  or a molar amount of $\pu{6.67 mol}$.

It is time to set these quantities into context.
$$ \mathrm{ratio} = \frac{ \mathrm{\mbox{new water}} } { \mathrm{\mbox{old water} }} =  \frac{ \pu{200 \times 10^{-6} mol} } {\pu{6.67 mol}} = \pu{29.99 \times 10^{-6}} \approx \pu{3.0 \times 10^{-3}\%} $$

This is like for every 33350 molecules of water present by the reagent solutions, there is one molecule generated by the neutralization reaction between the acid and the base.
The autoprotolysis of water at room temperature equates $K_w = [\ce{H+}][\ce{OH-}] = \pu{10^{-14} mol^2/L^2}$.  Thus, in neuter water at room temperature and pH = 7, the concentration of $[\ce{H+}] + [\ce{OH-}] = 2 \cdot \pu{10^{-7} mol/L}$. This predicts for $\pu{120 mL}$ of neuter water $\pu{1.8 \times 10^{-8} mol}$ of $\ce{H+}$, and additional $\pu{1.8 \times 10^{-8} mol}$ of $\ce{OH-}$.  Because this is even much less, than the quantity of water newly generated by neutralization of the lye by the acid, the initial assumption to neglect the autoprotolysis of water is not invalidated.


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What an electrolyte "really" does in electrolysis?
I am reading about electrolysis and I have trouble to understand why an electrolyte must be added in order for electrolysis to occur. I have read this question but still can't get it. The definition of electrolyte is a substance that when dissolved produces an electrically conducting solution.
Electrolytes in electrolysis
Why we need an electrolyte in order for electrolysis to occur? In electrolysis of water the water is consumed not the electrolyte. So why we need the electrolyte? Couldn't the reaction happen without the electrolyte?
After reading the comments I have modified the question. As I said before I can't understand the purpose of the electrolyte. For example in the electrolysis of water with sulfuric acid why we add it? Water molecules are in contact with both anode and cathode. So why we need to add the electrolyte?
What happens at molecular level when the electrolyte is added? How the ions move in the solution (if they move at all)?

You are confusing between electrolysis and electrochemical cell. Both are opposite processes. The bulb glows when electrodes are inserted in an electrochemical cell and not in a electrolysis vessel.

For reference read this:  Electrolytic Cells and Electrolysis https://chem.libretexts.org/@go/page/266 (accessed May 13, 2021).

I don't understand if you need the remark in the comments above or this one will suffices: in order for the electrolysis to occur, the current entering one side of the cell need to escape from the other one. This requires electrical conductivity, and this is why you need an electrolyte in solution (as per the same definition that you have mentioned).

The conduction of a steady electrical current through a medium requires that the medium contain charge carriers that can move in response to an applied electric field. Pure water is a relatively poor electrical conductor because it has only a very low concentration of charge carriers: 100 nM each of hydronium and hydroxide ions. So the electrical conductivity can be increased by dissolving appropriate ionic substances, the so-called “electrolytes” such as NaCl, in the water. The electrical conductivity generally increases with increased electrolyte concentration.

What I can't get is why water molecules don't oxidise/reduce if they are in contact with the anode/cathode? Why we need to add the electrolyte?

The answers are good, but maybe this is giving you pause: say you apply 9 V to a pair of platinum electrodes in pure water. The 9 V will be divided into three portions: 1) the potential required to oxidize whatever at the anode, 2) the potential required to reduce whatever at the cathode, and 3) the potential drop across the water. Pure water has high resistivity, so, assuming typical cell electrode separation and electrode areas, the water’s resistance is very high. Hence almost all the applied 9V is uselessly dropped across the water and resultant current flow is very small. Small current

means very little useful electrolysis happens. Adding an electrochemically inactive electrolyte greatly increases the solution conductivity, which is why it is called an electrolyte, so the voltage drop across the water is greatly reduced and the oxidation at the anode and reduction at the cathode are no longer strangled by the water resistance. The resistance of the pure water is just an obstruction that is greatly mitigated by adding the electrolyte, even if the electrolyte does no chemistry per se.

@EdV So by adding electrolyte we reduce resistance. But where the ions of the electrolyte go? How they move in the solution (if they mvoe at all)?

The electrolyte ions move: cations toward the cathode and anions toward the anode, hence their respective names. This bipolar ionic current flow is required to complete the electrical circuit. But only electro active species get involved in the electrode reactions. So, in neutral aqueous solutions, water gets oxidized and reduced at the anode and cathode, respectively.

Sometimes the electrolyte directly takes part in the chemistry, by being oxidized or reduced at an electrode. In those cases, it is easy to understand why the concentration would have an influence on the kinetics of the redox reactions (and with that, on the observed electrical current).
For water electrolysis, the "active" electrolytes are hydrogen ions and hydroxide ions. If you change the pH, you change their concentration, and would expect to affect the rate of reaction. You are asking about other electrolytes such as sodium sulfate, which don't participate in electron transfer.
To explain their role, take a look at this animated GIF. It shows electrolysis of water in the presence of universal indicator and sodium sulfate:

Source: CarolinaBiological
With the 9 volts applied to the graphite electrodes, you can see that the electrolysis causes local changes in pH (because one half reaction consumes hydrogen ions and the other hydroxide ions). It takes quite a bit of effort to mix the solution (with the plastic tube) to get the pH homogeneous again. This shows you that the diffusion of hydrogen and hydroxide ions takes a while. In the absence of other electrolytes, this would lead to local charge accumulation as well, stopping the reaction. With the electrolyte (at much higher concentration than the hydrogen and hydroxide ion), we get a pH gradient only.
This pH gradient also slows things down, and increases the voltage that needs to be applied (cell potential is 1.3 V at constant pH, but according to Nernst law, it would increase because of the pH differences). Because we are applying 9 volts, this still works.

What an electrolyte “really” does in electrolysis?

By moving in the appropriate directions, the ions of the "passive" electrolyte prevent large accumulations of charge, and allow the electrolysis to proceed at higher rates than without their presence. If you decrease the gap between electrodes down to sub-Debye length, they are no longer neccessary.

An electrolyte (in your case water) keeps charge carriers (in your case $\ce{Na+}$  and $\ce{Cl-}$) separate from each other and allows them to travel to the electrodes.  There is no constraint that this has to be a liquid phase, or a solid (e.g., a polymer).
I speculate when you say

«In electrolysis of water the water is consumed not the electrolyte.»

you accidentally skip the notion of the electrochemical window equally relevant e.g., to cyclic voltammetry.  This means, if the electrode potential applied is too high, or too low, the same compound acting as electrolyte  simultaneously reacts and may be oxidized / reduced.
Maybe the new form still is capable to be an electrolyte. Maybe the e.g., the electrochemical reduction of the electrolyte triggers the formation of bonds and a network representing an obstacle for the charge carriers to move; or products of reaction of the electrolyte leave the condensed phase (e.g., as hydrogen gas), which gradually depletes the electrolyte's ability about the charge carriers, too.

My main problem is that I can't get why in first place water molecules can't be oxidised/reduced without the electrolyte If they are in contact with both anode and cathode.

@ado sar They can be, but it's really slow.

In 1961, I generated Hydrogen in chem lab using a dry cell battery and water. No electrolyte needed.  Also in 1961, I generated hydrogen to fill balloons, used water, household products found at home: AL, NaOh,  and a pepsi bottle. Worked fine.  No electrolyte needed.  Reading about the catalysts used in the electrolysis process, the writers would have you believe that they are inventing new science comparable to turning coal into diamonds. In fact they are uncovering information that was known for 100 years at least.

You do not seem to answer the question.

If you used NaOH, you used an electrolyte. If you used anything other than very pure water, it probably contained impurities that acted as electrolytes.


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On basic mercury salts
$\ce{HgCO3.3HgO}$ and $\ce{HgO.Hg(NH2)Cl}$ are called basic mercury carbonate and basic mercury amidochloride, respectively. I found this thread on the former; I've only seen the latter as the product of a qualitative test in one of my textbooks, and Google yields nothing on it.
Is "basic metal salt" a general term for salts of the form $\ce{M_yB_x.M_2O_x}$(M the metal, B some anion with valency y, x the valency of the metal) or is it just for some compounds of mercury? If it's just for mercury, which ones more are there?
Are there any "acidic salt" analogues? If so, what general formula do they have?

Metal hydrogen sulfates MHSO4 are generally acidic

Also, see this: https://chemistry.stackexchange.com/questions/133364/why-does-hgcl2-react-with-hco3-but-not-with-co32

Iodide of Millon's base has the formula $\ce{HgO.Hg(NH2)I}$. So, your compound of "basic mercury amidochloride" should analogically be called chloride of Millon's base.

@NilayGhish: I actually do know about the terms in your first comment; I just wondered if there was a separate subclass or something, because I was seeing so many mercuric salts with HgO in them temed as such.

No, there is no separate subclass. Salts which contains hydroxide with other anions are called basic salts and ones which contains hydrogen are called acidic salts. But this is a very crude classification because the above compounds are also considered basic salts when neither of them contains hydroxide. I assume alkali compounds have been used in their making and/or the overall pH is basic.

@NilayGhosh: that makes sense, but there's also basic mercury sulfate, HgSo4.2HgO, made by reacting mercuric nitrate and the sulfate anion. That doesn't smack of a basic medium.

The technical name for that compound is mercuric subsulfate or mercury oxonium sulfate because of presence of polymeric oxonium cation $\ce{[Hg3O2]_{x}^{2+}}$ in a 2D infinite layer. The sulfate ion lies within the infinite mesh of cation lattice. So, we should try to avoid the word "basic" as it might create confusion. Try using the proper name. For ex: FeOOH is called iron oxide hydroxide. It also contains hydroxide but nobody calls it "basic iron oxide".

@NilayGhosh: just to be clear; so the earlier point about the basic pH isn't valid in every case?

It depends. One calls them "basic" if it satisfies one of the 3 criteria: (1) If it contains hydroxide in its formula (2) Its overall pH is basic (3)It is made using alkali compounds. But then again, this whole classification create confusion. One more example: aq. soln. of KMnO4 is slightly basic. Why? Because, since it is prepared in the presence of alkali, the product might contain some leftover alkali and hence the solution is basic But nobody calls it "basic" potassium permanganate.

@NilayGhosh: Great, that last comment was really comprehensive, thanks. I get the thing about not using this; it's way too fuzzy a definition and it's unnecessary. But it comes up multiple times in inorganic analysis, so i guess asking about it is warranted.

There exist plenty of basic salts. For example,

Copper produces a "simple" copper carbonate $\ce{CuCO3}$ as high $\ce{CO2}$ pressure. At lower pressures, the following basic copper carbonates are produced : $\ce{Cu3(CO3)2(OH)2}$ or $\ce{Cu2CO3(OH)2}$ (malachite), depending on the partial $\ce{CO2}$ pressure. Other copper salts produce basic copper salts like : $\ce{Cu2NO3(OH)3}$ , $\ce{Cu4(OH)6Cl2}$, $\ce{Cu4(OH)6SO4}$ .
Zinc produces the two following basic zinc carbonates, according to the chosen pH : $\ce{Zn5(OH)6(CO3)2}$ or $\ce{Zn2CO3(OH)2}$
Iron(III) produces also basic salts like $\ce{FeSO4(OH)}$ which is produced by air oxidation of iron(II) sulfate. Dilution of a ferric acetate solution produces a precipitate of basic acetate $\ce{Fe(CH3COO)(OH)2}$
Aluminum solutions are often hydrolyzed producing basic salts. In the case of acetate, it gives a precipitate of basic acetate $\ce{Al(CH3COO)(OH)2}$
Cobalt chloride solutions treated by cold NaOH solutions give a precipitate of basic chloride $\ce{CoClOH}$ When treated by carbonate solutions they produce a great variety of basic carbonates.
This list is not exhaustive.

Could you verify the criteria detailed in the comments above? That solves my problem. This answer is great, btw

@Harry Holmes. Nilay Ghosh proposes three criteria, which can be discussed. 1) A basic salt must contain a hydroxide compound. No ! It may contain oxide ion $\ce{O^{2-}}$ like the mineral goethite $\ce{FeOOH}$. 2) Its overall pH is basic. It does not make any sense, as a precipitate or any solid does not have a pH value. 3) A basic salt is made using alcali compounds. NO ! It may be made using Barium hydroxide. And the example of impure potassium permanganate is poor.  It is basic when insure may mean that it is not basic as a pure compound.

I see. I could make a point for the second criterion, though; the basic pH might refer to the pH of the salt in solution. Either way, could you specify an alternate list of criteria? I know there aren't hard and fast rules for these common names, but a rough sketch would be helpful.

@Harry Holmes. The pH of a solution made with an insoluble salt is a non-sense. It may happen that there is not even one atom of the insoluble salt per liter water. As to the possible rules about basic salts, I am afraid there are none.


text:
Pressure of two flasks connected to each other at different temperatures

Two flasks of equal volume connected by a narrow tube (of negligible volume) are at $\pu{27 °C}$ and contain $\pu{0.7 mol}$ of $\ce{H2}$ at $\pu{0.5 atm}$. If one of the flasks is then immersed into a bath kept at $\pu{127 °C}$ while the other remains at $\pu{27 °C}$. What is the final pressure?

Will the pressure of the two flasks which are now kept at different temperatures be the same or will they have different pressures?
In the solution I found of this question, the pressure of the two flasks at different temperatures are taken to be equal. Can someone please explain why this is so?

Because they are connected. You don't have the two gases equally divided per amount of particles.

The gist here is that in order for each flask to be at equilibrium then there must not be any net flow of gas through the tube which connects the two flasks. Thus the whole system must be at constant pressure.  That means that there must be a 100 degree C temperature gradient along the tube. Thus the gas is denser at the 27 C end than at the 127 C end.  But since the tube was deemed to be "of negligible volume" you can ignore the gas in the tube when calculating the solution.

The easiest way to prove the equality in pressure is by taking a cross-section of the narrow tube and finding the forces acting on it. By extending this, we can prove that the pressures are equal.
Take a cross-section area of the narrow tube. We assume this is at steady state and there is no average motion of particles in the entire system.
Now, since there is no net average motion, this means that there is no pressure gradient across any cross-section. Expanding this cross-section into the tube, we see that the ends of the tube have to be at the same pressure since there is no net average motion of the gas within the two flasks at steady state (which is where the parameters are asked to be measured).
Therefore $$p_\mathrm{left} = p_\mathrm{right}$$


text:
Effect of side reactions on galvanic cell using iron rod as anode and sodium chloride as salt bridge
We just made a galvanic cell on a chemistry lesson using iron, iron sulfate , copper and copper sulfate.
We used a sponge soaked in $\ce{NaCl}$ solution as the salt bridge. The issue is that we measured a voltage way below the expected value and it was dropping quite fast.
The iron sulfate solution was yellow which may have indicated that the iron ions were reacting with chloride ions.
I just can't see why this reaction would cause it to stop working from a charge perspective. What is happening here?

Iron sulfide solution doesn't exist it's insoluble in water may be you are seeing is iron chloride complex which is yellow

I'm sorry i meant iron sulfate not iron sulfide

Iron(II) solutions as easily and quickly oxidized by oxygen from the air, producing a yellow solution of basic Fe(III) ions, according to : $$\ce{4 Fe^{2+} + O2 + 2H2O -> 4 [Fe(OH)]^{2+}}$$

Since you state, "We measured a voltage way below the expected value and it was dropping quite fast," it indicates a few things that could lead to those observations.

How did you measure voltage? If you used a low impedance voltmeter, it would show much lower voltage because the impedance of the meter is low compared to that of the cell. Voltage is distributed in proportion to the impedance (here, the DC resistance) of the meter and the cell. For example a 1 volt cell with internal resistance of 20 kilohms, measured by a voltmeter with resistance of 20 kilohms, would show as only 0.5 VDC, because half the voltage is dropped inside the cell.
As a cell delivers current, the surface of the electrodes change. Corrosion, surface deposits and bubbles can cause current to drop. While the cell is being measured, observe the electrodes closely. Look for any change in appearance that progresses during measurement.
How well were the two half-cells isolated? Did the sponge effectively block liquid (as opposed to ion) flow between them? You can test that by putting some dye, such as food coloring or bromothymol blue, on one side and seeing if it migrates to the other. Flow might indicate the cell was "short circuited".


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What is the physical significance of the universal gas constant R?
While studying work done in isothermal processes I was told that the universal gas constant $R$ is the work done by the gas per mole per kelvin. I am sure I am missing something basic but I didn't find anything beyond that statement when I looked for an explanation.
This is what I found on Wikipedia:

The physical significance of $R$ is work per degree per mole.

In doing what, is 8.314 joules of work done by one mole of an ideal gas?

Charles' law says that at constant pressure the volume and temperature of an ideal gas are related as
$$\frac{V_1}{T_1}=\frac{V_2}{T_2}$$
If $V_2=V_1+dV$ and $T_2=T_1+dT$ then
$$\frac{V_1}{T_1}=\frac{V_1+dV}{T_1+dT}=\frac{V_1}{T_1}\left(\frac{1+dV/V_1}{1+dT/T_1}\right)$$
which can be rearranged into
$$\frac{dV}{dT}=\frac{V_1}{T_1}$$
But Charles' law says that the rhs in the previous expression is a constant ratio at constant pressure and amount of substance. And by multiplying both sides by $-p$ we have that
$$-p\frac{dV}{dT}=\frac{dw}{dT}=-p\frac{V_1}{T_1}$$
Then invoking the ideal gas law for one mole of substance we obtain
$$\left(\frac{\partial w}{\partial T}\right)_p=-R$$
This result could of course have been obtained without using such a roundabout way, by direct differentiation of the ideal gas law (holding composition and pressure constant) or by simple finite differences (again applying Charles' law):
$$w=-p\Delta V = -nR\Delta T$$
So $R$ is indeed a proportionality constant between work and temperature (a measure of thermal energy) change per mole of ideal gas.

In doing what is 8.314 joules of work done by one mole of an ideal gas?

Assume that during an expansion against constant pressure one mole of an ideal gas does an amount of work equal to $-R\cdot \pu{1 K}\cdot\pu{1 mol} = \pu{-8.3145 J}$. That is the amount of work that is done while the temperature of the gas increases by 1 kelvin (due to the heat that would need to be supplied to maintain constant pressure).

Yeah, that's a lot better than my answer. I don't think it's worth keeping mine around, luckily OP unaccepted it.

@orthocresol My answer is perhaps not the best way to think of R, but it's one. The question was surprisingly tricky to get a grip on.

As it turns out, thermodynamics is hard. :)

It may be helpful to look at a related value $k_\mathrm{B}$, the Boltzmann constant, which is widely used in thermodynamics.
These two are related by $R = k_\mathrm{B}N_\mathrm{A}$, allowing the ideal gas law to also be written:
$$PV = Nk_\mathrm{B}T$$ where $N$ is the number of particles, as opposed to the number of moles.
The units are $\pu{J\cdot K^{-1}}$.
It's a proportionality between energy and temperature. In the ideal gas law, it relates temperature on the right to the product of pressure and volume, which is an energy.
This general idea is frequently used in thermodynamics, as you will see factors of the form $\exp{(-E/k_\mathrm{B}T)}$, where the $k_\mathrm{B}$ allows the exponent here to be unitless.
As examples:

Planck's law, where the energy is in the form of quantum energy level spacing:
$$B_{\nu}(\nu, T) = \frac{2h\nu^{3}}{c^{2}}\frac{1}{\exp\left(\frac{h\nu}{k_{\mathrm{B}}T}\right) - 1}$$
Maxwell-Boltzmann distribution, where the energy refers to the kinetic energy of gas molecules:
$$f(v) = \left(\frac{m}{2\pi k_{\mathrm{B}}T}\right)^{\frac{3}{2}}4\pi v^{2}\exp\left(-\frac{mv^{2}}{2k_{\mathrm{B}}T}\right)$$

$R$ is a proportionality constant between the units we use to measure temperature (kelvins) and the units we use to measure energy (joules) and number of particles (moles). If you measure temperature in energy units and use the number of particles instead of the number of moles, the constant $R$ (and $k_\mathrm{B}$) disappears from your formulas.
$k_\mathrm{B}$ and $R$ exist for historical reasons, due to choices made in the definition of the Kelvin temperature scale and the Celsius temperature scale before it, as well as the fact that the link between temperature and energy was not understood at the time when the scales were conceived. The concept of temperature seems to predate that of energy, with attempts systematically at standardizing its measurement dating to 170 CE. The modern concept of energy didn't exist until Newton's mechanics (although Aristotle had a qualitative idea of potentiality). The earliest energy concept was the vis viva, which is proportional to kinetic energy and was proposed by Leibniz around 1676. The Fahrenheit scale, the first popular quantification of temperature, was invented around 1724, and was based on the boiling point of pure water and the freezing point of a salt water mixture. A scale similar to the modern Celsius scale based on the freezing and boiling points of pure water, was proposed in 1743. The fact that mechanical energy could be converted into a change in temperature was not understood mathematically until the experiments of Joule in around 1843.
The value of $R$ was deduced from studies of mechanical properties of gases and was known at least by 1856. Its development required the concept that gases were made of small particles and the relation between the number of particles and molecular mass. The first value of Avogadro's number was estimated by Lochschmidt in 1865. This is allowed the Boltzmann constant ($k_\mathrm{B}$) to be defined, linking temperature directly to the microscopic kinetic energy of particles in an ideal gas.
The modern understanding of the relation between temperature, energy, and entropy wasn't arrived at until the work of Boltzmann and Gibbs in the late 1800s:
$$\frac{1}{k_\mathrm{B} T} = \left( \frac{\partial \ln \Omega}{\delta U} \right)_{V,N}$$
Here $U$ is the internal energy, $\Omega$ is the number of microstates corresponding to the system's macrostate, and the derivative is taken with the volume of the system ($V$) and number of particles ($N$) fixed.
Some good statistical mechanics textbooks (for example, Landau and Lifshitz), present formulas with temperature in energy units (or equivalently, with energy in temperature units), so neither $R$ nor $k_\mathrm{B}$ appear in their formulas.

From kinetic theory the total kinetic energy of an ideal gas is $\frac{1}{2}mC^2 $ where $C$ is the root mean square velocity. This is also equal to $\frac{3}{2}RT$ which means that the gas constant $R$ is equal to two thirds of the total translational kinetic energy of a mole of ideal gas at $1$ degree kelvin.
In practical terms it was discovered (starting in 1870 ) that if the product $pV$ for a gas (e.g. H2, N2, O2 ) is plotted against $p$ from about $0\to 40$ atm. that the plot is largely linear, especially so at low pressures, and that the limit at $p\to 0$ is the same for all gasses at the triple point of water $pV\to 22.4$ litres and $30.6$ at the boiling point of water and so on.  In other words $pV$ is a constant at a given temperature, and the constant of proportionality is the gas constant $R$. Using statistical mechanics this can be given a molecular interpretation.


text:
Oxidation of sulfur dioxide to trioxide and its decomposition under same condition at higher temperature
I found that $\ce{SO2 + 1/2 O2 -> SO3}$ when catalysed by platinum and heated to $\pu{700 ^\circ C}$. On the other hand I found that $\ce{SO3 -> SO2 + 1/2 O2}$ with same catalysator and temperature about $\pu{800 ^\circ C}$. How depends these two cases in real reaction? What controls which reaction will happen?
I thought that one reaction is exothermic and other endotermic, and duration of reaction and temperature controls which reaction happens. But facts I found are different. Sorry for maybe too simple question. I would be pleased with some tips which chapters of chemistry to study to understand my problem more deeply.

https://chemistry.stackexchange.com/a/149516/106377 Similar concept applies here.

There is an equilibrium between $\ce{SO3}$ and $\ce{SO2 + 1/2 O2}$, at least at high temperatures and with an appropriate catalyst.  We could write the equilibrium as
$$\ce{SO2(g) + 1/2 O2(g) <=> SO3(g)}$$
Just from writing this reaction this way we can glean several things:

The reaction combines 1.5 gas molecules into one gas molecule, a net reduction in the number of molecules.  Since fewer molecules (but the same number of atoms) means there are fewer translational degrees of freedom, the entropy will decrease the more $\ce{SO3}$ "product" is formed.  Higher temperatures will thus disfavor the right-hand $\ce{SO3}$ side of the equilibrium and favor the left-hand $\ce{SO2 + 1/2 O2}$ side.

Total pressure will also affect the equilibrium but in the opposite way.  Higher pressure (crudely speaking) is a force that squeezes things together, so at high pressures the right-hand side will be favored.

The other principle factor that influences the equilibrium would be the partial pressure of $\ce{SO2}$, of $\ce{O2}$, and of $\ce{SO3}$.  So for example  if you had 99.9 mol % oxygen gas and only a trace of $\ce{SO2}$, that would favor the conversion of nearly all sulfur to the LH sulfur trioxide side.

The exact balance will depend on the precise enthalpy change of the reaction, the partial pressure of each species, the temperature, and the total pressure.  You could modify the system to favor one side or the other, and could model this accurately and quantitatively with appropriate numerical data.


text:
What happens to an enzyme after it breaks down a starch molecule?
I am a master distiller at a distillery and I am about to make some whiskey. First adding corn, cooking it and releasing it's starches. After that, will add in some malt, packed with enzymes to break down those starch molecules into sugars. My question is, what happens to the enzyme after it breaks down the starch? can 1 enzyme breakdown multiple starch molecules?

Pretty much nothing - that's the point of catalysis. Enzymes are molecular machines doing reactions.

So yes, multiple starch molecules

Enzymes are a form of catalyst. Catalysts do not get consumed in a chemical reaction. I believe in reality, enzymes do degrade overtime and you need to replenish them overtime to keep the reaction efficient.
Can one (1) enzyme break down multiple starches? Sure.
I did not find specific literature on fermentation, but I found one (1) carbonic anhydrase in our tissues and bloodstream can hydrate 1,000,000 carbon dioxide molecules per second. So your enzyme probably has a similar magnitude of catalytic rate.
Reference
Berg JM, Tymoczko JL, Stryer L. Biochemistry. 5th edition. New York: W H Freeman; 2002. Section 8.1, Enzymes Are Powerful and Highly Specific Catalysts. Available from: https://www.ncbi.nlm.nih.gov/books/NBK22380/

So the enzymes in our GI tract get reabsorbed and reused? They won't be used up like a chemical reactant or get excreted? Am I right?

Little clarification may needed. When you say enzymes in our GI get reabsorbed, you mean the enzymes are reabsorbed by intestinal lining? Enzymes do degrade eventually, so the body needs nutrients to make new ones.

Yeah I meant anywhere in our alimentary canal. I always thought that enzymes get used up completely everytime like reactants in a chemical reaction. It's useful to know that they are almost like catalysts which practically decompose after reasonable number of use.

So far I only find evidence for protein and hormone absorption by the gut lumen. Given digestive enzymes are proteins themselves, they should be reabsorbed into the epithelial cell too. I reference this article.


text:
Are these correct?
We were tasked to write the electronic configuration of the elements in the third energy level.
My answers are:
1s2 2s2 2p6 3s1
1s2 2s2 2p6 3s2
1s2 2s2 2p6 3s2 3p1
1s2 2s2 2p6 3s2 3p2
1s2 2s2 2p6 3s2 3p3
1s2 2s2 2p6 3s2 3p4
1s2 2s2 2p6 3s2 3p5
1s2 2s2 2p6 3s2 3p6
But I am not sure if these are in the third energy level cause my logic is new period in the periodic table = new energy level so I just wrote all electronic configuration of elements in the third period.

So you were asked to write the electron configurations for the elements in the third energy level? I am a little confused. I think elements do not reside on energy levels, electrons do. OR, were you asked to write the electron configurations for the elements in the third period of the periodic table? If this is case, your answer should be correct from sodium to argon. I am not aware of any elements violate the Aufbau Principle in the third period.
If you were actually asked to write the electron configurations for the elements containing electrons reside in the third energy level (n=3), I think you need to write the elements starting from sodium all the way to the end, Oganesson.
You can use this electronic periodic table to check your work. https://ptable.com/#Electrons


text:
Would gold have the negative charge in gold phosphides?
It should make sense that phosphorus has the positive charge,and gold itself
should have the negative charge in gold phosphide(and for any other gold phosphides) because phosphorus has a lower electronegativity than gold (phosphorus is 2.19, gold is 2.54),but then, they can both just share electrons. Is this true or not?

FWIW, two other phosphides are also known: $\ce{Au2P3}$ and $\ce{Au3P4}$

This source^ suggests it's alloy-like. Hard to tell what actual partial charges would be, but it's indeed possible that Au hard slight negative one.

Because their electronegativity difference is less than 0.4, I think the bond would have a lot of covalent character. I personally don't think it is a good idea to treat this substance as an ionic compound and assign a positively charged ion and a negatively charged.
According to a chemical supply company, $\ce{Au}$ is +3 and $\ce{P}$ is -3.  We typically write the cation first, so from the formula $\ce{AuP}$, you should get another clue that the gold ion is positively charged (sort of if you have to assign a positively charged ion).
I don't think you can assign charge signs solely on electronegativities. Atomic radius and position on the periodic table also play roles in determining which atom is more positively charged.
For electronegativity, more reference here.

The electronegativity of an atom can change in different environments. Gold as an isolated atom is less electronegative than fluorine, but when oxidized to the $+5$ oxidation state it can force a positive charge on a difluorine ligand. Phosphorus is apparently more electronegative than silicon, yet it becomes positively charged (by donating its extra valence electron to the band structure) when used as an $n$-dopant in silicon-based semiconductors.
So we cannot tell a priori how the charge will be distributed in a material with predominant covalent or metallic bonding from the electronegativities seen in an element chart. Unless we have detailed calculations to resolve this, the actual distribution of charges in gold containing phosphorus may be hypothesized as practically anything.


text:
Why is strontium(II) ion bigger than krypton atom?
$\ce{Sr^2+}$ is exactly the same as $\ce{Kr}$, in terms of electrons and orbitals.
The only difference between the two, is that $\ce{Sr^2+}$ has a couple of extra protons in the nucleus (and probably a couple of extra neutrons too, but these don't influence ionic/atomic size).
Considering the two extra protons in the nucleus of $\ce{Sr^2+}$, and the overall 2+ charge on the ion,
shouldn't this draw the valence electrons in closer to the nucleus in $\ce{Sr^2+}$ in comparison to $\ce{Kr}$?
(Krypton size= 88 pm, while Sr2+ is 132 pm.)

These are different, incomparable kinds of bigness.

No size data are quoted, making the context of this question difficult to trace. Do tell.

References for size data being considered would be very useful as well

Incomparable kinds of bigness? I see a textbook problem comparing the radius of an sodium atom to its ion (see Zumdahl & Zumdahl, 9th Edition, Ch. 7, Exercise 109). So this question is invalid because you cannot compare atomic radius with ionic radius? Notwithstanding my last sentence, I need to say that I never saw a question on a standardized test asks me to compare atomic radius with ionic radius.

Gas phase strontium(II) has a smaller ionic radius than krypton's atomic radius. The thing is that in a crystal lattice it isn't quite the same as in the gas phase.

Thankyou so much for all the useful comments and discussion! It's helping me understand

Sr2+ is exactly the same as Kr, in terms of electrons and orbitals.

I think the appropriate word for this relation is isoelectronic.
According to Ptable and LumenLearning, $\ce{Kr}$ has an atomic radius of 88 pm, while $\ce{Sr^2+}$ has an atomic radius of 132 pm. Although it is true that $\ce{Kr}$ and $\ce{Sr^2+}$ have the same electron orbital configuration and $\ce{Sr^2+}$ has two more protons than $\ce{Kr}$, $\ce{Sr^2+}$'s greater nuclear attraction is likely countered by the nature of metal ions being more willingly to lose electrons, which implies the electrons are more loosely held by the nucleus than the noble gas $\ce{Kr}$.
Another reason I conjecture is that $\ce{Sr^2+}$ has unfilled 5s orbitals to readily accept promoted electrons, while $\ce{Kr}$'s electrons have less tendency to fill the 5s orbitals because the property of being a noble gas.

Comparing Vander Waal's radius with atomic radius doesn't make sense. Krypton has an atomic radius of 88 pm which is less than 132 pm.

I thought they were the same thing. I will do some research and edit the answer accordingly.

You are comparing atomic radius with ionic radius. They are not the same thing.

Please see my comments under the question.


text:
How do you name a bicyclo compound with a fused benzene ring?
What is the IUPAC rule for naming a bicyclo compound with a fused benzene ring? Are you required to state the number of double bonds by adding something like 2,4,6-ene to the name or do benzene rings have a specific name ?
For example:

What will this compound be named? Will it be bicyclo[4.3.0]non 1,2,4 triene or is there some specific rule for fused benzene rings, as I stated before?

This is, derived from the stem compound, indene, $\text{2,3-dihydro-1}\textit{H}\text{-indene}.$

One follows the rules for fused rings. Years ago there were different acceptable approaches. And many structures can have traditional name (in the discussion at least, for identification, as it goes with data such as mp, H-NMR, elemental analysis,... perhaps one has to be stricter. But name as carbazole, naphthalene, indene, etc are OK.

Please don't answer in comments.

Also known as indane

Bicyclic compounds with a fused benzene ring may be named correctly in two or more ways in accordance with the several methods recommended by IUPAC.

If there is a retained name (e.g. naphthalene) for the parent ring components listed in Chapter P-25, this name is preferred.
If there is no retained name, a fusion name (e.g. benzo[8]annulene) is preferred for a benzene ring fused to another cycle of five or more members.
In the remaining cases, a von Baeyer name (e.g. bicyclo[4.1.0]hepta-1,3,5-triene) is used.

The following list shows examples for benzene rings with simple bridges from 1 to 6.

Beside systematic nomenclature about organic structures, there are other systems in use.  One of them is the by Hantzsch and Widman which departs from a number of stem structures.  Some of these archetypical motifs are named systematically, others (still) retain their traditional name which does say little  to you about the structure (e.g., pyridine) until you learn them.  One (incomplete) compilation of these names, for example, is this one.
This being said, you either look up, or over time (i.e., training) recognize that the parent structure of your molecule is indene:

except that the double bond in the five-membered ring of the structure you showed is completely saturated.
You remember reactions on benzene, and recall about the importance to denote the sites in organic structures.  The approach to use numbers equally exists for larger compounds, too; a few positions are labeled.  While there are cases where you may shorten 1H-indene to indene, this no longer is the case to describe unambiguously the compound shown by you, which is 2,3-dihydro-1H-indene.

I didn't say I apply Hantzsch-Widman here. I followed @Long's comment "Please don't answer in comments".


text:
How do steric effects affect inductive effect in compounds with tetravalent nitrogen?
I want to compare the electron density on methyl group between tri‐tert‐butyl(methyl)azanium (1) and triisopropyl(methyl)azanium (2), i.e. essentially the inductive effects of $\ce{-\overset{+}{N}R3}$ groups.
$$\underset{\large\textbf{1}}{\ce{Me-\overset{+}{N}(t-Bu)3}}\qquad\underset{\large\textbf{2}}{\ce{Me-\overset{+}{N}(i-Pr)3}}$$
The only difference I see in these compounds is that due to steric effects the bond angle would be greater in compound 1 than in 2. Based on this info alone I am not able to judge which one would have greater electronegativity on $\ce{\overset{+}{N}},$ or so as to say greater inductive effect due to the $\ce{-\overset{+}{N}R3}$ group.
How should I go about comparing the electron density on methyl group in both of these species?

An $\ce{sp}$ hybridized species, which has a bond angle of $180^{\circ}$ is more electronegative than an $\ce{sp^2}$ hybridized species, which has a bond angle of $120^{\circ}$ followed by $\ce{sp^3}$ which has a bond angle of $109.5^{\circ}$. So, as bond angle increases, electronegativity increases. So, as you said that 1 has a greater bond angle than 2, it should be more electronegative and hence it would show greater inductive effect on $\overset{+}{\ce{N}}$.

Are you considering a theoretical comparison or approaches to develop experimental approaches to produce measurements?

theoretical comparison

@LightYagami that seems like you are relating two phenomenon's but why does that happen. Even If ur relations are true, I don't really understand how can steric effects/repulsions lead to higher electronegativity

Since you are asking for a theoretical explanation, I believe a theoretical calculation might be helpful. It is possible to calculate the partial charges on atoms using electronic structure programs. I have calculated the partial charges at $\text{B3LYP/pcseg-1//GFN2-xTB}$ level (with GAMESS and xTB).
$\ce{Me-\overset{+}{N}(^iPr)3}$
We only need to look at the partial charges on the Me group to know the electron-withdrawing capacity of $\ce{-N+(^iPr)_3}$ group.
\begin{array}{cc} \hline
\mathrm{atom} & \mathrm{partial\;charge\;(e)} \\ \hline
\ce{C} & -0.4367\\
\ce{H} & \;\;\, 0.1954\\
\ce{H} & \;\;\, 0.1957\\
\ce{H} & \;\;\, 0.1966\\ \hline
\end{array}
Total charge on $\ce{Me} = 0.151$.
$\ce{Me-\overset{+}{N}(^tBu)3}$
Again, we only need to look at the Me group.
\begin{array}{cc} \hline
\mathrm{atom} & \mathrm{partial\;charge\;(e)} \\ \hline
\ce{C} & -0.3087\\
\ce{H} & \;\;\, 0.1436\\
\ce{H} & \;\;\, 0.1430\\
\ce{H} & \;\;\, 0.1414\\ \hline
\end{array}
Total charge on $\ce{Me} = 0.1193$
So, the partial charges would indicate that $\ce{-N+(^iPr)3}$ is more electron withdrawing than $\ce{-N+(^tBu)3}$.
Explanation?
I am not sure I have a definitive explanation for this. But it's possible to look at the differences between $\ce{^iPr}$ and $\ce{^tBu}$—

The cone angle formed by the three R groups for $\ce{-N+R3}$ would be larger for $\ce{R=^tBu}$ obviously, due to the steric bulk of that group. When the angles are the ideal $109.5^\circ$ each bond can be assumed to form from $\mathrm{N\;sp^3}$ orbitals. When N is $\mathrm{sp^2}$, the three bonds would be $120^\circ$ to each other. So, when the cone angle increases, the amound of p-character in those three N-R bonds can be assumed to go down. Conversely the amount of p-character in the 4th bond goes up which would mean less electron withdrawing capacity (p-orbitals are further away from nucleus). So, $\ce{-N+(^tBu)3}$ should be weaker EWG.

As there is no empty p-orbitals on N, so there should be no hyperconjugative effects for the compounds drawn. However, if instead of $\ce{Me}$, there is some other group that has a $\pi$ system attached to N, then there is a possibility of hyperconjugation.

The $\ce{CH3}$ groups have a very weak electron donating effect, so $\ce{-N+(^tBu)3}$ will be slightly weaker EWG than $\ce{-N+(^iPr)3}$. *

Notice that points 1 and 3 predict the same trend, which matches with the calculated trend. However, it is difficult to tell which factor is the most important (and whether these factors actually mean anything).

* Curiously, the partial charge on $\ce{C}$ of the side chain goes up when its $\ce{H}$ is replaced by $\ce{CH3}$, which would suggest that $\ce{CH3}$ is actually electron-withdrawing. This does not match with the standard notion of methyl groups being electron donating. I am not really sure if there is any explanation for this discrepancy. But note that there have been some evidence that $\ce{-CH3}$ groups can act as EWG's in some circumstances. The linked paper specifically mentions in the abstract—"That methyl groups attached to carbon atoms are electron donors must not be generally assumed."

Both N are sp3 hybridized. Both N are quaternary nitrogens with a positive formal charge. The tBu group is more electron donating (think vectors of the N-C-C geometries) than the iPr group, but there is more of a steric penalty due to its bulk; on the other hand, the N-C-H linkage in the iPr ammonium cation will also have a slightly larger dipole than the N-C-C. So, in answer to your question about electron density, I would say the [MeN(tBu)3]+ again because of the EDG effect of tBu > iPr.
Does this make sense? :)


text:
How to identify multiple molecules in a single xyz file?
I wonder if there exists a tool/script/algorithm to identify multiple molecules and/or atoms in a single XYZ or gaussian log file.
As an example, imagine one does a single point calculation for the ethanol-$\ce{Cl2}$ system, where the two molecules are placed next to each other at some distance. If one wants to detect all participating entities (be it molecules or atoms) what would be the best approach?
Could we use "standard" bond lengths between atoms and atoms allowed max number of bonds to do that? I suspect the problem is much more complicated than that.

Yes, you have stumbled upon the fact that there is no unique physical/mathematical definition of a molecule. The same goes for bonds. Nothing stops me from considering a dimer as a single entity. In the calculation it is even treated as such. But as you realized, you can make up your own definitions and check for them. If you know or assume that you have certain monomers that are retained, you can check for them. But  you need to define them in some way. Using a distance as bond length and checking all atom pair distances for it is a possibility. Many visualization tools work that way.


text:
What is the difference between lead tetraacetate and sodium periodate as reagents?
I recently came across this question which has confused me about the  usage of $\ce{Pb(OAc)4}$ and $\ce{NaIO4}$ reagent.
Both the reagents have nearly the same cyclization mechanism, but the answer to this problem says that $\ce{Pb(OAc)4}$ should be used for the conversion instead of $\ce{NaIO4}$.
The first thing that I noticed was that, the compound is a trans-1,2-diol, so $\ce{NaIO4}$ might not be the appropriate reagent, but I could not understand how $\ce{Pb(OAc)4}$ is appropriate for this conversation .
Please suggest the essential difference between these reagents.

I am not sure but the book from which I studied the mechanism, it is written that the cyclic mechanism was proposed by Criegee. Perhaps after investigation that mechanism was discarded.

Have a look at this: https://en.wikipedia.org/wiki/Criegee_oxidation . It gives several mechanisms. The radical one is for beta-amino alcohols so maybe I was wrong.


text:
Grignard reaction with cycloheptatriene
I'm having trouble with the following set of reactions and their products.

Product A is achieved by exploiting the unique aromaticity of cycloheptatriene cations. When A  is treated with the Grignard reagent $\ce{MeMgI}$, a mixture of B and C is obtained. Treating A with catalytic amount of $\ce{HCl}$ in $\ce{CCl_4}$ furnishes isomer A’, which is again converted into a mixture of B and C upon treatment with Lewis-acidic Grignard reagents.

For A I got the following structure, since I could not figure out how the aromaticity of the first structure is affecting the reaction.

This cannot be correct, since the grignard reaction would produce only one product and not a mixture of two.

Having done some reasearch online on the resonance structures of cycloheptatriene, I thought of the following structure for A. But this would not produce two different products either if I am not mistaken.

I think your initial structure for A is correct. Consider that MeMgI can attack A in a 1,2 mode or 1,4

Would that result in the products B and C being 1) the one below my structure for A and the other 2) being the same as 1) but with methyl group moved to the adjacent carbon in the cycloheptatriene structure?

correct  2) Me group in the benzylic position on benzocycloheptadiene ring - like your last structure but without the OH

The product A, shown below, is formed by first reducing the ketone to a secondary alcohol with $\ce{LiAlH_4}$ and then by adding a methyl group to form an ether using the reagent $\ce{CH_3I}$.
$\hskip2in$
Next the grignard reaction with product A will form two products. The grignard reagent $\ce{MeMgI}$ can attack A in a $1$,$2$ or in a $1$,$4$ mode, due to the unique aromaticity of cycloheptatriene, giving a mixture of products B and C, the structures of which are shown below.
$\hskip1in$


text:
Keto-enol tautomerism in a bicyclo compound
Question

Which one is more stable- the keto or enol form. The compound given is this:

Solution

The enol form would violate Bredt's Rule and would therefore be unstable. Hence the keto form is more stable

My doubt
But I am not entirely convinced. Can't we have alpha hydrogens migrating from either of the carbon atoms I've indicated by arrows? Probably they're expecting us to assume that the alpha hydrogen would migrate from the bridge-head carbon, but I just wanted to know if there's any loophole in my concepts. Any help would be appreciated.

Such tautomers are, like always, minor ones - it doesn't make a difference.

Please don't use the homework tag, it's depreciated.

@Mithoron Please elaborate. How did you decide that its a minor tautomer? I don't quite understand.

By now you should know that ketone is usually favored not enol.

Two arguments against this enol. 1) The atoms attached to the double bond do not lie in the same plane; 2) The p-orbitals of the double bond do not overlap with each other nor with those of the carbonyl group for that matter.

@user55119 OP understands that the enol shown is unstable. (S)he is asking why $\alpha$-H can't migrate from the marked positions.

@Ritam_Dasgupta: If the OP's question is does the equilibrium shown occur, the answer is no. If the question is can enolization occur at the ethano-bridge, the answer is yes.

@user55119 agreed.

Most of the times, the keto-tautomer is lower in energy than the enol-tautomer of a given compound. In all cases this is also very dependent of the actual conditions.
There are few cases, where the enol-tautomer can be dominant. One of the most common examples is malondialdehyde, and, of course, its derivatives.
I am fairly certain that the original question is trying to trick you into thinking this way; and testing your 3D awareness. Or something like this; I'm not sure.
With these structures it's always a good idea to visualise then. I recommend a molecule kit for hands on approaches, but clay and matches are great, too. Even chestnuts and toothpicks are great tools. If you are technically inclined, there are great programs, such as Avogadro (avogadro.cc), which can do basic modeling.
For the following, I have used gfn2-xtb as an optimizer; that is a semi-empirical level of theory, see xtb-docs.readthedocs.io.

You can clearly see that if you migrate the bridgehead-hydrogen, the resulting lone pair would not be able to form resonance with either of the carbonyl groups. That is known as the empirical observation named Bredt's rule.
On the other hand you are quite correct that you could form a tautomer from the other α-carbon. This carbon can easily distort to have planar coordination and the lone pair can be in resonance with the carbonyl groups. (Provided of course that the framework is localised bond orbitals; and that the model is a valid approach.)

On this level of theory, the enol form is about 64.5 kJ/mol higher in energy than the keto-form. So while there certainly is a theoretical possibility, the likelihood is very small.
(Just for giggles: the bridge tautomer is 246.6 kJ/mol higher in energy.)

Appendix

Keto-Tautomer
17
symmetry c1
C       -0.232194517     -0.455059284     -0.108971891
C        0.435695625      0.521858137      0.853516654
H        0.456616659     -1.075279085     -0.674274662
C       -0.652035075      0.924669592      1.832141805
O        1.572314637      0.893979605      0.826125495
C       -1.888025507      0.221717603      1.265101729
H       -0.393829692      0.553384752      2.825493466
H       -0.740240162      2.008868613      1.896308913
C       -2.240914528      0.890781562     -0.065758217
H       -2.719883618      0.149774125      1.962761267
C       -1.080374859      0.487832943     -0.956781929
H       -3.167688540      0.496107389     -0.485905106
H       -2.336524789      1.975006913     -0.012902862
O       -0.862082222      0.837168772     -2.079995835
C       -1.289181703     -1.103984849      0.787745832
H       -1.992066775     -1.719717994      0.228564710
H       -0.851574934     -1.695108793      1.590840632

Enol-Tautomer
17
symmetry c1
C       -1.252989400     -1.048025267      0.775079043
C       -0.235131476     -0.354975263     -0.135182749
C        0.316010839      0.694441373      0.817231157
C       -0.666608713      1.048338273      1.645393318
C       -1.895763330      0.268945971      1.228776635
C       -2.331880748      0.853259922     -0.128837161
C       -1.153613858      0.494907935     -1.014278443
O        1.568858497      1.186824174      0.741278311
O       -0.952399200      0.822318914     -2.150244908
H        0.493388572     -0.958265049     -0.672260873
H        2.056130434      0.748941273      0.034835696
H       -0.640184178      1.790739752      2.416980529
H       -2.686913718      0.203266237      1.970705846
H       -3.231277420      0.373221210     -0.518677687
H       -2.502755602      1.928070454     -0.108020068
H       -1.934290934     -1.708837720      0.239732785
H       -0.777205764     -1.591075191      1.590309569

Bridgehead-enol-tautomer
17
symmetry c1
C       -1.375174317     -1.103606241      0.843045628
C       -0.248454950     -0.489532946     -0.004463589
C        0.359288830      0.442230533      0.837984097
C       -0.605216725      0.791584850      1.950720322
C       -1.905351826      0.244045543      1.301925006
C       -2.166063721      0.867188742     -0.087057960
C       -0.935024925      0.450040319     -0.933532366
O        1.456979798      1.143062588      0.565484195
O       -0.594744149      0.929934770     -1.975783635
H        1.590334610      1.860425911      1.199389402
H       -0.422954467      0.294599984      2.906280842
H       -0.604773151      1.872841199      2.115837534
H       -2.773451234      0.269475098      1.961354663
H       -3.086355462      0.469282331     -0.513923137
H       -2.229917288      1.956721084     -0.092669863
H       -2.077588583     -1.677749691      0.243281460
H       -0.985680441     -1.728640075      1.644336401

I think you are correct, the α-H at the positions you marked would be able to migrate and form an enol tautomer. However, the keto form would still be more stable because in the enol form we do not gain aromaticity or conjugation, but lose out on the strong intermolecular H-bonding that the keto form was undergoing.


text:
Salt solubility and solvent dissociation
A textbook, describing a qualitative test for the $\ce{Al^3+}$ cation in which you use $\ce{NH4Cl}$ and $\ce{NH4OH}$ to precipitate $\ce{Al(OH)3}$, mentioned that the solubility of this precipitate decreases in the presence of the ammonium salt.
I sort of reasoned it out to the common-ion effect decreasing the $\ce{OH-}$ concentration before I wondered why that would have an effect at all. As far as I know, the solubility of a salt only varies with the solvent used, the pressure, and the temperature (and of course, on the common-ion effect if it's present).
I made up an explanation, that a more dissociated solvent wouldn't have to physically contort to solvate the salt ions; that when the bonds are broken, the ions of the solvent can more freely arrange themselves in optimal positions to stabilize the salt constituents.
Is that line of reasoning correct? If not, how can the relation between solvent dissociation and salt solubility be rationalized?

Water dissociation is not important here at all. It is the equilibrium between NH4+ and OH- that matters.

@IvanNeretin: yes, that equilibrium was what I meant. The common-ion effect effectively decreases OH- concentration by acting upon this equilibrium, as I understand it. But how does that affect the salt solubility?

Salt? What salt?

Common ion effect is the key.
A solution of $\ce{NH4OH}$ and $\ce{NH4Cl}$ is basically(no pun intended) a basic buffer solution, which has an almost  constant $\mathrm{pH}$ and thus a fixed concentration of hydroxide ions.
The solubility product of $\ce{Al(OH)3}$ as you say is also constant. So as to maintain the constant nature of both $[\ce{OH-}]$ and $K_{\textrm{sp}}$, there will be a fixed equilibrium concentration of $\ce{Al^3+}$ for a given $\mathrm{pH}$ above which it will start precipitating as $\ce{Al(OH)3}$. By increasing the hydroxide concentration we are forcing the $\ce{Al^3+}$ cation to precipitate earlier than it would have done in a normal, neutral aqueous solution.
This process of forcefully changing the $\mathrm{pH}$ of a solution to alter the solubility of different compounds is very common. In this case it was used to reduce the solubility and precipitate a substance, in other case it might be used to increase the solubility.
For example, during the same qualitative analysis schedule you mentioned, $\ce{HCl}$ is added alongside $\ce{H2S}$ during the detection of group 2 cations(like $\ce{Cu^2+, Hg^2+}$) to prevent precipitation of group 4 cations(like $\ce{Zn^2+, Mn^2+}$) in the form of sulphides. As the acidic conditions increase the solubility of sulphides, added with the fact that group 4 sulphides are not that insoluble prevents them from precipitating with the group 2 cation sulphides.


text:
How many boiling points does a mixture of liquids have?
Let's say I have a mixture of two miscible liquids. I want to get the liquids by fractional/pure distillation. My textbook says a mixture boils over a range of temperatures. On the contrary, another book I read says a mixture has only one boiling point and that the liquid with a lower boiling point adds more of itself to the mixture. I am wondering which of those is correct and will work with distillation.

The mixture boils over a range of temperatures. The lowest point of that range is called the boiling point.

There is too few information to anticipate if a separation of the two liquids by distillation may be achieved, or not.  Depending on their mutual interaction, perhaps they form an azeotrope and you only reach a maximal concentration of A in B (c /= 100%), e.g., HCl/water.

The actual mixture (at a given x) has one defined boiling point (at given P) but as boiling change composition the boiling occurs over a range of T.

There is no unique answer to this question. A mixture water-ethanol may boil at a temperature  between $78$°C and $100$°C, depending on the relative proportion of the constituants. A mixture of two non miscible liquids like oil + water boil at a given temperature, whatever the proportions of the constituants.

It would be nice to know which books you are reading.

My school science notes

I do use my comment above because the meaning of the books is, with little interpretation, clear and the answer is simple and don't require digging in theory of fractional distillation. So it would be pity to let the question unanswered.
A mixture as those discussed boils over a range of T in the sense that its boiling point BP depend on its composition.
The actual mixture, i. e. at a given molar ratio, has one well defined BP (at given P).
If you do not look at a given composition, but you actually conduct a distillation, as the boiling change composition the BP changes all along the process/column/plates, actually spanning a range of T.

so as the mixture boils, the composition changes and the boiling point also changes

I like @NinePlusTenEqualsTwentyOne revised answer, the boiling point changes as the substance is being boiled. But is it changing from the substance with the lower bp to the higher bp? Or no correlation?