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Predicting sign of enthalpy of allotropic transition of phosphorus
How to predict the sign of enthalpy during the transition
$$\ce{P(s, white) -> P(s, red)}?$$
$\ce{P(s, red)}$ is more stable than $\ce{P(s, white)}$ due to release in angle strain.
Since during formation of red phosphorus from white phosphorus bond breaking and bond formation both takes place, how to predict sign of enthalpy?

Does the downvoter care to give a chance to improve the question?

white P has 4 P atoms in a tetrahedral arrangement, so all bond angles are 60degrees. Like just about any three-membered ring, these structures are highly strained. Conversion to red phosphorus breaks two of the three membered rings open, relieving some of the strain with the entropic cost of polymerizing. Very similar to polymerization of ethylene oxide.

You can use the fundamental relation between energy and enthalpy. At constant temperature and pressure:
$$\Delta H_m = \Delta U_m + P\Delta V_m = \Delta U_m + PM (\frac{1}{\rho_{red}}-\frac{1}{\rho_{white}})$$
where the subindex m labels molar properties and M is the molar mass. Taking values of the density posted in the Wikipedia indicates that red phosphorus has a significantly higher density than white (~2.25 vs 1.82 g/mL). This means the volume decreases and the surroundings do work on the system during the reaction. From this data one can approximate the value of $P\Delta V$ at standard pressure (1 bar):
$$\Delta H_m=   \Delta U_m -\pu{0.32 J/mol} $$
In addition the difference in strain would indicate that $\Delta U_m<0$. This would mean that the enthalpy change should be negative, as can be verified by looking up thermodynamic tables. The Wikipedia page describing the allotropes of P (or alternately the CRC handbook) provides a value  of $\Delta H_m=\pu{-17.6 kJ/mol}$, so the correction for the volume contraction can be seen to be negligible (and does not influence the conclusion that $\Delta H<0$). Interestingly, even though white phosphorus is less thermodynamically stable it is considered the standard state of P.

It is assumed that strain in fact is a decisive determinant of the change in energy, although this is not something I know about, and irrelevant for the calculations, but if true it allows the logical conclusion that $\Delta H<0$.


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How to calculate electric force on valence electron?
In order to understand the periodic trends more fully, I wanted to calculate the net force exerted on the electron (with the highest principal quantum number and azimuthal quantum number) of an atom.
I decided to use the following procedure:

First, I found the effective nuclear charge of the electron in a table from Wikipedia.
Then, I found the calculated atomic radius of the respective atom (which tells me the distance from the nucleus to the electron) using another table from Wikipedia:
Lastly, I converted these units into Coulombs/meters and then used Coulomb's law and the charge of an electron to calculate the electric force.

This is the graph that resulted from these calculations.

For example, for lithium:

2s electron - effective nuclear charge: $\pu{1.279 e} \implies \pu{2.05e-19 C}$
atomic radius: $\pu{167 pm} = \pu{1.67e-10 m}$
Coulomb's law: $\displaystyle \frac{kq_1q_2}{r^2} = \frac{(\pu{8.99e9})(\pu{2.05e-19})(\pu{-1.602e-19})}{(\pu{1.67e-10})^2}$
Magnitude of force: $\pu{-1.06e-8 N} \implies \pu{1.06e-8 N}$

Is this a proper way to calculate the electric force exerted on the electron I specified?
If so, how does this graph relate with other atomic trends, like ionization energy and electronegativity?

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I am not sure your calculations make sense. I may be wrong. But the force changes constantly if the distance of the electron to the nucleus changes a bit. An electron has a certain energy on a given orbital, that does not change in time. But the attractive force changes as a function of the distance to the nucleus. it is a little like the vibrating pendulum. The driving force is maximum when the extension is maximum, and equal to zero when its speed is maximum. So what you do is obtaining a sort of intermediate value of the attractive force, which does not mean too much.

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Why does the water dissociation constant Kw remains constant?
At $\pu{25 ^\circ C}$ at equilibrium the value of $K_\mathrm{w}$ is $\pu{1\times 10^{-14} M}$ and the concentration of $\ce{H+}$ and $\ce{OH-}$ is equal i.e. $\pu{1\times 10^{-7} M}$.
Now if we add $\ce{HCl}$ to it the $\ce{HCl}$ will form $\ce{H3O+}$ and so the concentration of $\ce{H+}$ increases and so the equilibrium shifts towards the left to counteract the changes. As a result the concentration of $\ce{OH-}$ decreases but why does still the $K_\mathrm{w}$ stays the same. If we add too much $\ce{HCl}$ such as 1 mole then we know it will completely dissociate and would form $\ce{H3O+}$ whose concentration would be $\pu{1 M}$.
So now what will happen to $K_\mathrm{w}$? Does it still remains constant but why? Does still the concentration of $\ce{H+}$ can not rise above $1\times 10^{-14}$ but why?

$K_w$ is not really an absolute constant. It depends on the ionic strength of the solution as well as the temperature. In chemistry problems it is typically regarded as an absolute constant because (1) close is generally good enough for most chemistry problems and (2) if a viral expression was used the mathematics would get too complicated to solve by hand.

@ Moon. I do not understand your question. You ask "why the concentration of $\ce{H+}$ cannot rise above $10^{-14}$? " Well. The concentration can of course be higher than $10^{-14}$ . In pure water, this concentration is $10^{-7}$, which is much higher than $10^{-14}$ !!!

See Wikipedia's article Self-ionization of water.

I've rollbacked the edit to the previous version.Your newer edit is better asked as a new question as it negates the answers provided.. the question changed from explanation of ionisation constant to how addition of acids and bases affects the ionisation constant.

@Safdar okay  so i  will ask it as a separate question

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Your question reflects your knowledge in chemistry so I'll try to keep this explanation simple as possible.
Water conduct electricity because it contains $\ce{H+}$ and $\ce{OH-}$ ions by auto-ionization:
$$\ce{H2O <=> H+ + OH- \tag1}$$
It has been defined that this ionization is a constant at $\pu{25 ^\circ C}$. Thus by definition:
$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = 1.0 \times 10^{-14} \tag2$$
Accordingly, by definition for pure water, $ [\ce{H+}]=[\ce{OH-}]= 1.0 \times 10^{-7}$. And then, there is a widely accepted principle in chemistry called Le Chatelier’s principle:

Le Chatelier’s principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

Accordingly, increasing the concentration of reactants (species on left hand side of the reaction) will drive the reaction to the right (more products), while increasing the concentration of products (species on right hand side of the reaction) will drive the reaction to the left (more reactants). At constant temperature, even though concentrations of species (e.g., here they are $\ce{H+}$ and $\ce{OH-}$ in equation $(1)$) at new  equilibrium state is changed, the equilibrium constant remains same. In other word, the equilibrium constant is depends only on temperature.
According to Le Chatelier’s principle, in in equation $(1)$, if you add more $\ce{H+}$ or $\ce{OH-}$ (RHS), the equilibrium would be adjusted to reduce that amount by making more water. But since this addition is at constant temperature, $K_\mathrm{w}$ remains constant.

It can be said that at the given temperature, the rate of the water autodissociation is constant, but the rate of the ion recombination is proportional to $[\ce{H+}][\ce{OH-}]$. As it is proportional to probability two such ions meet each other, because the reaction rate is limited by diffusion. According to Wikipedia,

The inverse recombination reaction   $\ce{H3O+ + OH− -> 2 H2O}$ is among the fastest chemical reactions known, with a reaction rate constant of $\pu{1.3×10^11 M−1 s−1}$ at room temperature. Such a rapid rate is characteristic of a diffusion-controlled reaction, in which the rate is limited by the speed of molecular diffusion.

The kinetics of the [H+] change is like:
$$\frac{\mathrm{d}[\ce{H+}]}{\mathrm{d}t} = k_1 - k_2[\ce{H+}][\ce{OH-}]$$
where $$K_\mathrm{w} = \frac{k_1}{k_2}$$ and $$k_1 = k_{1\mathrm{a}}[\ce{H2O}]$$
implying $[\ce{H2O}] \simeq \pu{55 mol/L}$ is constant.
The consequence is the product of ion concentrations at equilibrium must be constant.
If $[\ce{H+}][\ce{OH-}] \gt K_\mathrm{w}$, then ions recombine faster then water dissociates until it is equal.
If $[\ce{H+}][\ce{OH-}] \lt K_\mathrm{w}$, then water dissociates faster then ions recombine until it is equal.
The ionic strength and activity coefficients, or major presence of other non-ionic compounds complicate the thing, but the above as the general principle of chemical equilibriums remains.

User Mathew Mahindaratne's answer is excellent, but I'd like to belabor a point. The equilibrium is given by the chemical equation:
$$\ce{H2O <=> H+ + OH- \tag1}$$
and the mathematical equation is normally reduced to:
$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = 1.0 \times 10^{-14} \tag2$$
However thinking about equilibrium expressions in general you can see that the mathematical expression for (1) should be:
$$K^*_\mathrm{w} =\dfrac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]}\tag{3}$$
The point here is that for dilute aqueous solutions that $[\ce{H2O}]$ is a constant and hence:
$$K_\mathrm{w} = [\ce{H2O}]\times K^*_\mathrm{w} = [\ce{H+}][\ce{OH-}]\tag{4}$$
Thus solutions with significant amounts of some miscible organic solvent are not "dilute aqueous" solutions, and mathematical equation (2) won't hold. This is not to say that mathematical equation (3) would be significantly better, but rather tying to create some understanding of why the simple equilibrium expressions fail.

That is why in solutions with negative pH the pH and pOH are not equal to 14 and their equilibrium constant value is also different. Please correct me if i am wrong .

@Moon - yes and no. For pH <0, or >14, you must have a very strong acid or base. Thus when dissolving HCl in water the most important effect isn't the change in the $[\ce{H2O}]$, but rather the increase in the ionic strength.

Yes MaxW we need very strong acid such as as 12 M of HCl has pH -1.08 . MaxW i want to ask you a question that in case of negative pH do we have an equilibrium  between water ,H+ and OH- ions ? And in the end you talked about ionic strength . Do you mean an increase in ionic strength of water i.e. the conectration of H+ and OH- ions become more than 1×10^{-14} M

@Moon - See wikipedia's article on ionic strength

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Do different alkyl groups affect chirality?
I am a bit confused on the definition of a chiral center;

A carbon atom is chiral if it has four different items bonded to it at the same time. source

However I am not sure whether different alkyl groups such as $\ce{CH_3}$ or $\ce{C_5H_{11}}$ on a single carbon count as "different" items, or do only different functional groups count?

See, you did not say "identical alkyl groups such as CH3 or C5H11". Why? Because you know they are different. Yes, that counts.

Also, keep in mind that it is not limited to carbon with 4 different groups. For example, if you look at other elements with different ligands in inorganic compounds, they can be optically active as well.

Quoting your source itself:

An object that is chiral is an object that can not be superimposed on its mirror image.

Let's think about this a little bit:
If we have a single carbon atom bonded to four completely different groups (be it functional or alkyl) we will achieve non-superimposability:

(Try re-drawing the structures replacing the $\ce{-COOH}$ group with $\ce{-C2H5}$ and check if you can achieve superimposability)
Achieving non-superimposability is all that optical isomerism is about. As long as we cannot superimpose the object on its mirror image, and can find no symmetrical elements, we can confidently declare that particular compound to be optically active.


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Effect of pressure on gas solubility in liquids
By Henry's Law ($K=\frac{x}{p}$), we know that increasing the partial pressure of a gas also increases its solubility on liquids.
Now a lot of sources (this, for instance) illustrate this by presenting "the bends", a condition that divers may experience when ascending too quickly due to the rapid decrease of the N$_2$ solubility on the blood.
So, my question is: why is the N$_2$ solubility greater deep on the ocean than on the surface?
Henry's Law says nothing about the total pressure, only about the partial pressure of the gas in question. When deep in the ocean, the pressure rises thanks to the water above, not due to a increase in N$_2$ partial pressure.
Bonus: does this video illustrate a greater CO$_2$ solubility in coke because of the greater total pressure? I feel like this is the same situation. Why would the CO$_2$ partial pressure be greater?

The pressure underwater increases by roughly 1 bar every 10 meters, that is, ambient pressure is doubled every 10 meters compared to standard atmospheric pressure.
When you breathe air underwater using a SCUBA tank, your lungs expand against the surrounding ambient pressure in order to increase their volume. As the lungs expand they fill with gas which props up the lungs (balances pressure within the lungs with the surrounding pressure). The SCUBA device adjusts the pressure of the delivered air so that it balances the ambient pressure. The partial pressure of the component gases in the gas mixture inhaled from the SCUBA tank is not the same at the water surface and when a diver is submerged. That means the pressure of the air in the diver's lungs is not ~1 bar, it can be far greater, within limitations imposed by the potential toxicity of high blood concentrations of oxygen and nitrogen.


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Nucleophilic allylic substitution via SN2

Since SN2 reaction is more favorable with primary substrates, do allylic primary halides undergo nucleophilic substitution reaction mostly on carbon 1 via "SN2". and in case of secondary allylilc halide would the reaction proceed mostly via SN2 pathway?

In case the nucleophilic attack occurs on carbon 3, would we obtain both configurations, R and S?

Doesn't this diagram show clearly enough that it's SN1 not SN2?

I made this diagram!
What I did not understand, is if SN2' reactions are possible when the allylic carbon is secondary since all sources I found use primary allylic carbons. In case it is possible, then how could I determine the stereochemistry of the final products since SN2 reactions gives the inversion product. Does nucleophile attack from above and beneath pi bound ?
Any thoughts?

Well, if you somehow don't got it yet, there's no SN2, but SN1 here!

You have not provided any conditions for these reactions so it is difficult to assess specifically how you are substituting a hydroxyl group for bromide. If you use water as a solvent, then an SN1 reaction would be likely but liberated HBr would have to be neutralized to avoid acid-catalyzed isomerization or dehydration of the resultant alcohols. If hydroxide is employed as the nucleophile, then elimination is an issue. The primary bromide will likely undergo SN2 displacement with hydroxide to give the primary allylic alcohol. As to the secondary bromide, elimination is a major concern. Ideally, acetate would be a better nucleophile to optimize substitution and repress elimination. Saponification of the acetate esters will liberate the alcohols.
The achiral primary bromide will afford achiral primary alcohol and racemic tertiary alcohol. There is little mechanistic information to be gleaned here. The secondary bromide, if a racemate, will provide racemic products. The secondary bromide is a more interesting case and one that can potentially provide useful information. Ideally, I would employ tosylates as opposed to bromides because they are formed from alcohols that can be prepared as their pure enantiomers. Your secondary bromide R,Z-1 would yield S,Z-2 by SN2 displacement. Syn SN2' addition affords R,E-3 while anti SN2' addition provides S,E-3. In the event that solvolytic conditions are employed, i.e., buffered aqueous acetone, racemic 2 and 3 would be expected barring solvent cage effects. SN1 reactions would proceed via cation transoid-4 and not cisoid-5, which has a syn-pentane interaction.


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Nomenclature of K2(PdCl4)
In the compound $\ce{K2[PdCl4]}$, the oxidation state of Pd is +2 so as per the  nomenclature rules, the name of the central metal inside the coordination is simple if it's a cation.
So if we name the compound it comes to be potassium tetrachloridopalladium(II) but in the answer it is tetrachloridopalladate(II).
Can anyone help me understand where I am going wrong?

It is an anion, not a neutral complex. If it was neutral, it would be IV, not II.

@poutnik can you please show how it's an anion ?

Any one of them works

Like if we say the O.N of Pd is x

Then +1(2)+x+(-1)4= 0 so x = +2

It's a cation then

@Poutnik I didn't get this

Imagine $\ce{Pd^2+}$ coordinates 2-4 $\ce{Cl-}$, having the overall charge 0 .. -2

As it has negative charge, its name ends -ate, not -ium.

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$\ce{K2PdCl4}$ consists of two entities: $\ce{K+}$ cation and $\ce{[PdCl4]^2-}$ anion. This anion is a complex anion and is named accordingly. The rules for formulating and naming coordination compounds are presented in Chapter IR-9 of the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book). An abridged version is included in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.
Names of anionic coordination entities are given the ending ‘ate’. The oxidation number of the central atom may be indicated by a Roman numeral in parentheses to the central atom name; alternatively, the charge on the coordination entity may be indicated in integers. While writing the complex, the cation is placed before the anion and no individual charges are shown.
The systematic name of the entire compound is formed by combining the name of the electropositive constituent, cited first, with that of the electronegative constituent, both suitably qualified by any necessary multiplicative prefixes. The two parts of the name are separated by a space.
So, the systematic name of the complex is potassium tetrachloridopalladate(II) or potassium tetrachloridopalladate(2-) . This roman numeral is the oxidation state of palladium which is +2 and when writing in terms of charge, the overall charge of the entity is given, in this case, it is -2. Also, do note that the ligand name ‘chloro’ is obsolete. According to current IUPAC recommendations, the correct ligand name is ‘chlorido’.
Read IUPAC name of a salt made of both cationic and anionic complexes for more information.


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Research-backed articles on cast iron seasoning
I've always seasoned my cast iron pans with vegetable oil because that was what I was taught. I'd done some research before, and even came across this StackExchange post on asking the best oil for the job.
To my dismay however, when I tried doing some research myself, all sources pointed to blog posts, particularly the Sheryl Canter one. There is not a single source on her post and she doesn't seem to be very near a materials scientist. The closest paper I could find to the topic was this one. Other than that, it was all anecdotal evidence, many people claiming many different things.
It felt almost absurd that I couldn't find a paper on something that seemed relatively common to me. This will bug me till the end of time unless I figure it out, so I need to know - does anybody have any real sources on cast iron seasoning or what factors play a part in the polymerization of different oils? Thanks in advance.

Google "Thermal Polymerization of Drying Oils" which should get you started. Most of the papers are behind firewalls and I don't have access.

There is this research on Chinese equivalent of seasoning process called "Kitchen God's Blessing" [1].
They argue that the non-stick property of seasoning comes by iron oxide $(\ce{Fe3O4})$ nanoballs.
If they are correct then the oil or its polymers is not exactly what is causing the non-stick property but the those nanoballs on the surface. Their seasoning method is somewhat different however. It uses temperatures up to 450 °C.
Reference

Gao, C.; Yang, N.; Li, C.; Wang, X.; Yu, X.; Zhang, L.; Wei, Z. Seasoning Chinese Cooking Pans: The Nanoscience behind the Kitchen God’s Blessing. Nano Materials Science 2020. DOI: 10.1016/j.nanoms.2020.06.001.

There is 'seasoning' to protect the pan, but on heating the pan, the chemical breakdown products are 'some' things I would not recommend to anyone for continuing good health for yourself or your family (see, for example, Genotoxic and carcinogenic risks associated with the dietary consumption of repeatedly heated coconut oil).
Unfortunately, many oils do breakdown even at low heat, and some at high temperatures, which includes even my favorite healthy (especially when acquired fresh from its harvest) Olive Oil!
However, a safer alternative path may yet exist to resurrect ourselves (but not likely literally).
Electrochemistry to the rescue!
In particular, try employing a sacrificial anode, like a small piece of zinc metal sitting in distilled water plus sea salt, covering your entire pan. Yes, to be explicit, my recommendation on how to best season your iron pan to guard it from future corrosion is zinc metal and saltwater free of oxygen, which unlike an oil coating, is not easily subject to removal or chemical disruption (from air/light exposure).
On the mechanics, your iron metal will no longer be the most anodic metal present, and the zinc may undergo limited anodic corrosion resulting in white ZnO/Zn(OH)2 product. However, even the latter can be consumed in low doses daily whereas, one must be more mindful of iron intake especially with small children. One has also, at low cost and safely, eschewed the dangers of associated heat generated cariogenics from various organics.
Aesthetically, I would also challenge you to compare the results from two iron pans, one stored in the manner I have suggested, to the other pan that was coated in whatever oil-based seasoning formulation you may believe is suitable, after washing both pans.

I think you miss the point. First the seasoning isn't mainly to protect the iron from rusting but rather to get a non-stick coating. Second, the oil polymerizes on the pan, thus it isn't the same as whatever oil you started with.

MaxW:  Is that the same, to quote per my source,  "polycyclic aromatic hydrocarbons (PAH), some of which have been reported to have carcinogenic potential. Consumption of these repeatedly heated oils can pose a serious health hazard. ", which sort of makes my point!

Interesting, I've seen conflicting evidence on the breakdown of oils. I have seen that repeatedly heating oil does release carcinogens, but this study and this one both suggest olive oil is one of the safer oils at high heats.

The dangers of cooking are greatly exaggerated and irrelevant to this question. Most every theoretical worry (PAH, nitrosamines, acrylamide...) has proved irrelevant in the real world. And the oil used in "seasoning" the pan is polymerised to an extent that means it does not contaminate anything (this is very different to repeatedly reusing the same oil to cook).

@matt_black proven where? and by whom? 'the oil used in "seasoning" the pan is polymerised to an extent that means it does not contaminate anything (this is very different to repeatedly reusing the same oil to cook).' high polymers contain free radicals (i.e. oxidized monomers) and can be digested by lipases, which means that yes, you are actually harming your cells DNA as those polymers are released on micro level almost during every cooking process, even more so during cooking with acidic ingredients.

@bodhihammer Did you know that you are exposed to many free radicals when you breathe? The oxygen in the air is a diradical. Are we all about to die because we breathe? My point it that your theoretical concerns don't show up in epidemiology stats. And it is upside down to demand that perfect safety is "proved"; the onus is on the people proposing theoretical worries to demonstrate harm (with epidemiology). They haven't, despite many theoretical worries.

@matt_black Of course the extent of radicals is important. In comparison to something like Teflon, cast iron pan seasoning releases of a lot  more of its coating, and that by daily use. It would have to be first properly measured. "My point it that your theoretical concerns don't show up in epidemiology stats." What epidemiology? There was no long-ranging study nor meta analysis conducted yet on that matter. It is simply an unexplored subject. We can only have some indirect evidence. It is nearly impossible to ascertain the structure of seasoning coating let alone draw any conclusions today.

@bodhihammer How do you know a seasoned cast iron pan releases more of its coating? Has it been properly measured? And do you even have any evidence that what is released contains anything harmful? Cast iron pans have been in use far longer than pans with teflon non-stick coatings. Are there studies showing people who cook using them have high rates of specific diseases?

@matt_black that is why I wrote that it hasn't been measured, because there was no research on the topic yet. "And do you even have any evidence that what is released contains anything harmful?" Absence of evidence is not evidence of absence. Do you know how scientific method works? "Are there studies showing people who cook using them have high rates of specific diseases? " Are there such studies? Can you name one? There hasn't been. The industry been interested mainly in Teflon, and that was successfully researched. Teflon is well-studied contrary to cast iron seasoning, which it's not.

@matt_black "How do you know a seasoned cast iron pan releases more of its coating?" By common observation, which is Empirically. Yes–it is only inductive kind of argument, but it is possible to chip off large swathes of the coating in a single cooking session using either acidic ingredients or a combination of ingredients.  This is not the case with cooking anything with the use of Teflon, unless you scratch the surface with metals but this is not that argument. This "Cast iron pans have been in use far longer than pans with teflon non-stick coatings." is also not a deductive argument.

Well, I came here in search of the same thing so it looks like we are the leading researchers in the field of pyrrolic polymerization on antique cookware. (Frowny face) I similarly reject 100% of the dot com heroes' wisdom available. It's garbage mixed with myth.
Here is what I know: Researching the chemical properties of cooking oil under pyrrolic polymerization is a start. It will teach you a lot but you could get a Ph.D. in the field with how much learning is out there.
I will do my best to summarize what I've learned in the process in the name of comparing notes.
There are three main types of oil we are talking about. Saturated, monounsaturated, and polyunsaturated
Saturated is a solid at room temp because the Vanderwall forces on long straight chains of carbons and no double bonds. Lard or coconut oil.
Monosaturated have one double bond so they have one kink. These a liquid at room temp. Olive oil
Polyunsaturated therefore are the kinkiest of the groups having more than one double bond. These polymerize very well but are very reactive and often aren't suitable for cooking due to their tendency to oxidize and go rancid at temperature. Grapeseed
The science says the presence of double bonds creates the possibility of covalent cross-linking which is a powerful and permanent bond. It's also impervious to corrosion under natural conditions.
The presence however of long uniform chains of carbon in unsaturated oils allows for uniform sheets of patina. So each has its place and none are "best"
Those are things I know. Here is what I've found to be the case from experimentation:
Lard makes a good low-maintenance patina but I find that it isn't as durable as I'd like in some cases. Particularly in the presence of tomato sauce. In some cases, I've found that I'm back to bare metal on the edges just through normal cooking.
Grapeseed creates an incredibly hard patina that I've seen chip under mild abuse. Worse, its polymer is lipophobic so it resists additional layers and creates a patchy, unreliable patina. The more layers, the worse the effect gets until the oil just beads up and rolls off preventing any additional layering. The science tells us that there are cross-links that are awesome but in practice, I've found this patina to be sticky and hard to cook with owing to the thinness of just one layer in many places. I tried hard to love grapeseed but I don't.
The best luck I have had is a mix of 1:1:1 unsaturated, mono, and polyunsaturated oils. So coconut, olive oil, and grapeseed seem to bridge the gap between the weaknesses of each oil and have given me the best results yet.
In theory, I am using the unsaturated fats to fill in the inconsistent holes of the poly and mono crosslinks and reduce the available oxygen for the more reactive oils to rancidify. I have also found that the unsaturated patina isn't nearly as lipophobic so I can make several layers and get a very smooth, hard, durable surface that food doesn't stick to.
Lastly, the temp I use has proven to be the most crucial part. Too hot, the patina cracks and falls off like dry paint chips, too cool and it's sticky. I go 5 degrees over the smoke point of the highest smoking oil. Then I leave it in until I notice the room is less smokey. This step has been a hard-learned one because there is a lot of crappy advice out there.
I would love to see some actual documentation on this so if you find it let me know. Good luck


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Lewis structure of dinitrogen trioxide
I have a confusion related to the structure of $\ce{N2O3}.$ I was taught 1 is the correct one, but I believe 2 is better since it has no formal charge on it. Why is 1 correct and 2 wrong?

https://chemistry.stackexchange.com/questions/72263/what-is-the-criteria-for-determining-structure-of-a-molecule?r=SearchResults ... https://chemistry.stackexchange.com/questions/91233/why-is-the-n-n-bond-length-greater-in-dinitrogen-trioxide-than-in-dinitrogen-tet?r=SearchResults

What really happens in nature has higher priority than our believe what happens or should happen. See Dinitrogen_trioxide with the structure, bond lengths and angles.
Be aware both $\ce{NO}$ and $\ce{NO2}$ are radicals with an unpaired electron.
$\ce{N2O3}$ being( $\ce{ON-NO2}$ ) as 2 paired radicals is the direct analogy
to the dimer $\ce{N2O4}$ being $\ce{O2N-NO2}$, both being formed at low temperatures.

So it’s just what it is?

@Rasputin Yes, it is what it is. Also keep in mind you're using a model, which is not the same thing as reality. Some times, models do not give the prediction they ought to give.

I think you assume that Nitrogen Sesquioxide is formed by a dehydration reaction of an acid of Nitrogen, hence necessitating the presence of a bridged oxygen. This isn't the case. It is created due to the addition of the radical on the Nitrogen of $\ce{NO}$ to the radical on the Nitrogen on $\ce{NO2}$. Hence why the structure is the way it is.


text:
Software to perform molecular dynamic simulation
I have got aggregate of 4 non-protein compounds I would submit for molecular dynamics simulation. The problem is I can't find software capable to do that. Amber, GROMACS, LAMMPS, all of them require structure to be built up from the aminoacids so the topology can be read. I've seen GROMACS has x2top that should be able to generate topology from non-protein structure. However, for my system it fails. What can I do?
Here are .xyz coordinates of the aggregate:
248
    Energy:    -128.8882289
N          2.94981        4.07926       -5.95549
C          1.55689        4.07222       -5.57866
H          0.95978        3.62638       -6.43298
C          0.93686        5.43505       -5.21651
C          0.87184        6.32909       -6.38603
C          1.67148        7.45272       -6.69880
N          1.24256        8.02487       -7.89049
C          0.19644        7.27018       -8.30869
N         -0.06225        6.22632       -7.42362
C          1.30522        3.12008       -4.38134
O          0.47546        2.19923       -4.49799
N          2.01969        3.35894       -3.21757
C          1.92802        2.65116       -2.01591
C          0.88945        1.74242       -1.73904
C          0.84547        1.06621       -0.51116
F         -0.18927        0.21946       -0.29679
C          1.82258        1.27431        0.49798
N          1.80287        0.69700        1.80611
N          1.39566       -0.45359        1.96381
C          1.34687       -1.02655        3.27192
C          1.52415       -2.43840        3.37979
F          1.81094       -3.17945        2.28364
C          1.42458       -3.09620        4.60909
C          1.13511       -2.37755        5.79381
N          1.02421       -3.08555        6.99322
C          1.04112       -2.55703        8.27589
O          1.18165       -1.33803        8.49824
C          0.83998       -3.56312        9.44452
H         -0.17494       -4.03912        9.27838
N          0.78281       -2.92516       10.73973
C          1.93934       -4.64304        9.37883
C          1.54272       -5.80248        8.55823
C          2.10713       -6.27545        7.34959
N          1.46038       -7.43692        6.94522
C          0.51317       -7.68025        7.88448
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C          0.94035       -0.98628        5.70659
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C          2.91419        2.87552       -1.02517
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H          2.51022        7.85384       -6.12959
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N          1.11328      -12.99119       -1.34556
C          2.03505      -10.97543       -0.74023
C          2.44727       -9.55973       -0.70751
N        -11.70170       -5.56231       -3.14774
H          3.57245       -9.50454       -0.72587
C        -10.39538       -5.93575       -2.82231
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H        -15.00747       -6.26815       -2.83416
C        -16.24979       -2.62698       -3.52327
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N         -5.88501       -7.91808       -1.55368
C         -3.87925       -9.32440       -1.64388
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F         -4.56957      -10.23766       -2.36676
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O          0.35611       -6.94553        0.32760
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F         -4.23242       -6.02796       -0.06245
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C        -10.08912       -6.84769       -1.79487
H          2.07812      -14.10928        0.27181
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C          1.90838       -8.82060        0.53509
N         -0.46010       -9.03051       -0.24672
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H         -2.12268      -10.56741       -1.65501
C        -14.40721       -4.19212       -2.54285
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H         -0.19493       -9.97937       -0.43360
N          2.86803       -7.88077        1.06569
H        -13.69657       -3.43595       -2.97421
H          3.09956      -11.82715        0.90716
N        -16.90981       -4.19498       -2.05957
H        -19.01808       -3.61296       -2.11480
O        -12.90229       -7.09102       -1.88528
H        -10.88203       -7.32042       -1.19068
H        -16.92146       -4.93886       -1.40693
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H          2.46898       -7.36611        1.82784
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H         18.09012        9.42357       -1.81836
F         16.48163        1.39387       -2.87539
H          7.96220       -4.06105       -3.51819
H         17.60268        3.68354       -2.56623
N         17.92323        8.43985       -1.73935
H         18.68774        7.93896       -2.14905
C          7.37441       -4.57902       -2.75953
H          9.42815       -6.32849       -3.18884
C         16.75252        3.51753       -1.88275
C         16.06761        2.30160       -1.96011
H         17.92140        5.71877       -1.46626
N          6.15726       -4.08817       -2.30354
C          7.67815       -5.80012       -2.11118
C          8.83234       -6.69634       -2.31074
N         17.09147        5.70197       -0.90756
H          8.45585       -7.72472       -2.57522
C         16.36409        4.50931       -0.94985
C         17.71875        8.06851       -0.35999
C         14.95253        2.01848       -1.11633
C          5.71401       -4.98725       -1.39019
N         14.35655        0.72269       -1.21801
H         21.61587        8.58107       -0.75362
H         17.19055        8.92234        0.16625
C         21.09462        9.17140        0.00018
C         16.74102        6.87002       -0.24821
N          6.60605       -6.04649       -1.24540
N         21.60136       10.37175        0.48346
C         12.54069       -0.70153       -1.10751
N         13.14748        0.58850       -1.03360
C         11.16073       -0.75363       -1.47075
H         10.79779       -8.38500       -1.72194
F         10.49608        0.37743       -1.80507
C         13.15822       -1.93853       -0.78035
C         10.45825       -1.96102       -1.50679
O         12.02403       -5.92934       -1.08952
C         12.45238       -3.15052       -0.80187
C         11.09434       -3.18059       -1.17037
F         14.45681       -1.97792       -0.40174
C         10.80784       -5.65638       -1.12525
C         15.27807        4.24224       -0.09560
H          4.78194       -4.91884       -0.81412
C         19.85964        8.88077        0.62505
C          9.73569       -6.78130       -1.06352
N         10.33617       -4.35316       -1.19072
C         14.58375        3.02684       -0.18691
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H          9.39657       -1.93873       -1.80486
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H         12.98913       -4.07196       -0.51983
H          9.35005       -4.23241       -1.32787
N         10.30574       -8.09865       -0.89693
H         19.56647        6.90777       -0.07566
H          6.50934       -6.82415       -0.64051
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H         20.77041       11.74227        1.97621
O         15.70174        6.99012        0.42661
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H         18.66052        7.30605        1.44630
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H         -9.18708       10.67465       -0.82679
N        -13.17375       13.38157        0.17397
H        -12.40898       13.52486       -0.45771
C         -7.26280       -2.38041       -0.19256
H         -4.92758       -1.46248       -1.31115
C         -9.45404        9.87889       -0.11042
C         -8.75922        8.66874       -0.17804
H        -10.77948       11.98996        0.22485
N         -8.11638       -3.21139        0.52310
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C         -4.67369       -2.10453       -0.42557
N        -11.12066       11.31524        0.88069
H         -4.00416       -2.93001       -0.79885
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C        -12.71357       13.14074        1.52107
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N         -8.22918        6.42421        0.60687
H        -10.97501       16.64039        0.56276
H        -13.61494       13.20121        2.20653
C        -11.72035       16.60868        1.35758
C        -12.18298       11.69437        1.68630
N         -5.96893       -3.74749        1.02909
N        -12.42268       17.73783        1.76153
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N         -8.66644        5.32269        0.93868
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H         -2.17063       -1.22446       -0.39021
F         -9.87124        2.84337        0.74802
C         -6.42644        4.11547        0.87959
C         -7.81553        1.69236        0.74875
O         -3.58659        1.12673        0.30641
C         -5.71718        2.90526        0.84970
C         -6.40005        1.67698        0.77865
F         -5.70416        5.25640        0.96629
C         -4.39806        0.21018        0.54285
C        -10.77381        9.02711        1.73940
H         -7.64584       -4.80706        1.94962
C        -12.12509       15.48849        2.12004
C         -3.92767       -1.26917        0.63595
N         -5.74810        0.44199        0.76341
C        -10.07810        7.81061        1.66405
C        -13.24784       17.32506        2.75528
H         -8.39188        0.75356        0.69354
C        -11.63630       14.09952        2.06136
H         -4.61552        2.94570        0.88616
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N         -2.49355       -1.41802        0.53990
H        -10.72396       14.07694        1.40613
H         -5.19146       -4.22413        1.41527
N        -13.09845       15.96346        3.00678
H        -13.95622       17.95292        3.31041
O        -12.70838       10.94446        2.52936
H        -11.55643        9.13140        2.50996
H        -13.58224       15.42900        3.68464
F        -10.40171        6.84946        2.56012
H         -4.21459       -1.62930        1.67173
H         -2.03679       -0.78457        1.17047
H        -11.33159       13.74988        3.08562

The MD software you mentioned were developed for biochemical problems. Other software, eg NAMD, DL-POLY, Desmond are able to perform calculations on other kind of systems, too. First I would figure out what kind of forcefield you need for your model and then check where they are implemented. (Maybe you want to briefly describe your problem. Most contributors here will not have time to download and visualize the xyz file just to figure out what you are talking about)

I told what I need - to achieve most probable relative positions of each of four compounds belonging to the one aggregate compound.

I personally think this question is on topic here, however, you might want to try a more specialist community to ask it: [MatterModeling.se]. (Since you have already gotten an answer, I can't recommend asking it there again and deleting it here. We generally discourage cross-postings though. But maybe you can browse their questions to find some more relevant information for you.)

In this case, the problem is not the software really, it's the topology of your molecules. This is a hairy problems, but there are some tools that you can use.
To generate topologies, look into acpype for Amber-based topologies and LigParGen for OPLS-AA-based topologies. However, note that the performances might vary, since these tools are very generic and do not guarantee solid results for all types of molecules. Nevertheless, these will allow you to run simulations (these tools offer topologies in formats for different softwares; GROMACS is free and relatively easy to use, so I would try that.)
If you want something a bit more solid and you have computational resources available (or a good desktop), try xtb, which is a semi-empirical tight-binding software. Its accuracy is somewhere between DFT and older semi-empirical methods, and it is fast. You can run molecular dynamic simulations or use their tool for conformational sampling called CREST.

There is at least one well-known third-party topology generation service for Gromacs and that is the PRODRG server maintained by the van Aalten group at the University of Dundee.
The Gromacs site though states a caveat:

The charges and charge groups generated by PRODRG are typically not consistent with functional groups that were derived more rigorously for the force field of interest. It is therefore the task of the user to validate any parameters obtained, and to modify them, if necessary, since the artifacts of not doing so are now published.  A thorough understanding of the underlying force field theory is recommended.  See the parameterization page for more details.


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Non-computational theoretical chemistry
Theoretical chemistry is nowadays almost synonymous with computational (and usually quantum) chemistry.
Are there any examples of recent (let's say post 2000) research in chemistry that is theoretical (i.e. non-experimental) but does not use extensive computations (i.e. HPC), does not aim to produce better computational methods, or does not investigate mathematical foundations of computational chemistry methods?
Something I imagine can exist are models or model Hamiltonians in quantum chemistry that are analytically solvable (or with minimal help from computers), or abstract mathematics like topology applied to chemistry.

Since practically no Hamiltonian can be solved analytically (ok, H aside), all you left is computers. Topology is a nice branch of mathematics, but I haven’t seen any relevant applications. Graph theory is your only shot, it may have new applications, mostly in chemoinformatics or such.

maybe something like https://aip.scitation.org/doi/10.1063/1.5002894?

@Greg my question was actually motivated by a graph theory question at the Matter Modelling SE https://mattermodeling.stackexchange.com/questions/1360/what-are-the-applications-of-chemical-graph-theory
So I was curious about other examples.
By exactly solvable Hamiltonians I mean 2-state models, BCS Hamiltonian... that reveal the essential physics even though they do not give qualitative answers and are popular with physicsts. I heard a joke that says "Chemists seek approximate solutions to the exact Hamiltonian while physicists seek exact solutions to approximate Hamiltonians."

@LukasK The theoretical physicists I know are spending considerable amount of time in front of HPCs and doing tons of numerical work. So that joke maybe were true 30 years ago, I am not sure it is true any more. Also, if a Chemist solves a toy Hamiltonian eg for spectroscopy, generally everyone (especially the Physisists) say it was Physics, just accidentally were applied by a Chemist.

I really don't mean to toot my own horn, but I'm curious whether something like this counts.

Maybe the quantum theory of atoms in molecules?

I think I can answer this in the affirmative. There are articles which find connections between abstract mathematics and chemistry, sometimes even bypassing physics altogether. Of course, these kinds of articles are considerably rarer, but they're sprinkled out there.
I can think of two articles which I'd love to discuss, but I literally do not have the necessary background, it just goes way over my head. The first one is Quantum Interference, Graphs, Walks, and Polynomials, Chem. Rev. 2018, 118, 10, 4887–4911. This is some rather pure graph theory which is not related to cheminformatics. In particular, I find it interesting that the connectivity in azulene is comparatively unusual, and this probably is deeply connected to its unusual photophysical properties. And then there's The Rouse Dynamic Properties of Dendritic Chains: A Graph Theoretical Method, Macromolecules 2017, 50, 10, 4007–4021, more graph theory with a little bit of physics, and whose content eludes me entirely.
Surely some amount of computer assistance was employed in deriving the results for these papers, because there's no reason to give up that tool. However, I don't think this counts as "extensive computations" in the sense you may have had in mind.

The answer is yes.
There is work in non-equilibrium thermodynamics that is purely theoretical, e.g.:
Jarzynski, Christopher. "Rare events and the convergence of exponentially averaged work values." Physical Review E 73.4 (2006): 046105.
https://journals.aps.org/pre/abstract/10.1103/PhysRevE.73.046105
Jarzynski, Christopher. "Fluctuation relations and strong inequalities for thermally isolated systems." Physica A: Statistical Mechanics and its Applications 552 (2020): 122077.
https://arxiv.org/pdf/1907.09604.pdf
You may recognize Jarzynski as the creator of the famous Jarzynski equality: https://en.wikipedia.org/wiki/Jarzynski_equality
And some of the work on phase transitions and criticality certainly meets your criteria.  Here you'll see papers whose principal purpose is to develop better theoretical models to explain these phenomena and, to the extent computation is used, it's to test those models.  But the heart of these papers is the development of the theoretical models themselves.  Here's one example:
Goodrich, Carl P., Andrea J. Liu, and James P. Sethna. "Scaling ansatz for the jamming transition." Proceedings of the National Academy of Sciences 113.35 (2016): 9745-9750.
https://www.pnas.org/content/113/35/9745.short
Finally, if you'll accept work on biological macromolecules as falling within the realm of chemistry, there is also work in theoretical chemistry/theoretical biophysics that meets your criteria, in which theoretical models are developed to predict and explain the properties of macromolecules, and the models are then tested either computationally or experimentally:
Yoffe, Aron M., et al. "The ends of a large RNA molecule are necessarily close." Nucleic acids research 39.1 (2011): 292-299.
https://academic.oup.com/nar/article/39/1/292/2409062
Chakrabarti, Shaon, Christopher Jarzynski, and D. Thirumalai. "Processivity, Velocity, and Universal Characteristics of Nucleic Acid Unwinding by Helicases." Biophysical journal 117.5 (2019): 867-879.
https://www.biorxiv.org/content/biorxiv/early/2018/07/10/366914.full.pdf


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What would the van der Waals parameters look like for gaseous metals?
So basically I'm trying to do some calculations on laser based machining, and it involved vaporizing a small mass from a target metal workpiece. I was assuming that a certain m mass of the metal vaporizes nearly instantaneously, and the resulting vapor expands from that initial volume.
I was using the van der Waals corrected gas equation, and wasn't sure how to quantify a and b for gaseous metals (take Al for example). I tried searching online, but couldn't find any values for gaseous metals.
Any help or resources I could refer to would be appreciated.

My first question would be why not treat it as an ideal gas and be done with it? The metal vapor is highly non-equilibrium to begin with, but will rapidly mix with the ambient and become dilute enough that any VdW parameters are irrelevant.

Not sure what you're trying to model, but the metal atoms will only be a gas for a fraction of second. As the hot gaseous metal diffuses into air it will probably oxidize and form  particulates.

@JonCuster I want the instantaneous pressure when the gas forms, because I'm using that to model the reaction force on the workpiece caused by the pressure exerted by the gas rapidly expanding. Mixing will indeed happen, but I'm attempting to see the part before that.

@MaxW Yeah I want to see the pressure during that fraction of a second for my model (since logically that should be the time when the pressure exerted by the gas is highest on the remainder of the workpiece)

Well, you get quite a nice shock wave just from rapidly heating a small surface volume with the laser without any vaporization, which complicates things further. That is a bigger effect than using VdW vs ideal gas parameters.

Aside from the metal vapor oxidizing, the pressure of gas over your weld is unlikely to be much different from one atmosphere whereas the critical pressure of almost everything is 10 atmospheres or (often a lot) more.  Generally gases become non-ideal when the pressure is a large fraction of the critical one; we are well below pressures where that occurs.  So the Ideal Gas Law works -- until the metal vapor oxidizes.

But wouldn't the instantaneous pressure of the gas be pretty high (Considering that it is in gaseous state, but within the volume of the solid state)? It will indeed expand rapidly and cause the pressure to equalize with the external environment, but I wanted the pressure at that instant of gas formation.


text:
Is there a difference between oxidation state and valency?
I just looked up peroxide on Google, and I saw that the oxidation state of oxygen in $\ce{R-O-O-R}$ is $–1$, and not $–2$ (like $\ce{H2O}$), when I could clearly see that each O atom forms two bonds. Did I get either of the concepts wrong?
I think that's probably the case but, can someone explain where I went wrong? To me, Valency dictates the no of bonds that an element can have. In the structure, I can clearly see that each oxygen forms $2$ bonds so, if I remove the other atoms, won't the imaginary charge left still be $-2$? Then how, does $-1$ come into play?

Does this answer your question? An introduction to "oxidation state"/"oxidation number"

The valency is the number of bonds. The degree of oxidation is the number of charges that are left on an atom once all valence bonds between different elements are replaced by an ionic bond. Oxygen has two valences in the molecule H-O-O-H. But if you don't touch O-O bonds, and replace OH bonds by charges, it produces a positive charge on H, and a negative charge on Oxygen. The oxidation state of Oxygen is $-1$.

See, where does the imaginary charge come from? It must be electrons stolen from some other atoms. What atoms?

@ Ivan. The imaginary charge comes from the bond. A bond between two atoms A and B is made of 2 electrons coming each from one atom (A and B). If the two electrons are not used for making a bond, but are fixed on the most electronegative atom (let's say it is B), A looses one electron, the valence bond disappears, and is transformed into a positive charge on A, and a supplementary negative charge on B.

@Maurice Sorry for not pointing out that my question was not for you, but for the OP.

I think it will help you understand if you know the historical context. The concept of valency existed well before chemical bond, electrons, protons, etc. were known. It is a primitive concept from late 1860s. It referred to the capacity of an element to bind with hydrogen or any other element.
For example, if we say carbon has a valency of four, all it means that it can combine with 4 hydrogen atoms. If we have say oxygen has a valency of 2, it means it will combine with two hydrogen, the famous H$_2$O! Does this make it clear? One would determine valency by pure elemental analysis. If I have FeCl$_3$, iron has a valency (=combining power) of 3, because it can combine with three chlorine atoms. I am avoiding the word ion, because ions did not exist then! The valency of oxygen in peroxide is still 2 because each oxygen atom is combining with two atoms via single bonds. In this case, the other atom is another oxygen itself.
Now read the Wikipedia article Valence. Note that I did not assign any sign to valency
Oxidation number or state is a newer notion which came about when electrons were very well recognized to play a role in bonding and the atomic structure become relatively well understood. I quote it's definition from the OED

a number representing the degree of oxidation of an element in a
particular ion, molecule, or substance, being a notional electric
charge equal to the number of electrons lost (or gained, if the number
is negative) by an atom of the element in forming the species or
substance, all bonding being assumed to be completely ionic

Oxidation state/ number has a sign associated with it. Sometimes, valency and oxidation number can match in terms of their absolute values.
NH$_3$
Valency of nitrogen= 3
Oxidation state of nitrogen = -3


text:
Ways to determine the concentration of alkyl halides in organic solvents
I have a volatile alkenyl bromide dissolved in hexanes. I have to use the halide in a Grignard reaction but separating it from hexanes is hard (the bromide is light sensitive and very volatile). I would like to use the solution of the halide in hexanes but I don't know how to determine its concentration. I tried taking an NMR of the solution but all the peaks interfere and I can't choose two characteristic peaks to take a ratio. Is there a titration technique or some other technique to determine its concentration?

It'd be very helpful, if you are able to provide the structure of your alkenyl bromide. Also, how accurate it should be for the determination of the concentration?

For NMR you only really need one distinct peak of the alkyl bromide and one distinct peak of the internal standard. The latter can even be chosen so that its peak has a resonance frequency far away from your sample’s peaks. Is even that not possible due to spectral overlap?

Change the solvent to cyclohexane?

You could form the Grignard and titrate it. The Grignard should be fairly straightforward to titrate (e.g. as described by Knochel). This also has the advantage of giving you precisely the concentration of your reactive specie. If you knew only the concentration of your alkenyl bromide, you would not know precisely the quantity of Grignard formed (as the reaction is not always perfect).

The alkenyl bromide is likely allyl bromide, which has a boiling point of $\pu{71 ^\circ C}$, very close to that of mixed hexanes: $\pu{65-70 ^\circ C}$. Yes, those would be difficult to separate by distillation. Higher alkenyl bromides have boiling points higher enough to enable easier separation.
Two other possibilities present themselves, with reasonable accuracies possible, if you are moderately careful. First, allyl bromide has a density of 1.4697; hexane has a density of 0.6606. If you determine the density of the mixed product to $\pm 2\%$, you can determine the concentration to about the same accuracy - assuming linear concentration vs density, which is likely to be true, since hydrogen bonding or other special attractions do not seem evident.
A second method would be to determine the refractive index, which can be found to 4 decimal places - therefore perhaps more accurate than $\pm 2\%$. Allyl bromide has a refractive index of 1.4697 while hexane has a refractive index of 1.375. Refractive index of mixtures has been linear in my experience, tho I have no literature reference for that.
It is good practice to determine the concentration of your reactive ingredients by a method without side reactions. Unfortunately, running a test reaction to see how much Grignard reagent is formed may have an undetermined amount of the side reaction:
$$ \ce{CH_2=CHCH_2Br + CH_2=CHCH_2MgBr}$$ -> $$\ce{CH_2=CHCH_2CH_2CH=CH_2 + MgBr_2}$$
so it may not be easy to quantify what you are going to do next.


text:
Three-membered ring pathway bromination of alkene
Bromination of alkene occurs by the mechanism as shown (source: Master Organic Chemistry — Bromination of Alkenes: The Mechanism):

Here $\ce{Br^-}$ attacks bromonium ion from a side which is more hindered to form Markovnikov product. But shouldn't it attack on the less hindered carbon as we do in case of reaction following SN2 mechanism to form anti-Markovnikov product?
For example, in ring opening of epoxide in basic medium attack by $\ce{CH3O^-}$ occurs on the less hindered carbon (source: Lumen's Organic Chemistry 1: An open textbook — 9.6. Epoxide reactions):

Why does the same doesn't happen in case of bromination of alkenes and oxymecuration of alkenes?

While the process involves the formation of a cyclic bromonium intermidiate, with positive charge residing on bromine, it may not be completely ignored that during the opening of the ring, a partial positive charge will develop on either carbon atom. Since, the partial positive charge is more stable on the -CH3 attached carbon, formation of that is favoured. Similarly, it is formed when attack happens on the more hindered sight. Moreover the rate determining step is the bromonium ion intermidiate formation. See Hammond Postulate.

@AdityaRoychowdhury the partial positive charge is also more stable on tertiary position in case of ring opening of epoxide then why the does the same does not apply to epoxides

https://chemistry.stackexchange.com/q/100788/16683 and generally the same explanation is invoked for acid-catalysed epoxide opening too as the transition state has cationic character. Base-catalysed / promoted epoxide openings are different.

Generall, epoxide opening occurs by acid and basic means. Since the nucleophile is neutral the ring openong step is the rate determining step and more stable carbocation forms like Sn1. But in basic medium the nucleophile is stronger negatively charged. Hence the rate determining step invoves simultaneous ring opening and necleophilic attack like an Sn2 reaction. Here steric effect prdominates. The bromination is completely different. In that the bromine is leaving. It is a much better leaving group and is also larger than Oxygen.

Comparing epoxide and bromonium is a bit ambiguous.

A very good exercise is to think about how one would even go about designing an experiment to show that the addition of bromide to a brominium ion proceeds with anti-Markonikov selectivity. Also, does it even matter given the nature of the products you will generate?


text:
Anomalous boiling point of "iso-" alkanes
I was attempting to compare the boiling points of iso-hexane and 3-methylpentane.
Boiling points of organic compounds depend upon three factors according to Master Organic Chemistry – 3 Trends That Affect Boiling Points. They are:

The relative strength of the four intermolecular forces is: Ionic > Hydrogen bonding > dipole dipole > van der Waals dispersion forces. The influence of each of these attractive forces will depend on the functional groups present.

Boiling points increase as the number of carbons is increased.

Branching decreases boiling point.

The number of carbons are same in both 2-methylpentane and 3-methylpentane and so that trend doesn't make a difference. Since there are no functional groups present, the only force acting between two molecules would be van der Waals dispersion forces and this depends upon the surface area of the molecule.
3-Methylpentane is more symmetric than 2-methylpentane and so would form a more spherical structure than iso-hexane. For a fixed volume, the surface area decreases the more spherical the shape becomes. Due to this, the more symmetric molecule (3-Methylpentane) should show a lower boiling point.
However, the boiling point of 3-methylpentane is $\pu{63 ^\circ C}$ whereas for isohexane it is $\pu{60 ^\circ C}$ which contradicts the above statement, according to which iso-hexane should have had the greater boiling point.
Thinking there was a flaw in my assumptions, I compared the boiling points of 3-methylheptane and 4-methylheptane.
This showed the same trend as my statement provided. 3-Methylheptane has a boiling point of $\pu{120 ^\circ C}$ and 4-methylheptane has a boiling point of $\pu{117 ^\circ C}$.
Going further, comparing 4-methylnonane($\pu{165.7 ^\circ C}$) and 5-methylnonane($\pu{165.1 ^\circ C}$), we see that the difference between the boiling points decreased to $\pu{0.6 ^\circ C}$ from $\pu{3 ^\circ C}$ but still can be predicted using the above statement.

Update
With reference to J. Am. Chem. Soc. 1929, 51 (5), 1540–1550, the values of boiling point and density measured for the isomers (only taking straight chain and single branched isomers) were as follows for:

Hexanes

\begin{array}{|c|c|} \hline
\text{IUPAC Name} &\text{Boiling point }(\pu{^\circ C})& \text{Density}(\pu{g ml-1)} \\ \hline
n\text{-Hexane} & \pu{68.95} & 0.6595\\
\text{2-Methylpentane} & \pu{60.2}&0.6542 \\
\text{3-Methylpentane} & \pu{63.2} &0.6647 \\ \hline
\end{array}

Heptanes

\begin{array}{|c|c|} \hline
\text{IUPAC Name} &\text{Boiling point }(\pu{^\circ C})& \text{Density}(\pu{g ml-1)} \\ \hline
n\text{-Heptane} & \pu{98.4} & 0.6836 \\
\text{2-Methylhexane} & \pu{90.0} & 0.6789 \\
\text{3-Methylhexane} & \pu{91.8} & 0.6870 \\ \hline
\end{array}

Octanes

\begin{array}{|c|c|} \hline
\text{IUPAC Name} &\text{Boiling point }( \pu{^\circ C}) & \text{Density}(\pu{g ml-1)}\\ \hline
n\text{-Octane} & \pu{124.6} & 0.702 \\
\text{2-Methylheptane} & \pu{116.0} & 0.6985 \\
\text{3-Methylheptane} & \pu{122.2} & 0.707 \\
\text{4-Methylheptane} & \pu{118.0} & 0.722 \\ \hline
\end{array}

Decanes

\begin{array}{|c|c|} \hline
\text{IUPAC Name} &\text{Boiling point }( \pu{^\circ C}) & \text{Density}(\pu{g ml-1)}\\ \hline
n\text{-Decane} & \pu{174.0} & 0.730 \\
\text{2-Methylnonane} & \pu{160.0} & 0.724 \\
\text{3-Methylnonane} & \pu{166.9} & 0.735\\
\text{5-Methylnonane} & \pu{166.2} &0.732 \\ \hline
\end{array}

In all these cases, it seems as though the 2-methyl isomer (the "iso-" isomer) has the lowest boiling point.
Therefore, it seems as though the answer for the question about comparing the boiling point of 2-Methylpentane and 3-Methylpentane lies in the reason for why an "iso-" alkane has the lowest boiling point for an n-carbon alkane.

Is there a reason for the "iso-" isomer having the lowest boiling point in the single branched methyl isomers?
Why is the comparison of the boiling points of 3-Methylpentane and 2-Methylpentane anomalous? (This would be answered since 2-Methylpentane is the "iso-" isomer of hexane)

Page 937 of this ancient paper and references therein might contain an answer.

Actually 2-methylpentane can be considered as symmetric as 3-methylpentane - both have plane of symmetry, it's just not perpendicular, but overlaps with longest chain.

@Safdar Actually I agree with Mithoron. While the molecule may have a more compact circular shape, it apparently has lesser waan der vaals's forces. But the molecule is more symmetric and as a result the surrounding molecules may fit more easily. Now though this serves as a more suitable reason for melting point, the symmetery helps better interaction and the boiling point is higher. Moreover, you may see that the difference is not so high, only 3 degrees so this may be the reason.

@Aditya How do you decide the "apparently less van der waal" forces? that is the whole question.

Is there a similar effect in the density of the alcanes ?

@ Safdar. I had a tendency to think that if the density of one of these isomers is lower than another one, it may mean less molecular interactions in the liquid phase. So it may boil at a lower temperature. Who knows ?

I doubt (after you added density) that it's either the "attractive forces" or the "surface area" which is creating a small difference in BP. But, I guess it's more of a data-based than a subjective problem. Let alone MCQ's.

@RahulVerma A mix of the two or are you saying that there are more factors in play?

Safdar, it's pretty clear from the density values that either the volume or the forces are different or both, and hence the density (as mass is same). But, I couldn't find any data for surface area / van der waals forces...

There is a flaw anyway. More spherical implies also more tight packing. It seem there is a contradictory passage in your question. What exactly does your statement in bold suggest?

@Alchimista I might be wrong here, but tight packing is more of a factor for the melting point than the boiling point. However, the reason for the actual values of boiling point might be a mix of the two factors-van der waal's forces and more efficient packing, since liquid phase is an intermediate between solid and gaseous phase.

Safdat I repeat my question. What exactly the statement in bold implies? Then I might think on this issue. And about packing, you are correct. But still a certain "solvatation" works the same way. Note the quote because we have one phase.

@Alchimista I state that using the fact that van der waal forces decrease with a decrease in surface area and that the boiling point  of a substance decreases when forces of attraction decreases. Now, if it gets more symmetrical then the body becomes more spherical (the more efficient packing if you will) and so less surface area is available for the force of interaction, thereby lowering boiling point. That statement regards only van der waals forces since that was the only factor I could think of at the time. I also assume that van der waal forces are the only intermolecular forces acting.

Is the passage that you made I am looking for. The bold statement what does implies in your opinion? That 3-methylpentane boils at lower or higher T? And what is observed? There might even be a entropy term seen in the liquid phase for the molecule with a long fragment.

@Alchimista According to my statement 3-methylpentane should boil at a lower T. However the 2-methylpentane shows a lower boiling point in reality. Also how does entropy work here?

@Safdar entropy might work by 2-methylpentane having higher entropy content in liquid phase owing to its longer branch. I might write an answer with time. Nevertheless I thought of it after the question.

Let us continue this discussion in chat.

I suggest that an entropic term explains the little difference observed.
While this aspect is normally important to justify what isomer melts at lower temperature, in principle it can be invoked to justify why an isomer having a longer - not too much - branch boils at a bit higher temperature.
In this case, which might be rare as for a balancing between enthalpy factors it is needed to make this entropic term of decisive importance, the $3$ carbons chain can access a considerable space - both in physical as well conformational terms - precluded to the $2$ carbons ones in liquid phase.
Considering $\Delta G$ of the liquid to gas transition would imply a slightly higher $T$ for 2-Methylpentane.

why just the entropic terms, can we not use the fact stated by the OP that the packing of 3me- pentane would be better than 2 me-pentane  having a slightly higher effect on the vander waals attraction counter to the claim that it decreases the surface area?

I have also proposed that fact in a comment. But it would be better called "effective surface". The point is that should not be anomalous anymore but seen in most analogue terms. I am convinced that the more compact shape of the 2 isomer as compared to the more flexible propyl branch is responsible for a higher entropy content in liquid 2-methylpentane... @I am Alita


text:
Comparing the solubility of glucaric acid and mucic acid
Why is glucaric acid soluble in water but not mucic acid? The chemical formulae seem to be the same so why is it that one is more soluble than the other?

One seems to be symmetric(mucic acid) whereas the other is not.. Also what have you attempted to find the solution so far?

Symmetry might actually answer it, I looked at the structures again and mucic acid has a plane of symmetry so that might explain why it isn't soluble. Thanks! I wanted to know about their solubilities because in my biochemistry textbook they said that you can identify galactose in samples by oxidation as it forms insoluble crystals. So I wondered why glucaric acid is soluble.

Wikipedia lists the melting points of the sugar acids as:

Glucaric (or saccharic) acid: $\pu{125-126 ^\circ C}$

mucic (galactaric) acid: $\pu{230 ^\circ C}$.

The very great solubility of all the sugary materials (glucaric acid: $\pu{63 mg/mL}$1, or perhaps $\pu{912 mg/mL}$2) is due to their hydroxyl groups, which are very compatible with $\ce{H2O}$. Mucic acid, even though, it has as many hydroxyls as glucaric, is soluble only to the extent of $\pu{3.3 mg/mL}$, although it was predicted from its structure to dissolve reasonably well ($\pu{63.1 g/L}$3). (Note the differences in units; I used the same numbers and units as in the original reference.)
The likely reason for the great difference in solubility and melting point is the way the mucic acid crystal packs. This has shown up in density and x-ray crystallography4.
Looking at the structures, or playing with molecular models (typically, one molecule at a time - you never have enough atoms to make a half a dozen molecules, nor the time to rotate them all and pack them, nor the straight-line vision to see how well they pack) doesn't give you the same sense of difference that the solubility and melting point give.
Maybe this is not terribly unusual: tartaric acid(s), a simpler diacid, shows similar differences in solubility and melting point just because of the chirality of the halves of the molecule. Melting points: D or L: $\pu{171-174 ^\circ C}$; DL $\pu{206 ^\circ C}$; meso: $\pu{147/ 165 ^{\circ} C}$. Solubilities (T=?): D or L: $\pu{1.33 kg/L}$; DL: $\pu{0.21 kg/L}$; meso: $\pu{1.25 kg/L}$ (Wikipedia).
References

https://www.drugbank.ca/drugs/DB03603
https://pubchem.ncbi.nlm.nih.gov/compound/Glucaric-acid#section=Experimental-Properties
https://hmdb.ca/metabolites/HMDB0000639
The crystal structure of galactaric acid (mucic acid) at −147 °C: An unusually dense, hydrogen-bonded structure by George A.Jeffrey, Richard A.Wood, Carbohydrate Research, Volume 108, Issue 2, 16 October 1982, Pages 205-211, DOI: 10.1016/S0008-6215(00)81790-5

Thank you so much for providing the links! I understood from ref. 4 that hydrogen bonding in the crystal state of mucic acid is the reason for its decreased solubility, but that begs the question as to why the crystal structure of glucaric acid doesn't have hydrogen bonding to the same extent as mucic acid?

@Sid: You're welcome! But I gave you the answer: Make a dozen molecular models...or maybe do a crystal x-ray analysis...or maybe do it in your mind. Exercise those brain muscles. Or just look in awe at Nature's ability to make things interesting, and go on to a simpler problem.

Ill definitely try that, thanks a lot!


text:
Thermodynamics of micellisation

Micellisation is found to be spontaneous i.e. $\Delta G < 0.$ It's found that $\Delta S > 0,$ which was intuitive since, solvated molecules are released. And, it may be most loosely started that $\Delta H > 0$ i.e endothermic at low temperature and exothermic ($\Delta H<0$) at higher temperatures. Guess why?
Hint, derive that $\Delta H^\circ=-RT^2 \dfrac{\mathrm d(\ln{\text{CMC}})}{\mathrm dT}$. And CMC may be defined as CMC as the point corresponding to the maximum change in gradient in an ideal property–concentration $(\phi$ against $S_\mathrm{tot})$ relationship $\dfrac{\mathrm d^3\phi}{\mathrm dS^3_\mathrm{tot}} = 0.$

(emphasis mine)
Why was most loosely stated term used in the text? Isn't it true? Excluding non-dressed micelle model, $\Delta G$ can be shown to be approximately $(1 + \beta)\ln{\text{CMC}}$.
Writing $\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ$ and plugging in the values I couldn't get the desired term. Is my process correct?
Also, what's the meaning of “ideal property–concentration”?

Deriving $∆H^\circ$ won't help in telling $∆H$ probably, author wanted that you deduce above inference during deriving process. 2. There's a typo it should be excluding dressed model, not non-dressed model.


text:
If a chemical reaction is repeated multiple times, will it produce the same result?
If a chemical reaction is repeated multiple(infinity) times, will it produce the same result?
For example, I put two $\ce{H}$ and one $\ce{O}$ together, and repeat it many (infinity) times within an exact same environment, will it produce $\ce{H2O}$ every time?
My chemistry teacher tells me it will. But I still wonder if it's true. Hope you will help me. Thanks.

You may also get a noble prize for being able to generate a molecules using atoms :-D. jokes apart, why do you think something different will happen?

If I remember correctly this is one aspect of selectivity or in catalysis. If two products can form in a reaction both have an activation energy. Even though your catalyst might give you one of two possible products only you are not allowed to call it 100% but you will always find 99% or so. That is because for reaction A to be 100% selective reaction B would require an infinately high activation barrier. If that is not the case then upon repeating it for infinate runs you will also get the second product. Please correct me if I'm wrong here but it's been a while.

@Safdar Emm, you must know Uncertainty Principle. I think the atoms move uncertainty and it may cause different results. Maybe I'm naive, but forgive me.

Some reactions generate side products, which are produced with less abundance than the primary product. perhaps that is what you are looking for.

However uncertainly they move, the only way a reaction happens is if they collide with each other. Here taking what you have, I would say that there would be other products $\ce{H + H -> H2}$,$\ce{H + H + O->H2O}$. This is assuming there are only 2 atoms of H and one atom of O. if you have a mixture of a very high number of H and O atoms other products like $\ce{H2O2, O2}$ can also form. Only issue here is that they do not exist as atoms and do not wish to remain so. So if you have a reaction conditions where only $\ce{H2O}$ can form, then only $\ce{H2O}$ will form for that reaction condition.

The uncertainty principle, you must know, is referred to the information arising for a single (or a few) particles. When studying many particles, the uncertainty principle becomes irrelevant, and macroscopic statistics gets it shape. What is a mole of "particles"?

@Safdar Ah,thanks. Moreover, if I have two $\ce{H}$ and two $\ce{O}$, it can get $\ce{H2O}$ and $\ce{O}$ or $\ce{H2O2}$. So there's two results?

They are possible not certain.. However for a given set of reaction conditions, you would get the same yield assuming you do everything perfectly.

@Safdar You mean, if I do every perfectly, it will only result the same?

Yes. Maintaing the same conditions, extracting the product in the exact same manner and so on would give the same product at the same yield.

@Safdar Thanks. This will happen in macrography or every time(include micrography)?

@ Sir Kay. The Uncertainty Principle states that no parameters related to a single particule can be measured with precision at the same time. For example, if you determine the position of a particule at time t, the speed cannot be known with precision at the same time. It has nothing to do with your problem

There are some kinds of reactions which display mathematical chaos even on a macroscopic scale (e.g. the Belousov-Zhabotinsky oscillator), and in a rough sense, could be said to never produce the same macroscopic result, even for reactions run with the exact same conditions (up to the uncertainty principle).

@Maurice Doesn't it, really? Tell what it says is not understanding consequences, and one is that you never know the state of system with absolute precision and therefore it's impossible to duplicate it with such precision and OP's question is moot.

This is a difficult question because some of the terms are not as well-defined as one might like.
What constitutes an experiment and what constitutes repeatable?
A experiment isn't just what you are doing but also requires an observable. It's unclear how repeatable certain setups are. You can try to eliminate variables, but you're limited in exactness. For the observable, you're limited by accuracy of detection and also by the fact that observations may change the system under observation.
You have to consider the quantum mechanical aspect as well. For example, if you try to measure the maximum occupancy of electrons in a given orbital, the answer is two, you will be unable to measure anything else because it is forbidden under quantum mechanics (assuming the model is correct). On the other hand, the uncertainty principle will make certain measurements and setups unreasonable, though you might be able to get around those facts with clever experiments.
There is also the issue that much of chemistry is driven by stochastic processes which we model fairly well with statistical mechanics. What are you supposed to get if you measure the temperature of a glass of water? Well, in a large enough sample, it will be fairly close to some value, but there are very small fluctuations around that value that are not easy for us to detect. But in the limit of infinite measurements over infinite time, you might see a random large fluctuation. Is that temperature measurement still repeatable? Practically yes, but in theory, not necessarily.

The answer depends on whether you are looking at the level of a single atom or a large population of atoms. Because of quantum effects, there is a lot of randomness in the behavior of single atoms. If there are multiple possible outcomes, we can fairly accurately determine the probability of each outcome, but which outcome occurs is not predetermined.
With a large population of atoms (such as we have in any normal laboratory experiment like the ones your teacher is talking about), the large sample size means that we get the same result every time if we control every relevant variable (which is actually quite hard to do).

In radical chemistry, one can obtain varying results with the same starting radicals. The products do, however, appear to follow a consistent probabilistic level of respective yields for each product.
To help clarify, a set of radical reactions around the ∙ClO dimer, for example:
$\ce{∙OCl + ∙OCl = (OCl)2 or (ClO)2 }$  (support on the existence of the latter, see, for example, Eq(14) here)
$\ce{∙OCl + ∙OCl = ∙ClOO + ∙Cl }$
$\ce{∙OCl + ∙OCl = ∙OClO + ∙Cl }$
And, as:
$\ce{ ∙Cl + ∙Cl = Cl2 }$
One can also find the cited reaction:
$\ce{∙OCl + ∙OCl = O2 + Cl2 }$
And even:
$\ce{∙OCl + ∙OCl = Cl2O2 }$
Source: See https://pubs.acs.org/doi/abs/10.1021/j100286a035?journalCode=jpchax .
where above ∙OClO is commonly cited as ClO2 (chlorine dioxide).
In general, an area of study where one experiences a range of products upon the photolysis of a compound commonly occurs in the field of photochemistry.
[EDIT] More completely, I would further cite cyclic reaction systems displaying periodicity. Examples include the action of H2O2 with HCl, and much more famously, Iodine Clock reaction systems, where one can visually see decisive variations in reaction products occurring over time.

The most essential thing of all sciences, that they apply to the laws of the nature, independently from space and time.
It is not even for chemistry. That burning $\ce{H2}$ in $\ce{O2}$ makes water (in the common pressure/temperature), is similarly so in our current best knowledge, than for example, the laws of the Newtonian mechanics.
If it would not be so, then the search for the more general laws, saying why, when and how the result of the repeated experiments differ, is started on the spot.


text:
Why do 1s and 2s orbitals intersect?
So I was going through the Schrödinger wave equation to determine the shapes of orbitals. We know that an orbital is nothing more than a region where probability of finding an electron is maximum.
Further, we know that the probability of finding a 2s electron in 1s orbital is just too low! Why do 1s orbital and 2s orbital intersect? I mean both the orbitals have the same shape and the radial function of 2s orbital is more than that of 1s which clearly means that 1s orbital lies inside 2s orbital!
But as it's impossible to find a 2s orbital electron in 1s orbital, shouldn't the shape of 2s orbital be something like a hollow sphere with thick walls instead of a complete sphere?

I suspect you are trying to visualize the orbitals stacked as a matryoshka doll. The thing is, this "atomic doll" isn't made of wood, it's more like a 3D model made in Blender consisting of a of bunch of intersecting meshes that are not baked together yet.

You have it all backwards. Any orbital intersects any other.

Electrons aren't running in circles.

Does this answer your question? How are subshells and shells arranged in an atom?

I read your question a few times, and I think you're confusing occupancy with spatial probability distribution.
Orbitals have a lot of spatial overlap. But if the orbital has two electrons “in it,” then it cannot describe the probability behavior of any other electrons in that atom, as other electrons must be in another orbital.
A 2s electron absolutely can go into a 1s orbital provided there is space. This is exactly what happens with atomic emission spectra when an electron in energy level $n=2$ transitions to $n=1$ and releases energy.

I would even dare to say that, generally, orbitals largely overlap each other. By other words, if there is nonzero probability of occurence of an electron in the particular quantum state, it does NOT mean the probability of occurence of electrons in other quantum states is in the same point zero.
The probability of occurance near kernel for 2s electron has a local maximum, as ns orbitals have (n-1) radial nodes. So no, 1s and 2s orbitals do not have the same shape, neither 2s and 3s. They just share the spherical symmetry.
Here is the ( local PNG copy of not pastable SVG) picture for 1s,2s,3s orbital from the page radial probability distribution function.

Note the quantity on the y-axis of the chart: $4 \pi r^2 \Psi^2$ what is probability an electron occurs in the infinitesimal spherical shell $r,r+\mathrm{d}r$
One can also check the radial term R(r) of wave function on chem.libretexts.org on Figure 8.2.2, describing orbitals 1s, 2s, 2p.
There is a close analogy with the Maxwell-Boltzmann distribution for the speed of ideal gas molecules. Even if the maximum probability of 1-axis velocity component is at zero axis velocity, following the Boltzmann energy distribution, for the speed is the maximum at nonzero speed.

The curves reported by Putnik show the probabilité of finding the electron in a given sphere around the nucleus. If the probability $\psi^2$ of finding an electron at position ($x,y,z$) were reported, the obtained curves would always show a maximum at $r$ = $0$.

Sure, that is why I have explicitly written the radial probability distribution functions. See also the A update.


text:
Freezing point of solution of Water, Ethanol and Sugar (aka, Limoncello)
I'm looking for a diagram of the freezing point of a water solution depending on its content in sugar and ethanol. More colloquially, I want to know how to balance alcohol and sugar so that my limoncello doesn't freeze in the freezer. I know that increasing both will eventually work, but that seems unsatisfying.
I've managed to find the appropriate diagram for solutions sugar-water and sugar-ethanol. There are some related research on emulsions, but the phase diagram there is done at fixed "high" temperature (7°C or above).
Is it possible to simply extrapolate from the data above ? Also, is there a formula for this ?

https://chemistry.stackexchange.com/questions/5946/does-freezing-a-solution-with-water-always-cause-the-water-to-separate-and-form https://chemistry.stackexchange.com/questions/73173/why-doesnt-frozen-sugar-solution-taste-sweet

That's sort of the point, I want the temperature range at which none of the mixture freezes. I agree that once it starts to freeze, things get a lot more complicated.

Wikipedia says it has 25-30 % - that's what you could get after freeze distillation IIRC, so it probably won't freeze at all.

What is the target range of your ethanol-sugar content?

Homemade Limoncello can go from 25% to not-legal-for-sale% alcohol (although it usually stops at 40%), and up to 30% sugar. Arguably, I could find out the data by myself, experimentally. :p


text:
Oxidation of hydroxyl groups of glucose
In all oxidation reactions of glucose, it seems that the aldehyde group alone gets oxidised and none of the hydroxyl groups. In one reaction with nitric acid, the aldehyde group and the terminal hydroxyl group get oxidised but none of the remaining hydroxyl groups. I was wondering why this happens. Are the hydroxyl groups less susceptible to oxidation for any reason? If so why would that be? And also is there a difference is ease of oxidation of the terminal hydroxyl and remaining hydroxyl groups?

When there are various pathways then what is oxidised depends more on the oxidising reagent - as Matthew mentioned, you can oxidise or reduced in various ways. Do you want to know about mechanism of oxidation with nitric acid?

@Mithoron Yes i wonder why nitric acid doesn't oxidize the secondary alcohols to ketones

Aldehydes are easier to oxidize via radical mechanisms because of the ease with which one can break the C-H bond. The resulting radical is stabilized via delocalization of an oxygen lone pair.
Such stabilization is absent in an alcohol. The alcohol shows a similar moiety, but I think the stabilization in the aldehyde is greater because of the shorter C-O bond, which improves orbital overlap.
In a laboratory setting, alcohols tend to be more stable to air, whereas aldehydes can be prone to air oxidation.

Thank you so much for the answer! But why wouldn't the hydroxyl group be able to stabilise a carbon radical attached to it because it too has a  lone pair on it, would the attached hydrogen to the oxygen in hydroxyl affect it ability to stabilise the radical? Of so why? Also, what if the oxidation reaction is based on an ionic mechanism, how would difference in ease of the two groups be explained in this case?

That's a really good question. I think added an edit. I would think that the two-electron mechanisms more readily oxidize an alcohol (many reagents stop at the aldehyde, e.g., DMP). However, I would think that nitric acid operates via a two-electron mechanisms though I could be wrong here. If it were a two-electron mechanism, my guess is that the steric factors involved with being terminal versus non-terminal were at play...

The steric factors and orbital overlap concepts make so much sense! Thanks a lot for the answer!

OP's statement of "In all oxidation reactions of glucose, it seems that the aldehyde group alone gets oxidised and none of the hydroxyl groups" is not entirely true. For example, the step by oxidative cleavage with periodic acid ($\ce{HIO4}$) has been successfully applied in the constitutional analysis of sugars. The presence of several pairs of vicinal diols such as those in glucose during oxidation with $\ce{HIO4}$ can lead to the formation of complex product mixtures:

Accordingly, the oxidation reaction of D-glucose with 5 equivalents of $\ce{HIO4}$ gives five equivalents of formic acid and one equivalent of formaldehyde (Figure A). An analogous degradation of D-fructose yields three equivalents of formic acid, two equivalents of formaldehyde, and one equivalent of carbon dioxide (Figure B):

This oxidation is believed to be step-wise. For example, glucose (a hexose) contains a terminal aldehyde group ($\ce{C}$1) adjacent to a hydroxylated carbon. This setup tends to be oxidized to formic acid, $\ce{H-C(OH)2-CH(OH)R -> H-C(=O)OH + (O=)CHR }$, and an another aldose, which contains one less carbon atom than original sugar (here, it is the corresponding pentose). This formation can be explaned by a $\ce{C-C}$ cleavage followed by each $\ce{C-OH}$ oxidation to $\ce{C=O}$:

Now, this corresponding pentose also contains a terminal aldehyde group adjacent to a hydroxylated carbon. Thus process continues until $\ce{C}$6. The $\mathrm{R}$-group of $\ce{C}$6 is $\ce{H}$. Thus, final product is $\ce{R-CHO \ # \ H-CHO}$, which is formaldehyde. Additional description of this oxidation can be found in here.

Nice answer to some other question but it doesn't address the OP's question about nitric acid.

@ user55119 : True. But I am concentrating OP's concern about not oxidizing hydroxyl groups (the title and the first sentence of OP's text). So, I think my answer is appropriate as well.

I still don't understand why the aldehydes are easier to oxidise than the hydroxyl groups with weaker oxidising agents like say Benedicts reagent, and also why terminal hydroxyl group alone is oxidised when aldehyde is protected and not the other hydroxyls? And what about nitric acid, why the selectivity of oxidation of only aldehyde and terminal hydroxyl groups? Is it because it isn't too strong an oxidising agent?