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Choosing electrodes for corrosive environments
Background
I'm building a watering robot for my household plants. For the purpose of measuring the moisture content of the plant I intend to use a resisitive approach where I pass an electric current through two electrodes in the soil.
I initially used a cheap off-the-shelf probe with a metal I don't know what it is but it corroded very quickly in the soil. I have learned since then that soil is quite corrosive to metals. I'm aware of a capacitive method of measurement that doesn't require the metal to be in contact with the soil but I haven't found a lot of information on how that works and I've already built most of the project around the resistive approach.
I'm not good with chemistry, and I'm trying to figure out what material I should use for these electrodes. As the plants I'm growing are eventually going to end up for personal consumption I am concerned with toxicity and if these materials will react with the soil and produce some thing that can be absorbed by the plant's leafs and fruit.
Research Done
I read that Graphite, Palladium, Platinum and Gold are the three best materials from a corrosion resistance point of view with Silver and Titanium coming in close. [1]
My understanding is that Platinum and Gold are noble metals and they are inert and are normally found in their pure form in the earth so they would not react with the soil at all under normal circumstances. So these metals would be my first choice if availability and cost was not a concern.
I managed to find some information on titanium since we use it in medical implants, it's not directly toxic invivo. Although I did find this site claiming you can have titanium poisoning (also claiming platinum poisoning). But to be honest I'm not sure if I consider that page a credible source, it seems to be fear mongering designed to sell 'home tests'. I found some research papers detailing that lab rats had adverse effects on lung development when exposed to TiO2 nano particles, but no studies to show any actual adverse effects on humans, some recommended caution however.
As for Graphite I found that it really depends on the alloy and basically I felt like it would be too hard for me to make an informed decision so I discounted that, also I'm kind of afraid that it might be too brittle for the application. But this may also be lack of knowledge on my part.
I know you can get platinum plated titanium mesh for electrolysis applications (but I only found alibaba online and that seemed sketchy) but they are too big and unsuitable for my application. And if I could find platinum coated titanium rods, I guess that would be optimal but I have had zero lucky finding such things (2-4mm thick, 7cm long or there about)
Questions
So I turn to you guys: Gold and Platinum are apparently the best choices. But would titanium be good enough? Is it safe? Should I still consider Graphite? Can the soil react with Titanium to produce TiO2? Do you know of a place online where I can purchase these things in Europe?

I would search the patent  and marketing literature, I would think this has been done many times before.  Using AC excitation is a common way to avoid the galvanic corrosion.  With chloride ions present in the soil it may be possible to electrolytically corrode even gold if the measurement voltage is high enough.

@KalleMP Yes, I'm doing AC excitation at 5V which helped a lot compared to the off-the-shelf thing that was always on DC 5V... Will try searching, can you suggest some keywords?

I would start with the search below and also check the image results for interesting ideas that might lead somewhere.  -  https://www.google.com/search?q=soil+moisture+electrodes+corrosion+patent

Unless you are using lead, cadmium etc. it is unlikely that your plants will take up toxic amounts of any electrode material, I expect the problem to be electrode lifetime and not soil toxicity.  I would start testing with stainless steels unless literature has better ideas.  Capacitive sensing is a common alternative in industry but not universal.


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Solvents for monoalkylations?
Just beginning to grasp this. Betaphenylethylamine to n-methyl-b-pea. Can I use water instead of DMSOesque solvent in regards to cesium carbonate/hydroxide monohydrate? Also are molecular sieves necessary? As well as 1-bromobutane?
Bottle + water + pea + cesium carbonate/hydroxide = NMPEA?

Your phenethylamine and bromobutane/methyl iodide are not going to be soluble in water but will dissolve in each other and react. If you use aq base and stir, possibly with gentle heating for bromobutane, then you will get your product. No need for sieves or drying agents. This is essentially a phase-transfer reaction but without a second solvent. As with any attempted amine alkylation, the danger is over reaction to give the quat. salt. If you find over reaction to be a significant problem then consider reductive amination.

Water for the solvent... and then heating and stirring to replace the need for bromobutane? I appreciate your initial response sir!

You'll still need haloalkane (MeI or BuBr) or there's nothing for the phenethylamine to react with. Water for solvents as the two reactants dissolve in each other, the initial product (which will be a salt) goes into aq phase where it is neutralised, then returns to the organic phase.

A little over my head. Cesium hydroxide monohydrate or cesium carbonate + phenylethylamine in a water base devoid of anything else besides stirring and heating? One pot / shake and bake style betamethylphenylethylamine? Forgive my crude explanation, all due respect to the professional chemist.

Phenylethylamine + cesium hydroxide or carbonate in water does nothing. If you want N-methylphenylethylamine then you must introduce a source of that methyl. Methyl iodide will do fine.

Well appreciated. Hope to apply in the near future.

If this doesn't work, consider using acetonitrile and potassium carbonate

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Metal reactivity with hot water VS HCL
Does being able to react with hot water take more metal reactivity than being able to react with HCl? To me, it seems like a metal would need to be more reactive to be able to react with hot water since HCl is a very strong acid and so should theoretically be able to react with more metals, but I am not so sure.
Edit: Context: (Question) Al reacts with hot water (boiling pt), Mg does not react with hot water but reacts with HCl, which is more reactive? According to the reactivity trends, Mg is more reactive than Al, but according to the information given in the question, it seems logical that Al is more reactive since it reacts with hot water but Mg does not. So is Mg or Al more reactive if we were to assume that everything in the question were true?

What is the temperature of "hot" water? 300 K? 2,000 K , well above the critical point?

Boiling so 100 degrees Celsius

How does magnesium not react with hot water?

you might as well check the reactivity series.There is no unique and fully consistent way to define the reactivity series.The reactivity series is sometimes quoted in the strict reverse order of standard electrode potentials, where it is also known as the "electrochemical series":according to the series which has been defined there is no reason behind it.It is experimentally defined and according to the reactivity series magnesium is placed above aluminium.You might as well check this link for more clarity https://en.wikipedia.org/wiki/Reactivity_series#Reaction_with_water_and_acids.In your question it is given that magnesium doesn't react with hot water but it does actually vigorously.
 Magnesium  reacts very slowly with cold water, but rapidly
in boiling water, and very vigorously with acids.Aluminium reacts with acids and steam but less vigorously than magnesium.Magnesium DOES react with hot water even more vigorously than aluminium . Please check the link given.
Yes another factor is the formation of oxide layer on the surface of aluminium which reduces its reactivity

Al (not AI, @Jon), forms an adherent oxide on the surface under some conditions which greatly reduces its apparent reactivity. In dilute HCl, it reacts slowly, but in NaOH solution, the oxide is penetrated and Al reacts rapidly.

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Atkins' Physical Chemistry 10th edition Question on steady-state approximation
Please help me with Example 20E.2 of Atkins' Physical Chemistry 10th edition textbook on using the steady-state approximation. I have attached my attempt but am just not sure how to get to the final answer.

Your equations 1 , 2 and 3 are correct so your method is correct but you have not finished and need to substitute for NO3 in your final line then you will have your answer.

exactly. $\ce{NO3}$ is an intermediate so it can't appear in the rate law.


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Why is the boiling point of para hydroxybenzoic acid lower than that of ortho hydroxybenzoic acid?
As per the literature at 760mm of mercury, p-hydroxy benzoic acid has lower boiling point ($\pu{336^\circ C}$ - ChemSpider) than o-hydroxy benzoic acid ($\pu{372^\circ C}$ - ChemSpider). What is the reason behind this although for the former we can see intermolecular H-bonding?

One consideration:  the para compound can have a "bimolecular" hydrogen bond structure like acetic acid.


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Berry Pseudorotation
I was at a chemistry competition, and one of the questions they asked was on Berry pseudorotation
Now the question gave 4 molecules, namely, $\ce{IF5, PF5, PCl5, and XeF5-}$, and asked which of these molecules is unable to undergo Berry pseudorotation. I immediately eliminated $\ce{PF5 and PCl5}$ as they have trigonal bipyramidal structure, so that left me with the other two molecules.
At this point, I am unable to determine which of the remaining molecules should be the answer, so I take a guess and select $\ce{IF5}$, as my reasoning is that it has lone pairs so it is unable to interconvert. (However this also could be used to explain for why $\ce{XeF5-}$ is a possible answer too)
After reading the Wikipedia article above, it turns out $\ce{IF5}$ is able to undergo such pseudorotation, which leaves me surprised, but in that case, could someone explain why $\ce{XeF5-}$, as opposed to $\ce{IF5}$, is unable to undergo such rotation? (I think it has something to do with the lone pairs on $\ce{XeF5-}$, but I'm not sure)

It would appear that $\ce{XeF5^-}$ has all 6 atoms co-planar (D$_{5h}$) so would not fit the criterion needed for pseudo-rotation. (See JACS 1991, vol 113, p 3351)

As per classic literature, PF5 shows identical bond lengths of all P-F bonds which are supported by data from NMR and IR spectroscopy which led to the development of berry-pseudo-rotation. But data's from electron diffraction shows that there are eventually two types of bond lengths (P-F (axial)> P-F(equatorial)) which supports the fact that a naive might predict the hybridisation to be sp²d and hence T.B.P. Even in JD lee, it has been clearly stated that this fluctuation may be called the "Paradox" of Chemistry


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More effective fire extinguisher liquid than water
I join robot fire fighting contest.  One of the mission is to extinguish candle by spraying it with liquid (water is default option). For spray mechanism I use pump with Venturi effect principle.  Is there any liquid substance that is more effective extinguishing fire than water by Venturi spraying mechanism? Homemade is preferred.

You could consider an aqueous film forming foam (AFFF) system, this is water with some additives to form a foam. Originally foam was something intended for class B fires, but there are now foams which are intended for both class A and class B fires. They are even considered to be acceptable for use on gas fueled fires where there is no risk of explosion (class C). Thus there are ABC foams.
A word of warning some ABC foams such as those based on PFOS are horrible and harmful. It might be best if you stick to a water spray system. The PFOS foams are considered to be problem when they enter the environment, if I recall correctly they are very harmful to fish and the perfluoro compound PFOS only breaks down very slowly in the environment.


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What does unactivated, activated and deactivated aryl chloride mean in cross-coupling reaction?
Hartwig et al. [1] used unactivated aryl chloride for Buchwald–Hartwig amination. What does unactivated, activated and deactivated aryl chloride mean in a cross-coupling reaction?
Reference

Hartwig, J. F.; Kawatsura, M.; Hauck, S. I.; Shaughnessy, K. H.; Alcazar-Roman, L. M. Room-Temperature Palladium-Catalyzed Amination of Aryl Bromides and Chlorides and Extended Scope of Aromatic C−N Bond Formation with a Commercial Ligand. J. Org. Chem. 1999, 64 (15), 5575–5580. DOI: 10.1021/jo990408i.


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Can air inside a high temperature (1300C) Kiln cause an explosion?
I saw many kiln designs that lack an opening for hot\pressurized air to come out , or any pressure valve.
But when air is heated to such a high temperature (1300C) inside a closed chamber (the kiln), won't the pressure inside increase enough to explode the kiln ?
(These kilns usually operate for several hours under this high temperature)

How's that a chemistry question? Sounds like physics to me. Now to the point: maybe it would, maybe it wouldn't. Also, consider the possibility that these kilns might not be hermetically sealed.

@IvanNeretin I wanted to make sure i am not missing any chemical reactions inside the air when it reaches high temperatures, this is why i thought it would be good to ask it here, i can remove it if it's not relevant. and about the hermetically sealed, i didn't think tiny gaps between the kiln door or lid  would be sufficient to release pressure from the kiln, and these doors are usually heavy

The tiny gaps might pretty well be sufficient. After all, a kiln does not get hot instantly. As for the reactions, why, there surely may be some, but why would that be important?

@IvanNeretin I don't know , i thought about bulling a tiny kiln for ceramics, and i didn't want to explode in the process ;-) ,  So you are saying the  reactions would be negligible ?

I think in general the physics would make it unlikely that the air pressure would build up faster than it leaks out, attempting to seal a furnace is unlikely to happen by accident.
A bomb calorimeter or pressure vessel made to specifically seal the inside from outside does not have a loose fitting lid over a cast refractory edge, it is made to much finer tolerances.
Chemical reactions are not likely unless you have a significant mount of vaporisable fuel in the charge and this is contained until at some point it lifts the lid and lets vapour out and air in with glowing elements to ignite the mixture, a flame may result, explosion still unlikely as volume small and not contained any longer.
In practice most kilns have a hole for temperature sensor that is not airtight and would let pressure equalise as they are heated.
Unless a specific atmosphere (inert, reducing, oxidising, carburising) is required in a kiln there is no real need to make them air tight and they rarely are, even then it is often achieved within a protective wrapping of the load and supporting chemistry inside the wrapping like charcoal, cyanides etc).


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Phosphonate reaction with oxides and their stability?
I have read some papers stating that $\ce{SiO2}$ surfaces can bind phosphonates, but in water they hydrolyse. However, for $\ce{Si}$ surfaces, they are  stable in water. In both cases, the reaction takes place through $\ce{OH}$ groups that are present on the surface.

What I think is that on $\ce{SiO2}$ surfaces, there could be some steric effects which make hydrolysis "more convenient".
Another answer which seems less reasonable is that maybe the phosphonate group is electron withdrawing, and as the Si in the oxide already lacks electron density, this makes $\ce{Si-O-P}$ less stable, which is not the case for pure Si.

Related to this, why is the $\ce{Al-O-P}$ bond more stable in water than $\ce{Si-O-P}$?

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Is sodium acetate a neutral or a basic salt?
I read somewhere that salts which do not contain a replaceable hydrogen or hydroxyl group are classified as normal salts. But I also know that strong base-weak acid salts are basic in nature.

Normal salts do not have to be neutral, and vice versa. There are two different features of a salt: (1) containing a replaceable hydrogen, and (2) making an acidic solution in water. They are not necessarily related. For example:

$\ce{NaHSO4}$ has both.
$\ce{NaHCO3}$ has the first and not the second (in fact, being a salt of a strong base and a weak acid, it has to be basic in nature, as you correctly recall).
On the other hand, $\ce{CuSO4}$ is partially hydrolyzed in water solution and hence has the second feature, but not the first.
To make the picture complete, $\ce{NaCl}$ has neither.

Likewise, the opposite features of (1) having a hydroxy group and (2) making a basic solution in water are even less related. For example:

$\ce{Cu2(OH)2CO3}$ has the first feature, but not the second (for it is not even soluble in water).
$\ce{CH3COONa}$ is partially hydrolyzed in water solution and hence has the second feature, but not the first.
$\ce{MgCl2}$ and $\ce{Mg(OH)2}$ form several basic salt phases with both features (J. Chem. Eng. Data 2017, 62 (4), 1384–1396).  They are only sparingly soluble but still form measurably alkaline solutions, especially when warmed.
$\ce{MgSO4}$ has neither.

What about the particular case in the question?

Sodium acetate is hidden somewhere in my answer.

Okay so I get it...a salt is only neutral when it follows neither 1 not 2. Thanks :)

No, a salt is neutral when it doesn't follow 2, and normal when it doesn't follow 1. These things may or may not coincide.

I find it very hard to get the idea. You are defining those "features", but what follows from them? What do you mean CuSO4 "hydrolyzes" in water? If you mean the sulfate ion, it's a rather weak base, with pKa of $HSO_4^-$ of ~2. How should I know the acetate ion is basic, without knowing the pKa of acetic acid?

Well, then chemistry is hard. Now to the point: sulfate ion does not hydrolyze in water, since it is a very weak base, like you said. As for the acetate ion and its basicity: without knowing the pKa of acetic acid, you wouldn't know that.

I still don't get what you mean by "CuSO4 hydrolyzes", and how that makes the solution acidic. Copper sulfate dissolves in water, right. And now? $Cu2+$ hydrate is notably acidic? That's not what you wrote, and I don't even immediately know where to look it up.

When we say that a certain salt hydrolyzes, it isn't clear without context whether we mean the hydrolysis of the cation, or anion, or both. In the case of CuSO4, it is the cation.

Since sodium acetate is the salt formed from sodium hydroxide (a strong base) and acetic acid (a weak acid) a solution of the salt should be alkaline.


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Do we consider the dissociation of water when dealing with ion concentrations of a solution?

One mole of hydrated copper(II) sulfate, $\ce{CuSO4 . 5 H2O},$ is dissolved in water. How many moles of ions does the solution contain?
1) 1
2) 2
3) 6
4) 7

The correct answer is 2. Specifically, I'm confused about why the $\ce{OH-}$ and $\ce{H+}$ ions aren't considered.

The solution will remain neutral with pH=$7$, thus the concentration of OH$^-$ and H$^+$ will be $10^{-7}$ M which is negligible compared to the copper and sulphate of one molar each.

It dissociates into 7 parts; copper and sulfate ions and five moles of water.
Water is also the solvent, so these don't count.
However the answer "two" is indeed not perfectly right, because the sulfate ion would (partially) react with water, to form
$$\ce{SO4^2- + H2O <=> HSO4- + OH-}$$
A sulfate ion is so to say the opposite of sulfuric acid, which reacts
$$\ce{H2SO4 + 2 H2O -> HSO4- + H3O+ + H2O <=> SO4^2- + 2 H3O+}$$
$\ce{HSO4-}$ is not strongly acidic anymore $(\mathrm{p}K_\mathrm{a} = 2),$ so the sulfate ion logically must react somewhat basic. (OK, $\mathrm{p}K_\mathrm{a} = 2$ is quite acidic, acetic acid has $\mathrm{p}K_\mathrm{a} = 4.8.)$
So the correct answer is slightly larger than two. But from the given possibilities in this multiple-choice test, two is by far the most reasonable.
(Update: The hydrated $\ce{Cu^2+}$ ion might well be acidic enough to overcompensate the basicity of the sulfate. In which case you'd still have more than two equivalents of ions, just it'd be a small amount of $\ce{H3O^+},$ not $\ce{OH^-}.$ And of course there is always the autoprotolysis of water.)

@Karl: Do you mean $\ce{Cu^{2+}}$ ions? And also 7 parts, copper ..?


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Term for compound of carbon, hydrogen, and oxygen
Is there a term for a compound of carbon, hydrogen, and oxygen?
There is hydrocarbon for a compound of carbon and hydrogen.  There is also carbohydrate but that requires the ratio of hydrogen to oxygen to be 2:1.
As well as the carbohydrates, there are many other compounds of these three elements: alcohols, aldehydes, ketones, esters, etc.  Yet, I cannot find a term for the whole group.  Have I missed something or just no need is felt for such a term?

There is no general term for this group. It's quite arbitrary to categorise organic compounds just by the elements they contain, and not useful.

@Karl Thanks.  Of course, it is not an important point but given that the term hydrocarbon is quite common, it seemed a little surprising that there was no term for this group.  I guess that hydrocarbons share a lot more properties than this group.

"hydrocarbons" (!= carbohydrates) is also only useful with an additional adjective (aliphatic, olefinic, aromatic). The group is too big and diverse, nobody talks about just plain "hydrocarbons". "Carbohydrate" is a synonym for sugars. Now that's a useful category.

@Karl Thanks again.  Just "hydrocarbon" is common outside of chemistry but I can see that it is too broad in a more serious setting.  Of course, my group with oxygen is huge.  I am fairly familiar with carbohydrates as my son studies biochemistry.

There is no term for this group as it is too broad and diverse.  Even the narrower term hydrocarbon is usually qualified e.g. aliphatic, olefinic, aromatic.
With thanks to Karl.


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Confidence in conformation search results from molecular mechanics
Conformational searches using molecular mechanics can be either systematic or 'random' using Monte Carlo methods (or other non-systematic methods) to sample conformational space.
In general, my understanding is that Monte Carlo methods perform just as well as systematic searches # assuming that the right parameters are used. This is often described in (organic) computational papers along the lines of :

A sufficient number of conformations were explored such that all low-energy conformers within  a X kj/mol window of the global minima were found at least X times.

There is a lot of ambiguity over how many times a low energy conformer needs to be found in order to 'guarantee' that the conformational space has been fully explored. Certainly, in my own experience I can generate an ensemble that fulfils this criteria but is clearly non-complete (if I run the same search from a slightly different starting geometry I generate another ensemble with some different low-energy conformers).
My question is two-fold:

Is there any mathematical basis for the statement about finding a low energy conformation x times, or is it largely a case of convenience that works well empirically in the majority of situations ?
Is there really any way to know that a search is comprehensive without systematically exploring the entire PES ?

#: There was a report that Ken Houk lost a bet due to his belief that systematic sampling was the only true way to explore a potential energy surface, though I cant find the paper where this was discussed.

"A sufficient number of conformations were explored such that all low-energy conformers within a X kj/mol window of the global minima were found at least X times." Do you mean for the first X to be different from the second X, meaning that the second one should be called something else, such as Y?

Note that both systematic and random search can only quarantine finding global minimum for a certain probability. The question is not if random sampling is guaranteed correct, the question is which of the two methods are better in finding the answer with high probability in the shorter time.

We've been interested in this question - and are currently generating sets of exhaustively-sampled minima.

I'd guess that the answer is "it seems to work well" based on an estimate of statistics - if you're seeing the same conformer X times, the probability of missing minima can be hoped to be small.

We have published² a probabilistic treatment of this question, but for the case of MD simulations (and not MC). The principal idea is to use Good-Turing statistics to estimate the probability of unobserved species (ie conformations that have not -as yet- been observed). I have no idea whether this treatment can be transferred to MC, but there is an open source program³ -based on R- that you could try to use.
² https://pubs.acs.org/doi/10.1021/ci4005817
³ https://github.com/pkoukos/GoodTuringMD

That's an interesting approach


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Mechanism of preparation of methylene blue
Methylene blue is prepared by reacting p-amino N,N dimethylaniline with strong acidic solution of $\ce{Na2S}$ and later adding aqueous $\ce{FeCl3}$. In this reaction does the mechanism begin with the addition of proton on amino which then undergoes substitution with another of the substituted aniline molecule?
Also how does sulfur attach to the ring?


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Do theoretical plates required to obtain a pure product decrease under vacuum distillation?
For this question I am using Ethanol as an example. The following is the vapor-liquid equilibrium diagram for Ethanol under atmospheric pressures.

Under reduced pressures do the number of theoretical separations decrease for a given system? Ethanol for example needs approximately 14 separation "plates" to achieve the azeotrope.
From my understanding wouldn't this only break the azeotrope and allow separation to a higher purity? Ex. Vacuum operation doesn't increase the efficiency as far as theoretical required separations go. (Vacuum Operation Plates == Atmosphereic Operation Plate)
Note: For the purposes of this question efficiency is in reference to the number of theoretical plates, and not the total heat required to obtain the pure product.

The azeotrope is alway only reached for an infinite number of plates, so this question makes no sense in the way it is written. At different pressure, the two curves will look differently, making it easier or not to reach a certain composition. Without data at other pressures, that's hard to say.


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Reaction of Grignard reagent with primary amines
What does the reaction $\ce{CH3CH2NH2 + CH3MgBr ->}$ form?  Does it form a secondary amine or a tertiary amine or an alcohol? I can't find what a Grignard reagent does to a primary amine.

Deprotonates the NH2 group to form methane plus magnesium salt.

Here's the thing about Grignard reagents.  Though we typically want them to do nucleophilic addition to something, these powerful bases tend to go for protons -- a faster reaction -- if there is just about anything that might make the proton vulnerable.  Attaching the proton to an electronegative atom like nitrogen is enough, so primary and even secondary amines will react like that with Grignard reagents.  You just deprotonate the amine, the hydrocarbon part of the Grignard reagent ends up an an alkane (or an arene if there are aromatic rings), then on acid workup you get just the amine (or ammonium ion) back.  Yup, this is not a very productive synthesis.
Choose your substrate carefully with Grignard reagents to avoid this deprotonation and get the addition you want, or consider less strongly basic alternatives like an organocadmium compound (which itself can be made from a Grignard reagent plus a cadmium salt).


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Molecule alignment
I'd like to align ~2000 molecules consisting of 8 atoms of carbon and evaluate their similarity with RMSD (all vs all molecules).
What open software can do it in a day or less? I'd like it to work on Ubuntu, and if possible that it calculates the alignments in parallel.
My files are xyz: number of atoms, element+coordinates of each atom.
Thank you.

Assuming you know basic coding and are able to write a simple script doing the input and writing the output for all the 4 million calculations you can use for example this script I found some days ago for doing the alignment and RMSD calculation. Alternatively PyMol can do this too (align function).
Not quite sure about the timing*, even if one would take a second, I expect it to be much faster, that's still about two days if run in sequence. If you are doing this only once it might not be worth the effort to optimize this and I'd rather wait some hours or days or run several of those jobs in parallel on a cluster.
*I've made a similar script and reading the xyz files, calculating the RMSD and even saving the aligned structures takes about 5 ms, which means it's done in less than 6 hours even without running it in parallel.

The rmsd package is nice, yes.

A good idea to get this working in parallel would be to write a Python script that calls rmsd with multiprocessing, rather than using xargs or GNU parallel from the command line.


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What might be the reason for the longer stem found in some separation funnels?
Most separation funnels have a stem around 50 to 70 mm long (often less if fitted with a ground taper joint) that is used to direct the output into a receiving flask.
While the stop-cock has a small bore matching the small end of the funnel to reduce dead volume the stem tube is generally short and has an internal diameter around 6 to 10 mm to prevent a valuable liquid column being retained in the stem.
Having received a separation funnel c.1970 with a stem of around 200 mm (below the stop cock) I was curious what benefits it could offer over the disadvantages of more difficult clearing of the liquid column in the stem and possible much more likely mechanical mishaps when waving the thing around.  I went looking and found that long stem variants are available from numerous vendors but without any explanation of what their purpose might be.  I have not seen any demonstrations or descriptions that use or call for a long stem separation funnel so I am curious how they are used.
Could the long stem relate to the occasional presence of long stems in regular 60 degree filter funnels and thistle funnels. Could it have some application for faster draining with the addition of the weight of the liquid column?
P.S. The funnel in question is not the Gilson type with an intermediate diameter cylindrical section between the bulb and stopcock. The stem has a larger bore than the stopcock so not a capillary.  It has a typical ground glass stopper.

The long stem would speed up draining (esp. at the end), but without a filter paper, that's not a big issue anyway. I find a long stem helps against splattering (as does an undamaged end of the stem). But then you have to be a bit more careful with it.

@Karl While  Hmm, if the top layer was viscous and you wanted to remove it you could pour it out the top. I mention this capillary feature of filter funnels as it was new to me and novel but the funnel I acquired has a stopper (ground glass joint) and no way to use as a filter.  Perhaps it is still useful for draining viscous liquids faster but how to break the liquid column after closing the tap, or does one leave the top product in the stem? Risky I think.  I would be interested in a catalogue or experimental procedure that calls for such a separatory funnel and perhaps mentions the reason.

If the stem has the right dimensions, it empties when the switch is closed. Otherwise I'd want my money back.


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Finding Gibbs energy at different temperatures
How can I find $∆G$ at a higher/lower temperature if I am given $∆G°$ and $∆H°$ $\pu{298 K}$, and a reaction equation (solubility)?

Are you familiar with the Van't Hopf equation?

You might know that Gibbs free energy is given as:
$$\Delta G=\Delta H-T\Delta S$$
This implies that:
$$\Delta G^\circ=\Delta H^\circ - T \Delta S^\circ$$
Which can be rewritten as:
$$\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T}$$
Often it is a good approximation to assume that $\Delta H$ and $\Delta S$ are temperature independent. Now that you have found the entropy you can calculate Gibbs free energy at different temperatures using the first equation.
If you want to find the equilibrium constants, you might have seen this equation before:
$$\Delta G=-RT\ln\left( K_\text{eq} \right)$$


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Electron Probability Density Functions
I am a mathematics and chemistry double major and I just finished my statistics requirement and a great deal of the class was spent on probability distributions (e.g. Binomial, Poisson, Geometric, Gamma, etc.) and so I was wondering since electrons have a "probability" of being in a certain location according to the Heisenberg Uncertainty Principle, what sort of probability distribution would this represent?
As in, could we treat electrons and their positions as discrete or continuous random variables? And if so, what sort of distribution could we use for an electron being in a particular cloud?

As a math and chemistry double major, you must have heard of Schrodinger equation and its solutions. These are the distributions you are asking about.

Yeah that's what I figured, I was hoping there would be some strange connection to between stats and chem.

There is nothing strange about it (other than the deep inherent weirdness of quantum mechanics, that is). You solve the equation, you get the probability distribution. It is as simple as it gets.

I suppose what I would like to know is what the probability distribution the Schrodinger Equation most closely resembles and whether or not another method to predict electron locations is possible using your average statistical models other than the Schrodinger Equation.

Well, the distribution is known and even expressed in elementary function; look and see what it resembles. As for another method, I've never heard of such. Let's wait for someone else's opinion.


text:
Reactivity and selectivity in free radical halogenation of alkanes
I read about the free radical halogenation of alkanes with $\ce{X2}$ and I have the following questions:

Given that the first step is the homolytic breaking of the $\ce{X-X}$ bond, and the homolytic bond dissociation enthalpies of halogens (page 23, 31, and 34 of the PDF - I think it does not have data on $\ce{I-I}$ - but I'll use values from my textbook) are in the order $\ce{I_2 < F_2 < Br_2 < Cl_2}$, why is the order of reactivity of halogens $\ce{F_2 > Cl_2 > Br_2 > I_2}$?
My book said that bromine is very selective when reacting with alkanes - it will prefer a tertiary carbon greatly over primary or secondary ones. Chlorine is less selective and fluorine is not selective at all - it's happy to react with all types of carbons. Why is this so?

PS: I am using the textbook 'Organic Chemistry For JEE' by Solomons, Fryhle, and Snyder.

Related: What determines the high degree of selectivity in free radical halogenations, Why is it possible to form bromoalkanes but not iodoalkanes from free radical substitution? and What is the major product on chlorination of 2-methylbutane?

Your order for bond dissociation energies correct (DOI: 10.1039/TF9625800633)

Also related: Why does radical chlorination and bromination of propane occur at different positions?


text:
How does activated carbon catalyse oxidation?
Context: I am doing an experiment on the quantitative analysis of ascorbic acid in a blackcurrent syrup using fluorescence spectroscopy. In order to do so, ascorbic acid was first oxidised into dehydroascorbic acid. Emission reading was then recorded.
An aliqout of the sample was taken and boric acid was added to form a complex that would be fluorescent inactive to account for other impurities that are fluorescent active and would interfere with the results. The loss in emission between the original sample and the sample with boric acid was used to deduce the amount of ascorbic acid present.
The reagents used were: acetic acid and sodium acetate as buffer, metaphosphoric acid, and activated carbon. In this scenario, how does activated carbon aid in the oxidation of ascorbic acid?
Also, I've done some searching online and found that the purpose of metaphosphoric acid was to prevent the atmospheric oxidation of ascorbic acid. Why would we want to prevent the atmospheric oxidation of ascorbic acid if what we wanted to achieve was to oxidise it anyway?


text:
What's the identity of the crystalline solid in the following reaction?

A mixture of toluene, benzoic acid, aniline and phenol was dissolved in diethyl ether and extracted with 5% $\ce{NaHCO3}$ solution? This extract was acidified until a white crystalline compound appeared. What is the identity of this compound?

I had this question in my exam today. The answer is benzoic acid but I wanted to know how.

Well what you are doing here is trying to extract it by making it slightly alkaline using water, as water and diethyl ether seperate anything that is only soluble in alkaline water will be extracted from the organic phase. Of course toluene and aniline won't do much if $\ce{NaHCO3}$ is added. For phenol and benzoic acid it's just about the base you are using. $\ce{NaHCO3}$ is not strong enough to deprotonate the phenol. For benzoic acid, the acid is strong enough to be deprotonated by $\ce{NaHCO3}$. When it is acidified the benzoic acid is formed again and precipitates.
The problem here is that many people forget how poorely soluble benzoic acid and salicylic acid really are. I made the same mistake many years ago. Although they only feature a quite small organic fragment they are insoluble in water although they are acids. Their ionic salts however are soluble. So adjusting the pH value either dissolves or precipitates them in water.
Edit:
With the same method you could also probably extract the aniline using acidified water (instead of alkaline) as a soluble anilinium forms.


text:
Reverse rate coefficient for thermolecular and thermal dissociation reactions
I am an astrophysicist working in exoplanetary atmosphere chemistry right now, in particular modelling the chemical kinetics taking place within the atmosphere. Based on this IOP article, we have that given a certain chemical reaction,
$\ce{\alpha A + \beta B <=> \gamma C + \eta D}$
In which its forward rate coefficient is $k_{\mathrm{f}}$ and the reverse counterpart is noted as $k_{\mathrm{r}}$ the equilibrium constant let us to compute the reverse reaction rate coefficient as
$$\frac{k_{\text{f}}}{k_{\mathrm{r}}} = K_{\text{eq}}\left(\frac{k_{\mathrm{b}}T}{P_{0}}\right)^{\Delta \mu}$$
$$K_\text{eq} = \exp\left(-\frac{\Delta H^{0}-T\Delta s^{0}}{RT}\right)$$
The previously cited paper is pretty clear at this statement and it is relatively easy to find this information to me in other sources. Nevertheless I came up with a certain kind of reactions, so called "Combination
and Dissociation Reactions" in which a third body is involved, in my case representing the atmospheric bulk, for example molecular hydrogen for a Jupiter-like planet and it is represented by a letter "M". For example we could have
$\ce{A + M ->  C +  D + M}$
For these kind of reactions the rate coefficient is parametrized based upon on to limiting to zero and infinite pressure, $k_{0}$, $k_{\inf}$ and an "F" coefficient leading to the equivalente rate coefficient
$$k = \frac{k_{0}M}{1+\frac{k_{0}M}{k_{\inf}}}F$$
My question is, is this kind of reactions reversible?. Can we reverse the reaction using the previous formalism just exchanging $k_\mathrm{f}$ by $k$?

While not a direct answer you should look at the 'Collision theory of Unimolecular reactions' also called 'Lindemann Theory'  (in most under-graduate phys. chem. textbooks). These reactions have the form $\ce{A + M  <=> A^* + M; \quad A^* \to products }$. You should be able to generalise from this:)

Thank you a lot for the reference!. I will have a look at it :)


text:
Work done in an isothermal irreversible process
In calculating the work done in an isothermal irreversible process, why do we consider the external pressure only (using formula: $W_{\text{ext}} = P_{\text{ext}} \mathrm{d}V)$ but not the internal pressure? When the system expands, the internal pressure does the positive work, but even in this scenario, we use the equation as mentioned above. Why is this happening?

The ideal gas law (or any other equation of state) can only be applied to a gas at thermodynamic equilibrium.  In an irreversible process, the gas is not at thermodynamic equilibrium, so the ideal gas law will not apply.  The force per unit area exerted by the gas on the piston is comprised of two parts in an irreversible process:  the local pressure and viscous stresses.  The latter depend, not on the amount that the gas has been deformed, but on its rate of deformation.  Of course, at thermodynamic equilibrium, the rate of deformation of the gas is zero, and the force per unit area reduces to the pressure.  In this case the ideal gas law is recovered.
So, you are correct in saying that, for a reversible process, the internal pressure is equal to the external pressure.  But, for an irreversible process, even though, by Newton's 3rd law, the force per unit area exerted by the gas on its surroundings is equal to the force per unit area exerted by the surroundings on the gas, the force per unit area exerted by the gas on its surroundings includes viscous stresses; thus, the force can cannot be determined by simply applying the ideal gas law.  In this case, our only alternative (aside from solving the complicated gas dynamics equations for the gas, which include the viscous stresses) is the explicitly specify the external force per unit area imposed by the surroundings.

in irreversible isothermal expansion, formula for work done is    W=P(external)x change in volume. here why we are not considering  internal pressure? here expansion is caused due to internal pressure of gas  against external pressure and so work is done .please clarify me irreversible process formula

For an irreversible expansion process, a crude approximation to the force exerted by the gas on the piston  (where the work is done) can be provided by the equation $$\frac{F}{A}=P_{ext}=\frac{nRT}{V}-\frac{k}{V}\frac{dV}{dt}$$where $P_{ext}$ is the force per unit area of the piston on the gas and the constant k is proportional to the gas viscosity.  So the gas force per unit area is equal to the external force per unit area $P_{ext}$, but there is an extra term related to the rate of volume change.

@ChetMiller I have a doubt which I think I can ask here , as we know that irreversible processes are like sudden jerks or quick changes then in that case how can we call the process isothermal because we cannot specify temperature uniquely at each point inside system as piston moves out, I mean to call it isothermal the temperature at each point should be same but this is not the case during irreversible expansion

@ChetMiller or for irreversible isothermal process only initial and final temperature need to be same ?

@Lalit It depends on how you define an isothermal process.  It could be that only the initial and final temperatures are the same, or it could be that the system is in contact with a constant temperature reservoir at the same temperature as initially (and finally).  For an irreversible process, this still doesn't say anything about how the temperature is varying inside the system (only at its boundary with the reservoir).

The key word here is irreversible. Irreversible processes are more like sudden jerks or a quick change in the system and therefore the ${P_{\text{ext}}}$ is not same as the ${P_{\text{int}}}$ throughout the process.

Why do we NOT use the internal pressure (${P_{\text{int}}}$) to find the work done?
  It is very difficult to find the ${P_{\text{int}}}$ exerted by the gas in the container since ${P_{\text{int}}}$ is varying with time. For this, we would have to use integration. Something like this:$${W_{\text{int}} = P(V)_{\text{int}}\mathrm{d}V}$$
  (where $P(V)_{\text{int}}$ is a function of $V$)
  But we know the value of ${P_{\text{ext}}}$ and this is constant throughout the process. This is why we use ${P_{\text{ext}}}$ instead of troubling ourselves with integration. Something like this: $$W_{ext}= P_{\text{ext}} \mathrm{d}V$$

So if a system undergoes a decrement in volume suddenly, it is the external pressure that does the positive work throughout this decrement. The equation would be: $$W_{\text{ext}}= P_{\text{ext}}\mathrm{d}V$$ (However we can use ${P_{int}}$ as well, please read the NOTE at the bottom)
And if a system undergoes an increment in volume suddenly, ${P_{\text{int}}}$ is doing the positive work. However, to find the $W_{int}$ we would have to use integration since ${P_{\text{int}}}$ is varying with time. It is much easier if we just find the negative work done by ${P_{\text{ext}}}$ and since we know that $${W_{\text{int}} = - W_{\text{ext}}}$$ we can find the work done. This is why we consider the ${P_{\text{ext}}}$ and according to the situation in the question, we apply the signs.
NOTE: Magnitude of work done by the internal pressure and external pressure is always same. We only use ${P_{\text{ext}}}$ because it is constant whereas ${P_{\text{int}}}$ varies with time. If we use integration to find ${W_{\text{int}} = P(V)_{\text{int}}\mathrm{d}V}$, where $P(V)_{\text{int}}$ is a function of $V$, we will get the same magnitude as we would from $W_{\text{ext}}= P_{\text{ext}} \mathrm{d}V$. Only signs are opposite.

In irrreversible expansion,expansion is caused due to internal pressure and force is exerted by gas on piston causing expansion.External pressure in not responsible for expansion and then why do we  consider?

@MouryanPalla could you add your comment to your question as well so others can understand better?

@dr.drizzy if we have the initial and final pressure of the system, we can use the formula Pavg*(Delta V) ... So why don't we use that?

Expansion
Work is always done against the external pressure, because that's what opposes the expansion of the gas (and not internal pressure). It's just that in case of a reversible process, the internal pressure (or pressure of the gas in other words) is equal to the external pressure - so we can use it to find the work done. This is not the case for an irreversible process - the gas does not get time to make its pressure equal to the external pressure since the process is not being performed slowly. So, we need to consider the external pressure.
Compression
In this case, the opposing force is the internal pressure. But, since it's an irreversible process, I think we can't know the internal pressure. But, we know the temperature, so I'd use the FLOT to indirectly find the work done.
PS: I'm not sure about the compression part; it'd be nice if someone good at thermodynamics clarified this in the comments. :)

In irrreversible expansion,expansion is caused due to internal pressure and force is exerted by gas on piston causing expansion.External pressure in not responsible for expansion and then why do we  consider?

Have a look at my solution. I think the compression part is covered there. Let me know if it's helpful


text:
Valence shell expansion and sulfur
I've been searching some things regarding the ability of sulfur to accomodate negative charge and reached the conclusion that it's due to its size and its ability to expand its valence shell.
I can't fully understand the expansion part though. It is mentioned in various sites and books that sulfur uses the 3d orbitals due to their similar energy level with 3p orbitals in order to achieve this "expansion". However, 4s are supposed to be filled first. I even came across this question: How does expansion of electron shells work? but the answer given didn't really helped me. So why do we consider the expansion to happen with 3d orbitals, while 4s is filled first?

I have a feeling we should make some banner next to "top questions" list saying  "Octet expansion is outdated and deprecated theory"

I've also seen that being mentioned. So what is accepeted regarding this phenomenon?

That there's no such phenomenon?

How do we justify the ability of sulfur to create more bonds then? And regarding its ability to accomodate negative charge better than oxygen?

https://chemistry.stackexchange.com/a/5242/9961 I suggest to ignore rest of answers there.

I hadn't found this thread. Thank you. So, the only accpepted justification of sulfur "controlling" a negative charge better than oxygen, is simply it size?

I really hate linking to that post, but @Mithoron is right, that one answer is ok. (There are two more that are not awful.) The top answer is just terrible. Anyway, I don't know what caused Oscar to delete his answer, but he linked to a paper that you might want to check out: Alan E. Reed and Frank. Weinhold (1986). "On the role of d orbitals in sulfur hexafluoride". J. Am. Chem. Soc. 108(13), 3586–3593. https://doi.org/10.1021/ja00273a006

authors: Αντώνιος Κελεσίδης


text:
Formation of strontium peroxide
Wikipedia states:

Strontium peroxide is produced by passing oxygen over heated strontium oxide. Upon heating in the absence of $\ce{O2}$, it degrades to $\ce{SrO}$ and $\ce{O2}$.

I found some literature about what temperature is needed but I did not find any literature about the condition of oxygen pressure. If I do annealing of $\ce{SrO}$ at $\pu{800 ^\circ C}$ in $\pu{0.5 atm}$ oxygen, will I form $\ce{SrO2}$?

Posted from Physics.SE

There are a few method to synthesize $\ce{SrO2}$ from $\ce{SrO}$ and $\ce{O2}$. One of them uses $\ce{KClO3}$ as the source of $\ce{O2}$ [Ref. 1]. The paper is in German but has English abstract which states:

Abstract. Single crystals of $\ce{SrO2}$ have been oblained after high
  pressure/high temperature reaction of a $\ce{SrO}$/$\ce{KClO3}$ mtxture at $\pu{20 kbar}$, $\pu{1400 ^\circ C}$. ...  From a refinement of the site occupation factor for oxygen a composition $\ce{SrO_{1.95(2)}}$ has been found for the crystal investigated.

Note that $\pu{20 kbar}$ pressure was inserted by the $\ce{O2}$ released by the following reaction at high temperature (Ref. 2): $\ce{2KClO3(s) -> 2KCl(s) + 3O2(g)}$.
Experimental information on the $\ce{Sr-O}$ has been reviewed in Ref. 2 and stability of $\ce{SrO2}$ and phase-diagram of $\ce{Sr-O}$ is graphically presented (see Figure 1 & 2, respectively). Accordingly, $\ce{SrO}$ under $\pu{0.5 atm}$ of $\ce{O2}$ pressure at $\pu{800 ^\circ C}$ would not produce $\ce{SrO2}$ (see Figure 1: Only $\ce{SrO}$ exists under the conditions, $\log_{10} p = \log_{10} 0.5 = -0.30$ and $\frac{1000}{\mathrm{T}} = \frac{1000}{1073.15} = 0.93$).

References:

K.-J. Kange, F. Rau, U. SchieBl, U. Klement, Verfeinerung der Kristallstruktur von $\ce{SrO2}$ (Refinement of the Crystal Structure of $\ce{SrO2}$), Z. anorg. allg. Chem., 1994, 620, 879-881 (https://onlinelibrary.wiley.com/doi/pdf/10.1002/zaac.19946200521).
D. Risold, B. Hallstedt, L. J. Gauckler, The strontium-oxygen system, Calphad, 1996, 20(3), 353-361 (https://doi.org/10.1016/S0364-5916(96)00037-5).

as I read on this phase diagram (left) at room temperature (3.66 [1000/T^-1]) and athmosperic pressure (0(zero)  [log(P[bar]] SrO-2) is stable. am i right?

From atomistry.com:

Very large pressures are required in order to make strontium oxide
  combine directly with oxygen. Combination under a pressure of 76
  atmospheres of oxygen is possible, but the maximum yield, 15-16 per
  cent, of peroxide, is obtained at about 400° C. under a pressure of 98
  atmospheres.


text:
How to compare dipole moment in thiophene, furan, and pyrrole?
This is what I found (source):

I wonder about several things here.
First, why is tetrahydrothiophene's dipole moment bigger than that of tetrahydrofuran, when considering that tetrahydrofuran has an atom of stronger electronegativity?
Secondly, why thiophene has weaker dipole moment when the S atom has larger p obitals, and therefore, has weaker conjugation ability than O to aromatic ring. The result seems that S in thiophene has better conjugation effect. (that's why it has less stronger dipole.) and I don't seem to explain why.
So what are the reasons for these dipole moment orders?

Related: Dipole moments of pyrrole and furan

As another (downvoted) answer points out, the key to this comparison involves the heteroatom donating an electron pair to the ring in the aromatic compound.  This shifts the negative charge away from the heteroatom's corner into the ring.  Thus we either do not get as much of a negative end at the electronegative heteroatom's corner, or with a stronger interaction we can reverse the dipole.  In the examples shown, pay attention to the directions as well as the numbers of the given dipole moments!
Clearly nitrogen has this stronger interaction, while oxygen and sulfur give the weaker effect, with a reduced but not reversed dipole moment.  Oxygen is so electronegative that even in forming the aromatic ring it donates an electron pair only weakly.  The large sulfur atom has lower electronegativity, but overlaps poorly with the smaller carbon atoms. (In the nonaromatic tetrahydrothiophene, the larger size of the sulfur atom makes the dipole longer and thus may give a larger dipole moment with less charge separation.)  Nitrogen in pyrrole has the moderate electronegativity and strong overlap which, together, enable the stronger interaction.

Thanks for your answer but my question is still not solved. I'll make it easy. Tetrahydrofuran has dipole 1.73D and furan has dipole 0.7D. So furan's overlaping effect offset 1.73-0.7=1.03D. On the other hand, Thiophene offset tetrahydrothiophene's dipole 1.9-0.51=1.39D. Isn't this mean thiophene donates electron better than furan? By the way, I checked furan is better at EAS than thiophene, which means furan donates electrons better. But what happened to dipoles??? (and thanks for explanation of tetrahydrothiophene's bigger dipole. That makes good sense.)

1.03 vs 1.39 vs ... 3.39, for pyrrole.  It is pyrrole that srands out for dipole alteration, the others are just small potatoes.

It depends on the extent of which the heteroaton can donate its lonepair into the ring and as N does it the most the dipole moment is upward and similar way it can be determined for the other two


text:
Is there anyway to derive the number of water molecules in a hydrated salt, given just the formula for the anhydrous salt?
Is there anyway to derive the number of water molecules in a hydrated salt, given just the formula for the anhydrous salt?
That's pretty much it. If not, why?
For coordinated water, my guess is counting the number of unfilled pairs in the metal ion or something like that would give us an indication of the number of water molecules.

No.$\mathstrut$

For a given salt there may be lots of hydrates.

@Mithoron Come on! Science can't be that arbitrary. There must be something

https://en.wikipedia.org/wiki/Magnesium_chloride#Structure,_preparation,_and_general_properties

I disagree that this question is too broad, other than the fact that it is of general interest, even if the answer is not much more complicated than "no".

@pentavalentcarbon This matter is much more complicated. One can't just derive it with simple equation, but there can be guesses similar to what OP proposes, sometimes it would work other times not. One could use various computational methods to estimate enthalpies of hydrates one by one to find which probably exist. No doubt there's middle ground between these approaches, but it seems like putting whole book summary into few posts.

@Mithoron "No doubt there's middle ground between these approaches" If such an approach exists, I have never heard of it, and it would make a great answer.


text:
Under what circumstances can the stereochemistry at nitrogen atom be faithfully retained and identified?
Under what circumstances can the stereochemistry at nitrogen atom be faithfully retained and identified?
Because in most cases nitrogen atoms tend to undergo stereorandomisation by flipping due to its lone pair. I know of one example which has a three membered ring where nitrogen flips very slowly due to the high s character in the lone pair which requires significant activation energy to flip. But I wanted to know of the other cases where this is possible (probably in a stereo locked system inhibiting flipping).

Kill that lone pair, and the problem will be gone. I believe tetraalkylammonium cations don't flip.

The TS for nitrogen inversion involves the lone pair being in a p orbital.  Putting more s character in the orbital holding the lone pair moves it further away from the TS and increases the inversion barrier.  Electronegative substituents will have this effect (Bent's Rule) on the lone pair orbital.  $\ce{NF3}$ has a calculated inversion barrier of 78.5 kcal/m

You could try to lock the conformation of structure. For example, derivatives of a  1-azabicyclo[2.2.2]octane system could have a chiral nitrogen, and the fact that the substituents are tied back prevents inversion of the nitrogen. Quinine has this kind of structural moiety.
This example isn't exactly what you're looking for, but it does it by using diastereocontrol to limit the amount of undesired isomer: J. Org. Chem. 2004, 69 (3), 714–718.
As Ivan suggested, you can hold onto the lone pair with something. Here's an example where metal coordination prevents inversion: Inorg. Chem. 2012, 51 (15), 8516–8523.
There are plenty of others. I just did a quick search on the JACS website for stereogenic nitrogen.


text:
Why does a Beckmann Rearrangement not occur in the in nitrosation of cyclohexanone?
In my textbook Organic Chemistry, 2nd edition by Clayden, on pg. 464 it states the addition of $\ce{NO2-,HCl}$ to cyclohexanone to 2-nitrosocyclohexanone, which hydrolyses to 1,2-cyclohexadione. Why does the Beckmann rearrangement not occur in 2-nitrosocyclohexanone instead?

First, reaction conditions are important here. Second, do no assume that the nitroso is the same as the oxime.

Because the oxime hydrolysis is faster than the Beckmann rearrangement (after the nitroso has isomerised)


text:
Question about the composition of a 1902 non-polarisable electrode diagram by Jagadis Bose
In his work, "Response in The Living and Non-living" (1902); page 6, Jagadis Bose describes a method for detecting nerve response. He explains that he uses iso-electric contacts with no current in the galvanometer and diagrams a non-polarisable electrode. The following are his diagrams:

On Diagram C, one will note that he uses a strip of saltwater soaked cloth attaching to a node that contains, what appears to be, a Zinc Sulfate solution with a Zinc rod resting in this ZnSO4 solution.
Below the ZnSO4, there's a seemingly unspecified area that acts as an intermediary between the ZnSO4 and the "strip of cloth."  I have highlighted it in yellow in the diagram below.
Could someone please explain what this medium may be or what could be used in it's place?  It appears to be some kind of solid or interstitial porous material (or another compound that he does not mention). Or is it simply a generic attachment or some kind of neutral absorbent material (perhaps sand?) that acts as a medium between the Zinc Sulfate, the "node" and the moistened cloth?
Here is Diagram C enlarged and highlighted for viewing convenience:


text:
What does borax do in and to the washing machine?

Does borax actually do anything to make clothes cleaner? If it does ....
Does it do something with the detergent to remove more dirt from clothes?
Does it make it easier to rinse detergent and dirt out of the clothes?
Does it work only if the wash cycle uses hot water?
Does it un-dissolve in the cooler rinse cycle water and end up building up in the machine?

I honestly don't think this is a broad question. It's just a few closely related questions bunched up together. It certainly doesn't make much sense to ask all five questions in five new questions, does it? Thoughts?

This is just a list of super specific yes/no questions - the correct answer will likely not help anyone. I'd like to see this rewritten - the header question is much better than a list off suggestions and a yes/no question. In addition, a short description of the background of the question and the result of a preliminary investigation (google has lots of suggestions, for example) on the author's part will help attract good answers.


text:
Why does Xe have a higher boiling point than Ne?
I am certain that the reason involves intermolecular forces, but since both $Xe$ and $Ne$ are noble and non-polar gases, shouldn't these forces have a much smaller effect or negligible on boiling point?

You'd be surprised, dispersion forces are much more important than high school chem makes them out to be.

The main interaction between noble gases is Dispersion. (inert gas that's where the name comes from)
Dispersion is larger in heavier atoms. (Considerable larger, 18+18+8 more electrons difference is massive)
an example:
When Interaction energy passes zero point and thermal energy you can have the solid phase. This is why you will not have solid helium.


text:
Prediction of enantiomeric excess through kinetics

Suppose we have a reaction such as the one represented above. Is there a method to estimate enantiomeric excess of the product if the rate constants are known and the product does not isomerize (and the other stereocenter doesn't either)?
To simplify, let us suppose that $a=100$, $b=900$, $c=50$, $d=30$.
We will begin with a racemic mixture (of the first two compounds). Methods that I've tried so far seem inadequate to approximate enantiomeric excess.

Did you try chiral GC? Guessing from cyclopropyl amine (b.p.: ~$\pu{50^\circ C}$), these compounds shhould be fairly low boiling. Alternatively, if you suspect they'd recemize faster at higher temperature,  try chiral HPLC.

This is a computational study (with a different compound, I am just using cyclopropylamine as a model), I was wondering if there is a way to predict enantiomeric excess with rate constants.

Just model it numerically. A simple python script can give you the answer within minutes. Btw. what do you mean by racemic mixture of the first two compounds? You can just look at the series of one enantiomer in each case just like drawn.

Modeling using python worked well, thank you.


text:
Compare stability of transition metal cations in aqueous solution
I have to compare the stability of the following transition metal cations: $\ce{Co^3+, Fe^3+, Cr^3+, Sc^3+}$ in aqueous solution. (Source: Joint Entrance Exam (JEE) 2013 Mains India)
The first thing I looked at were the ions' electronic configurations:
$$
\begin{align}
\ce{Co^3+}:&\ce{[Ar] 3d^6}&\\
\ce{Fe^3+}:&\ce{[Ar] 3d^5}\\
\ce{Cr^3+}:&\ce{[Ar] 3d^3}\\
\ce{Sc^3+}:&\ce{[Ar] 3d^0}\\
\end{align}
$$
It suggests that $\ce{Fe^3+}$ should be the most stable, but the answer key given for increasing stability instead is $\ce{Co^3+ < Fe^3+ < Cr^3+ < Sc^3+}$. This seems to suggest that lower number of d-electrons leads to more stability, but I am not sure why that is true.
I also thought of the fact that more water of hydration would probably lead to higher stability, and that depends on charge/radius ratio. However, that would suggest $\ce{Sc^3+}$ to be the least stable, but that isn't the case here.
So, what factor is at play instead? And why are the stabilizing factors of half-filled d-subshell and greater water of hydration ignored?

Update: As pointed out in the comments below, to narrow down the cases, we may look at the enthalpies of formation, Gibbs free energy of formation, or the standard electrode potential - whichever one of these is the most applicable to ions in aqueous solution - to compare their relative stability.

"Stability" is an ambiguous term, which can encompass a variety of things. This question is very vague.

@TanYongBoon Can you give some examples of the "variety of things"?

Stability could be either thermodynamic stability or kinetic stability. Even defining the term in this manner would still have some ambiguity, the stability could also be seen as how easy it is for the species to undergo particular reactions. Those reactions would then also need to be specified.

@TanYongBoon Correct me if I'm wrong, but thermodynamic stability and kinetic stability were defined when a reaction actually occurs (example). In my question instead I'm just comparing the stability of these free ions in the aqueous solution.

Yes. These terms refer to the relative ease of the chemical species undergoing reaction. How else could you define stability if you do not use a reaction as a yardstick for comparison?

@TanYongBoon I guess by measuring the Gibbs free energy of formation? The lower the value of Gibbs free energy change for the formation of a particular aqueous ion, the more stable it is.

@GaurangTandon Actually, stability is related to enthalpy of formation not gibbs free energy of formation.

@Abcd Okay, enthalpy then.

It is written in NCERT Chemistry class 12 "In a medium like water, d3 is more stable than d5 (see CFSE).
CFSE stands for Crystal Field Stabilisation Energy. Basically in water there are h2o molecules which complexate the ions. Assuming octahedral geometry the d orbitals split into d3 (lower in energy compared to barycentre) and d2 (higher in energy compared to barycentre). If you only fill the d3 then you would actually have a more stable configuration as compared to d5 situation where the e.g. levels would also be occupied.
This is a more prominent factor than half filled orbital stability. In $\ce{Co^3+}$ there are no stabilizations involved and hence it is least stable.
$\ce{Sc^3+}$ has a noble gas configuration and this dominates all effects in terms of stabilizing the ion.


text:
Why does the structure of haemoglobin change when the ferrous ion changes to ferric ion?
It is said that $\ce{Fe^2+}$ can bind oxygen while $\ce{Fe^3+}$ cannot. Why is that so? $\ce{Fe^3+}$ has an extra electron, it could bind more easily to the oxygen. And how and why does the structure of heme protein change with the ferric ion?

"Fe3+ has an extra electron" that's probably a typo. Did you mean Fe2+ instead?

The $\ce{Fe^{II}}$ ion is held in the centre of the heme's porphyrin ring and is also attached to a nitrogen atom on a nearly histidine group which is part of the protein. When there is no oxygen present the Fe ion is in a high spin state. The Fe has two electrons in the $e_g$ orbital and is 5 coordinate. The other 4 bonds are to the porphyrin N atoms making the geometry square pyramidal. In this state the Fe is just a little too big to fit into the porphyrin ring which becomes a distorted, bending towards the histidine. The Fe is 0.07 - 0.08 nm out of the plane and the Fe-N distance 0.22 nm (to the imidazole N on histidine F8). When an oxygen molecule binds the Fe becomes low spin, (the $e_g$ orbitals are now empty) and 6 coordinate, and  has a slightly smaller radius and so fits into the porphyrin ring which becomes planar. As the Fe has moved on binding oxygen, the histidine is also moved and is pulled towards the porphyrin ring by $\approx$ 0.075 nm. This motion is transmitted and amplified through the protein to the other three binding sites which affects their ability to bind. This is called an 'allosteric' interaction as it is 'non-contact' between the sites.
The picture (b) is self explanatory. The structure (a) shows the protein backbone as cylinders and the porphyrins and Fe at atoms. The Fe is yellow.

(Picture from 'Protein Physics' A. Finkelstein & O. Ptitsyn, publ. Academic Press.)


text:
In the addition of HBr to isoprene why does the less stable carbocation form?

According to me, the answer is B as the tertiary allylic carbocation should be more stable. But the answer given is D and in the given solution, the tertiary allylic carbocation rearranges to primary allylic carbocation to yield Option D.
My question is why does rearrangement from tertiary allylic carbocation to primary allylic carbocation take place?

Source: JEE Advanced 2015 Paper 1 Q31

have you hearrd about 1,2 addition and 1,4 addition in conj dienes

https://www.youtube.com/watch?v=JcvQXSm341A this might be helpfull

Steric hindrance in the Br attack step, maybe?

The carbocation will be formed as per stability rules but as it is conjugated system the cabocation resonates and as the less stable carbocation is more reactive  it is attacked by Br-


text:
Does 2-chloro-2-methylpentane react with NaI faster in acetone or water?
Does 2-chloro-2-methylpentane react with NaI faster in acetone or water?
I thought that since it is a tertiary halide it will proceed via SN1 reaction. So it must be faster in water as it is protic solvent.
But the answer given is Acetone. Why?

It  is simple this is the famous finkelstein reaction.

As you can see in the image NaCl is formed which is insoluble in acetone hence it precipitates and the reaction cannot go reverse.

But if we take water the reaction will shift in backward direction as NaCl is soluble in water.

 That is the reason we take acetone.

plus 2-Cl-2-Me-pentane is not soluble in water so that hinders the reaction further to the point where I doubt it would proceed at all


text:
Biological contamination from ground
I took samples of tapwater in order to examine concentrations of Escherichia coli and enterococcus in a rural area. I was told, from an analytical lab, to take samples in four sterilized vessels bought from pharmacy.
The problem is that at the completion of the last vessel, its cap fell frοm my hand down into the soil and some grains of sand were stuck on the inner surface of the cap. As fast as I could I picked it up and washed it off totally with the water of the sampling tap. After that I closed my last sampling vessel with it and this is the completion of the sampling.
How large is the possibility my sampling has a serious contamination of these twο biological pollutants after this incident?

The point is not how large the possibility is.
There is a possibility that the flask has been contaminated, so you should have taken a new flask.
I suppose you did not have a spare one with you.
Then, the best should have been to sterilize again that flask, but that is not easy in the field.
I mean, it is much easier to take more sterile flasks with you than to sterilize them in the field.
Now, if this sample shows that there are no bacteria, you can conclude that there are no bacteria in the water.
But if there is some concentration of bacteria, you can not know if they come from the water or from the soil or how much come from each part.


text:
Camera obscura chemical inquiry
I’d like to make a rather large scale camera obscura and use the light reflected inside of it to recreate the first process for a permanent image so I need a sheet metal as well as a chemical (the recommended is bitumen of judea), to expose the sheet metal to create a heliograph image, it is then said that Niepsce used oil of lavender to wipe off unexposed bitumen(I’m assuming it’s a solvent so I’d also need one to pair with whatever chemical you’d recommend for exposing). The end result is going to be rather large scale so I’d like to use a lens that won’t let in as much light as a glass lens because I believe that this will cause the image to be overexposed I was curious as to what chemical I should use to apply to the metal sheet, and what type of metal to use to get the best result that will be the most light sensitive and result in a clear final product. If this doesn’t make sense I apologize and I’d be glad to try and make as much sense of it as possible, any help is good help thank you!

Glass lens?! Camera Obscura a.k.a. pinhole camera doesn't use any lenses - just a tiny hole in the side opposite to the screen/film. This also causes very little light to enter, making for its primary drawbacks: long exposure time, need for bright scenes.

@SF. As addendum: The lens in the camera obscura is the aperture; the pinhole. If the aperture vs image size and depth of field necessary is adjusted correctly, there is no need for a focusing lens. All incoming light arrives (nearly) perpendicular to the image plane.

I believe the original photo plate was made of silver. However, having seen the result with my own eyes (it's currently kept in a museum a few miles from my place) I'm not sure you'd want to replicate his procedure exactly. Granted the plate has aged quite a bit, but it takes most people a minute or two to be able to see anything at all.


text:
How to electron count bridging ligands?
Although I am okay with rationalising bridging halides and $\mu$-2 CO I can't seem to figure out others, including $\mu$-2 hydride, CN- and NO.
When considering the electron count of bridging ligands, this is my rationale so far:
-Neutral bridging ligands contribute to the overall complex the number of electrons they would have when terminally bonded. These electrons are given in any proportion. This means that $\mu$-2 bridging CO gives 1 electron to a metal centre, for example.
-Ionic counting gives even numbers of electrons for ligands; I am unsure if this is still the case for bridging ligands - I suspect not?
Considering NO, I would therefore believe it to be a 2 electron donor to one metal and a 1 electron donor to the second metal.
Considering hydride, more specifically the dimer form of titanocene, I have been taught that each Ti centre is 16 electrons. I have been told, separately, that each hydride contributes half an electron in neutral counting, and 1 electron in ionic. This would lead to a 16 electron complex for "Titanocene" if there is a Ti-Ti bond, however this source: https://link.springer.com/content/pdf/10.1007%2Fs40828-015-0010-4.pdf seems to imply that it is in fact an 18 electron complex, with each individual hydride contributing more than its total electron count? Is this source correct (and my lecturer wrong) in how you would count this complex?
Considering CN-, this is similar to hydride I believe, and should be treated in the same way - is this correct?

Are you using L or X?

Sorry, I'm not quite sure what you mean?

If you're counting electrons, you're using the L method (counting ligands as neutral) or the X method (counting some ligands as anionic). Which are you using?

I'm using predominantly the L method, although I am also considering the X method for hydride and CN-.

The above compound obeys the EAN rule, and has 36 electrons in total, with each Titanium center having 18 electrons. I will leave the link here. It is clearly specified that according to the covalent model, the electron-donation for μ-H is 1 and 2 for the first and second metal centers respectively.

This means that each μ-H donates two electrons to the first metal center and one electron to the second metal center simultaneously.
ie, while taking the total electron count, assign 3 electrons to each bridging hydride.
Parallelly, you can count the electrons around one metal center also. Each of the Titanium metal centers would have one μ-H (3e-), two Cp ligands (10 e-), and a half metal-metal bond (1 e-). There are 4e- in Titanium metal. This adds to a total of 18e-.
This solves the problem.

In the covalent (L) method, $\mu$-hydrides should be treated as a normal hydride (1-electron ligand) for one metal. The $\ce{M-H}$ bond itself should be treated as a 2-electron ligand for the other method. This explains the two entries for hydride in the link you provided:

Cyano-ligands are isoelectronic to carbonyl ligands, so I'll leave that to you as an exercise to figure out.

My issue with cyano ligands is that although they are isoelectronic to carbonyls, they have a negative charge. Therefore my treatment of carbonyls are: 1) Terminal ligand, 2 electron donors, 2) mu-2 bridging, 1 electron donor to each centre in both L/X methods. However, with CN-, I treat terminal, L method as a 1 electron donor and X method as a 2 electron donor, therefore shouldn't the bridging considerations still be different? Charge notwithstanding it would also behave as a 1 electron donor per bridging centre however it is similar to hydride in how you treat the terminal ligand?

Therefore wouldn't that suggest that you could treat the cyano ligand reasonably as how you treat the hydride in the bridging case?

That would be fine. Alternatively, by analogy to CO is perfectly fine as well. Consider a complex with 2 bridging cyano ligands. Change these to bridging CO ligands and increase the charge of the overall complex by 2.

The problem is we are assuming pairwise sharing of electrons.  That may not be true.  Think of diborane -- it has two hydride moieties that share their electrons in three-center bonds and thus contribute two electrons simultaneously to both boron centers.  Thereby each boron atom has a full complement if eight valence electrons around it.
The bridging hydrogens in ditatanocene could do the same thing.  Add the electrons around titanium from other sources and you find that the titanium atoms, like the three-center bonded boron atoms in diborane, have fully filled valence shells.


text:
Dextro or laevo-rotatory compound conversion
Will applying an external electric field to a racemic mixture will convert the optically active compound(mixture of both Dextro and laevo-rotatory) into a particular dextro or laevo-rotatory compound?
I thought that if I apply an external field, then all the chiral molecules will arrange in a particular direction i.e their spatial arrangement in the solution aligns in a specific direction, resulting in a particular dextro or laevo-rotatory compound.
Is my thinking right or I am misunderstanding the topic?

You are misunderstanding the topic. Electric field isn't chiral, hence the result of its action isn't either.

@IvanNeretin it can be, for instance circularly polarized light has chirality.

This question is unclear as written because "arrange" is not well-defined enough. What do you mean to say that they arrange? Are you inverting centers? Are you causing reactions?

@Tyberius What "it"? We were talking about an electric field, not about circularly polarized light.

@IvanNeretin maybe I'm misunderstanding, but circularly polarized light is a chiral electromagnetic wave. Thus I would think you could apply a chiral electric field.

@Tyberius Electric field is a vector; how can it be chiral? Well, it kinda can, if you would rotate it with high frequency, at the same time moving pretty fast through the media (that's what light does, after all). But that would be an awfully broad interpretation of the words "apply an external field".

@Zhe They are just two enantiomers of each other. Do even the $(\pm)$ enantiomers react with each other? Also, the OP has said "their spatial arrangement in the solution aligns in a specific direction". What more clarity is required?

@IvanNeretin, I am a high-grade student, so I couldn't understand the point when you said electric field isn't chiral, does it mean electric field around a particular molecule is symmetric? by the way, I mentioned that I applied an external field, which I thought would orient all the molecules in a particular direction such that when I pass  a PPL through the solution, the polarimeter needle aligns in a particular direction unlike when I pass through a racemic mixture...

@IvanNeretin en.m.wikipedia.org/wiki/Circular_polarization. Circularly polarized light is used fairly commonly in chiral spectroscopy, like circular dichroism. Electric field is a vector with a polarization, which can vary with time such that the vector moves in circle (or more generally an ellipse).

@GaurangTandon If there is interconversion, how does it take place? Photochemical or a reversible equilibrium? If the former, it might be possible to generate a large enough gradient. Via the latter, you will have no chance because the system is in equilibrium.

@PrathamArya Yes, that's precisely the case: the situation with molecules in a static electric field is symmetric. But if the field isn't static and moves in a fashion I've described earlier (as it does in the circularly polarized light), or if you add some other influence that would break the symmetry, then yes, chirality might emerge.

@ivanneretin so can I conclude that if I managed to apply a static electric field, all the molecules of racemic mixture are arranged in a fashion like either dextro or laevo-rotatory at that particular moment..


text:
NMR - coupling of chemically equivalent protons
In Klein's Organic Chemistry 3rd Edition page 671, it states

This observation, called the n + 1 rule, only applies when all of the neighboring protons are chemically equivalent to each other.

However, in the following page, it states

Equivalent protons do not split each other

The first statement infers that for the n + 1 rule to be valid, the protons must be chemically equivalent - inferring that equivalent protons can couple, while the second statement directly contradicts this.
Could my misunderstanding be clarified?

The "n+1" rule refers to a situation where you have a proton of type A with $n$ protons of type B next to it. Proton A's signal will be split $n+1$ times by the B's. However, none of the B's will split each other because they are equivalent.

Note:  the Bs won't split each other if they are 'magnetically equivalent', which is a stronger condition than chemical equivalency.

The chemically equivalent proton energy levels  are split by the static and local magnetic fields but nmr transitions out of some of these levels are forbidden by selection rules and those signals that remain are only the allowed transitions that give the appearance  as if splitting has not happened.
The case for AX spectra is shown in the figure where the spin-spin coupling moves degenerate levels apart. The signals that are observed all have the same energy so appear as if nothing has happened to them.
(There are 2 spins labelled as $\alpha$ or $ \beta$, $J$ is the spin -spin coupling constant and $h$ the Planck constant. The energy $A$ is given by $\gamma (1-\sigma)B_0/(4\pi)$ where $\gamma$ is the magnetogyric ratio, $\sigma$ the shielding constant and $B_0$ the static magnetic field).


text:
How to convert from ppm to ug/m3 in case of PMx?
For $\text{PM}_{10}$, $\text{PM}_{2.5}$ sensors, sensitivity, minimum and maximum recordable concentrations are generally expressed in $\pu{ppm}$ (parts per million) while legislation uses $\pu{\mu g / m^3}$ or $\pu{mg / m^3}$.
For gases you usually use the molar mass to convert from one to the other.  Since PM is composed by heterogeneous particles how can I convert $\pu{ppm}$ to $\pu{\mu g / m^3}$? Is there an approximate weight for each type of PPM?


text:
How does salt push out the water from vegetables, meat, etc.?
Where can I find out, how does this process work? Have you noticed, that when you salt mushrooms, eggplant... or add sugar to lemon - some "juice" comes out.
Any explanations? Or maybe name that reaction?

I'm not 100% certain on this, but my guess would be osmosis.

It's more like pull out, than push out

It is called "Osmosis" and the applications of this are many. Since you are a self-identified "noob of chemistry" I'll not go too far into the specifics of why this happens but try to explain on a laymans basis.
As you probably know, water is a solvent. It can dissolve (some) solids so that no solid remains, only liquid. This usually has some effect on the solute (water in this case). Lets say you have two glasses of water - one of which has some salt dissolved in it. Nature would very much prefer if they were at the same concentration of salt. Nature wishes for 2 glasses of water with $\frac{some}{2}$ salt in each.
So, when there is an apple, with water inside and some salt outside, nature would like there to be about the same concentration of salt inside and outside. So she (nature) will provide a driving force for taking water out of the apple and into the salt. The skin of the apple is a semi-permeable membrane: only water can pass. So nature will drive the water through the skin to the salt. And she will keep at it until the apple is so empty of water that it is so hard to find water in the apple that the forces are in balance.
The same thing applies to sugar. Sugar is also one of these compounds that can dissolve in water and will affect the water in the same way.
For a more detailed explanation I'd google "Osmosis". This interesting phenomenon is responsible for many things, wonderful and useful. You can actually pump water with it. Nature will "pull" water against the force of gravity in order to balance out the concentrations. (This is basically how a salt power plant works)


text:
Substitution reaction rate for tertiary halide substrate with adjacent carbonyl

In the book "Organic Chemistry" by Clayden et al. p 343 it is stated that this substrate reacts very slowly through both the SN1 and SN2 pathways. It is shown side by side with neopentyl chloride as an example of notoriously unreactive compounds.
I get that being tertiary might slow down SN2 to counteract the effect of the adjacent carbonyl. But why would SN1 not still be good?
My question is: why does this substrate react slowly through both SN1 and SN2?

You should consider what the transition state of the SN1 reaction looks like.


text:
Percentage purity of Ba(MnO4)2 sample that completely reacts with given H2O2?

$\pu{40 g}$ $\ce{Ba(MnO4)2}$ (mol. wt. = 375) sample containing some inert impurities in acidic medium is completely reacted with $\pu{125 mL}$ of $\pu{3 M}$ $\ce{H2O2}$. What is the percentage purity of the sample?

I actually found this question in a book, I found its solution but I cannot properly understand it. This is the first equation given:
Since milliequivalents of $\ce{Ba(MnO4)2}$ = milliequivalents of $\ce{H2O2}$
$$(w/375)\times10\times1000 = 3\times125\times2 $$
I understand the RHS as $3\times125$ gives number of millimoles which multiplied by n-factor gives milliequivalents. But where does the LHS come from? And what is $w$? In the next line it is given
$$ \text{percentage purity} = (w/40)\times100 $$
where the value of $w$ is taken from the first equation. Would someone please explain this to me?

The n-factor of $\ce{Ba(MnO4)2}$ is $\mathrm{10}$ in the above reaction. And $w$ is the mass of pure $\ce{Ba(MnO4)2}$ in the impure sample which is what we are required to find inorder to get the % purity.
Thus we have the equation,
$$\pu{milli-equivalents of \ce{Ba(MnO4)2} = moles*n-factor*1000}$$
which in your case is
$$\pu{Meq.of \ce{Ba(MnO4)2} = \frac{w}{375}*10*1000}$$

Thank you! I know this is a silly doubt but n-factor is 10 when MnO4- gets oxidized right? How can it get oxidized if Ba(MnO4)2 is reacting with H2O2? I had actually assumed that n-factor of compound is 2 since valency of Ba is 2, would you please tell me what was wrong with my assumption?

@Hema No, MnO4- in acidic medium is always reduced to Mn2+(n-factor =5). Since one mole of the compound contains 2 moles of MnO4-, the n factor is 2*5=10.

The question does not demand to solve it using "equivalents," I'm going to attempt to solve the problem in universal way, using moles. As OP correctly suggested, this reaction of $\ce{Ba(MnO4)2}$ and $\ce{H2O2}$ is a redox reaction. Since the reaction has taken place in acidic medium and has completely reacted (assuming observation is made by appearance so assuming it was judged by clear solution), the two half reactions should be:
$$
\begin{align}
\ce{MnO4- + 8H+ + 5e- &-> Mn^2+ + 4H2O} & E^\circ &= \pu{1.507 V} \\
\ce{H2O2  &-> O2(g) + 2H+ + 2e-} & E^\circ &= \pu{-0.695 V}
\end{align}
$$
Thus, total redox reaction can be written as:
$$\ce{2MnO4- + 6H+ + 5H2O2 -> 2Mn^2+ + 5O2(g) + 8H2O} \quad E^\circ_\mathrm{rxn} = \pu{0.812 V}$$
The positive $E^\circ_\mathrm{rxn}$ means the reaction is spontaneous. And it also shows that you need $\pu{5 mol}$ of $\ce{H2O2}$ to completely react with $\pu{2 mol}$ of $\ce{MnO4-}$. Since $\pu{1 mol}$ of $\ce{Ba(MnO4)2}$ contains $\pu{2 mol}$ of $\ce{MnO4-}$, it is correct to say that $\pu{5 mol}$ of $\ce{H2O2}$ would completely react with $\pu{1 mol}$ of $\ce{Ba(MnO4)2}$.
Suppose that $\pu{40 g}$ of impure sample contains $x~\pu{g}$ of $\ce{Ba(MnO4)2}$. Then, amount of of $\ce{Ba(MnO4)2}$ in the sample is
$$\frac{x~\pu{g}}{\pu{375 g\cdot mol-1}} = \frac{x}{\pu{375 mol}}.$$
To completely react with that amount, you need
$$\left(5\cdot\frac{x}{375}\pu{mol}\right)~\text{of}~\ce{H2O2}.$$
Thus,
$$5\times\frac{x}{375}\pu{mol} = \pu{3\frac{mol}{L}}\times\pu{125 mL}\times \pu{10^{-3} \frac{L}{mL}} = 3\times \pu{0.125 mol} \label{eq:1}\tag{1}$$
$$\therefore x = \pu{\frac{ 3\times 0.125 \times 375}{5} g} = \pu{28.1 g}$$
Thus,
$$\text{percentage of $\ce{Ba(MnO4)2}$ in the $\pu{40 g}$ of sample} = \frac{28.1}{40}\times 100 = 70.2$$
Note that, the equation $\eqref{eq:1}$ is exactly the same as yours (with minieq.)

I'll upvote your answer since you also included the chemical equation. // I'll also point out that milliequivalents is a depreciated concept.


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The structure of 1-(butan-2-yl)-4-tert-butylbenzene is displayed below. What is -yl on this compound and when do I use it?
I can't figure out how to use -yl on organic nomenclature and how do I use it in nomenclature. Please explain everything about the use of -yl in nomenclature.

Think of the ending -yl as defining an adjective. The term (butan-2-yl) is an adjective modifying the noun tert-butylbenzene. No different than 3-methylhexane. [3-adjective-noun].

The "yl" refers to removing a hydrogen atom from a molecule making a univalent group.
Butane is
$\ce{CH3-CH2-CH2-CH3}$
To make butan-2-yl take a hydrogen atom off carbon 2:
$\ce{CH3-CH-CH2-CH3}$
Then butan-2-yl X would be
$\ce{CH3-CH(X)-CH2-CH3}$.


text:
Major product in nitration

My approach was that when the carbon attached was a secondary one, it would have a greater inductive effect compared to the primary one.
Why is the answer 3? Is it due to steric reasons or am I missing some other factor?

i think hyperconjugative effect is seen in such cases not inductive

You can also discuss it here : https://chat.stackexchange.com/rooms/74807/jee-high-school-chemistry-problems

Read this first: Why methyl group is 2,4-directing?
From that answer, we understand that methyl groups (or alkyl groups in general) are o/p directing because of hyperconjugation, and not inductive effects! In fact, as you may also know, inductive effects are generally overshadowed by hyperconjugative effects.
The aromatic ring on the left is more activated because it has two alpha hydrogens - implying more number of hyperconjugative structures - as opposed to only one for the right aromatic ring (a small difference, but good enough for elementary level questions). Thus nitration takes place at the para position of the left ring.


text:
Lone pair on carbon in CO2?
CO2 has sp hybridization and no lone pairs on carbon.

For some reason calculation with VSEPR gave sp2 hybridization and lone pair on carbon.

Actually, from where did you read that "It has 1 lone pair in reality"? The picture you've attached shows no lone pair either.

Why do you think the hybridization is $sp^{2}$?

The carbon atom in $CO_2$ has zero lone pairs of nonbonding electrons.
[By the way, the carbon has $sp$ hybridization, as it requires the bonding orbitals to have the same lobe sign on each side to make two identical $\sigma$ bonds. The $2p_z$ orbital alone is not sufficient to make two identical bonds, because the molecule would then lose the reflection symmetry it ought to have.]
Let's construct the Lewis structure ourselves:

Carbon has four valence electrons. We accept that.
Each oxygen has six valence electrons. We also accept that.

Therefore, the total number of valence electrons is $4+6+6=16$. Carbon is the larger atom, so this is a more usual structure with the largest atom in the center. The skeletal structure so far is:
$\text{O}-\text{C}-\text{O}$
We place three lone pairs of electrons on each oxygen next, to account for $2+2$ bonding and $2\times3+2\times3=12$ (currently) nonbonding electrons, for a total of $16$ as we needed.
Lastly, we use one lone pair from each oxygen to make a $\pi$ bond. This only redistributes electrons, without adding any extra or losing any. That results in:
$:\ddot{\text{O}}=\text{C}=\ddot{\text{O}}:$
And are the formal charges (FCs) minimized?

Carbon owns 4 valence electrons and came in with 4 valence electrons, so $FC = 0$.
Oxygen owns 6 valence electrons and came in with 6 valence electrons, so $FC = 0$ as well.

This must be the actual Lewis structure, and it also is in the correct image you have placed in your question.

SORRY I PUT 6 VALENCE ELECTRONS IN THE VSEPR FORMULA WHEN IT'S 4. AND IT'S HYBRIDISATATION IS sp. And it has no lone pair as L=H-X-D =2-0-2=0. Sorry for the mistake.  @mithoron

If you can delete the original question it would be better as the question means no sense now @mithoron

@user187604 Well, now it's a bit trickier to delete it. I can flag it for mods to consider it, but it would also mean deleting the answer.

@Mithoron does the answerer lose the reputation for deleting the question? If that does happen don't delete it. If not please do.

@Mithoron if I even edit the question to give correct values that also now means no sense!

@user187604 I might just have asked, "How do you determine the Lewis structure of CO2 from scratch? I tried this and got one lone pair on carbon for some reason." I can edit my answer to reflect that. No one is trying to embarrass anyone here.

@timaeus222 I never complained ever about your answer! Did I? Actually it's very disrespectful to me to post a wrong question


text:
All roads lead to S4N4
Why does thionyl chloride $\ce{(SOCl2)}$ when it reacts with liquid ammonia form a trace of $\ce{S4N4}$. What is the mechanism of the reaction forming $\ce{S4N4}$.
See discussion of sulfur chemistry for details.

Are you sure this is thionyl chloride ($\ce{SOCl2}$) and $\ce{NH3}$ reaction? Or is it sulfur(I) chloride  ($\ce{S2Cl2}$) and $\ce{NH3}$ (balanced equation: $\ce{6S2Cl2 + 16 NH3 -> S4N4 + 8 S + 12 NH4Cl}$)?

The problem is that while a the reaction of SCl2 with ammonia gives a nice and high yield of S4N4, the reaction of thionyl chloride with ammonia is reputed to always give traces of S4N4. The problem is what is the mechanism of the reaction.

What is the intended product (or major product) of the reaction?

The intended main product is H-N=S=O which can deprotonate to form the N=S=O anion.

Yes, you are indeed right in the comment section that the reaction between thionyl chloride and ammonia will give trace amount of tetrasulfur tetranitride(1) and the intended product is thionyl imide ($\ce{HNSO}$)(2).

From the 'thionyl chloride' entry from the same site you used
in the question:

Thionyl chloride reacts with ammonia giving rise to various complex
  products, the nature of which varies with the conditions.

As said, the main product from the reaction in gas phase is thionyl imide and trace amount of thiazyl chloride($\ce{NSCl}$) and sulfur dioxide. The reaction are as follows:

$$\ce{SOCl2(g) + 3NH3(g) -> HNSO(g) + 2NH4Cl(s)}$$
$$\ce{HNSO + H2O -> SO2 + NH3}$$
Ref. : Inorg. Chem., 1977, 16 (1), pp 216–217
DOI: 10.1021/ic50167a050

If we work out oxidation states, we find $\ce{S^{IV}}$ in thionyl chloride, and of course$\ce{N^{-III}}$ in ammonia. Thus formation of tetrasulfur tetranitride would involve the combination reaction
$\ce{4 S^{IV} +4 N^{-III} -> S4N4}$
But we see that to balance charges four electrons must be added, so sulfur tetranitride is actually a reduction:
$\ce{4 S^{IV} +4 N^{-III} + 4e^- -> S4N4}$
Hence a redox reaction must be occurring.  Assuming the quite plausible notion that the oxidation half-reaction is additional sulfur being oxidized to $\ce{S^{VI}}$ ($\ce{S^{IV} -> S^{VI} +2e^-}$)*, we would form two moles of $\ce{S^{VI}}$ per mole of sulfur tetranitride.  Then we could have something like this:
$\ce{20 NH3 + 6 SOCl2 -> S4N4 + 14 NH4^+ + 2 SO3NH2^- + 12 Cl^-}$
*This reference reports the formation of sulfate as well as sulfamate from thionyl chloride and ammonia, both of these species having $\ce{S^{VI}}$.


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Why is ΔH (enthalpy)=Σreactants-Σproducts and not Σproducts-Σreactants?
I was told that ΔH=Σreactants-Σproducts in regards to finding the enthalpy, given average bond enthalpies.
Why is the case? It seems counterintuitive, because of the delta, which would imply ΔH=Σproducts-Σreactants.
Is this the case, because of the fact that for breaking the bonds in the products energy is required, while for forming the bonds in the products, releases energy and that enthalpy is equal to the heat released?


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Conversion of an alcohol to a methyl group
I'm only interested in the first conversion: D to E.
I've been thinking about this for a little bit. My first thought was to use $\ce{TsCl}$ to make a good leaving group then a $\ce{Me}$ nucleophile. This didn't seem like it would work since $\ce{Li-Me/MeMgBr}$ would just displace the leaving group. My second thought was to convert the alcohol to a halide, make it a Grignard reagent, then react it with $\ce{MeBr}$. But that also wouldn't work. Any suggestions?

Consider going via ketone.

You might also consider an organocuprate as a nucleophile. Ordinarily, I wouldn't suggest a secondary electrophile for SN2, but this is also benzylic so...

I considered ketone, that would add the methyl easily using Me-Li then an OH would be left. Add TsCl then a weak reducing agent, NaB(CN)H3, but would that work to remove the OH?

Oxidise to ketone, Wittig, then hydrogenate

@MalcolmHall It should, but there is a more reliable way. The resulting alcohol should dehydrate really easy. In fact, it might dehydrate during acidic treatment of Grignard reagent adduct.

Ionic hydrogenation with Et3SiH/TFA will reduce the benzylic OH very easily

As @orthocresol mentions in the comments, the better way to undertake this synthesis would be to first oxide the alcohol to a ketone, perform the Wittig reaction and than hydrogenate the product.

Another possible reaction sequence might be first convert the alcohol into a ketone, use a Grignard reagent and than reduce the alcohol.