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Intermolecular Forces of Attraction
Imagine a bottle 85% filled with water and the rest is just air, and the cap is closed. Here are now my questions.
Since the cap is closed completely, the air bubble is formed. Is that air bubble compressed? In other words, since air has the least force of attraction between its molecules, they are more free than the molecules of water. Now since water is creating a pressure against the air bubble, does it mean that the molecules in the bubble are now closely packed than usual?
Is it possible for the air bubble, if compressed a lot, to mix with water?
If the air bubble is able to hold itself even in water, why do we say that air has lesser force of attraction that water?

Please explain what you mean when you say : "the air bubble is formed" ! Which air bubble ? How and why is it formed ? What do you mean when you say that" the molecules in the bubble are closely packed" ? In a gas the molecules are not packed at all ... Explain !

If you compress it, then it will be compressed, otherwise no.

Assuming that by air bubble you're referring to the pocket of air now sealed in the bottle above the water surface:
Immediately after sealing the bottle, the air pocket is not at all compressed: it's at the same pressure as the surrounding environment (the air outside the bottle).
However, because you have a closed container, some water will evaporate, until an equilibrium is reached between the water vapor in the air pocket and the water in the liquid phase. This occurs when the chemical potential of the water in the two phases are equal.
Due to that vaporization, the total pressure in the air pocket (which now contains a higher amount of water vapor than the surrounding atmosphere outside the bottle) will be larger than the surroundings. This is a prominent effect when it's hot outside (have you ever left a sealed water bottle in a hot car?).
While this effect is usually small, as gases mix very well, it is non-zero and so yes, the air pocket is slightly compressed due to presence of water vapor.
As for whether the air pocket can mix with water: Assuming our container can tolerate such pressures and we were to externally compress it: liquids are much less compressible than gases, and so the air would (given high enough pressure) condensate, mixing into the liquid phase (but you wouldn't achieve this by hand!)
To address your final point: let's say we invert the bottle, so the air pocket travels as a bubble through the liquid phase to re-emerge at the other end of the bottle. One could ask, "why doesn't the air bubble fragment into little bubbles as it progresses upwards?"
To be fair, you do see a few bubbles splinter off, but for the most part, the bubble is intact. The air bubble stays largely intact due to the viscosity of water (the intermolecular attractions between water molecules is larger than the entropic benefits of mixing with the air bubble). So, to answer your question, the fact that the air bubble doesn't mix is not due to its intermolecular attraction, but rather it's due to water's intermolecular attraction.

thanks a lot for the detailed answer. I have another doubt now. Let’s imagine the same bottle with water and the air pocket. Now picture the molecules of air in the pocket (freely enclosed) and the molecules of water in the bottle (closely packet). Now since the air pocket has a defined boundary, I want to know if there is a force acting in between the two different forms of matter (air and water) that holds the air pocket together and does not let it mix with water. Also, the molecules part, what is present in between the molecules of air?

Assuming you just closed the bottle in typical surroundings, the initial composition of the air pocket will be identical to the surrounding air (4 parts N2, 1 part O2 by volume, with some water vapor in there too), and then the composition will start to become more humid (more water vapor) as equilibrium is reached between the water in the vapor phase (in the air pocket) and the water in the liquid phase.

There are quite a few effects acting on the air and water to prevent them from mixing: 1) gravity makes the more dense water “want” to occupy the region of the bottle with the lowest gravitational potential. While the air wants to do this too, it has a much lower density, so the water would push the air bubble out of the way (to the top) to minimize its energy. 2) As for why they don’t just mix: water has an intricate network of hydrogen bonds, giving it viscosity. This means the entropic effect of mixing with the air is overruled by the favorable water-water attraction.

as for why the air doesn’t mix into the liquid phase so that they both get to occupy the bottom of the bottle: that would leave a very negative pressure at the top of the bottle, which would exert a much larger force on the system than gravity.


text:
What does HClO2 aqueous solution decompose to in an open environment?
I looked up the decompositions of diluted $\ce{HClO2}$ aqueous solution (used in a disinfectant) to know if it is safe to use on some substances, such as metal, wood, leather or skin. However, the chain I found leading to $\ce{HClO4},$ which is a very strong acid.

$$
\begin{align}
\ce{5 HClO2 &→ 4 ClO2 + HCl + 2 H2O}\\
\ce{4 HClO2 &→ 2 ClO2 + HClO3 + HCl + H2O}\\
\ce{3 HClO2 &→ 2 HClO3 + HCl}\\
\ce{2 HClO2 &→ 2 HOCl + HClO3}\\
\ce{HClO2 &→ HCl + O2}
\end{align}
$$

Then there is

$$\ce{3HClO3 → 2ClO2 + HClO4 + H2O}$$

(Source: Inorganic Chemistry by Garg and Singh [1], Google book preview)
The $\ce{HClO2}$ solution is diluted (I do not know the exact number, but maybe similar to mild bleach), sprayed to substances' surface, at room or body temperature.
Can the balances in this case lead to $\ce{HClO4}?$
On the other hand, $\ce{HCl}$ recombines to make $\ce{Cl2}$:

$$
\begin{align}
\ce{HClO2 + 3 HCl &→ 2 Cl2 + 2H2O}\\
\ce{HOCl + HCl &→ Cl2 + H2O}
\end{align}
$$

So, can the remaining $\ce{HCl}$ be concentrated enough to corrode metal or cause harms to skin?
From a consumer's point of view, $\ce{HClO2}$ seems to be a complete disinfectants for all of water, surface (by producing $\ce{HOCl}$), and air (by producing $\ce{ClO2}$). However, the issue left for concerns is the byproduct acids.
Updates: I found a list of patents about producing $\ce{HClO2}$ solutions to be used as disinfectants, which are likely the one I am talking about in this question. The patents claimed that the disinfectants are safe to human, can be used to clean food-processing utilities, and even as food additives. The exact reasons for the claims are not clear to me.

https://patents.google.com/patent/US9516878B2/en
https://patents.google.com/patent/AU2013205834B2/en
https://patents.google.com/patent/EP2999490A2/en
https://patents.google.com/patent/US20160106106A1/en
https://patents.google.com/patent/US20160113282A1/en

References

Garg, R.; Singh, R. Inorganic Chemistry; McGraw-Hill Education: New Delhi, 2015. ISBN 978-1-259-06285-8.

$\ce{HClO2}$ is a very disagreeable substance to use, as it quite easily produces plenty of unbearable and toxic gases like $\ce{ClO2}$ and $\ce{Cl2}$ if the solution is made acidic. The danger of $\ce{HClO4}$ being formed in the solution may be considered as nearly negligible, compared to the danger of these gases.

Is ChemiDay a reliable enough source (for inorganic reactions) to be cited on our site? (spoiler: no, not really)

@andselisk Sorry, I'm new here. Can you refer me to a more suitable source for this site? Google show ChemiDay at the top of search page.

@Maurice How is it compared to other chlorine-based substances for disinfection? Maybe for disinfectant, those toxic gases are unavoidable. For example, HOCl and ClO2 aqueous solutions are also used as disinfectant.

@THN No need to apologize, the thing with ChemiDay is that they very rarely cite any sources, and even if they do, these are of questionable quality. You can use any source of information as long it's trustworthy and the claims are backed up by reputable sources, such as textbooks, articles, established databases etc.

@Maurice Actually I'm not making it, I asked from a consumer's point of view.

@Maurice I think you are right that toxic gases is the main issue, as I found some patents about HClO2 solution disinfectant which aim to reduce ClO2 production. Can you take a look and say about the if or why the solution can be used on human and household?

Per Wikipedia on Chlorous acid to quote:

The pure substance is unstable, disproportionating to hypochlorous acid (Cl oxidation state +1) and chloric acid (Cl oxidation state +5):
$\ce{2 HClO2 → HClO + HClO3}$

Also, here are some interesting comments on the decomposition of chlorous acid in "Kinetics and Mechanism of the Decomposition of Chlorous Acid" from J. Phys. Chem. A 2003, 107, pages 6966-6973, to quote:

Of the many mechanistic models tested, the one that fit best included the following reactive intermediates: HOCl, Cl2O2, Cl2O3, •ClO, •OH. The stoichiometric ratio of ClO2 produced to Cl(III) consumed varies with pH and [Cl-]. Reaction of Cl2O3 with Cl(III) yields chlorate exclusively. Reaction of Cl2O3 with Cl- favors ClO2 over chlorate, but does not entirely exclude chlorate, because it is produced by hydrolysis of Cl2O2. Invoking Cl2O3 explains the variation in stoichiometric ratio as well as the maximum observed in the initial rate of ClO2 formation as a function of pH. The kinetics of chlorous acid decomposition cannot be quantitatively fit through the last stages of the reaction without postulating a first-order decomposition. Scission of chlorous acid to give short-lived hydroxyl and chlorine-(II) monoxide is a plausible route for this process [...]
Several groups of investigators[5-7] have found
  that in the absence of chloride ion the stoichiometry of the
  decomposition of chlorous acid is given by reaction A:
$\ce{4 HClO2 -> 2 ClO2 + ClO3- + Cl- + 2 H+ + H2O}$ (A)
The stoichiometry of the decomposition of chlorous acid in the
  presence of chloride ion is given by reaction B:
$\ce{5 HClO2 -> 4 ClO2 + Cl- + H+ + 2 H2O}$  (B)

Also, to quote:

Earlier studies,[9,18] in agreement with our present results, have also found the formation of more chlorate than predicted from reaction A. Reaction C
$\ce{3 HClO2 -> 2 ClO3- + Cl- + 3 H+}$  (C)
also plays a role in determining the stoichiometry at higher $\ce{HClO2}$ concentrations."

So, several intermediate products and depending on chloride presence possibly $\ce{ClO2}$, which is a problematic explosive and toxic gas, along with the strength of the $\ce{HClO2}$, which can introduce chlorate at higher chlorous acid concentrations.
Not a particularly good path to an acid as you are also inquiring about. As a disinfectant, the transient creation of the powerful disinfecting HOCl and associated radicals, may actually make it weaker, albeit more stable (considering end products) than hypochlorous acid, in my opinion.
[EDIT] To answer a comment question, as to whether is ClO2 safe, here is  a statement from the CDC, Public Health Statement for Chlorine Dioxide and Chlorite, to quote:

If you are exposed to chlorine dioxide or chlorite, many factors will determine whether you will be harmed. These factors include the dose (how much), the duration (how long), and how you come in contact with them. You must also consider any other chemicals you are exposed to and your age, sex, diet, family traits, lifestyle, and state of health.

Interestingly, ClO2, which is a stable free radical, has found commercial application as an odor removal agent, likely due to its interaction with organics to create volatile organic chlorides (I would suspect that the presence of light would be catalytic). These VOCs are much more of a long-term health danger (as in carcinogenic), in my opinion.

OK, so by your quotes, it seems that in a solution, HClO2 will result in mainly ClO2, HCl, and HClO3, all nasty products.

I found a list of patents about HClO2 solution disinfectant (listed in the updated question). Can you take a look and say about the if or why the solution can be used on human and household?

Thank you for the expositions. I think I will stick to the good old NaOCl/HOCl solutions for now.

It is my understanding that the main decomposition product of HClO2 is ClO2. Commercial ClO2 generation systems typically incorporate a membrane used to separate the gas from the other components, such as the solids and undesirable side products. This helps create a highly pure solution of ClO2 in water.
Here are some examples:
https://www.dioxide.com/systems/chlorine-dioxide/
https://selectivemicro.com/about-us/#smtdiff
From a consumer's point of view, as you state, I would look into how the disinfectant solution is generated. Typically, you can react sodium chlorite (NaClO2) with an acid such as HCl or citric acid to generate HClO2 and therefore ClO2. Does the product use membrane technology to separate the gas? You mentioned HCl. Based on the reactions described, don't think concentrated HCl is a concern  (unless HCl is the acid used in the reaction). With Na ions present, you would more likely form NaCl.
Also "...water systems using chlorine dioxide for disinfection or oxidation must monitor their system for chlorine dioxide and chlorite." (https://www.evoqua.com/en/brands/municipal-services/Product%20Information%20Library/MSPOTABLEAP.pdf)

As you said that the production process of HClO2 is important. I found a patent about producing HClO2 solution for disinfectant (https://patents.google.com/patent/US9516878B2/en). Maybe this is the exact disinfectant that I am concerning. Can you take a look at it?

I found a list of patents about HClO2 solution disinfectant (listed in the updated question). Can you take a look and say about the if or why the solution can be used on human and household?

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text:
Can glass be deionized?
Is there a chemical treatment that could remove sodium and calcium ions from the surface of soda-lime glass to turn it into quartz glass, increasing the hardness?

You can indeed "harden" glass by exchanging ions on the surface, but not because it turns into quartz glass.
Quartz glass is mostly so robust because it has an extremely small thermal expansion coefficient, and is therefore mostly free of internal stress due to uneven cooling. Ordinary glass already gets a lot more durable when you temper it, thereby relieving those internal stresses.
Generally, glass breaks under a very small extensional strain. It has a huge $E$ modulus and can take a lot of stress, but when you bend it, so the extensional stress and deformation is largest at the surface, where also you always have small scratches etc.), it shatters.
The trick is to put the glass surface under a large compressive stress. That way it can take a lot more pull before the critical strain (deformation) is reached and breakage occurs.
There are basically three ways to do that: Firstly, you can coat the hot glass with another glass that has a lower thermal expansion coefficient. That way the surface gets compressed during cooling. Tricky, and expensive.
Then, the most common way, is to rapidly cool the still nearly molten surface by e.g. blowing cold air on it. The outer shell is already cold, the inner part takes longer to cool and still shrinks, thereby compressing the surface.
The last, chemical trick (making supremely solid glassware) is to exchange sodium ions in the surface layer (100nm or so i think) by potassium ions. This was done in the eighties in the GDR for household glassware, under the brand name "SUPERFEST", by soaking the glasses in molten potassium nitrate (at 400 °C) for twenty minutes or so. The ions are larger, but the silica network lattice does not yield below at least 500°C, so the surface gets compressed by the exchange.
You know you had a glass that was treated in one of these ways when it does not break into large shards, but basically crumbles into small pieces or even dust. The classic example for the latter is Prince Rupert's drop, of the former any car windshield. (It seems the Na/K exchange does not (necessarily) do that, see your mobile phones "Gorilla glas".)

GDR = East Germany? I haven't heard that term in decades. It reminds me of my dad (in the UK) bringing home a chunk of the Berlin wall from a business trip. And the songs: wind of change, 99 red balloons...

@LevelRiverSt I hope that was a long time ago. Taking home chunks of the Berlin wall is frowned upon, now that the remaining pieces are basically memorials and not something they're trying to demolish.

Chemically hardened glass will not necessarily shatter into dust or small pieces - it depends on the type and degree of treatment.  We routinely perform drop-ball testing on K+ hardened glass lenses and they break into typical big chunks when they fail.

@Karl No it doesn't, on human or any other timescale.

@Karl - Matt's correct. The variation in thickness in old panes of glass if due to the manufacturing process, which inherently produced glass of uneven thickness (see crown glass). With the advent of float glass production in the 20th century, more uniform sheets of glass were produced. The "flowing glass" story is scientifically inaccurate and is a myth.

@ToddMinehardt I know of course, being a bloody rheologist and wanted to point out exactly that. Total language failure. Tnx.

Why, it has had a wiki article for a long time. ;) https://en.wikipedia.org/wiki/Chemically_strengthened_glass What am I doing here ...

@LevelRiverSt Jup, a small gem in Erichs Lampenladen. You can ask Corning for quote for a batch of "Gorilla" pint glasses .... ;)

@Karl - Ha! You should be running for President :)

@user253751 Don't worry too much, missing pieces are what makes it interesting. Of course the Eastside Gallery is a work of art, not a wall now. And you're right for parts that are still in their place.

@user253751 you can still buy those pieces in souvenir stores. Some even come with a certificate for their originality.

You could dealkalize the surface of a borosilicate glass to get rid of alkali ions, but I'm not aware of a method that allowed the extraction of alkali metals from the bulk of the glass.
There are probably quite a few ways to modify the surface, but here's a few reported by Yashchishin and Zheplinskii [1]:

Sulfur anhydride or Freon is the most widely used reagent for this purpose. The method consist in that the gaseous reagent is fed into the firing furnace or mixed with the air used for blowing hollow articles (USSR Inventor's Certifi- cate 647273). An interaction between the gaseous reagent and the glass surface occurs in the processes of blowing and firing.
The chemical resistance of glass containers can be increased by means of special treatment by the products of pyrolysis of ammonium sulfate or chloride. This increases the chemical resistance of glass containers by almost a factor of 10. The method consists in that granules of $(NH_{4})_{2}SO_{4}$ or $NH_{4}Cl$ are introduced inside the items before the beginning of the annealing. In the furnace, the salt decomposes and the resulting gas mixture interacts with the glass, removing alkalis from its surface.

One more note about the hardness: Although fused quartz is thermally more stable than borosilicates, its hardeness on the Mohs' scale is comparable to the latter (5.3 to 6.5 and 6, respectively).
References

Yashchishin, I.N., Zheplinskii, T.B. Improving the chemical resistance of glass containers by thermochemical treatment with a reagent solution. Glass Ceram 53, 135–137 (1996). https://doi.org/10.1007/BF01166432

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text:
Which is a stronger nucleophile in aprotic solvent: water or hydrogen sulfide?
Which one is the stronger nucleophile in aprotic solvent: $\ce{H2O}$ or $\ce{H2S}?$
According to my notes, nucleophilicity order is directly proportional to basicity order in an aprotic medium, and is inversely proportional to basicity order in protic medium.
And $\ce{H2O}$ is a stronger base than $\ce{H2S},$ so it should be the strong nucleophile, but the answer is given as $\ce{H2S}$ is the stronger nucleophile. What is the reason for that? Am I missing something?

Related: Nucleophilicity with respect to the solvent

H2S is a stronger base then H2O, not that it matters much. More importantly trialkylthionium ions are much more stable then oxonium analogues.


text:
Reason behind more Reactivity of Unsaturated Hydrocarbons
My textbook says that unsaturated hydrocarbons are more reactive than saturated ones. But double and triple bonds are stronger than single bonds.
What is the explanation for this?

Double and triple bond have greater electron density that make them reactive to electrophiles

Double and triple bonds as a whole are stronger than single bonds, but the additional pi bonds are easier to break than a sigma bond.

The unsaturated regions are electron-rich and typically allow for electrophilic attack, such as by halogens or protons. Saturated hydrocarbons usually react only with radicals (radical halogenation). C-H activation is often hard in saturated hydrocarbons.
While double and triple bonds are stronger than a single bond when looking at complete cleavage of a bond, one notices that the bond strength of a double bond is less than twice the bond strength of a single bond (with other factors being comparable, there are surely some exceptions). When a double bond reacts, it typically is changed into a single bond, not fully cleaved. Similar arguments hold for triple bonds.


text:
Screw axis in P21c space group
I am having some trouble with understanding glide axis. The following is a question asking to reconstruct the full unit cell:

And the following is the answer to the question:

I have highlighted an element which I want to focus on. I understand the rotation of 180 degrees and translation of 1/2 of a unit cell. However, I don't understand why the element undergoes a change from 1/2 + to - . My understanding of 1/2 + is that it is located 1/2 above the left element following a glide plane (c) transformation. But I don't understand why it would be transformed into a "-" after the axis screw. I also don't understand what 1/4 near the screw axis means, I'm guessing it could be that it is an axis at c = 1/4, but the transition that the highlighted element undergoes still doesn't quite make sense if that is the case.
Please provide any guidance with this question, I've been struggling to understand what is going on for a long time.
Many thanks

Well, the 1/4 near the screw axis means exactly what you're guessing. Also, it changes + to -, because, er, well, that's what all horizontal axes do. What were you expecting of a 180 degrees rotation? Just a shift along X and a reflection in Y, with no change in Z? That would be a plane.


text:
Is an epoxide opening more favorable than a chloride leaving in the same molecule?
In general, we've learned that alkoxide groups are awful leaving groups, unless we're talking about epoxide rings whose ring strain allow the alkoxide to act as a leaving group in a ring-opening reaction. However, if a structure contained both an epoxide and a relatively good leaving group like Chloride, what would predominate if a strong base/nucleophile was introduced: Epoxide opening or the substitution reaction? An example is provided below of what I'm referring to:

If I perform, the epoxide ring opening, I get something unusual:

I left the stereochemistry ambigious because I honestly did not know how to assign it since a new stereocenter is formed.
If I perform the Sn2 substitution, the reaction is completed quickly:

Notice that both of my proposed mechanism reform the epoxide ring, but do generate entirely different structures.

https://chemistry.stackexchange.com/questions/73398/how-to-rationalise-the-major-product-formed-in-a-nucleophile-promoted-epoxide-cl

I did not realize someone had posed a similar question. Thank you for the link.

Chloride isn't really that great a leaving group, either. The summary of this question is that two fast steps are collectively faster than one slow step.


text:
Guidelines for mixing gases?
I want to create several mixtures of oxygen and nitrogen gas, such as 0% oxygen, 10% oxygen, 20% oxygen, 30% oxygen... 100% oxygen, but I know very little about working with gas cylinders and regulators. Is it as simple as connecting the two cylinders with a y-shaped hose and setting the flow rates to 10% oxygen and 90% nitrogen? Or will I need a more complicated setup?
This is for dissolving the gas mixture in water so I can test methods for determining dissolved oxygen levels. I'm not filling SCUBA tanks or anything dangerous, but I would still like the mixtures to be fairly accurate so I can get good data.

I think you should speak with your compressed gas vendor to get an understanding of what is required once you have your experiment properly defined. Working with pure oxygen and nitrogen is dangerous and shouldn't be done without proper safety guidance.

@J.Ari - I'd agree that working with pure oxygen can be dangerous because of the flammability hazard. The only problem that I can think of for pure nitrogen would be asphyxiation which seems unlikely.

@MaxW Since the OP indicated a lack of experience working with compressed gas sources and technologies, it's my opinion that no hazard (e.g. over pressure, flammability, asphyxiation, leakage) should be considered unlikely until proper system design and construction expertise is consulted and a documented hazard review is completed.

@J.Ari - You are correct. I handled gas cylinders for so many years that I forgot about even basic considerations like strapping the cylinder. Breaking the stem off would of course turn the cylinder into a missile.


text:
Effect of ionization on dipole moment
Suppose we have a non-dipolar molecule such as carbon dioxide. If carbon dioxide is ionized by some means in the gas-phase, will it develop a dipole moment? I am asking with reference to microwave spectroscopy, because the selection rule requires the presence of a permanent dipole moment. In the interstellar space,  $HCO_2^+$ has been detected and it does have a microwave spectrum, because it does have a dipole moment. However, carbon dioxide's dipole moment is zero, so its ordinary microwave spectrum does not exist. There are very weak lines due to its quadrupole.

So, you ask if CO2 cation could be detected?

Indirectly yes, but the main idea is can we make non-polar molecular as dipolar (having a non-zero dipole moment) by ionizing them?

You should propose an electronic structure for the radical cation in question.

Yes there can be a dipole due to the fact that a molecular ion is not a point charge but an unbalanced distributions of charges.

@Alchimista I don't think that is true. Whether a species possesses a dipole moment is ultimately down to its symmetry. Is there a good reason why a molecular ion can't belong to (for instance) a $D_{5h}$ point group? Cyclopentadienyl immediately springs to mind. Such a species will not have a permanent dipole moment  - a charge and quite possibly a quadrupole, but not a dipole. My guess is the loss of electrons distorts the structure into a point group which can display a dipole moment, but that is purely a guess.

@Ian Bush Sure, it can be that a ion has enough symmetry to have zero dipole moment. I did not say that one must have it. In fact, you said the same as in my comment.

@Karl, I do not know enough quantum mechanics to propose the electronic structure of ionized carbon dioxide. What a microwave spectroscopist told me is that sometimes ionization may decrease or sometime it may increase the dipole moment. He was not sure either. May be physicists know this better.

Well, ionization can change geometry - I wouldn't be surprised if CO2+ was bent instead of linear.

I think geometry change would be likely. This seems more like a complex problem, because bent carbon dioxide will not rotate in the same way as linear carbon dioxide (different moment of inertia). Interesting perspective!

Actually it seems to me that this particular case is simple as for symmetry likely has a dramatic change. But what I wanted to say above is that a dipole moment can result because a distribution of n negative charges and n-1 (or n+1) positive charges is not a single charge. Forget molecule for a while and see it from a basic pov. The distribution orients itself in an electrical field.

In addition to a possible geometry change, ionization of a closed-shell molecule results in an unpaired electron and a (large) magnetic dipole moment. This can facilitate the observation of magnetic-dipole transitions in the rf and microwave range of the spectrum.


text:
Best bags to contain chemical odours?
I have noticed that if you put household or other chemicals in bin liners or freezer bags the plastic doesn't keep the chemicals in. Rather they travel through the bag material and out into the air. However I have noticed that some bin liners have stronger material and are better than others.
Perhaps it will depend on which chemical is in the bag. However I guess I am asking about pollen, pesticides,washing detergent and general cleaning chemicals.
Is there a specific type/grade of plastic bin liners that I can get that will, generally speaking. keep odors in? Actually it doesn't have to be pvc any bag material which does the job will do however I guess PVC bags are most easily accessible so would like to know what options there are. See through bags are preferable.

As you state, the ubiquitous polyethylene (or polythene) bag allows many chemicals to slowly seep through. Pollen and dust particles themselves are far to large to get through, but water vapor, for example does.
For that reason, many foods are sold in plastic (e.g., mylar and polypropylene) sacks layered with aluminum (ah, aluminium for you polythene-sayers) to block evaporation, oxygen (which leads to fats and oils becoming rancid) and loss of flavor.
For storing small amounts of food, reuse these aluminized bags -- which also are effective shields to block RF tags in passports and licenses. They're available commercially, e.g. this list.
These bags are comparatively expensive, though, and last millenia in landfills, and impractical for general use waste bin bags.
Non-chemical work-around: store food wastes and other potentially odorous items in paper or plastic bags that would have been discarded anyway, and that bag inside the bin liner, to delay odor migration.

What you are seeking is a material with very low permeability.  That's the search term you should use if you want to look for this on your own.
There are modern plastic composites with very low permeabilities, but I don't know if any of them are commercially available as trash bags.
Among non-composite plastics, PVDC (polyvinylidene chloride), which is the plastic that used to be used in Saran Wrap (before it was reformulated in 2004 to instead use polyethylene, a non-chlorinated plastic), is known for its especially low permeability (see:  https://www.britannica.com/science/polyvinylidene-chloride).
You would need to search the internet to determine if trash bags that are either made of, or lined with, PVDC are commercially available.  And even if you could find them, you might not want to use them, for environmental reasons: Plastics that contain chlorine, such as PVC and PVDC, have fallen out of favor, especially for use in disposable products, in part because they release the chlorine when burned in commercial incinerators.
You might also try contacting a speciality polymer company (e.g., https://www.aaapolymer.com/can-liner-materials/ ) and ask them for their recommendation.


text:
What exactly are the graphs of orbitals?
In my QM class we are finding the wave functions of the hydrogen atom. In spherical coordinates, these wave functions are functions of 3 variables: $r,$ $\theta,$ and $\phi.$ My professor stated that "the wave functions of the electron are called orbitals".
Say, we are interested in the $l = 0$ (or s-orbitals). Since $l = 0,$ $m = 0.$
$$\psi_{n00}(r,\theta,\phi) = R_{n0}(r)Y_{00}(\theta,\phi) = \left(\frac{1}{4\pi}\right)^{1/2}R_{n0}(r)$$
So, we can set a specific $n$ value. Say, $n = 1,$ and we get the the corresponding 1s orbital. Since we set an $n$ value, the wave function reduces to a function of the radius $r.$ During the lecture, he said that this image:

is an isosurface (or level surface) of the wave function at a constant radius.
I interpreted that as finding the isosurface of a 4D graph, our wave function, at some radius. This isosurface is the the sphere in 3D. What confused me was when he stated that the graph was an "isosurface of the full orbital".
What does he mean by the "full orbital"? Is this "full orbital" something in 4D space?

No 4D! The spherical surface is mathematically 2 dimensional: it is the boundary of the 3 dimensional “ball” within. Note that ordinary language usage is not always the same as the mathematical language. So the spherical surface is embedded in three dimensional space.

Isn't an orbital a probability distribution?

@124c41 Someone can correct me but from what I understand, an orbital is described by a wave function. It is a region of space where an electron can be found. The probability density is the square of the wave function.

Yes it is correct. But why being the orbital a probability distribution causes your question? Graphs of the orbital can be graphs of that function (the orbital) or related functions, such as a probability density function, its square. If you want to visualize how a s changes radially, a section along a plane is better suited. A common way is to plot 2D in 3D as such that the probability to find an electron within exceed 90 percent or more. Tough usually these plots are used qualitatively l

For isocontours or isilosurfaces you just set a value to be kept. In your example is not n, but a value that lead finally to a meaningfully high probability to find the electron

@Alchimista I agree with everything you said. So I think what it boils down to is that the 'best' we can do to visualize orbitals is via these density plots (or alternatives graphs). I don't have a problem with this, but I think it is still important to make certain distinctions. Though density plots are one of the best ways to visualize, I don't want to equate it with orbitals.

As you may have surmised by now: An orbital/wavefunction is defined for all values of $r$, and in that way exists through all space (though the wavefunction has negligible value except for near the origin). The isosurface graphs are produced by picking a constant value of $r$, so that the "shape" of the orbital can be visualized intuitively. However, keep in mind that mathematically the orbital extends past the isosurface, with exponentially decreasing value. You will see this if you look at the radial factor of the wavefunction $R$, which will have an exponential decay term.

@JamesBond Strictly speaking: No, an orbital is NOT a region of space. Also, the orbital is not described by a wave function, instead it IS the wave function (of a single electron). Therefore, the probability density is actually the absolute square of the orbital. Since the wave function is never really zero, the electron could be located anywhere, and that region of space would span the whole universe. Saying an orbital "is a region of space where an electron can be found" is a simplified visualization, but imprecise!

@Feodoran Since the wave function spans the entire universe, and the wave function IS the orbital. Would it be correct to say that the orbital spans the entire universe? (Strictly speaking)

@JamesBond By that definition, yes. But I don't know whether quantum mechanics is still valid when talking about the universe as a whole. It is a theory made for the small things, not the macrocosms.


text:
1H NMR of 4-Methylanisole
I have been pondering the assignment for the 1H NMR peaks observed. I can't really make sense of the observations from what I know.
https://www.chemicalbook.com/SpectrumEN_104-93-8_1HNMR.htm
My question is the assignment of the signal between 6 and 7.5ppm. Unlike the way it is currently assigned, I would imagine Ha to be the most upshifted peak (near 7ppm) because methoxy group is slightly electron withdrawing and therefore de-shield the protons of Ha more than Hb (Hb is near methyl group which is slightly electron donating).
Any idea of how you would interpret the peak assignments?
Data retrieved from here: https://www.chemicalbook.com/SpectrumEN_104-93-8_1HNMR.htm

The methoxy group actually interacts with the aromatic ring via two competing mechanisms.
It is indeed electron withdrawing via the sigma bond interaction between the O and C of the aromatic ring. However the lone pair on the oxygen conjugates through the delocalised pi network and increases the electron density at both the ortho and para positions. If you draw out the conjugation pathway you will see the electrons can sit on the ortho and para carbons. This is the more significant contributor to the overall electronics and, therefore, chemical shift. So, the protons ortho to methoxy are far more shielded than those ortho to the weakly electron donating methyl group.
This effect can be generalised across all aromatic directing groups, and the size of chemical shift difference from benzene (7.3ppm) is directly correlated to the strength of the directing group. So, aniline has ortho chemical shift of about 6.5ppm (-NH2 is a strong EDG), while the ortho protons in nitro benzene are up at about 8.5ppm (-NO2 is a strong EWG).


text:
Spin Operator algebra
I am trying to teach myself some QM.
In Christopher J. Cramers textbook Essentials of Computational Chemistry: theory and models, in Appendix C, he goes over Spin algebra.
I am unable to calculate myself that $S^2\alpha = \frac{1}{2}\left( \frac{1}{2}+1 \right) \hbar^2 \alpha$ as well as that $S^2\beta = \frac{1}{2}\left( \frac{1}{2}+1 \right) \hbar^2 \beta$
Where $S^2$ is the spin squared operator and $S = S_x + S_y + S_z$.
This seems like it should be trivial. It certainly is to him, he says that  repeated application of the following equations lead to the above
$S_x\alpha = \frac{1}{2}\hbar\beta$
$S_x\beta= \frac{1}{2}\hbar\alpha $
$S_y\alpha = \frac{1}{2}i\hbar\beta$
$S_y\beta = -\frac{1}{2}i\hbar\alpha$
$S_z\alpha = \frac{1}{2}\hbar\alpha$
$S_z\beta = -\frac{1}{2}\hbar\beta$
my understanding is that $S^2=S_x^2 + S_y^2 + S_z^2$. However, when using this operator on $\alpha$ or $\beta$ I am not sure what the eigenvalues would be..., relatedly, I am not sure even how to operate just $S_x^2$ on $\alpha$.

The operators work one after the other just as if you are differentiating and the order of doing this matters, so if $S_x^2$, say, operates on $\alpha$  then $S^2_x\alpha\equiv S_x(S_x\alpha) = S_x(\hbar/2) \beta=(\hbar/2)^2\alpha$.  The $S^2$ operator has 9 terms as $S_xS_y$ is not necessarily the same as $S_yS_x$.

I have to go back a step. Particle in a box has mathematical eigenfunctions that can clearly be differentiated. I am not seeing what the mathematical form for $\alpha$ is. Hence, I can't differentiate anything. I am missing a pre-requisite piece of info

$S$ is not equal to $S_x+S_y+S_z$, but rather $(S_x,S_y,S_z)$ or a vector of the spin in each direction. $S^2$ is actually the dot product of this vector with itself, which is why you get $S^2=S_x^2+S_y^2+S_z^2$. Generally, we don't worry about the mathematical form of $|\alpha\rangle$ or  $|\beta\rangle$ and just treat the equations you listed as rules for how they behave. So porphyrin is right that you should just apply your operators one after another and follow the rules you have listed.

I answered it myself. It is trivial, I was starting on the wrong foot when I looked at it.

I would put your solution as a self answer rather than editing it into the question.

I feel like a clown doing that though.. I feel like a clown for having the question to start with ;)

Solution
$S^2\alpha = \left(S_x^2 + S_y^2 +S_z^2  \right)\alpha= S_x(S_x(\alpha)) + S_y(S_y(\alpha)) +S_z(S_z(\alpha)) $
from the ''rules", where for instance $S_x$ operating on $\alpha$ gives $\frac{1}{2}\hbar\beta$, we make the three replacements to get
$=S_x(\frac{1}{2}\hbar\beta) +S_y(\frac{1}{2}i\hbar\beta) +S_z(\frac{1}{2}\hbar\alpha) $
Now take the constants outside so that we have
$=\frac{1}{2}\hbar S_x\beta +\frac{1}{2}i\hbar S_y\beta +\frac{1}{2}\hbar S_z\alpha $
We have our familiar Operator_$\phi$ where $\phi = x,y,z$ and we can follow the rules above for how each operator deals with the respective eigenfunction.
This leads to
$=\frac{1}{2}\hbar (\frac{1}{2}\hbar \alpha) +\frac{1}{2}i\hbar (-\frac{1}{2}i\hbar \alpha) +\frac{1}{2}\hbar (\frac{1}{2}\hbar \alpha) $
This can be simplified to (remembering that $i = \sqrt{-1}$ so $ii = -1$,
$=\frac{1}{2}\hbar^2 \alpha +\frac{1}{2}\frac{1}{2}\hbar^2\alpha  $
and finally
$S^2\alpha = \frac{1}{2} \left(\frac{1}{2}+1 \right)\hbar^2\alpha$
So, it is indeed trivial once you see it in the correct light.

One possible way:
As you say
$${\bf{S}^2} = S_x^2+S_y^2+S_z^2$$
Thus
\begin{align}
{\bf{S}^2} &= \left( \frac{\hbar}{2}\right)^2
\left[\begin{pmatrix}
0  & 1 \\
1  & 0 \\
\end{pmatrix}\begin{pmatrix}
0  & 1 \\
1  & 0 \\
\end{pmatrix}
+
\begin{pmatrix}
0  & -i \\
i  & 0 \\
\end{pmatrix}
\begin{pmatrix}
0  & -i \\
i  & 0 \\
\end{pmatrix}
+
\begin{pmatrix}
1  & 0 \\
0  & -1 \\
\end{pmatrix}
\begin{pmatrix}
1  & 0 \\
0  & -1 \\
\end{pmatrix}
\right] \\
&=
\left( \frac{\hbar}{2}\right)^2\left[\begin{pmatrix}
1  & 0 \\
0  & 1 \\
\end{pmatrix}+\begin{pmatrix}
1  & 0 \\
0  & 1 \\
\end{pmatrix}+\begin{pmatrix}
1  & 0 \\
0  & 1 \\
\end{pmatrix} \right]\\
&= \left(\frac{3}{4}\right)\hbar^2
\end{align}

This is correct, but doesn't answer the question. I appreciate the effort though.


text:
Drug design methods - docking software, binding affinity
Lately I've seen videos and some articles explaining drug design. I would also like to specifically cite this guy, who isn't the most credible guy in the planet, but some of the stuff in the video makes sense to some degree, and he just confused me further.
Now I work on a completely different field on my day job, but I would like to ask a few questions, out of curiosity, because the techniques described in the video really piqued my interest.

Is it the case that when designing drugs (i.e. these proteins) we attempt to optimise binding affinity (maximise?)
How well does Autodock Vina compute this affinity? Is it opinionated? I've seen a few docking software, it looks like there are different methods to compute docking score? (can you recommend further reading for me in this regard)
Is there a linear correlation between binding affinity and drug efficacy? I.e. does it mean that higher binding affinity directly means we have a better drug, or is there some sweet point kinda thing?
Why can't we use very complex molecules, more than 300 characters in SMILES length to combat viruses for example? I noticed everyone was trying to minimise ligands?
How can software like AutoDock tools prepare your proteins for docking, is there a standard method (they add hydrogens, add Gasteiger charges). Is that guaranteed to be accurate in every case?

To clarify, I'm not looking for detailed answers, since I am completely off the domain, I just want to have a basic idea for these questions, and perhaps recommendations on further reading from you guys.
Thank you, and stay strong!

This is drastically too broad.

I mainly work with simulation (MD), so take these answers with a grain of salt.

Is it the case that when designing drugs (i.e. these proteins) we attempt to optimize binding affinity (maximize?)

Computational docking usually employes two methods: A force field based estimation of the free energy of binding, and exploration of the conformational space for the ligand.
After a docking run you will end up with set of different ligands, a set of conformations (or poses) for each ligand, and a corresponding score value for each conformation. The score is usually based on the binding constant and Gibbs free energy (which is based on various numbers of physicochemical parameters, and may or may not include parameters like desolvation and entropic effects). Different software use different scoring functions and methods to estimate the free energy.

How well does Autodock Vina compute this affinity? Is it opinionated? I've seen a few docking software, it looks like there are different methods to compute docking score? (can you recommend further reading for me in this regard).

As far as I can tell, it is hard or impossible to make general statements about which piece of software is the best, since it depends a lot on what you are trying to use it for. Some of the limitations of the AutoDock suite (implicit hydrogens, spherically symmetric H-bond potentials etc.) are given in this publication.

Is there a linear correlation between binding affinity and drug efficacy? I.e. does it mean that higher binding affinity directly means we have a better drug, or is there some sweet point kinda thing?

Not necessarily. A docking run gives you an (rough) estimation of the ligand affinity for the receptor you chose. It might be that the ligand also binds strongly to a lot of other proteins, making it less effective in an in vivo setting.

Why can't we use very complex molecules, more than 300 characters in SMILES length to combat viruses for example? I noticed everyone was trying to minimise ligands?

You can, but not necessarily with the AutoDock suite. Large compounds have too many degrees of freedom, and the software is not capable of properly exploring the conformational space. Usually these large compounds are split in to smaller fragments which are then docked individually.

How can software like AutoDock tools prepare your proteins for docking, is there a standard method (they add hydrogens, add Gasteiger charges). Is that guaranteed to be accurate in every case?

Again, depends on the software. You listed some of the features of AutoDock, and you can find the details of the receptor preparation scheme from their website. It is definitely not accurate in every case. As far as I know, most docking runs are performed on protein structures obtained with X-ray methods, and there is no guarantee that the conformation of the protein in crystal is a good representation of the conformation in the solvent phase.

Thiis is perfect, thank you! Just one clarification, when you say "... ligand also binds strongly to other proteins... less effective in an in vivo setting" does this mean that the ligand could have side effects, i.e. bind to proteins necessary for the proper functionality of a healthy organism, therefore, rendering more harm to the organism rather than providing alleviation?

Yes pretty much. The ligand could have unwanted cross reactivity. Other factors such as bioavailabilty, rate and path of metabolism could also mean that a ligand with great binding may not make a good drug.


text:
If all bond angles in AX₃ are the same, then which of the following are correct conclusions about AX₃?
Question

If all bond angles in $\ce{AX3}$ are the same, then which of the following are correct conclusions about $\ce{AX3}?$
(A) $\ce{AX3}$ must be polar.
(B) $\ce{AX3}$ must be planar.
(C) $\ce{AX3}$ must have at least 5 valence electrons.
(D) $\ce{X}$ must be connected to the central atom via single or double bonds.

My Attempt
This is clearly a case of $\mathrm{sp^2}$ hybridisation. So, I thought about molecules like $\ce{BF3}$. As the bond angle needs to be same, it must not be polar. $\mathrm{sp^2}$ has trigonal planar structure, so (B) must be correct.
But my book says that the answer is (D) only (this question was of single correct choice type).
Why are (B) and (C) not the answers and why is (D) the answer?

Bond angles in $\ce{NH3}$ are equal, but the molecular structure is not planar. That's a good place to start.

@ToddMinehardt Yes I see that rules out (B). But what do we mean by the option (C)? What does it mean when it says that the valence electrons of AX$_3$ is 5? Is it referring to the central atom or the whole molecule's number of valence electrons?

@ToddMinehardt Also, isn't NH$_3$ polar??

Yes, it is polar. As for option (C) - it's unclear to me and I think it's poorly-worded - all you need to do to disprove it is come up with one example of a molecule that has neither 5 valence electrons on the central atom or in the whole molecule. Maybe $\ce{BH3}$?

@ToddMinehardt Boron having 3 valence electrons and 3 electrons, 1 for each in hydrogen would make the molecule have 6 valence electrons wouldn't it?

Correct. And there you have disproved option (C), which leaves you with (D). This approach to problem-solving - find a counterexample to inform you yes/no - is generally useful for multiple choice questions such as this.

@ToddMinehardt But how would that be disproving option (C)? I simply got that there are 6 valence electrons in the particular molecule of BH$_3$. How do I know that there isn't some molecule with 5 valence electrons possible? Because it says "at least".

You're right, I'm wrong. Anyway, find a counterexample and there you go.

(A) and (B) are answered well in the comments by Todd Minehardt and Techie5879. So, I'll summarise their points and clarify (C) to justify the answer (D).
(A) $\ce{AX3}$ must be polar.
Consider the case of $\ce{BH3}$ to disprove statement (A), since it is a non polar molecule with a trigonal planar structure (equal bond angles).
(B) $\ce{AX3}$ must be planar.
Consider the case of $\ce{NH3}$ to disprove statement (B), since it is has a trigonal pyramidal shape and isn't planar, but has equal bond angles.
(C) $\ce{AX3}$ must have at least 5 valence electrons.
Here, the valence electrons of the central atom is referred and not the total number of valence electrons in the molecule because, $\ce{AX3}$ has 3 bonds, thus, 6 electrons are involved in bonding at the very least. If 4 valence electrons are considered, that could equate to only 2 bonds, hence contradicting the existence of $\ce{AX3}$.
With that mentioned, consider $\ce{BH3}$, it has 3 valence electrons, less than the 5 stated in the question. Hence (C) is disproved.
With (A), (B), and (C) disproved, (D) is the correct option by elimination.
(D) $\ce{X}$ must be connected to the central atom via single or double bonds.


text:
Easy way to derivatize silsesquioxanes
I have a polymer end-capped with a polysilsesquioxane (POSS) like this one $(\ce{R} = \ce{i-Bu})$:

I haven't found much info on what type or reactions these POSS groups can undergo. Are there any procedures to modify the POSS moiety and convert it into an amino, carboxyl or triethoxysilane group, for example?
Also, are those $\ce{-R}$ groups easy to replace? How stable are the $\ce{Si-i-Bu}$ bonds?


text:
why is lithium oxide so different from water
Take water and replace hydrogens with the next element down in the periodic table, and you get a white solid, nothing like water.
https://en.wikipedia.org/wiki/Lithium_oxide
Lithium itself is very different from hydrogen, but it's isoelectronic and only has a few more nucleons. You'd think if there was any sense to this periodic table thing they'd be similar...

Take water and drop below 0C. You get a often white solid.

Periodic table only accounts for chemical and some very basic physical properties, it, by no means, takes into consideration the phase in which a compound is which is formed when an element combines with another.

Hydrogen is not isoelectronic to lithium. According to wikipedia, for example, $\ce{CO, CN−, N2, C2^2-, and NO+}$ are isoelectronic because they have the same electronic structure. Typically, elements in the same group are not described as isoelectronic (like the noble gases).
Moreover, hydrogen is less of a typical group 1 element than, for example, fluorine is a typical group 17 element. Looking at the electron configuration, hydrogen has a single electron in the outer shell (like group 1 metals), has its out shell half-filled (like carbon), and is one electron away from noble gas configuration (like group 17 elements).
You don't even have to discuss compounds and ions formed to see that hydrogen behaves very differently from lithium. It is sufficient to compare the elements. Hydrogen is a diatomic gas (like many non-metals) while lithium is a ductile metal.

I see. It's far more complicated than being in the same column.


text:
What is the trend of thermal stability of group 1 oxides?
The thermal stability of most compounds of Group 1 elememts (hydroxides, carbonates, nitrates) increases down the group due to decrement in charge density of the cation.
Now, according to one of my study sources, thermal stability of oxides is as follows:
normal oxide(that of Lithium)>peroxide(that of Sodium)>superoxide(that of Potassium, Rubidium, Cesium).
However another source exactly says:
"The stability of peroxides and superoxides increases as the size of metal ion increases."
I can not understand whether these two statements contradict each other or not. If yes, then which statement is true and what is the actual trend of thermal stability of oxides down a group?

The stability of "normal" oxides decreases, while that of superoxides increases going down the group. This can be reasoned as such - The charge density of metals at the top is higher than those at the bottom, so they attract the "normal" oxide ion with a much higher force which doesn't give it enough time to combine with another oxygen atom to form a peroxide, so they prefer to form normal oxides, but when we go down the group, like in sodium the charge density becomes lower so it doesn't attract the oxide ion with as high force and thus allows it to combine with another oxygen atom. This is why sodim prefers to form a peroxide over a normal oxide. When you look at the bottom metals, you will observe a similar pattern, seeing Rubidium or cesium forming superoxides.
You can also see the stability of oxides by decomposition of the corresponding carbonate of the metal ion i.e more stable the carbonate, more will be the heat required by it to be decomposed and hence lesser stable will it's normal oxide be.

can you establish the general relationship of stability of between these oxides? As in which is more stable than which among the oxides I mentioned above?


text:
Log(Q) is zero, therefore the effect of temperature change is unaccounted for by Nernst Equation?
I have an internal assessment on Nernst Equation. The experiment is seeing how changing temperature effects cell potential. The solution are zinc and copper sulfate with each having concentration of 1M.
I have read online on many sources that Q in the Nernst equation is calculated based on the initial concentration values. In my case that is 1M for both the product and the reactant, hence Q=1 and log(Q)=0. If log(Q)=0, then the whole left part of the Nernst equation(E=E°-RT*log(Q)/nF) is equal to zero (E=E°-0) and therefore E=E°. However, I have changed the temperature and my experimental results do show a change in cell potential with changing temperature, but how can I calculate that change in cell potential with changing temperature theoretically.
Thank you.

The standard potential applies when the temperature is the standard temperature. So changing the temperature changes the standard potential. Now you need to research that and best of success!

The full answer was given by @orthocresol here. Sorry I did not find this last month!


text:
Concentrating Isopropyl alchohol using non-iodized salt
I am using 99% isopropyl alchohol to clean soldering flux residues for my electronics projects. Due to a certain viruse running amok in the world today, I am not able to get from my usual supplier. I am running low on supplies. So i got this idea that i would concentrate the more common (but still not so easy to get in my area) 40% or 70% ispropyl alchohol.
Searching the internet on guides to do it, it would seem that there is an easy way to do it. According to this instructables; add non-iodized salt to the solution until its past its saturation point. By this time the heavier water should sink and the lighter isopropyl should float. Extracting it would now be easy with a syringe.
For my question. Water and especially salt is an enemy to any metal. I would like to ask if this process mixes tiny amount of salt to the isopropyl? if so by how much %? Is NaCl even soluble to isopropyl or are these just tiny particles floating around and if i leave the solution to settle for a longer period of time those particles would sink down. up until what percent can i bring the concentration to?
As for the Salt to use what other ingredients should i look out other than iodine( or other close sounding).
P.S. I am no chemist any way, the only chemistry background i have is during my high-school days. It was a time that i was still not able to appreciate class lessons. So go easy on me

I've never heard of adding salt to create two phases between water and isopropyl alcohol, but if you want to create concentrated isopropyl alcohol I'd suggest distillation. The alcohol should distill off before the water.

Anhydrous magnesium sulfate would do what you want

Or calcium sulfate; see https://en.wikipedia.org/wiki/Calcium_sulfate

@CH3M i have considered distillation at first since isopropyl boil at lower temp than water, but i found that guide on instructables which is a lot easier to do.

@Waylander and DrMoishePippik only easily bought items please :/ or do they have a laymans name :)

Waylander & DrMoishePippik - Both of those salts are hygroscopic. That doesn't seem to be a good thing on a circuit board.

Simple distillation won't work. From Wikipedia "Isopropyl alcohol forms an azeotrope with water, which gives a boiling point of 80.37 °C (176.67 °F) and a composition of 87.7 wt% (91 vol%) isopropyl alcohol." // So you could  dry over a salt, then distill to be sure that there were no particulates in the liquid.

Not a bad question, but all in all there is no "simple" way for a ordinary person without a lab to purify isopropyl alcohol. The chemicals and equipment needed will be a much greater expense than the isopropyl alcohol. Without a proper lab there is also a safety concern. Isopropyl alcohol is flammable.

Enough Mag Sulfate salts the water out from IPA to give a two phase system. It is more effective than NaCl

@Jakequin The typical path to dry isopropanol is 1) distilling this mixture of constant boiling temperature (aceotropic mixture) at 80C which is of 91% isopropanol which than 2) is boiled under reflux for hours with about 200g/L of Calciumoxide before this is distilled (Perrin, Purification of Laboratory Chemicals). However, according to https://www.all-electronics.de/wp-content/uploads/migrated/document/197287/infodirectkyzen.pdf -- written in English, with contact emails -- some modern flux residues are removed by dish washer and water, which could liberate you from isopropanol at all.

I did not read the pdf yet but, are removed by dish washer and water that does not sound instinctively right. although i can see it happening, since dishwashers mostly use hot water which could be enough to melt the flux and get removed by the water pressure.

Common table salt can have additives, so I would recommend working with high purity Epsom Salts (magnesium sulfate hydrate). Heat to remove water of crystallization at 100 degrees Celsius for two hours.
Add your isopropyl alcohol with the MgSO4 in excess and wait until there is a separation. I would decant and allow for a safety margin above the visible separation line.
Here is a reference.

would the isopropyl contain any traces salt after the process?


text:
Disinfectant production
One of my friends says that he made a disinfectant. These are the materials:

83% methanol
isopropyl alcohol
propylene glycol
and the others are: citral, geraniol, linalool, dlimonene, l alpha terapineol...

Does this really work? Is this dangerous for health if I use it only for surfaces?

Methanol is poisonous. While it will evaporate over time this puts methanol vapour in the air, and before it evaporates it is a hazard to touch the surface it is on.

I think we should nix all disinfectant questions as medical advice.

No, it is not a good disinfectant.
To quote from the CDC on chemical disinfectants 'Guideline for Disinfection and Sterilization in Healthcare Facilities (2008)' available here:

Methyl alcohol (methanol) has the weakest bactericidal action of the alcohols and thus seldom is used in healthcare 488.

i forgot to say, he says he added Colloidal silver to this but i dont know how much.
and he also said one liter of this solvent should be mixed in 200 litters of water, and mixing this much of water with the solvent, make the methanol bad effects to go away,
now what?


text:
222nm HARMLESS UV lamp for protecting from virus covid-19
I did asked before , but apparently providing a link to a science report of harmless for humans (both skin and eyes) far-UV light was not enough to make people understand the significance of the solution to the raging COVID-19.
So this link (please read a bit of it):
https://www.nature.com/articles/s41598-018-21058-w
Says that the UV light of wavelength 222 nanometers (the photons bounces off the skin and the eyes) is absolutely harmless and do not make neither skin cancer or cataract. So this light is deadly for viruses and is not commercially available to anyone like ordinary humans (perhaps only some labs got it but sure cannot by on amazon or ebay or so). So if you wish to protect people you ideally need something to kill the virus in mid-air exactly as this lamp does.
Cite: "By contrast, we have previously shown that far-UVC light (207–222 nm) efficiently inactivates bacteria without harm to exposed mammalian skin." - this line is at the beginning of the link above.
So my idea is to help ordinary people that have some DIY experience to make a lamp and disinfect the clothes and themself like skin and face and face mask and so.
So a guy good in chemistry of similar that ca help with his knowledge and experience to help the people and get rid of the virus.
The research I did in the internet says that the UV has ability to kill virus in 10 seconds.
So having a lamp emitting harmless 222nm UV light instead the cancer deadly 254nm UV germicidal lamps will be able to protect from the diseases.
The question is how to make it.
so I suppose you can type a good way to do it and save the half of the western world population (the old people like your grandma and grandpa)?
(Nobody needs glass (or quartz) blowing of a lamp. Neither skills beyond the ordinary DIY - well it might be a need of experienced DIY-er.)

Well I’ll be damned. 222 nm radiation doesn’t seem to affect humans. However, it would be immensely difficult (dare I say impossible) to DIY a lamp of 222 nm radiation. Getting that out of typical radiation sources (blackbody) would require insanely high temperatures. To do so through chemical means is beyond the ability of any DIY.

I find the claim that 222 nm radiation does not cause damage, just because it only penetrates the skin by a small, amount quite extraordinary. The energy is still absorbed and still has the ability to break bonds so how can it be safe?

@porphyrin - because the top layer of skin cells are dead. So, if they absorb it all there is no harm. Just don’t scratch a lot...

The main problem is not making a lamp which emits at 222 nm. It is to be sure that this lamp does not also emit at 256 nm.

The paper claims that 222 nm UV does "not [...] penetrate through [...] the outer tear layer on the outer surface of the eye". That outer tear layer consists in good approximation of water. So what about viruses or microbes under a thin layer of water?

@Jon Custer yes I agree if they absorb it all, but that's the whole point.

https://www.who.int/images/default-source/health-topics/coronavirus/myth-busters/mythbusters-31.png

@porphyrin the original paper postulated that the layer of dead skin that encases the living dermal layers absorbs the UV light, so unless those dead cells become bioactive again, or unless toxic byproducts of photon absorption arise, or unless coverage in dead cell matter is incomplete, damage to living cells could be attenuated to a safe level by virtue of this dead matter shroud.

A deuterium lamp produces a continuous spectrum concentrated around these wavelengths, but it also emits UV in the harmful portions of the spectrum. Given the experiments, it may be impossible to say whether wavelengths outside of 206 and 222 nm are safe, but one can purchase a filter which confines the wavelengths nearly (NOT exactly) to this band: https://www.knightoptical.com/stock/optical-components/uvvisnir-optics/filters/band-pass-filters/interference-bandpass-filters-uv-range-214nm-320nm/214nm-bandpass-filter-125diax10nmbw/ Continuous spectrum is an untried method and could be unsafe.

@pygosceles yes the paper postulated that dead cells absorb all the 222 nm radiation, this is not the same by any stretch of the imagination that they actually do , or even if they do absorb some radiation that some does not get through and so potentially cause cancer.

UVC is very harmful to people and requires specialist equipment to create and use safely
You are not going to be able to create or use lamps that emit 222nM light without specialist equipment, and even if you could it would almost certainly not be safe.
Far UV lamps are widely commercially available for germicidal applications and these usually use mercury vapour fluorescent discharge tubes with no phosphors so have a strong emission around 250nM from a line in the mercury emission spectrum. They are very dangerous to people and have caused significant harm when used accidentally (see this video from BigClive describing what happens when they are used accidentally instead of the much less harmful "black light" tubes commonly used for discos and events).
Another important fact to bear in mind is that UVC lamps of this type have to use specialist glass in manufacture because common types of glass strongly absorb UVC wavelengths. So, while they are readily available, they are not easy for an amateur to make.
You report a claim from a Nature paper than a tightly controlled UVC light of around 222nM wavelength is not harmful to people. I would be far more cautious as the Nature paper did not do extensive safety studies on the actual effect on people and given what we know about wavelengths not much longer, the claim would need some extensive validation.
But you have bigger problems than this in your pursuit of an easy-to-make solution accessible to amateurs. First, you cannot easily get around the problem of the glass used absorbing the wavelengths you want to see: you will need specialist material. Second, the actual excimer lamps as used in the nature paper are sophisticated devices that cannot easily be made by non-specialists (they are far more complex than the relatively conventional discharge tubes used in commercial UV lamps which are essentially the same as conventional visible fluorescent lamps apart from the need for specialist glass). Apart from anything else the lamps have a strong emission at around 258nM, exactly the wavelength known to be very dangerous to people from simple UVC mercury vapour lamps. The NAture paper reported using specialist narrow band filters to eliminate those wavelengths. These are unlikely to be easy to obtain. Third, creating the excimer lamps is a sophisticated process requiring specialist pulsed electronic drivers to get the lamp to work (most likely, according to this reference in the Nature paper, you need to fine tune the gas mixture and the pressure and the driving electronics for the lamp which is far more complex than a simple driver for a fluorescent tube).
And, as for "Nobody needs glass (or quartz) blowing of a lamp", I'm afraid you are delusional. You need specialist glass to let the right wavelengths through and specialist filters to kill the harmful wavelengths. This is not "simple DIY".
You have seen an interesting paper in Nature and assumed it is easy for someone to make in their shed. This is not true. Specialist skills are required and attempting anything without those skills is likely highly dangerous (and that is ignoring the issues of handling the chemicals and vacuum technology needed to create the lamp: both are dangerous).
And, perhaps more importantly, the best way to avoid contamination of others with covid-19 is to wash your hands. There are no easy technological silver bullets here, though, perhaps, professional engineers and chemists can build something safe based on the Nature observation in the future.

Interesting and per your cited source, to quote:

[...] far-UVC light generated by filtered excimer lamps emitting in the 207 to 222 nm wavelength range, efficiently inactivates drug-resistant bacteria, without apparent harm to exposed mammalian skin 13,14,15.

Also, more details:

Far-UVC lamps
We used a bank of three excimer lamps containing a Kr-Cl gas mixture that predominantly emits at 222 nm 25,26. The exit window of each lamp was covered with a custom bandpass filter designed to remove all but the dominant emission wavelength as previously described 15. Each bandpass filter (Omega Optical, Brattleboro, VT) had a center wavelength of 222 nm and a full width at half maximum (FWHM) of 25 nm and enables >20% transmission at 222 nm.

This is a technical area and to be honest, this may not be the best forum on how to make a lamp with a Kr-Cl gas mixture with a custom bandpass filter.
I do get the feeling that they may be an associated patent application somewhere, that may provide more details. And, yes there is at least a patent pending upon performing a patent search.  To quote from the company's website:

The New Columbia University Study again confirming the effectiveness and safety parameters of our patented wavelength (222nm) referred to by Columbia as Far-UVC.
These results confirm studies that we have been doing for the last 12 years. We have patented 222nm for air, liquid and surface disinfection.
Our patents cover all forms of skin and eye disinfection. We are in the process of getting FDA approval for a number of applications.
We call this wavelength Far-UV Sterilray.


text:
1H NMR of cyclohexanecarbaldehyde
Can someone please explain how to assign the 1H nmr peaks of substituted cyclohexane group? I am more familiar with interpreting substituted benzene.
I get that the signal near 10ppm refers to the aldehyde group. The rest of the signal refers to the hydrogens on the cyclohexane group. I also understand that the assignment of Hb (see the picture below).
However, I don't know how I would compare the relative chemical shift between the remaining hydrogens. Since there are multiple peaks, the "Hd"s are chemically distinct. That is also observed when we use do the NMR analysis with higher resolution (See the second spectrum below).
Indeed, I have 1H NMR results with slightly better resolution. For "Hc" and "Hd"s, we can indeed resolve 4 peaks corresponding to signals at ~1.9, ~1.8, ~1.7 ~1.3ppm. The integration for these peaks are 2:2:2:4.
How would you assign the chemical shift for those 4 peaks?

the data is coming from http://www.hanhonggroup.com/nmr/nmr_en/B21091.html
Here is spectrum with a better resolution.

The picture is retrieved from https://spectrabase.com/spectrum/LEvG58hCD1J?a=SPECTRUM_LEvG58hCD1J

How would you assign the chemical shift for those 4 peaks?

Since there is overlap between some of the resonances I would run a COSY experiment to resolve the peaks and find correlations, and make assignments on that basis. A number of things can complicate assignment. First, the CHO group can be in axial or equatorial positions, leading to two overlapping subspectra under slow exchange. In addition you have different shifts for geminal Hs, in equatorial/axial positions, separated by ~0.5 ppm (equatorial further downfield), with resolution into two peaks depending on the exchange time (e.g. temperature). The equatorial/axial H are coupled with a relatively strong J-coupling (~12 Hz). The chemically equivalent hydrogens are not magnetically equivalent, but this is not very important since long distance couplings are small.


text:
Practical considerations for electrolysis of water
I'm interested in electrolysis of water to form $\ce{O2}$ and $\ce{H2}$, as pure as possible given the constraints of a "household chemical" setup. I've read other questions on here (in particular, Best Settings for Electrolysis of water, Does the electrolysis of water produce chlorine gas?, Why would water, and not saltwater, undergo electrolysis? and others), but still have some open questions, in particular concerning the used electrolyte and the material of which the anode/cathode are.
Electrolyte
I'll be using tap water for my reaction that is not chlorinated. However, unspecified salts/impurities might be present that I cannot closer define, but we're talking about infant-safe drinking water (with little Na present). To have an efficient reaction, I will need to dissolve an electrolyte. There are many choices:

$\ce{NaOH}$ (lye): As far as I can tell, the "ideal" candidate for my requirements, but might not be present in a household and might be difficult to obtain. Should form pure $\ce{H2}$ and $\ce{O2}$ if I am not mistaken.
$\ce{NaCl}$ (table salt): Present in every household. Greatest drawback is that as far as I read chlorine gas is formed during the electrolysis. My question would be, is the chlorine emitted at the $\ce{H2}$ or $\ce{O2}$ electrode?
$\ce{NaHCO3}$ or $\ce{KHCO3}$ (baking powder): Present in every household. Does not form chlorine gas. However, as far as I have read, forms $\ce{CO2}$, but again I'm unsure at which electrode.

Electrode material
We can forget platinum and gold right off the bat for price reasons. Stainless steel would be a good choice, but I've also read that pure nickel sheets should be working well. My main concern is corrosion and passivation of the electrode material which then slows down the overall reaction. Would Nickel sheets be preferable over stainless steel? If we use some material that does corrode easily (e.g., regular steel or just run-of-the-mill hardware store items that are some combination of galvanized iron), would the corrosion and solution of the metal in the water affect the gas quality? If so, how?

What is the purpose of this home water electrolysis ?

Stainless is good, no corrosion, conductive, won't affect gas quality. Graphite is good too.

The best choice is sulfuric acid, or sodium sulfate. With NaCl, you obtain mainly chlorine gas at the anode. This gas will get partly dissolved in the water, then will react with NaOH produced at the cathode, producing sodium hypochlorite. With sodium sulfate and sulfuric acid, you obtain H2 at the cathode and O2 at the anode.

Use magnesium sulfate heptahydrate, also known as Epsom salt. It is inexpensive, available in grocery and drug stores (people soak their feet in solutions of it), safe, and works well. Also, sodium bicarbonate is baking soda, not baking powder (a mixture that contains some baking soda). Potassium bicarbonate is just that. For electrodes, use stainless steel, nickel or graphite. You can buy graphite rods very inexpensively on eBay.


text:
Why does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site?
I am currently studying the textbook Mass Spectrometry, third edition, by Jürgen H. Gross. Chapter 2.4.3 Bond Dissociation Energies and Heats of Formation says the following:

The heat of formation of organic radicals and positive ions decreases with their size and even more importantly with their degree of branching at the radical or ionic site. A lower heat of formation is equivalent to a higher thermodynamic stability of the respective ion or radical. The corresponding trends are clearly expressed by the values given in Tables 2.2 and 2.3, This causes the fragmentation pathways of molecular ions proceeding by formation of secondary or tertiary radicals and/or ions to become dominant over those leading to smaller and/or primary radical and ionic fragments, respectively (Sect. 6.2).

Why does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site?
I would greatly appreciate it if people would please take the time to explain this.

What about hyperconjugation comparing primary, secondary, tertiary radicals; or for the row of phenyl, phenoxyl, benzyl radicals?

@Buttonwood What are you referring to? I'm not sure how this relates to my question.

hyperconjugation is one of the forms stabilizing cations and radicals if a $\sigma$ orbital may interact with an adjacent empty or partially filled orbital, e.g., $\pi*$. (https://en.wikipedia.org/wiki/Hyperconjugation).  One methyl group may not offer much stablization (two or three are increasingly better) which eventually lowers the enthalpy of formation for example of tertiary radicals over the one of primary ones.

@Buttonwood I have thought about your comments and posted a related question here https://chemistry.stackexchange.com/q/129076/75460

There are essentially 2 ways to explain this, the modern and possibly more coherent way via „hyperconjugation“ as the other commentators mentioned, the other via a small electron donating effect of alkyl/aryle groups due to EN differences in C and H. QM I think it’s even wrong, but it still offers a nice model for many cases.

Here is a rather simple structure explanation presented on this site. To quote, where 'pieces' refer to an organic molecular with a non-symmetric branching structure:

You’ll notice something – the simpler the pieces are, the easier they are to stack together, which provides a tighter fit with fewer spaces. Here, by putting a kink in the block, we make them harder to stack.
What’s this got to do with chemistry?
When compounds freeze, the process is a lot like stacking bricks. The more symmetrical the molecules are, the easier it will be  and the fewer spaces there will be between the molecules. Fewer spaces = better stacking. Hence, when you compare hexane to its structural isomer, 2-methylpentane, hexane has a much higher melting point due to the regular arrangement of its structure.
Better stacking, higher melting point. Case closed. Right? Not quite.
It’s also about surface area [...]
It looks like as we increase branching, we’re increasing melting point and decreasing boiling point. What’s going on? [...]
Going from “branched” to “highly branched”  makes a molecule more compact and sphere-like. As the surface area of the molecule decreases (remember that spheres have the lowest surface area/volume ratio of any shape) they will become more compact and thus easier to pack. This explains the melting point phenomenon [...]
Boiling point is related to the forces between molecules, which in the case of hydrocarbons is Van Der Waals interactions. If you’ve ever seen microscope images of a gecko’s feet – which allow it to climb walls – you’ll see that there is no adhesive but the pads contain a tremendous amount of surface area. It’s all about the Van der Waals interactions.
As we decrease surface area, we are going to decrease intermolecular Van Der Waals interaction and therefore decrease boiling point.

The summary relationships are presented as follows:

linear versus branched —> higher melting/boiling points due to better stacking and surface area contact.
highly branched vs. branched —> more sphere-like –> better stacking –> higher melting point
highly branched vs. branched  —>more sphere-like – -> lower surface area —> lower boiling point.

The above connects boiling point to branching structure, and this paper "Hypothetical Thermodynamic Properties: The Boiling and Critical Temperatures of Polyethylene and Polytetrafluoroethylene" found here links BP to thermodynamics. Per the opening reference statement: "A lower heat of formation is equivalent to a higher thermodynamic stability".
This completes the argument as to why the heat of formation decreases for organic radicals with decreased size and degrees of branching.


text:
What is the true caloric impact of Alcohol?
I've only had 1 year of basic Chemistry at school, so I am waay too uneducated to answer this question myself:
What is the true caloric impact of ingested alcohol (ethanol) in the human body?
Here are a few things I presume to be fact:

When testing, how many calories a substance contains, it is done by setting that substance on fire, then measuring the missing oxygen, maybe the resulting CO2 and the heat.
Alcohol is poisonous to the human body, and cannot be used directly as an energy source
the human body has to expend energy to turn alcohol (and starch ect) into some usable form. (by oxygenating or hydrogenation or something alike)
every energetic substance we consume will be turned into adenosine triphosphate (ATP), which is the only 'energy carrier' our body uses when powering muscles

Conclusions I draw:

When we test things for calories, we get a good estimate of how much 'energy' it would serve us
yet the testing is just a guideline, as human digestion and internal biochemical processes are a bit more convoluted than the simple 'set it on fire' combustion
When we have two different substances with the same caloric value, consuming them might yield different energetic value to the body

So, how to formulate this mathematically:

tcv is the tested caloric value
ecv is the effective caloric value for the human body
tcv >= ecv
for those sugars, that can be used directly (e.g. converted into ATP): tcv = ecv
for most things, where internal biochems have to turn it into usable form : tcv > ecv
alcohol: tcv > ecv

So how energetically effective is alcohol to the human body, compared to sugars?

on a pure chemical level: ecv/tcv=? 50%? 70%?
considering neutralizing / storing / discarding unwanted byproducts

Ethanol does not bring any energy to our body. it just allows arteries to be a bit enlarged, so that the blood gets quicker to the capillaries, and gives an impression of more heat being produced.

@Maurice Ethanol does have calories; it is converted to acidic acid and funneled into the citric acid cycle: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4338356/

The method is a bit more sophisticated, but still a rough estimation: https://www.mcgill.ca/oss/article/nutrition/how-caloric-value-food-determined

@Maurice Just wrong. But there is an interesting question as to whether conventional ways of measuring food calorific value apply to alcohol as alcohol metabolism doesn't go through exactly the same rout as carbohydrates. It is both an interesting and important question.

@KarstenTheis hey man, thanx a lot for the links! Quite interesting to read. Yet they basically state that the full impact on our metabolical digestion has by far not been understood or measured, that there are only preliminary thesises and estimations, especially not on the chemical level. Also cleared up my assumption about how caloric testing is achieved, but the article is still a little vague on that. Thanx also to matt_black

An interesting study, 'Can calories from ethanol contribute to body weight preservation by malnourished rats?', to quote:

In terms of caloric content, ethanol is a unique drug whose oxidation by alcohol dehydrogenase (ADH) produces 7.1 kcal per gram of ethanol oxidized. However, it is not known to what extent these calories are converted into body weight. There is no controversy about the malnutrition produced by chronic ethanol intake in alcoholics. In addition to the known toxic effects on absorption and synthesis of essential nutrients (1), some important metabolic changes occur. Pirola and Lieber (2) reported that activation of the microsomal ethanol oxidizing system (MEOS) in hepatocytes of alcoholics may consume one third of the caloric content of ethanol and impair ATP synthesis. Lands and Zakhari (3) suggested the existence of a futile cycle in ethanol metabolism which would account for a loss of six ATPs per gram of ethanol oxidized. They proposed that ethanol may be oxidized to acetaldehyde, consuming three ATPs, and acetaldehyde may be again reduced to ethanol, consuming three extra ATPs.
  However, there is some controversy about what happens to ethanol calories when consumption is moderate and not related to the metabolic changes of the liver that characterize chronic ethanol intake. It has been suggested that calories derived from ethanol may function as do calories derived from carbohydrates, producing a thermogenesis value of about 10% (4,5). This thermogenic value may be higher than that of fatty food (3%) and lower than that of proteins (about 20%) (6).
The contribution of calories provided by ethanol to body weight gain has also been studied but there seems to be no consensus about the data from animal and human studies.
Moderate ethanol consumption seems to favor an increase in fat storage and could result in weight gain, especially in overweight individuals (7). In an epidemiological study involving 89,538 women and 48,493 men, Colditz et al. (8) reported a strong negative association between alcohol intake and body weight index for women and the absence of this association for men.
Animal studies are consistent in reporting a decrease in the body weight of rats receiving ethanol solutions as the only source of liquids. Concentrations of ethanol as low as 5% (v/v), which are similar to the ethanol content of a Brazilian beer, or as high as 40% (v/v), solution similar to spirit drinks, are related to decreased body weight gain (9). Similar results have been reported for 20% (v/v) ethanol solution (10).
Different results have been obtained for malnourished animals. Da-Silva et al. (11), studying rats which had been treated with ethanol for 90 days, reported a significant weight gain by malnourished rats (50% food restriction) drinking a 20% (v/v) ethanol solution when compared to malnourished rats drinking water. A more recent study (12) reported improvement in somatic and motor development and a decrease in the mortality rate of the offspring of malnourished rats drinking low doses of ethanol (5%, v/v). These data suggest that malnourished rats can benefit from ethanol calories.
In summary, in spite of the large number of studies on the effects of ethanol in well-nourished animals and humans, there is still controversy about how well ethanol-derived calories can be utilized. Fewer studies are available about special physiological conditions such as malnutrition. Over the last few years, scientific research has mainly focused on obesity, an increasing problem in developed countries, which led us to the false belief that malnutrition was no longer a problem worth investigating. However, there are still 800 million malnourished people in the world (13). The decreasing interest of the scientific community in problems related to malnutrition has left many questions without an answer. Ethanol consumption and its consequences on the malnourished organism are among them.
In view of the importance of malnutrition in Brazil - 22% of the population or 40 million people are malnourished (14), as well as ethanol consumption and alcoholism - 11% of Brazilian population are alcoholics (15), the aim of the present study was to assess the use of ethanol calories in a dose/effect model by evaluating body weight before and after the installation of malnutrition.

My take, ethanol is apparently somewhat unique and there remains some controversy about how well ethanol-derived calories can be utilized. Study classification factors include well-nourished vs. malnourished, low doses (5%) vs high doses (20%), producing significant variations in effects.

This is quite a good article. Thanx for that. The lower part that you did not share here is also quite interesting. Thanx a lot!

@JayC667 see also: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3484320/figure/F1/


text:
Boiling points of ethylene, formaldehyde and oxygen
The boiling points of ethylene, formaldehyde and dioxygen are $\pu{-103.7 ^\circ C}$, $\pu{-19 ^\circ C}$, and $\pu{−183 ^\circ C}$, respectively. I expect formaldehyde to have the highest boiling point of the three because of dipole moment mostly due to the carbon-oxgyen bond. However, I don't know why ethylene has a higher boiling point than dioxygen.
Different from this similar question, where the atoms involved are of very different size, molar mass, and polarizability, the structure and electronic structure of ethylene and dioxygen are more similar. They both have a single conformation, but they have different symmetry, which might have same bearing on the entropy change during phase transition. I know that both the $\ce{C-C}$ bond in ethylene and the $\ce{O-O}$ bond in dioxygen have a bond order of 2, but dioxygen is paramagnetic while ethylene is not.
The latent heat of vaporization is 6.8 kJ/mol for dioxygen and 13.7 kJ/mol for ethylene, so enthalpy seems a major contribution to the difference in boiling point.
What are the differences in intermolecular interactions in liquid $\ce{O2}$ compared to liquid $\ce{H2C=CH2}$ underlying the difference in normal boiling point?

Et tu Brute... Second answer you linked answers that.


text:
Sphalerite type structure
There are binary diamond type structures for example S and Zn form sphalerite. I thought the reason this formed was because of the big difference in the radii of the atoms that allows the smaller atom to fit into the tetrahedral hole that the bigger atoms form. However GaAs also forms sphalerite type structure, their radii are not that different. Why do Sphalerite type structures form?


text:
What did Richter mean regarding an acid base reaction?
Recently I've been reviewing concepts belonging to the history of chemistry. But I came stuck at trying to understand a passage which I read from wikipedia entry (and which it seems has been mentioned in different sources). This is related with the law of reciprocal proportions.
It states as this:

It took 615 parts by weight of magnesia ($\ce{MgO}$), for example, to
  neutralize 1000 parts by weight of sulfuric acid.

There's an existing answer which explains a similar concept (law of multiple proportions) but not specifically to Richter's.
From what I understood in the law of reciprocal proportions there is a set of three elements. Two of them react with a fixed amount of a third element. The ratio of these two elements is the same when they combine between.
For example:
When $63.5~\rm{g}$ copper combines with sulphur produces copper sulphide
$\begin{array}{lllll}
\ce{Cu}&+&\ce{S}&\rightarrow&\ce{CuS}\\
63.5~\rm{g}&&32~\rm{g}&&95.5~\rm{g}\\
\end{array}$
Conversely when those $63.5~\rm{g}$ of copper combines with oxygen produces cupric oxide.
$\begin{array}{lllll}
2\ce{Cu}&+&\ce{O_2}&\rightarrow&2\ce{CuO}\\
63.5~\rm{g}&&16~\rm{g}&&79.5~\rm{g}\\
\end{array}$
Therefore the proportion between the mass of sulphur and oxygen masses is:
$\frac{m_{\ce{S}}}{m_{\ce{O_{2}}}}=\frac{32}{16}=2$
When sulphur and oxygen combine together they make sulphur dioxide.
$\begin{array}{lllll}
\ce{S}&+&\ce{O_2}&\rightarrow&\ce{SO_{2}}\\
32~\rm{g}&&32~\rm{g}&&64~\rm{g}\\
\end{array}$
Then the proportion between these two becomes:
$\frac{m_{\ce S}}{m_{\ce{O_{2}}}}=\frac{32}{32}=1$
The latter is in proportion which is half of the first ratio. Hence follows the Richter law.
But how can I use this information to understand what it was mentioned in the paragraph from above? How is exactly related with equivalent weights?
The second part of the question arises from the fact, that would it be okay to state this?
In an hypothetical reaction between $A$, $B$ producing $C$ and $D$.
$\begin{array}{ccccccc}
aA&+&bB&\rightarrow&cC&+&dD\\
\textrm{1 eq gram of A}&&\textrm{1 eq gram of B}&&\textrm{1 eq gram of C}&&\textrm{1 eq gram of D}\\
\end{array}$
Becoming in
$\textrm{1 eq gram of A}=\textrm{1 eq gram of B}=\textrm{1 eq gram of C}=\textrm{1 eq gram of D}$
Would it be accurate to put it in that way?
or?
$\textrm{number of eq gram of A}=\textrm{number of eq gram of B}=\textrm{number of eq gram of C}=\textrm{number eq gram of D}$
where:
$\textrm{1 eq gram A} =\frac{\textrm{grams of A compound equal to  the equivalent weight of A}}{\frac{\textrm{formula weight of A}}{\textrm{number of electrons transfered}}}$
which is different when considering moles:
$\begin{array}{ccccccc}
aA&+&bB&\rightarrow&cC&+&dD\\
\textrm{a moles of A}&&\textrm{b moles of B}&&\textrm{c moles of C}&&\textrm{d moles of D}\\
\end{array}$
where:
$\textrm{a moles of A}=\textrm{b moles of B}=\textrm{c moles of C}=\textrm{d moles of D}$
So, is what I wrote correct? How can I relate Richter's law of reciprocal proportions to the equivalent weights? How is it related with the paragraph from above? Perhaps does it meant to say that there is a definite amount of reagent which can react with a certain amount of sulphuric acid? Then why isn't it related more with the concept of limiting reagent rather than the equivalent weight? Can someone guide me on that?

I am afraid you are mixing all modern and old concepts. Moles and the concept of limiting reagents did not exist in Richter's time.

It took 615 parts by weight of magnesia (MgO), for example, to neutralize 1000 parts by weight of sulfuric acid

This relatively famous statement has nothing to do with law of multiple proportions but it illustrates the concept of equivalents. All this observation shows is that it always took the same amount of X to react with Y. Hence the ratio of X/Y remained constant.
If we take 1000 parts of sulfuric acid as a standard, then 615 parts by weight of MgO will neutralize it. It will take 672 parts of ammonia to neutralize the same amount of acid.
Note carefully this concept does not even need the knowledge of atomic weights.
Addendum
Also note that Richter's values are incorrect as shown in the comments i.e. if we use moles. Most likely the reason was purity of his magnesia and sulfuric acid. The inaccuracy of his values was realized in the late 1860s but we should respect his idea of constant ratio of reactants. I don't know enough German to check his entire three volumes of books nor have access to them. However a table from 1860s shows the comparison of corrected values (which are still slightly off):
An Introduction to Chemical Philosophy, according to the Modern Theories, by Adolphe Wurtz, 1865.

"It will be noticed that for many bodies Fischer's numbers are widely different from the theoretical figures, and consequently the analyses of Richter from which
they are calculated were not nearly so accurate as those of Wenzel. The inaccuracy of these analyses, and the obscurity of a perplexed explanation, have not been much noticed in giving Richter credit and some authority to
his works."

I totally agree that moles are a more appropriate measure. I can't figure out the reaction however. I'm expecting $$\ce{MgO + H2SO4 -> MgSO4 + H2O}$$ However let's take the parts as grams. 615 grams of $\ce{MgO}$ is 15.3 moles, but 1000 grams of sulfuric acid is 10.2 moles.

Thanks for the additional information to the answer. I'd say that definitely clears up the discrepancy.

@M.Farooq Interesting, but this doesn't clear up the last part of my question. Is it okay to state that the number of equivalents for each reagent and product to be the same? as indicated in my first example given with $A$, $B$, $C$ and $D$?. So overall the statement given by Richter was merely obtained from an experiment not calculated or anything related with that?. What was him referring with parts of?. Was it grams?, liters?, percentage by weight?. or was just any of those?. How did you obtained  that it will take 672 parts of ammonia to neutralize the same amount of acid.?

@M.Farooq I was just checking with the table that you provided. This value was also obtained from an experiment right?. An all in all he compared it with reaction with sulphuric acid?. Did he used other acids to make a table?. Can you please attend the latter questions from my earlier comment?.

@ChrisSteinbeckBell, I used the value from the table. Yes those numbers must have come from the experiments of acid base neutralization. I am not a native English speaker either, however, " 1000 Parts of Sulfuric Acid by Weight" is an old English terminology. By current standards, it would simply mean 1000 [weight units] of sulfuric acid react with X [weight units] of a base B. "Parts of" is just a filler of old language style.

@ChrisSteinbeckBell, Actually, I can explain you equivalents, but I really do not understand your question about equivalents. Please clarify.

@M.Farooq Sorry about any confusion. What I intended to say is (if you look at my first equation which has $A+B\rightarrow C+D$) that you can conclude (from reading what Richter mentioned), that the number of equivalents of $A$ is equal to the number of equivalents of $B$ and the number of equivalents of $C$ and the number of equivalents of $D$.?. Given that the number of equivalents is equal to $\frac{\textrm{mass of A}}{\textrm{equivalent weight}}$. Where $\textrm{equivalent weight}=\frac{\textrm{molar mass of A}}{\textrm{numer of electrons transfered or hydrogens displaced or OH ions}}$.

@M.Farooq In my question I did mentioned 1 equivalent for all, where, let's say for oxygen would be $8,g$. I understand that from the original perspective when this law was given. Neither electrons or atoms were known but from looking it from a modern view. Would it be okay to say what I mentioned above?. So far I only recall that the number of equivalents is $\frac{\textrm{mass}}{\textrm{equivalent weight}}$. Then $\textrm{equivalent weight}=\frac{\textrm{molar mass of the substance}}{\textrm{electrons or hydrogens displaced or OH}}$. But I don't know how could this be drawn as a conclusion.

@M.Farooq I don't know how that can come as a conclusion from what Richter mentioned. Perhaps is it better understood now?. Btw I noticed that parts in that given context were referring to an old terminology.

Richter's statement has nothing to do with equivalents. So the answer is no, you cannot draw equivalent information from Richter's table.  Recall, this is 1790s, so atomic weights and electrons did not exist. So there is no need to draw more information from less. All Richter is saying that for a fixed amount of acid, a fixed weight of a base will neutralize it.

Your understanding of equivalents is correct, otherwise.