text:
Understanding the octet rule
I just started selfstudy with an organic chemistry textbook and I'm not quite getting the motivation behind the octet rule.
Per Bruice's textbook, the octet rule reads

an atom is most
  stable if its outer shell is either filled or contains eight electrons and it has no electrons
  of higher energy

Why are atoms with a filled valence shell most stable?
What is the special case with 8 electrons? I understand that nobles gasses have full valence shells, but why aren't there noble gasses with filled valence s- d- or -f valence shells --- without higher subshells being filled?

There's already dozens of posts about that and this looks like one of broadests.

Its a quantum effect. E.G. Pauli exclusion principle has electrons (fermions) in pairs, and there is room for 1 pair in an s orbital, 3 in a p orbital (5 in d and 7 in f... notice odd numbers).

Atoms surrounded by 8 electrons look like noble gases (elements from the last column). And noble gases are particular stable, since they are very reluctant to make bonds with any other atom. And when they accept to make a molecule with some other atom, this molecule is highly unstable, and is decomposed at the next opportunity, sometimes with explosion.
The goal of the octet rule is to explain why Na reacts so violently with many other substances like Chlorine to make NaCl for example, which produces a huge amount of heat. The same for C burning in O2 to make CO2. The flame of C produces  a lot of heat. The octet rule explain why NaCl and CO2 are stable : they don't react so often, and when they do react, the reaction does not liberate much heat.


text:
How is Henry's constant affected by change in temperature
According to me Henry's constant $K_h$ should be directly proportional to the effect of temperature.
It is because $K_h$ is inversely proportional to solubility and as we increase temperature solubility decreases and hence $K_h$ value should increase.
But the correct answer is value of $K_h$ also decreases with increase in temperature.
Can anybody explain how?

Wikipedia answers your question well. Also, solubility doesn't always increase with temperature.

It is an application of the Le Chatelier principle. The dissolution of a gas in a solvant is exothermic. If you heat a system in equilibrium, where an exothermic reaction occurs, the system reacts by favoring the endothermic reaction. So at high temperature, the gas quits the liquid : the gas becomes less and less soluble in the solvant. The solubility of a gas in a solvant decreases with increase of temperature.

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text:
Can ascorbic acid be neutralized and keep it's elastic properties in dough?
Ascorbic acid (C6H8O6) is commonly used in dough to give it strength and elasticity. However, I need this dough to be alkaline, or at the very least neutral. I've used sodium carbonate (Na2CO3) before for this purpose but never at the same time with ascorbic acid.
If I use vitamin C in dough to make it elastic, as well as sodium carbonate to it neutralize it, will the dough retain the elastic properties caused by the ascorbic acid? What ratio of ascorbic acid to sodium carbonate creates a neutral solution? Thanks!

Never heard of this use of ascorbic acid before, interesting. I am not sure if this question is a little off-topic, though.

Ascorbic acid is vitamin C, which I am considering adding to my dough mix in powdered form. I included this context in original post because the way I intend to use it relates to one of its properties (elasticity), which I want to keep while neutralizing it.

sodium ascorbate is a different substance than ascorbic acid. Why should it keep the acid properties ?

I DON'T want the acidic properties; I want the elasticity that ascorbic acid adds to dough. Thing is, it also lowers the pH level. This can be countered by adding a base, like sodium carbonate. I wanted to know if having ascorbic acid and sodium carbonate in the same solution would cause any unwanted effects or lack thereof, such as no longer contributing to dough elasticity.

I have written acid properties ( = properties of the acid ), not acidic properties. Salts do not have behaviour of their respective acids.Ascorbic acid and sodium carbonate cannot exist in the same solution. One of them will disappear. You may need to use a different substance, giving the desired improving properties without being acidic. Unless it is directly related to being acidic. I suggest you experimenting with the VitC/soda ratio to reach acceptable trade-off between not being much acidic and still providing dough improvement.

https://bakerpedia.com/ingredients/ascorbic-acid/ Here is an explicit question about using calcium o magnesium ascorbate. the answer is "No. We highly recommend using ascorbic acid". But as usual amount is 15-20mg per 1 kg of flour, I do not see acidity as a real issue, unless you write why you do.

@Poutnik Sorry, my bad for misunderstanding. As for ascorbic acid and sodium carbonate not being able to exist in the same solution, I meant adding those to things to a solution (i.e.: the dough). For example, having dough comprised of flour and water, adding vitamin C to strengthen the dough and give it more elasticity, then adding sodium carbonate to counter the acidity of vitamin C. My question is whether that final step (adding sodium carbonate to increase the pH level of the dough) would also affect the increased elasticity due to the vitamin C in any way.

As for the reason I need it to be alkaline, pH affects the yeast's fermentation, color, as well as some other physical properties. This article explains a few of these regarding ramen noodles specifically: https://www.ramenchemistry.com/blog/2015/1/26/ramen-1014-everything-else-noodles-oils-and-toppings.

There is high probability the desired properties of ascorbic acid are directly related to acid being acidic and in the free acid form, i.e. not in the form of the ascorbate.
It is hard to advise, not being familiar with this part of the applied food chemistry, it the effect of ascorbic acid happens in the cold dough conditioning stage and/or in the stage of the later thermal processing, in case of noodles.
If the sodium carbonate is added later to the dough, it is unclear, how well would be the dough mixed to neutralized the acid, and what the sideeffect would be.
Bakerpedia.com  explicitly advices against using ascorbic salts.
I highly advice to experiment, as theory will probably not provide clear advice.
I am not sure, if the acid is applicable to the noodle dough as much as for bread dough.

“Baker baker, baking a cake / Make me a day, make me whole again / And I wonder what's in a day / What's in your cake this time…”  (Tori Amos – Baker Baker)

I am a medical student and in my book, it is stated that L-ascorbic acid is destroyed by alkalis but the mechanism is not mentioned

It is not directly destroyed in mild alkalic solution, but it rather rapidly oxidated. But the point is, even before destruction of the ascorbate, the ascorbate itself may not be active in the bakery context, if the acid form is required.

@Poutnik Thank you for your help. Once I get the chance I'll experiment with adding ascorbic acid and sodium carbonate to dough. I'll also measure their pH levels to see what the end result is and update this post with my findings as well. Cheers!

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Can a reaction have carbocation resonance followed by a hydride shift?
Consider 5-methylcyclopentadiene:

If HX attacks at carbon 1, we have an option of 1,2 or 1,4 (carbocation resonance) addition. In the case of 1,4 addition, since the carbocation (secondary) will be one bond away from a tertiary carbon, will there be a hydride shift, thus yielding in 1,5 addition?
In essence, can you have both carbocation resonance and a hydride shift in one reaction? Or is it that once a carbocation is resonance stabilized, there is no need for stabilization via hydride shift?

Substituted cyclopentadienes are an unique molecules, which are not stable at room temperature (recall dimerization). For example, 1-methylcyclopent-1,3-diene (1) or 2-methylcyclopent-1,3-diene (2) or 5-methylcyclopent-1,3-diene (3; diene in the question) would rearranges to a mixture of all 3 positional isomers if leave at room temperature for 1-2 days by hydrogen migration (Ref.1 & 2). The abstract of Ref.1 states that:

A number of substituted cyclopentadienes display characteristic properties which are representate of all members in the series: (a) the endocyclic double bond isomers are in the thermodynamic equilibrium; the process of establishing the equilibrium being retarded; (b) the transformation of isomeric cyclopentadienes proceeds by steps, i.e. a proton from the methylene group (5th position) passes mainly to the adjacent C-atom (1st position). The proton-cyclopentadienyl anions are probably transitional states in the transformation of isomers. It is assumed that this transformation is a first order reaction. As a result of this investigation it has been shown that the structure of the substituted cyclopentadiene (or the composition of the isomer mixtures) depends mainly on the character of the substituent and not on the method of synthesis.

The ratio of 1 : 2 : 3 would be 45:54:1. Kinetically, the process is first order and activation energy is $\pu{20.4 kcal/mol}$ over temperature range of $\pu{298-313 K}$ (Ref.3). The kinetic studies by NMR has been suggested that rearrangement is due to 1,2-hydrogen shift (Ref.4).
Yet, these cyclopentadienes are also suseptible to protonation by strong acids (Ref.5). Thus, it is very possible that protonation at $\ce{C1}$ 1,5-addition after 1,2-hydride migration. However, as explained above, 5-methylcyclopent-1,3-diene (3) is relatively the least stable positional isomer. At room tempeature, it would stay mostly as 1-methylcyclopent-1,3-diene (1) and 2-methylcyclopent-1,3-diene (2) with $\approx 1:1$ ratio. Therefore, addition would be either 1,4- and 1,2-.

References:

V. A. Mironov, E. V. Sobolev, A. N. Elizarova, "Some general characteristic properties of substituted cyclopentadienes," Tetrahedron 1963, 19(12), 1939-1958 (https://doi.org/10.1016/0040-4020(63)85008-5).
Sigmund M. Csicsery, "Methylcyclopentadiene Isomers," J. Org. Chem. 1960, 25(4), 518-521 (https://doi.org/10.1021/jo01074a009).
Raymond K. Crandall, “Investigation of the electronic and thermal rearrangement properties of spiro[2,4]hepta-4,6-dienes,” M.S. Thesis, Rochester Institute of Technology, Rochester, NY, 1982.
S. McLean, P.Haynes, "Hydrogen migration in cyclopentadienes," Tetrahedron 1965, 21(9), 2329-2342 (https://doi.org/10.1016/S0040-4020(01)93887-6).
Ronald F. Childs, Maung Zeya, “Diprotonation of 5-Acyl-1,2,3,4,5-pentamethylcyclopentadienes,” Can. J. Chem. 1975, 53(22), 3425-3430 (https://doi.org/10.1139/v75-490)


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Why does the colour of the same compound (e.g. magnesium nitride) differ?
When I am searching the colour of $\ce{Mg3N2}$ I came across photos showing different colours: greenish-yellow on Wikimedia,
and some other just yellow on some Google Images results.
I am eager to know what causes some samples to be bit greenish.

Maybe due to impurities, the color differs.

Particle size and polymorphism would also affect the color of a solid.

Privacy note: be very cautious when you post a link to the Google search results. Not only the address string looks messy, it also contains private info such as your device and browser names in plane text. There are several browser extensions that fix that and create a short nice-looking URL without using services like tinyurl; at the moment, I personally use ClearURLs.

Minor differences in color are likely either due to impurities or due to intrinsic crystal defects such as f-centers. For example, $\ce{ZnO}$ is often a shade of yellow, rather than pure white, due to those defects. In alkali halides, h-centers are similar defects giving rise to colors. You can easily create color centers in $\ce{NaCl}$ with a spark coil in just a few minutes.
Another possibility is that there are mixtures of different allotropes, or polymorphic forms. An extreme example is transparent diamond vs. black graphite. BTW, diamonds having varied colors due to impurities (e.g. nitrogen) and to intrinsic defects are useful in studying quantum phenomena, as well as gems.


text:
Which has a greater dipole moment: methylamine or methanol?
Which has a greater dipole moment: methylamine $(\ce{CH3NH2})$ or methanol $(\ce{CH3OH})?$

It's difficult to compare the dipole moment from the structure of both the molecules. I guess you should go for experimental values.

As per experimental value dipole moment of methylamine is 1.31D and of methanol is 1.69D.

The bond dipole moment of a $\ce{H-C}$ bond is $0.3 D$, that of a $\ce{C-N}$ bond, $0.4 D$, that of a $\ce{H-N}$ bond, $1.3 D$, that of a $\ce{C-O}$ bond, $0.7 D$, and that of a $\ce{H-O}$ bond, $1.5 D$, each in the direction of the second atom designated.
Locate the carbon atom of each molecule at the origin of a three-dimensional ($r$, $\theta$, $z$) coordinate system with the nitrogen atom of the $\ce{CH3NH2}$ molecule and the oxygen atom of the $\ce{CH3OH}$ molecule on the $z$-axis of each coordinate system.
To find the dipole moment of the $\ce{CH3NH2}$ molecule, we add the vector dipole moment of a $\ce{H-C}$ bond (with the carbon atom above the hydrogen atom),
$0.3 D \cdot u_z$
the vector dipole moment of the $\ce{C-N}$ bond (with the nitrogen atom above the carbon atom),
$0.4 D \cdot u_z$
and the vector dipole moment of a $\ce{H-N}$ bond (with the nitrogen atom below the hydrogen atom),
$1.3 D \cdot (-u_z)$
obtaining as a net vector sum
$(0.3 D + 0.4 D - 1.3 D) \cdot u_z = -0.6 D \cdot u_z$,
the magnitude of which is $0.6 D$.
To find the dipole moment of the $\ce{CH3OH}$ molecule, we add the vector dipole moment of the $\ce{H-C}$ bond (with the carbon atom above the hydrogen atom),
$0.3 D \cdot u_z$,
the vector dipole moment of the $\ce{C-O}$ bond (with the oxygen atom above the carbon atom),
$0.7 D \cdot u_z$
and the vector dipole moment of the $\ce{H-O}$ bond (with the oxygen atom below the hydrogen atom and the $\ce{C-O-H}$ bond angle approximately $104.5^\circ$),
$1.5 D \cdot [\cos(180^\circ-104.5^\circ)\cdot u_z + \sin(180^\circ-104.5^\circ)\cdot u_r]$,
the net vector sum of which is
$0.3 D \cdot u_z + 0.7 D \cdot u_z + 1.5 D \cdot [\cos(180^\circ-104.5^\circ)\cdot u_z + \sin(180^\circ-104.5^\circ)\cdot u_r]$
= $[1.0 D + 1.5 D \cos(75.5^\circ)]\cdot u_z + \sin(75.5^\circ)\cdot u_r$
= $(1.0 D + 1.5 D \cdot 0.25038000405)\cdot u_z + 0.96814764037\cdot u_r$
= $1.37557000608 D \cdot u_z + 1.45222146057 \cdot u_r$
the magnitude of which is $2.00 D$ (rounded).
One might thus conclude that methanol's dipole moment (experimental value being $1.69 D$) is greater than that of methylamine (experimental value being $1.31 D$).


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Is electron density a good measure of similarity between molecules?
I am looking for a good similarity measure between molecules in order to use it with machine learning algorithms. I found a paper, Carbó, R., Leyda, L. and Arnau, M. (1980), How similar is a molecule to another? An electron density measure of similarity between two molecular structures. Int. J. Quantum Chem., 17: 1185-1189. doi:10.1002/qua.560170612 where they propose a similarity measure based on the electron density $\rho$:
$$\varepsilon_{AB} = \int_V|\rho_A-\rho_B|^2\,dV$$
This can be expressed as the sum
$$\varepsilon_{AB} = \int_V\rho^2_A\,dV+\int_V\rho^2_B\,dV-2\int_V\rho_A\rho_B\,dV$$
The authors then retain only the third term of this sum because only this term refers to the difference between the molecules A and B, and then they normalize it arriving to their final similarity measure
$$r_{AB} = \frac{\int_V\rho_A\rho_B\,dV}{\left(\int_V\rho^2_A\,dV \right)^{1/2} \left( \int_V\rho^2_B\,dV \right)^{1/2}}$$
which is zero for completely different molecules, and one if A = B.
My question is: why the electron density $\rho$ is appropriate here? A large portion of the electron density is due to the core electrons that don't participate in chemical bonding. They spend most of the time near the nuclei, and don't influence the rest of the molecule very much.
Instead of using the total electron density $\rho$, wouldn't it be better to use the density of only valence electrons: $\rho_{val} = \rho-\rho_{core}$?
Elements along the same group of the periodic table have similar chemical properties but have very different $\rho$ because they have different number of core electrons. However, their $\rho_{val}$ will be similar.
Is my thinking valid or I am missing something? What is the influence of the core electrons besides screening a part of the nuclear charge and preventing the valence electrons to come closer to the nuclei? Are there any papers where the density of the valence electrons is used as a descriptor of the molecular structure?

TLDR: Defining the similarity between two molecules is an open field of research, but is fundamentally a subjective question which depends on your context. If you are interested in the similarity of molecular structures, then a graph-based measure might be a logical choice. If you're interested in similarities of reactivity, then a measure incorporating the electron density or valence electron density might make sense. Ultimately, only you can know which measure makes the most sense for the problem you are trying to solve.

This is a good question and one which does not (and likely never will) have a definitive answer. That is, there will never be a physically correct way to answer how similar two different molecules should be, or even what type of mathematical object that answer should be (i.e. a number, vector, matrix, etc.). Nonetheless, some measure of the similarity between systems is clearly a useful thing to know about, and it is becoming more practical as machine learning becomes more common in chemistry.
From what I have seen recently, many of the most commonly used similarity measures are derived from a representation of a molecule as a graph. I think there are two main reasons for this. First, most people who are interested in molecular similarity are also interested in machine learning. Hence, they are working with a relatively large dataset. Computing a graph representation of a molecule and performing operations on this graph is almost always significantly cheaper than computing some kind of quantum mechanical property for every molecule in your dataset. Second, machine learning projects in chemistry tend to be very collaborative and there many applied mathematicians and computer scientists that know a lot about graph theory and are hence eager to apply this knowledge in a new context.
I have found a review from 2003 which discusses some more chemically motivated methods of defining molecular similarity and some methods based on a graph representation of the molecule[1]. They also discuss a similarity measure based on the overlap a field. The most natural choice for the field being the electron density, though one could choose something like an electrostatic potential or even some kind of magnetic vector potential. What you choose would just depend on what type of similarity you're thinking of.
As an example, you propose using the electron density due to valence electrons. I would be surprised if this hasn't been done before, but I couldn't find any references where this is done. This would be very natural if one is interested in similarities with regards to reactivity. On the other hand, if you were interested in predicting how similar the x-ray absorption spectra of various compounds would be, then it would be stupid to leave out the core electrons from the density as those are the electrons which are most relevant. This further illustrates that the similarity between two molecules is unanswerable because it is not a well-defined question. That is, this similarity will inevitably be connected to some way in which two molecules produce a similar outcome in some physical situation. If you're interested in predicting a molecule which will behave similarly to some other molecule in a reaction, then valence electron density could be a great descriptor, but this descriptor might be entirely useless if you're actually interested in finding molecules with similar x-ray absorption spectra.
Now, chemists have a serious habit of trying to make a connection between everything a molecule does and the structure of that molecule. Very often, this works out extraordinarily well, so chemists keep doing it. I think this is another reason why graph based measures of similarity have become so popular. Virtually all methods of determining molecular similarity based on a graph begin by computing the adjacency matrix of the molecule based on connectivity along covalent bonds (or maybe hydrogen bonds). Then, some operation is performed on a pair of graphs which results in a number or vector or somethings else which is interpreted as a measure of similarity. Or, an operation is performed on each graph individually, and then some measure of distance is used to define the similarity of these graphs. These operations are usually called kernels. A review of some of the kernels which have been used to define graph similarity can be found here[2].
There are all sorts of tricks which can be played with the adjacency matrix to give some more physical meaning to the measure of similarity. For instance, you might weight each of the edges by the atom-atom distance defining that connection. You might also weight each of the nodes with an atomic charge or polarizability and incorporate this into the adjacency matrix. You might also choose to do this and then work with the laplacian of the graph rather than the adjacency matrix. Anything at all can be done really. You just have to choose what makes sense for your context.
More recently, people have come up with measures that also incorporate similarities of the subgraphs of two graphs[3]. This seems like it has some potential advantages if functional groups, etc. are important to you.
References:
[1]: Nikolova, N., & Jaworska, J. (2003). Approaches to measure chemical similarity–a review. QSAR & Combinatorial Science, 22(9‐10), 1006-1026.
[2]: Rupp, M., & Schneider, G. (2010). Graph kernels for molecular similarity. Molecular Informatics, 29(4), 266-273.
[3]: Kondor, R., & Pan, H. (2016). The multiscale laplacian graph kernel. In Advances in Neural Information Processing Systems (pp. 2990-2998).


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What are the properties of COₓ where x ≥ 3?
$\ce{CO_x}$ are part of the oxocarbon molecule class. Commonly known are $\ce{CO}$ and $\ce{CO2}.$ However, there are molecules in which $x$ is equal to or larger than $3.$ Those molecules are barely known, other than the fact that they're unstable.
$\ce{CO3}$ decomposes slowly enough that it may come into contact with other molecules (including biomolecules), but nothing is known of its possible oxidizing potential.
$\ce{CO4},$ $\ce{CO5}$ and $\ce{CO6}$ are very unstable, and could possibly be strong oxidants (though it may never be known since they're too unstable).
Other than that, any information?

Its hard to give (m)any solid facts about molecules which barely exist. ;) What would CO5 even look like? A ring structure?

What are the properties of two-headed people that live on an island beyond the ocean?

The compounds do exist, however, they aren't stable above really cold temperatures (around 120 K for carbon tetroxide, to 106 K for carbon pentoxide and then 60 K for carbon hexoxide).

Adding to previous comment, logically, carbon trioxide should be stable until an even higher temperature than carbon tetroxide. In the gas phase however, as noted, it still decomposes slowly enough to potentially encounter other molecules.

At least the trioxide $\ce{CO3}$ exists, and it has even been detected in at least two isomeric forms.  Wikipedia reports that it forms as a transient species in reactions between carbon dioxide and atomic oxygen or ozone, in irradiation of solid carbon dioxide, and even in oxidation of carbon monoxide by $\ce{O2}$.

If you have said A, you must also say B: $\ce{CO4}$; $\ce{CO5}$; $\ce{CO6}$. Don't leave Wikipedia half-milked:)

Thus, "at least" the trioxide.  Feel free to add more info!


text:
Stability of 4-pyridone vs 4-pyridinol
In kryptomerism assignment, there was a question to compare more stable tautomeric form between 4-pyridone (1) and 4-pyridinol (2):

Answer: 4-pyridone.
My friend gave two plausible reasons:

$\ce{C=O}$ is more stable than $\ce{C=N}.$
$\ce{-N=}$ will have greater repulsion between the lone-pair and the double bond than the same in $\ce{-O=}.$

Though, not being convinced by his explanation, I have two questions. First, is the answer correct? And second, what's the reason if it's correct?

In my opinion, the aromaticity gained by 4-pyridinol would offset the energy considerations mentioned by @Zenix's friend, making it more stable.

@AniruddhaDeb both are aromatic

@Zenix researched and answered accordingly. You're pretty sharp for someone in eighth grade :)

I don't think this is an either this or that situation. As with most of the equilibria we know it depends very much on the circumstances we look at it. Many examples of keto-enol-tautomerism are heavily influenced by the solvent. Due to time and resource restrictions I won't be able to go in as much detail as I did in explaining whether enolates get protonated at the carbon or oxygen atom, but I ran a few calculations showing some of the controversy.
The calculations were carried out using Gaussian 16 Rev. B.01 at the DF-B97D3(BJ)/def2-TZVPP level of theory, and thermal corrections were obtained from normal coordinate analysis (in the harmonic approximation) at the optimised structures at the same level of theory using $T=\pu{273.15 K}$ and $p=\pu{1 atm}$ (because that is the G16 standard, sorry for the non-SI). The polarisable continuum model was employed except for the gas phase. All energies are in $\pu{kJ mol^-1}$ for reaction \eqref{tauto}, which means that negative values indicate that 4-pyridinol is more stable than 4-pyridone and vice versa for positive values:*
$$\ce{4-pyridinol <=> 4-pyridone}\tag1\label{tauto}$$
\begin{array}{lrrr}
\text{solvent} & \Delta E_\mathrm{el} & \Delta E_\mathrm{o} & \Delta G \\
\hline
\text{gas}     &  -3.4  &  -4.2  &  -3.7  \\
\text{water}   &  21.0  &  18.7  &  18.9  \\
\text{benzene} &   6.9  &   5.2  &   5.5  \\
\hline
\end{array}
From this we can imply (although there are too few points) that in condensed phase 4-pyridone is the predominant species, while in vacuum it would be 4-pyridinol. Unfortunately, at this time I cannot offer a satisfactory (and easy) explanation for this. The values above have not been calibrated, they are a mere snapshot. I don't want to wander deep into hand-wavy territory and propose a rule of thumb based on bond energies and/or similarities to other molecules.
It is probably true that water is able to stabilise a possible negative charge at the oxygen and at the same time stabilising the positive charge of the proton bonded to nitrogen, but that's about as much of an educated guess I would like to give up.
I want to caution everybody with expressions like the aromatic form when referring to a resonance contributor. Aromaticity is a concept which applies to a (sub-) structure and as such to all resonance contributors. This is why both tautomers are aromatic regardless of the depiction chosen.
In a similar way I want to caution against the argumentation with bond energies. Whenever resonance comes into play, we will have fractional bond orders, which deviate significantly from the conditions under which the tabulated values have been measured (or calculated).
The above notes apply directly to the two reasons your friend offers:

My friend gave two plausible reasons:

$\ce{C=O}$ is more stable than $\ce{C=N}.$

$\ce{-N=}$ will have greater repulsion between the lone-pair and the double bond than the same in $\ce{-O=}.$

There is neither an isolated $\ce{C=O}$, nor a $\ce{C=N}$ bond, as they are both part of the same π-system. The lone pairs cannot (directly) interfere with the π-system as they are orthogonal.
I wish I could offer you a better explanation, but with most chemistry there is nothing easy about simple systems.

Notes:

There is a more extensive theoretical study available, which compares THF, water, and methanol and focuses more on the solvent model. However, obtained values are in a similar region:
Nagy, P. I.; Alagona, G.; Ghio, C. Theoretical Investigation of Tautomeric Equilibria for Isonicotinic Acid, 4-Pyridone, and Acetylacetone in Vacuo and in Solution. J. Chem. Theory Comput. 2007, 3 (4), 1249–1266. DOI: 10.1021/ct6002252.

* For those unfamiliar with the notations: $\Delta E_\mathrm{el}$ refers to the electronic energy in the Born-Oppenheimer approximation; $\Delta E_\mathrm{o}$ refers to the observed energy, i.e. it is the former including the zero-point energy correction; $\Delta G$ refers to the Gibbs energy.

When I have more time, I might come back to this and add some more calculations. And if I find a nice way to look at the solutions and analyse them I will add that, too.

Thank you for correcting me and cautioning me not to jump into hasty conclusion. Do you have access to ACS papers mentioned by HahaHaha in his answer.

@Zenix I wasn't really correcting you, merely pointing out a different view. And after all you have had some doubts about your approach that needed clearing up. It's a good instinct to question things as they don't always are as easy as it seems, or less so: obvious. I currently do not have access to the publication mentioned by Haha; I have my doubts about how comparable this molecule is with the ones we're looking at.

Ik that two compounds aren't same that I am particularly interested in, but reason provided there for stability might be of some help. Further it isn't that important, so don't purchase the membership of ACS for this answer.....

The solvent dependence is pretty interesting. Do you have dipole moments of the two molecules? Joule/Mills, Heterocyclic Chemistry 5th ed. agrees with your findings (pp 141–2): "Under all normal conditions, [2- and 4-hydroxypyridine] exist almost entirely in the carbonyl tautomeric form [...] the hydroxy-tautomers are detected in significant amounts only in very dilute solutions in non-polar solvents [...] or in the gas phase. The polarised pyridone form is favoured by solvation."

& the references it cites: https://doi.org/10.1021/ja00417a027, https://doi.org/10.1016/S0040-4020(01)83237-3, https://doi.org/10.1021/jo01296a002, https://doi.org/10.1016/S0040-4039(01)86850-7. Feel free to use all, any, or none; I'm too lazy to read them properly.

@orthocresol The calculated dipole moments for pyridone in the order gas, water, benzene: 6.7502; 9.6483; 7.9517. And for pyridinol (same): 2.7196; 3.7804; 3.1573. I probably have neither access, nor time, nor the patience to go through the material, but I might add it as a note, if (or when) I come back to this post. Thanks for digging anyway.

If I find the time I might write an answer based on the polarity / solvation link, but that can only rationalise the trend going from gas-phase -> benzene -> water... it is certainly not trivial, perhaps impossible, to explain why gas-phase pyridinol is more stable than pyridone to begin with.

Okay i gotta give this one totally to you, great answer, i learnt a lot from it. The answer could be in the grey zone is what i failed to understand. Thanks for the inputs to my answer and the great answer, and my apologies for getting too excited with my answer @Martin-マーチン i guess being anew user i got too excited by the data evidences i was getting for what is the answer.

In a similar way I want to caution against the argumentation with bond energies. Whenever resonance comes into play, we will have fractional bond orders, which deviate significantly from the conditions under which the tabulated values have been measured (or calculated). I was just curious if this is correct: if I add all the bond energies not considering the delocalisation, and then subtract the resonance energy, then I will get the resultant bond energy. (Of course, my answer will not hold, as I do not even know the resonance energies), but is what I said right now correct @Martin-マーチン?

@Zenix the bounty ends in 4 hrs, imo you should accept martin’s answer

@HahaHahaha I guess accepting answer won't give Martin bounty, further reasons that I was looking for aren't given in this answer. Answer seems even more mystified after this. It's in the hand of Andselisk to award bounty, not in my hand... I will accept the answer as soon as all my queries are answered.

Idk to whom @andselisk will award bounty...

@HahaHahaha That approach might be valid, if you actually know these values as independent data points. It could well be that you are running in some kind of circular reasoning, or self-fulfilling prophecy, for the cases where the one is derived from the other. I generally only wanted to caution, not say a rigorous no to the approach; models have limitations, even the one I have used. If you know them well enough, you can apply the models, otherwise you might get a false positive (or explanation).

@Zenix The answers you are looking for probably does not exist; at least there won't be a simple explanation, which you can summarise in two keywords. I agree with orthocresol that any explanation of the solvation-stability relation won't be trivial; I share the fear that it might be impossible to explain the gas phase at all. I don't know whether there is enough experimental data to correlate the findings to (I believe I read somewhere that they decompose before being able to measure them in the gas phase). I hope you won't be too disappointed

@Martin-マーチン: Pls see this also.

This Article states the following about 4-pyridone in its abstract:

Aromatic resonance energies for 4-pyridone and its analogues are similar to those for the 2-pyridone series.

This means that in order to find out how 4-pyridone is so stable, all we have to do is take a look at it's sister, 2-pyridone. 2-pyridone and it's tautomeric form, 2-pyridinol have the following resonance structures respectively, as shown in this question.

The major reason of the stability of 2-pyridone is that Its aromatic form has the negative charge on the more electronegative element and the positive charge on the more electropositive element. More about the tautomerism of 2-pyridone can be found on Wikipedia, however for the purposes of this explanation I'll restrict myself to the above concept.
Applying this concept to the resonance of 4-pyridone and 4-pyridinol respectively, we obtain the following structures:

This explains why 4-pyridone is more stable than 4-pyridinol
References:

Cook, Michael J.; Katritzky, Alan R.; Linda, Paolo; Tack, Robert D.: Aromaticity and tautomerism. Part II. The 4-pyridone, 2-quinolone, and 1-isoquinolone series. J. Chem. Soc., Perkin Trans. 2, 1973, 1080-1086
Forlani L.; Cristoni G.; Boga C.; Todesco P. E.; Del Vecchio E.; Selva S.; Monari M. (2002). Reinvestigation of tautomerism of some substituted 2-hydroxypyridines. Arkivoc. XI: 198–215.

I have a doubt, isn't in the question asked here,OP make fundamental mistakes in counting no. of π electrons. Also your answer is contrary to what answer says, and acc. to you 2-pyridinol is more stable, which is not the case

Sorry I fixed a couple typos above. As for the answer, the $\ce{\pi}$ electrons seem in order. Nitrogen’s orbitals in 4-pyridinone are sp3 hybridised and the lone pair electrons in the sp2 orbital do not contribute to aromaticity

"On the 2-pyridone resonance forms (right) we have on the left side of the equilibrium a non aromatic (8pi e−) uncharged molecule and on the right side an aromatic (6pi e−) charged molecule", this is quoted from above link. How are there 8π e- and aren't both aromatic?

Both are not aromatic. The nitrogen having two lone pairs does not conjugate it's lone pairs with the pi-electron system in the resonance structure as it's orbitals are sp3 hybridized.

There are no sp3 nitrogens in any of these structures. Mind you, it's not possible for hybridisation to change between two different resonance forms of the same thing.

@orthocresol aren't both aromatic? Also can you please check my first comment on this answer and verify or correct me...

The bottom image uses the wrong arrows, these symbolise tautomerism, but the correct form would be $\ce{<->}$ for resonance (or mesomerism [outdated]).

This answer does seem right. In the reaction I have drawn here, a similar logic prevails, you would be shocked to know that the equilibrium shifts to the right here in an ethanolic solution. The reason simply is the very high $\ce{C=O}$ bond energy. $\ce{C=O}$ has a bond energy of more than $\pu{100 kJ/mol}$ than the $\ce{C=N}$ and $\ce{C=C}$ bond energies (data from here). So, both these compound being aromatic, the debate of resonance energy dies, so it is a perfect explanation.
Here is a solid ACS paper[1] which says the same, the source for my claim even if you cannot access the paid version if you are not an ACS member, the first page is free, you can read clearly, that the mixture exists largely in the quinoid form.
On page 72 in the paper by Suzanne Slayden and Joel Liebmann titled The organic thermochemistry of hydroxylamines oximes hydroxamic acids and their derivatives, there is proof as well as justification of the same.

References:

Anderson, L. C.; Yanke, R. L. The Tautomerism of Quinoneoxime-para-nitrosophenol Systems1. J. Am. Chem. Soc. 1934, 56 (3), 732–735. DOI: 10.1021/ja01318a062.
Rappoport, Z.; Liebman, J. F. (eds.): The Chemistry of Hydroxylamines, Oximes and Hydroxamic Acids, Volume 1. John Wiley & Sons, 2008. (See a preview on books.google.co.in.)

I am sorry for the weird screenshot, I'm a new user and not aware how you guys draw reactions

ChemDoodle is fine, but it would be nice to crop the image so that it's not bloated with the irrelevant web interface elements. Also, there is a brief software community Wiki. If you are interested in tips on formatting, feel free to visit this page, this page and this one on how to format your posts better with MathJax and Markdown.

One more thing: cite the original source whenever possible. The web page states the data is "Referenced from Chemistry by Zumdahl (5th ed.) Table 8.4 on page 373", so it's a good idea to cite the textbook as a primary source and append URL for convenience.

Thanks for the help andselik, but what should I cite when the reaction was given by my teacher in-class notes? Is citation avoidable in such situations, because i do not have many books at my disposal to search from where the reaction was given. Thanks for the software I will try.

Regarding citing the reaction by your teacher: it's a gray zone. Most likely, the teacher adapted it from somewhere else, so ideally that original source should be cited. If finding this out is deemed impossible, then post it as is (e.g. citing your teacher's private lecture notes), but be prepared for the critique for the lack of credibility of such source.

As said by @andselisk, here comes your first critique. Your answer seems correct. And have a look what explanation my friend gave.... But bounty was placed to cite the source...

found it @Zenix I guess ACS is trustworthy enough

Thanks for the research, so the answer given is correct. But I am afraid if first page provide reason for the same? No, I do not have the paid membership of ACS. In case you have, what's the reason that this article provide, and on which page..

Nah, nor do I, I'm pretty much a starter in organic chemistry. Ill try and find something else

I searched a lot, all I could find is NMR proof on the first page of the acs link I showed you which says the answer is correct. Why is c=o bond energy not acceptable??/

Please do not use edit statements, rewrite your post as necessary to form a continuous text. New readers will not care about the timely progression of writing; for those interested there is the history available. I have cleaned up your post a bit also using the mhchem package (look here and here. Also note that not all content on google books is licensed everywhere; it would be better to quote the relevant portions and leave a link as reference.

For the chemistry at hand: I do think that the molecules you cite a fundamentally different from what was asked in the question, and I would not be too sure to make this leap in argument.

For the re edits, sorry i wont do that again, as for the the chemistry, nature loves the $C=O$ bond over aromatic hetero rings of nitrogen, as this is illustated in my example, the question itself, in the struture of guanine, in the structure of cytosine, in the structure of thyaime, in the structure of uracil, etc. So i do not agree with you that the chemistry in different. In my opionion its a perfect analogy. If there was exact data easily available, then Zenix wouldnt ask this and Andselisk wouldnt bounty it. @Martin-マーチン

In this case, the dominance form of the compound depends on which phase it is in. Ref.1 states that:

In the gas phase, there is a preference of the hydroxy form in both 2- and 4-hydroxypyridine, while in a nonpolar solvents such as cyclohexane and chloroform, the two tautomers exist in comparable amounts. However, the two equilibria are shifted entirely to the right in favor of the oxo-form, 2- and 4-pyridone, respectively in polar solvents as well as in the crystalline state.

In general, pyridones (e.g., 4-pyridone) predominance strongly over their tautomeric hydroxy forms (e.g., 4-hydroxypyridine) can be rationalized by considering the mesomerism of the alternative forms. The charge-separated form of 4-pyridone (far left; negative charge on more negative oxygen), being aromatic, is greatly preferred over the nonaromatic, charge-separated form of 4-hydroxypyridine (far right; negative charge on relatively more positive nitrogen).

There are compilations about this type of keto-enol tautomerism.   The freely available Tautobase for example is provided as searchable database in DataWarrior.
References:

Jiali Gao, Lei Shao, "Polarization Effects on the Tautomeric Equilibria of 2- and 4-Hydroxypyridine in Aqueous and Organic Solution," J. Phys. Chem. 1994, 98(51), 13772–13779 (https://doi.org/10.1021/j100102a051).
Oya Wahl, Thomas Sander, "Tautobase: An Open Tautomer Database", Chem. Inf. Model. 2020, 60, 1085–1089 (doi 10.1021/acs.jcim.0c00035).

@Buttonwood: Thanks for addition of good reference.


text:
Does the concentration of water accounts for determining the rate of a reaction?
I found this problem in my book and I'm not sure how to assess this question.
It is as follows:

Regarding the relationship between the rates of the chemical species
  in the following reaction. Which of the following alternatives is the
  correct one?

$\ce{2H2 + O2 -> 2 H2O}$
$\begin{array}{ll}
1.&v_{H_2}=2v_{H_{2}O}\,, v_{O_2}=2v_{H_{2}}\\
2.&v_{O_2}=2v_{H_{2}O}\,, v_{H_2}=2v_{O_{2}}\\
3.&v_{H_2}=2v_{O_{2}}\,, v_{O_2}=v_{H_{2}O}\\
4.&v_{H_2}=v_{H_{2}O}\,, v_{O_2}=\frac{1}{2}v_{H_{2}}\\
5.&v_{H_2}=v_{H_{2}O}\,, v_{O_2}=v_{H_{2}}\\
\end{array}$
In the original source (which doesn't list a specific authorship) doesn't indicate the state of matter for each species. So I'm assuming perhaps that the reagents are gases and the products is liquid?.
Anyways what I've attempted to do is to establish the rate law as follows using the law of mass action:
$\ce{2H2 + O2 -> 2 H2O}$
For Hydrogen gas:
$v_{H_{2}}=-k[H_{2}]^2[O_{2}]$
For Oxygen gas:
$v_{O_{2}}=-k[H_{2}]^2[O_{2}]$
From this it can be established that:
$v_{H_{2}}=v_{O_{2}}$
For water (since it is on the products it must be established as if it were a reagent):
$v_{H_{2}O}=k[H_{2}O]^2$
And that's where I'm stuck.
Supposedly the answer is
$v_{H_2}=2v_{O_{2}}\,, v_{O_2}=v_{H_{2}O}$
But I don't see any way of how to get there. Can somebody point me exactly what did I missed?.

Rate of consuming of reagents ( or production of products) must be in ratio of their stoichiometric coefficients ( for case of a single step, or a pseudo single step reaction)

@Poutnik I recall that if water is a solvent then it can be left out of the rate law. But in this case it is a product. By comparing to what I attempted to do I placed the stoichiometric coefficients in the rate law for each species, were you referring to this?.

Hm, unless it is in context of some fuel cell, water will be hardly in liquid state.But for fuel cells, different reactions would occur.

@Poutnik You're right about fuel cells. But in a general context as (I think it was intended in this question) does it exist a way to relate those three rate laws as presented in the alternatives?.

Rate law vs. definition of rate
For this problem, you don't need the rate law. And, just for the record, you can't get the rate law from the stoichiometry - it depends on the mechanism and has to be determined empirically.
Definition of rate of reaction
The rate of reaction is based on the change of concentrations. For a homogeneous reaction (i.e. all species in the same gas phase, or all species part of the same solution), the definition is the following:
$$\mathrm{rate} = \frac{d[\ce{X}]}{\nu_X \cdot dt}$$
where $\nu_X$ is the stoichiometric coefficient of the species $\ce{X}$. This does not apply to heterogeneous reactions. For one, if a species is a pure solid or liquid, its concentration will not change. Also, the stoichiometry of a reaction refers to amount of substance - only if all species are in the same solution (or the same gas phase) does this translate into easy relationships of concentration changes (or partial pressure changes).
Definition of "rates of the chemical species"
This is not a technical term I am familiar with. In order to be able to answer the question with answer 3., you would have to define it as:
$$v_\ce{X} = \frac{\mid dn_\ce{X} \mid}{dt}$$
Once you define it such, it does not matter what the physical state of reactants and products is, it does not matter whether a species is a reactant or a product, the only thing you need to related the different rates are the stoichiometric coefficients.
So for example, the ratio of the amount of hydrogen and oxygen used is equal to the ratio of stoichiometric coefficients:
$$\frac{\Delta n_\ce{H2}}{\Delta n_\ce{O2}} = \frac{\nu_\ce{H2}}{\nu_\ce{O2}} = \frac{2}{1}$$

The answer according to the answers sheet in my book says it is the answer $3$ which would be the same as answer $4$?. From your answer the idea is that the rate of reaction by definition is the inverse of the stoichiometric coefficient?. Then why in literature is it always presented with the law of mass action instead?. Did I missed something. Can you explain this part or the relationship between these two concepts?. Btw I was aware that the rate law depends on the mechanism.

@ChrisSteinbeckBell My mistake, answer 3 is the one that matches the stoichiometry. I edited my answer - thanks for paying attention.

You're welcome but there's still one of my doubts unnatended which is regarding on why in the books rate law is defined by the law of mass action. I mean $aA+bB\rightarrow cC$ being for $A$ rate law $\dfrac{dA}{dt}=-k[A]^a[B]^b$ and not by $\textrm{rate (A)}=\dfrac{d[A]}{adt}$?. You used the word rate but isn't rate the same as $\frac{d[X]}{dt}$ why is the stoichiometric coefficient in the denominator?. why is it the inverse?. Is it due a definition?

@ChrisSteinbeckBell You want to have a single rate for a reaction even if not all species have 1:1 stoichiometry. So the reaction rate is defined with the stoichiometric factor in the denominator. No matter which species concentration you monitor, you will find the same rate.

@ChrisSteinbeckBell "In the books, rate law is defined by the law of mass action"? The rate law usually starts with rate =, and uses coefficients m, n, q etc. that are distinct from the stoichiometric coefficients, e.g. https://openstax.org/books/chemistry-2e/pages/12-3-rate-laws. If your textbook says otherwise, ask a specific question showing the portion of the textbook.

The fourth possibility is the only correct answer, because the numbers it contains are proportional to the stoichiometric coefficients, whatever the physical state. The fourth possibility is the only one where O2 is consumed at a lower rate than H2. And this is exactly what the equation says : when 1 H2 is consumed, only 1/2 mole of O2 is destroyed.
And by chance the other relationship in 4th answer is also correct. The number of H2O produced is always legal to the number of H2 consumed, whatever the physical state.

Im still confused about what you mentioned about velocities proportional to their stoichiometric coefficientes, can you put this as a relationship for a generic situation so i could see it better? What ive attemped to do is follow the definition for rate law based on law of mass action but as you can see i could not get to an answer. There's still my question about water, should it account for rate law? How should it be wrote that velocity in terms of the other species? Is what i did correct?


text:
When aluminium is ignited in atmosphere, will it form only aluminium oxide or also aluminium nitride?
We know when lithium is burnt in air, it forms lithium nitride and lithium oxide. I am wondering if this applies to aluminium (other metals) also?

Magnesium also forms the nitride as well as the oxide when burning in air.

According to the paper by Gromov and Vereshchagin [1], it is indeed formed when burning aluminium, but you could have predicted this simply by considering the standard enthalpy of formation of aluminium oxide $\Delta_\mathrm{f}H^\circ (\ce{Al2O3}) = \pu{−1675.7 kJ/mol}$ which will be liberated upon ignition and this energy helps in breaking the strong nitrogen-nitrogen triple bond (bond energy is $\pu{−945 kJ/mol})$ and finally this would form aluminium nitride, which also produces some energy (standard enthalpy of formation is $\pu{−318 kJ/mol}),$ but the combustion here is not a simple one it requires high temperature as mentioned in the paper, I think this is to sublimate $\ce{Al2O}$ which will react with nitrogen and give rise to aluminium nitride.
As HYDR0GEN stated, this combustion isn't simple like throwing a match on aluminium sheet, it needs high temperature. According to the work by Trunov et al. [2, p. 38], as the size of aluminium particles decrease the needed temperature for combustion decreases and that's why researchers who wrote the paper used superfine aluminium powder:

Figure 5. Experimentally determined temperatures of aluminum
  ignition as a function of the used sample size.

References

Gromov, A.; Vereshchagin, V. Study of Aluminum Nitride Formation by Superfine Aluminum Powder Combustion in Air. Journal of the European Ceramic Society 2004, 24 (9), 2879–2884. DOI: 10.1016/j.jeurceramsoc.2003.09.020. (PDF)
Trunov, M. A.; Schoenitz, M.; Dreizin, E. L. Ignition of Aluminum Powders Under Different Experimental Conditions. Propellants, Explosives, Pyrotechnics 2005, 30 (1), 36–43. DOI: 10.1002/prep.200400083.

When citing third-party resources, make sure to provide a complete human-readable reference and not just "this link" to make your post robust against link rot. Also, always cite the original publication, not a reproduced rip-off of a questionable quality where there is typo in the axis name ("Diamter") and illiterate numbers formatting. I took a liberty to replace the illustration with with the one from original paper.

@andselisk thanks for help, it's my first answer

When the aluminum [powder] is ignited, it will cause a reaction between the aluminum and oxygen in the atmosphere. The required temperature for the aluminum oxide to be formed is almost 3000+ °C.
$$\ce{4 Al + 3 O2 -> 2 Al2O3}$$
In addition, aluminum nitride can't be formed when the aluminum powder is ignited in atmosphere, because the aluminum nitride starts to decompose into aluminum and $\ce{N2}$ at a high temperature under atmospheric pressure.

Aluminium needs high temperature to be ignited but that's why the researchers who wrote the paper used ultra fine powder of aluminium to be ignited easily . See this link :https://www.researchgate.net/figure/gnition-temperatures-of-different-samples-of-aluminium-samples-reproduced-from-Trunov-et_fig3_240497487


text:
A question about nicotinamide adenine dinucleotide
In the structure of NAD+, why does the nitrogen of pyridine bind covalently to the the first carbon of glucose while the anion stabilizing the nicotine amide ion is the phosphate group? In other words, why didn't NAD+ exist as an ionic salt between nicotine amide ion and phosphate ion in the  nucleotide?


text:
Dipole Moment of the following organic compounds

Question: Which of the following compounds will show highest dipole moment and why?
I figured out that, (4)th compound is very symmetric - so net dipole moment on the compound must be null.
How should I go about finding the compound with the highest dipole moment? Any help or tips are welcome!
[Question borrowed from the JEE - Main]

“The image and the title are very self-explanatory.” – Please don’t write such things. Write your question in the Body box. Your question should be clear even without the title. Especially, do not write your whole question in the title and then put something like “title says it all” in the body. Write your question out in full; do not post a picture of the question.

It is all about aromaticity.

@IvanNeretin well.. none of the compounds are aromatic out-of-the-box. Perhaps you're hinting that I should alter the compounds a bit and look for aromaticity?

Well, yeah, sort of.

@IvanNeretin well.. (IV) is out of the question for highest dipole moment. For aromaticity; one needs conjugation and 4n+2 $\pi$ electrons - this doesn't seem possible for (III). Only options left are (I) and (II). For (II), conjugation could be achieved if I performed some form tautomerism... however condition for 4n+2 electrons fails. Speaking about option (I), thinking of tautomerism doesn't make sense since conjugation of double bonds is impossible as this ring is too small. The least sketchy option is (I) - but I'm not happy with the way I got the answer. How do I make (I) look aromatic?

By thinking of tautomerism, even if it's making you uncomfortable. Yes, the ring is pretty strained, so what?

@IvanNeretin I'm sorry - I didn't understand what you wrote..

Imagine a form of (I) with a minus on O and plus on the ring.

@IvanNeretin I get it now. Thanks. That structure you just spoke about is clearly aromatic, and is very stable. So it would be prone to stay like that, hence the big dipole moment it will have, thanks once again.

Great! You may write an answer yourself, this is allowed and even encouraged.

@McSuperbX1 if we just switched to looking for orbital I think the carbon atoms attached to the carbonyl at III are sp3 hyperidized so it will the easiest to strip electron cloud off. Right?

For further studies you may search for quasi-aromaticity.

Im not sure why this question is closed. What effort could I have possibly shown? I hit a dead end. If Ivan had not given me an idea about aromaticity, I would've remain stuck. This question is from an engineering college entrance examination, intended for 17/18 year olds. It would be helpful if someone could explain how I could've possibly shown more effort. Thank you.


text:
Way to confirm dichromate ion presence where other ions are messing with the color
I'm currently working on an electrolytic setting where a stainless steel (430 series, mainly made of iron and chromium, but no nickel) anode is dissolved in acidic environment (sulfuric and phosphoric acids at moderate concentrations).
My concern is the hypothetical creation of dichromate ions, since they are carcinogenic. But I can't confirm only color-wise because of the presence of other ions that mixes all the colors.
The experiment runs for long (3 hours). At first an orange color appears, which suggests dichromate ions (but could be Fe III). But after only a little while, the solution turns to a very dark and deep green, almost blue depending on the light. This suggests chromium III ions.
Several things could be happening, such as reduction of Cr VI to Cr III at the cathode, or just creation at the anode of Cr III whose color overwhelms that of Cr VI, or even Fe II or Fe III messing with the color.
Does anyone have an idea about a means to confirm or infirm the presence of dichromate ions despite not being able to judge by the color ? And if so, how could I determine the concentration of it ?
Thanks a lot in advance.

The best reagent for detecting Cr(VI) in aqueous solution is the diphenylcarbazide, which produces a red color in presence of Cr(VI). The sensibility limit is as low as 0.2 ppm Cr(VI). This method has been developed one hundred years ago. See
1) M. A. Moulin, Dosage du chrome, Bull. Soc. chip. 31, 295 (1904).
2) H. Crossley, Diphenylcarbazide. An indicator for the titration of iron with dichromate. Analyst 61, 164 (1936).
The reaction with alcohols to produce aldehyde and acids does work. But it is not very sensitive. It needs at least some millimoles of chromate to be detected. The reaction with diphenylcarbazice is about 1'000'000 times more sensitive.

Thanks a lot, I guess I'll order some and make the test.

I think, the other way to detect the presence of dichromate on a solution is when the primary alcohols are converted into aldehydes while the secondary alcohols into ketones.
On the other hand, the presence of aldehydes and ketones can be analyzed by various confirmatory tests.

Oh thanks, indeed I remember this property ! So adding ethanol and heating should get me all the way to ethanoïc acid, which has a caracteristic vinegar smell... That should be enough to confirm.

@MrFrederic. I think, you can determine the concentration of dichromate by means of titration.


text:
What is Meant by Mole Percent?
I am attempting to interpret an experimental section on a scientific paper. The paper calls for 0.51 mmol of 3,5-dimethoxy phenol, 1.0 mmol of 3-methyl-3-butenal, 5 mol% of phenylboronic acid, 20 mol% of benzoic acid, and 2.5 mL of heptane. I am trying to figure out what is meant by mole% in this case. There are many different interpretations of this I have seen online, including mole percent being based on a fraction of the limiting reagent (the phenol in this case), the mole percent being based on the fraction of all of the total moles of reactant and catalyst (most fitting of the definition of mole percent), and the mole percent being based on all of the moles of reactant, catalyst, and solvent. It makes most sense that the mole percent for this protocol would be based on the limiting reagent, but I included the calculations based on the total moles of reactants and catalysts for reference. Does anyone know what the term "mole percent" would appropriately based on in this case?
0.51 mmol (phenol der.) + 1.0 mmol (aldehyde) + x mol PhB(OH)2 + y mol PhCOOH = z total mol
0.2 z (total mol) = y mol PhCOOH
0.05 z (total mol) = x mol PhB(OH)2
0.51 mmol (phenol der.) + 1.0 mmol (aldehyde) + 0.2z + 0.05z = z total mol
0.75z = 1.51
z = 2.013
0.2 (2.013) = 0.40 mol benzoic acid
0.05 (2.013) = 0.10 mol phenyl boric acid
If the mole percent was based on the moles of the LR, the moles of catalyst would be as follows:
0.10 mol benzoic acid and 0.026 mol of phenyl boric acid

In the absence of any other information, I would always assume that it refers to the limiting reagent. This is by far the most common usage (in fact, I don't think I've ever seen your other definitions before). But you know what would be really helpful - if you provided a link to this paper...

The information I provided is about all the paper provides in terms of the experimental section, however, here is a link the the paper in case anyone is interested: DOI: 10.1039/C6OB01026A

Eli, you should have read the paper very carefully before asking questions. I think you are very intelligent person, so don't choose easy way out. Now, to answer your question, about mol%. It is right in the paper. As you already figured out and also Orthocresol pointed out in his comment, The amount equals to given mol% is $LR \times \frac{given \ mol\%}{100}$. It is in the General procedure A of given reference:

Condensations of phenols and naphthols with $\alpha,\beta$-unsaturated ketones. An oven-dried $\pu{25 mL}$ Schlenk tube equipped with a teflon-coated magnetic stir bar was evacuated and purged with argon. The phenol ($\pu{0.2 mmol}$), pentafluorophenylboronic acid ($\pu{8.5 mg}, \ \pu{0.04 mmol}, \ \pu{20 mol\%}$), diphenylphoshinic acid ($\pu{8.7 mg}, \ \pu{0.04 mmol}, \ \pu{20 mol\%}$), $\alpha,\beta$-unsaturated ketone ($\pu{0.4 mmol}$) and heptane ($\pu{1 mL}$) were added to the tube under argon.....

Accordingly, The phenol used: $\pu{0.2 mmol}$ ($LR$), pentafluorophenylboronic acid used: ($\pu{0.04 mmol}$ meaning $0.2 \times \pu{20 mol\%}$), and diphenylphoshinic acid ($\pu{0.04 mmol}$ meaning $0.2 \times \pu{20 mol\%}$).

Thank you for pointing this out. I did not think to read general procedure A since it was not referred to in the sub-procedure I was following; however, I see how valuable information can be held within every aspect of the paper.


text:
Suitable reagents for nitration of thiophene
I know that for nitration of benzene we can use concentrated nitric and sulfuric acid. As furan and pyrrole are less aromatic than benzene, they can be protonated more easily and undergo polymerisation. We can use $\ce{AcONO2}$ and $\ce{AcOH}$ instead.
Give the order of aromaticity: benzene > thiophene > pyrrole > furan, I'm not sure whether thiophene is suitable for use with concentrated nitric and sulfuric acid.

According to this reference by Katritzky et. al (2005) (see below):

Thiophenes  are  easy  to  nitrate  compared  to  other  five  membered  heterocycles.  They  react  with  mild  nitrating  agents  such  as  copper  nitrate.

The reference further reports the nitration of thiophene to 2-nitrothiophene by nitric acid/Trifluoroacetic anhydride in 78% yield.
The classic $\ce{c HNO3/c H2SO4}$ conditions are too strong and likely to result in substrate degradation.
Reference:
Alan R. Katritzky, Eric F. V. Scriven, Suman Majumder, Rena G. Akhmedova, Novruz G. Akhmedov, Anatoliy V. Vakulenko, "Direct nitration of five membered heterocycles," ARKIVOC 2005, (iii), 179-191 (https://doi.org/10.3998/ark.5550190.0006.320).

Suitable reagents for nitration of thiophene have already been addressed by elsewhere here. Therefore, I'd address why concentrated nitric and sulfuric acid are not suitable for the task, since you are not sure whether the traditional mixture is suitable or not for use to nitrate thiophene.
Because of thiophene is much reactive towards electrophilic substitution, compared to benzene, the traditional mixture of concentrated nitric acid and sulfuric acid is not the ideal nitrating reagent for thiophene (Ref.1). That's because it is essential that the nitration of thiophene should be carried out in the absence of nitrous acid as it can lead to an explosive reaction. A part of the abstract of Ref.1 states that:

".....the mechanism of reaction appears to be the same as that for benzene compounds. However, it is not a suitable preparative procedure owing to the high reactivity of thiophen. Nitration of thiophen, with a solution of nitric acid in acetic acid, proceeds with explosive violence owing to the easy nitrosation of thiophen in an autocatalytic reaction, leading to a variety of unidentified products. If nitrous acid is removed by addition of urea, nitration of thiophen at low concentration is of the first order in thiophen and nitrothiophen is obtained in high yield. However, at high thiophen concentration nitrosation again becomes the predominant reaction. The reason for this is discussed. As has been known for some time, the most successful reagent for the nitration of thiophen is nitric acid in acetic anhydride. Acetic anhydride does not greatly increase the rate of nitration but appears to remove complications due to nitrosation."

In addition to use of acety1 nitrate (nitric acid in acetic anhydride; Ref.2) as the nitrating reagent (which prevent the formation of nitrous acid), nitronium tetraflouroborate is also a satisfactory nitrating reagent for thiophenes (Ref.3):
$$\ce{ArH + NO2+BF4- -> ArNO2 + HF + BF3}$$
Like nitration of substituted benzenes, nitration of thiophene also gives at least two mono-substituted products: The major product is 2-nitrothiophene, is accompanied by approximately 10% of the 3-nitro isomer (Ref.4).

References:

Anthony R. Butler, James B. Hendry, “Electrophilic substitution on the thiophen ring. Part IV. Kinetics and mechanism of the nitration of thiophen,” J. Chem. Soc. B 1971, 102-105 (https://doi.org/10.1039/J29710000102).
V. S. Babasinian (Checked by Roger Adams and A. E. Knauf), “2-Nitrothiophene,” Org. Synth. 1934, 14, 76 and Organic Syntheses, Coll. Vol. 1943, 2, 466 (DOI: 10.15227/orgsyn.014.0076).
G. Oláh, S. Kuhn, A. Mlinkó, “Aromatic substitution. Part II. Nitration of aromatic compounds with nitronium tetrafluoroborate and other stable nitronium salts,” J. Chem. Soc. 1956, 4257-4258 (DOI: 10.1039/JR9560004257).
Börje Östman, “Preparation of 2-Nitrothiophene of High Isomeric Purity,” Acta. Chem. Scand. 1968, 22(5), 1687-1689 (DOI: 10.3891/acta.chem.scand.22-1687).


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What is Meant by "Eluting with 1-2% EtOAc/pentane" in Terms of Flash Chromatography?
I am reading the experimental section of a paper, and it states

[compound of interest] Isolated as a pale yellow oil after flash chromatography on silica gel, eluting with 1-2% EtOAc/pentane...

This is all this paper states in regards to the work-up procedure. I fully understand everything in this section except what is meant by 1-2% EtOAc/pentane. Does this mean that in total only 1-2% by volume of ethyl acetate is used compared to the amount of pentane used in the flash chromatography? This could also mean that a solution of 1-2% ethyl acetate in pentane was used throughout the entire run (which I find most likely). Which of these interpretations is most likely? Would this mean the author of the paper did not change the polarity of the solvent during chromatography and only use a 1-2% EtOAc in pentane solution?
Here is the DOI to the paper: 10.1039/c6ob01026a

This looks to me like a gradient starting at 1% EtOAc in Pentane and finishing with 2% EtOAc in pentane was used. Must be a very non-polar product.

In another section of this same procedure, however, it states "eluting with 2% diethyl ether/hexanes". Would this suggest that the gradient starts at pure hexanes and is gradually increased to 2% in this case. In the case of my question above, would it mean we start with pure hexanes and gradually increase the gradient to 1-2% EtOAc/pentane?

It is going to have the same outcome whether you start with neat hexanes or pentane and gradient to 2% EtOAc/Ether or start at 1%. This is a very shallow gradient for a very non-polar material. If in doubt start with neat hexanes/pentane and run a slow gradient to 2% EtOAc/ether and expect to collect a lot of fractions.

As Waylander say, there shouldn't be any different outcomes operating according to your various interpretations. That is probably why they do not mention any gradient.

@Waylander How do you apply a gradient to a flash chromatography? That sounds very cumbersome to me, unless you have a machine that mixes and applies the solvent automatically.

By making up 100ml volumes of eluent with succesively more polar solvent so you elute with for example 100ml neat hexanes, then 100ml 1% EtOAc  then 100ml 2% EtOAc. It is time-consuming!

Based on OP's question, I tentatively assumed OP is a novice to the isolation of organic compounds using column chromatography. Thus I'd explain this question with as basis as possible.
This paper OP sighted is for the synthesis of 2$H$-chromenes (2$H$-benzopyrans). They can be non-polar to polar compounds based on the substitution on benzene nucleus. The paper describes a series of relatively non-polar chromenes. When you do organic reaction, you monitor the progress of reaction with TLC analysis. During that analysis, you can determine that what solvent should be the most suitable for the particular compound elution. That's how the authours have determined several solvents such as 1% ethyl ether/hexanes and 2% ethyl acetate/pentanes, etc. for different different chromenes.
Some of these chromenes have been isolated using just one solvent. For example, evidently, they have mentioned isolation of 7-Methoxy-4-methyl-2-propyl-2$H$-chromene, using flash chromatography on silica gel, eluting with 2% diethyl ether/hexanes solvent system (this means solvent made by 2 parts of diethyl ether and 98 parts of hexanes, by volume). Note that it would also mean that they might have used certain volumes of following systems: (1) pure hexanes; (2) 1% diethyl ether/hexanes; (3) 2% diethyl ether/hexanes; etc. However, expected compound has come out with only fractions eluted with 2% diethyl ether/hexanes (remember, you supposed to analyze all the fraction by TLC), which may have combined and further analyzed by spectroscopy to confirm the structure.
As in another data set, they have mentioned isolation of 5,7-Dimethoxy-2,2-dimethyl-2$H$-chromene, using flash silica gel column, eluting with 1-2% EtOAc/pentane solvent system.It would definitely mean that they should have used certain volumes of few solvent systems: (1) pure pentane; (2) 1% EtOAc/pentane; (3) 2% EtOAc/pentane; (4) 3% EtOAc/pentane; etc. However, their expected compound has come out with fractions eluted with 1% and 2% EtOAc/pentane (concluded by the fractions analysis by TLC).
Later Note: As commented by Waylander, these type of separation is called gradient elution. By the scale of reaction, total elution must not go over $\pu{100 mL}$ of solvent for each compound.

Just to clarify, when I actually carry out the gradient elution for my compound I should gradually increase the polarity ABOVE what was mentioned in the paper. So in my case gradually increase the polarity to something like 3.5% EtOAc/pentane. Since I will be using about 28 fractions of about 12 mL each, this would mean that I should start with ~40 mL of pure hexane and increase the EtOAc concentration by 0.5% each run until I reach 3.5% EtOAc. I will, of course, be checking the TLC each time, but this will give me a good estimate of where I should expect my coupound.

It you are using a fraction collector, you can program the gradient with volumes.


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How does a double displacement reaction work?
When a double displacement reaction occurs with compound AB and CD, what determines that B and C are going to bond instead of A and C?

Does this answer your question? What determines whether a double displacement reaction will occur?

Double displacement reactions involve ionic compounds, i.e. salts formed from positive cations and negative anions. Consider the following setup,
$$\ce{\color{blue}{A^{+}}\color{red}{B^{-}} + \color{blue}{C^{+}}\color{red}{D^{-}} -> \color{blue}{A^{+}}\color{red}{D^{-}} + \color{blue}{C^{+}}\color{red}{B^{-}}}$$
where the cations are indicated in blue and anions in red (and all charges are $\pm 1$ to reflect your question). All salts are comprised of cations and anions, hence $\ce{\color{blue}{AC}}$ wouldn't form since both are cations. $\ce{\color{blue}{C}\color{red}{B}}$ would form instead.


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What are some software packages that allow visualization of molecules in 3D?
Could anyone suggest a good application or a software that helps visualise molecular structures in 3d?
I have tried a few applications like King Draw and molecular constructor but I want to know if there are better ones out there.

Avogadro is free and is compatible with almost any standard chemical format, plus it has some force fields for structure optimization.

Possible duplicate of How do I make 3d molecular graphics similar to those shown on Wikipedia?; Freely available software for generating 3D coordinates of molecules.

Some examples are: MOLDEN, Avogadro, JMol and VMD.
But there is a very thorough list (along with comparisons) here: https://en.m.wikipedia.org/wiki/List_of_molecular_graphics_systems.


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A question about thiamine
How does the the nitrogen of thiazole in thiamine acquire a positive charge without being stabilized by another negative charge or by being a salt of an anion?

Even if it's not always reported, the thiamine you can buy or isolate is a salt: the thiamine molecule is a cation, and a counterion is present (typically Cl- or NO3-)

But this molecule doesn't enter my mind as being simply a cation, was the nitrogen atom neutral first then the hydrogen of methylene group attached to nitrogen was striped off by a strong alkali pushing the nitrogen to form a pi bond with the deprotonated group and acquiring a positive charge?

Thiamine (Vitamin B1) has a positive charge on the nitrogen of thiazole ring, because that nitrogen is tetravalent. Thus, it should be neutralized by a counter ion. Usually, the counter ion is chloride ($\ce{Cl-}$) ion. Over-the-counter Vitamin B1 usually supplies as hydrochloride salt of thiamine chloride (simply called thiamine hydrochloride), which is very soluble in water.
According to Wikipedia:

The salt thiamine mononitrate, rather than thiamine hydrochloride, is used for food fortification, as the mononitrate is more stable, and does not absorb water from natural humidity (is non-hygroscopic), whereas thiamine hydrochloride is hygroscopic. When thiamine mononitrate dissolves in water, it releases nitrate (about 19% of its weight) and is thereafter absorbed as the thiamine cation.

The structures of two compounds are depicted in following diagram:

But how would nitrogen acquire this charge did a base deprotonate  the methylene group  attached to the nitrogen causing the nitrogen  atom to form a double bond and acquire a positive charge?

In the biosynthesis, the neutral thiazole portion has a lone pair on the N that is perpendicular to the ring (ie not involved in aromaticity). It makes a nucleophilic attack on an electrophilic carbon, knocking off a negatively charged group and acquiring a positive charge because it now has four bonds. See https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6039189/

It is not uncommon in biosystems with tetravalent nitrogen (with a positive charge), e.g., nicotinamide adenine dinucleotide (NAD+). Andrew has given an excellent answer for OP. Thanks Andrew.


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Electronegativity of elements on pauling scale
When we say the electronegativity of Nitrogen is 3.04 ( on pauling scale ) what is the hybridisation of Nitrogen? Similarly for other elements?

I don't know why this question has been down voted. I was once in a room with several chemists and physicists all pretending that electronegativity of C is one fixed value - basically because periodic tables gives a single value for each element :( Perhaps the downvote is due to the fact that OP could have digged a bit more. The Wikipedia article seems rather good and answers to the question https://en.m.wikipedia.org/wiki/Electronegativity

Firstly thanks for your guidance . I read the article and as much I understood is that the value of electronegativity for any generic element E is calculated by considering it's bond H-E as we know the formula for difference of electronegativity. This is ok for F and Cl as we get the gases H-F , H-Cl but what about C ? We can't get the molecule C-H ? Please clarify me.

Is it in there. There is also a specific paragraph on C and its differently hybridised states.


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How can the formula of WD-40 remain a trade secret?
Wikipedia says

WD-40's formula is a trade secret

How can it remain a secret today? Can't mass spectrometers, NMR and other equipment be used to determine the structure of the compounds that go into it?
To put the value of the secret into perspective, the WD-40 company (whose flagship product is its namesake) makes \$76M a year and is currently worth \$2.7B

"How can it remain secret?". Trivial answer is - it can't (it isn't).
WD40 is a Fast Moving Consumer Good (FMCG). It is also a reasonably cheap product - as an industrial chemist this suggests at least two things to me:

It doesn't contain anything rare or expensive
It isn't difficult to manufacture

As an ex-petroleum chemical analyst, with access to a reasonably complete lab, checks to do :

Get the Safety Data Sheet (below is composite information from a number of WD-40 Penetrant SDS)
cas        %
64742-47-8      50-70    isoparaffins petroleum hydrotreated HFP
64742-56-9      30-35    paraffinic distillate, light, solvent-dewaxed
61789-86-4      <5       calcium petroleum sulfonate
111-76-2        1-3      ethylene glycol monobutyl ether
Not Available    1       fragrance
Check the SDS info using physical/chemical testing
If you have the gear available :
Density
Distillation
FTIR
GC (GC-MS if you have it)
AAS/ICP - metals (ICP multielement would be best)
CHOSNP analysis for hetero atoms
Rebuild the product from data. Check to see if it performs the same.
Improve on the product (believe me - WD-40 is NOT the be all and end all of penetrating oils - technology has improved since its formulation).

No one is asking for the precations, read the question carefully

What are "precations"? The questioner asked why the product's formulation couldn't be determined by spectroscopic/chemical means. Trivial answer is - it can (it has been). The process is trivial if you have the right equipment and background.I included a schema to follow to derive the formulation - it completes the answer.

A fundamental problem in analytical chemistry is that we never analyze a complete unknown mixture simply because of time and money constraints. Finding each and every component of a complex unknown mixture is quite impossible. NMR is good for pure compounds only, mass spectrometer is "blind" to several analytes because they do not ionize very well. It is unable to distinguish the masses of isomers. You may ask what can't we make synthetic gasoline? The reason is that it literally contains thousand and thousands of compounds and people are even discovering more by using million dollar Fourier transform ion cyclotron resonance mass spectrometers. So back to the question, is it worth analyzing a completely unknown mixture?

Having said that, WD-40 is rather far from being completely unknown.

No idea if it's worth it (It's big business AFAIK). How much do you think it would cost to analyze it?

Several thousand dollars at the minimum. See here, someone attempted gas chromatography https://www.wired.com/2009/04/st-whatsinside-6/

WD-40's annual revenue is $381M and income is $76M (they have other products too though)

Hmm, NMR can handle mixtures even if it does have a somewhat high detection limit and all those overlapping alkanes might prove challenging (if not impossible) to disentangle. Not push-button technology exactly. You can also use hyphenated techniques (GC-NMR) in parallel with GC-MS which would be a step up from the quick analysis in the linked wired article.

@BuckThorn, I am sure coupling GC with NMR must be a very expensive set up. It is amazing that it is possible.  The caveat is that one must use a preparative GC.

@M.Farooq You can run it in-line with special microvolume probes. NMR probe coil sensitivity increases with smaller size which makes up for the decreased mass of sample. The technique is pretty mature but I don't know about expense, certainly the NMR instrument will be the most expensive part. See eg https://www.bruker.com/products/mr/nmr/probes/probes/flow-nmr-probes.html

We can make synthetic gasoline - in fact pretty much all gasoline from a modern refinery is "synthetic". The molecules expressed in the final product don't start in the original crude - alkylation, isomer rebuilding, cracking, distillation all modify the product profile - we do it all the time. An d that doesn't even mention things like the "gasohol" conversion process (ExxonMobil). Balancing the final product is the job of a fuels chemist / technologist - correct IBP/FBP, Octane #, evaporation profile, benzine content, RON/MON, etc. We literally do this all the time.

Bruce Heath, My point is that gasoline whether synthetic or natural is very hard to analyze. No one can do a total and complete analysis (= identifying each molecule and isomer therein) of such complex mixtures without spending years or dedicated service.

Sorry, I disagree - gasoline is basically just a series of hydrocarbons, occasionally with some oxygenates or additives. GC-MS makes both qualitative and quantitative evaluation of the mixture not only possible but relatively trivial. Sure, it's a $1MM dollar piece of equipment, but it's also a relatively common piece of equipment in any large commercial analytical laboratory. Couple that to a reasonable background in petroleum chemical manufacturing and you'll get the "formulation" of gasoline +/- a few %. Nobody is going to "build" gasoline from discrete components - but analysis is trivial,

I'm reminded of a formulation I was asked to produce about 40 years ago by someone who had taken a sample of an anti-corrosive spray to a University for evaluation. The chemists at the uni had GC/MSd the product and done ICP to determine metals present - the customer ended up with a 45 page list of "ingredients" and asked me to reproduce the product - I looked at it and the analysis and said - "Bitumen  + wax+ barium sulphonate + solvent". "But what about all these other bits" - :"That's the bitumen - that's the wax - that's the solvent" - "pure" chemists just don't "get" real world products


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Can Silicon be used as anode for Li-Ion batteries
Can Si be used as anode. I assume you could store much more energy in Si since the atom is much larger than C, but then Si seems not to like to bond with itself. Can there be a solution?

Yes, it can and was initially used as Li-Si alloy (also look for the modern alloy "SiliconX"), later as a thin film (approx. 500 nm), nanowires, or being incapsulated in graphene. Pros, cons and the history are summarized in Wikipedia.

Hmm, interesting, never heard of it.

Generally, heavier atoms means lower specific stored charge per mass.
Specific energy per mass depends of the cell redox systems, but also on the above.
Silicium, in contrary to carbon, does not form hexagonal planar structure with delocalized electrons like graphite, but prefers the diamond-like structure.
The Li-ion cell anode is formed by lithium-graphite intercalate, for which does not exist Silicium analogue. If it had existed and the cell had been possible to create, the major difference would have been being heavier for the same mAh capacity.
Silicon anodes (thanks @andselisk  for the link) are possible, but not as analogue of the graphene structure  of graphite. The huge problem with them is material swelling in the charge state, limiting the anode life to few cycles. Few percents of silicium together with carbon can be used to increase somewhat the graphite anode capacity.

Pure silicon is not conducting like metals. It is an electrical insulator. But it becomes a semi-conductor if it is doped. And it is doped if a tiny amount (less than 1%) of some impurities is present in the crystal of silicon. So pure silicon cannot be used as an electrode, if it is pure. But it can with doped silicium.


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What happens to air at 1000 degrees Celsius
I was looking at the graphs for the density of air at different temperatures (link below) and I was wondering if air would not react with itself at high temperatures? I read that oxygen and nitrogen react at 1600 degrees Celsius, but even below that average temperature (because of the Maxwell Boltzmann distribution) I’m guessing the reaction takes place? Also would it just catch fire with contact with anything in the environment like carbon / almost anything? I’m wondering about the hypothetical use in hot air balloons with very high temperatures and low volumes.
https://www.engineeringtoolbox.com/air-density-specific-weight-d_600.html

Radiation loses increase proportionally to the fourth power of the temperature (Boltzmann law), so your baloons would be extremely inefficient

Are you aware 1000 Deg C hot surface thermally radiates about 150 kW/m2 ? Even if surface itself was 500 Deg C, it would be 20 kW/m2. If we consider the mass 500 kg, and lift about 1 kg/m3, it would have radius 4.9 m and surface 305 m2. That is 6.1 MW for 500 deg C surface. It would need impossibly huge power source and the useful load would be baked in a huge grill.

For just 200 Deg C surface, it would be cca 2.8 kW/m2 and 866 kW.(orvsone 10-15% less after subtracting background radiation) Still very mighty power. Some light aircraft would be more energy efficient.

At $\pu{1000^oC}$, oxygen and nitrogen will not react substantially or probably even noticeably if at all. Besides that, whatever organic material you are using for a hot air balloon will most definitely ignite immediately. The average temperature that most organic materials char and eventually ignite is $\pu{270^oC}$. At almost double the kinetic energy at $\pu{1000^oC}$, most things will almost instantaneously ignite. Note that in air, temperature diffuses rather quickly so even if the fire or whatever you are using to get to $\pu{1000^oC}$ will probably be significantly colder when it reaches anything flammable. In fact, most fuel used in normal hot air balloons burn at over $\pu{1000^oC}$, but it diffuses very quickly that the average temperature of the air in the balloon is around $\pu{110^oC}$

Perhaps to make clearer use of terminology it is better to say "Note that hot air diffuses rather quickly so even...."?


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A general definition of bond order in molecular orbital theory?
I stumbled upon this problem upon learning organic chemistry after learning the foundations of MO theory in physical chemistry. It seems that VBT and MOT can coexist with each other in many scenarios. One problem that especially confuses me about their compatibility is that how can we define a general notion of bond order in MOT. This question arises naturally as we assign integer values to bonds between atoms in VBT but this is definitely not true in MOT.
It's often taught in general chemistry course that bond order is 1/2 times number of bonding orbitals minus antibonding orbitals, but this definition only works for diatomics. I wonder if there's some more general definition that works for each adjacent(perhaps this is ill defined) pair of atoms, perhaps in the spirit of Mulliken population analysis that assigned partial atomic charges based on MO coefficients of AOs.

One example for pi bonds: Coulson defined pi bond order within Huckel MO theory as $\Sigma_i n_ic_{ri}c_{si}$. See Levine, Quantum Chemistry 4th edition p. 567

Wiberg bond indices and Mayer bond order may be starting points.


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Why Frenkel defect does not impact the density of the solid?
The following text is from Chemistry Part I - Textbook for Class XII (NCERT), chapter "The Solid State", page 23, topic "1.9 Imperfections in Solids", sub topic "1.9.1 Types of Point Defects - (iii) Frenkel Defect":

This defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location.
Frenkel defect is also called dislocation defect. It does not change the density of the solid.

(Emphasis mine)
Earlier I've learnt that a vacancy defect results in decrease in the density of the substance, and an interstitial defect increases the density of the substance. I could understand, due to the presence of vacancy or interstitial defects, the density varies locally or in other words density of the substance is not uniform throughout.
According to the quoted text, Frenkel defect creates a vacancy defect at one location and an interstitial defect at a different location (not at the same location), so I think the density will again have local variations and so the density changes. But it is given Frenkel defect doesn't alter the density. What is the reason for this fact? Why does the same logic fail here?
The Wikipedia article on "Frenkel defect" gives the following statement:

Frenkel defect does not have any impact on the density of the solid as it involves only the migration of the ions within the crystal, thus preserving both the volume as well as mass.[citation needed]
Source : Frenkel Defect - Effect on density

If density is plainly defined as the ratio of mass and volume it doesn't seem to change, but in reality I think there will be variations within the crystal having this defect. But why is this fact not specified in many sources?

The sources may not explicitly say this, but I suspect that they consider mass density to be a macroscopic property, to be measured from at least milligrams and cubic millimetres.
If the volume considered is small enough, then yes, one would see local variations of the density. However, if you make the volume even smaller, you would see huge variations in the density as additional nuclei are included. (Similar statements will be true for Frenkel defect concentrations.) On this level, the definition of mass density simply breaks down and is no longer very useful.

Yes, there will be local variations in density as the density would decrease where there is a vacancy defect (as volume is the same and mass becomes less) and it would increase at the point where there is an interstitial defect (as volume is the same and mass becomes more). But since there is no loss of ions in the compound, the overall density of the compound remains the same (as the ions just move to an interstitial site).


text:
How does acetone remove pencil marks?
In another question it was noted that acetone is a useful solvent for removing pencil marks from wood surfaces.
(Source: https://woodworking.stackexchange.com/a/1870/763)
I'm used to acetone being used as a solvent for things like spray foam, nail polish, etc. but was surprised it could 'dissolve' pencil lead. I take it pencil lead is a mixture of graphite & clay substances. What enables the acetone to remove pencil marks?
Perhaps another way to think about this is, why does pencil lead adhere to surfaces like wood or paper in the first place? This is discussed in How are pencil marks "adhered" to the paper? but without mention of using solvents to remove it. How does the acetone reverse or interfere with the adhesion?

Pencil lead writes on rough surfaces by abrading into tiny particles and getting pushed into surface irregularities. If the surface isn't rough, you would get no visible residue of graphite. A pencil might sometimes even leave a scratch on dense wood that is so smooth that it has almost no blackness from graphite.
Wood is hydrophilic and even contains some water (unless bone dry). Graphite is not so hydrophilic or polar. Acetone wets the wood surface and either swells it by diffusing into watery regions or simply gets between a graphite particle and the wood surface, loosening any van Der Waals forces holding the graphite to the wood. While the surface is wet, the graphite can be wiped off (or smeared out so much that the original mark is not visible any more). Then the acetone evaporates quickly and the wood looks clean(er). Ethanol or isopropanol might do as well - even water (especially with a surfactant to coat and disperse the graphite) might work.
A grease pencil would be much more difficult to remove because so much material is pushed into crevices that a solvent can't easily loosen it without spreading it.

I'm vaguely recalling that acetone dissolves some of the chemicals in wood.


text:
How does one tell (or conclude) if a substance is carcinogenic?
It's common to read statements like:

Tetrachloroethene ($\ce{CCl2=CCl2}$) was earlier used as a solvent for  dry cleaning, but since it contaminated ground water and is a suspected carcinogen, it has been replaced by ... .
Various hydrocarbons harms plant tissues, shed leaves, flowers and twigs. Many of them are also carcinogenic.

How do we tell or suspect one compound to be carcinogenic? And are all carcinogens organic compounds?

Look up "Ames test"

It can be told by extensive, expensive and time consuming studies, and even than, some conclusions may not be decisive.

Please see the applied tags as somewhat of a better place-holder than [tag:everyday-chemistry]. As far as I know there is no tag (yet) that deals with carcinogenicity.

How do we tell or suspect one compound to be carcinogenic?
As written in the comments to the question, this the result of large studies on the human population, correlating blood or urine levels of some substance to instances of cancer. Alternatively or concurrently, studies on animal models, such as mice, rats or even in-vitro studies may be carried out. As a result, regulatory bodies may include a substance in lists, which may preclude them from use in consumer products, manufacturing procedures etc.
Are all carcinogens organic compounds?
No, this does not have to be the case: plain arsenic and basically all inorganic cadmium compounds have been classified as suspected or definitively carcinogenic.

Now arsenic has been mentioned and interestingly, inorganic arsenic compounds interfere with the basic mechanism of DNA repair, which apparently increases frequency of chromosomal aberrations. So this is chemically predictable.
Cadmium has also been mentioned, but what is little known is "even tumor formation [52] can be suppressed by the simultaneous Zn treatment" cited in this work. So chemically predictably treated.
For organic compounds, I suggest a simple process to determine the POTENTIAL carcinogenic ability of an organic compound (I believe this is what you are, in effect, asking in being able to 'tell').
In the lab or reported in the research literature, is it apparent that the compound readily undergoes decomposition creating (directly or otherwise) radicals or groups in an environment of aqueous saltwater containing transition metals, CO2, oxygen and naturally produced HOCl or other reactive oxygen species (ROS) at pH levels found in the human body?
Some generated radicals, like the hydroxyl radical and the carbonate radical, for example, can attack DNA as it would any large organic compound to extract a hydrogen atom.
Damaged DNA can lead to cancer cells.
While, for example, short-lived hydroxyl radicals are good in breaking down problematic pesticides in natural waters, they are not recommended in large amounts in the human body.
Note, I am not recommending the direct testing of the decomposition of DNA,  but just the expected (and conveniently measured) side products created with the associated breakdown of the organic compound in question. This may be suggestive of potential DNA damage, and at best, a predictive tool. Varying dosing levels may also be quite informative. This is categorically distinct from expensive short-term animal studies (examining tumors) and also investigations of long term low dose exposure and reported incidents of cancer in human populations with inferences drawn from correlation analysis.
If a compound displays a chemical potential to be carcinogenic, per above, you may still want to await the opinion of say Germany, England, or other countries with socialized medicine,.., to verify. But, in my opinion, I would just avoid/limit exposure and consume over the counter anti-oxidants (Vitamin C,...), especially if chemically related suspect compounds are now publicly believed to be carcinogenic.
Thanks Andrew for the reference to the Ames test! It, however, apparently employs "several strains of bacteria (Salmonella, E.coli) that carry mutation" which could limit the accessibility of the test.
DISCLOSURE: I have no financial interest, or am I employed, in the health supplement industry or the health care industry.

While stressing cells with radical species is a mechanism for generating cancers, it is not necessary (and possible also not sufficient). Consider DNA intercalation compounds like several polycyclic aromatic hydrocarbons, or other molecules with planar aromatic moieties - no chemical reactions are involved. Another mutagenic source is thymine photodimeriaztion by UV light, independent of UV formation of radicals.

I do agree on the UV exposure issue. I read about certain drugs where sunlight exposure is to be avoided. However, in the lab, UV is employed in photolysis assisted radical pathways, so not exactly always a direct (or indirect?) non-radical path to DNA damage.

Alkylation is another common non-radical mmechanism

My use of the term 'radical' in 'creating (directly or otherwise) radicals" can be inferred to apply to more generically to encompass anions and such. Also, per Wiki on 'Alkylation' transfer of an alkyl group is via an alkyl carbocation, a free radical, a carbanion or a carbene.  I view both comments as supportive of my assertion that one can apply research/knowledge and NOT await a massive human trial to 'tell' if a compound is potentially problematic.

This answer assumes that cancer can be attributed to DNA damage, which is the standard explanation but by no means exhaustive, and that DNA is damaged by free radicals, which is just the explanation du jour for DNA damage but I know several others (ionizing radiation, any poison that inhibits DNA repair, anything that inhibits the apoptosis mechanisms). The answer is also alluding that cancer can be predicted, which isn't even remotely true.

@toolforger All of those examples you listed ultimately lead to DNA damage except things that "inhibit apoptosis mechanism" which doesn't necessarily mean cancer because a cell can be free of apoptosis but not proliferate due to lack of growth factors. And the reason why cancer can't be predicted is because most DNA damaging agents are random in nature via the bases they damage. The only class of agents that you probably can't predict as carcinogenic are chemicals that aren't reactive but get converted to reactive intermediates in the liver. Like why benzene is carcinogenic but not toluene.

@Cell Actually you can't predict much. You can make informed guesses, but you still need to test. (If we knew, we wouldn't need things like studies.)

@toolforger That was kind of my point. Anything that leads to increased DNA damage in vivo is carcinogenic. But not everything directly reacts to cause DNA damage. Like benzene if I recall correctly is carcinogenic because a breakdown intermediate in the liver is reactive with DNA. So an in vitro experiment with benzene and pure DNA in solution wouldn't tell you anything or with asbestos etc. The studies are necessary to uncover indirect carcinogens or weakly reactively ones.

Compounds that are radiation/UV sensitive induce radical creation. One apparently must  believe that one can have 'suspected' carcinogenic agents and not 'potential' based on tests. Really?