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SN1 products for (1S,2R,4R)-1-chloro-2,4-dimethylcyclohexane
Question

Total number of SN1 products of given compound are
A) $3$
B) $4$
C) $5$
D) $6$

Answer

 (C) $5$

My attempt
I can find only $4$ products. The two stereoisomers formed from the carbocation, when chlorine leaves, and two stereoisomers formed when the hydrogen from the neighbouring carbon centre migrates (i.e. rearrangement occurs) and nucleophilic attack takes place from both sides.
I still am unable to find the fifth product. Moreover, I don't suppose the $\ce{-CH3}$ opposite to the point where the carbocation forms can interfere with the product. What would be the fifth product?

https://chemistry.meta.stackexchange.com/questions/4808/a-concise-template-for-homework-questions. When asking problem statements, could you try to conform to this template? You seem to be someone who would stick through SE for an extended period of time and we would be glad to keep it that way

@Safdar I started actively today in this forum. Hence I am unaware about the etiquette. I shall keep the code of questioning in mind.

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

Is there no other information given in this question?

I don't think a 5th product should be possible here

@Safdar No, nothing more is provided in the question.

A very weird proposition, a hydride shift to the other side since the number of alpha hydrogens increases from 3 to 4, the other shift is more stable since it goes from 3 to 7, but still is possible. This makes the answer 6 though. @YusufHasan

Are stereoisomers considered? (I'm guessing they are)

@AniruddhaDeb they have to be since you only have 3 possible locations for the carbocation to achieve.

Since total number of products is more than 2, we have to counter initial secondary carbocation survived enough time to give two of other products. If that so, 1,2-hydride transfer can happen in both side to give the fifth (and sixth?) product. If the answer is expected to be number of theoretical products, then there are a lot of possibilities to have more than 5 products. That means, the question is misleading.

@Safdar Yeah, if we do the shift towards the other side, I guess we can end up with 6(not 5) products. As Mathew says, this is probably a misleading question

This is a bad question. The nucleophile should be specified. What if we used a mildly nucleophilic, non-basic source of hydride?

@Safdar I thought about the hidride shift proposition too. But, the ring strain generated during migration to opposite side will be immense.

Yes, maybe its an incomplete question or everything is not sufficiently provided.


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Why is the molar entropy of hexane lower than that of butane?
Entropy is dependent on the complexity of the molecule.
Due to this, I expected entropy to increase as the complexity of the molecule increased.
According to p 918 of Atkins' Physical Chemistry (9th ed.), I noticed that the standard molar entropy of n-hexane at is $\pu{204.3 J mol-1 K-1}$, whereas that for n-butane is $\pu{310.23 J mol-1 K-1}$.
How can butane have a higher standard molar entropy, even though it is a smaller molecule?

Thing is butane is gas!

I should also point out that standard molar entropies must be specified with a temperature, as the word "standard" by itself does not indicate a temperature. If your book does not point out that these entropies apply to 298 K (or whatever temperature they used), then it is doing you a disservice.

Also, annoyingly, there's some disagreement over the value of the molar entropies in the sources I have on hand (neither is given in the CRC Handbook), but I think the general idea that butane > hexane appears to be valid. For now I've stuck with values which I can actually cite; if anybody has better values please go ahead and update it.

There are two factors to consider. Certainly, the dominant factor is that the degrees of freedom available for the molecule to disperse energy into increases as the carbon chain extends. Thinking along these lines, one would expect a monotonic increase of entropy with carbon chain length.
However, we must also consider phase. As the carbon chain increases in length, the equilibrium phase at RTP changes from gas to liquid. This occurs between butane and pentane. At this point, there is a drop in the standard entropy, which then proceeds to increase monotonically from this point (pentane). The cause for this drop in entropy upon phase-change is due to the motional degrees of freedom being more constrained by proximity to other molecules in liquids compared to gasses.
See this link, where you can see some nice plots of standard entropy vs. carbon chain length.

The next question that arises is why a gas and a liquid would have so little difference in entropies  but I guess hexane being much more complex would compensate it.

It is a problem of phase.
The standard entropy of butane is $\pu{310 J mol-1 K-1}$ in the gaseous phase.
The standard entropy of hexane is $\pu{296 J mol-1 K-1}$ in the liquid phase. But it is $\pu{389 J mol-1 K-1}$ in the gaseous phase. So the entropy increases from butane to hexane in the gaseous phase, as expected.

Where did you get the number 389 from?

The entropy of gaseous n-hexane is given in the reference : D.W. Scott, Correlation of the Chemical Thermodynamic Properties of Alkane Hydrocarbons, J. Chem. Phys. 1974, 60, 3144 - 3165.

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Comparison of basicity of o-phenanthroline and ammonia
I was trying to compare the basicity of o-phenanthroline and ammonia.
According to me, the factors that are seen in o-phenanthroline are:

o-Phenanthroline has two nitrogen atoms at the 1,10 positions.
It has a free lone pair on one nitrogen that is not delocalized due to resonance.
The nitrogens are $\mathrm{sp^2}$ hybridized which is more electronegative compared to the $\mathrm{sp^3}$ lone pair on ammonia.

However, would there be hydrogen bonding in its conjugate base? If so, then how do we compare its basicity with a base like ammonia?

A quote from here explains and give a good answer to your wondering:

Amines are the most basic of the common organic functional groups, but are still fairly weak bases. Protonation occurs on the non-bonded electron pair exclusively. The basicity of amines is directly dependent on the “electron density” at the nitrogen atom. Both inductive and resonance effects can alter the basicity of a nitrogen atom.
Hybridization on the $\ce{N}$ also affects basicity. An increase in $\mathrm{s}$ character on an atom increases the electronegativity of that atom which favors acidity and therefore disfavors basicity. Hence $\mathrm{sp^3}$-hybridized nitrogen is more basic than either $\mathrm{sp^2}$ or $\mathrm{sp}$ hybridized nitrogen.

The availability of this non-bonding lone pair is a factor of basisity. The 1,10-Phenanthroline is a pyridine derivative. Thus, both lone pairs of two $\ce{N}$ atoms in the ring system are contributed to the system's aromaticity. This make them not 100% available to incoming protons while the lone pair in ammonia is 100% available. Therefore,  in aqueous solutions, the basicity of $\ce{NH3} \ (\mathrm{p}K_\mathrm{a} = 9.3)\gt$ basicity of 1,10-Phenanthroline $(\mathrm{p}K_\mathrm{a} \approx 4.9)$. This is similar to $\mathrm{p}K_\mathrm{a}$ of pyridine, which is $5.2$ in water (comparison with piperidine is depicted in following scheme):

Keep in mind that there are some other factors are effecting the basicity of amine as well. For example, the ring sizes of cyclic amines (Ref.1):

The order is 5-membered $\ge$ 4-membered $\gt$ 6-membered $\gg$ 3-membered.
References:

Scott Searles, Milton Tamres, Frank Block, Lloyd A. Quarterman, "Hydrogen Bonding and Basicity of Cyclic Imines," J. Am. Chem. Soc. 1956, 78(19), 4917–4920 (https://doi.org/10.1021/ja01600a029).

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Intuitive Explanation for why more negative charge on ligand does not imply more splitting effect?
I have studied coordination chemistry as part of my school curriculum and we covered Crystal Field Theory as our final theory, and the book says that Ligand Field Theory and Molecular Orbital Theory are used to explain the limitations of CFT but are beyond our scope of learning.
One of the limitations of CFT according to the book is that if the ligands are point charges then it should follow that an anionic ligand should exert more splitting effect but that is not the case as we see in the spectro-chemical series.
A good example would be $\ce{S^2-}$ < $\ce{F-}$ < $\mathrm{EDTA^{4-}}$ < $\ce{CO}$
Also we can see that size is not the determining factor as then $\mathrm{EDTA^{4-}}$ wouldn't be so high as compared to $\ce{F-}$
Is there some intuitive explanation or logic for this which I can follow with my current knowledge till CFT without going too deep into MOT or LFT?

One of the major flaws in your assumption is that you are using absolute negative charge as the cause for higher splitting. The truth is, it is not the absolute charge, but the overall charge density of the approaching ligand the determines the splitting. Ligands with higher charge density will often cause more splitting than those with low charge density. That explains the positioning of sulphide and fluoride. Chelating ligands also in general, cause more splitting. CO is a special case where synergic bonding strengthens the metal to ligand bond, and hence causes even more splitting

Thanks for the explanation if you can write that as an answer it would be better @YusufHasan

It would be good if you can add some data showing the variation in different cases you mentioned. Also is this the only factor if not what other factors come to play?@YusufHasan

You already know the answer to your question, as evidenced by your first paragraph. I don't think there's much of a shortcut to it. The charge density thing only accounts for a portion of the spectrochemical series. The rest can only be explained by MO theory.

Okay understood that @orthocresol I've learnt MOT for diatomic species having upto 20 electrons, Can that knowledge help me here?

It's a start. The polyatomic case is a bit more complicated. But if you keep at it I'm sure one day you'll reach it. Suffice it to say that things like CO, CN-, PPh3 etc have strong metal-ligand covalent bonding which is best described using MOT.

Crystal Field Theory assumes that all of the interactions occuring are purely electrostatic and only concern d-orbitals.  This is clearly not true when dealing with some ligands which create a covalent bond.
Most limitations of CFT all stem from the inability to explain non-electrostatic interactions.  For this reason, it works well when the electrostatic aspect is the dominant factor but it completely fails for some cases such as neutral CO ligands.  While you could easily argue electrostatic forces still exist in this situation, clearly they are not dominant and do not explain any observed behavior while still having an effect.


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I want to purposefully dry up (polymerize) an entire bottle of ethyl cyanoacrylate (Super Glue). How to do it?
So, I want to do some tests with the solid form of ethyl cyanoacrylate.
I have a small 20g bottle which was very cheap so I don't mind.
How can I cause the entire contents of the bottle to set and turn solid?
I know CA glues require moisture to polymerize, and that after some time it will eventually dry out...
Will just leaving the bottle open in air work? Can I do something to speed it up?

Polymerization of cyanoacrylate in contact with wet air should last a couple of minutes. No need to speed it up.

My problem is usually keeping a once-opened bottle of superglue un-polymerized...

All right then. Will just keep then cap off and wait for moisture and air to do their thing.

Sorry. I meant "The" cap off...

Exercise caution while working with cyanoacrylate. A part of population is shown to have allergic reaction to it on dermal exposure. Also, cyanoacrylate fumes can sometimes be toxic.

You probably don't want to polymerize the CA in the bottle, because then you will have to peel/pry the bottle material away, and the interior of the polymer mass could still be fluid if you do not wait long enough.
First, prepare a plan. For instance, arrange some barriers on waxed paper, say, birthday candles or some disposable sticks, or put waxed paper into a small box (as if you were preparing to bake a cake) to make a rectangle. Then pour the CA glue onto the waxed paper and let it harden in air. The thickness of the layer will determine how long you have to wait. After hardening, you can peel the waxed paper away, leaving a coupon which you can cut or test in various ways.
You really need good ventilation when spreading CA out; allergies and sensitization are common.
You can pour the CA onto glass fiber or sawdust or sand to make a composite material. Some of these combinations react exothermically, and emit much vapor/smoke; therefore the need for good ventilation to avoid breathing the vapors. Sounds like fun. Other polymerizable materials have literature available and commercial products will come in larger sizes and might be good to work with too.


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Synthesis question: Using reductive amination to form pyrrolidine
What molecule results from treating 4-oxopentanal with $\ce{NH4Cl}$ and $\ce{NaCNBH3}$? Why? Is this a good, high yield reaction?
I believe that pyrrolidine is formed because amination will convert one carbonyl to an amine that reacts with the other carbonyl. However, I am unsure how reasonable this is and I'm unsure what happens to the original carbonyl oxygen.

Reductive amination is OK. Succindialdehyde is harder to come by than pyrrolidine.

The carbonyl oxygens are lost as water during imine formation


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Effects of evaporating the solvent of a supersaturated solution
What happens to the solute if you heat up a supersaturated solution? Since a supersaturated solution is already more than saturated with solute and unstable, my theory is that as you heat the supersaturated solution, the solubility limit will have been increased enough to allow all of the solute to dissolve. So technically, the solute does not precipitate out if you heat it enough.

Is the solvent volatile? It depends on how you set up the experiment. If more solvent evaporates then the concentration of the solute will increase.

Molten sugar is miscible with water. NaCl in water at 100°C not. NaCl in water at 800°C (under high pressure) possibly, you´ll have to check.

Two caveats:

If you sufficiently disturb a supersaturated solution, the solute will spontaneously come out solution.  So you would need to heat it without sufficiently disturbing it.

The solubilities of some solutes decrease with temperature.  So let's assume your solute's solubility increases with temperature (as is the case for most, but not all, solids dissolved in a liquid).

So if we allow for the above then, as you heat the solution, you will eventually reach at point at which it is no longer supersaturated.
Indeed, the reverse of this is how supersaturated solutions are typically prepared:  A solution is prepared at a high temperature with a concentration of dissolved solid just below the saturation point (you don't want any undissolved solid left, since that will act as a nucleation site and thus prevent supersaturation).  Then it is carefully cooled.  So, by heating it sufficiently, you would simply be moving it back towards the starting point.
Finally, someone might argue that, if the solution is supersaturated near the boiling point, it might not be possible to heat it up sufficiently to get below the saturation concentration.  I suppose that's possible in theory, but I don't know how you would prepare such a system, since it couldn't be created using the typical cooling process.

The solubility of all solutes decrease with temperature.

@JonCuster The OP is talking about the solute preciptating out, so clearly he or she is referring to a solid dissolved in a liquid.  If the solvent is water, solubilities of solids typically (but don't always) increase with temperature. And even if the solute were a gas, it's not generally the case that the solubilities of gases decrease with temperature.  In water they generally do, but in organic solvents one can see the opposite.


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Why sugar solution/sugarcane juice becomes sour when kept for certain time?
I stir-dissolve some sugar in a cup of water and then keep it open for certain time (say a few hours), when I taste it I feel that it has became sour.
I have observed the same thing many times when I went to juice shop(s) and asked especially for sugarcane juice when I drank it, it tasted a bit sour (sometimes too sour to drink) and I just realized that the sugarcane (or other fruits) juice was extracted some hour(s) ago because I know that a fresh juice doesn't taste sour until kept (open) for a certain time.
From all this experience one thing I know is that the sugar solution or sugarcane (or other fruits) juice becomes more or less sour when kept for a certain time after extraction. It's most probably due to acid (which usually tastes sour) formation but I don't know what chemical changes turn a sugar solution or (sugarcane) juice sour (acidic) when kept for a certain time?
Any explanation will be greatly appreciated. Thanks.

Have you read about fermentation? Recall the actual mean of acid, it means "sour".

acetic (adj.)
1808 (in acetic acid), from French acétique "pertaining to vinegar, sour, having the properties of vinegar," from Latin acetum "vinegar" (properly vinum acetum "wine turned sour;" see vinegar), originally past participle of acere "be sharp; be sour" (related to acer "sharp"), from PIE *ak-eto-, suffixed form of root *ak- "be sharp, rise (out) to a point, pierce."

A warm and humid land, like India, and a sugary solution, is an ideal breeding ground for natural microorganisms (yeasts) that can turn sugar into acids. Try an experiment, take fresh sugar cane juice and cool it in a fridge. Does that taste sour? Don't drink fermented juice, make sure it is always fresh.

@M.Farooq: Thank you for your comments and giving important information. I would like to know the mechanism or chemical changes that actually turn sugary solution sour. It will be good if someone elaborates the things to exactly explain the reason.

https://en.wikipedia.org/wiki/Ethanol_fermentation#:~:text=Biochemical%20process%20of%20fermentation%20of%20sucrose,-A%20laboratory%20vessel&text=Alcoholic%20fermentation%20converts%20one%20mole,H5OH%20%2B%202%20CO

@Karl: You are very right but I like to know the ideas of others also. This also helps me learn a lot of things over here. Anyway thanks.

This is not a complete answer but it should give you a head start.

I suspect two things: Fermentation and Caramelization.
In any fermentation process, sugar will turn into acid. It is observed from this answer from cooking.SE that at a temperature of 35-45 °C, fermentation occurs. If any sugar laden food is kept at this temperature, bacterial or yeast colony would flourish giving you the fermentation effects and making your food acrid. Fermentation process is a two-step process. First, glucose is turned to ethanol by Saccharomyces cerevisae, also known as Brewer's yeast. This ethanol is then converted to acetic acid by Acetobacter aceti, giving you the sour effect. This two-step process is called aerobic fermentation.

Caramelization generally occurs at higher temperatures, at 150-170 °C where sugars are broken down to simpler products and turning sugar into a brown-black mixture of products. This reaction is also called Maillard reaction. At this temperature, the reaction happens almost instantly but at lower temperature (35-45 °C), this reaction will takes hours, days or even months to finish it but it will speed up if there is moisture where it will take a few hours. Note that caramelization process is very complex as there are various steps. The finished product is a brown-black goo which is actually a mixture of several products. But if put it in a simplified way, sucrose breaks into glucose and fructose which further breaks into acetic acid, maltol, furan and many other products. This acetic acid contributes to the sour taste. So, in a hot-humid day, sugar can turn sour.


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How to find number of moles of HNO3 behaving like acid in a redox reaction?

In the reaction $(1)$, how to find the number of moles of $\ce{HNO3}$, which is behaving like acid?

The reaction is given as follows:
$$\ce{Sn + HNO3 -> Sn(NO3)2 + NH4NO3 + H2O} \tag1$$
To find number of moles of $\ce{HNO3}$ behaving like acid, I first balanced the reaction using oxidation number method which gave me:
$$\ce{4Sn + 10HNO3 -> 4Sn(NO3)2 + NH4NO3 + 3H2O}$$
But I could not proceed any further. I am unable to find any hint. How should we proceed to reach the answer? Answer mentioned in book is 9.  This is a question from chapter "Redox Reactions".

Try splitting into half-equations to find how many moles of HNO3 are accepting electron pairs and acting as Lewis acids.

I also tried Half reaction method, but I couldn't observe the solution.

@Sahil in the future, enclose chemical equations and formulae in $\ce{}$ : it typesets chemistry automatically.

https://chemistry.stackexchange.com/questions/67911/how-can-i-prepare-snno32/67938#67938 ... https://chemistry.stackexchange.com/questions/103349/sn-hno3-when-will-the-metastannic-acid-h2sno3-appear ... https://chemistry.stackexchange.com/questions/8748/make-tin-nitrate

Welcome to Chemistry SE! When asking problem statements, could you try to conform to this template? You seem to be someone who would stick through SE for an extended period of time and we would be glad to keep it that way.

It is always easy to balance redox equations by considering two half reactions. The equation to be balanced is:
$$\ce{Sn + HNO3 -> Sn(NO3)2 + NH4NO3 + H2O}$$
Thus, the two relevant half reactions to this redox equation are:
$$\ce{Sn (s) -> Sn^2+ (aq)} \\
\ce{NO3- (aq) -> NH4+ (aq)}$$
First balance elements other than $\ce{O}$ and $\ce{H}$. In this case they are $\ce{Sn}$ in first equation and $\ce{N}$ in second equation. Then, balance $\ce{O}$ and $\ce{H}$ in both equations with water and $\ce{H+}$, respectively. This is because the reaction is in aqueous acidic medium:
$$\ce{ Sn (s)  -> Sn^2+ (aq)}\\
\ce{NO3- (aq) + 10 H+ (aq) -> NH4+ (aq) + 3 H2O (l)}$$
Finally, balance the positive charges by electrons so that each equation has no net charges:
$$\ce{ Sn (s)  -> Sn^2+ (aq) + 2e-} \tag1$$
$$\ce{NO3- (aq) + 10 H+ (aq) + 8 e- -> NH4+ (aq) + 3 H2O (l)} \tag2$$
These are your mass and charge balanced oxidation ($(1)$) and reduction ($(2)$) half-reactions. Now, you can add these two equations in order to cancel electrons. To do so add $4 \times (1) + (2)$:
$$\ce{ 4 Sn (s) + NO3- (aq) + 10 H+ (aq) -> 4 Sn^2+ (aq) + NH4+ (aq) + 3 H2O (l)} \tag3$$
This is your balanced ionic equation. As evidence, you need nine $\ce{NO3-}$ ions to balance as counter ions in right hand side, so you need to add $\ce{9NO3-}$ to both side to keep the mass and charge balance.:
$$\ce{ 4 Sn + NO3- + 10 H+ + 9 NO3- -> 4 Sn^2+ + NH4+ + 9 NO3- + 3 H2O} \tag4$$
When simplify, the equation $(4)$ would look like:
$$\ce{ 4 Sn +10 H+ + 10 NO3- -> 4 Sn(NO3)2 + NH4NO3 + 3 H2O} \tag5$$
Since $\ce{ +10 H+ + 10 NO3- = 10 HNO3}$, your complete balance equation is:
$$\ce{ 4 Sn +10 HNO3 -> 4 Sn(NO3)2 + NH4NO3 + 3 H2O}\tag5$$
Therefore, your answer is correct. Answer in your textbook must be a misprint or accidently for got that one $\ce{HNO3}$ is needed to be reduced. Other $\ce{9NO3-}$ is to balance the products (as counter ions). All $\ce{10H+}$ ions needed to complete the redox reaction.

While balancing,
$$\ce{Sn + HNO_{3} \longrightarrow Sn(NO_{3})_{2} + NH_{4}NO_{3} + H_{2}O}$$
first balance oxidising agent and reducing agent only, we will get $\ce{4Sn}$ and $\ce{1HNO_{3}}$ , after balancing them, write number of extra moles of acid obtained while balancing other atoms. That will give us the answer.


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Extrema in free energy of a solution?
I am a high school student reading about the second law of thermodynamics, and one of the equations given was the free energy change in dilution.
$$\Delta G_{\text{dil}} = -RT \log(C_1/C_2)$$
This implies that as you make a solution more dilute, the free energy decreases. However this equation does not take into account the entropy change in breaking a solid crystal, which I would expect to be highly positive as the crystal is a highly symmetric arrangement. Also it assumes ideal solution.
If we take into account the entropy change in breaking up the solid crystal, we see that the free energy increases while diluting, and also increases in breaking up the solid , that is, concentrating. So I would expect a minima in the free energy vs concentration curve. My guess is it might be the saturation point?
Is my reasoning correct? If it is, then is it possible to calculate the minima?

I think that the use of dilution in the description of the equation implies that you start with a solution, and then dilute, rather than starting from a solid. Starting from a solid, you would need to calculate the energy change for dissolving the solid first. Once the solid was completely dissolved, then you could use this equation for diluting it further.

Okay, but even without the equation my logic still holds,

If entropy in dilution is positive, then according to $\Delta G = \Delta H - T\Delta S$, the free energy decreases, does it not?

@ManitAgarwal-Elpsycongroo:  Sol is correct.  You can see this for yourself by substituting numbers into your expression. C1 is the initial concentration, and C2 is the final concentration.   Suppose C1 =20 and C2 = 10 (this would be a two-fold dilution; the final concentration, C2, is half the initial concentration, C1).  Then $\Delta G = -RT \ln (C1/C2) = -RT \ln (20/10) = –RT \ln 2 = -0.693 RT$. And since $RT>0$ , $\Delta G < 0$.  I suspect you got confused because you were thinking more dilute means a higher C, when in fact it's the opposite.

For Gibbs Free Energy, the standard equation is ΔG=ΔH−TΔS. The overall change in Gibbs Free Energy will depend, at least in part, on the enthalpy of dissolution for the specific substance. As for entropy, increasing entropy by breaking up the crystal decreases Gibbs Free Energy. The saturation point is where ΔG for the dissolution equals zero: where ΔH = TΔS. As ΔH and ΔS vary depending on the particulars of the solute and solvent, the saturation point also varies.
The equation you cite is specifically for when you have two solutions, so the dissolution of the solid is not described by it. Since the saturation point depends on the enthalpy of dissolution, you can't use the log equation to find it.


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Change in water vapor with change in pressure and volume
Given a closed (air tight) container (Volume $V_0$) held at the room temperature ($T_0$) and pressure ($P_0$). The container contains 100% saturated water vapor with no air (at $T_0$ and $P_0$).
Now assuming the container is compressible (say a balloon whose volume can be changed by applying uniform external force/pressure how does the water vapor inside the container change when we apply external pressure on its surface ($P_1$) and its volume changes ($V_1$).
Assuming the Ideal Gas Law I assume none of the water vapor would condensate with any such change in external pressure and volume or i am missing something?
Edit 1: the container contains no air and no liquid water either. Just water vapor at 100% saturation (whatever its value is at initial room temperature and pressure i.e. $T_0$ and $P_0$).

"Assuming the Ideal Gas Law I assume none of the water vapor would condensate with any such change in external pressure and volume or i am missing something?" Can you explain this assumption? You also need to indicate whether T is maintained constant during compression or allowed to rise. Look at a phase diagram of H2O to figure out how much pressure is needed to liquify the vapor at a given temp.

since the change in volume etc. happens in the open air and the effective external environment temperature won't change significantly we can assume the initial and final temperature of container to be effectively equal.. so yes the T is effectively constant.

I am not a chem student so I might get things wrong (phase diagram suggestion)..  can you help me assuming the temp as reasoned above doesn't change much..

I think the more interesting would be “ehat would happen if the chamber were insulated?”

If water is introduced in an empty container, the water will slowly evaporate. The pressure $p$ will slowly increase from $0\ \mathrm{Pa}$ to a maximum which is $3167\ \mathrm{Pa}$ at $25\ \mathrm{^\circ C}$. This corresponds to a molar concentration of water vapor equal to: $${c = p/RT = \frac{3167\ \mathrm{J\ m^{-3}}}{8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}\times298\ \mathrm K} = 1.278\ \mathrm{mol\ m^{-3}}}$$
Expressed in gram per cubic meter, the concentration of water is:  $ c = 23.0\ \mathrm{g\ m^{-3}}$.
This means that between $0$ to $23\ \mathrm{mg}$ water can be evaporated in a $1$ liter container at $25\ \mathrm{^\circ C}$, and that a maximum amount of $2.3\ \mathrm{mg}$ water can be evaporated in a $0.1$ liter container. As a consequence, if the gas volume of your water vapor is compressed from $1$ liter to $0.1$ liter without changing the temperature, $23\ \mathrm{mg} - 2.3\ \mathrm{mg} = 20.7\ \mathrm{mg}$ water will be condensed as a liquid on the inner wall of the container.

This neglects the small volume occupied by the liquid water(20 cc).

@ChetMiller 20 g of liquid water would be 20 cc or 20 mL. Here, it is a thousand times less.

There will be no water vapor in the container if you apply "room pressure" on it (~1 atm, applied with for instance a piston) at room temperature.
If you want to have coexistence between liquid water and its vapor you can set T or p but the remaining intensive variables (including density) are then fixed.
The reason is the requirement of chemical, thermal and mechanical balance between phases, which leads to Gibbs phase rule:
$$f=c+2-p$$
where f is the number of degrees of freedom (intensive variables such as T or p which you get to set), c is the number of components (here c=1, there is just water), and p is the number of phases (here p=2, because we want coexistence of liquid and vapor).
Therefore f=1. This means you get to set one intensive variable. The other ones depend functionally on the first, describing a coexistence line.
For a single phase, the application of a specific pressure implies that the density is no longer under your control at a fixed temperature. For two phases , at a specific {T, p} set on a liquid-vapor coexistence line the total volume can be changed independently (up to a point) but the density of both liquid and vapor will be constant).

I should add a caveat: on the coexistence line at a specific T, p set on the line, the total volume can be changed independently (up to a point) but the density of both liquid and vapor will be constants.

Assuming that $P_0$ corresponds to the equilibrium vapor pressure of water at $T_0$ and the temperature does not change from $T_0$ (i.e., equilibrates thermally with the room air), when you try to compress the saturated vapor, the pressure won't change significantly from $P_0$, but some of the vapor will condense to liquid water at $T_0$ and $P_0$.


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Mechanism of addition of H-X to alkynes
Vinylic carbocation is unstable as the carbocation is bonded to a $\mathrm{sp^2}$ carbon that has high electron density, and pulls the electrons from the carbocation, which is already electron-deficient (empty $\mathrm{p}$ orbital)
However, the mechanism for the addition of $\ce{H-X (X = Cl/Br/I)}$ to alkynes involves the formation of vinylic carbocation, which to me, should not be plausible due to extremely high activation energy (the cation is very unstable).
Is the mechanism wrong or my concept on the stability of vinylic cation is wrong?

If I'm not wrong, Carey & Sundberg's Advanced Organic Chemistry has a section on these reactions and the available information regarding their mechanisms (and associated literature references). My suspicion is that it need not necessarily follow the "standard" mechanism that is taught for alkenes. I don't have the time to read and summarise it now, but it may be worth looking in there.

If you look at the acid catalysed hydration of alkynes the intermediate is indeed a vinylic carbocation. See here for details  https://www.masterorganicchemistry.com/2013/05/14/hydroboration-and-oxymercuration-of-alkynes/

The exact same reaction is explained here https://www.masterorganicchemistry.com/2013/05/24/alkyne-reaction-patterns-the-carbocation-pathway/

"In general, alkynes are somewhat less reactive than alkenes toward many electrophiles. A major reason for this [...] is the substantially higher energy of the vinyl cation intermediate [...]" (Carey/Sundberg A, 5ed, p 539) They go on to say that addition can occur either via the cation like in alkenes, or in a termolecular fashion which doesn't involve a discrete vinyl cation intermediate - the proportion seems to depend on substitution pattern - although in the link @Safdar posted there is an argument that vinyl cations are very rarely true intermediates

https://youtu.be/d7qdeCYq3UY?t=2340, in this lecture, around 39:40, the lecturer says vinyl cation can ONLY be formed with the addition of alkyne. However, according to Hammond's postulate, the activation energy of this step should be extremely high that the reaction cannot proceed, which I think @Safdar's post is right. So may I ask for the rate law of this reaction?

https://pubs.acs.org/doi/pdf/10.1021/jo01292a023 The preview of this article talks about the dependance of rate law on atmost three different constants depending on reaction conditions

However that is for bromination, for addition of Hydrogen chloride, according to The Chemistry of the Carbon-Carbon Triple Bond, Part 1, Chap 8, p 281. It is stated that It would be second order for addition of H-Cl to acetylene.. I do not have access to the book itself so I can't say more. this was using google search in books to find out. Someone else who could access would be able to tell you more

@234ff That is not the correct application of the Hammond postulate. It is perfectly reasonable to argue that a transition state is higher in energy than the intermediate it leads to without any reference to the Hammond postulate.


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Comparison of carbocation stability in bicyclo compounds
Question

Which is the most stable carbocation ?

Answer

 3.

My attempt
I answered as 1 as the first one had +M of -O- and hyperconjugation effect as the positive charge is at para position. Why is the answer 3? It just has +I and +H of the second chain and +M is stronger than +I

When asking problem statements, could you try to conform to this template? You seem to be someone who would stick through SE for an extended period of time and we would be glad to keep it that way.

As for the answer, a hint would be the strength of oxygen in direct extended conjugation is much stronger than that of carbon. think on that front.

@Safdar okat ill stick to the template. about the hint, i dont understand what direct extended conjugation means. Could you please use a simpler term ?my chem level is that of a high school student

if you try to draw resonance structures, youll find that 3 has the most number of those.

I think the third option as greatest number of resonance structures can be alluded to it.
The following may suffice:

In the first structure, the carbocation is isolated except for the presence of a single double bond in conjugation beside it. The oxygen and the double bond beside it play no role in stablizing it.
In the second structure, the lone pair on oxygen, the double bond and the carbocation are in conjugation. Hence, it is somewhat stable.
In three, similar conjugation like two is present, but is it extended over another double bond i.e. the lone pair of oxygen, two double bonds and the carbocation are all in conjugation and an extended resonance comes into play. Thus, it is more stable than two, since more number of resonance structures may be drawn for it than 2.
The fourth one is the least stable among them, as no conjugation or resonace structures may be constructed. Moreover the I- effect of the neighbouring oxygen tends to destablise the carbocation.

Hence, the order of stablity:
3>2>1>4
An arrow-pushing representation of resonance in the third option can be made as follows:

Hope it helps.


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Hydrolysis of magnesium boride
Does magnesium boride react with water to form boranes?
My attempt:
Magnesium nitride reacts with water to form ammonia and magnesium hydroxide.
Similarly, does magnesium boride also react with water to form a borane and magnesium hydroxide?

Hydrolysis of magnesium boride has previously been attempted1. Two hydrolysis products were observed. The main hydrolysis product was found to be $\ce{Mg3B2(OH)6}$ but  at $\pu{-10 ^\circ C}$, a different product is formed whose composition was found to be $\ce{H3B2(MgOH)3}$ which immediately converts back to the main product if the temperature were slightly increased. The corresponding reactions were:
$$\ce{Mg3B2 + 6H2O -> Mg3B2(OH)6 + 3H2}$$
$$\ce{Mg3B2 + 3H2O -> H3B2(MgOH)3}$$
$$\ce{H3B2(MgOH)3 + 3H2O -> Mg3B2(OH)6 + 3H2}$$
No boric acid, magnesium borate or even boranes were formed although these products were expected from the hydrolysis reaction.
Reference

The hydrolysis of magnesium boride by Rames C. Ray  and  Prafulla C. Sinha , J. Chem. Soc., 1935, 1694-1695, DOI: 10.1039/JR9350001694 (PDF)

The compound $\ce{Mg_3B_2(OH)_6}$ seems to be made of $3$ $\ce{Mg(OH)_2}$ + $2$ atoms of elementary Boron. It looks as if Boron had been oxidized from $-3$ to $0$. Strange reaction ! Everything looks as if both Mg and B are at the oxidation number of zero in magnesium boride !!! Never seen such a strange reaction !

Are there other borides which gets hydrolsed to form boranes?

@NevilleVempeny Not that I know of, but action of dilute acids on magnesium boride gives various types of boranes especially $\ce{B4H10}$. See here and here for reference.

@Maurice I think there is something even more strange.  We are speaking of $\ce{Mg3B2}$, but does this even exist?  Magnesium-boron phase diagram I've seen show no compound with stoichiometry $\ce{MgB_x}$ for any $x<2$.


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Regioselectivity in coupling reactions of α-naphthol
I have seen in the following reaction of azo coupling where coupling takes place at position 4 of $\alpha$-naphthol.

Now the electron density accumulation is similar at both position $4$ as well as $5$.

However, there are more resonance structures that can be drawn for position 5. It should also be sterically easier to attack position 5.
The only reason I can think of for position 4 over 5 is that $\ce{=OH+}$ has a -I effect, which may somewhat stabilize the carbanion. But at the same time due to large distance this may be very weak and may not efficiently stabilize the negative charge. Then why is it in the para and not the peri position that coupling preferentially occurs?

The product at the para-position is more stable as it allows the compound to exist in the quinoid form, rather than only in the benzoid form:

This is important as in your case, if the substituent $\ce{X}$ is $\ce{H}$, then depending on the solvent, as much as 89% of it may exist in the quinoid form ($\bf{2}$).
References:

Satoshi Kishimoto, Shinya Kitahara, Osamu Manabe, Hachiro Hiyama, “Studies on coupling reactions of 1-naphthol. 6. Tautomerism and dissociation of 4-arylazo-1-naphthols in various solvents,” J. Org. Chem. 1978, 43(20), 3882–3886 (https://doi.org/10.1021/jo00414a020).


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Are moles of the solution equals to moles of the solvent + moles of the solute?
I'm a little bit confused here, does the mole of a solution equals to the mole of the solvent + mole of the solute?
Any help would be appreciated. Thank you.

Welcome to Chemistry SE! What efforts have you made to find the answer on your own? could you add that in your question?

A mole of solution has no meaning. Moles are applied only to pure substances, not to mixtures or solutions.

Maurice is wrong about moles only being applied to pure substances. Consider a flow of air that goes into a process. It is perfectly acceptable to say that there 10 moles of air per min in flow. Air is a mixture. It contains about 79% nitrogen gas and 21% oxygen gas. Knowing this, I can now tell you that because 10 moles of air are flowing, this also means that 7.9 moles of nitrogen are flowing and 2.1 moles of oxygen are flowing!
By this same logic say I have a solution of household white vinegar. It IS in fact a solution. Acetic acid is the solute and water is the solvent. Regular white vinegar usually has about 5% acetic acid by weight. So say I had 100 g of white vinegar. 5 g of that solution would be acetic acid and the other 95 g would be water.
So to calculate the average molar mass of a solution using mass fractions, the formula is 1/(avgMW) = sum(massfrac of component i/MW of component i)
So in the vinegar case above, the molecular weight of a 5% solution of household white vinegar would be  18.95 g/mol. I can now use this to tell you how many mols are in 100 g of 5% white vinegar (it's 5.28).
I hope this let's you see that moles are not additive in the manner that you described in your question. If you're still not understanding, think about the fact that in a chemical reaction, mols are not always conserved, but mass always is. Mols are really just a convenient stoichiometric tool. Like there are 12 eggs in a dozen, there are 18.02g in a mol of water.
Hope this helps :)

@ cqferrier. You are right in stating that one mole air has a meaning, because air has a constant composition. I am not so sure about the meaning of one mole vinegar. Vinegar has not always the same composition. 5% is an average value. Some vinegars may have up to 6% acetic acid,

By other words, the particular composition of a mixture must be supposed.

A mole of vinegar solution has meaning if the concentration is given. However, it has no useful meaning, because that mole of solution is not an amount that is used for a stoichiometric relationship.


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What happens when you burn salt in paraffin
My 9 year-old son and I tried to do some extremely amateur flame testing...
We lit a paraffin burner [a little tin of paraffin jelly that came with a marshmallow 'smores' kit!] and sprinkled on about a teaspoon of salt. We got a brief yellow flame (as expected), which dimmed to blue. Then we simply let it burn for a while outside.
After a couple of hours, when it had all cooled down, we discovered that the salt had coalesced into a kind of mass. We removed it and found that it had the consistency of chewing gum. What was left behind was a clear, non-viscous liquid which smelled of paraffin.
We hoped to replicate the flame experiment, using more salt and the 'leftover paraffin' liquid. However, we found that the liquid is non-flammable!
I wondered if anyone could explain this. What exactly did the salt absorb that turned it into 'gum'? What liquid was left behind? And why is that liquid non-flammable?
(I'm delighted that we happened upon something strange and unexpected - much more interesting kitchen-chemistry than an Alka-Seltzer bottle rocket!)


text:
How is the temperature in which vapor pressure is equal to atmospheric pressure the boiling point of the liquid?
I was just learning about vapor pressure and got very confused. Vapor pressure was defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system, which was the pressure the vapor exerts on the closed container when it reaches the dynamic equilibrium. However, this is merely a calculation AFTER it reaches equilibrium, and I don't understand how vapor pressure can also be interpreted as the measure of the tendency of a material to change into the vapor state. Also, if it is the lack of enough vapor pressure, or "tendency of the liquid to change to vapor" compared to the atmospheric pressure that's keeping the liquid from vaporizing, how can evaporation occur with lower pressure than the atmospheric pressure(because if the vapor pressure is higher than atmospheric pressure it is already boiling), since evaporation is supposed to be the molecules on the surface of the liquid randomly gaining enough energy to overcome the atmospheric pressure, implying that it has greater pressure than the atmosphere. I feel like I am overthinking the topic, but I didn't just want to give up thinking about it, so please help if possible!

Vapour pressure is not merely calculation, it is easily measureable.
Evaporation is due molecules leaving liquid not by overcoming atmospheric pressure, but by overcoming cohesive intermolecular forces. The saturated vapour pressure is a measure of the rate of this massive leaving of molecules.
Boiling is just a special evaporation case. Above boiling point, vapour is always unsaturated and liquid evaporate at its surface regardless of gas composition. Additionally, random processes create miniature bubbles ( or these may be bubbles of formerly dissolved gasses ), that have the same or higher pressure then external pressure and expands themselves.
The result is more or less massive and turbulent evaporation with creation of vapour bubbles.
Liquids can be frequently overheated above their boiling point, if there are absent "bubble seeds". Once available, e.g. by strirring or repulsion of dissolved gas, creation of vapour bubbles quickly cools liquid down to its boiling point.


text:
What happens when you put a battery in saltwater?
If somebody puts a battery in a salt solution, what path do the electrons take?

It depends on battery voltage.

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected the author elaborates the topic in the question by at least basic own topic review and writes what he/she understands and what is the eventual stumble stone. The "quick questions" without explicitly expressed particular effort are not very welcome, and may be closed.

Electrons do not take a path, it is the charged ions which act as messengers of current though an electrolyte as in an electrolytic cell.

Salt dissolves in water to left behind positively charged sodium ion and negatively charged chloride ion. once you put battery in salt water, the sodium ion migrate towards the "negative tank" and chloride ion migrate towards the "positive tank". Salt water is much more conducive than regular water, so the battery would discharge more quickly and the current will break down the salt by electrolysis, producing hydrogen and chlorine gas.


text:
What is the relationship between current flow and low AC voltages within electrolyte solutions
I have recently been trying to find references or textbooks which may explain the principles of electrical current conductance within an electrolyte solution when low AC voltages are induced. Does anyone have specific expertise or knowledge into this area and may be able to explain or direct me to the right resources or authors? Thanks!

Search for molar conductivity and ion mobility. https://en.wikipedia.org/wiki/Molar_conductivity

@Poutnik, molar conductivity is measured at really high frequencies. I think the OP might benefit from electrochemical impedance spectroscopy. Search this term and you might useful info.

@M. Farooq I do not suppose really high frequencies, as high frequency phenomena start to occur, like reactance of electrodes and different ion behavior in dielectric layer, compared to the bulk solution.


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Formation of Goethite
Probably a simple question: When you're mixing potassium hydroxide ($\ce{KOH}$) with iron nitrate ($\ce{(FeNO3)3}$) to form ferrihydrate, does the temperature have to be below $\pu{60 ^\circ C}$?

TL;DR  Yes, the reaction is feasible at $\pu{60 ^\circ C}$. Just extending my comment into an answer.

Goethite and ferrihydrite(not ferrihydrate) are two different things. They are polymorphs of iron oxyhydroxide ($\ce{FeOOH}$). Goethite is $\ce{α-FeO(OH)}$ while ferrihydrite is amorphous/nanocrystalline hydrated iron oxide, officially formulated as $\ce{FeOOH•1.8H2O}$ but also exist in many hydrated states and so formula is theoretically indeterminate as its water content is variable.

Reaction between iron(III) nitrate hydrate in alkaline solution has been studied by Atkinson et al.1 at different concentrations and amounts. One case is: 50g of $\ce{Fe(NO3)3}$ salt dissolved in 825 ml of water is added to $\ce{200 ml 2.5 N}$ potassium hydroxide solution. At pH 12 and maintaining the temperature at $\pu{60 ^\circ C}$, it took 24 hours to complete the reaction. They also noted that in this goethite synthesis, ferrihydrite is the initial precipitate that results from the rapid hydrolysis of Fe(III) solutions, before aging to goethite at higher temperature.
Pure ferrihydrite synthesis is difficult to isolate because it requires careful control of pH and temperature and its composition can changes if any of this parameter changes. Also, ferrihydrite is the only polymorph which is called nanomineral because its crystal size is in the range of 2 nm to 5nm. Nonetheless, its synthesis has also been discussed in the same paper where ferric nitrate is dissolved in potassium hydroxide but the pH is maintained at 7-8. Vigorous stirring and centrifugation yields the ferrihydrite.
An alternative reaction where the pH is not needed to maintain strictly is by adding $\ce{Fe(NO3)3.9H2O}$ and $\ce{NH4HCO3}$ in a 1:3 mole ratio at $\pu{80-100 °C}$ 2 or iron(II) salt or iron(III) salt with silicon or germanium at pH 5-11 3.
Notes and References

Laboratory synthesis of goethite and ferrihydrite of controlled particle sizes by
Milton Villacís-García, Mariana Ugalde-Arzate, Katherine Vaca-Escobar, Mario Villalobos,
Rodolfo Zanella, Nadia Martínez-Villegas, Boletín de la Sociedad Geológica Mexicana
Volumen 67, núm. 3, 2015, p. 433-446 (PDF)
Novel Synthesis and Structural Analysis of Ferrihydrite
Stacey J. Smith, Katharine Page, Hyunjeong Kim, Branton J. Campbell, Juliana Boerio-Goates, and Brian F. Woodfield
Inorganic Chemistry, 2012 51 (11), 6421-6424
DOI: 10.1021/ic300937f
SYNTHESIS OF FERRIHYDRITE AND FEROXYHYTE, NAGANORI YOSHINAGA and NOBORU KANASAKI
Faculty of Agriculture, Ehime University, Matsuyama 790, Japan, Clay Science 9, 43-51 (1993), (PDF)
Simple hydrolysis of iron(III) nitrate yields $\ce{\alpha-Fe2O3}$. Fusion of potassium hydroxide and iron(III) oxide ($\ce{Fe2O3}$) in air produces potassium(VI) ferrate but in laboratory, $\ce{K2FeO4}$ is prepared by oxidizing an alkaline solution of an iron(III) salt with concentrated chlorine bleach.


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What is the hybridisation of the bridging oxygen atoms in the borax anion?
Below is a neat looking borax anion.

While exploring the molecule in a computer simulation (Gaussian 16, level of theory: HF/3-21G), I found that the oxygen atoms in the 8-membered outer ring had sp2 hybrid orbitals. My curiosity led me to question the presented data. Using Natural Bond Orbital analysis, I found out that:
BD(1)B1(O2)
1s(20.94%)p3.78(79.06%)
BD(1)O2(B1)
2s(10.48%)p8.54(89.52%)

BD(1)O2(B3)
2s(23.60%)p3.24(76.40%)
BD1(1)B3(O2)
3s(36.38%)p1.75(63.62%)

LP(1)O2
s(17.50%)p4.71(82.50%)
LP(2)O2
s(48.39%)p1.07(51.61%)

Here BD denotes a 2-centered bond with atom(i)atom(j) and LP(i) denotes the lone pair and the s,p characters follow. My question is as follows:
Given the chaotic data, how does one analyse or interpret the harmony and claim that the oxygen[2] is indeed ~sp2 hybridised?
What is the correct hybridisation of the oxygen atoms not connected to a hydrogen?

NBOA was done using Engine provided as Gaussian, theory: Hartree-Fock in basis set 3-21G

Oh, than that wouldn't be borax, because it's a salt of such anion and you made another mistake in simulation. not to say this title is particularly vague.

@Mithoron Your suggested title was misleading, this atleast is in direct relation in what I want to know. And yes, Borax anion...that has been changed

First, it's important to understand that the simple integer hybridization labels sp, sp2 and sp3 are gross approximations and only accurately describe the electron density of simple symmetric molecules like $\ce{BeH2}$ and $\ce{CH4}$. Atoms in more complex molecules (even something relatively simple like $\ce{H2O}$) cannot be described accurately using these labels. Instead, the contribution of atomic orbitals to localized bond orbitals must be described by non-integer combinations.
That said, the calculations you did are remarkably consistent with an sp2 hybridization of oxygen [2]. If I understand the notation correctly, it says that the bond from O[2] to B[1] is ~90% p (roughly a pure p orbital), while the other bond to B[3] is ~75% p (i.e. sp3). The two lone pairs on the oxygen average 67% p, perfectly sp2.
We are able to average the two lone pair orbitals because electrons are completely interchangeable and are not localized to a specific lone pair. Any hybrid orbital description that captures the combined electron density of the four electrons in the two non-bonding orbitals is acceptable.
So the oxygen can be described as having two sp2 nonbonding orbitals, an sp3 bonding orbital and a nearly pure p bonding orbital. That's pretty close to sp2 hybridization, where we expect three sp2 orbitals and one pure p. The only deviation here is the shift of ~ 10% s from one of the sp2 orbitals to the p orbital, so that we end up with an sp3 and a not-quite-pure p instead.

Your interpretation was correct, Thanks for the explanation, can please explain how did you average?...say for lone pair, I did (5.78/2) ....and what you did was adding the percentages and obtaining a ratio. Also, last paragraph implies that as long as we donot have nearly 100% p character, we cannot say the bond is $sp^2$ am i correct?

@AnindyaPrithvi - The reason you cannot do 5.78/2 is that the 4.7 and 1.1 p contributions aren't normalized. They are only relative to an s value of 1. To give them equal weight, you need to add 4.7/(4.7+1) + 1.1/(1.1+1), where the denominator represents the total s + p contribution. These normalized values are the percentages shown in parentheses in your calculation. For your second question, I guess you are talking about the B-O bond that is sp3? Yes, it's sp3, but the original statement is that the oxygen atom is approximately sp2 hybridized, which is roughly true.

This answer is extrapolating from incomplete data on a very, very crude level of theory. It's probably also using an NBO version years deprecated and the data which is actually not meant to be interpreted. It's not even clear that the structural data is sound. Sure, this is one way or interpreting it, but in my opinion, this is a more misleading interpretation than just guessing hybridisation from the structure with Coulson's theorem (given a crystal structure of something).

@Martin-マーチン Supposedly as the website states, Gaussian 16 is used, and Gaussian 16 uses relatively newer version of NBO as indicated in the answer. Structural data is sound? (structure of NBO or the compound)....and Apply coulson's theorem on the most optimised geometry of Borax( the B.angle is 109), you get sp3 as opposed all over the internet

@Martin-マーチン - fair point about the theory. I was approaching the question as being specifically about why the calculated result did not match OP's expectation and simply pointing out that it does match. Whether or not the expectation and the calculation correctly describe the situation is a different question.

@pipiedipie You do not cite any details for the calculation. While Gaussian is certainly able to interface to newer versions of NBO (i.e. 6 or 7), it only has implemented version 3.1, which is deprecated for more than a decade now. Even if the older version works for this molecule, there is still the issue of the crude level of theory and whether you are actually interpreting data from a true (or even the global) minimum. On another note, I don't see much benefit in describing binding within the constraints of hybridisation for this molecule.

@Martin-マーチン yes right, should have cited. Also, if you meant least energy state by global minimum, it supposedly should be, because I tried tera-chem G.O. and OCRA G.O. before Gaussian (All on same theory).. And also as Andrew stated, i wanted to know how does this work, so might say a fault of the question


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Reaction of carboxylic acid with phosphorus trichloride
The reaction of carboxylic acid $\ce{RCOOH}$ with $\ce{PCl5}$/ $\ce{SOCl2}$ / $\ce{PCl3}$ yields an acyl chloride $\ce{RCOCl}$. With $\ce{PCl5}$ and $\ce{SOCl2}$ the reaction is quite simple utilizing one mole reactants and yielding products.
But in case of $\ce{PCl3}$ the balanced reaction is as follows:
$$\ce{3RCOOH + PCl3 -> 3RCOCl + H3PO3}$$
First I thought the reaction would proceed with $\ce{PCl2−}$ attacking the oxygen from which the oxygen-hydrogen bond cleavage happened but then I realized that it would form product $\ce{POCl}$ which is not a real compound I suppose so I am stuck how would the reaction proceed.
So what is the mechanism for this reaction?

What do you think it is? Extrapolate from the PCl5/SOCl2 reactions and give us your view.

@Waylander , first i thought the reaction would proceed with $\ce{PCl2}-$ attacking the oxygen from which the oxygen-hydrogen bond cleavage happened but then I realized that it would form product $\ce{OPCl}$ which is not a real compound i suppose  so i am stuck how would the reaction proceed.

The mechanism is same for $\ce{PCl3}$ as it is with $\ce{PCl5}$ ans $\ce{SOCl2}$. Only the side products are different.

Edit your question to include this. It shows you've made an effort and the question will not get closed as homework

@NilayGhosh if the mechanism is the the same so why are are using 3 moles in this case?

A plausible mechamism of this reaction is discussed in a paper by Xiao et. al1 where they tried to react benzoic acid with $\ce{PCl3}$ to get acyl chloride. They noted the following mechanism:

Although the detailed mechanism was not clear, a possible mechanism is
proposed [...] the first step should be a reaction of the acid with
$\ce{PCl3}$, probably via a four-center transition state A, with
carbonyl oxygen attacks the electron-deficient $\ce{P(Cl)}$ while $\ce{Cl}$
attacking the electronically deficient carbonyl carbon. Formation of a
$\ce{Cl-C}$ bond and $\ce{P-O}$ bond (A’), further $\ce{C-O}$ bond cleavage and
hydrogen transfer took place to give the products B and C.
Product C then reacts further with the acid to give the
corresponding chlorides and $\ce{HP(O)(OH)2}$

This product C will again react with carboxylic acid two times so as to replace all the chlorine atoms and ultimately forming phosphorous acid (actually phosphorous acid technically exist as $\ce{HPO(OH)2}$ called phosphonic acid) So, the final reaction would be three moles of carboxylic acid reacting with $\ce{PCl3}$ to form three moles of acyl chloride and phosphonic acid thus fully utilizing the three chlorine atoms in $\ce{PCl3}$. The major advantage of this reaction is that there is only one side product i.e phosphonic acid which is non-toxic and easily removed by filtration whereas reaction with $\ce{PCl5/SOCl2}$ gives two side products making the reaction more volatile.
Reference

Atom-efficient chlorination of benzoic acids with PCl3 generating acyl chlorides, Jing Xiao and Li-Biao Han, Journal of Chemical Research, 2019, Vol. 43(5-6) 205–210 (PDF)

The structure of H3PO3 has a hydrogen bonded to the central phosphorus. How can we account for this hydride transfer? It is actually the deviation from the standard reaction of PCl5 or SOCl5 with alcohols or carboxylic acids.

@ADITYAROYCHOWDHURY I am not sure if I am interpreting your comment correctly. 1. Are you sure there is a hydrogen bond in H3PO3, IIRC, hydrogen bond only occurs with oxygen, nitrogen and fluorine. 2. H3PO3 is the minor tautomer. The major one is actually HPO(OH)2 called phosphonic acid and this form exist almost all the time. As you can see, this acid is subject to tautomerization  involving hydrogen transfer between the oxygen and phosphorus atoms. My organic chem is rusty so if you are having doubts with my answer, feel free to post a new answer sharing your idea.

@ Thank you I was asking about the tautomerism part. Not  hydrogen bonding.


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Regioselectivity in Electrophilic Aromatic Substitution of Pyrrole and Indole
If we look into the structures of Pyrrole and Indole, they are very similar. Only Indole has one extra Benzene ring attached to it, i.e. if we attach one Benzene ring to Pyrrole, what we'll have is nothing but Indole.

But if we delve into the preferable positions of Electrophilic Aromatic Subsitutions for these compounds, it's pretty well known that Pyrrole undergoes EAS preferentially at the $2$-position, whereas for Indole it's the $3$-position which is favoured.
A typical example of an Elcetophilic Aromatic substitution reaction where this distinct feature between Pyrrole and Indole can be clearly visible is the Vilsmeier-Haack Reaction. The major outcomes of the reaction in the two cases clearly show the difference.

Now, the preference of $2$-position in case of Pyrrole can be justified easily by considering the greater stability of the $\sigma$-complex intermidiate formed in case of substitution at $2$-position due to more number of resonating structures as clearly been described in the answer to the question Regioselectivity in Electrophilic Substitution of Pyrrole
Now my question is, applying the similar logic Electrophilic substitution at the $2$-position should be even more favoured in case of Indole, as the $\sigma$-complex in that case should be further more stabilised due to resonance with the Benzene ring as well. Therefore, the preference of substitution at $2$-position should be exceptionally higher than at $3$-position in case of Indole. But in reality, it's actually the opposite. Why is this anomaly observed in case of Indole ?
The resonating structures in case of two different substituent positions in Indole are shown below,

P.S. Thoughtful answers are highly appreciated. Any kind of Quantum Mechanical Calculations or Computations or Orbital Analogy are also welcome.

See Indole Elctrophilic Substitution in Joule and Mill's Heterocyclic  Chemistry

An abstract from the book referred above by Aditya, under Reactions with Electrophilic reagents for Indoles: Reactions and Synthesis: "By NMR and UV examination, only the 3 - protonated cation (3 H - indolium cation) is detectable; it is the thermodynamically stablest cation, retaining full benzene aromaticity (in contrast to the 2 - protonated cation) with delocalization of charge over the nitrogen and α - carbon". It seems like the argument is that EAS at position 2 disturbs the aromaticity, hence position 3 is preferred

The traditional explanation for the reactivity of indole at C–3 is that attack at C–3 does not disrupt the aromaticity of the benzene ring in the cationic intermediate (see any organic chemistry book; for heterocycles a good book is Joule and Mills as has been suggested in the comments):

You have drawn many of the resonance structures for the intermediate, including the two shown above. So I should mention that the major resonance contributor should be the one where the positive charge is borne by the nitrogen. In this case, the positively charged nitrogen has four bonds and a complete octet; whereas if the positive charge is on carbon, that's formally a six-electron, electron-deficient carbocation. Many thanks to Buttonwood for pointing this out.
To a first approximation, one could consider the reactivity of indole to be as if it were a benzene ring plus an enamine, and enamines are very, very happy to react β to nitrogen and not α. This is of course not an enamine, but the idea is the same.
For pyrrole, there is no "other benzene ring" to worry about, so this consideration does not apply.

Equally to note, back to the original question: For a substitution at C2-position, the intermediate state was drawn including mesomers breaking the aromaticity of the the benzene ring; the analogue for the substitution at C3 is missing.  As if the OP recognized the additional price to pay to break aromaticity (but this is a speculation).

@Buttonwood thanks for pointing out! I think the missing point which I failed to include was that the major resonance structure should be one where the positive charge is on nitrogen (octet complete) vs a 6-electron carbocation. I will edit that in.


text:
Is hybridization of the cyclopropyl anion sp5?
I have read some answers about the $\mathrm{sp^5}$ hybridisation of carbon atoms in the $\ce{C-C}$ bond in cyclopropane as well as the $\sigma$-bond resonance in methylcyclopropyl cation.
Overall, the answers say that the $\ce{H-C}$ bond is due to overlap of $\mathrm{sp^2}$ orbital of carbon and $\mathrm{1s}$ orbital of hydrogen, while the bent bonds of the ring are formed by overlap of $\mathrm{sp^5}$ hybrid orbitals.
But, overall true separation of orbitals in such cases is not defined i.e. we cannot say that some orbital is completely $\mathrm{sp^2}$ hybridised.
Similarly, if the loss of a proton occurs from cyclopropane then the following:

In such an ion, will the hybridization change and $\mathrm{s}$ and $\mathrm{p}$ characters of the orbitals be redistributed?
My assumptions for this are:

In the $\mathrm{sp^2}$ bond the lone pair due to loss of proton will pose increased steric hinderance as it occupies greater volume. This will cause increased repulsion and decreases stability. Moreover, shouldn't the $\mathrm{s}$-character increase to accommodate the increased negative charge?

Accordingly, wouldn't the $\mathrm{sp^5}$ bonds will now have greater $\mathrm{p}$-character to ease even greater strain produced?

Then, what will be the change in the hybridisation in the overall molecule on formation of this ion?
References:

Sp5 hybridization in cyclopropane?
Are the triply-bonded carbons in pyridyne (and benzyne) sp-hybridised?
Non-integer hybridization
https://www.ch.imperial.ac.uk/rzepa/blog/?p=14548

Initially, I thought about an accidental claim of an carbon atom orbital thought to be hybridized $sp^5$ (with a five) instead of $sp^3$ (with a three), because the hybridizations of carbon atom orbitals taught in school are $sp$, $sp^2$, and $sp^3$.  Because the body of your question repeats the claim of $sp^5$ more than once, may you either indicate a reference for this unusual concept, or consider to revert (back) to $sp^3$ (because, the cyclopropyl anion still is a cyclopropane)?

@Buttonwood I have included some references.

I tried to check your claims, it seems that, C-C bond (without anion) came out to be $\ce{sp^{3.5}}$ and the anion ones are $\ce{sp^{2.6}}$ towards, $\ce{sp^{4.06}}$ away and the lone pair was in $\ce{sp^{1.64}}$. so i think there wont be sterric hinderance,due to the s, but p character decreased against your claim

@AnindyaPrithvi Yes, so it does not account to any whole number value in either contribution. What do we do in such a case?

I do not have a clear answer about that, because it is an aromatic molecule as well...so there should be an average $\ce{sp^r}$ . My calculated values corresponds to a static state

@AnindyaPrithvi It doesn't have unsaturation. Should it be aromatic?

Any further information regarding this compund will be greatly appreciated.

@ADITYAROYCHOWDHURY yes, I read it in a book...I will share the name as well (can't find it right now)....I think it was written that it will do intra-molecular proton transfer to delocalize the charge

@AnindyaPrithvi Thank you, will wait.

Michael B. Smith- March's advanced organic chemistry_ reactions and mechanisms and structure-wiley 2020

@AnindyaPrithvi Thank you, will definitely give it a read.

I was reading it, I couldnt find it though....Also, I maybe $5%$ wrong about aromatic nature...I was just going through the huckel MO, and found that there is a small break in the orbitals plotted at -16.266eV (in the negative sign of wave region) (It might be due to a computational fault)

It's reasonable to ask when you've pushed a model like hybridization farther than it can be pushed to be useful.

@Zhe So hybridization cannot be used to explain this?

I don't even know what you are saying!

Background
Charles Coulson was the originator of Coulson's Theorem, a useful tool for the chemist.  It allows you to make "back of the envelope" estimations of hybridization if bond angles are known. Conversely, if one knows the hybridization from, say, $\mathrm{p}K_\text{a}$ or $J_{C^{13}-H}$ data, then the bond angle can be estimated.
The key equation is $$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

See these earlier answers for interesting examples where the theorem is applied:

estimating the hybidization in cyclopropane
angle of attack of a nucleophile at a carbonyl
the lone pairs in water are non-equivalent - the resultant hybridization of water

The Question
The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114°.  From this, and using Coulson's theorem
$$1 + \lambda^2 \cos(114^\circ) = 0$$
where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Using the equation
$$\frac{2}{1 + \lambda_{\ce{C-H}}^2} + \frac{2}{1 + \lambda_{\ce{C-C}}^2} = 1$$
(which says that summing the "s" character in all bonds at a given carbon must total to 1), we find that $\lambda_{\ce{C-C}}^2 = 3.74$, or the C–C bond is $\mathrm{sp^{3.74}}$ hybridized.
Now, if we remove a proton from cyclopropane and generate the cyclopropyl anion, we move from a situation where we had a pair of electrons shared between carbon and hydrogen in a $\ce{C-H}$ bond to a situation (the anion) where we have a pair of electrons residing entirely in an orbital on the carbon atom.  In other words, we have increased the electron density in this carbon orbital.  Bent's Rule tells us that the molecular geometry will change so as to lower the energy of this pair of electrons and that it will lower the energy of these electrons by increasing the s-character of the orbital they are in.  In order to increase the s-character in this orbital, we will take some s-character away from the 2 $\ce{C-C}$ bonds and 1 remaining $\ce{C-H}$ bond.  Hence, $$\lambda^2_{\ce{C-C}}>3.74$$ $$\lambda^2_{\ce{C-H}}>2.46$$ and $$\lambda^2_{\ce{C-electron pair}}<2.46$$
Said differently, the interorbital (not internuclear) $\ce{C-C-C}$ angle at the anionic carbon will decrease and the $\ce{C-C-H}$ angle will decrease making the substituents about the anionic carbon appear more puckered.

I don't get the leap that s character will be removed from the bonds. I would assume the lone pair to be in a p like orbital, increasing s character in the bonds. Could you clarify that please.

@Martin-マーチン Just like in ammonia where the molecule is approximately tetrahedral
with the lone pair in an approximately $\ce{sp^3}$ orbital; this being a lower energy geometry than the molecule being planar with the lone pair in a p orbital.

Yes, you're right, I somehow got my reasoning the wrong way around.


text:
Using Beer-Lambert law to calculate light intensities
Supposing the mean absorption coefficient of seawater in the visible region is $\pu{0.8 m-1}$, calculate the depth at which a diver will experience half the surface light intensity.
I know that you have to use $A = -\log \left(\dfrac{I}{I_o}\right) = \epsilon cl$ but I'm not sure how to proceed

The answer depends on whether you use log to base 10 or base e, i.e whether you are taught $I/I_0=10^{-\epsilon c l}$ or $e^{-\epsilon c l}$. As the absorption coefficient is $\epsilon c=0.8 $ (which you can see by looking at units) you can calculate the answer  by making $I=I_0/2$

In my opinion, Beer–Lambert law is no use. I would rather write
$$\frac{\mathrm dI}{I\,\mathrm dl} = \pu{0.8 m^-1},\tag{1}$$
which leads to the integrated formula
$$\ln\left(\frac{I_0}{I}\right) = 0.8 l.\tag{2}$$
And, if $\displaystyle\frac{I}{I_0} = 0.5,$ the corresponding depth is
$$l = \frac{\ln 2}{0.8} = \pu{0.846 m}.\tag{3}$$


text:
Why do electron-rich aryl groups have greater migratory aptitudes?
In the following reaction there are three possible aryl groups that can migrate: an unsubstituted phenyl group, a p-methoxyphenyl group, and a p-nitrophenyl group:

Why does the electron-rich p-methoxyphenyl group migrate instead of the phenyl or p-nitrophenyl group, especially when it leads to a carbocation that is less stabilised (since it will no longer be conjugated with the stabilising p-methoxyphenyl group, cf. the crossed-out arrow)?

Probably due to an increase in tendency to stablize the positive charge. But this assumption is not true always as there are reactions where the contrary occurs.

This is a shift which follows a migratory aptitude which differs slightly depending on the reaction you are talking about.  In your case, the migration will take place with the most donating ring to stabilize the cation by a sort of "neighboring group participation" .See this for more details

Also see ron's answer here  for a general discussion of migratory aptitude

But why not think of it in another way, if the nitrobenzyl migrated, we will get a more stable carbocation with the NO2 being further away and (OCH3Bn) being closer to the carbocation, which means closer electron donating group and further electron withdrawing group = more stable carbocation.

@Katia Did you check this link that I gave above? Your carbocation will be stabilized by a similar "neighbouring group participation" by the formation of a phenoium cation ispecies. While this migration is occuring, the positive charge of the carbocation will be delocalized over the migrating ring. Now try the migrating ring was nitrobenzyl, the -NO2 would in fact highly destabilize the transition state by it's electron withdrawing effect

@YusufHasan indeed, this would certainly be the kinetic product. Any thoughts on thermodynamic product? Presumably, the other structure Katia proposed would be the thermodynamically favored product (but it would have a hard time getting there!)

@jezzo Yes, I believe that Katia's proposed product might fit the bill of thermodynamically controlled product, as I can see that the product #2 shown in the diagram has the positive charge delocalized over the anisole ring. But again, the activation energy barrier to get to this product would be quite high due to the destabilized transition state, hence under normal conditions, I would propose the kinetically controlled product(that is, product #1 in the diagram) unless condition are specified for thermodynamic control

That is because migratory aptitude depends on electron density.Here, -m groups withdraw electrons from benzene ring making it slightly less electron dense.

The answer varies slightly depending on whether the migration is one-step or two-step via an intermediate.
In either case, the respective transition state or intermediate looks like this:

You can see that the middle structure would be stabilized most by a group that is able to stabilize the ring via resonance, for example, an alkoxy substituent.
If the mechanism is single-step, then that substitution would stabilize the transition state and make this migration the fastest.
If the mechanism is two-step, then we can invoke the Hammond postulate to say that the transition state leading to the higher energy intermediate will resemble it in structure and thus also be stabilized by the same kind of substitution, leading to the same conclusion.
We therefore suggest that the propensity of migration is X=OMe > X=H > X=NO₂.

Thank you so much! any reference? where did you learn this?

I have no idea where or even if I learned this. You gave an observation, and I used what I know about principles of organic chemistry to provide a rationale. This is a core skill for a chemist.

This should be covered in a first- or second-year organic chemistry course, @Katia. Likely not this specific reaction, but the general skill of drawing resonance structures for intermediate products and assessing their relative stabilities in order to predict the most likely products of the reaction. Any college-level organic chemistry textbook should have a treatment of this in general terms, although it likely won't be a specific page or even chapter. As Zhe said, it's more of a general or core skill.

@Katia if the answer has helped you, please consider accepting it (click the green check mark near the upvote/downvote buttons). See: What should I do when someone answers my question? for more information.


text:
Change in internal energy for isothermal process

A gas expands isothermally against a constant external pressure of $1\ \mathrm{atm}$ from a volume of $10\ \mathrm{dm^3}$ to a volume of $20\ \mathrm{dm^3}$. It absorbs $800\ \mathrm J$ of thermal energy from its surrounding.
The change in internal energy is:

I came across this question and my doubt is, for isothermal process, $\Delta U=0$ ( change in internal energy) for ideal gas is zero right? Then how can we calculate $\Delta U=0$ here, assuming the gas to be ideal?
Is the error with the question or am I missing something?

$\Delta U=0$ for an isothermal process on an ideal gas.

So the gas in question is not ideal?

The process is irreversible, see the P-V graph, ext. P is at 1 but volume changes by 10 units implying a straight line. After this observation, proceed with dq=dw+du

@Anindya does it mean this is not a isothermal process?

No, it is isothermal, just irreversible....

There are two types, one is reversible process and one is irreversible, it can be for adiabetic process as well, i.e. irreversible adiabetic process and reversible adiabetic process exist seperately.

refer here https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Physical_Chemistry_(Fleming)/03%3A_First_Law_of_Thermodynamics/3.03%3A_Reversible_and_Irreversible_Pathways

I suppose the question didn't specify the gas is perfect, so delta U need not be 0. If you compute for w and the delta U turn out to be 0 then the gas is perfect.

It must not be an ideal gas.  This is irrespective or whether the process is reversible or irreversible.   That is the only possible conclusion.  Otherwise the amount of heat received would adjust  until it was equal to the work done.


text:
Determination of percentage of Methanol in an Alcohol based hand sanitiser
Note: I am looking for advice on a chemical test - not a sermon.
I expect that this is entirely on topic in this forum, downvotes and sermon comment notwithstanding.
More than about 5% of Methanol in alcohol based hand sanitisers indicates that corners have been cut during manufacture and that the product should be avoided.
I'm looking for an easy test to determine the approximate percentage of Methanol in an Ethanol / Methanol hand sanitiser mix. Many hand sanitisers may contain small percentages of Methanol (typically in the 1% - 5% range) and this is acceptable. Anything over 5% I'd like to avoid. Much over 5% (say 10% +) I'd like to be very clearly distinguished. If actual approximate percentage can be indicated, so much the better.
Ideally I'd like a test with either low cost, simple to use, readily available non-hazardous reagents (one can hope) or failing that, something that can be prepackaged for easy use by the unskilled.
In web searches I found a range of tests for Methanol but most were not very quantitative and most were not intended to work in an ethanol-methanol mixture.

Background: Added:
Some guy y'know said:

Using disinfectants outside of a heathcare environment is still nonsense in the year 2020. Using disinfectants of questionable provenience is simply hazardous. Don´t do it.

The use of hand sanitisers is well covered in recommendation documents from regulatory authorities in many administrations and by internationally relevant bodies. Sanitisers are sold under regulatory guidelines which if followed would guarantee their safety when used appropriately. In the US many out of specification brands are being sold which are indistinguishable from those which comply. Several of the largest US sellers have been selling out of spec product. There is no way to determine provenance based on labelling.
While the use of soap and water is often a good or even superior solution it is incorrect to suggest that the use of sanitisers in general applications is nonsense.
In this instance I am interested NOT in their efficacy or application but re their specific bonafides with respect to a crucial component which is known to be used in excessive and illegal amounts in some cases.
I'm an engineer. Sanitisers are sold by a wide range of manufacturers and repackagers. I read labels, note sources, note claimed contents and viscosities and residence times and more.

Using disinfectants outside of a heathcare environment is still nonsense in the year 2020. Using disinfectants of questionable provenience is simply hazardous. Don´t do it.

There's no such test. Methanol is very similar to ethanol. Perhaps it could be selectively oxidised, but strong oxidisers aren't particularly safe or freely available.

You could conceivably use mass spec or NMR, but neither method is conveniently available outside of labs. But to be fair, methanol is perfectly fine, provided you're not inhaling large amounts, swimming in it, or drinking it.

@Zhe Methanol is readily absorbed into the body through your skin. No need to drink it to get sick.

@Zhe For information. Some santisers in the US have been found to have only Methanol or high percentages of Methanol. These are reported to all or mostly have been manufactured in Mexico but I imagine labels are unlikely to reflect this. Reports indicate that regular use over 2 days can start to cause vision effects and blindness. This is also reportedly reversible if use is then stopped. It would be very useful if such products could be reaily detected in advance. My interest covers not only NZ where I live but other countries who may be being taken advantage of in the present circumstances.

@Karl Yes, but methanol also evaporates quickly. We're not talking about bathing in it.

@RussellMcMahon If you're talking the composition being primarily methanol, that is very different from having some fraction of methanol. I think the main point is that the analysis you're talking about is non-trivial...

Noted: Closed. This question needs to be more focused. It is not currently accepting answers.
Update the question so it focuses on one problem only. This will help others answer the question. You can edit the question or post a new one.

@RussellMcMahon I have served you up a fresh green (as in vegetation) embodiment to possibly ascertain CH30H presence (and even perhaps relative concentration) for virtual (so far) consumption. Here is the ScienceDirect link: https://www.sciencedirect.com/science/article/abs/pii/S1671292708600863#:~:text=The%20results%20revealed%20that%20photosynthesis,maintained%20for%203%E2%80%934%20days.

Russell McMahon - I saw the question early. Sorry I didn't give a comment then. You did ask a simple straightforward question. However there isn't a corresponding  simple straightforward answer. There are really two nested problems. First you need a qualitative analysis to figure out what is in the product. Then you need to analyze to determine how much of each component.  With a GC/MS setup the qualitative analysis to identify methanol would be trivial. To get a quantitative value would be a little more work, but still easily doable. But such an analysis just isn't possible with a dipstick.

Russell, agree with your down voting frustration but it is a common problem of any public forum. The bad news is that "simple" distinguishing test do exist, but they generate more toxic products (including a carcinogen) in a so-called iodoform test. Good for school labs only. The problem is not in distinguishing methanol versus ethanol, the problem is the matrix (as analytical chemists call it), i.e., these alcohols are present in a complex mixture.
The only reliable test test to determine methanol in a complex mixture like a hand sanitizer is gas chromatography. You need to consult a professional certified service to ask them to analyze the sanitizer. There is no home-kit or simple test to differentiate the two in a mixture.

DV has now been followed by question closure :-). I've been active on SE for 9+ years, I'm in 50+ stack exchange groups where I have more than the basic 101 points and another 30 with ~= 101 rep. I'm a moderator of an active SE site. I have a reasonable idea of how to ask good questions on SE. I find the reactions here 'interesting'. I'm aware that in some groups it seem obligatory to give newcomers a hard time (and do not know if this is the case here), but it's an interesting experience. | I'm astounded that anyone thinks they could not perceive a focused question in my original post.  :-)

Russell McMahon, I am one of the major opponents of this unnecessary down voting issue in chemistry. I guess, it gives some people a dopamine boost. Chemistry also receives low quality queries from students as well because chemistry teachers at many schools are not that great. Anyway, you should not care on the other hand, because faceless and nameless people down voting a decent question have zero importance outside the internet (equivalent to generating a storm in a tea cup). ResearchGate disabled down voting option altogether.

I disagree with this absolute negative response and I plan on testing the burning splint variation (testing actually for oxygen) with a methanol/glycerine mixture on leaves versus ethanol/glycerine. For CH3OH, expect less CO2 and more pure O2. It will be fun and further fascinating if this biological-based experiment (based on a ScienceDirect article) that is performed on three plants (one with ethanol, and two with different doses of methanol) can actually roughly estimate the CH3OH loading.

@AJKOER Any practical results that you achieve would be of interest.

Not sure if I can achieve practical results. One obstacle that I have already been confronted with (source: a younger member of my family), is that it is advisable to talk to the plants. Perhaps I should have listened to my esteem colleagues.

It is likely that without sophisticated equipment that one cannot accurately determine methanol concentration.
That being said, my speculation on a possible process, for possible use outside of the lab, could be based on this source outline of relevant chemistry, to quote:

Methanol influences the oxidizing capacity of the atmosphere through reaction with the hydroxyl radical (Jiménez et al., 2003), its main sink, leading to the formation of water vapour and either CH3O or CH2OH radicals, which both react with O2 to give HO2 and formaldehyde (H2CO) (Millet et al., 2006 ). The photo-oxidation of formaldehyde, a key intermediate in the oxidation of numerous volatile organic compounds , leads to the formation of HO2 radicals and carbon monoxide (CO). ...

As such, consider an experiment consisting of placing a test sample of the hand sanitizer in an expandable transparent plastic bag along with some B-12 mixed with 3% H2O2 (which was shortly before made alkaline by adding Na2CO3 until CO2 evolution ceased) in direct sunlight. Periodically shake.
My expectation: The B-12 is a photocatalyst that turns clear with sunlight exposure. Some hydroxyl radicals, along with CO and O2 are produced in the reaction from the breakdown of the H2O2 and importantly, the CH3OH (methanol) in direct sunlight in the presence of B-12. The alluded to HO2 radical above at pH>5 (from the added alkaline Na2CO3) is now in the form of gaseous superoxide ($\ce{.O2-}$).
Perform a parallel run employing ethanol mixed with an ethanol-based hand sanitizer. The chemistry on photolysis differs and is discussed here.
Comparatively, I would expect a faster reaction with methanol over ethanol (accelerated loss of the B-12 coloration) and more/faster gas creation. If I am wrong on the latter point, not material, as it is still likely quantitatively different.
You may be able to use the measured timing of the volume of gas along with the timing of coloration lose, to assess relative methanol presence.
Note: Unfortunately, there is more than alcohol added to hand sanitizer mix (like a moisturizer, perhaps glycerine) which may make the comparative analysis even harder.
[EDIT] Now, some may have suspected that my suggested photolysis experiment was a bit far afield from classical analytical chemistry. Wrong, this experiment is definitely more so, but even greener, likely easy to implement at home, and decidedly fun. Select two (or more) small nearly identical green plants, treat the leaves of one with a glycerine/methanol mix and with the other substitute ethanol. Place each plant in a container which has been fulfilled with a fixed amount CO2 gas, but still should have some remaining air (plants apparently need to breathe O2 at night, but during daylight, on net, produce much more oxygen from CO2) where the very top part of the plant should be exposed to air. Place the open mouth vessel with the plant in sunlight with a thin transparent plastic seal. The plant demonstrating a greater consumption of CO2 and corresponding formation of pure oxyen can be tested by placing a glowing splint into the vessel. Record the seconds and/or the splint's position in the vessel until the burning bright oxygen flame is reduced/extinguished. This may produce a relative metric to judge the relative presence of methanol given its apparent photocatalytic acceleration of a leaf's ability for oxygen formation in sunlight.
Here is a supporting Science Direct source, to quote:

The results revealed that photosynthesis was greatly improved by methanol, as indicated by higher photosynthetic rates and stomatal conductance. The enhancement effect of methanol on photosynthesis was maintained for 3–4 days. Different methanol concentration treatments also increased intercellular CO2 concentration and transpiration rates.

As the methanol treatment is effective for days, repetition of the oxygen generation/sunlight treatment, after replenishing the CO2 supply, may be viable. Also, for very low added amounts of CH3OH, conducting a four-day trial run is recommended as time, assuming a constant rate of photosynthesis, is a multiplicative factor.
Finally, as to whether this can actually work, in practice, and could quantify variations in the concentration of CH3OH, remains to be demonstrated, AND I PLAN TO TEST IT.
By the way, the inspiration of my original photolysis path was based on my acquaintance in the photolysis literature of the reputed abilities of methanol, so the latter photosynthesis reported results with CH3OH should not be too surprising to some.

Sorry, this is complete non-sense. That is not how analytical chemistry works.

True, I added the (obvious) qualifier, outside of one's lab. However, I have tested the relative strength of H2O2 with NaOCl at home via O2 evolution on several occasions. My photolysis suggested experiment is based on my experience working with B-12 and shares a similar gas evolution comparative path. Conducting classical analytical chemistry experiments is a luxury for many, and I am opened minded to at least consider improvising, when necessary, outside of the lab. I also suspect M. Farooq has never performed any fun experiments with direct sunlight and home photocatalyst like ZnO or B-12.

Ajkoer, I don't your age group, but 20 years ago, I have had all this fun which excites you now ;-). What boggles my mind, is how conveniently you drag in free radicals and bleach in almost all your answers.

Actually, HOCl (in a Fenton-type reaction) and H2O2 (Fenton-like or Fenton with iron) engaging with transition metals manages to show up in disease proliferation (in our lungs and elsewhere) all by themselves (no need to drag them in), or somewhat extensively in AOP (advanced oxidation processes) to remove pollutants even, at times, using light, or more enjoyably use NaOCl (or HOCl) in a bleach battery (electrochemistry).  I regret to inform all that photolysis, as the source above notes, is apparently all about radical formation and subsequent interactions (especially with O2).

M. Farooq, for my first suggested photolysis based approach, your comment was 'non-sense'. As I have now have gone even more 'green' (as in vegetation) with my suggested second embodiment, I eagerly await your new depiction.


text:
The color of the most sulfides of p- and d-block elements
My teacher said that many of the sulfides of d-block and p-block are black. Why?

https://chemistry.stackexchange.com/questions/32019/why-are-hgo-and-cds-compounds-colourful

In my opinion nobody knows why $\ce{FeS, CoS, NiS, CuS, Ag_2S, PbS}$ are black.

There, there, colors of complexes and inorganic compounds seem to be well studied and there was lots of posts about them already. Still, it's not like this knowledge is particularly useful or common, even among chemists...

Fro example https://chemistry.stackexchange.com/questions/98915/black-green-nickel-oxide-supplemental

The band gap between conduction band and the valence band plays a role in color of these compounds. The good example is $\ce{SnS2}$, which is bronze color (Image $\bf{A}$; $\ce{SnS2}$ is used in decorative coating where it is known as mosaic gold) and have a band gap of $\pu{2.18-2.44 eV}$ (Image $\bf{B}$; Ref.1). When an electron in the conduction band falls down to the valence band, it loses energy and the lost energy is emitted as a photon ($h\nu$). This photon has an energy equal to the band gap (Image $\bf{C}$). In mathematics, it is: $E_\mathrm{gap} = h\nu$. The crystals of $\ce{SnS2}$ and its calculated band gap is shown below:

The compounds is colored when the band gap is between $1.8$ and $\pu{3.0 eV}$ (visible range). The compounds with larger band gaps $(\gt \pu{3.0 eV})$ are either white or transparent crystals.  Those with smaller band gaps $(\lt \pu{1.8 eV})$ give  black color crystals. For example, the band gap of $\ce{ZnS}$ is $\pu{3.54 eV}$, the color of which is white. Meantime, the color of $\ce{CdS}$ (Pigment name: Cadmium yellow) is yellow because its band gap is $\pu{2.45 eV}$. On the other hand, the colors of $\ce{CdSe}$ and $\ce{CdTe}$, which have the same structure of $\ce{CdS}$, are black because their band gaps are $\pu{1.74 eV}$ and $\pu{1.49 eV}$, respectively. Also, the color of $\ce{PbS}$ is black because its band gap is $\pu{0.37 eV}$.
For some crystals, the band gap change with temperature (Ref.1). Hence, for those, the color depend on temperature.
References:

Lee A. Burton, Thomas J. Whittles, David Hesp, Wojciech M. Linhart, Jonathan M. Skelton, Bo Hou, Richard F. Webster, Graeme O'Dowd, Christian Reece, David Cherns, David J. Fermin, Tim D. Veal, Vin R. Dhanak, Aron Walsh, “Electronic and optical properties of single crystal SnS2: an earth-abundant disulfide photocatalyst,” J. Mater. Chem. A 2016, 4, 1312-1318 (https://doi.org/10.1039/C5TA08214E).

@ Mathew. So a compound is black if the band gap is smaller than 1.8 eV. OK. So the real question is : why is the band gap smaller than 1.8 eV in FeS, CoS, NiS, CuS, Ag2S and PbS ? And why not in ZnS ?

More effective nuclear charge on Zn lowers the conduction band of ZnS, maybe?

I would leave a comment, but the basics of the answer are as follows. It's hard to give an answer that's complete because the chemistry is very diverse for the different compounds.
What you're basically seeing is the transition corresponding to breaking the metal-sulfur bond (i.e. the bonding orbital/antibonding orbital) or electrons moving between the metal and sulfur. The colours tend to be black for sulfides because sulfur orbitals are relatively high energy (compared e.g. to oxides) and form weak bonds so the energy of the transitions is small.
A complete answer would require more detail about the band structures of the relevant compounds. Some of the oxides are also black.


text:
Number of moles of acidified potassium dichromate required to liberate 6 moles of iodine from an aqueous solution of iodate ions
I came across this question in a recent exam I had taken:

How many moles of acidified $\ce{K2Cr2O7}$ is required to liberate 6 moles of $\ce{I2}$ from an aqueous solution of $\ce{I-}$ ?

So this is how I approached it (by writing the corresponding equations):
$$
\require{cancel}
\begin{align}
\ce{ Cr2O7^2- + 14 H+ + \cancel{6e^-} &-> 2Cr^3+ + 7H2O} \\
\ce{6 I- &-> 3I2 + \cancel{6e^-}}\\
\hline
\ce{Cr2O7^2- + 6I- + 14H+&-> 2Cr^3+ + 3 I2 + 7H2O}
\end{align}
$$
Clearly, every mole of $\ce{K2Cr2O7}$ liberates $3$ moles of $\ce{I2}$ from an aqueous solution containing $\ce{I-}$.
So, I concluded that $2$ moles of $\ce{K2Cr2O7}$ are required to liberate 6 moles of $\ce{I2}$.
But the answer key given to me says that the correct answer for this question is $1$. Am I going wrong somewhere?

Yes you are right, it should be 2 mol dichromate leading to 6 mol iodine. Can you balance the equation first (hand written), iodide ion is missing on the left hand side?

Oh yes. I'll edit it @M.Farooq. Thanks for pointing it out :)

Your answer is right, but only for the reaction you have written. There could be other reduction half-reactions that you might select, though based on the statement of the question, you have selected the correct oxidation half-reaction.

Be prepared that any task description can contain an error. If you underestimate your knowledge, overtrusting autority, you may not notice it.

@Zhe could you please tell what other reactions are possible?

Try consulting a table of half reactions. Judging by the expected answer, you're looking to reduce the dichromate down to elemental chromium, which, in all honesty, seems like a worse choice than what you've selected.

For the record, should be iodate not iodide. Here are cited reactions in the Briggs-Rauscher reaction with H2O2 replacing K2Cr2O7  in the presence of iodate:
Eqn. 1-4  5 H2O2 + 2 IO3- + 2 H+ -> I2 + 5 O2 + 6 H2O and
Eqn. 1-5  5 H2O2 + I2 -> 2 IO3- + 2 H+ + 4 H2O