text:
How can chemists distinguish pure chemical element specimens that look almost "the same" as well as what deposit is what in a multimineral mined rock?
As a non chemist I am most often charmed when visiting Wikipedia articles of chemical elements and see images of very pure specimens of element after element, proton by proton, and often also metal cube specimen made from smithing similar pure deposits.
The wiki article Periodic table allows me to do so easily; here are some elements I found looking almost the same and don't think I personally could distinguish between them without some instrument:

molybdenum and manganase
titanium and chromium
rutenium and cadmium
sodium and aluminium
silicone and germanium

My problem
Generally all examples are of synthetic specimens of about 99.9% purity but that me to wonder what test can a chemist make in a natural environment, say inside a natural cavemine, to determine which natural deposit contains what after the mining and positioning a seemingly multimineral natural deposit;
Of course, I understand that with natural deposits of mainly carbon or copper, or iron (water exposed and rusty) natural deposits, let alone with very pure specimens of these elements, it will be easy because of color, but I believe that generally or mostly color by itself won't be enough due to various color biases and I assume that today there are in-place very-accurate tests that can be done, especially for different parts of a multimineral deposits.
My question
How can chemists distinguish pure chemical element specimens that look almost "the same" as well as what deposit is what in a multimineral mined rock?

Most elements are not mined pure. Those which are (say, gold and silver) are relatively easy to tell apart.

As you enjoy reading Wikipedia this entry may interest you https://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy

I am quite sad this question got a downvote ; in my opinion, images of very pure deposits of elements might be one of the best ways to do public understanding of science ; I don't even call myself a chemistry amateur although I really like learning new things about chemistry in accessible terms for general audience of chemistry knowledge and here I tried to help others and myself get one of the most fundamental understandings about chemistry.

I’ll give you an upvote for making the point about facilitating public understanding of science! Maybe this link will be of interest: https://theodoregray.com/PeriodicTableDisplay/ . One minor thing: what you call deposits are usually called specimens. Also, manganese, ruthenium and silicon are mispelled.

@EdV thanks ! By specimen did you mean to any "a whole, mined, multimineral deposit" or just to a particular "sub deposit cutted out of a mined multimineral deposit"?

No. I mean element specimens such as the machined rods, arc-melted pellets, etc. shown at, e.g., this link: https://elementsales.com/ . Go there and look at cobalt or hafnium or whatever element you want. These are not mineral specimens and, like Ivan said, are not mined as pure elements.

Thanks @EdV I have just made the question more helpful for future readers (hopefully) by accessing the important distinction you presented, in the simplest words I could.

What you're describing is essentially the field of "qualitative inorganic analysis". That's a very broad subject, but a short description is that prior to the advent of advanced instrumentation, chemists developed extensive series of tests specifically to determine the elemental composition of samples.
For example, one would start with basic physical properties like color, density, and hardness, then look at reactions with heat, acids, bases, oxidants, etc. If the compound could be solubilized in any way, then one can look at how the solubility/color/etc is affected by addition of known cations and anions. If it was a geological specimen, then the location and the types of surrounding minerals would help inform the analysis as well.
A well trained inorganic chemist would have standard flow-charts of tests and could fairly confidently identify the elemental composition of common minerals and metals.

The qualitative inorganic analysis is a science in itself. To know what sort of elements you have in your sample, you have to first make it react with an acid like concentrated nitric acid $HNO3$. At the end of the reaction, you obtain a solution and an insoluble substance. You filtrate the residue. And now you get a solution S1 and a deposit D1.
You add some other solution, say $HCl$, to the solution S1. This will produce a precipitate D2 and a solution S2, that you separate by filtration.
Simultaneously you treat the deposit D1 by a new reactant, like $NaOH$. Part of the deposit is dissolved. You obtain a new solution S3 and an insoluble part D3, that you separate by filtration.
Afterwards, you treat each solution and deposit by a new reagent, to obtain a new solution and a new deposit. It is a long job. And I cannot describe to you how to separate all different metals of the periodic table. It takes months of lab study to master this technique. But the principle is the same overal. At the end you obtain a precipitate which contains only one element. By weighing it, or by titration, you can determine its weight. You may need a lot of different reagents to carry out the nth separation.


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Is the concept of a "self cleaning" banknote plausible?
Although in 2020 there is a big movement of bankers marketers to lean the public towards cashless society, cash in the form of banknotes is still used in many if not most countries.
Cash banknotes can be very dirty in some occasions, especially old banknotes in humid countries and furthermore in humid storage areas.
Is the concept of a "self cleaning" banknote plausible?

No,  but there are currency launderers.

Something like idrophobic and self cleaning paints and coatings might be a possibility, like TiOx nanoparticles.

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HCP unit cell c-axis
Today I was asked to show that the perpendicular height, $H,$ of a regular tetrahedron is related to the length of a side, $L,$ of the tetrahedron, by $H = \sqrt{(2/3)} \cdot L$. I did this no problem.
I was then asked:
Explain how this is relevant in determining a relationship between the unit cell parameters, $a$ and $c$ of a hexagonal-close packed crystal structure: $c = 1.633 · a$.
Could anyone possibly help me out here?

Well, if you know what an HCP structure is, then you could have guessed that the said ratio of 1.633 was not put there on someone's whim.

No I realise it’s twice of a - I think I know what’s going on but I don’t know what they’re asking for here

Imagine the whole construction, with spheres and everything. That 1.633 is not some random decimal number. It must have been derived geometrically, you know, with square roots and stuff.

A picture is worth a thousand words (see the two tetrahedra heights making up the vertical c-axis?):

The HCP packing has layers of closest-packed spheres in an ABAB... pattern. The distance between layers is one height of a tetrahedron, and you need two layers to get a repeating structure along c, i.e. the unit cell parameter.
(picture found at https://chemistry.stackexchange.com/a/111921)

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Is levomenthol superior to Menthol as a pain killer in humans and what are the ways to determine this?
I desire to learn whether levomenthol is superior to Menthol as a pain killer in humans.
I didn't find a wiki article about levomenthol and the article Menthol doesn't mention this molecule.
The chemical formulas are the same but I understand that the only difference is the spatial position of chemical bonds between the atoms, as can be learnt in pubchem:

https://pubchem.ncbi.nlm.nih.gov/compound/Menthol
https://pubchem.ncbi.nlm.nih.gov/compound/l-Menthol

My problem
I don't know what are the main ways for chemists to decide which one should work better in humans but hope for a simple explanation for non chemists.
I thought about this after reading that Strepsils, the famous British sore throat lozenges, contain levomenthol.
I think about this because I know some other medicines are "levo" versions of others (for example, cetrizine and levocetrizine).
My question
Is levomenthol superior to Menthol as pain killer in humans and what are the ways to determine this?

Oookay, I can see you try there, but menthol isn't exactly a painkiller and l- (levo) is just a description of isomer, another being d-  (dextro)  https://en.wikipedia.org/wiki/Dextrorotation_and_levorotation

@Mithoron Here is a starting point to read on the analgesic properties of menthol, and the channel it binds to: https://www.ncbi.nlm.nih.gov/pubmed/29524352

Another paper: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6600640/

@KarstenTheis Tell that to House M.D. Strepsils isn't exactly Vicodin.


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Why does hydrogen have a lower ionization energy than fluorine?
I found here that the ionization energy of hydrogen is $\pu{1312kJ/mol}$ and for fluorine, it is $\pu{1681kJ/mol}$. Now clearly, from the data, we can see that hydrogen has a lower ionization energy when compared to fluorine.
I did not expect this because I saw here that fluorine is twice as big as hydrogen (I am referring to the empirical values of the atomic radii). This means that it would be significantly easier to remove an electron from a fluorine atom than from a hydrogen atom. That's why I expected hydrogen to have a higher ionization energy. But I found that it's the other way round. Why is it so? What is wrong with my reasoning? Is this an exception?

@SirArthur7 btw, you don't need to typeset everything in MathJax, so you can save yourself some effort; 1681 kJ/mol is typographically just as correct as $\pu{1681 kJ/mol}$, and H2O (try it on a post) is just as correct as $\ce{H2O} $\ce{H2O}$. The most important thing is to be consistent. In this case, it arguably looks better without the Jax, because then it fits into the surrounding text better. See: https://chemistry.meta.stackexchange.com/a/2935/16683

@orthocresol You know what? Your linked post was for long an ideal doubt of mine, Thanks a lot for clearing out this, this'd surely save myself from needlessly editing units, but if it's really not a problem could I use \ce for $\ce{H2O}$, is it the buffering time that you want me to avoid? If so then H2O could be fine.

Fluorine is bigger, but there are $9$ protons in the nucleus, compared to $1$ in case of hydrogen. So, the outermost electron in fluorine experiences more attraction than what hydrogen's single electron experiences, in spite of larger size of fluorine.
However, the outermost electron in fluorine doesn't experience attraction from all the $9$ protons (nuclear charge). Due to improper shielding, the effective nuclear charge experienced by one $2p$ electron of fluorine becomes -
$\mathrm{ [9 - {(6 \times 0.35) + (2 \times0.85)}]= 5.2}$.
In comparison, the single electron in hydrogen experiences effective nuclear charge of $1.0$
Note: For hydrogen, there is no other electron to shield nuclear charge. So, effective nuclear charge = actual nuclear charge.
Source: To know how to calculate effective nuclear charge visit this Wikihow link.

But then the ionization energy should increase down the group due to increasing nuclear charge. Why does that not happen?

According to Coulomb's law, attractive force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
So, it is basically a competition between the two as Ionization Energy estimation involves calculation of attractive force between the nucleus & the outermost electron.

In the same group, the size factor dominates the charge factor. And that is pretty regular in most cases. But the comparison is tricky when you compare between different groups & periods.
(Note: If you calculate the Effective Nuclear Charge for each members of a group, you will get a clearer idea about why the size factor is getting dominated.)
Li has lesser ionization energy than H. But, along a period Ionization energy increases gradually, with some exceptions, and becomes the maximum at the Noble Gas element.

There is confirmation of the ionization values being for Flourine $1680$ $\pu{kJ/mol}$ and Hydrogen $1312$ $\pu{KJ/mol}$ from the Theoretical Chemistry Textbook (online link).  Theoretically, if you have a chemical like $\ce{HCl}$ it is very understandable that it forms an ionic bond, because the the out-most electron is held with less force for Hydrogen than with Chlorine which takes the electron from Hydrogen in an ionic bond. And Florine is in the same column in the periodic table as Chlorine, so its behavior is similar.
At ValenceElectrons.com the Aufbau electron configuration of Florine is listed as  $1S^2$ $2S^2$ $2P^5$:

It is then important to understand the Aufbau principle further as it represents hybrid energy levels because of a combination of Theoretical Hydrogen charge distributions.  The $S$ $orbitals$ and $P$ $orbitals$ are somewhat combined in atoms aside from Hydrogen, and their energies can be shown in the picture as hybrid $P$ $Bonds$:

In the referenced diagram, just above, due to the Aufbau principle and hybrid orbitals, the energy of the $2$ $S$ $Orbitals$ are lowered, making the energy-well deeper because of the hybrid bonds from the $2$ $S$ and $2$ $P$ levels.  That means that going across a row in the periodic table (with $S$ and $P$ levels to it) that generally it becomes more difficult from left-most (easiest) to right-most (most-difficult) element to remove the outer-most electron, as a general trend across the row.
In the far-field, the $2$ $S$ electrons are not fully shielded and are attracted with a partial $+2e^+$ charge, resulting on a tighter hold of these electrons and a greater bonding energy then normally expected for Hydrogen which in the far-field has no-shielding and only a $+1e^+$ to hold the electron in its shell.
The Theoretical Chemistry Textbook further explains:

The Effects of Electron Shells on Ionization Energy
Electron orbitals are separated into various shells which have strong
impacts on the ionization energies of the various electrons. For
instance, let us look at aluminum. Aluminum is the first element of
its period with electrons in the 3p shell. This makes the first
ionization energy comparably low to the other elements in the same
period, because it only has to get rid of one electron to make a
stable 3s shell, the new valence electron shell. However, once you've
moved past the first ionization energy into the second ionization
energy, there is a large jump in the amount of energy required to
expel another electron. This is because you now are trying to take an
electron from a fairly stable and full 3s electron shell. Electron
shells are also responsible for the shielding that was explained
above.

For Florine, before ionization, it has seven outer shell electrons.  Before ionization, the outer shell structure can be described as s s p p p p p (since the inner $1$ $S$ shell stays basically the same in before and after ionization). The shell has more of a s characteristic then after ionization.
Before ionization, the outer shell structure of Florine can be described as s s p p p p p.  After ionization, the outer shell structure can be described as s s p p p p which has a greater s influence.
The general trend versus row position can be viewed at the below diagram:

I have calculated this effect of the ionization energy of Helium being greater than that of Hydrogen, with some accuracy.  The calculations for Helium are referenced my the free online-book here. The same concept of incomplete shielding for Florine is what is dominating its higher biding energy than that for Hydrogen also.

Your intuition is spot on : the issue is with the definition of atomic radius, which according to that same Wikipedia page "is not a well-defined physical entity, for which there are various non-equivalent definitions". This ambiguity finds its roots in the quantum mechanical description of the atom.
If you use the column with the "calculated" value, you'll see that the radius for hydrogen is 53 and for fluorine 42, i.e. hydrogen is bigger than fluorine. Correspondingly, the ionization energy is smaller, roughly by a factor $\frac{42}{53}$, in line with your reasoning.
Mills* classical model of physics allows us to calculate these radii and ionization energies fairly precisely, as it describes electrons as having fixed distances from the nucleus. The classical Bohr radius (52.91pm) for hydrogen, and 0.7807 times the Bohr radius (41.39pm) for fluorine. The approximate ionization energy, in Rydberg atomic units, is then simply given by $\frac{Z}{r}$ i.e. 1 for hydrogen and $\frac{Z-8}{r}$, or 1.281 for fluorine, with each electron shielding exactly one proton. (Multiply by 1313 to convert to kJ/mol)

*Randell L. Mills, The Grand Unified Theory of Classical Physics, 2018 edition (ISBN: 978-0-9635171-5-9) see equations 10.182 and 10.183


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What do you mean by "periodic" in the periodic table?
Modern periodic law states:

“The physical and chemical properties of the elements are periodic
  functions of their atomic numbers”.

But I don't think this is so! (Forgive me for my stupidity. But please try to understand what I am saying and if you find any mistakes then please let me know).
If I consider the property of Li then I find that after 7 element there comes an element Na which have similar properties like Li. Now according to periodic law I must get another element which has similar properties to that of Li after 14 elements. And indeed I found one named K . Now again the same thing must happen (no?) ie, there must be an element which have similar properties to Li and is 21 elements away(Mn). But I don't find such element! Why?
The properties of Element is periodic so why then this "periodicity" doesn't hold in case of Li?

There is a pattern, but the period is not a constant.

@orthocresol but the word PERIODIC means : occurring or recurring at regular intervals [ https://www.merriam-webster.com/dictionary/periodic ]. Why then scientist are using such word which can led confusion in a law?

Well, https://www.lexico.com/en/definition/periodic suggests "appearing or occurring at intervals" as their definition, so not as specific. Yes, English can be a bit weird, but the fact remains that technical definitions and uses may diverge from standard definitions and usages. Language is not static.

I think most people understand that when someone says they go to the doctor periodically, that does not imply exactly equally-spaced intervals.

@JonCuster Science starts from words used in daily life and then make changes to it for it's own use. Indeed. But in this case a better word is "intermittent": .... stopping and starting repeatedly or with periods in between. [ https://www.google.com/amp/s/dictionary.cambridge.org/amp/english/intermittent ]

@Idk, As you can see in the answer, periodic concept is derived from mathematics. Hope you understood the answer. Periodic functions must be taught in grade 10,11 or 12 in the context of trigonometry. Intermittent has a very different context. Intermittent means "ceases for a time; coming at intervals; operating by fits and starts." The properties of elements don't have this fits and starts or ceasing of properties.

contd...One can have intermittent fever, which comes and goes but one cannot have periodic fever which will imply that fever comes at exactly the same time and goes away at exactly the same time, every day. Imagine a situation of fever starting at 8:00 am, ending at 8:00 pm, then starting at 8:00 am and then ending at 8:00 pm.

Science does not start with words from daily life. Science uses words which may be incorporated into daily life. Take "quantum mechanics." In daily life, no one knows what "quantum" is, and "mechanics" fix cars.

@Zhe This is even written in Einstein's book The evolution of Physics "But science must create its own language, its own con cepts, for its own use. Scientific concepts often begin with those used in ordinary language for the affairs, of everyday life, but they develop quite differently.
They are transformed and lose the ambiguity associated with them in ordinary language, gaining in rigorousness so that they may be applied to scientific thought. " https://b-ok.cc/book/2626393/ada93b  .

First you need to understand what a periodic function is in mathematics. For example, if you are standing at any point P on a circle, you start walking on the circumference and count the degrees you completed. Once you complete 360 degrees, and you will find yourself on exactly the same point P. Sines, cosines and many other functions are periodic. This term existed before the chemists started using the term periodic law in the early 1800s or perhaps well before that. If $f(x)$ is a function with period $T$, then $f(x+T)= f(x)$ for a periodic function in a defined range of $[a,b]$. The period $T$ may change to $T'$, when x is in a different range say, $[c,d]$. This is  an example of a discontinuous periodic function. As you will see, periodic properties of the elements form a discontinuous periodic pattern in a loose sense.
Now come to the periodic law: "The law that the chemical elements, when listed in order of their atomic numbers (originally, atomic weights), fall into recurring groups, so that elements with similar properties occur at regular intervals." [unabridged OED].
Instead of $x$ for a mathematical function, the independent variable is the atomic numbers Z (Zahl for a number in German) for the chemists. So this periodic law has a very loose connection with mathematical periodic function, because none of the values are actually repeated. Each element is unique, but in a loose way, their properties or even a better word is trends "repeat" themselves after a certain number of elements. See this chart for example. Atomic radii vs. Z, do you see a loose periodic pattern?

Once you have got the idea of periodic function in a loose way for the chemical elements, get ready to the see the physicists style of periodic table. It explains why some properties loosely repeat themselves.
Each element in the column, corresponds to filling of a certain orbital, for example look at the column $s^1$. All elements falling in that column correspond to the filling of $s^1$ electrons. Then move to the next column $s^2$, all elements listed there corresponding to filling of $s^2$. Can you see a pattern? In this sense, one calls it periodic function in a modern way, that certain elements repeatedly fall in $s^1$ column, some elements repeatedly fall in $s^2$ column.


text:
Is there a way to create natural Red food coloring from Turmeric in the house kitchen?
As a non chemist I have read in the Wikipedia article Curcumin that:

Curcumin becomes bright red when it interacts electrostatically with
  phospholipid film.

This text appeared without source under this image in the article:
https://en.wikipedia.org/wiki/Curcumin#/media/File:Curcumin_thin_lipid_film.jpg
As an amateur of cooking Indian dishes (which often have bright red "Tandoori" color):
Is there a way to create natural Red food coloring from Turmeric in the house kitchen?
That is to ask; what edible supermarket products contain phospholipid films and how should I use at least one together with Turmeric (fresh-grinned or dry-powdery) to get a bright red solution?

Update
I should note that I didn't intend to ask this question as a "how to make Tandoori colored seasoning paste" question;
About six years ago, I was told by the owner of a large Indian restaurant in Israel, what is the secret (or at least one of the secrets) and the following seems to be that secret from what I understood then and did and replicated several times:
1) Take about 500 gram of dried Kashmiri peppers; clean them from pedicel, placenta and seeds.
2) Put the dried Kashmiri peppers in a metal pan and pour boiled water upon them.
3) Let the dried Kashmiri pepper boil and cool ("water-marinate") about 3-6 hours but don't cook them.
Mere boiling, without cooking, enhances pigmentation but doesn't destroy some pigments as in cooking.
4) Remove the water and let the peppers dry.
5) Grind the processed peppers into a paste with a bit of water and cooking oil and add a tablespoon of that paste as a basis for any Tandoori paste you are making (along with salt, lemon, garlic, ginger and possibly other spices).

Turmeric turns distinctly red at a pH above ~7.8 (we had this experiment in school). Considering that you're cooking tandoori, you would probably use salt and lemon in the seasoning, which would cause the medium to be acidic and revert turmeric into its natural yellow colour.

@AniruddhaDeb in great plea ; please publish this in an answer such as "adding on what Dr. Pippik said"; as you probably know, comments might get deleted with time from various reasons, and yet this is very important. If you choose to do so --- please say if potassium would be a seasoning alternative to salt to prevent yellowization.

Alright will do! Im in the middle of exams so I'm a bit squeezed on time but I'll try to write an answer tomorrow

@AniruddhaDeb if you still have the will (and hopefully if you have spare time), about yesterday the question was reopened and you are now available to publish an answer; kindly,

Yep I'm working on an answer right now. Some practical stuff is involved (and I want to take pictures) So I'll post in around a day from now.

@JohnDoea You just wanted reddish color, what about beet juice? It will impart reddish pink color to foods

@M.Farooq I thought about it a few years back but didn't find any way to implement it to make Tandoori color - perhaps the powder of some species would be helpful.

@JohnDoea, The tandoori color on chicken is mostly a food dye in all ethnic restaurants because most Westerner would not like a very spicy chicken. The reddish color in real tandoori chicken comes from adding dried red chillies to the $masala$ (all species). and sometimes tomatoes. When you fry red chillies in oil or tomatoes, the oil develops a reddish color. It is beyond my tolerance as well. Another source of nice organe color is saffron  and as you know genuine saffron is extremely expensive. https://tasty.co/recipe/homemade-chicken-tikka-masala

@M.Farooq yes, and there is also the Ratan Jog / Ratin Jot (Alkanna tinctoria root) common in Vindaloo; but indeed, some real, non-lousy Indian restaurants use a paste based on the kashmiri boil-marinate paste I described in the update; but it is indeed rare to find in comparison to artificial dye.

@JohnDoea, Interesting I wasn't aware of ratan jog, but I would say it is extremely rare in typical South Asian household foods. Apparently there are side effects.  https://punampaul.com/kashmiri-rogan-josh/

Curcumin(the active ingredient in Turmeric) can turn red in two ways: Either by acting as an acid-base indicator or by forming a complex with a phospholipid film, as you stated.
Colour change due to Acid-Base action
The curcumin molecule is a tribasic acid in basic solution, where it exists as its enol tautomer. The loss of one or more protons from the phenolic hydroxyl groups convert the benzene rings into their quinonoid form, which have a larger absorption/emission wavelength than the benzenoid form, giving a red colour.

This indeed works experimentally: a paper highlighting this can be found here. A simple home experiment would also suffice

The left one is a suspension of turmeric in water; the right one is the same suspension, but with a few drops of soap added to it, which make the medium alkaline and cause the turmeric to turn red. Unfortunately, around my kitchen, I could not find an edible compound basic enough to cause turmeric to change colour. Lime is a possible compound that I could test with but I don't have any foodgrade lime powder lying around. Another issue with using this colouring is that any acidic medium will revert the colour back to yellow: this means that sprinkling lemon as garnishing on your dish would change the colour from an appetizing red to a dull yellow.
Colour change due to Phospholipid complex
A lot of papers have been released on this topic, such as this one and this one. The second paper mentions the following synthesis method to prepare curcumin liposomes (curcumin surrounded by a spherical membrane of phospholipid)

The MFGM phospholipids/soybean lecithin was dispersed in chloroform. Curcumin was dissolved in anhydrous ethanol and thoroughly mixed with the MFGM phospholipids/soybean lecithin dispersion in a 250-mL round-bottomed flask. The mixture solution was evaporated on a rotary evaporator at 35°C, until a thin film was formed on the flask. The film was washed with PBS and completely hydrated for 2 h. After the hydration of thin lipid film with PBS, the lipid dispersion was sonicated for 4 min, and finally the liposomes encapsulated with curcumin were obtained.

For cooking food, this above synthesis method seems unreasonable and overkill. I urge you to experiment with a few methods of forming the red curcumin-phospholipid complex yourself, with soy lecithin/eggs. I would have tried some methods out but I don't have these ingredients at home. see edit.
The advantage of having curcumin in a liposome is that it is much more stable: unlike the acid-base action, curcumin in a liposome remains stable across a wider range of pH's, as shown in this graph. This method does show promise, but as I mentioned, you would have to give this a try yourself.

Other practical solutions
Seasoned Advice would probably give you would get a better answer in this field, but I'll drop in my two cents. Generally, chilli powder is used to provide the red colour to Tandoori. More specifically, Kashmiri chilli powder is specifically used for tandoor preparations, such as tandoori chicken and chicken tikka. This powder is applied as a dressing (in combination with oil, rice flour etc after cooking the food) and during marination (chicken is marinated with curd, red chilli powder and turmeric).
Happy cooking!
References:

Kuntal Maiti, Kakali Mukherjee, Arunava Gantait, Bishnu Pada Saha, Pulok K. Mukherjee; Curcumin–phospholipid complex: Preparation, therapeutic evaluation and pharmacokinetic study in rats, International Journal of Pharmaceutics, Volume 330, Issues 1–2,
Sam-ang Supharoek, Kraingkrai Ponhong, Watsaka Siriangkhawut, Kate Grudpan;
Employing natural reagents from turmeric and lime for acetic acid determination in vinegar sample, Journal of Food and Drug Analysis, Volume 26, Issue 2,
Hong-Hao Jin, Qun Lu, Jian-Guo Jiang; Curcumin liposomes prepared with milk fat globule membrane phospholipids and soybean lecithin, Journal of Dairy Science, Volume 99, Issue 3,
Research gate: Why does curcumin have a red colour in the alkaline condition
Wikipedia: Phospholipids, Curcumin

EDIT: I tried to make a phospholipid complex with turmeric and egg yolks in the kitchen. Around a teaspoon of turmeric was mixed with an equal amount of egg yolk and heated in a water bath: An orangish colour was observed, which was in contrast to the bright yellow of the egg yolk. (I scraped the top layer away which had a larger amount of turmeric to reveal the bottom layer, which had more egg yolk)

A possible conclusion that can be drawn from this experiment is that the complex is indeed forming, but the concentration of phospholipids/curcumin is too small to impart the magnitude of red colour required for food colouring. This pretty much conclusively proves that no easily made derivative of turmeric can be used as red food colouring.

Many foods have high phospholipid content. Eggs and soybean have comparatively high amounts, for example. Phopholipids are essential components of all cell membranes, and are in virtually all living things. They function as emulsifiers, one side of the molecule adhering to water (hydrophilic), and the other to fats (termed lipophilic or hydrophobic).
You can buy food-grade lecitithin, derived from various sources, e.g. sunflower seeds, soy beans, eggs, fish, etc., for about US$7.00/100 g.
So I suggest you to try mixing the turmeric with eggs or soybeans or plain lecithin and to let us know if the color changed to an appealing red! Or, as others stated, use some alkali, such as baking soda, though that certainly would change the taste.

Hello Dr, thank you ; just one question - about using baking soda and turmeric, must it be in an oily solution right? Is there a best oil or group of oils for such solutions (refined/cold-pressed)?

As stated by Aniruddha Deb, the pH is the cause of the color change. See https://en.wikipedia.org/wiki/PH_indicator . Read, and learn.

I assume Dr. Pippik won't be mad if I give the green check mark and 15 rep points to Aniruddha Deb who worked very hard to construct his answer which included direct testing.

Some common materials you might have at home are baking soda (sodium bicarbonate) or washing soda / soda ash (sodium carbonate). Adding them to hot water + turmeric makes a solution that stays bright red for quite a while (as others have said, it will revert over time, but in my experience with dying fabric it's a matter of several hours, so it won't affect food you use in the space of a meal). Avoid using lemon or anything acidic because that will send it back to yellow (even if there's baking soda in it). If you add a small spoonful of baking soda, the flavor won't really be influenced (especially if you're using other spices with a strong taste).

Do you use dying fabric when cooking? Maybe you want to make this part a little more clear.


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Why is iso-octane different from other iso-s?
While learning nomenclature of organic compounds I found the common  names to be a bit difficult to learn than the IUPAC ones. One of these common name stuff is the iso-,neo-, sec-, etc. So I searched for the ways to differentiate between these on the YouTube and found this video (note that the anchor of the video uses hindi for communication so it might not be intelligible to most people here). To summarize what the anchor said:

Prefix-iso: is used when all carbon atoms except one form a continuous chain.
$$\mathcal {OR}$$
They have a structure like this:

Now in the same video the anchor says that iso-octane is an exception to this rule, as it has the following shape:

Before this came in the video I formed the view that these are just the ways of naming compounds with similar structure but now I think that that isn't the case. So:

Why iso-octane has shape different than the other iso-s?

The anchor says that the structure of iso-octane was determined experimentally. This implies that there is some way to identify these compounds without knowing their structures beforehand so:

How do we know that whether some given compound is iso-, neo-, etc?

What might be the most general guidance for using iso-, neo-, sec- and tert- prefixes?

Do not try to learn chemistry from youtube videos of questionable background.

"secondary" and "tertiary" (+"primary", "quarternary") are clearly defined (by IUPAC) structural descriptors. "iso" and "neo" are not.

@Karl thanks for the suggestion. Actually this was the first time I watched a video from that channel and thought that it might be a legitimate one given that there are 100k+ subscribers. Though you are right and I do try to avoid such things but for now I wasn't able to find any other resources online so I had to reside for that video.

This would be one that is reasonably reputable: https://en.wikipedia.org/wiki/Alkane#Nomenclature

Looks like a slang term that got commonly recognized.  It's definitely unpleasant to the eye, much like $`` \mathrm{C} 0{}_{2} ".$

The usage of non-standard prefixes such as "iso" has never been defined with hard, unambiguous rules such as those in the IUPAC recommendations. Instead, their usage has often more been "by convention". In Organic Chemistry (1st ed.) by Clayden et al., the authors write (p 315):

Strictly speaking, if the standard meaning of ‘iso’ is followed, the name isooctane should be reserved for the isomer 2-methylheptane. However, 2,2,4-trimethylpentane is by far the most important isomer of octane and so, historically, it has been assigned this name.

(This phrase is also found at Wikipedia, where it is copied wholesale without any indication that it's a direct quote. Please don't do this in your work. Even if you provide a citation, you should either paraphrase the information, or clearly indicate that it's a quote, e.g. by enclosing it in quotation marks.)

If I remember right from ~25 years ago quoting my chemistry teacher "Iso-" [equal] indicates a mixture of carb-molecules where the longest chain is the name-giver .. as in iso-octane being an 8-carbon chain. .. Other possible molecules use side-chains of 1 or 2 carbons length.


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Can density functional approximations including dispersion correction perform worse than without it?
I am trying correlate some experimental properties of a molecule to its calculated equilibrium geometry. To do so, I calculated the geometry and energy of some conformers.
Now, I obtain mixed results when performing the same energetic calculation with dispersion-corrected (D3) and not corrected functional (B3LYP): when no correction is applied, a conformer has lower energy, while with D3 correction another conformer has lower energy.
The idea of using D3 looked reasonable, as the molecule has two aromatic and two aliphatic wings, which from the D3-uncorrected geometry seemed to promote to some degree an aliphatic-aromatic stacking interaction.
The point is: my intuition drove me to the use of the dispersion correction, and indeed the D3-corrected result would agree wonderfully with spectroscopic results. But is there a case in which a D3-uncorrected result would be more accurate than the corrected one? Or, alternatively, is there a way to understand quantitatively if the uncorrected model does not describe correctly the system?
I saw a lot of reviews talking about the "benefits" of DFT-D3, but could it do worse than DFT alone?
Edit, to answer Marvin's questions:
Yes, the molecule is organic. It all started with a "routine calculation set-up": b3lyp/6-31g*, but it later became evident that more refinement was required, so the current level might be seen as an expansion of the original calculation. Beyond the 6-31+g** level, I don't stress on Pople's basis anymore, and if more refinements will be required, Alrichs basis sets will be employed. Yes, I omitted it, but B-J damping was employed. At the moment I have optimized one structure with D3(BJ), the other one is being optimized. Of both the geometries optimized with B3LYP, the D3BJ-energy was calculated, together with the uncorrected one.
The two molecules arise indeed from a semiempirical conformational search, followed by a fast and inaccurate hf/3-21g* optimization on 100 geometries. A dozen of low energy conformers was optimized with hf/6-31g*, and the two outliers/low lying conformers were subject to further study with the methods discussed. The molecule contains 90 atoms, and the calculated energy difference is less than 0.5 kcal/mol, but as NMR experiments show a net predominance of one geometry, it's likely that the low energy difference is coupled with an high isomerization barrier, which I still haven't calculated: at the moment I am focusing on the equilibrium geometry.
As soon as the "new" geometry calculations will complete, I will probably do more calculations with a functional better suited for dispersion corrections, thank you very much!
Note: an NMR calculation is planned, but as good results require time and a strong basis set, it will probably be the last thing I will do, starting with better geometries.

Very, very unlikely. Especially since version 3 it should very generally yield improvements only. Now the real issue may well be the functional you are using. While B3LYP is popular, its many failures have thoroughly been reported. You have not been very specific about the differences either, that may very well be within the general error of the method. Another thing to consider is basis set. But again: DFT-D3 will very likely be the smallest source of error.

Thank you! I used B3LYP because originally I didn't consider implementing it with D3; as to the basis set, at the moment the basis set is 6-31G(d,p) for the geometries, and energy calculations were run on some geometries at the slightly higher 6-31+g(d,p). The molecule is neutral, non ionic and with no radicals, so the basis set looked to me like a good compromise in terms of performance and time, but I might consider improving it in the next days

I'm not a big fan of Pople basis sets, but the set-up makes sense. I guess your molekel is organic, so it is surprising to see b3lyp fail. Can't make reasonable statements without knowing more specific things. General recommendation: use many functionals for initial benchmarking. If you need something fast, go with M06L (or later, no dispersion needed), BP86-D3BJ, PBE-D3BJ, TPSS-D3BJ. Hybrids: M06, PBE0-D3BJ, B3LYP-D3BJ, etc... You'll probably find an outlier within those... Are you implementing the DFA yourself?

More questions: Did you use Becke-Johnson damping? You have optimised the structure with/without D3, then calculated the energy on the opt. struc.? What is the energy range we're talking about? Another recommendation: Try xtb and crest (https://github.com/grimme-lab/xtb/releases) for automatic conformer search; it's semi-empirical, so it's incredibly fast, produces very good structures.

tldr; I'll check the detailed results, but the overall answer is that dispersion correction is absolutely worth it
We just finished a benchmark over ~700 molecules with up to 10 conformers, so ~6700 single points at the DLPNO-CCSD(T) / def2-TZVP level. I will post a link to the preprint when it's available soon.
In the meantime, here's a quick summary for the effects of dispersion corrections:

Comparing un-corrected PBE, B3LYP, and ωB97X single-point energies to DLPNO-CCSD(T) illustrate a significant effect. The uncorrected median R2 values drop by ~0.12, and the median Spearman correlations drop by ~0.08. For example, the median R2 of B3LYP / TZ drops from 0.920 to 0.706 without the D3BJ dispersion correction.

The metric here is the correlation between a method (i.e. B3LYP) with the DLPNO-CCSD(T) energies.
I'll look to see if there's ever a case where the dispersion correction does worse. But there are many, many papers illustrating that conformer energetics are dominated by non-bonded interactions .. which are improved with dispersion-corrected density functional methods.
Your methodology seems solid, although if you want good energy evaluations of different conformers, I'd consider either RI-MP2 or DLPNO-CCSD(T) single-points. Our timings in the paper show that RI-MP2 can be about as fast as hybrid DFT methods, but more accurate at ranking conformers.

Any consideration for F12 computations at your CC level of theory?

AFAIK, there’s not a DLPNO-CCSD(T)-F12.. the ORCA manual says these will be available “in the foreseeable future.”

When they are available, see how F12 affects any of your structures (if any) that have 2nd row main group elements, particularly P and S.  I've found F12 to DRAMATICALLY improve relative energies with these types of compounds, particularly when 2nd row elements are bonded to elements in the 1st row.


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Discussion about Warren-Cowely Parameter
The Warren-Cowely Parameter (WCP) for short range solution order is defined as:
$$\mathrm{WCP}_{ij} = 1 - \frac{Z_{ij}}{x_j \cdot Z_i}$$
where $Z_{ij}$ is the partial Coordination Number (CN) of  j atoms around i, $Z_{i}$ is the total CN around i, and $ x_{j}$ is the nominal composition of j.
Consider a two component solution of $\ce{A}$ and $\ce{B}$.  If $\mathrm{WCP}>0$, like neighbors are preferred, If $\mathrm{WCP}<0$, unlike neighbors are preferred.
If both of $\mathrm{WCP}_{AA}$ and $\mathrm{WCP}_{BB}$ are positive, which one is preferred?

Could you clarify what $Z_{ij}$ and $x_j$ are in your equations? Does your definition match the one in this paper: https://deepblue.lib.umich.edu/bitstream/handle/2027.42/29021/0000051.pdf;sequence=1

Dear @Tyberius I have updated my question, and added the definitions

@Tyberius do you not have any idea about this question? I checked the parameter with the definition in the paper you introduced, they are identical

I do not know for sure. It could be that $WCP_{AB} is what actually matters.$WCP_{AA}$ and $WCP_{BB}$ both positive seems to me like it implies that both components prefer like neighbors. I had hoped clarifying your question and introducing some terminology for people who aren't familiar (like myself) would encourage an answer.


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Morse interatomic potential
I am working with Morse interatomic potential, I searched through the Internet but I could not find any Morse parameters for Zirconium element. Is there any table, handbook or something like these which list these parameters? or Is it possible to calculate these parameters for hcp metal elements?

Since the Morse potential was constructed to model diatomic bonding, it seems a bit of a stretch to easily generalize it to a crystalline potential, be it hcp, fcc, etc. So, it seems unlikely that one will find a readily available table.

In principle, one can perform quantum chemical calculations on the zirconium crystal structure and distort it to obtain Morse potential parameters. However, this is a lot of work even for an expert and may even provide the answer OP is looking for, presuming they do not care for the parameters but want to find some property using it.


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How does gefitinib affect the EGFR(ecptor) and why is there relapse
Gefitinib has been shown to be an effective tyrosine kinase inhibitor in a fraction (~$10$%) of non-small cell lung cancer patients.
These patients are characterized as having a mutation (usually a deletion or missense) in the EGFR gene affecting the ATP-binding site.
I have two questions:
-- How does gefitinib impede the proliferation and apoptosis pathway? Does it in effect "ace out" the binding of EGF? If this is correct, how does it accomplish that? If not, I would appreciate a brief explanation as to what the successful mechanism is.
-- It seems that (similar to imatinib, another tyrosine kinase used to treat CML) successfully treated patients tend to relapse after a period of time. Why is this?

I suspect this question would fit better on the Biology or Medical Sciences SE pages.

How does gefitinib impede the proliferation and apoptosis pathway? Does it in effect "ace out" the binding of EGF? If this is correct, how does it accomplish that? If not, I would appreciate a brief explanation as to what the successful mechanism is.

No, EGF binds on the extracellular side (to the ectodomain), and gefitinib binds on the intracellular side (to the tyrosine kinase, purple in the figure). The inhibitor interferes with ATP binding (see below, middle panel - the part boxed in green has structural similarities to adenine in ATP). The ATP would otherwise phosphorylate the tyrosines of other signalling proteins. So even if EGF binds to the ectodomain, there is no downstream effect because the tyrosin kinase domain is inhibited.

Source of images: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6523254/
I recommend section 3.2 to learn about how various inhibitors work.

It seems that (similar to imatinib, another tyrosine kinase used to treat CML) successfully treated patients tend to relapse after a period of time. Why is this?

Additional mutations might directly prevent the drug from binding, or change signalling pathways in a different fashion to render the drug inefficient.


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Is it correct to put a methyl group on each end of a skeletal formula?
On my organic chemistry exam there was a skeletal formula that had, supposedly one methyl group at each end of the skeletal formula and one somewhere in the middle.

But is it correct if you draw a formula like that and then put those methyl groups on each end? I feel like it's very misleading because I thought that those were just the ends of the main chain, so not methyl groups. I lost a lot of marks on that question because of this confusion and I just really want to know if it's even correct to write it that way.

Please just answer this question, because if this is an incorrect way of using the skeletal formula then I could have the possibility to pass this exam.

Do you think that the drawing shows something different than 3-Methylhexane?

But the ends of the main chain are methyl groups.

Here, all three terminal carbons are also methyl groups.
Carbon labels may be omitted or displayed, e.g. for the straight angles or implicit hydrogen atoms.
Both structures 1 and 2 below are identical and represent 3‐methylhexane:

From IUPAC Recommendations 2008 [1, p. 290]:

GR-1.4 Terminal single bonds
As discussed in GR-2.1.2, unlabeled atoms are assumed to be carbon atoms, and so terminal single bonds are assumed to represent methyl groups. Unlabeled bonds should not be used to represent unspecified or variable attachment points (see GR-9), as such diagrams are extremely prone to misinterpretation.

Further [1, pp. 300–302]:

GR-2.1.2 Labeling of carbon atoms
[…]
It is acceptable to add labels for terminal carbon atoms connected to unlabeled carbon atoms, but only when it is possible to do so without overlapping other portions of the diagram.

When ethane, ethene, ethyne, and related molecules are drawn with only one explicit bond, both terminal carbon atoms must be labeled explicitly to prevent the molecule from being interpreted as a stray line or set of lines.

References

Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). Pure and Applied Chemistry 2009, 80 (2), 277–410. DOI: 10.1351/pac200880020277. (Free Access)


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Understanding the use of Le Chatelier's principle in the industrial production of ammonia
I understand why the reaction
$$\ce{N2 + 3 H2 -> 2 NH3} \qquad \Delta H = \pu{-92.2 kJ mol-1}$$
isn't carried out in the lowest possible temperature (even though it seems that it should be as it favors the right side of the reaction). As the rate of reaction would be very slow, the amount of $\ce{NH3}$ produced would be lesser than the amount that would've been produced if the reaction was carried out at the optimum temperature in a given time span.
Why isn't then the reaction carried out at the highest possible temperature since the rate of reaction would be very high and the yield would be very high consequently as well?

There is an optimal temperature, for given production line setting, where the effective yield is optimal, as reaction thermodynamics and kinetics have the opposite effects, wrt temperature.

Why isn't then the reaction carried out at the highest possible temperature since the rate of reaction would be very high and the yield would be very high consequently as well?

The rate constant would be high, but that does not mean that the net forward rate is high. If you start without product, the initial rate would be high, but would drop to zero very quickly as equilibrium is reached. The equilibrium concentration of product would be very low at high temperature, as you said, so the yield would be very low.
If you had a very cheap and highly efficient process to remove product directly from the reaction vessel, you could run the reaction at high temperature. However, the product removal is slow and expensive (cool down the reaction mixture, remove ammonia, and reheat the remaining reactant to feed back into the reactor). Because this is costly, the reaction is run at a compromise temperature where at equilibrium, about 15% of the reaction mixture is product.

I think the simplest way to conceptualize this would be to pretend the heat produced by the reaction is a "product" (it's not really correct to say that, but it may help you understand it). Since the reaction is exothermic, you could rewrite the reaction as: N2 + 3H2 ⟶ 2NH3 + Heat
If you raised the temperature, you'd add heat, which is a "product".  As per Le Chatelier's Principle, increasing the concentration of a product shifts the reaction equilibrium left, which is why the reactants predominate at a higher temperature.
If you're familiar with Gibbs Free Energy, I think that would be a better (and more accurate) way to look at it. The forward reaction converts four moles of gas (3 H2 + 1 N2) to two moles of gas (2 NH3). This reduces the total entropy of the system.
Now, we know that: ΔG = ΔH - TΔS
We also know that since entropy is lowered by the forward reaction, ΔS < 0, which makes the term -(TΔS) positive. Lastly, we know a reaction will be spontaneous only if ΔG < 0.
If the temperature were very high, the (TΔS) term would also be very high, making ΔG positive. For the reaction to be thermodynamically possible, the temperature needs to be very low, otherwise the decrease in entropy would offset the favorability added by the loss of enthalpy. This is why there's an "optimal temperature", and why you can't just try and maximize the temperature during ammonia synthesis.
There may be some other ways to look at it, but that's how I would conceptualize it.


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Why is the ionic product of water $K_w$ the same in acid, water, and alkali?
Seen on the web:

The concentration of [H+] and [OH−] both vary based on the the composition (acid/alkaline) of the solution, but the remarkable thing is that their product does not. When [H+] goes up, [OH−] goes down, in the same proportion.

The word "remarkable" was used to describe the ionic product of water (Kw). Ideally, it should not seem more remarkable than anything else, there should be simple explanations that make sense. Why is Kw constant?

You seem to be implying that something that has an explanation that makes sense can't be remarkable?

Also you should give where you found the quote. "On the web" isn't helpful nor a valid way to give credit.

"Bohr was inconsistent, unclear, willfully obscure and right. Einstein was consistent, clear, down-to-earth and wrong. "  John Bell on the Bohr-Einstein debates.  I'm afraid in real physics "simple explanations that make sense" don't always work, even for genius.

@AaronStevens I'm implying that explanations and descriptions are different (but related) things. Keppler observed, for example, his third law, empirically. Surface tension on water, another empirical observation. That it is a result of the gel-phase of water, or "ionized ice", still not very broad consensus, socially.

Another way to put that, people tend to conflate cause and effect. They'll observe an effect, and, a bit lazily, ascribe it as the cause. Lazy thinking. Common in the history of ideas, "generatio spontanea".

The equilibrium reaction for the auto-dissociation of water is:
$$2\text{ H}_2 \text{O}(l) \leftarrow \rightarrow \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$
The associated equilibrium constant $K_w$ is:
$$K_w=[\text{H}_3\text{O}^+]\times [\text{OH}^-] \approx 10^{-14}$$
(Strictly speaking the expression is:
$$\frac{[\text{H}_3\text{O}^+]\times [\text{OH}^-]}{[\text{H}_2\text{O}]}$$
But because $[\text{H}_2\text{O}]=\text{constant}$ for dilute solutions we can use the first expression.)
The reaction rates for the reactions are, where $k$ are the rate constants:
Forward reaction:
$$\text{H}_2\text{O}(l) \rightarrow \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$
$$v_f=k_f[\text{H}_2\text{O}]$$
Reverse reaction:
$$\text{H}_2\text{O}(l) \leftarrow \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$
$$v_r=k_r[\text{H}_3\text{O}^+][\text{OH}^-]$$
We can now show easily that:
$$\boxed{K_w=\frac{k_f}{k_r}}$$
Because in our case the rate constants $k$ are only dependent on temperature (see Arrhenius) that means that $K_w$ is a temperature (only) dependent constant.

One can also say that any equilibrium constant $K$ is related to the change of free enthalpy ${\Delta G°}$ of the particular reaction :
$${\Delta G° =  RT lnK}$$  And this ${\Delta G°}$ does not depend on any other compound present in the solution.


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Is it correct to say that active carbon is an enhancer of either acid or base?
As a non chemist I tried to read to understand if active carbon acts in solutions as an acid or a base but I understood that it can act as both, depends on the material which used to make the solution.
For example, mixing a sodium bicarbonate with active carbon in a solution will make that solution with higher PH (hence a stronger base, per my understanding).
Is it correct to say that active carbon is an enhancer of either acid or base?

Could you please provide evidence, i.e., measured pH values, for the assertion in the second sentence?

@EdV did you mean the second passage For example, mixing a sodium bicarbonate with active carbon in a solution will make that solution with higher PH (hence a stronger base, per my understanding).?

Yes. I see no reason the pH would change unless adsorption was taking place. But then why? So if measured pH values, with and without the active carbon, displayed a significant difference, then an explanation would be warranted. But a small difference might just be experimental error.

@EdV I don't know where to start - I don't have the equipment to measure the PH if I mix (the amount of) one capsule of Active carbon with say half a teaspoon of baking soda which I bought both in the supermarket; I can do that but don't have the equipment and will not purchase it for this question - if the question cannot be answered regarding this or a similar solution than I guess the question should be closed.

Understood. Best to just wait and see how the answers shake down.

Active carbon is neither an acid, nor a base. It cannot. It does not react directly in water and in acidic or basic solutions.  But it may have a catalytic action on the solutes. It may accelerate the rate of chemical reactions. That is probably what you have observed when mixing sodium bicarbonate in water. Without knowing it, you must have increased the rate of the forward reaction : $$\ce{HCO_3^- <=> CO_2 + OH^{-}}$$ and this reaction is favored if the air in the room contains a low concentration of $\ce{CO_2}$, according to Le Chatelier's principle. As a consequence, the pH of the solution has slightly increased, due to this supplement of $\ce{OH^-}$ ions.

If one is already employing microwave (MW) assisted preparations, in this regard, per this 2007 source, 'Generation of hydroxyl radical in aqueous solution by microwave energy using activated carbon as catalyst and its potential in removal of persistent organic substances', by Xie Quan, et al, then one may, indeed, be able to increase the respective power of acidic reactions.
As to mechanics with bicarbonate at low pH in the presence of any created hydroxyl radicals (from MW irradiation of aqueous activated carbon):
$\ce{HCO3(-) + .OH = .HCO3 + H2O}$
the powerful hydrogen carbonate radical.
Similarly, working with a weak acid (for example, dilute acetic acid) and H2O2, accelerated activity via the formation of the more acidic hydroperoxyl radical from the action of the hydroxyl radical acting again on hydrogen peroxide:
$\ce{H2O2 + .OH -> .HO2 + H2O}$
Source: 'Radical-Enhanced Acidity: Why Bicarbonate, Carboxyl, Hydroperoxyl, and Related Radicals Are So Acidic’, referring in particular, to the hydroperoxyl radical, •HO2. To quote from the abstract:

Comparison of accepted pKa values of bicarbonate, carboxyl, and hydroperoxyl radicals, with those of models having the unpaired electron replaced by H atoms, implied the acidity of the radicals was greatly increased. A Density Functional Theory computational method of estimating pKas was developed and applied to a set of radicals designed to probe the phenomenon of radical-enhanced deprotonation (RED-shift) and its underlying causes. Comparison of the computed pKa values of 12 acid radicals to those of the corresponding model acids confirmed the intensified acidity of the title radicals..

More generally, in the realm of electrosynthesis (which relates to the chemistry of chemical synthesis of compounds in an electrochemical cell) and, in particular to the alkaline 'bleach battery', replacing the copper metal with more noble carbon, will also accelerate the reaction. Source: see comments and reference cited here.


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Why is the boiling point of ethyl fluoride lower than that of hydrogen fluoride?
The book, Solomons' Organic Chemistry (for JEE Mains and Advance), contains the following question:

Hydrogen fluoride has a dipole moment of $\pu{1.82 D}$; its boiling point is $\pu{19.34 ^{\circ} C}$. Ethyl fluoride ($\ce{CH3CH2F}$) has an almost identical dipole moment and has a larger molecular weight, yet its boiling point is $\pu{-37.7 ^{\circ} C}$. Explain.

Now while explaining the (very)  concept of boiling point and why it differs for different substances the book considers:

Dipole-dipole forces
London dispersion forces
Molecular weight

But when I tried to explain the phenomenon stated in the question, I found it hard to explain the lower melting point. Following is my approach on the explanation:

Since both have the same (similar) dipoles therefore there would be very negligible difference in their boiling points.
London dispersion forces increase in magnitude as the surface area of the molecule increases. So contrary to what was said $\ce{CH3CH2F}$ should have stronger dispersion force interaction (due to larger surface area)  and hence higher boiling point as compared to $\ce{HF}$ (given other properties were same).
Given that $\ce{CH3CH2F}$ has higher mass than $\ce{HF}$, therefore it should lead to higher boiling point (as is stated in the question itself).

Now it can be clearly seen that the above stated arguments are against the phenomenon, so I believe that I am missing something in my explanation.
So

What is (if any) wrong/missing in my explanation?
Why is the boiling point of $\ce{CH3CH2F}$ lower than that of $\ce{HF}$?

Hint: Recheck how the dipole moment is defined.

@Karl the said book only talked about the surface area while explaining such phenomenon. It says "and because the
surface areas of larger molecules can be much greater, intermolecular dispersion attractions can also be much larger".

they are in aqueous condition, then hydrogen bonding can be one of the reason in HF both hydrogen and fluorine are able to form hydrogen bond, where in ethyl fluoride only F which can form intermolecular bond.

The three reasons that you present are not sufficient to explain the high boiling point of pure $\ce{HF}$. There is a fourth reason, that you have not mentioned : the Hydrogen bond. Molecules $\ce{HF}$ are interacting with one another through $\ce{H}$-bonds, exactly like $\ce{H_2O}$ molecules attract one another because of $\ce{H}$-bonds. In the liquid state, $\ce{HF}$ behaves as if it is made of heavier molecules, something like $\ce{(HF)_n}$, with $\ce{n}$ being between about $4$ and $7$. The boiling point of $\ce{HF}$ and of $\ce{H_2O}$ is extremely high when comparing to substances of nearly the same molar weight, like $\ce{CH_4}$ or $\ce{Ne}$. The boiling point of $\ce{H_2O}$ and $\ce{HF}$ does not depend significantly on the van der Wals forces. $\ce{H}$-bonds are much more important than van der Wals forces.
And there are practically no $\ce{H}$-bonds in ethyl fluoride, because $\ce{H}$-bonds happens only  when an $\ce{H}$ atom is attached to a strongly electronegative atom.

The last sentence is of course completely and utterly wrong. Just look up ; but these are at least an order of magnitude weaker. Therefore they hardly play a role in the comparison of OP.

@ Martin. Please Martin ! Keep cool ! I perfectly know that some extremely weak H-bonds exist with C-H hydrogen bonds, and even between a lot of other X-H bonds. They can even be measured. But they are so weak that they do not have any visible effect on the boiling point.

Maurice, a technical note, adding a space between @ and the username like you did will not notify the person you're pinging. Just type @user and not @ user.


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Why is o-nitroaniline not more basic than aniline?
We know from SIR effect (Steric Inhibition of Resonance) that in o-nitroaniline, the nitrogen's p-orbitals will change their plane, causing the conjugation/resonance of its lone pairs to stop. So, the only effect acting on it now is −I effect from the $\ce{NO2}$ group.
Comparing with just aniline, the $\ce{NH2}$ lone pairs are in constant resonance with the benzene ring making them unavailable for donation. Thus, aniline should be less basic than o-nitroaniline. But it isn't as per experimental values. So what's going on? What's the logic behind this?

Oh, do we know that? No, it's your misunderstanding - in such cases it's ortho-effect not SIR.

Yes it is true that SIR is observed in o-nitro aniline so it will not show -M effect.
But I think you're confused in the resonance of the lone pair of NH2 group with the benzene ring which in fact occurs in both cases. The thing that makes o-nitro aniline less basic is the -I effect which makes the lone pair on - NH2 group less available for donation .
In fact the complete order of basic strength of nitro substituted anilines is:
[Aniline>m-nitro aniline>p-nitro aniline>o-nitro aniline ]

How can resonance of Lone Pairs of Nitrogen of$NH2$ occur in ortho case?? Resonance completely stops due to SIR effect.

I think it the nitro group goes out of plane


text:
Antoine law and ideal gas law
I have the following question:

A $100\ \mathrm{ml}$ sealed flask with $60\ \mathrm{ml}$ acetonitrile is heated to $140\ \mathrm{^\circ C}$. The boiling point of acetonitrile is $82\ \mathrm{^\circ C}$. What will be the pressure in the flask?

1) I understand I can use the Antoine equation, $\log{p} = A-B/(C+T)$, to calculate the vapor pressure of acetonitrile, but that equation only applies if there is some liquid acetonitrile left in the flask. How do I know if this is the case, or instead all acetonitrile has boiled and is in the vapor phase?
2) The total pressure is the sum of the partial pressures of acetonitrile and air. I can calculate the air pressure inside the flask with the ideal gas law, PV=nRT. However, I'm not sure what is the volume? Is it the total volume minus the acetonitrile volume ($100\ \mathrm{ml}-60\ \mathrm{ml}=40\ \mathrm{ml}$), or is it $40\ \mathrm{ml}+x$, where $x$ is the additional volume due to the conversion of liquid acetonitrile into vapor?
Any ideas are welcome.

To know the partial pressure of the air in the flask following heating you need to know how many moles of air you have, which you can calculate from the partial pressure of air at the start.But to determine this, you need the initial temp. of the flask, and the conditions under which it was sealed (e.g., was it sealed such that tthere was 60 mL acetonitrile after saturating the air with it, or was it sealed before the acetonitrile vaporized, in which case the partial pressure of the air would be 1 atm)?  Or did it have any air to start?  You need this info. to begin answering the question.

Ad 1) Remember the pressurized cooking. The water is filling the comparable fraction of the volume, and temperature is raised above the standard boiling point by comparable amount. No cook is afraid of total vaporization.

Thank you for your comments. I will consider your points.

Assume that the initial temperature is room temperature and neglect the initial amount of acetonitrile in the gas phase.  Calculate the equilibrium vapor pressure of acetonitrile at 140 C and compare it with the partial pressure you would calculate from the ideal gas law if all the acetonitrile had evaporated (so that its volume is 100 cc).  If the latter is greater than the former, then there will still be some acetonitrile remaining as liquid.  The partial pressure of the acetonitrile in the gas phase will then be equal to the equilibrium vapor pressure.  Assuming that x gm of acetonitrile have evaporated, you can then calculate, in terms of x, the new mass and volume of the liquid and the new volume of the head space.  You can then determine x from the requirement that the total amount of acetonitrile has not changed.

Thank you, that was very helpful. As the partial pressure calculated with the ideal gas law is greater, I can use the Antoine equation and solve the problem.

EDIT I fouled up my original answer. So as a matter of pride I have fixed it. The OP was right on his idea #1. The answer to the question hinges on the calculated vapor pressure assuming all of the acetonitrile evaporates. I have no idea what coefficients the OP might have had for the Antoine equation.

Let's go down the rabbit hole! Given no other information you have to make several assumptions.

Does the sealed flask initially contain $40\ \mathrm{ml}$ of air or not?

For instance the apparatus could contain two chambers. One $60\ \mathrm{ml}$ to be completely filled with liquid acetonitrile and a $40\ \mathrm{ml}$ chamber which is evacuated. Then when the experiment starts the chambers are first connected and then heated.

Alternatively the apparatus could contain one $100\ \mathrm{ml}$ chamber into which 60 ml of liquid acetonitrile is poured, the other $40\ \mathrm{ml}$ being part air and part acetonitrile vapors.

What is the starting temperature?

This is more important if the flask initially contains $40\ \mathrm{ml}$ of air.

Can the partial pressure of acetonitrile at the starting temperature be ignored?

If yes, then just state the assumption.

If no, then you'll need the partial pressure of acetonitrile at the starting temperature, and you'll have to assume that the $40\ \mathrm{ml}$ gaseous phase is in equilibrium with the liquid phase.

What is the density of liquid(?) acetonitrile at the starting temperature?

Since the problem states "$60\ \mathrm{ml}$ acetonitrile" you'd be led to assume that the acetonitrile was initially a liquid.

I'd assume whatever temperature I could find a density for liquid acetonitrile.

Since $140\ \mathrm{^\circ C}$ is above the atmospheric boiling point, which does the acetonitrile do?

Totally vaporize?

Form gas and a liquid phases?

Turn into a supercritical fluid?

Does gaseous acetonitrile act as if it is a perfect gas?

If yes, then $pV=nRT$ will work.

If no, then some more "advanced" model needs to be chosen and the additional constants that it needs to better fit the gas phase behavior.

Number of significant figures for the answer?

It would seem to be 2, so this feeds back into assumptions 2, 3, 4, and 6.

Given the above assumptions, the problem should be easy to solve using $pV=nRT$.

Let's assume:

The sealed flask initially contain $40\ \mathrm{ml}$ of air.

The starting temperature is $\pu{68 ^\circ F = 20 ^\circ C = 293 K}$

The partial pressure of acetonitrile is $\pu{73 mmHg at 68 ^\circ F}$

Density of liquid acetonitrile is $\pu{0.787 g/ml at 68 ^\circ F}$

Since $\pu{140 ^\circ C = 413 K}$ is above the atmospheric boiling point, the acetonitrile totally vaporizes.

The gaseous acetonitrile acts as if it is a perfect gas.

2 significant figures are needed for the answer.

OK, let's check some our our assumptions. For the acetonitrile:
$$m_{\text{acetonitrile}} = \pu{0.787 g/ml}\times\pu{60 ml}=\pu{47.22 g}\tag{1}$$
$$n_{\mathrm{acetonitrile}}\dfrac{\pu{47.22 g}}{\pu{41.053 g/mol}} = \pu{1.150 mol}\tag{2}$$
One mole of a gas at STP occupies $\pu{22.711 l/mol at $0\ \mathrm{^\circ C}$ and $100\ \mathrm{kPa}$. Since the assumption is that only two significant figures are needed for the answer, the approximately $0.040\ \mathrm l$ of the gas phase (air + acetonitrile) can just be ignored. A more careful check confirms the notion.
$$n_{\text{air}}=\frac{pV}{RT}=\frac{\pu{100 kPa}\times\pu{0.040 l}}{\pu{8.314 l\cdot kPa}\cdot\pu{K^{-1}}\times\pu{293 K}}=\pu{0.0016 mol}\tag{3}$$
$$ n_{\text{liquid acetonitrile}}\gg n_{\text{air}}\gt n_{\text{gaseous acetonitrile}}$$
Now assuming that all of the acetonitrile vaporizes, calculate the pressure assuming that the acetonitrile behaves as an ideal gas equation:
$$p=\frac{nRT}V=\frac{\pu{1.150 mol}\times\pu{8.314 l\cdot kPa}\cdot \pu{K^{-1}}\cdot\pu{mol^{-1}}\times\pu{413 K}}{\pu{0.100 l}}\\
=\pu{39 487 kPa}\approx\pu{3.9\times10^4 kPa}=\pu{3.0\times 10^5 Torr}\tag{4}$$
On the Wikipedia data page for acetonitrile gives a formula for the vapor pressure (temperature range from $\pu{229.32 K}$
to $\pu{545.50 K}$) where $p$ is the pressure in $\mathrm{mmHg}$ and $T$ is the temperature in $\mathrm{K}$:
$$\ln p=\ln{\dfrac{760}{101.325}}-3.881710\cdot \ln T-\dfrac{4999.618}T+41.05901+3.155956\cdot10^{-6}\cdot T^2$$
$$\therefore p=\pu{3 379 Torr}\approx\pu{3.4\cdot10^{3} Torr}$$
Since the vapor pressure of acetonitrile in equilibrium with the liquid at $\pu{413 K}$ is less than the vapor pressure that would be created by all of the acetonitrile evaporating, all of the acetonitrile will not evaporate.  The relative volumes of the two phases won't matter for the pressure.

It is one 100 ml chamber. 2. I agree, so I have to assume the starting temp. 3. I don't know. 5+6. That are the problem that concerns me most, because I don't know it.


text:
Odd looking carbocation stability order
I am studying introductory organic chemistry, suddenly this one started looking odd,
The carbocation stability for three compounds as given by my book is shown below,

Let me call those as compounds A>B>C as given in image. It is very well known that 'N' and 'O' are electron withdrawing group(shows -I effect) and they can also show +M effect( some texts call it as Resonance effect ) by delocalising their lone pair over the vacant p-orbital of the carbocation.
Now all i want to know is, How to decide whether +M effect or -I effect will operate in this case? If I consider only inductive effects, the order would be C>A>B   which is wrong in this case.
One part of my text book says that +M effect > -I effect (in more than 99% of cases). But then too the next question arises, why +M effect of 'N' is more than that of 'O'?
I got stuck in this one and i am now asking whether such questions are feasible without conducting experiments.
Explanation with the principles which are familiar to my level would be nice.

You already stated (correctly) mesomeric effects and inductive effects may occur simultaneously, and sometimes one of them dominates the other.  Think a moment about rules about electrophilic aromatic substitutions on benzene: For the entry of a second substitutent, you have substitutents directing preferentially either a) toward ortho- and para-substitution, or b) toward meta-substitution -- where both the mesomeric influence typically is more important. And c) the halogens where both mesomeric effects (free electron pairs) and inductive ones ($\Delta{}EN$) affect the reaction.

If hetro atom with lone pair is neighboring to carbocation, then that lone pair can be donated to empty orbital on carbocation (see figure below). These orbitals are said to be in conjugation. Conjugation leads to delocalization of electrons resulting in resonance structures.

Quoting from "Organic Chemistry" by T.W. GRAHAM SOLOMONS , CRAIG B. FRYHLE .SCOTT A.SNYDER
12edition:

Structures in which all the atoms have a complete valence shell of electrons
  (i.e., the noble gas structure) are more stable.

The third structure (your structure C) does not have conjugation. Hence it is comparatively less stable then 1st and 2nd structures.
Among 1st and 2nd structures , due to lower electronegativity of Nitrogen compared to Oxygen , Nitrogen has greater electron pair donating tendency. Therefor the order of stability is as follows.

How to decide whether +M effect or -I effect will operate in this case?

The extent of stabilizing effect follows the order: $\ce{\text{Mesomeric} > \text{Hyperconjugation} > \text{Inductive}}$.
In general, this order is based on extent of $\ce{e-}$ transfer. In mesomeric effect, $\pi$-bonds are in conjugation which completely transfers $\ce{e-}$ density to carbocation. On the other hand, in inductive effect, induction of charge takes place, which partially transfers $\ce{e-}$ density through $\sigma$-bond. Hence, it is weaker.

why +M effect of 'N' is more than that of 'O'?

As $\ce{N}$ has lower electronegativity than $\ce{O}$, hence it has greater $\ce{e-}$-donating tendency, therefore greater $\ce{+M-\text{effect}}$

According to you, when one answers theoretical questions like this, our thought process should be like , 1) +M > -I . 2) if (1) is true, COMPLETELY IGNORE INDUCTIVE EFFECTS. 3)find the strongest +M ??? Does this always work? I hope you understand that ignoring the inductive effects pesters me a bit.

@MukunthA.G: 1) It's not feasible to cover every detail of the answer, unless you've asked in the post itself. 2) You asked for "find the strongest +M?" - I would suggest you to either search via google for previously asked question on this site, or maybe ask a new question. 3) You can read ron's answer in: Competing resonance and inductive effects in a substituted benzene...

...Also, you should prevent asking multiple questions here. You may've gone through the how to ask a good question page earlier, but it would be nice to have a look at it, if you feel like it.

I'd try to give my explanation with the principles simple enough so that it would familiar to your level of education:
Rahul Verma has given simple enough explanation to answer most part of your question. I said here 'simple enough' because of your subsequent comments. However, I have to point out that mesomeric effect ($M$) almost always predominate inductive effect ($I$), even when the distance between two concerning centers are neighbors like in the case in hands where inductive effect is in its highest level.
When consider structures (A) and (B), the electron-withdrawing inductive effect ($-I$) of (B) is lager than that of (A) simply because electronegativity of $\ce{O}$ is higher than that of $\ce{N}$. However, the electron-donating mesomeric effect ($+M$) of (A) is lager than that of (B), which can be simply explained by following reason you have learned in general chemistry classes:
Consider $\ce{H2O}$ and $\ce{NH3}$ for their affinity for electron donating:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
The solution is always basic ($\mathrm{p}K_\mathrm{b}$ of ammonia is about 4.75) because ammonia donate its lone pair to grab a proton from water. It'd never be other way around like: $\ce{NH3 + H2O <=> NH2- + H3O+}$. By this phenomenon along, we can assume electron donating ability of $\ce{NH3}$ is comparatively larger than that of $\ce{H2O}$. Hence, one can consider that $(+M)_\ce{N} \gt (+M)_\ce{O}$.
Based on these two comparisons, it is evident that $(-I+M)_\ce{N} \gt\gt  (-I+M)_\ce{O}$ when compared to net positive value. Therefore, one can conclude that (A) is stable than (A).


text:
Temperature Programmed Disorption

For second part of the question I'm wondering if the peaks on the right denote the chemisorbed layer, while the peaks on the left denote the physisorbed layers. Since chemisorption results in a stronger covalent bond, the energy to remove adsorbates than the physisorbed layers

What is plotted on the vertical scale ? It is 1000 or 2000 what ? What units ?

The Y axis is rate


text:
Molecule or molecular ion that spontaneously breaks up in two different ways
I teach physics, and I'm looking for simple, easy examples from chemistry to use in order to illustrate basic ideas about quantum mechanics for my students. A conceptually simple example is the $\text{He}^-$ ion, which has a metastable state with a mean lifetime of 0.36 ms. This is for the totally isolated ion, i.e., we're not talking about a gas or plasma in which there can be collisions or reactions. (How would chemists refer to this? "In vacuo?") This is easy to relate to in chemical terms (a noble gas needs an electron like a fish needs a bicycle) and provides a simple example of quantum mechanical concepts like exponential decay.
What would be more fun for my purposes would be an example of a metastable molecule or molecular ion that would spontaneously break up in two different ways. I would imagine that for the ions, one could measure the half-life and branching ratios by storing them in a circular accelerator. So we could have something like the decay channel
$$\text{XYZ}^+ \rightarrow \text{XY}^+ + \text{Z}$$
competing with, say,
$$\text{XYZ}^+ \rightarrow \text{X}^+ + \text{YZ}.$$
We can then discuss things like the probabilities (branching ratios) and the fact that if the ion starts out in a well-defined state, presumably its ground state, then we can only predict these probabilities, not the actual outcome.
Part of what's making it hard for me to find examples is that I don't know the terminology to google on. Is this "autodetachment," or is that only used for the loss of an electron? "Autodissociation?"
Note: I really am looking for the process described above, not something else like breakup induced by a collision, nuclear decay, electromagnetic decay of an electronic excitation, electron emission, or chemical reactions that require two reactants to come in contact. Those other examples might be fine for some educational purposes, but they're not what this question is about.

The fragmentation patterns seen in mass spectrometry might be partly what you are looking for: https://www.chemguide.co.uk/analysis/masspec/fragment.html. Not sure how much the rates of decay have been studied, but the patterns of relative abundance are the core of the technique.

Phosphorescence would obey an exponential decay curve.

Any number of radioactive isotopes decay in multiple ways.

x-ray excitation of atoms can decay by fluorescence or Auger electron emission.

many chemical reactions can branch to form multiple products. Copper and nitric acid is one such example.

Almost any aromatic molecule, benzene, naphthalene, amongst many many, others,  show fluorescence from the excited singlet state, this state also decays (intersystem crossing)  to a triplet state at lower energy and then this triplet decays by  phosphorescence, which, like intersystem crossing, is a forbidden transition partially allowed by spin orbit coupling.

This happens all of the time in mass spectrometry following electron-impact ionization. Following ionization, each molecular ion decomposes (or fragments, to use mass spectrometry lingo) into other radicals and ions, usually following multiple pathways. As the pressure inside a mass analyzer is usually near vacuum, each fragmentation event is independent. No collisions or reactions between particles occur. The mass analyzer collects all of the fragments and generates signal that computer software converts to a mass spectrum - a frequency bar graph of the masses of species generated. For some very simple molecules, only one fragmentation pathway may predominate, but for most organic compounds, there are multiple fragmentation pathways.
For example, let's consider the fragmentation of but-3-en-2-one radical cation with the formula of $\ce{C4H6O+}$ and the structure shown below. The mass/charge ratio (m/z) is 70 for this compounds

There are at least two different fragmentation pathways.
The first fragmentation pathway is a loss of a vinyl group ($-\ce{C2H3}$, -27 m/z):
$$\ce{C4H6O+ -> C2H3O+ + C2H3 }$$

The second pathway is a loss of a methyl group ($-\ce{CH3}$, -15 m/z):
$$\ce{C4H6O+ -> C3H3O+ + CH3}$$

Each of these fragmentation pathways is independent. Here is a copy of the mass spectrum from the NIST Chemical Webbook, which is a frequency bar graph of all possible fragments. You can see in addition to the two pathways mentioned above, there are many other possible fragments.

We can estimate the relative rates of these processes from the frequencies of the peaks in the mass spectrum. The first pathway occurs with a relative frequency of 83.19 (based on the height of the m/z = 43 peak) , and the second pathway occurs with a relative frequency of 1.000 (from the height of the m/z= 55 peak). We can use these as proxy for the relative rates of decomposition. We can also see that the parent ion survives this process with a relative frequency of 0.57779 (from the m/z = 70 peak), which indicates that the parent ion has at least some stability.

I was hoping to see a mass spec answer. It seems to fit with what the original post is looking for, with multiple fragmentation pathways and fairly well characterized ratios between those paths based on stability of given types of fragments.

This is nice, thanks. It would be cool to have an example in which there was a known state in the parent ion, and a measurable lifetime.

There are many compounds that will break down into different products when involved in collisions with inert molecules. For example, $\ce{F3-}$ can separate into $\ce{F2 + F-}$ or into $\ce{F. + F2-}$. But atoms or molecules that will breakdown without being involved in a collision are rarer (simply because their instability means they are harder to observe).
One example, though, is the radioactive decay of the bismuth-212 isotope. It decays ~64% of the time by $\beta-$ decay, producing polonium-212: $\ce{^{212}_{83}Bi ->  ^{212}_{84}Po + e- + \nu_e}$. The other ~36% of the time it decays by $\alpha$ decay to thallium-208: $\ce{^{212}_{83}Bi -> ^{208}_{81}Tl + ^4_2He}$.
More info can be found here: http://nucleardata.nuclear.lu.se/toi/nuclide.asp?iZA=830212

Thanks, but this doesn't answer the question. The question is about spontaneous breakup, and it's a molecular physics question, not a nuclear physics question.

The key operation is to make a metastable particle that has at least two decomposition pathways and a short lifetime. Fragmentation is probably the key word you were looking for. Pyrolytic fragmentation of complicated molecules (1-methoxycyclopropylamines) is one way https://www.sciencedirect.com/science/article/pii/S0040403900741402. Photolytic fragmentation of acetaldehyde, a relatively simple molecule, has been investigated https://aip.scitation.org/doi/10.1063/1.4878668 , and many others have been investigated with lasers. Another well known area is mass spectrometry fragmentation https://en.wikipedia.org/wiki/Fragmentation_(mass_spectrometry).  For example, toluene is bombarded with electrons, loses one to form a

cation, then decomposes to form a benzyl cation or a phenyl cation.
An advantage of using mass spectrometric data is that there is a lot out there, and the idea can also be hypothetical, in that a molecule can be selected on the basis of its simplicity and likely decomposition products to allow theoretical calculations, e.g., for educational purposes, even in the face of insufficient actual data.

This looks interesting, thanks. Is the parent ion in a defined state before it breaks up? Is there a measurable lifetime and measurable branching ratios?

Original state for volatile compounds is just the ground state; branching ratios will be measured as output; lifetimes will be more difficult, not really sure how that's done now. I did mass spec work as a grad student, many moons ago.

This is not really a complete self-answer, but I thought it was close enough that it was worth posting. I would be happy if others could come up with better examples.
The ions $\text{N}_2^{2+}$ and  $\text{CO}^{2+}$ have been studied experimentally and theoretically in Pandey, Bapat, and Shamasundar, J. Chem. Phys. 140, 034319 (2014). Pandey's thesis also has a helpful table on p. 122. Their experimental setup is only sensitive to a breakup in which both fragments are charged. They do a total kinematic reconstruction of the two fragments and infer the kinetic energy release (KER). Their resolution is not quite good enough to resolve contributions from individual vibrational states in their KER spectrum, but you can see structures in them which they explain as arising from those vibrational states. They didn't store the fragments, just let them fly through a spectrometer, so they could only see decays that happened within a few milliseconds. With these two dications, you don't get tunneling on these time-scales when the initial state is the vibrational and electronic ground state. They do observe it from the vibrational ground state coupled to an electronic excitation and vice versa. They don't actually seem to have measured any lifetimes, and they don't give references to anyone else who has. There are no branching ratios to measure, because their molecules were binary, and their detector needed to see two charged fragments.
Eland et al., https://arxiv.org/abs/1908.11441.pdf  (2019), have studied the breakup of $\text{HNCO}^{2+}$ and $\text{HNCO}^{3+}$. There are cases where different breakup channels compete, and there seem to be some lifetimes on the order of microseconds. But the experimental setup basically seems pretty messy, and it doesn't seem like they got a lot of detailed data.
Relevant keywords seem to be "dissociative ionization" and "vibrational autoionization."
These are both pretty recent, and I was kind of surprised that the state of the art was so crude. To me as a physicist, these seem like obvious things for a molecular physicist or quantum chemist to want to study. No doubt my ignorance of the experimental techniques is my reason for not understanding why this kind of experiment is so hard. I wonder if there are any systems that happen to be particularly easy to study and that have been better characterized than these.

It's not clear to me how one knows that the cations are all in the ground state before dissociation. Couldn't the breakup channel be dependent on the initial state produced by the ionization collision (eg angle, energy, etc)? I think that's the challenge of finding a system of the sort you are looking for. One needs to have molecules that are unstable enough that they fall apart, but stable enough that they have time to relax to a common ground state as the starting state before dissociation.

@Andrew: In the Pandey experiment, none are in the ground state. See Pandey's table on p. 122, and the text of my answer starting at "With these two dications..."


text:
Polycondensation of PMS

This is the reaction that I'm interested in. I reckon this is what occurs:

But how does the reaction proceed from there, there is a lone pair on the oxygen in right side of my picture, but how do you get rid of the CH3 group on that oxygen?
Although I do see a problem here that the structure of my final product was due to my aforementioned reaction scheme would still have RO-(PMS)-OR, where R is methyl group, instead of RO-(PMS)-R. I don't know what to do...

The more common reaction of this type eleminates acetic acid instead of methanol. Silicon rubber during curing smells like vinegar. I´ll have to look up the specifics ...

"Cationic polymerization can occur with the use of strong or Lewis protic acids. The mechanism of this reaction is yet unknown." https://www.ukessays.com/essays/chemistry/polysiloxanes-synthesis-structure-2201.php
Thereby, we are freed to speculate!
It is more likely that silicon goes pentacoordinate rather than tricoordinate. If a proton attaches to an oxygen on the siloxane, the silicon goes positive, which will attract an oxygen from: 1) a water, 2) a methanol, or 3) a methyltrimethoxy silane. The second and third options can only reform the original reactants. If water attaches after protonation, forming $\ce{H2O-Si-OHCH3}$, methanol could leave, followed by a proton from the water, to give $\ce{CH3Si(OCH3)2OH}$. This becomes a fourth attacking option. If this attacks the silicon of a protonated siloxane thru the $\ce{Si-OH}$ to give $\ce{CH3O(H+)-Si-OHSiCH3(OCH3)2}$, the pentacoordinated silicon can lose methanol, then lose the proton from the oxygen between the silicons, to form $\ce{CH3(OCH3)2Si-O-SiCH3(OCH3)2}$, and that's the first step, a dimer.
Because there are two more alkoxy groups on the silicons which can leave, more reactions can occur, and likely will produce a three-dimensional gel, or even a solid, if you cook it.


text:
Titrating iodine starch solution with sodium thiosulphate - Colour change
I investigated two mixtures with different solvents, one with water and one with n-heptane. Both contained iodine $\ce{I2}$ as a solute. To both solutions I added a bit of starch.
As I remember this resulted in a colourchange. So the solution turned from yellowish to dark blue (if I remember correctly!).
Now according to wikipedia starch and iodine indeed form a structure which has a dark blue colour. But it only forms in the presence of $\ce{I^-}$.
This leaves me wondering, why do I remeber the solution to be dark blue, eventhough I think there was no $\ce{I^-}$ present? Could it be the solution turned dark blue only after I added some sodium thiosulfate? Because in the next step I did a titration with $\ce{Na2S2O3}$.
In this case I don't see which reaction could have produced the $\ce{I^-}$ though. I thought only $\ce{NaI}$ is produced after adding the sodium thiosulfate.
$$\ce{I_2 + 2Na_2S_2O_3 -> 2NaI + Na_2S_4O_6} \tag{1}$$
So at which point did the solution turn dark blue and where did the $\ce{I^-}$ come from, that was needed for the formation of the starch-iodine-compound? Could it be there is an intermediate step to (1) in which $\ce{I^-}$ is formed and this $\ce{I^-}$ was used to produce the dark blue starch-iodine compound?

I don't think your memory is serving you right. That is why we write everything in the notebook, especially color changes.
I think you are doing distribution experiments where iodine is distributed between aqueous layer and an organic layer. When we add indicator for titration, it is not a solid starch but starch which is boiled in water. So when you added starch $solution$ to heptane which contained iodine, I would not be surprised if the starch solution turned blue.
Remember that iodine is strong oxidizing agent as well. A very small fraction of it can easily convert into iodide. You really really need a trace of the triiodide ion to form a dark blue iodine complex.

Your assumptions are correct. This is my first chemistry lab. And yes I should've wrote everything down more carefully. One question for clarification: You think the Iodine interacted with the sodium thiosulphate, forming some $\ce{I^-}$ which then lead to the reaction $\ce{I^-}+\ce{I_2}+\textrm{starch}\leftrightarrow\textrm{dark blue starch}$? And when adding more and more thiosulphate all of the $I_2$ and consequently all of the dark blue starch reacted to the colourless $\ce{I^-}$?

Right, this is what I think happened in your case.

The molecular iodine $\ce{I_2}$ is poorly soluble in water : maximum $0.0011$ M. If starch is added to this solution, the iodine will react with starch and the solution is dark blue. In the lab, this experiment is rarely done with simple $\ce{I_2}$ solutions, because the solutions to be titrated are usually more concentrated than $0.001$ M. Usually $\ce{I_2}$ is dissolved in $\ce{KI}$ solutions, producing $\ce{KI_3}$ or $\ce{I_3^-}$ ions.$$\ce{KI + I_2 <=> KI_3}$$ The "solubility" of $\ce{I_2}$ as combined in $\ce{KI_3}$ is at least $1000$ times higher than $\ce{I_2}$ in water. So in the presence of $\ce{KI}$ in solution, more $\ce{I_2}$ can stay in solution. And if some starch is added to a $\ce{KI_3}$ solution, it will produce a dark blue-black color, due to the small amount of free $\ce{I_2}$ in the $\ce{KI_3}$ solution.
When titrating either $\ce{I_2}$ or $\ce{KI_3}$ by adding thiosulfate  ions $\ce{S_2O_3^{2-}}$, the free $\ce{I_2 }$ is consumed. $$\ce{I_2 + 2 S_2O_3^{2-}-> S_4O_6^{2-} + 2 I^-}$$
$\ce{I_2}$ is consumed by adding $\ce{S_2O_3^{2-}}$. But as the equilibrium $\ce{KI + I_2 <=> KI_3}$ is rapid, new $\ce{I_2}$molecules are continuously regenerated from $\ce{KI_3}$, so that the starch solution stays dark blue up to the end of the titration. Apparently, the titration proceeds as if the solution of $\ce{KI_3}$ is a solution of $\ce{I_2}$.


text:
Why is the viscosity of water and ethanol lower than that of a water-ethanol mixture?
We found in an experiment using Ostwalds method that
ethanol is more viscous than water, which
is already surprising as water has stronger intermolecular bonds than ethanol as @MaxW pointed out.
The mixture
of both substances furthermore is significantly more viscous than both individual substances. This
is interesting. We did not come up with a
satisfying explanation for this behaviour.
I know that one shouldn't expect that a mixture behaves the same way or proportional to the concentration of the components, but I would appreciate an explanation of this effect on a molecular level.

Ethanol boils at $\pu{78 ^\circ C}$ and water boils at $\pu{100 ^\circ C}$. I'd say that water has stronger intermolecular bonds.

@MaxW You're right.. I corrected this. Thank you.

This answer just popped up when I searched for viscosity of water-ethanol mixtures, so I just copied the abstract:
Viscosities of Binary and Ternary Mixtures of Water, Alcohol, Acetone, and Hexane
https://www.tandfonline.com/doi/abs/10.1080/01932690802313006?journalCode=ldis20
"This articles studied and determined the viscosities of the binary mixtures of water–methanol, water–ethanol, water–propanol, water–acetone, acetone–ethanol, methanol–ethanol, and acetone–hexane and the ternary mixtures of water–methanol–ethanol and water–ethanol–acetone at 20°C. It is shown that the mixing of water with the alcohols and acetone resulted in a positive deviation of viscosity, which reached the maximum value at the water mole fraction x 1 ∼ 0.7 for water–methanol, x 1 ∼ 0.72 for water–ethanol, x 1 ∼ 0.74 for water–propanol, and x 1 ∼ 0.83 for water–acetone binary mixture. This viscosity deviation can be mainly attributed to the formation of micelles of alcohol or acetone molecules in water because of the hydrophobic attraction between the hydrocarbon chains. The micelle surfaces are surrounded by hydration layers, leading to the positive viscosity deviation in the liquid mixtures because the water in hydration layers has a much higher viscosity than bulk water. Also, the contrary observation was found in the binary mixtures of acetone–ethanol and acetone–hexane, having a negative viscosity deviation."