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Why is the rate inversely proportional to the square root of temperature in Grahams Law of Diffusion?
In my book it is given that:

The general form of the Grahams Law of Diffusion can be stated as follows when one or all of the parameters are varied:
  $$\text{rate} \propto \frac{PA}{\sqrt{TM}},$$
  where $P$ - pressure, $A$ - area of the hole, $T$ - temperature, $M$ - Molecular weight.

How can rate be inversely proportional to the square root of temperature? I feel that on increasing temperature it must increase, but how can the rate decrease?
Is the formula in my book correct or incorrect?

It helps to rewrite
$$\text{rate} \propto \frac{PA}{\sqrt{TM}}$$
by assuming the ideal gas law holds, as follows:
$$\text{rate} \propto \frac{RA}{V_m}\sqrt{\frac{T}{M}}$$
where $V_m$ is the molar volume or inverse of molar particle density.
Written this way it is clear that if the particles occupy the same volume, increasing their temperature increases the rate at which they effuse through the apperture.


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Is there any mathematical or logical proof that carbon with 4 different groups would be chiral?
In chemistry; chirality is generally defined in 2 ways.

Lord Kelvin's definition: "I call any geometrical figure, or group of points, chiral, and say it has chirality, if its image in a plane mirror, ideally realized, cannot be brought to coincide with itself" (source: 1. chirality.org , 2. Biochemistry book by Lubert Stryer.)

"If a tetrahedral (sp3) carbon atom contains a different chemical group on each of its 4 hands, the central carbon atom is called chiral. Though there are some exceptions where we do not found any chiral carbon still some compounds can be chiral. Chiral molecules cannot be superposed with their 3-dimensional mirror image" (common classroom-definition for chirality).

But it seems the first definition is the basic definition; and the second definition seems to be a corollary or application of the first definition.
Now I want to know; is it possible to 'prove' or 'derive' the second definition (better to say second statement or second description) from the first (Lord Kelvin's) definition mathematically or geometrically or logically? and if possible then how to prove the second definition from the first definition?
This question is related but different from this question

Build two enantiomeric models out of matchsticks and chewing gum, and try to superimpose them. If that isn't enough of a proof to you, then what is?

The two definitions are not trying to define the same thing, to begin with. (1) refers to the chirality of a molecule, and (2) refers to chiral carbons, and is not directly related to the chirality of the entire molecule (that's why there is a disclaimer saying that there are some exceptions). So you need to be clear about what you're asking: you cannot prove the definition of A using the definition of B.

@IvanNeretin i think that is more of trial-and-testing (than a proof) that if there are 4 different groups the carbon would be chiral.

@orthocresol please consider chiral carbon.

If you want mathematic proof, take a seat and ingest two semesters of linear algebra. The keyword is point groups. Unfortunately most chemistry faculties dont care. ;)

@Karl lovely. it answers my question

Also in biology university level classes the math was taught to me in a repulsive and non-visual way.

I dont have the age and score-credential to get a scope to "take a seat" to learn linear algebra in any academia. Maybe online will help.

If you find a good online course that teaches group theory, please post the link. I would love to get a refresher.

@AlwaysConfused — Geometrically speaking, the two rules are equivalent. It is demonstrable that if a point (i.e. the (a)chiral center) two identical groups, there always is a type of symetry (point, plane or rotation symetry). As a consequence, any chiral tetrahedral point MUST have 4 different groups.

Exhaustive trial and testing is a totally valid mathematical method to prove sth. The possibilities are very limited, after all.

Related : How can I prove that the angles of the tetrahedral structure is 109.5∘ with calculus. I could do it with geometry.

@SteffX No, the two rules re notequivalent. There are far more types of chirality than those generated by tetrahedral carbons.

@matt_black I am totally agree. The carbon tetrahedron's is only a small subtype of broader definition by Lord Kelvin

Yes, with a couple of simplifying conditions, you can easily show that a tetrahedral carbon with 4 different chemical groups (let's call them 1, 2, 3, 4) is chiral. Let's call the atoms directly bound to the carbon L1, L2, L3 and L4.
So the first condition I will impose is that none of the groups have another tetrahedral carbon with 4 different chemical groups attached.
Chemistry
In a tetrahedral carbon, the groups can't swap places. If that happens, we will call it a chemical reaction, and we are not allowing any chemical reactions.
Geometry
Tetrahdral should be taken loosely: For the plane defined by L1, L2 and the carbon atom, L3 is on one side of the plane and L4 on the other. The exact angles don't matter.
Showing that the carbon is chiral
If we put a mirror in the plane defined by L1, L2 and the carbon atom, the mirror operation swaps L3 and L4 (L3' on the side of L4, and L4' on the side of L3 - not sure we need a formal proof for that). I cannot superimpose that mirror image with the original (L1, L2 and the central carbon are already in the same position as the mirror image. I would have to rotate the molecule to get L3 and L4 superposed with L3' and L4', but then L1 would move away from L1', and L2 away from L2' - not sure we need a formal proof for that). Thus, the carbon atom changes configuration when taking the mirror image.
3D illustration
see https://proteopedia.org/wiki/index.php/R/S_nomenclature (you have to turn the molecules a bit to have L1 and L2 point in the same direction as L1' and L2')
Is this trivial?
No, not at all. For trigonal bipyramids there is sometimes pseudorotation, and for ammonia derivatives the trigonal pyramid can "umbrella flip", so these are typically not chiral. If there are four ligands on a central atom and the geometry is square planar, there is no chirality. So a couple of things have to come together to get a chiral center.

Another way to show this is to first define improper torsion angles, and then show that they switch signs upon reflection in a mirror.

I don't think there is any mathematical way to prove it.It's only experimental.
See Chirality is defined for the compounds whose mirror image doesn't superimpose on each other.

-source : LG Wade Orgainc chemistry.
if you want more explanation then you can take help of LG Wade Or Solomons Organic Chemistry(both are available free on net)

Yes empirical observations demonstrate it, but the proof is nothing to do with that: it is pure geometry. A rigid tetrahedron with 4 differently labelled vertices cannot be the same as its mirror image.

source: Stack Exchange

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Why is Thorium (At. no. 90) an f-block element at all?
Being an exception of the Aufbau principle, Thorium has an electronic configuration of $\ce{[Rn]}\mathrm{7s^25f^06d^2}$ instead of the expected $\ce{[Rn]}\mathrm{7s^2 5f^2 6d^0}$.
Two other elements, Lanthanum and Actinium, also showed such an anomaly as $\mathrm{6s^2 4f^0 5d^1}$ and $\mathrm{7s^2 5f^0 6d^1}$, respectively. But we placed them both in the d-block of the periodic table.
Since, even in Thorium, the last electrons occupied d-orbitals only (disobeying Aufbau), then why is it placed in the f-block of the periodic table. I mean, it should be placed in the d-block, shouldn't it?
Edit: Note that here, I don't intend to ask or know how or why the 2 electrons disobeyed the Aufbau, and occupied the higher energy 5d instead of lower energy 4f.
Rather, I want to know only that why it is kept in the f-block when the last orbitals which the electrons occupied were of d-subshell.

I don't think it violates the Aufbau principle, but instead it fails the generalised prediction patterns that have been derived from observation, like $n + \ell$ 'rule'. About the placement in d-/f-block, I don't think there is consensus on that either. There recently have been different news about how the elements should be displayed, even turning the PT upside down https://doi.org/10.1038/s41557-019-0253-6

“…even turning the PT upside down doi.org/10.1038/s41557-019-0253-6”: this would be a decent article for a design journal, but how on Earth did this pave its way to Nature (judging from the name of the first author I think I know the answer, so this is more like a rhetorical question)?

Related (or probably duplicates): Anomalous Electronic Configuration of Thorium; Outermost electronic configuration of f block elements.

As one gets more and more electrons, their interactions really start messing with simplistic expectations (e.g. Aufbau) based on hydrogen-like energy levels. But, really, to what purpose does one call something an 'f-block' vs a 'd-block' element - you look at the actual filled levels and nearby levels and try to figure out the chemistry from there. For crystal solids, Thorium is just at the edge of some really weird crystal structures popping up because of all those electrons (like Pu with the first 3 allotropes being specific to it).

@andselisk I edited my question. The first link you related my question to, is demanding the reason why/how it happened, which I'm not. So not a duplicate or related. The second link, which was also my doubt but I later accepted it as one of the abundant errors of chemistry and thus decided not to include in my doubt, is also neither a duplicate of nor related to my question.

The bulk metal actually does involve the $f$ orbitals, distinguishing it from what would be its congeners in Group 4 of the $d$ block.  From Wikipedia:

Despite the anomalous electron configuration for gaseous thorium atoms, metallic thorium shows significant 5f involvement. A hypothetical metallic state of thorium that had the [Rn]6d27s2 configuration with the 5f orbitals above the Fermi level should be hexagonal close packed like the group 4 elements titanium, zirconium, and hafnium, and not face-centred cubic as it actually is. The actual crystal structure can only be explained when the 5f states are invoked, proving that thorium is metallurgically a true actinide.[1]

Cited reference
1.
Johansson, B.; Abuja, R.; Eriksson, O.; et al. (1995). "Anomalous fcc crystal structure of thorium metal". Physical Review Letters 75 (2): 280–283. https://doi.org/10.1103/PhysRevLett.75.280, erratum at https://doi.org/10.1103/PhysRevLett.75.3968

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DNA pairs (adenine-thymine, guanine-cytosine)
Well I was wondering why adenine pairs with thymine and cytosine pairs with guanine.
From those pictures:

Thymine has the lowest acidity and adenine has the biggest acidity.
So it is logical that the intermolecular bond will be strong.
Adenine and Cytosine have pretty much the same acidity so why is it energetically favorable for guanine to make an intermolecular bond with cytosine?
Is it due to the three combinations this (adenine-thymine, cytosine, guanine) has the lowest energy?

Simplest explanation in that only in G-C there can be 3 hydrogen bonds - 3 beats 2. BTW your reasoning is based on false assumptions as there's no such correlation.

Thymine is the least acidic since it contains 2 O bases.

Adenine doesnt contain any 0 bases so it is the most acidic.

I looked, but I disagree. If you want to have someone look at your comment, use their username prefixed by @, as in @Software_Player.

@Karsten Theis Why do you disagree? O base is less acidic than N base...

That is a new question. Maybe use examples with less functional groups, like acetone vs. 2-amino propane. Acetone has no acid/base chemistry in the pH range of 0-14, and 2-amino propane is a base. Neither is an acid.

@Karsten Theis Amino-propane  is a base indeed.

@KarstenTheis and acetone is a base as well due to the molecule being polar.

@Mithoron there is also repuslion from a partially charged negatively oxygen to a partially charged negatively mitrogen .

In Principles of Nucleic Acid Structure, W. Saenger argues that hydrogen-bonded bases contain at least two hydrogen bonds (forming a "cyclic" pattern). Often, there is a tautomeric form possible that also makes two hydrogen bonds, with two covalent bonds turning into hydrogen bonds, two hydrogen bonds turning into covalent bonds, and double bonds moving around so that each atom still has an octet. In the OPs first example (A:T pair), for example, the ketone would turn into an enol, and the protonation of the amino and imino groups would change.
Saenger goes on to say that 28 pairs are possible.

Many of these are observed (the ones that are labeled are so common that they were given names), in structures of DNA, RNA, and in crystal structures of dinucleotides.

I was wondering why adenine pairs with thymine and cytosine pairs with guanine.

They pair because they make strong hydrogen bonds with a geometry of the base pairs that allows base stacking and is compatible with base pairing of the other nucleotides in a DNA strand. As the discussion above shows, many other base pairs are theoretically possible, and some of them are actually observed in nature.

Thymine has the lowest acidity and adenine has the biggest acidity. So it is logical that the intermolecular bond will be strong.

I'm not sure about the premise, and the conclusion is incorrect, as other base pairs also form. As long as a group has the right protonation state, the pKa does not directly determine hydrogen bond strength.

Hydrogen bond strength  (any intermolecular bond strength)  is determined by the polarity of the molecules.

@SoftwarePlayer OK, but is acidity directly correlated with polarity?

In Hydrogen bonding , Hydrogen is charged partially positive , so this makes the molecule acidic.By definition an acid is a molecule which can donate a proton so yes.


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Acetone basicity
Why doesn't acetone have high basicity?

The C−O bond is polar, O becomes partially negatively charged and it should attract protons, becoming a strong base.

Related: Are hydrogen atoms in acetone molecule acidic?; Why is acetone less acidic than cyclohexanol?; Why is the pKa of a ketone higher than of water?

Andelisk im not asking about the acidity of hydrogens in acetone , just why the oxygen cant become a proton acceptor.

Inspired by (I think) comments to this question: https://chemistry.stackexchange.com/questions/119921/dna-pairs-adenine-thymine-guanine-cytosine/119924#119924

Do you have a good explanation why acetone isnt a base?

I don't know the reason, but I can describe the trend (source: http://cactus.dixie.edu/smblack/chem2310/summary_pages/pka_chart.pdf):

Protonated ketones have a pKa of about -7. Protonated ethers have a pKa of about -4 and protonated primary alcohols of about -2. Comparison to hydronium in water is difficult, but in DMSO, hydronium is more acidic by about 2 pKa units (source: comment by DavePhD below https://chemistry.stackexchange.com/a/27278/72973).
So it seems water is a little bit basic. If you replace one of the bonds to hydrogen with a bond to carbon, it becomes less basic (alcohol). If you replace the other bond to hydrogen with a bond to carbon as well, it becomes even less basic (ether and ketone).
A hand-waiving explanation would be to say the electronegativity difference between hydrogen and oxygen is higher than that between carbon and oxygen. I am not sure what to say about the difference between ethers and ketones. I think of ketones as the better hydrogen bond acceptors, and as more soluble in water (all other things being equal), so I don't have a good rationale why it would be easier for an ether to accept a proton.

And what do you mean by" protonated ketones"?

I would consider the geometry of a ketone vs an ether to perhaps explain the answer. I am pretty sure the real reason lies somewhere in the MO of the two groups (i.e. the HOMO of a ketone sits lower than an ether). However, I am not sure if it would be too rigorous to dive into that though.

@SoftwarePlayer A protonated ketone is the product of a ketone acting as a base, when a proton (or H+) attaches at the oxygen.

@SoftwarePlayer "extra oxygen atom"? Alcohols, ethers and ketones all have oxygen atoms - there is no extra oxygen.

yes you are correct i thought you meant something else


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Chaos in chemical reactions
I recently read about Chaos Theory and was wondering if a chemical reaction results in or shows characteristics of chaos (I found a few examples of such reactions here)

Chaos theory concerns deterministic systems whose behavior can in principle be predicted. Chaotic systems are predictable for a while and then 'appear' to become random.

If the reaction becomes unpredictable towards the time of attainment of equilibrium we can't possibly determine how the reaction will proceed during the time frame so will the reaction attain a state of equilibrium and will it be at the time it should've without chaos being in the picture?
If the equilibrium shifts can we quantify by how much and how much more or less time will it take to attain it in general or we need to take specific reactions and analyze them?
Also after attaining equilibrium can the system again show Chaotic behaviour and if it does what can se say about the state of equilibrium?

Indeed, there is quite a bit of chaos in chemical reactions. I wonder, though, what's the point of your question.

@IvanNeretin I was curious about chaos in Chemical Reactions and the concept of equilibrium which pretty much says that we can predict the future and determine the concentration of different reagents involved and hence I'm asking this

Well, can you tell someone what they'll do 20 years in the future? Can you tell if someone will die at all? Reaction like life is unpredictable, but inevitably ends.

@Mithoron it does and will as a system stays in chaos only for specific time intervals but I'm asking is will the reaction attain equilibrium at the same time as it was supposed to and will it stay in the dynamic equilibrium or move away from it

I still don't get your point... When you die, you stay dead, people might somewhat postpone death, though.

@Mithoron your analogy isn't helping. I am simply asking whether the reaction will attain equilibrium at the time when it would've if there was no chaos or not? It's a binary question also if it doesn't then why and how can we find out the deviation at the point of time equilibrium and also when it does will it ever re-enter a chaotic state and when it does will the equilibrium shift?
Sorry if my previous (deleted) comment was not friendly it was a mistake keeping the caps lock on

My comment did answer that: chaotic regime does somewhat alter time of completion of reaction and when it's complete, it's complete.

An Introduction to Nonlinear Chemical Dynamics: Oscillations, Waves, Patterns, and Chaos.

@Mithoron and can we quantity by how much it deviates?  and also provide the sources for your claims

What Mithoron says is based on chemical common sense. The very basics of thermodynamics and kinetics. If you want to quantify anything, you should be a lot more specific about the system.

@Karl I want to know the general way in which these things are quantified I don't have any specific example or reaction in mind

I think the point of chaotic (nonlinear) behavior is precisely that prediction becomes difficult and not amenable to standard linear methods. However you can probably bracket the expected response such as the time to equilibrium (provided the system is not continuously driven), as well as say something about the duration of intervals during which the system may behave "chaotically". You may want to consult a physics or math source for more details, or read more about the individual reactions you cite.

@BuckThorn I've updated the question making it more clear to understand hopefully

Equillibrium does not shift, unless you actively change the composition or the temperature or pressure. Never. You should revise your highschool natural science.

@Karl I got your point I thought maybe chaos can shift it as Le Chatelier's Principle did not account for it

A homogeneous reaction mixture can hardly show macroscopical chaotic behaviour. Well known cyclic counterexamples exist, and practically any reaction which proceeds faster than diffusion or mechanical mixing can homogenise it again (e.g. any reaction that generates a lot of heat!) is liable to show some intermediate chaotic concentration gradients. Its usually hard to notice, and most reactions are perfectly reproducible, if you set them up in the same way (educt impurities!).
If your reaction vessel contains inflows of different chemical composition, then you easily get chaotic behaviour. Think about a tube reactor. I believe chemical engineers are spending a lot of their time trying to control and avoid chaos.

Thanks for the info but this doesn't exactly answer my question

@StackUpPhysics Then I also do not understand your question.

you have given a qualitative analysis however you didn't answer the things like how much more time will it take to attain equilibrium and will the system oppose this equilibrium if chaos kicks in again or not

Just FYI I am not the downvoter

no I'm asking that after equilibrium can it again become chaotic

@StackUpPhysics Are you seriously asking wether a system can stay chaotic when it has reached thermodynamic equillibrium? And of course a chaotic system will take a different amount of time to reach equillibrium than a reaction that just progresses straigthforward. It is a different system, after all!

One hint, perhaps that is behind your misunderstanding: Chaos is by no means a state of maximal entropy. Chaotic systems usually show selfsimilarity! Thermodynamic equillibrium IS the point of maximal entropy, and there is no way out of it.

yes I'm asking the first question you mention I think it can but I'm not sure as this isn't my area of expertise

Why do you think it is called equillibrium? When have you ever seen that something that was in motion and has come to rest suddenly starts rolling uphill again?

it is in a dynamic equilibrium hence I thought maybe chaos could disturb the equilibrium as it's impossible to predict how the processes proceed during that time frame

@StackUpPhysics In a dynamic equillibrium, any local excess concentration that occurs must favour the reaction that makes it vanish again. Le Chateliers principle. Otherwise you had a perpetuum mobile. Chaos depends on self–amplification. Heat production speeds up the further reaction, self–catalysis only occurs when the catalyst concentration is still very low, etc.

What I think: A macroscopic chaotic system is a system whose "components" influence each other, and the macroscopic outcome depends on the status of the components. This happens for the two-arms pendulum, for air particles in cigarette smoke, and for particles in wind. Each microscopic variation can be propagated between particles, and the final status of the system can vary a lot, as varies the status of the particles (eg: temperature, position).
Typical chemical reactions, at the typical macroscopic level have different features: they show "average" properties of the microscopic systems of which they are made (that's the base of statistical thermodynamics, and of Boltzmann's law).
Anyway, if we look closer, we might find chaos, after all: fluctuations of thermodynamic properties arise from the microscopic cause of macroscopic properties (for example, if the temperature of a liquid was mesured with extreme precision every picosecond, it would show variations arising from the way the particles hit the sensor). The variations arise from complex interactions between the microscopic components, and they are non-predictable. That is chaos, albeit not as "macroscopic" as the behavior of cigarette smoke or of the double pendulum.

Chaos means that you have variations at least on a mesoscopic lenghtscale, and it absolutely needs an energy source, unless you have a completely lossless system. Its not to be confused with high entropy. The surface shape of a cloud is highly structured, selfsimilar and chaotic, fog (or the inside of the cloud) isn't. https://en.wikipedia.org/wiki/Belousov%E2%80%93Zhabotinsky_reaction#/media/File:The_Belousov-Zhabotinsky_Reaction.gif


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Reaction between chlorine gas and oxygen at 600 degrees Centigrade
Chlorine gas will not normally react with oxygen gas directly at STP. However, would the two react at standard pressure and 600 degrees Centigrade?

No.$\mathstrut$

Why always 600 degrees Celsius? There are several questions here that involve that specific temperature.

AP chem, apparently it's a common temp in our teacher's lectures and on the class practice sheets.


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Chloride Process TiO2 recovery
In the chloride process (the process used to create titanium oxide from titanium ore), how is TiO2 collected? What prevents the TiO2 from building up on the walls of the reactor chamber?
It's my assumption that the solid titania produced in a TiCl4+O2 reaction (assuming that the TiCl4 and O2 are all gases at 200 C) would plate out on the walls of the reactor chamber (since it is the one solid in the reaction). When manufacturing titania, how do manufacturers prevent the titania from plating out on the walls? Also, how would they collect it?

Can yoe be a bit more specific about where you see the problem, because at the moment this looks like a homework question.

@Karl Sure thing, it's my assumption that the solid titania produced in a TiCl4+O2 reaction (assuming that the TiCl4 and O2 are all gases at 200 C) would plate out on the walls of the reactor chamber (since it is the one solid in the reaction). When manufacturing titania, how do manufacturers prevent the titania from plating out on the walls? Also, how would they collect it?

Its a fine powder with very high melting point. I dont see it sticking to the wall, and would collect it with a cyclone, like those filterless vacuum cleaners, the brand name of which I will not mention. ;)


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How to convert Mn2O3 to Mn3O4?
I was preparing $\ce{Mn3O4}$ but it was when characterized by XRD, a $\ce{Mn2O3}$ phase existed. Now how can I convert $\ce{Mn2O3}$ to $\ce{Mn3O4}$?

You may need a good lab oven.
Manganese(II,III) oxide on Wikipedia

Preparation
$\ce{Mn3O4}$ formed when any manganese oxide is heated in air above 1000 °C. Considerable research has centred on producing nanocrystalline $\ce{Mn3O4}$ and various syntheses that involve oxidation of $\ce{Mn^{II}}$ or reduction of $\ce{Mn^{VI}}.$

As @Poutnik has answered, heating manganese oxides at elevated temperature results in forming of $\ce{Mn3O4}$, research paper suggest that the reaction gets affected not only temperature but the partial pressure of oxygen and further heating at 1400 °C results in formation of $\ce{MnO}$. Here is an excerpt from a research paper [1] (emphasis mine):

TG and DTA curves of $\ce{γ-MnO2}$ have been obtained in nitrogen, air and oxygen. The reactions $\ce{MnO2 → Mn2O3}$ and $\ce{Mn2O3→ Mn3O4}$ were observed in the ranges 450° to 600° and 750° to 1100° respectively, and the decomposition temperatures are affected by the partial pressure of oxygen. The endotherm at ~1200° is unaffected by the atmosphere and is not accompanied by weight loss. It is, therefore, due to a polymorphic transformation of $\ce{Mn3O4}$ rather than formation of $\ce{MnO}$.

Another process is carbothermic reduction [2], i.e. reduction of manganese oxides with the help of carbon or carbon monoxide. It requires high carbon monoxide/carbon dioxide pressure and precise change of pressure w.r.t. to temperature. The following reactions depicts how the standard energy of formation varies with temperature:
$$
\begin{align}
\ce{3 Mn2O3 + C &->[$T = \pu{1100 °C}$] 2 Mn3O4 + CO} &\quad ∆G^\circ &= -0.25 - 0.17T\\
\ce{3 Mn2O3 + CO &->[$T = \pu{1100 °C}$] 2 Mn3O4 + CO2} &\quad ∆G^\circ &= -170.71 - 0.004T\\
\end{align}
$$
References

Tinsley, D. M.; Sharp, J. H. Thermal Analysis of Manganese Dioxide in Controlled Atmospheres. Journal of Thermal Analysis 1971, 3 (1), 43–48. https://doi.org/10/bvwj2v.
https://www.alliedacademies.org/articles/practical-reduction-of-manganese-oxide.pdf
Puerta, M. C.; Valerga, P. Thermal Decomposition of a Natural Manganese Dioxide: A Laboratory Experiment for Undergraduate Students. J. Chem. Educ. 1990, 67 (4), 344. https://doi.org/10/fqg3kp.


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Will Proving or Disproving of any of the following have effects on Chemistry in general?
I am working on a project relating to-
https://en.wikipedia.org/wiki/Millennium_Prize_Problems
I wanted to list the effects of them being proved or disproved in different aspects of science and maths. It was fairly easy to find such results on physics and maths as the questions primarily come up from them however I couldn't find any reference to effects it would have on our understanding of chemistry.
Is it because there will be no effect at all and that they are the least concern to chemists (in terms of applicability) or if there are then what are they?

To understand just how absurdly far-fetched this may come across, ask yourself the same question: what effect will it have on your life? Say, you get up, brush your teeth, have your breakfast, then BANG! $P\ne NP.$ So what?

I suspect that two problems (p-np and Navier–Stokes existence and smoothness) would have repercussions, in defining some problems as either bounded or unbounded. Guaranteeing that you can find a solution to a certain level of accuracy in a given amount of time is akin to guaranteeing that you will come out winning in a lottery if you buy enough tickets. However I admit to speculating - I am not directly familiar with the problems.

@IvanNeretin I know it wouldn't be of any importance to me in my general life but i wanted to know it's impact on a discipline or a field of study such as Chemistry

@IvanNeretin I was reading the Code of Conduct for this website. Maybe you might find the Friendly and Unfriendly part helpful :-) https://chemistry.stackexchange.com/conduct

All right, let's sum it up.

P vs NP, if solved the way that pretty much everybody expects (that is, $\rm P\ne NP$), will have no far-reaching consequences, because that's what we were thinking for quite a while now. If it would miraculously happen to be otherwise (that is, $\rm P=NP$), that would be quite a shock to many fields, especially to computer science, and by extension, to computational chemistry. But I don't believe in miracles.
Navier–Stokes (no matter how it will turn out) will have consequences in fluid dynamics, and by extension, maybe in some relatively narrow areas of chemical technology.
Yang–Mills will cause repercussions in the areas of physics which are the most distant from chemistry.
The rest of the problems, as far as I understand, have no bearing whatsoever on our material world.

"The rest of the problems, as far as I understand, have no bearing whatsoever on our material world." - Not true. The Hodge conjecture, at the very least, would impact string theory (https://mathoverflow.net/questions/221892/what-are-the-ramifications-of-the-hodge-conjecture-to-mathematical-physics), and the Birch-Swinnerton-Dyer conjecture and Riemann hypothesis have cryptographics ramifications, which would affect money, which is material (or at least is convertible into something material). The Poincare conjecture is also relevant for certain GR spacetimes.

String theory has no bearing on our world either, and won't have for a while. Poincare... well, if our universe is closed (which we are not sure about), then it is applicable and may have some consequences, though I don't see any. As for the rest, well, point taken: Riemann and probably BSD will affect the world, if only through computer algorithms. But then again, it is much like P vs NP: if what we think true is proven true, then some tentative estimates become exact, and things continue running as usual. And if it is proven not true, that would be a miracle.

... and possibly the worst version of $P = NP$, a completely nonconstructive proof of that equality.  So then we know that many problems are "easy", but we're ignorant of the easy way.

@IvanNeretin If string theory has no bearing on our world, then Yang-Mills surely doesn't either. Likewise, if P vs. NP affects our world, then Riemann and BSD do too. I'm just confused about the distinction between the things you say do have some bearing on "our material world" and those that you say have "absolutely no bearing on our material world."

That's why I said "point taken". Admittedly, Riemann and BSD do have some bearing on our world, just not on chemistry.

@EricTowers Actually, there is an algorithm that is guaranteed to be a P-time solution to an NP-problem if and only if P=NP.  (It is, however, exponential in the size of the algorithm if implemented naively, and it does not tell you if it found the algorithm, but it does tell you if it found the solution to a given case)


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Electrolytic refining of precious metals — why the anode rather than the cathode?
In electrolytic refining, metals like platinum and gold are deposited as anode mud.
Why aren't they deposited on the cathode since they have higher reduction potential?

For these metals to deposit, they would have to be present in solution such that they can come into contact with the cathode and be reduced. However, this would first require them to be oxidized, so that they dissolve in solution. Unless you are actually electrolytically refining for gold, the voltage will not be enough to oxidize it.


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Force on an electron in a given orbital
I was wondering whether there is any quantum mechanical  equation to describe the force on a given electron in a given orbital of an atom (lets say, for the simplest case, for hydrogen atom). Also if it exists then what is the use of effective nuclear charge?

The force concept exists in the Bohr's model. However, it is more "natural" to talk of angular momentum (recall it is related to force) in the quantum world which have no classical analogs. It is better to keep the ideas of Bohr model separate from QM.

In QM model, there exist the potential in the Schroedinger  equation, that is in direct relation to the electrostatic force.

Your question doesn't make much sense. If you think that your electrons go in circles then get it out of your head expeditiously.

No I don't think of electron moving in a circle.

Also asked on Physics. Please don’t cross post like that.

Well, electron described by any orbital can be found in virtually any place in atom so why you ask about a "given" orbital?

An orbital is not an orbit. Also for hydrogen atom an electron in its lowest energy state resides in 1s orbital.


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Decomposition of Metal Nitrates
An entrance JEE Main & Advanced AIEEE 2003 exam question:

Which one of the following nitrates will leave behind a metal on strong heating?
A. Copper nitrate
B. Manganese nitrate
C. Silver nitrate
D. Ferric nitrate

I looked up the answer and it was given to be silver nitrate, and I took it up as something to be remembered. However, when I checked online for the decomposition of other options such as ferric nitrate I found many research papers discussing the same which is not something one asks students who have just studied chemistry up to 12th grade.
Also, is there a way to decide which one will, because if it says strong heating, shouldn't most metals leave their salts and exist as free metals if we go on increasing temperature?

Most metal nitrates decompose on heating to give metal oxides. Decomposition of silver nitrate is different as it yields elemental silver.
$$\ce{2AgNO3 -> 2Ag + O2 + 2NO2}$$
Usually the metals which lie very low in the activity series decompose to give their elemental form.
I don't think it's possible to extract metals from just any nitrate by strong heating. They will yield their respective oxide which won't give the element rather they'll just fuse (melt) just like stable metal oxides like $\ce{Al2O3}$ does.


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Which of the following carbocations is more stable?

Which of the following carbocations is more stable?

I know two major factors in determining carbocation stability: resonance and the number of carbon atoms neighboring the positive carbon. Does the double bond in the  substituent of the first molecule participate in resonance?

The one with the aliphatic C=C bond may be more stable since there may neighbouring group participation, allowing dispersal of positive charge

@TanYongBoon There is an $\mathrm{sp^3}$ carbon in-between which breaks resonance so the shorter chain allows for more hyperconjugation.

@Jan I am not talking about resonance. When I mean "neighbouring group participation", I am referring to the $\pi$ electron density being donated directly to the positively charged carbon atom, forming a cyclopropyl or cyclobutyl structure

@TanYongBoon Oh, I think I see what you mean; analogous to a metal-ethene complex? If that’s the case, I’m not sure whether the single $\ce{CH2}$ group provides enough flexibility but I would have to build a model.

@TanYongBoon that mechanism would require a "six membered ring". In term or charge stabilisation/relative stability the smaller carbocation should be favoured.

@TanYongBoon actually now I feel stupid. The cation on the top is just the linear form of a cyclopropylmethyl cation … *head meets table*

@Jan Yes, that is one possible stable structure that can be derived from the "neighbouring group participation" I suggested

Are we talking about: What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?

The second one will be more stable.
We usually first check for resonance as it is more effective.
But here it has to be explained on basis of hyper conjugation because the resonance due to the ring is same in both. But the second one will have more hyper conjugating structures as greater number of alpha hydrogens are present.
More about hyper conjugation concept:
https://en.wikipedia.org/wiki/Hyperconjugation
Edit :  No the double bond wont participate in resonance.

The Second One Will be More Stable as it has one of the resonating structure with octet configuration that is when the the double bond of the ring comes out and participates in the formation of double bond with the carbocation which is out of ring in no case the double bond from the first molecule will be involved in resonance ....


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Classic/killer/easy application for the deuterium kinetic isotope effect?
I'm a physicist. While visiting someone's o-chem class I learned about the existence of the deuterium kinetic isotope effect. This seems like an extremely sweet example of zero-point energy, and I would like to use it as an example when teaching the basic quantum mechanics that we teach in freshman physics at a community college.
In this context, I would like to be able to give a good example where we actually care about the deuterium kinetic isotope effect. Ideally this would be an example that would be at least somewhat understandable to people who (like me) have no clue about organic chemistry. Something like:

Without the deuterium kinetic isotope effect, we wouldn't be able to synthesize x (where x is nail polish, fullerenes, methamphetamine, ...)
Historically, there was intense controversy about how this mysterious chemical process in x worked (where x is learning in slime molds, work-hardening of bronze, ...), but then they injected their (slime mold, ...) with deuterium and used the deuterium kinetic isotope effect to prove that ...

Related: Why is a C–D bond stronger than a C–H bond?

Involvement of water in enzyme-based catalysis, I'm sure there are examples.

The effect can be smaller that you imagine as the energy at the transition state may also be lowered so the barrier height need not change that much. In reactions involving many steps, particularly with enzymes a small, say 10%, change at each step becomes important and results in a big overall effect.

You might read about heavy water toxicity to find an overall impressive one. For instance an aquarium of heavy water can host no fishes, at least after a certain time.

@porphyrin "Big" is relative. I dont think there is a natural reaction pathway that can actually accumulate deuterium in a specific chemical function to twice the concentration in the surrounding water or so. Much less, usually.

The site specific hydrogen (and carbon) isotope ratios tell you if the alcohol in your wine was actually made by fermentation of fructose (instead of saccharose, maltose, ...), and even if the fructose was actually in grapes and not apples. https://en.wikipedia.org/wiki/Isotopic_analysis_by_nuclear_magnetic_resonance.

@Karl, Sure, 'big' is relative, but I meant only that if hydrogen was replaced with deuterium then the effect would be accumulative and so much slower.

@porphyrin A hydrogen atom can only be replaced by deuterium once. ;) Where du you see an accumulative slowing effect?

@Karl, quite so but in a sequence $A \to B \to C\cdots$  each step is slower so slower overall.

@Alchimista: Your comment sounds like it would make a good answer.

An expensive but possible demo would be to make two candles, one out of perdeuterated wax, and another out of regular wax, and to burn them both at the same time.  The rate at which the candles burn should be significantly different.


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Real Gas Behaviour

For a fixed amount of a real gas when a graph of $Z$ vs $P$ was plotted, then at a very high pressure slope was observed to be $\pu{0.01 atm^-1}.$ At the same temperature and pressure if a graph is plotted between $PV$ vs $P,$ then for $\pu{2 mol}$ of the gas the $y$ intercept is found to be $\pu{40 atm L}.$ Calculate excluded volume in litres for $\pu{20 mol}$ of the real gas.

Kindly note that Z is the compressibility factor of the gas.
I am not sure whether or not this problem involves differentiation in order to find the slope. Maybe the excluded volume is actually the product of the van der Waals constant $b$ times the amount of substance (in this case, $\pu{20 mol}).$
So, how can the value of $b$ be determined? Also, why is the slope between $PV$ and $P$ has its units in terms of atm-litre? Shouldn't it be just in terms of litres?
The answer was given to be 4. But I wish to know how it was calculated.

Can you define $Z$? It's not obvious from the statement of the problem what that is.

Z = PV/nRT, is the compressibility factor of the gas under the given conditions.

The van der Waals equation is
$$\left(p+\frac{a}{V_m^2}\right)\left(V_m-b\right)=RT$$
which for $p \gg a/V_m^2$ $^\ast$ can be rewritten as
$$pV_m=RT+bp$$
or
$$pV=nRT+nbp \tag{1}$$
The problem says that the tangent of this curve evaluated at the same $T$ has an intercept (the value of $pV$ when $p$ goes to zero) of $\pu{40 Latm}$ when $n=2$, which means that $RT=\pu{20 Latm/mol}$. But equation (1) can be written as
$$\frac{pV}{nRT}=\frac{pV_m}{RT}=Z=1+\frac{bp}{RT}$$
which means that
$$\left(\frac{\partial Z}{\partial p}\right)_T=\frac{b}{RT}$$
The problem states that this slope is $\pu{0.01atm^{-1}}$ so that $b=\pu{0.2 L/mol}$ and so that when $n=20$, the excluded volume is $nb=\pu{4 L}$.

$^\ast$ More precisely, assuming ideal gas behavior, the inequality is satisfied when $$\frac{(RT)^2}{a}\gg p$$ Looking up typical values of a (typically $\pu{1-20 atmL^2/mol^2}$) means that for the given problem this equality holds if $p\ll\pu{20 atm}$.

Hmm, But for the pressure term of the van Der Waals equation, for small and medium pressures, a/Vm^2 grows faster then pressure. The simpification is applicable, when gas is not well compressible anymore and the  equation is not well applicable.

@Poutnik good point, I will add a comment to my answer.

Just complementing the first answer of @Buck Thorn :
Full rewriting of the van Der Waals equation in terms of $Z=f(p,V_\mathrm{m})$:
$$\left(p+\frac{a}{{V_\mathrm{m}}^2}\right)\left(V_\mathrm{m}-b\right)=RT$$
$$pV_\mathrm{m} - pb + a/V_\mathrm{m} - ab/{V_\mathrm{m}}^2=RT$$
$$pV_\mathrm{m} \left( 1 - \frac{b}{V_\mathrm{m}} + \frac{ a}{ p{V_\mathrm{m}}^2} - \frac{ab}{p{V_\mathrm{m}}^3}\right)=RT$$
$$Z=\frac{pV_\mathrm{m}}{RT}\\ =\frac{1}{1 - \frac{b}{V_\mathrm{m}} + \frac{ a}{ p{V_\mathrm{m}}^2} - \frac{ab}{p{V_\mathrm{m}}^3}}\\=\frac{p{V_\mathrm{m}}^3}{p{V_\mathrm{m}}^3 - pb{V_\mathrm{m}}^2 + aV_\mathrm{m} - ab}\\
=\frac{p{V_\mathrm{m}}^3}{(pV_\mathrm{m}^2 + a)(  V_\mathrm{m} - b)}\\
=\frac{1}{(1 + \frac{a}{pV_\mathrm{m}^2})(  1 - \frac{b}{V_\mathrm{m}})}\\
\overset{p\gg \frac{a}{ {V_\mathrm{m}}^2}}
=\frac{V_\mathrm{m}}{V_\mathrm{m}-b}$$
$$\overset{b \ll V_\mathrm{m}}=1 + \frac{b}{
V_\mathrm{m}}$$

There is an error in equation 4, if I understand correctly what you are doing. When you correct that you get $Z=1+b/V_m$. Also, I am not sure the series in Z is useful.

@Buck Thorn Well,I guess you are right, it should be reciprocal. More eyes see more. :-)

I think you'd agree that in general $b<V_m$ and at lower pressures, $b \ll V_m$, meaning that $(1-b/V_m)^{-1} \approx 1+b/V_m$

Yes, sure. :-) But the point is, for low pressures is Z often < 1, as the cohesive pressure often overtakes non ideal compressibility.

But your inequality is OK.


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Is it wrong to allege that lemons and limes are alkaline?
Please ELI5; I've never studied chemistry. I read about  alkalinity v. basicity here and on Sciencing.com.
Lemons and limes are doubtless acidic.

Lime: The pH range of lime is 2.00 to 2.35.
Lemon: The pH range of lemon is 2.00 to 2.60. This makes lemon slightly less acidic than lime.

'The pH levels of citrus fruits can cause damage to tooth enamel. Typically, the lower the pH, the more damage is done to the teeth. Oranges have a higher pH level, at 4.35, and are less damaging to teeth than lemons at 2.75, limes at 2.88.'

So isn't it wrong for Yuri Elkaim BPHE, CK, RHN

and this unidentified author to allege lemons and limes to be alkaline?

https://chemistry.stackexchange.com/questions/105503/sugar-solutions-have-a-neutral-ph-in-themselves-but-it-makes-your-body-acidic

Yeah, I've seen such claims too and they always leave me puzzled. I heard also that it's good for the humors to eat a balance of wet and dry foods.

Whole lotta pseudoscience stuff from people who never took a chemistry course, and great marketing to the crowd who believes that they can "alkalize" their body with the snake oil being sold to them.

You're right to be confused. Limes and lemons are indeed acidic, not alkaline.

Grouping of food items into various classes is a very old idea of Ancient Greek, Alchemists & Ancient Indian School of Medicine (Aryuveda). These ideas are perhaps thousands of years old. From a scientific perspective these classifications don't mean anything today. For example, in those schools of thought, fish has a very "hot" effect on the body but yogurt is "cold". Similarly, citrus fruits are "cold" but dried dates are "hot". Now nobody can define what hot or cold means and what is the criterion?
Similarly, the chart shown in the original posts are all wrong. It is just a modern version of these ancient schools of medicine. I think the major source of confusion in calling limes and lemons is that citrates (salt of citric acid) are indeed alkaline. Many antacid concoctions use solid sodium bicarbonate and citric acid. The moment you add a spoon to water, it starts to fizz vigorously. The pH of such solutions are slightly alkaline. Perhaps whoever made those charts had these ideas in mind.

There might be some confusion between lime as in limestone and lime as in lime tree. Even though the current spelling is the same, the former has indo-european roots while the latter has Arabic roots. One is alkaline and the other acidic.

Very interesting, I did not know that the word lime (fruit) was derived from Arabic. In Urdu/ Hindi/ lemon is lee-moo, very close to Arabic (< Arabic līma). The calcium hydroxide (lime) is from Latin  *lîmo- = Latin līmus mud as per the OED.


text:
Osazone crystal
How does maltose form its osazone crystal? If osazone is to be formed, then there must be aldehyde form for phenylhydrazine to react, but maltose is formed after hemiacetal formation. How do we address this?

What happens if you tautomerize the hydroxyl proton on the far right to the nearby ring oxygen?

Then you have to break the ring.

And you then get what?  Sugars easily do this tautomerization.

The hemiacetal is in equilibrium with the aldehyde unless it reacts further to form a glycosidic bond (like the glucose unit on the left). When analyzing carbohydrates, you can distinguish between reducing sugars (aldehyde group is available, will react with Benedict's reagent like glucose unit on the right) and non-reducting sugars (aldehyde group is not available, like glucose unit on the left).
"Reducing" sugars will react to form osazone derivatives while "non-reducing" will not.


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What is the Lewis dot structure for ozone?
Currently I studying about the Lewis dot structure for different compounds and my textbook says that the Lewis structure for ozone may be drawn as(as shown in the image):

And then it goes on to calculate the formal charges on different oxygen atoms in it(which are shown in figure with $+/-$ signs.) In justification of the structure it says:

Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species. Generally the lowest energy structure is the one with the smallest formal charges on the atoms. The formal charge is a factor based on a pure covalent view of bonding in which electron pairs are shared equally by neighbouring atoms.

Now, my question is why can't the structure  be like this:

Clearly, if this isn't a possible structure then the flaw in my understanding arises because of not understanding the term 'smallest formal charge'(as I found out that the structure I gave has all oxygen atoms with zero formal charge )
Please suggest what does it mean by the smallest formal charge ? Also why energetically the structure isn't stable.
(For now please exclude any discussion about resonance hybrid of the ozone)

The reason the cyclic structure doesn't exist is due to the bond angles. The 60 degree bond angles puts the electrons in the covalent bonds too close to each other. The electron-electron repulsion makes the structure unstable.
Rarely will you find bond angles less than 90 degrees, making cyclic structures with less than 5 in a ring uncommon.

Why don't you post it as an answer.

I can't post as an answer, it's on hold.


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Acid-base titration: Unknown base in burette or flask?
While finding the concentration of $\ce{NaOH}$ via titration with $\pu{0.05M}$ Oxalic acid solution (in our lab) we put NaOH in the burette whereas added phenolphthalein to oxalic acid in titration flask. But in this video the demonstrator does just opposite ( i.e., $\ce{NaOH}$ in flak with phenolphthalein). I am confused in deciding which one might be correct. My view is that either would do the job but I need some clarification.

Both are perfectly correct methods as long as the volume measurements are accurate.

@M.Farooq - I disagree. It is easier to detect the change from colorless to colored. So, if possible, the titrant should be a base when phenolphthalein is the indicator.

No matter what you do students will always come to you holding the flask asking if this is the right end-point :-) For a skilled analytical chemist, both techniques should be fine and of course colorless to color is easier to detect. One is less painful than the other.

The base should be in the burette, and the acid in the flask. With this technique, the point of equivalence is obtained with a rather high degree of accuracy.

This is not a universal practice in analytical chemistry. For example, if we wish to determine NaOH contaminated with some sodium carbonate, one has to use HCl in the buret, and two use two indicators, NaOH solution is in the flask.


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Reaction with tin(IV) iodide and hydrogen peroxide
Today I reacted tin(IV) iodide $\ce{SnI4}$ (dissolved in water) with hydrogen peroxide. This produced a yellow/brown colour (I'm guessing due to formation of $\ce{I2}?).$
I then added some petroleum spirit which gave a bi-phasic system which was a clear/pink/ light purple colour on top and an orange-brown yellow underneath.
I'm struggling to find out what reactions have occurred here, can anyone please help?

I propose the following equation :
SnI4 + 2 H2O2 --> SnO2 + 2 H2O + 2 I2

$\ce{SnI4}$ quickly hydrolyses in water, so likely there will be hexahydroxostannic acid that hydrogen peroxide actually encounters.

Ahh ok, this would make sense, thank you for your quick reply! Would this explain the colours in each layer?

how would this account for the colours?

Clear/pink color is likely due to Mie scattering of white $\ce{SnO2}$ particles (sol formed from $\ce{H2[Sn(OH)6]}$), and orangey brown color is probably iodine from the reaction between HI (product of hydrolysis of $\ce{SnI4}$) and $\ce{H2O2}$. Do you need references from reputable literature sources to back this up? If so, feel free to add reference-request tag, I'll try to look them up and maybe post an answer with reactions.

Just in case, are you sure the brown yellow phase was underneath, and the clear pink one was on top and not vice versa? Also, are you sure it was an ordinary petroleum ether? If you have a photo, it would be nice to see it.

Note: this is a preliminary description, which I think answers the question, but disagrees in one detail (see asterisk).
Tin(IV) iodide is completely hydrolyzed by water [1, p. 120] to colorless hexahydroxostannic acid:
$$\ce{SnI4 + 6 H2O <=> H2[Sn(OH)6] + 4 HI}$$
Hydrogen peroxide can not oxidize $\ce{Sn^4+}$ further (in fact, hexahydroxostannate(IV) is used as a stabiliser for hydrogen peroxide), and is likely going to oxidize hydroiodic acid to iodine (which attributes to the brown color) instead:
$$\ce{H2O2 + 2 HI -> I2 + 2 H2O}$$
Hexahydroxostannic acid only exists in solution and tends to form cloudy/milky gel over time [2, p. 1021]; this can be simplified as formation of colloidal particles of tin(IV) oxide [hydrates]:
$$\ce{H2[Sn(OH)6] <=> SnO2 · n H2O + $(4-n)$ H2O}$$
When petroleum ether is added, iodine is extracted to organic phase (on top due to lower density of the ether). Tin(IV) oxide sol remains in aqueous phase and attributes to its slightly pink color due to Mie scattering of colloidal solution.*
* Phase order doesn't match OP's description which suggests opposite coloring of the phases (additional clarification on the type of petroleum spirit or a photo of the binary phase mixture would help).
References

Inorganic Syntheses; Bailar, J. C., Ed.; Inorganic Syntheses; John Wiley & Sons, Inc.: Hoboken, NJ, USA, 1953; Vol. 4. https://doi.org/10.1002/9780470132357.
Holleman, A. F.; Wiberg, E.; Wiberg, N. Lehrbuch der anorganischen Chemie, 102nd ed.; Fischer, G., Ed.; Walter de Gruyter: Berlin New York, 2007. ISBN 978-3-11-017770-1. (in German)

Thank you so much for this description, although yes the top layer was slightly pink but I suppose I should have described it as more clear? I’m not quite sure what’s going on @andselisk

@andelisk I should have taken a picture, I didn’t think at the time! I guess the top layer was just off a light purple, not a milky colour

@MarinaCalder Hmm, it's indeed hard to discuss colors in the absence of a photo. Purple color might be due to iodine remaining in aqueous phase. And by "milky" I didn't mean plain white, but a cloudy-white colloid solution, something like this (randomly picked from Google Images).

Which iodide is extracted to organic phase? Would it not be the iodine - I don't understand how the iodide would prefer a non-polar solvent.

@StianYttervik You are absolutely correct, this was a typo. Thank you!

Thank you so much for your help, I appreciate it so much!:) @andselisk


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Last step of proposed cubane synthesis
I was watching an online lecture on decarboxylation by soda lime and the first few examples were straight forward but after a while a felt that the questions couldn't simply have one effect and maybe other factors should also come into consideration as complexity of structures increases.

In the above reagent soda lime was added and the teacher said that cubane will be produced however though it might be straight forward based on the mechanism as a carbanion is produced in the intermediate is it correct to directly say that the mechanism should be followed as expected by one pathway without considering any other possibilities such as attempt to relieve angle strain?
Over the course of studying organic chemistry I have realized things often aren't as simple as they appear to be as many factors come into play such as ring closing or bond breaking to relive angle strain.
According to the video the product is Cubane
I know that cubane itself is kinetically stable (primarily due to lack of availability of pathways for decomposition) but the intermediate formed can it not provide any pathways for decomposition?

What was the yield on the reaction according to the video?

Sodalime forms the dianion of the carboxylic acid group and SN2 ring closure occurs. Decarboxylation of the anion of cubane carboxylic acid is facile given the kinetic acidity of cubane.  https://doi.org/10.1021/ja00001a051

@user55119 i didn't get your point about dianion. Isn't a simple anion formed? Also I would appreciate if you elaborated a but more on the kinetic acidity point as it is a relatively new term for me. Maybe formatting as an answer would be a good choice

Sodalime conditions are quite vigorous. The acidic hydrogen is removed from -CO2H but the alpha hydrogen next to the carboxylate is also deprotonated, thus a dianion. The carbanion displaces chloride. The smaller the C-C-C bond angle the more acidic the C-H bond of the central carbon. The authors claim that cabane is ~63K times more acidic than cyclohexane. Also, cyclopropane is more acidic than methane.


text:
Aluminium Oxide as Dehydrating agent
According to this paper on dehydration by $\ce{Al2O3}$,

Dehydroxylation causes a surface defects as Lewis acidic centers (unsaturated $\ce{Al^3+}$ ions) and basic Lewis centers ($\ce{O^2-}$ ions) creation. Water molecules deactivate Lewis centers and create strongly acidic Brönsted’s centers. Alcohols dehydration reaction performs on Brönsted’s active centers where hydroxyl groups disconnect proton from alcohol molecule. During reaction alcohol molecule adsorbed on aluminium oxide surface reacts with proton.

But I am not able to understand what they are trying to convey here.

I have revised my question and added other points also. It is now clearer than before.

Regarding the flag you've raised on your own question as to why it has been closed as HW, I think the reason is obvious if you read the information in the reason for closing. The question itself is all right, but you are supposed to add your thoughts and/or demonstrate the research done prior to asking.

Simply stating "I don't understand X" isn't constructive. Something like "I've read this paper [ref] and it claims X, whereas my textbook [ref] says Y, but I think it's Z because of this, that and the other" would be much better and help others help you since we would have better understanding what exactly you are having difficulties with.

Alumina absorbs water to give aluminium oxide hydroxide, a solid which removes the water from the reaction environment.
According to the paper by Pethrick et al. [1]:

Aluminium oxide exists in several forms which on addition of water give a complex range of oxide-hydroxide; boehmite, bayerite and gibbsite.

References

Pethrick, R. A.; Hayward, D.; Jeffrey, K.; Affrossman, S.; Wilford, P. Investigation of the Hydration and Dehydration of Aluminium Oxide-Hydroxide Using High Frequency Dielectric Measurements between 300 KHz–3 GHz. J Mater Sci 1996, 31 (10), 2623–2629. https://doi.org/10/chcqs8.

Is there any kind of mechanism involved in the reaction?

There is a reaction between the alumina surface and free water with water adding to the aluminium atom, but afaik that is as much as can be said.


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What does higher electron gain enthalpy mean here?
While talking about ionic bonds my book states :

Higher the value of electron gain enthalpy of the atom, greater the ease of formation of the anion from it, i.e., other atom should have high value of electron gain enthalpy.

Now I think it should say more negative than just higher (as negative enthalpy means release  of energy and hence more stability). So what should it be? Why the chemistry books (I have to read)  seem to always avoid the term magnitude(Is there some hidden assumption which are generally in the field of inorganic chemistry  ?

I take for granted that what you call "electron gain enthalpy" is generally described as "electron affinity". This is the energy gained (or lost) when an electron is added to an atom, ion or molecule. In general the electron affinity is a measure for the tendency of elements to form anions. The trouble is that you may find tables giving these numerical values as positive or negative values. For example, for fluorine $\ce{F}$, the electron affinity is $\pu{338 kJ/mol}$. But you may find books where the same affinity is given as $\pu{-338 kJ/mol}$. It depends on the sense you consider the reaction $\ce{A + e- -> A-}$.


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Are these numbers Madelung constants?
I have no idea of chemistry, just a little bit what I know from school, but I found through some search some connection between entropy (Shannon entropy) of the divisors of some natural numbers and the Madelung constants.
My question is this:
Are these numbers Madelung constants?
1 -0.000000000000000
2 1.38629436111989 ( http://oeis.org/A016627 )
3 1.38629436111989
4 3.31573052232910
5 1.38629436111989
6 5.54531864369184  ( http://oeis.org/A257872 )
7 1.38629436111989
8 5.57849454296371
9 3.29975689063759
10 5.54521210457048
11 1.38629436111989
12 10.7943529563000
13 1.38629436111989
14 5.54518969515687
15 5.54519368543309
16 8.08924328274917
17 1.38629436111989
18 10.7599543952466
19 1.38629436111989
20 10.7921021686465

Thanks for your help!
Related question:
https://mathoverflow.net/questions/346239/entropy-magnitude-diversity-of-finite-metric-spaces-in-number-theory
Edit:
I think I found an explanation for the numbers ( entropy ). It is approximately:
$\tau(n) \log(\tau(n))$, where $\tau(n)$ is the number of divisors of $n$.
So the $1.386$ and $5.545$ are just coincidences with Madelung constants I think.

1.386 is 2log2 but what about 5.545 the closet approximation is 7log7 which is still 5.91. Am i missing something?

$5.545=8 \log(2)$ see the oeis link which relates this number to one Madelung contant

Ok i was mistaking that both the numbers inside and outside the log must be the same

I don't know if this would be very helpful or not but as i had some free time on hand I got a list of Madelung Constant Values from the Internet
In the list most of values were reported to 3-4 decimal places so I have Compared only the first decimal places as once the second decimal place was used none of the values matched. Maybe this would be a helpful start.
$$M_a$$ of ZnS (Wurtzite Hexagonal) is $10.7153$ (nearby the 12th and 18th entry in your table)
As Oscar Lanzi has commented $1.386$ is approximately $$2\log(2)$$ which is the Madelung constant for a one-dimensional line of alternating charges
As the question has been edited to add further details maybe the second one is a coincidence.
Only Nearby Matches So Far

The $1.386$ figure is $2\log 2$, which is the Madelung constant for a one-dimensional line of alternating charges.

@OscarLanzi should i add this into my answer or are you going to write one on your own?

I say add it in.

@OscarLanzi Done


text:
Representation of bromelain molecule in two dimensional diagram
If I'm not wrong, I learned in 2011 introduction to neuropsychology course that all enzymes are proteins.
As far as I know, any protein, not even the smallest of the TAL group (which I understand to be the smallest currently known protein group), can be represented on a common-size two dimensional diagram (say, A4 page size) simply because it's too big in comparison to other molecules (such as methylcobalamin).
If bromelain is an enzyme which is larger than, for example, TAL proteins, how come it is represented on a common-size two dimensional diagram in PubChem?

Well, just like that. Look at that 2D diagram, it is all before you. Which part of it you don't agree with?

@IvanNeretin hello. Sadly I don't understand your comment - if it is sarcasm, this is a case in which I misunderstand the purpose of the sarcasm...

No, it is a genuine incomprehension on my part. You seem to be seeing some kind of mistake or contradiction where I see none.

@IvanNeretin thanks, I have edited the question to hopefully make it better.

I was forced to look at the wikipedia entry for enzyme, which states:

Enzymes /ˈɛnzaɪmz/ are macromolecular biological catalysts that accelerate chemical reactions. [...] Most enzymes are proteins, although a few are catalytic RNA molecules. The latter are called ribozymes. [Enzymes' specificity comes from their unique three-dimensional structures.]

The definition encompasses a rather broad range of molecules (macromolecular biological catalysts), is more of a functional rather than a structural definition, and requires clarification of what is meant by macromolecular. Also, not everything the wikipedia says should be accepted as consensus, of course.
This is an interesting question. For one thing, it challenged my biased view of enzymes as proteins (even if I know of ribozymes). I don't usually think of glycoconjugates as enzymes, but there you go, you have presented such a case. However, this is the structure shown in pubchem:

I see a couple of amide bonds, but mostly I see sugar rings, consistently with the description of the molecule:

Alpha-L-Fucp-(1->3)-[alpha-D-Manp-(1->6)-[beta-D-Xylp-(1->2)]-beta-D-Manp-(1->4)-beta-D-GlcpNAc-(1->4)]-D-GlcpNAc is a branched amino hexasaccharide consisting of a GlcNAc residue at the reducing end with a Man-alpha(1->6)-[Xyl-beta(1->2)]-Man-beta(1->4)-GlcNAc moiety attached via a beta-(1->4)-linkage and a Fuc residue attached via an alpha-(1->3)-linkage.

And yet the wikipedia says that "Stem bromelain (SBM) (EC 3.4.22.32), is a proteolytic enzyme". That's because I also missed the fine print: the pubchem entry is for MUXF, carbohydrate moiety of bromelain (hat tip to Ivan). Bromelain itself is a glycoprotein. And I am still looking for an experimental 3D structure to go with the molecule. Heading down another rabbit hole, the best I could do for now is find a link to a theoretical structure (Ref. 1):

A polysaccharide, attached to a folded protein, is shown in the lower right of the image.
So you opened up your post with a false premise, that all enzymes are exclusively polypeptides. Some of them, such as bromelain, are glycoconjugated proteins. And some such as ribozymes lack a peptide component altogether.
Finally, as to why the hexasaccharide is presented in 2D rather than as protein enzymes usually are, as a folded 3D structure, pubchem explains (see under 3D structure):

Conformer generation is disallowed since too many atoms, too flexible

Reference

Banerjee, N., Banerjee, A.. A three dimensional structural model of stem bromelain: a critical comparative analysis of glycosylation torwards functional structural stability. DOI: 10.5452/ma-c3x5z

I think I got it. The diagram in PubChem is not bromelain molecule, but a tiny fragment of it. This thing is indeed too small to be an enzyme. Also, it has nothing to do with proteins. Look at the formula, there is not a single amino acid residue here. Indeed, scroll down to 3.4.2 and read: "carbohydrate moiety of bromelain". That's what it is, and not the entire enzyme.
If you want to know what the rest of it is made of, then google "bromelain amino acid sequence". It seems to be known for quite a while. No nice pics, though.
On a side note, there is no such thing as "the smallest protein" or even "the smallest known protein", for the border between proteins and polypeptides is rather arbitrary. You are right, though, in that most enzymes are huge proteins, too big to fit on a 2D diagram. So is bromelain.
So it goes.

There is a recent crystal structure of a bromelain precursor protein.
PDB Entry 6U7D   http://www.rcsb.org/structure/6U7D
Usually "precursor" means that the active form of the protein is formed by cleaving off a piece or pieces. I do not know if that is the case for this protein.  However the protein used to grow these crystals was expressed in E. coli, so it does not contain the glycosylation characteristic of proteins produced natively in eukaryotic cells. If you are interested in the sugar residues of the native form, this crystal structure will not help you.


text:
How are London Dispersion Forces generated?
While talking about the gaseous state of matter we came to the topic of London Dispersion Forces caused by the generation of a dipole in one atom which induces a dipole in another. While talking about the cause of such interactions I stated that, since Schrödinger's equation for multi-electron atoms is time-dependent, at some moment a dipole may get generated (due to the increase in the probability of electrons in that region). Though now I am thinking that what I said was quite vague and may not be the reason for it.
So what is the reason for the generation of dipole moment in one atom to cause London forces?

Let me crash the party here.
TL;DR: The classical explanation of induced dipole attractions from electron densities "evading" each other does, by itself, not adequately or intuitively explain the actual charge density patterns that arise in these situations.
(I know that this must seem like an outrageous statement, and surely downvote fingers are itching now. Bear with me.)

Based on the Hellmann-Feynman theorem, it is known that the forces acting on a nucleus arise from two purely coulombic sources: Its attraction to its surrounding electron distribution, and its repulsion with other nuclei. Hence, the observation that e.g. the two atoms in a rare gas dimer are attracted to each other immediately implies that there is a concentration of electron density in between the nuclei, so that the resulting net forces pulls them "inwards" towards each other. As Feynman put it himself in 1939 (emphasis in the original):

... the charge distribution of each is distorted from central symmetry, a dipole moment of order $1/R^{7}$ being induced in each atom. The negative charge distribution of each atom has its center of gravity moved slightly toward the other.  It is not the interaction of these dipoles which leads to van der Waals' force, but rather the attraction of each nucleus for the distorted charge distribution of its own electrons that gives the attractive $1/R^{7}$ force.

This is the complete opposite picture of the momentary effects in the induced-dipole explanation, where the electron densities "evade" each other by simultaneous displacements in the same direction to create attractive dipole interactions. The Feynman explanation is not very popular, but there are in fact some authors who have picked it up and commented on it, most notably perhaps Richard Bader (of Atoms-In-Molecules fame). Politzer and Murray have written a nice article on the topic.

But, is Feynman actually correct? If so, then we should be able to observe some accumulation of charge density between two neutral atoms that are bound by dispersion interactions, right? Indeed we do. The below image comes from a publication by Thonhauser et al. and shows the difference in charge density that arises when the dispersion interaction between to Ar atoms is "switched on" (highlight mine):

So, this may hold true for atoms, but what about entire molecules? Luckily, Hunt has shown in a very laborious 1990 paper that Feynman's picture holds true in that case as well:

... Feynman's statement concerning forces between well-separated atoms in S states generalizes to interacting molecules A and B of arbitrary symmetry (with A and B in their ground electronic states). To leading order, the dispersion force on each nucleus I in molecule A results from the attraction of I to the dispersion-induced change in polarization of the electronic charge cloud on molecule A itself.

Obviously, the "inwards polarization" effect must seem counter-intuitive at first. Why would the negatively charged electron clouds actually want to approach instead of evade each other? Thankfully, a straightforward rationalization of this effect for a rare gas dimer comes from another paper by Clark, Murray and Politzer:

What causes the polarization of the electronic densities toward each other (...)? This can be easily understood when it is noted that the electrostatic potential produced by the nucleus and electrons of any free ground-state atom is positive everywhere; the effect of the nucleus dominates over that of the dispersed electrons. This positive potential is what each atom “sees” of the other atom (...).

Of course, the astute reader may also voice another point of protest: "The fluctuating dipoles are variable in time, whereas Feynman's deformed charge density explanation is entirely static. How do we even compare the two? And what is the effect of the fluctuating dipoles when averaged over time?"
As it turns out, the two explanations are apparently consistent with each other, as detailed by several authors; Hunt herself in the paper mentioned above acknowledges fluctuating dipoles as a possible starting point, and a paper by Clark dedicates a full paragraph to the seeming dichotomy of the two pictures. At its core, however, the suggestive "electron-evading" nature of this explanation is very much misleading in light of the observable static effects of the "inwards" charge redistributions -- which, again, are actually required to create the resulting attractve interactions.

It requires some thought to follow your argument but it certainly shouldn't be down voted.  My impression of VDW forces is simply on close contact the attraction of the nuclei for electrons on the "bond axis" is slightly greater that the repulsion of the perturbed outer electrons.  These slightly inelastic collisions might also account for blackbody radiation.

@jimchmst Thanks for your comment. Yes, the nuclei primarily follow the electron density, which directly implies its accumulation between the fragments. I've added an image that demonstrates this. -- After some more literature research I also have to concede that, in the limit of full mathematical rigour, the fluctuating-dipole picture is apparently consistent with the "Feynman" one. Still, I consider its simultaneous electric dipole attractions to be utterly misleading, and it also doesn't offer any straightforward way to deduce the observable static charge shifts that ultimately arise.

You are correct that simply stating that the Schrodinger equation is time
dependent does not really answer the question.
In fact, time-dependence is not a component of the derivation of the dispersion forces. At a surface level, they just represent how much the ground state energy of a system of polarizable molecules is lowered by their interactions.
It's the same idea as any of the other intermolecular forces (coulomb, dipole-dipole, etc.) where some interaction between particles leads to a lower energy state than the free particles.
Generally, the London Dispersion interaction potential is not derived in a general chemistry class because it is purely quantum mechanical and rather involved. These slides provide a clear and thorough (albeit advanced) derivation, but I will explain the main points below.
Derivation Explanation
Let's imagine for this model that a polarizable atoms can be thought of as a simple harmonic oscillator with a positive charge ($+q$) on one end of a spring and a negative charge ($-q$) on the other end. The spring has an equilibrium length of 0, so that the atom has zero average dipole.
This model of an polarizable atom captures the physics we want because applying a force of some kind to the spring will induce a dipole in this atom  ($\vec{p} = q \vec{d}$).
So far we have only considered one of these atoms. What if we move to consider multiple of them together?
Remember we're really interested in the energy difference between two non-interacting atoms and two interacting atoms, so let's start by finding the non-interacting ground state energy of two atoms. We'll let the two atoms have two different spring frequencies $\omega_1$ and $\omega_2$ respectively.
Setting $z_1$ and $z_2$
as the displacements of the springs from equilibrium for atoms 1 and 2 respectively and $m$ as the mass of an atom,
we can write the total potential energy of the system as:
$$
U_{\text{non-interact}} = \frac{1}{2}\left(m \omega_1^2z_1^2 + m \omega_2^2z_2^2\right)
$$
Solving the Schrodinger equation for this potential yields the ground state energy for the two non-interacting harmonic oscillators:
$$
E = \frac{\hbar}{2} (\omega_1 + \omega_2)
$$
Now, we turn on the interaction and the problem becomes harder more interesting because the positions of the springs affect the magnitude of the interaction. Physically, because the dipole-dipole interaction can lower the energy, we expect that the interacting ground state of the two atoms will be a balance between minimizing the stretching of the springs (which increases the energy) and maximizing the amount of dipole-dipole interaction (which lowers the energy).
Setting $z$ as the distance between the two atoms, with some algebra one can show that the interaction energy between these two dipoles is given by
$$
U_{\text{interact}}(z,z_1,z_2) = \frac{q^2}{2\pi\epsilon_0}\frac{z_1 z_2}{z^3}
$$
Now, our total potential energy for the system is given by:
$$
U_{\text{total}}(z,z_1,z_2) = U_{\text{non-interact}}(z_1,z_2) + U_{\text{interact}}(z,z_1,z_2)
$$
While intimidating looking, this potential can also be solved exactly by changing variables to $z_+ = (z_1 + z_2)$ and $z_- = (z_1 - z_2$) and rearranging terms to make it look like the non-interacting potential energy. Once that is done the ground state energy can be found just as easily as in the non-interacting case.
With this done, we can find that the (one dimensional) energy stabilization is given by:
$$
U(z) = \frac{-\alpha^2 I}{2}\frac{q^2}{(4\pi\epsilon_0)^2}\frac{1}{z^6}
$$
Where the final step was to replace some of our less measurable parameters with their atomic counterparts: polarizability ($\alpha$) and ionization potential ($I$)


text:
Mechanism for reaction of NOCl (Tilden reagent) with primary amines
According to The Pearson Guide To Organic Chemistry For The IIT JEE by Atul Singhal, the reaction between $\ce{NOCl}$ and $\ce{CH3CH2-NH2}$ is as follows:

$$\ce{CH3CH2-NH2 + NOCl -> CH3CH2Cl + N2 + H2O}$$

But how exactly does the reaction proceed?
The first step (what I think) could be that $\ce{NOCl}$ dissociates into $\ce{NO+}$ and $\ce{Cl-}$ ions and then the $\ce{Cl-}$ ion attacks the amine group and replaces it by $\mathrm{S_N2}$ mechanism.
But according to paperplane.io (via the Internet Archive), a diazonium product is produced. Even though a diazo product is formed, how does it produce an alkyl halide according to my textbook?

If you know how to form a diazonium salt with NOCl (or HONO/HCl) then you are an SN2 reaction away from the answer.

Please cite the textbook (as in 'according to my textbook'), or leave out that statement if it is not relevant. Always cite the sources you are using.

Why would the chloride ion attack the electron rich amine? The amine is the better nucleophile in the system, and will attack the $\ce{NO+}$. This gives a $\ce{RNH-NO}$ nitroso species which is then protonated on oxygen, loses water, and gives the diazonium cation. The resultant diazonium cation loses nitrogen rapidly and get attacked by chloride to give the primary alkyl chloride.
More can be read about this process here (which is where I sourced the reaction scheme) or by searching the term "nitrosation of primary amine"

I don't understand how amine is a better nucleophile than chloride ion.

@praig_ Well, what you don't understand? You have some serious misunderstandings there. Probably wanted to get too much while not knowing necessary facts. This is such fact. You can get amine from halide and ammonia, not otherwise (unless it's silyl amine)

@user85683: You may need to read this to start.


text:
Finding the wavelength of maximum absorption and Crystal Field Splitting of [Ni(H2O)6]2+
(I haven't been here for a long time, so tell me if my formatting is fine)
Given

The $\ce{[Ni(H2O)6]^2+}$ complex appears green in solution from the transmitted light. Estimate the wavelength of maximum absorption, and also the crystal field splitting $Δ_0$.

I got 1.59ev for the CFS, but I want to verify if I'm doing this right.
I wasn't sure if I also need to know the wavelengths of light and spin pairing energy for this problem. So I just looked up the average wavelength of red light (650nm) and plugged it in (and ignored spin pairing energy since my prof never talked about it)
My Method
Since H20 isn't charged, so Ni is 2+ to make this complex 2+.
Since $\ce{Ni^2+}$ is $[Ar]3d^8$, and water makes a weak ligand field, I basically did Crystal Field method and filled the bottom row (t2g) with 6 electrons and put the remaining 2 electrons in eg.
I then used stablization energy: $ΔE = -6(2/5)Δ_0+2(3/5)Δ_0=-6/5Δ_0$
And then $ΔE = hv = hc/λ= 6/5Δ_0$ (I think I can exclude that negative sign since I'm just finding the splitting)
to get $Δ_0 = 5hc/(6(650nm)) = 1.59ev $

Absorption of a photon (loosely speaking) causes the transition $\mathrm{t_{2g} \to \mathrm{e_g}}$. The energy needed for this is (by definition) exactly $\Delta_O$, which doesn't actually have anything to do with the CFSE or the electron configuration, so you've gone one step too far.


text:
Simple sandbox software for an interactive molecular dynamics simulation
I'm not sure if this is the right site to ask a such question.  But I'm getting to learn the world of the small.
I know there is software out there like NAMD, and LAMMPS.  But these tools are very non-intuitive, having complex interfaces that seem to hide the underlying mechanics.
I'm looking for something that can display live molecular simulations in 3D, with some kind of inbuilt editor to simply place, or remove atoms.
Again, sorry if this is the wrong place for a such question, I'm sure from what you can tell, I am extraordinarily new to this area of knowledge.  I would greatly appreciate any feedback.  I fear although, I will have to create my own intuitive frontend to NAMD or equivalents to have essentially a 'quantum sandbox'

Albeit not as complete as other packages can be, Ascalph is what you describe http://www.biomolecular-modeling.com/Ascalaph/index.html

Also: I have not tried it myself, but it should be possible to set up VMD so that it "streams" the output of MD packages like LAMMPS

I strongly recommend openmm. It is the easiest of the packages to interact with, followed by LAMMPS. I believe Macsimus also has video options, but I have never used it

Avogadro has a live minimizer when editing - pulling, adding, deleting atoms, changing bond orders, etc. The next version (v2) will support full MD.

Since this question has been up for a couple weeks and still doesn't have an answer, perhaps it might be something that a dedicated computational chemical physics community would be interested in. There is a proposal for a Stack Exchange for Materials Modeling for example: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=ODVjZmM2YzJmOTc0MzBkZDVkMTZmMTIzN2EyNmZmNTQyOWYyY2Y0YWFmNGYxNTQ2OTQzNmMwOWIyYjhhOTZjMhwblBomlTMtPu8EgXeAKtp-8dFsxghojkD0vBSe1wZl0. Perhaps you might consider supporting it if you think it would be useful for you! there's already MD questions there.

@The_Vinz that's if you can get VMD to compile for you. With Catalina, Apple decided to up the VMD compilation difficulty level to boss-mode.