text:
Who came up with the box and arrow diagrams for electron configurations?
In general chemistry texts, the electron configurations are sometimes shown in boxes with up and down arrows to show the concept of paired spins. My impression for years was that Hund used such diagrams. However, flipping through his book Linienspektren und Periodisches System der Elemente which used the s,p, d, f notation in a modern way, there is no such use of box diagrams with up and down arrows. Does anyone know who introduced this so-called box notation? Thanks.

This image is from Atomic Spectra and Atomic Structure by Gerhard Herzberg, John William Tranter Spinks - 1944. It does not really have boxes (just boxes of a table), but spin +1/2 and spin -1/2 are shown as arrows. Of course, I am not claiming this is the first usage, just that in 1944 it existed.

Thanks Karsten, I think then it must be somewhere between 1944 (before Herzberg's) and 1925 (after Pauli). I will also check Sommerfeld's book. It is available on Internet Archives.

As an aside, the German wikipedia uses parallel and anti-parallel spin (with respect to the z-axis) instead of spin up and spin down. If that was common, you might not use an upward arrow for spin +1/2. On the other hand, if you say "spin up" or consider the direction of angular momentum, an upward arrow would be a natural notation. Also, notice that Herzberg uses arrows while nowadays harpoons are more common.

Could you share the German wikipedia link? I am curious to see that version. What is traditionally taught in German, up and down or the German wiki style? Harpoon is rather new. I would be looking for arrows.

I don't teach in German, and I forget the nomenclature we used when I took the course. Here is the Wikipedia article section: https://de.wikipedia.org/wiki/Spin#Spinoperator_und_Basiszust%C3%A4nde_f%C3%BCr_Spin_%C2%BD

Somebody is downvoting this answer without any reason. Please don't spread your internal and anonymous negative energy like a poltergeist on good scientific discussions.

@M.Farooq - Downvotes are allowed, even without stating a reason. Maybe P.A.M. Dirac is downvoting because he came up with the notation, never published it and now feels bad that he is not mentioned in this answer.

The problem with anonymous forums like these is that there is no check. It could be a high school dropout or Dirac's spirit or a future Einstein downvoting your answer and my question too. You never know. That is why downvote should have a reason box for a healthy discussion. ResearchGate deleted the downvote option altogether.


text:
Why is a alkyne anion more stable than a benzene anion?

I would think the benzene anion (A) is more stable b/c the compound aromatic. Yes I understand that sp hybridized carbons are more stable than sp2 for negative charges, however I thought that aromaticity is a more important factor in determining stability.

Why do you think aromaticity has anything to do with an anion that's not delocalized into the ring?

S-orbitals are lower in energy than p-orbitals (Aufbau Principle). Hybrid orbitals follow similar behavior. An sp orbital is 50% s; an sp2 is 33% s; and sp3 is 25% s. The more s-character, the more stable the anion and the more acidic the conjugate acid

Aromaticity enters the reckoning only if the excess electron associated with the negative charge enters the conjugated pi electron system.
In your case that does not happen.  The negative charge is on a formally nonbinding pair on the carbon like other carbanions, but in a conjugated ring when you do not have a ligand on that carbon the electron pair takes the place of the ligand and isn't in the pi system.  So not stabilized by aromaticity.  Here you don't have the ligand, the top carbon in A where they put the negative charge has no hydrogen atom.  If a hydrogen atom were there the molecule would be neutral benzene.
You can have a negative charge stabilized by aromaticity in two ways represented by the benzyl anion and the cyclopentadienyl anion.
Benzyl anion:  This has the formula $\ce{C6H5CH2^-}$, a phenyl ring attached to a methylene group.  The methylene group acts as the necessary ligand that's missing in your $\ce{C6H5^-}$, so the excess electron pair enters the pi system and conjugates with the ring.  Note that the $4n+2$ rule for aromaticity does not work because with that extra pi-bonded carbon you don't have all the conjugated electrons in a ring anymore.  But you do have a highly de-localized structure that inherits the stability of the ring.
Cyclopentadienyl anion:  this ion, $\ce{C5H5^-}$, has the negative charge and all the ligands on the ring atoms, so the extra electron pair and negative charge are already in the pi system without having to conjugate with anything else like the case above.  You have a pure aromatic ring with the negative charge built in.  Pyrrole, furan, and thiophene are a variation on this theme where the ring is negative but neutralized by a positive formal charge on the nitrogen, oxygen or sulfur atom.

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text:
Is activation energy directly related to the reaction rate? (keeping the temperature, concentration, catalyst, particle size and pressure constant)
For different chemical reactions, does the activation energy of that reaction directly relate to the reaction rate? In other words, would a reaction always be slower if it had a higher activation energy than another reaction?
Of course, one of the reactants will be kept constant and the other reactant will differ for each reaction.
If pressure, catalyst, temperature, concentration, particle size, and other factors affecting reaction rate is kept constant, would the activation energy for each reaction be inversely proportional to the reaction rate?

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text:
What chemistry lab experiements can I do based on the bicarbonate buffer system?
I want to do a chemistry experiment, based on the reactions present in the bicarbonate buffer system. It needs to be heavily chemistry based, rather than biology based-- any suggestions  preferably based on a factor other than pH for the independent variable (pH can be dependent variable)? Especially preferable if they don’t involve the use of a pH meter, and can be clearly tied to chemical principles. Thanks!


text:
Is Br sp³-hybridised in HBr?
$\ce{HBr}$ formation can be explained by simple orbital overlap as well as by $\mathrm{sp^3}$ hybridisation of $\ce{Br}.$ Can't we use dipole moment to find if the lone pairs on $\ce{Br}$ are mutually perpendicular or $\mathrm{sp^3}$-hybridised?

Hybridization is a convenient rule, but more modern molecular orbital theory does not use it.  For HF see here, for example.

I don't understand how the dipole moment will help at all. The molecule has $C_{\infty}$ symmetry. There is only one direction the dipole moment, if it exists, can point.

If you add the three lone pairs when they would be in sp³-hybrid orbitals, it would be just the same as if they were in s and p orbitals. Hybridisation is just a mathematical trick. If we are employing any of this theory (and including molecular orbital theory), then the orbitals are always chosen orthogonal to each other. A note on language: Atoms cannot be hybridised, only orbitals can. Therefore it is incorrect (or at least very sloppy) to speak of sp³ hybridisation of bromine.

From a theoretical viewpoint we would expect the Br to be sp3 hybridized, because in the HBr molecule the Br atom has 4 pairs of electrons around it (the 3 lone pairs plus the bonding pair), and symmetry (plus the repulsion of electrons for each other) tells us to expect that 4 pairs of electrons arrange themselves in a tetrahedron, and as you know we explain a tetrahedral arrangement of electron pairs around an atom by arguing the s and p atomic orbitals form sp3 hybrid orbitals.
It's the same valence shell electron pair repulsion (VSEPR) logic that leads us  to expect the 4 bonding pairs around the C in CH4 should be arranged tetrahedrally, and, therefore, the C atom is sp3 hybridized.  (We would also expect that a more careful and precise determination would give us some deviation from this prediction for HBr, since unlike CH4 the 4 electron pairs around the Br in HBr are not equivalent.)
But you are interested in the question of whether there could be some experimental verification of this expectation.  I don't think the experimental dipole moment would work, because it's too crude a picture of the electron distribution, it just tells us whether the electrons spend extra time near the H or near the Br, and we already know it will be the Br because of its much stronger effective nuclear charge for valence electrons.  What you want to know is something like whether the valence electrons spend significantly extra time on the "far" side of the Br, away from the H-Br bond, as opposed to the "near" side, so we can conclude they're in sp3 hybrid orbitals that point away from the H-Br bond.  Offhand I can't think of an experiment that would unambiguously test this.
You could certainly calculate the electron density in the HBr molecule, with a good quantum chemistry package, and for a molecule this small the results should be quite reliable.  That would tell you whether it makes more sense to consider the Br lone pairs as in sp3 hybrid orbitals or atomic p orbitals.


text:
Is prefixes such as di-, tri-, etc. considered for alphabetical order in IUPAC nomenclature?
The correct IUPAC name of a compound was: 4-(1,1-dimethylethyl)-5-ethyloctane.
my question is why not 4-ethyl-5-(1,1-dimethylethyl)octane?
Are prefixes such as di-, tri-, etc. considered for alphabetical order?
and what about prefixes such as cyclo, spiro? Are they too taken into consideration for alphabetical order?

Related: Nomenclature of organic compounds containing complex side chains

Also note that the preferred IUPAC name actually is 4-tert-butyl-5-ethyloctane.

I understood sir, thanks for the help

We don't consider di,tri ,etc Just ethyl or methyl ,etc are to be considered


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Why is molar mass of a substance and atomic mass number usually the same?
I recognize what atomic mass represent and what molar mass represents by why is it that a substances molar mass and it’s atomic mass are usually the same number
Ex: chlorine has an atomic mass of 35.453 u and a molar mass of 35.453 g/mol

I like that tinge of incertitude ("usually"). Indeed, the skill of not making overly broad generalizations is rather important in a scientist. Now to the point. Long story short, this is because the units (u and g/mol) are not facts of nature; they were invented by humans, and chosen so as to be convenient.

The term mole is a short form of the German word Molekül which was coined by Ostwald in 1890s. At that time, the "atomic mass unit" did not exist. The common term at that time was gram atomic mass or gram-molecule. The original meaning of so-called was defined to be equal one gram molecule i.e, the mass of a molecule expressed in grams. Atomic mass unit came into use by the 1950s.
These values match by definition not by coincidence. Here is how Ostwald coined the concept of a "mole". In those times molecules weights were determined by colligative properties. This is a snapshot of Findlay "Principles of Inorganic Chemistry" (translation of Ostwald's work)

What is the unit of gram atomic mass? Gram or Gram/mol? I think it is gram, but I asked many people and they say it is gram/mol, but gram atomic was even used before mole (I guess).

In older times, it must be just grams as the name indicates.


text:
Are all lone pairs considered as π-electrons?

According to me, in methylenepyran 8 π-electrons are present as there are 3 double bonds. Therefore, 2 × 3 = 6 π-electrons and 2 π-electrons as one of the two lone pairs of oxygen will participate in resonance.
But the correct answer is 10 π-electrons. How? Are both lone pairs of oxygen are being considered? If yes, then why?

Your reasoning is fine, and 10 is nonsense.

I thought the same.

Can you quote the question statement along with options, or link the source, saying the other way.

I am quoting from http://www.hutters-online.de/publikationen/dmchdo.html.

One requirement that is basic to this concept and is agreed upon by most authors is that there have to be two double bonds that are in conjugation with a third one, but not in a linear arrangement. The p electronic system thus forms a bifurcation. So the simplest cross-conjugated hydrocarbon would be 3-methylene-1,4-pentadiene (I) that can be viewed as an ethene disubstituted geminally with two vinyl groups:

Based on the above the following resonance structures are possible. These have 8 $\pi$ electrons in conjugation. You are correct.


text:
How many ether metamers are possible for C4H10O?

How many ether metamers are possible for $\ce{C4H10O}?$

According to me, the answer should be 2 but it was given 3. I know 3 structures can be made:
$$\ce{CH3—O—CH2—CH2—CH3}\tag{1}$$
$$\ce{CH3—CH2—O—CH2—CH3}\tag{2}$$
$$\ce{CH3—O—CH(CH3)2}\tag{3}$$
Structure 2 and 3 are metamers, but 1st and 3rd structures are positional isomers(?), then how the answer can be 3?

Funny that my first answer was down voted because I right gave ethers as example of metaners apparently out of black. Lately I see more people comfortable with that. By the way I do not see much, if any, use of this particular class of isomerism which is covered by the other classifications.

Metamer, really? Im glad to be able to say that none of my teachers or professors has ever bothered to introduce this new metacategory.

Can anyone point me to a IUPAC definition of the term "metamer"? Google was unsuccessful for me, en:wp has nothing but a one -liner saying the term is obsolete, nothing on de:wp.

Metamerism definition (3rd bullet) includes position isomers!

"Metamerism" is an extremely old, obsolete, and ill-defined property. It is also not educationally or synthetically useful. An additional, and pointless, way to categorise chemical compounds.
See this answer to the question "What is metamerism": https://chemistry.stackexchange.com/a/104576/17952
If you have to answer this in a test, do your best with whatever definition your teacher has provided. And tell him that he should stop teaching such nonsense.

Whilst the 1st and 3rd are positional isomers, I think you can also class them as metamers as well - n-propyl and i-propyl will have some significant differences. For instance n-propyl alcohol can be easily oxidized all the way to a carboxylic acid, whereas i-propyl alcohol can only easily be oxidized to the ketone. If the alkyl chain was a pentyl, for instance I think you can class the 2-pentyl and 3-pentyl radicals as positional isomers and a different metamer to 1-pentyl.

Structural isomerism is categorised into 5 main types; which are listed from highest priority to lowest as:

Tautomerism

Functional isomerism

Metamerism- Metamerism is a type of structural isomerism where the carbon-side chain/ alkyl group attached to a polyvalent functional group differs( here ether is the polyvalent functional group).

Chain isomerism

Positional isomerism - a type of structural isomerism where the position of the functional group differs.

According to this, we check for metamerism before positional isomerism.
Here, since there are different alkyl groups on the right side of the oxygen atom in ether*( propyl group in structure (1), isopropyl group in structure (3))* we consider them to be metamers.
Only if the 2 compounds were not metamers, we would look for positional isomerism.
This youtube playlist is a useful way to understand structural isomerism https://www.youtube.com/playlist?list=PL_A4M5IAkMac-tIcsX-GmJB6Jfv5wjVTp
(the 7th and 8th video are specifically on metamerism)


text:
Does the activation energy depend upon the stoichiometric coefficient?

For the decomposition of $\ce{N2O5(g)}$ it is given that
$$
\begin{align}
\ce{2 N2O5(g) &→ 4 NO2(g) + O2(g)} &\quad\text{activation energy} &= E_\mathrm{a}\\
\ce{N2O5(g) &→ 2 NO2(g) + 1/2 O2(g)} &\quad\text{activation energy} &= E'_\mathrm{a}
\end{align}
$$
then
(1) $E_\mathrm{a} = 2E'_\mathrm{a}$
  (2) $E_\mathrm{a} > E'_\mathrm{a}$
  (3) $E_\mathrm{a} < E'_\mathrm{a}$
  (4) $E_\mathrm{a} = E'_\mathrm{a}$

In the definition of activation energy, it doesn't say if that is the minimum energy required for 1 mole of reactant or otherwise. Can anyone help me understand this?

It will become obvious as soon as you look at the reaction rate formula.

The reaction 1 and 2 are the same reaction... So, the activation energy of both must be equal.

@IvanNeretin In the reaction rate formula, the rate constant changes with different stoichiometric coefficients. So, going by the arrhenius equation, if rate constant changes, the activation energy should change too. But that's wrong according to the answer given. Can you explain further?

https://chemistry.stackexchange.com/q/38167/81005 Here is an answer I took as reference to my previous comment.

Don't look at the rate constant per se. (Of course it will change, but then again, it's not the activation energy, is it?) Look at its temperature dependence.

@IvanNeretin I'm sorry, I don't understand what you're trying to say. If we keep the temperature constant, and the stoichiometric coefficients change, then the arrhenius constant is the same, the rate constant changes and so does the activation energy. I'm sure I'm going wrong somewhere, but I can't figure it out.

You are totally right. Now what if we don't keep the temperature constant?

@IvanNeretin If we don't keep the temperature constant, the activation energy is constant but the rate constants are different. But, the question I was trying to solve only changes the stoichiometric coefficients of the reactions, so why are we considering the temperature to not be constant?

By considering the temperature to not be constant, we may deduce the character of the rate's dependence (or independence) on the stoichiometric coefficients.

Activation energy does not depend on number of moles of reactant(s) as it is defined for per unit mole of reactant. Thus the answer should be (4)

No, the activation energy does not depend of the stoicheiometric coefficients. The activation energy is determined by the energy of the electrons and nuclei as the reaction proceeds to its products.
The rate constant is an experimentally measured quantity and knowing the species involved we invent a possible mechanism, as shown by the two reactions you give. Another reaction could be $\mathrm{N_2O_5\to N_2O_3+O_2}$, you can invent more. The molecule will only react by one mechanism at a given temperature and we generally use our intuition to guess what happens (i.e. we write down a scheme by guestimation) but always many experiments have to be done to work out exactly what happens. In other words we finish up by writing down the reaction scheme after experiments show what happens, not start with it. Once you have a scheme you should use the rule $
\displaystyle rate =\frac{1}{a}\frac{dA}{dt}=\frac{1}{b}\frac{dB}{dt}\cdots$ so that we know how stoicheiometic factors $a, b$ etc. are treated.

Rate measurements, while measuring the overall reaction, are concerned only with the rate determining step. This is usually an elementary reaction in the overall mechanism so the overall stoichiometry has no obvious effect and until one looks real closely the orders work out to be small whole numbers. That is until low pressure gas reactions, enzyme reactions, polymerization reactions, photochemical reactions are studied. Then elaborate rate laws show up.


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Why carbon monoxide reacts with hydrogen to give different products in the presence of different catalysts
My books says how carbon monoxide will react with hydrogen to give different products with different catalysts, but how exactly do different catalysts yield different products. Do they provide different pathways for entirely different mechanisms, or is the formation of some products is better thermodynamically and for that reaction to occur some catalysts provide the necessary activation energy and the others cant.

All true what you say, it all depends on the temperature, (partial) pressures, activation energies, mechanistic pathways, catalyst morphology, ...

So its difficult to pinpoint what exactly is different specifically?

The thermodynamics are always the same. If you go from carbon monoxide to methane, there is only one final enthalpy, and if you have the choice of going to methane or methanol the thermodynamically preferred product is always the same.
The difference indeed less within the different pathways and the activation barriers needed to get from one intermediate to the next. This isn’t really thermodynamics, it is mainly kinetics (although you can argue about thermodynamically preferred intermediates). The pathways are often very complex and entire papers revolve around how one reaction is catalysed on such a solid state catalyst, so please forgive me for not looking them up; but the general point still stands: one catalyst will lead to intermediates where the $\ce{C-O}$ bond is easily broken (and thus give methane), the other will not (and thus give methanol).

Wow, two answers in a raw. It's great to see you back, Jan:)

@andselisk Only briefly visiting on my tablet ;)


text:
Mixing ink with alcohol, MEG and water
First of all, I'm not a chemist and my knowledge in chemistry is not that extensive. However, I would like ask the following question: are there any chemical reasons that would explain that mixing a blue ink with alcohol and MEG, lets say 3% as part of a volume of 1 L, gives a ways stronger blue than mixing in a big tank 3% of the same ink on a volume of 30000 L? Being in that last mix a greenish colour the result.

How do you measure the intensity of the blue color, exactly? Given the only variable that is different is the volume of solution, this is unlikely a chemical question and may have to do with the perception. There is a small chance that higher volume suggests greater surface area and quicker oxidation by oxygen from the air, but that's to be answered by an organic chemist once you provide additional info what the chemical composition of the blue ink was.

the intensity of the colour is measured by the sight of eye, the difference is clearly spotted. We know that sun light may dim the colour of the mixture but we rule that one out because the mixture happen in a closed tank, oxidation inside of the tank? I'll propose that as a possible explanation, the mixture though remains in the tank for only about 40 món.again I rinsed very well the mixing tank, check for leaks in pipes, also the flash point between tank and manual mixture and nothing..so I thought that there is some chemical reaction going on that escape to my knowledge.

If I find out what's going on I'll post it here. Thanks for your answer thought, for taking some time to try to help us. Highly appreciated


text:
“T” in subscript in formulas
Does a “T” in a formula subscript indicate all possible numbers of atoms? $\ce{FeO_T}$ is the specific one I came across (in the "bulk chemistry" table here: https://sciences.ucf.edu/class/simulant_lunarmare), it seems to make sense in context but I've not done chemistry since school (tried googling already) so thought I'd better check.

Never saw this one and wouldn't believe it even if I did. Little "x" would do, though.

http://www.geokem.com/glossary.html. See the 10th entry.

@IvanNeretin extremely common in geochemistry. For a visible example, see the Earth Wikipedia page, in the chemical composition table.

Interesting question.
It looks like this is a rather common notation in geology and refers to the total iron(II) and iron(III) oxide (e.g. ferrous and ferric) content.
For a chemist this is probably somewhat confusing as “total” is normally denoted via subscripted abbreviation “tot”.
Flipping through numerous sources, it appears there is no standardized notation and alongside with $\ce{FeO_T},$ $\ce{FeO_t},$ $\ce{FeOt}$ and $\ce{FeO^\mathrm{t}}$ are being used.
An introductory textbook by Shikazono [1, p. 21] explicitly defines $\ce{FeO^\mathrm{t}}$:

$\ce{FeO^\mathrm{t}}$: Total iron $(\ce{FeO} + \ce{Fe2O3})$

Reference

Shikazono, N. Introduction to Earth and Planetary System Science: New View of Earth, Planets and Humans; Springer: Tokyo; New York, 2012. ISBN 978-4-431-54058-8.

I will add to the correct answer by andselisk that the reason for this notation is that in most geochemical analytical methods, oxygen is not measured.
Instead, all metals are measured and oxygen is calculated by stoichiometry assuming (often very reasonably) that all metals are as oxides.
As the vast majority of major elements in geological materials have only one oxidation state throughout all geologically reasonable conditions, this works out very well.
Iron is an exception: it is a major element that commonly exists both as divalent  and trivalent. Because the oxidation state is much harder to analyse than the iron contents, iron is commonly reported as either all 2+ as FeO(t) or all 3+ as Fe2O3(t).

This seems similar to, for example, how a fertiliser can have its phosphorus content measured in "percent weight as $\ce{P2O5}$" or "percent weight as $\ce{PO_4^{3-}}$. I've always found it confusing and opaque; why perform the completely unnecessary assumption that the element is found as a particular substance, especially in cases which clearly do not reflect reality, such as $\ce{P2O5}$? Such easily avoidable problems...

@NicolauSakerNeto that's mostly a matter of opinion and convention. To me, as a geochemist, using $\ce{PO4^3-}$ is misleading, because it adds oxygens that do not "belong" to the phosphorus. It is also not a neutral thermodynamic component, which personally makes me cringe. On the other hand, $\ce{P2O5}$, or $\ce{PO_{2.5}}$, is perfectly fine as it tells you exactly how much phosphorus (and associated oxygen) there is in the compound, without any assumptions.


text:
Which has the highest enol content?

Which of the following compound has highest enol content?

The answer is an option (c) and I think the reason behind it is that there are two phenyl groups attached to it, hence there will be a good conjugation. Also, since phenyl is an electron-withdrawing group, it will provide stability to the structure (as the $\ce{C-C}$ double bond formed after tautomerism will have π-electrons). But I am not quite sure about my logic and I don't know the reasons to eliminate other options.

The extent of enol content can be explained on the basis of acidic hydrogen, intramolecular hydrogen bonding.

The stability of the carbanion formed is dependent on its stability after it is formed. In compound $\ce{(c)}$, the carbanion is stabilized by extended conjugation due to presence of phenyl group. Such stabilization does not occur in $\ce{(d)}$. In $\ce{(d)}$, methyl substituent at carbanion increases negative charge via inductive polarization. The resulting carbanion is unstable.

After enolization , enol is stabilized via intramolecular hydrogen bonding, which is a common feature in all examples provided.
The enol in compound $\ce{(c)}$ is stabilized via extended conjugation.

Taking the above factors into account, $\ce{(c)}$ has highest enol content.

Well, to comment on the enol content we have to check the stability of the carbanion formed during the mechanism process and also look at factors like hydrogen bonding, extended conjugation, aromaticty formation.
So, here we can comment the answer is c because the intermediate carbanion is stabilized by resonance with the carbonyl and phenyl.
Note: the intermediated carbanion will be formed at the site of the most acidic H. Here it's connecting $\ce{CH2}$ between the two carbonyls.


text:
Is it possible to have an allotrope of carbon as a long unbranched chain of itself?
Is it possible to have an allotrope of carbon with formula $\ce{(=C=)_n}$?
Well, let's actually leave alone the separate question like from which "monomer" would one produce such a thing (can one actually consider it being a polymer? Or a covalent crystal?), because it seems like the most suitable candidate would be dicarbon.
All in all, might it be (under some conditions) energetically favorable for carbon to form a long unbranched chain of itself?

It's been all over the news recently: http://www.ox.ac.uk/news/science-blog/making-and-imaging-ring-18-carbon-atoms-first-time although it's $\ce{C#C-C#C-C}$ rather than $\ce{C=C=C=C=C}$.

Thanks for the link, it was a pleasure to read! However, these guys made cyclocarbon, if I got it correctly. My question though is more about a linear form, reminiscent of a polymer. On the other hand, they tell "After the successful creation of the linear carbon segments ..." - which, maybe, means, that they can make out of them not a ring, but just keep going with linear chain?..

I seem to remember sth. about polyacetlyene, which gets first brominated and then dehydrohalogenated. Whether the end product is a cumulene or a polyyne is imo just a question of what end group(s) you have. Cant find a reference now, though.

it depends what you mean by "to have". If you mean that the structure is remotely stable then the answer is NO; if you mean that it is possible to create something, then the answer is still only maybe (and the structure of the bonding might not be what you think).

Well, this is a rather obvious development (in retrospect, that is). We had an $sp^3$ carbon with tetrahedral bonds, and it made diamond. We had an $sp^2$ carbon with trigonal bonds, and it made graphite. What if we just had a linear $sp$ carbon?
Such thoughts have been around for more than quite a while. Theorists swarmed around that non-existent tree. No wonder that people started trying to actually obtain that linear carbon.
Not knowing anything about you, I still can claim with a good deal of confidence that the first reports of its synthesis are quite probably somewhat older than you (or me, for that matter). Then they were proved... not quite reliable. Some groups agreed with that, some disagreed, a great deal of confusion abounded, then more reports came and things became hairy. So they stayed for the past few decades and will stay for a while, I guess.
Now to the Q&A.

Does this compound exist?

· Define "exist", otherwise no meaningful answer would be possible.

Is it stable?

· Define "stable", otherwise no meaningful answer would be possible.

Is it thermodynamically stable at any P and T?

· Most researchers agree that it isn't.

Is it electrically conductive along the strands?

· Most researchers agree that it is.

Is it made of double bonds or of alternating single and triple bonds?

· The difference is less than you seem to think, to the point that it might not be a question worth pursuing.

What is its crystal system?

· Nobody knows. Chances are there are polymorphs with different crystal systems.

Is it super strong?

· Nobody knows, as with any non-existent compound.

Should it be called a polymer?

· About as much so as graphite and diamond, which means: No.

Great answer, but I have to disagree on your last point: A linear macromolecule of undefined lenght and a repeating structure, thats usually a polymer.

I disagree, but I understand that I might be in a minority on that one.

Whats your reasoning for the "non–polymer" verdict? Oh, and you avoided the initial question: "Is this a carbon allotrope?" I also wouldnt want to judge on that. My inclination is no, however. ;)

Of course this thing would be a carbon allotrope (conditional on its existence, that is). As to the polymer question, see here: https://chemistry.stackexchange.com/questions/38238/is-graphene-a-polymer/38253#38253

When someone says "we found a new carbon allotrope", Im inclined to retort "no, you just devised a synthetic route for a new molecule which happens to be carbononly". By common definition, that is of course an allotrope. Just feels weird for me. ;)

@Karl Just when did that last happen?


text:
Calculate the potential of each electrode, $\ce{Mn(OH)2}/\ce{Mn}$ ox/red couple
So the exercise goes like this:
We consider the chemical battery(i have no idea what is its actual english name since i'm translating), $\ce{Mn_{(s)}/Mn(OH)_2//Cu^{2+}/Cu_{(s)}}$ Where $\ce{Mn}$ is submerged in a saturated solution of $\ce{Mn(OH)_{2}}$ of $\ce{pH}=9.86$
first is to calculate the constant of solubility $\ce{kS}$ and $S$
this i think i can do, since we have the concentration of $\ce{OH^-}$, and we have the relation $\ce{2S=[OH^-]}$ then we can get the $S$ then the $\ce{kS}$ which is equal to $\ce{kS=[OH^-]^2*[Mn^{2+}]}$ with $\ce{Mn^{2+}}=S$
Then comes the question that I couldn't do, it says to calculate the potential of each electrode, for the electrode with the Copper element i can do, but for the electrode with the $\ce{Mn(OH)2}/\ce{Mn}$ couple I'm confused.
We have the demi-reaction of Oxidation(Since the standard potential of the Copper couple is higher than the Manganese one):
$\ce{Mn +2OH^-\leftrightarrow Mn(OH)_2+2e^-}$
Now for the electrode potential:
$\ce{E=E^0+\frac{0.06}{2}log(\frac{1}{[OH^-]^2})}$
This is the one I'm not sure of at all.
Another question is to deduce the potential of the oxred couple $\ce{Mn^{2+}/Mn}$ from the previous data we have.
I think this one I have to equal the potentials of the couple $\ce{Mn(OH)_2/Mn}$ and $\ce{Mn^{2+}/Mn}$ since they coexist at the same electrode(I'm not sure of this either and I need confirmation)
Note: The potential standard of $\ce{Mn(OH)_2/Mn}$ is a given.
This isn't a homework, just me trying to understand the oxred batteries better and getting stuck.
Thanks for your help!

@OscarLanzi That is my bad hahaha


text:
Diborane reduction
I am stuck at a step in reduction of alkene using $\ce{B2H6}.$ I assume that step 1 is the addition of borane to alkene. After that a water molecule attacks $\ce{-BH2},$ it being a Lewis acid. Subsequently, the boron will depart and oxygen will remove a proton which gets attached to carbanion formed when boron exited.
This is what I think is happening, the sole reason to ask it here is because I am unable to find a mechanism online or from a book. Mostly, I stumble upon hydroboration which is of no use to me. Please correct my mechanism or guide me to a reliable source.

When you don't find good results, it often means that the search keywords are not appropriate. Use Google Scholar in such cases. I typed "mechanism of alkene reduction by diborane" in Google Scholar, and the second result is a >20 paged review. I don't have an answer but you can certainly start from there. Anything which is in a textbook is often 20-30 years old research. Here is the link to the review: https://pubs.acs.org/doi/abs/10.1021/cr60304a005

What are you talking about? You want to make triple bond diborane? With what base? This is at least weird...

@Mithoron I was talking about reduction by diborane

Aditya, I am not an organic chemist so I don't know the answer. My point was that the answer must be out there if you use the right keywords. Search this phrase "mechanism of hydroboration of alkenes". I see several papers right away some of them can be downloaded. It is definitely complex because I see lot of papers on this topic. Are you in a university of college?

I'm still in school , last year

For the time being I would say accept the reaction as a fact as shown in your textbook. Those research papers may be too advanced or complicated at this moment. All I can feel is that the mechanism is complex and perhaps undecided as yet. You can see nobody posted an answer.

You can find a simplest mechanism and more in here: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/11%3A_Alkenes_and_Alkynes_II_-_Oxidation_and_Reduction_Reactions._Acidity_of_Alkynes/11.6%3A_Addition_of_Boron_Hydrides_to_Alkenes._Organoboranes

Addition of boron hydrides to alkenes or alkyne is a valuable reaction in organic chemistry. The reaction is called hydroboration that gives organoboron compounds, which are a valuable group of intermediates for other organic synthesis (Chem.libretexts.org). The simplest borane,  $\ce{BH3}$, exists as the dimer, $\ce{B2H6}$, or in complexed form with certain ethers (e.g., $\ce{H3B}$-THF) or sulfides (e.g., $\ce{H3B-S(CH3)2}$). To my understanding, your expectation is to find a mechanism for hydroboration of an alkene by borane followed by its oxidation to corresponding alcohol. I found a suitable mechanism for that online and depicted in following diagram. First step is common hydroboration.

No I was looking for a mechanism that converts alkene to alkane , instead of oxidation as the second step if an acid is used instead in aqueous medium then the mechanism follows the process I described  in my question, fortunately I could find it in the link given by you itself , so thank you.


text:
Why are B-O-H and B-O-C angles in B(OH)3 and B(OCH3)3, respectively are higher than normal?
In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 77 under the topic "Back Bonding" it is given:

The $\mathrm{2p}\pi$-$\mathrm{2p}\pi$ back bonding in $\ce{B(OH)3}$ and $\ce{B(OCH3)3}$ explains why  $\ce{B-O-H}$ and $\ce{B-O-C}$ angles are $112°$ and $113°$, respectively.

Based on the above statement I arrived at the following:
Due to back bonding in $\ce{B(OH)3}$ and $\ce{B(OCH3)3}$ there is a partial double bond character in the bond between boron and oxygen in both the compounds. The oxygen in both compounds is $\mathrm{sp}^3$ hybridised and ideal bond angle is $109°28'$. But due to the partial double bond character and as we know the repulsion between a double bond and single bond is greater than between two single bonds the bond angles $\ce{B-O-H}$ and $\ce{B-O-C}$ are larger than the ideal $109°28'$.
Kindly tell whether my understanding is correct or not. If not, kindly give an explanation for the observed bond angles and their deviation from ideality.


text:
What is the biggest known difference between rₑ and r₀?
What is the biggest known difference between $r_e$ and $r_0$? Where $r_0$ is the average bond length of the lowest vibrational state, and $r_e$ is the location where the potential is lowest.
For a perfect harmonic oscillator, these are the same (notice that the peak probability is exactly at the minimum of the potential):

What is the largest known difference between $r_e$ and $r_0$ in a real molecule?
This is important because the CRC Handbook on Chemistry and Physics almost always gives $r_0$ because that's what can be determined more easily from experiment, but ab initio geometry optimization almost always gives $r_e$, yet many publications do not even say which type of bond length they are reporting.

I think what you are asking is, in effect, 'what is the most anharmonic bond?' as this will give the greatest ratio of lengths.

@porphyrin: I thought of asking that too, but as a separate question because I don't think it is guaranteed to have the same answer. Some bonds have a double-well potential, whose solution to the Schrödinger equation (eigenfunction/wavefunction) is not so simple anymore. Likewise in polyatomics the picture is much more complicated than in this diagram.

The largest anharmonic constant $\omega_e\chi_e$ in the CRC book for a diatomic is 121.34 cm$^{-1}$  (for H2).

Keeping things simple, if the average bond length ($r_0$) for $n=0$ is calculated for a Morse potential then the largest deviation from $r_e$  seems to be  proportional to the ratio  $ D_e / \omega_e$.

@porphyrin: Where do you get that from? So you are saying the anharmonicity $\omega_e \chi_e$ does not enter at all? To maximize $D_e/\omega_e$ we just have to find the heaviest molecule (meaning small $\omega_e$) that is bound strongly (large $D_e$).

The reason $x_e\omega_e$ is big for hydrogen is because the frequency is big so this product is misleading. A low frequency and a big dissociation energy seems to mean a big anharmonicity in the bond. For example PbBr, or SiTe. The question was about calculating the difference between $x_e$ and $x_0$ the latter I took to be the average bond length $\int \psi^_0\psi_0rdr/\int\psi^_0\psi_0$

@porphyrin: Actually you're wrong and you're right at the same time. You were right originally by mentioning the ratio $D_e/\omega_e$ and wrong about $\omega_ex_e$ not being related. See Eq. 3.16 !! http://scienide2.uwaterloo.ca/~rleroy/c209/Primer.pdf.

@porphyrin: also please see footnode 6 on the same page!

Good question +1. @porphyrin I think that it is not the same that anharmonicity. I would separate the contributions due the potential as $\mu(V)$ (with $\mu=r(V)-r_e$) into the sum of symmetric and assymetric contributions. The expected value of the symmetric part will be 0, so only the assymetric part contributes to $r_0-r_e$

Let me start by saying that this is a very difficult question to answer because it's not very hard to come up with pathological cases which I will address at the end. Also, as you point out, it is somewhat non-trivial to calculate $r_0$ from theory and also somewhat non-trivial to determine $r_e$ from experiment. That being said, I think I have a good first suggestion that I am guessing will be hard to beat. I will just answer with theory because this data is more readily available.
$\ce{F^-(H2O)}$:
As has been pointed out, this question is very closely related to the anharmonicity of the vibration in question. The $\ce{O-H}$ stretch of $\ce{F^-(H2O)}$ is the most anharmonic known vibration for a water molecule. This has an experimentally observed vibrational frequency shift of about $2000~\mathrm{cm^{-1}}$ which I'm almost sure is the largest known vibrational frequency shift in a molecular complex, where the molecule in question doesn't dissociate.
Now, that being said, this will vary based on the method used, but at the MP2/AVTZ level, $r_e=1.0653$ angstroms while $r_0=1.1248$ angstroms resulting in a difference of $0.0595$ angstroms.
This $r_0$ value is calculated from VPT2, which is probably the easiest way to calculate this quantity, but this can fail for certain pathological cases like those I'll discuss below.
$\ce{H5O2^+Ar}$:
Ok, now I know that 0.06 angstroms doesn't seem like that huge of a difference, but generally both of these numbers will increase together, so it's hard for them to be all that different. Now, another way you could consider finding a very large difference would be to look for a double-well system where the zero-point energy of the vibration in question is larger than the barrier height.
Now, the zundel cation is a proton shared symmetrically between two water molecules. This is a single well system, however, so $r_e$ and $r_0$ will be the same and they have a calculated $\ce{O-H}$ distance of $1.202$ angstroms at the CCSD(T)/aug-cc-pVDZ level of theory.
If, however, we disturb this system ever so slightly, we will break the symmetry and might get the desired behavior. This can be done by tagging the complex with a rare gas atom, which is often done in experiments as it reduces the noise in spectra for reasons I won't get into.
$\ce{H5O2^+Ar}$ has a hydronium like complex with an elongated $\ce{O-H}$ bond length of $1.108$ ansgtroms. It is very likely that the barrier to proton transfer in this case is very small because there is no barrier in the absence of an Argon atom. Therefore, it is likely that the proton will be close to symmetric after vibrational averaging, so if it were perfectly symmetric, we would get an $r_0$ of $1.202$ angstroms again. It is likely less than this, but it is possible this would lead to a difference greater than $0.06$ angstroms which we got from $\ce{F^-(H2O)}$.
I should point out that calculating $r_0$ for this system is pretty difficult because using VPT2 will almost certainly give nonsensical answers because the harmonic reference is a terrible one, so the perturbation theory will break down. The best option is probably to do diffusion monte carlo and extract the average from the wavefunction directly. It's possible this has been done, but a cursory search of the literature didn't turn anything up.
Also, the data for the first part comes from the SI of ref.[1] and the data for the second part comes from ref.[2].

References:
[1]: Boyer, M. A., Marsalek, O., Heindel, J. P., Markland, T. E., McCoy, A. B., & Xantheas, S. S. (2019). Beyond Badger’s Rule: The Origins and Generality of the Structure–Spectra Relationship of Aqueous Hydrogen Bonds. The journal of physical chemistry letters, 10(5), 918-924.
[2]: Park, M., Shin, I., Singh, N. J., & Kim, K. S. (2007). Eigen and Zundel forms of small protonated water clusters: structures and infrared spectra. The Journal of Physical Chemistry A, 111(42), 10692-10702.


text:
Is coupling reaction performed in acidic or basic medium?
Why is coupling reaction between aryldiazonium chloride and aniline carried out in a mild basic medium and not under acidic medium?
According to me, it can be performed in both mediums.


text:
Why is trichloromethane more acidic than trifluoromethane?
In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 78 under the topic "Back Bonding" it is given:

$\ce{CHCl3}$ is more acidic than $\ce{CHF3}$ which is explained as follows:
$$
\begin{align}
\ce{CHCl3 &<=> H+ + ^-CCl3}\\
\ce{CHF3 &<=> H+ + ^-CF3}
\end{align}
$$
The lone pair on the $\ce{C}$ atom gets delocalised through 2pπ–3dπ bonding in $\ce{^-CCl3}$ which is not possible in the case of $\ce{^-CF3}.$

My doubt is, fluorine being the most electronegative element exerts a greater negative inductive effect (compared to that of chlorine) which stabilises the conjugate base formed. So I feel trifluoromethane must be more acidic than trichloromethane. But the reverse order is observed.
Please explain why is this so. Should we want to prefer the conclusion from back bonding over inductive effect?

Related: How to prove that chloroform is more acidic than fluoroform?

@Loong, Thanks for sharing the link. The question is similar to that of mine but the accepted answer's author is not confident about the answer. But his reasoning is understandable. I thought of starting a bounty for that question but that is marked duplicate addressing a completely different question. Anyway thanks for spending time in helping me. Hoping for a better answer.

Also related: Which molecule is more acidic: CHF3 or CHBr3? I think that is more believable than some ideas of d-orbital involvement (which has fallen out of favour over the last few decades). But I also wonder what has been written in the primary literature.

@orthocresol, Thank you for the link. Initially, I thought that was a different question but it cleared my doubts.

Since OP mentioned the question answered theirs, I have no regrets dupehammering.

@Jan, That's fine. I was thinking of deleting the question, but there are some useful comments for this question which I would wish to keep.

We know that fluorine ($\ce{F}$) is more electronegative than chlorine ($\ce{Cl}$). Therefore, -I effect (Inductive Effect) of $\ce{F}$ would be more and trifluoromethane ($\ce{CHF3}$) would be more acidic.
But this is not the case. As -M (Mesomeric Effect) effect also plays an important role in determining the acidity. We also know that -M is considered over -I (Inductive Effect).
Now, If we compare the conjugate base of both the given molecules, $\ce{Cl3C-}$ and $\ce{F3C-}$, Chlorine has vacant $\mathrm{3d}$ orbitals whereas $\ce{F}$ does not have any vacant $\mathrm{d}$ orbitals (in fact, it does not even have $\mathrm{d}$-orbitals). Therefore, $\ce{Cl}$ can exert -M effect due to availability of vacant $\mathrm{d}$ orbital but $\ce{F}$ cannot exert the -M effect. Generally, -M effect stabilises the negative charge and increases acidic strength. Therefore, trichloromethane ($\ce{CHCl3}$) is more acidic than trifluoromethane ($\ce{CHF3}$).


text:
Ionization energy of neon vs its cationic counterpart

Which requires more ionization energy: $\ce{Ne}$ or $\ce{Ne+}?$

It seems to me like it should be neon because of noble gas configuration, but the answer given is $\ce{Ne+}.$ Does this anything to do with size of the atom shrinking?
Is this because cationic counterparts always have more ionization energy than their neutral counterparts because it becomes successively more difficult to remove electrons from atoms on successive removals?

Does this anything to do with size of the atom shrinking?

Yes, precisely; this is one way of putting it. Another way is to say that because one electron has been removed, the remaining electrons are less shielded from the nuclear charge: therefore, the effective nuclear charge increases, and the remaining electrons are harder to remove.

Is this because cationic counterparts always have more ionization energy than their neutral counterparts because it becomes successively more difficult to remove electrons from atoms on successive removals?

Yes. This is true for every element in the Periodic Table, with no exceptions, and it doesn't matter what the electronic configuration is.
If all else were equal, then it would indeed be harder to remove an electron from a noble gas configuration than a non-noble gas configuration. However, that is a very big if! Between $\ce{Ne}$ and $\ce{Ne+}$, the electron configurations are indeed different, but all else is not equal: as discussed previously, the effective nuclear charge in $\ce{Ne+}$ is greater than that in $\ce{Ne}$.

If the second ionization potential were less than the first, a certain amount of (limited) ionization would doubly ionize half of the neons and leave the other half unionized. You could never observe a significant concentration of singly ionized neon; two singles would collide to give a neutral and a double.
Although neon is far more complex than a hydrogen, you could consider Ne+ and Ne++ to be positive attractors for an electron. The attraction of Ne++ would be about twice as great as from Ne+ (21.5 eV vs 41 eV, CRC Handbook).
What is interesting is that the next 6 ionizations require between 63.5 and 239.1 eV, but the last two require 1195.8 and 1362.2 eV. Big jump in energy required to get the last 2 s-electrons off!


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Theory behind Experiment: Extraction and Identification of DNA
A few days ago, in my chemistry lab class, I performed an experiment aimed at extracting and identifying DNA. The class is mostly practical oriented, so I thought of asking my theoretical doubts on ChemSE.
First of all, let me explain the experiment:
We used onion extract, sodium dodecyl sulfate (SDS) solution, 4% $\ce{NaCl}$ solution, papain extract, ethanol, TE buffer solution, and diphenylamine reagent. I shall ask my questions while explaining the procedure followed-

I took $\pu{10ml}$ of the onion extract in a boiling tube, and added $\pu{1.5 ml}$ of SDS solution to it, following which the mixture was allowed to stand in ice for nearly $\pu{10 minutes}$.
What does onion extract + SDS do? What's the chemical reaction if any? Why is low temperature (ice bath) important?
I added 5-6 drops of papain extract to the mixture, and stirred gently. A quick Google search says Papain, also known as papaya proteinase I, is a cysteine protease enzyme present in papaya. I wonder how it 'reacts' with the mixture prepared in step 1?
Next, I slowly poured nearly $\pu{25ml}$ of ice cold ethanol down the wall of the test-tube, so that ethanol formed a separate layer on top of the mixture. Note that this won't happen if ethanol is poured in rapidly, as that disturbs the mixture at the bottom of the tube. Why so, though? Difference in densities should make sure separate layers are formed, even if so after a long time? What's wrong here?
After allowing the tube to stand straight in a test-tube holder for ten minutes, I noticed some stringy white substance collect in the alcohol layer. We were told that that substance was DNA! How did this happen?
Using a hooked glass rod, I spooled out whatever DNA I could from the top of the test-tube, very carefully. This was used to carry out the following confirmation tests:

Diphenylamine Test:
Some crude DNA was collected in a test-tube, to which $\pu{2ml}$ of 4% sodium chloride solution was added. Next, $\pu{2ml}$ of diphenylamine reagent was added and the mixture was placed in a boiling water bath for an hour. The colour of the mixture changed to pale blue. How does this work? What are the reactions involved? Why the colour change?
UV Absorption:
DNA was dissolved in $\pu{2-3ml}$ of TE buffer solution (why?), and ratio of absorbances was determined at $\pu{260nm}$ and $\pu{280nm}$. Furthermore, our instructor told us that this ratio should be less than 2 (I guess?), for a confirmation. I wonder how that number was determined. (Also, some of my classmates got the ratio slightly greater than 2, due to some error. What error could they possibly have made?)
I'm trying to make sense of what I did practically, with the help of theoretical inputs, and would appreciate any help with the same. No doubt, the experiment was really fun to do!

1) Sodium dodecyl sulfate (SDS) is a solid anionic detergent. It can solubilize proteins and lipids that frame the cell membranes by degrading the cell (the proteins from the cell membrane get damaged and cell gets broken) and nuclear membranes that protect the DNA. This will help the cell membranes to separate and expose the chromosomes that contain the DNA. SDS can also release the DNA from histones (a group of basic proteins found in chromatin) and other DNA binding proteins by denaturing them. Breaking the nuclear membrane exposes the DNA to reactive and possibly dangerous chemicals in the cell. To slow down these chemical reactions, which might damage or break up the DNA, you must use cold ingredients and add cold alcohol. Also, it is also known fact that the colder the alcohol is, the more DNA it will extract.
2) Once the DNA has been released, adding papain extract helps untangle and unfold the DNA from the other parts of the cell. The enzymes in papain break down protein attached to (or tangled with) DNA molecules, making them free from proteins.
3) DNA is not soluble in absolute alcohol but soluble in water. If you pour alcohol rapidly it will mix with water and you may not have DNA precipitation. Therefore, you should slowly pour ethanol (or isopropanol) down the inside of the test tube with a Pasteur pipette, so alcohol layer build up on the extract.
4) DNA is soluble in water, but not in salty water. All the other cell contents are also soluble in water and salt water. This difference in solubility allows you to separate DNA from the rest of the cell material. When the DNA leaves the salty water mixture, it comes into contact with the alcohol but cannot dissolve (DNA is not soluble in alcohol either). When DNA comes in contact with alcohol, it uncoils and precipitates. All the other cell materials stay in the salty water solution. Note that an uncoiled piece of DNA can be $\pu{2.8 inch}$ ($\pu{7.2 cm}$) long, and visible when few of them tangled each other.
5) Diphenylamine will react with deoxyribose sugar in an appropreate condition to give blue-colored complex. This is te principle underlying estimation of DNA using diphenylamine. This is called Dische Diphenylamine Test For DNA:

One of the acid degraded products, $\omega$-hydroxylevulinylaldehyde, under acidic conditions, reacts with diphenylamine to produce a blue-colored complex that absorbs at $\pu{595 nm}$. The intensity of the blue color of this complex is proportional to the concentration of DNA.
About using UV-vis-Spectroscopy to determine the purity: The fact is that nucleic acids have an absorption maximum at $\pu{260 nm}$ UV wavelength. However, this absorbance suffers from interference from othe molecules such as single nucleotides, proteins, and single-stranded nucleic acids (as well as RNAs). Thus:
DNA concentration (in $\pu{μg}$ DNA/$\pu{mL}$ Solution) $= (A_{260} - A_{320}) \times 50\times$ dilution factor where $50$ is DNA extinction coefficient, $A_{260}$ and $A_{320}$ are absorbance of DNA at $\pu{260 nm}$ and $\pu{320 nm}$, respectively. In general, since the DNA has limited absorbance at $\pu{320 nm}$, this is simplified as:
DNA concentration (in $\pu{μg}$ DNA/$\pu{mL}$ Solution) $= A_{260}  \times 50\times$ dilution factor
Yet, the DNA purity is determined with the help of software. This is done by measuring the absorbance of the DNA against protein:
$$\text{DNA Purity} = \frac{(A_{260} - A_{320})}{(A_{280} - A_{320})}\approx \frac{A_{260}}{A_{280}}$$
If $\frac{A_{260}}{A_{280}}$ is in the range of $1.7-2.0$, the content of the solution is considered to be pure DNA. In other words,, constructing an absorbance ratio between these two absorbance wavelengths can provide an estimate of sample purity. As a general rule any preparations with a $\frac{A_{260}}{A_{280}}$ ratio greater than approximately $1.7$ is called pure.
Sources:
http://www.biotech.iastate.edu/publications/lab_protocols/DNA_Extraction_Onion.html
https://openlab.citytech.cuny.edu/bio-oer/chemistry/biologically-important-macromolecules/nucleic-acids/nucleic-acids-dna-extraction-and-disches-diphenylamine-test-activity/
https://www.news-medical.net/whitepaper/20140203/DNA-Analysis-with-UV-Visible-Microspectroscopy.aspx
https://www.bmglabtech.com/fileadmin/06_Support/Download_Documents/Application_Notes/AN263.pdf


text:
How Does Benzyl Alcohol Produce a Peak at m/z = 79?
I have understood the mechanism for converting ion 1 (benzyl alcohol) to ion 2 (m/z = 107) but not the mechanism for the conversion of ion 2 to ion 3 (m/z = 79).
Please elaborate the mechanism for the conversion of ion 2 -> ion 3.

This is similar to the phenol fragmentation. In mass spectrometry, benzyl alcohol fragments with two mechanisms, the one in your question and by a simple loss of the OH(91m/z). The 79m/z is the fragment with the most intense peak.
I have added below two possible mechanisms for step 2 to 3.
First option:

Second option:

EDIT: in the second option, the homolytic cleavage step is meant to be a summary of these steps:

The homolytic cleavage step in Option 2 is a doubt. Nevertheless, satisfied with the concerted step in the first option.

Option one feels more natural to me, but I added the second option after reading about a possible radical intermediate in the phenol fragmentation. That homolytic cleavage is in fact 2 steps: 1)hemolytic cleavage with one electron to the carbonyl carbon and one electron to the carbon bound to 2 H. 2)lone pair of the oxygen forms a triple bond, and at the same time the unpaired electron moves to the adjacent carbon with positive charge(+) to avoid a 5 valence state and because it is attracted to the positive site

I have added all the homolytic cleavage steps in the edit section at the bottom of my answer

Okay.. more understandable now

Interesting question.
The cycloheptatriene - norcaradiene tautomerization is a common equilibium.  Usually the norcaradiene is present in small amounts, but as long as some is present the reaction can rapidly proceed through this intermediate.
In the diagram below, the norcaradiene can undergo hydrogen shift with concomitant decarbonylation to produce the m/z = 79 peak in the mass spectrum.

Is the mechanism for this reaction the same as what blu potatos has shown in his answer?

@TanYongBoon I don't see anything in his mechanism(s) that looks like a one-step chelotropic extrusion of CO from a norcaradiene-like structure.


text:
Reducing agent in electrolysis
I would like to make colloidal silver to kill mold. I think $\pu{100ppm}$ (weight) would be nice. I read a paper on it and in it they used citric acid + distilled water as fluid, two silver plates as electrodes, $\pu{300v}$ ($\pu{50Hz}$ Ac to reduce oxide buildup on electrodes), $\ce{N2H2}$ as reducing agent and a surfactant to encapsulate the uncharged $\ce{Ag}$ particles to prevent them from clotting and building aggregates.
I would like to make something similar. So I have a couple of questions:
About the electrolysis: Is there a problem with making the distance between the electrodes larger? In the paper, there was a distance of $\pu{2cm}$ between the electrodes. I'm planing on having perhaps $\pu{20-30 cm}$. What would be the impact?
Suggestion on alternative reducing agent for $\ce{Ag+}$ ions: Preferably something that is nontoxic and does not promote mold. I have heard of tannic acid and ammonia. Could it work? In the best of world, I would like a reducing agent that evaporate from the solution. It needs to work in a electrolysis environment of $\pu{240-300V}$ AC without contaminating the electrodes. Is there a way to calculate which reducing agent would work?
Why do they use citric acid in the water if it is to allow conductivity in the beginning? Could I omit it? If all the $\ce{Ag}$ particles are reduced to be without charge then the water will not conduct? I think I need it. Correct me if I'm wrong.
About surfactants: Could I use ordinary soap? But I do not think I need it since I'm not after a specific size and I think the $\pu{100ppm}$ solution is not saturated and hence the particles do not grow that much. Is there a way of measuring the $\pu{ppm}$ of uncharged species in a non lab environment or the best way is trough weighing the plates before and after electrolysis?

#1: You mention a paper you read.  May you add the literature reference to your question?  While the citric acid will protonate portions of the hydrazine, the relative amount mentioned in the paper's experimental section allows to infer the extent of this step.  It will show the relative ratio of reducing agent vs. Ag used and set the procedure into context of other publications to achieve the task in mind, too.  #2: Maybe starting from $\ce{AgNO3}$, $\ce{NaBH4}$ and ascorbic acid is much more convenient (and less toxic) than working with hydrazine hydrate; e.g. DOI 10.1007/s13204-014-0371-9.

#1: The paper is "Synthesis of Ag nanoparticles using an electrolysis method and application to inkjet printing
Jin Min Cheona, Jin Ha Leea, Yongsul Songb, Jongryoul Kima,∗"

Thanks for the reply.
The paper you are referring to they produce 24ppm solution of AA-Ag. Do you think I could get a stable solution of 75-100ppm?
Normally without AA you can not get a stable solution at [Ag]colloidal = 1.72x10-4 M = 18.6 mg L-1 = 18,6ppm (DOI: 10.17628/ecb.2013.2.700-705).
Would 75ppm AA-Ag have the same anti-mold properties as 75ppm colloidal Ag?

I do not know the answer to this subsequent question. While I attribute the anti-mold properties to the colloïdal silver, the mere increase of the concentration of silver (ppm or mg/L) needn't be the complete story.   By the very nature of colloïds I think the ratio of mass over exposed / accessible surface area of these little particles may play a significant role, too where again the ascorbic acid (as an stabilizer of the particles) may interfere, too. So in short: either someone already published this (-> library), or it needs to be determined experimentally (-> laboratory).


text:
Why does the d orbital size decreases on addition of electrons?
In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), on page 80 under section 3.7 "The Extent of d-orbital Participation in Molecular Bonding" it is given:

A second factor affecting the size of d orbitals is the number of d orbitals occupied by electrons. If only one 3d orbital is occupied on an S atom, the average radial distance is $2.46$Å, but when two 3d orbitals are occupied the distance drops to $1.60$Å.

I wish to know the reason why the size of the d orbital drops when additional electrons are added. I think this is counter-intuitive since on addition of an extra electron the interelectronic repulsions increases and confining them to a small volume increases overall energy of the system thereby making it unstable. So, I think the orbital size must increase, but it is happening the opposite way.
Kindly explain the above-mentioned concept.

Jan's answer below is spot on but I think the author borrowed the same logic from why atomic radii decreases as you move across a period.

There is not much to explain because the concept is wrong. There is no sulphur compounds in which the 3d orbitals (remember: they have an energy level similar to 4s) are involved in bonding in any noteworthy way, shape or form.
From what I believe I may remember about the outdated concept, ‘one d orbital’ refers to cases like $\ce{SO3^2-}$ with sulphur in the $\mathrm{+IV}$ oxidation state while ‘two d orbitals’ would refer to e.g. $\ce{SO4^2-}$ or an oxidation state of $\mathrm{+VI}$. In the Lewis formalism, the answer is easily deduced if you draw octet-abiding structures: $\ce{SO3^2-}$ will include a formal single positive charge on sulphur while in $\ce{SO4^2-}$ sulphur has a formal double positive charge. The higher an oxidation state, the more contacted an atom becomes.
A more correct answer would elaborate on ($\mathrm{sp}^n$-type, formed only by sulphur’s s and p atomic orbitals) molecular orbitals and the increasing number of electronegative substituents that need to bond to the central sulphur, reducing its charge density and leading to the same effect.

Thank you for your answer. In the first paragraph, you have said the concept of d orbital participation in sulphur compounds is incorrect. I understood the reason you have mention in the same paragraph. In the last paragraph, you have said (or what I understood based on your answer and what was given in my book as the first reason for the participation of d orbitals in bonding) increase in electronegative substituents causes d orbitals to contract in size thereby causing a fall in energy similar to that of the orbitals participating in bonding.

If the concept is incorrect why must it have an explanation? Sorry to irritate you by this large comment. Kindly reply when you find time.

@Intellex You misunderstood me. The calculated molecular orbitals for these types of compounds are made up of over 99 % s and p contributions. With these $\mathrm{sp}^n$-type orbitals, you can do the electronegativity discussion and reduce their energies/contact them. While the outer d orbitals as well as any other highly virtual orbitals will also contact and be lowered in energy, this has no effect on the bonding situation as the difference between the energies remains almost the same. The charge density mentioned is sulphur’s overall charge density across all (including core) orbitals.


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

@StackUpPhysics Then I also do not understand your question.

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

Just FYI I am not the downvoter

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

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

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

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

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

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

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

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

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


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anionic surfactants as salt water induced rust inhibitors
There are a multitude of product that purport to protect boat engines from the adverse effects of salt water.  These anionic surfactants claim to rinse away salt buildup.  I do not understand how a surfactant would improve water's ability to dissolve NaCl.

A surfactant won't improve water's ability to dissolve NaCl. Its ability to protect your engine (which might or might not be true) stems from other sources.

Sounds suspiciously like a fake, but where do you apply these products? Wouldnt they be washed away instantly with the stream of cooling water?


text:
Can an alcohol oxidation to ketone happen with hydrogen peroxide?
As far as I know, usually $\ce{CrO3}$ and $\ce{KMnO4}$ are used for this scope, but can other oxidizers such as $\ce{H2O2}$ be used?

Not directly but by the intermediacy of catalytic oxidants e.g. rhenium compunds.  https://pubs.acs.org/doi/pdf/10.1021/ic9806784


text:
Bonding in Sulphur dioxide
In Concise Inorganic Chemistry by J.D.Lee (Adapted by Sudarsan Guha, Fourth Edition), on page 82, under section 3.8 - Types of Covalent Bonds (Sigma and Pi Bonds) it is given (for Sulphur dioxide molecule):

The $\pi$ bonds do not alter the shape, but merely shorten the bond lengths. The bond angle is reduced from ideal $120°$ to $119°30'$ because of the repulsion by the lone pair of electrons. Problems arise when we examine exactly which atomic orbitals are involved in $\pi$ overlap. If the $\sigma$ bonding occurs in the $xy$ plane then $\pi$ overlap can occur between the $3p_z$ orbital on sulphur and the $2p_z$ orbital on one oxygen atom to give one $\pi$ bond. The second $\pi$ bond involves a d orbital. Though the $3d_{z^2}$ orbital on sulphur is in the correct orientation for $\pi$ overlap with the $2p_z$ orbital on the other oxygen atom the symmetry of the $3d_{z^2}$ orbital is wrong (both lobes have a + sign); whilst for a p orbital one lobe is + and the other -. Thus overlap of these orbital does not result in bonding. The $3d_{xy}$ orbital in sulphur is in the correct orientation and has the correct symmetry to overlap with the $2p_z$ orbital on the second oxygen atom, and could give the second $\pi$ bond.

I am unable to understand the part which has been typed in boldface.
We know $d_{xy}$ orbital lies in the $xy$ plane. We also know that $p_z$ orbital lies along the z-axis. Then how can these two orbital participate in $\pi$ bonding?
The following image is my attempt to show bonding in sulphur dioxide. The oxygen atom under consideration is the second one marked with a $\bigstar$.
In the following image the lone pair present in the s orbital is not shown for the sake of simplicity. There will be some distortion in the bonds and sulphur dioxide would be bent and will not be linear. But here it is shown to be linear.

Clearly, for the second oxygen atom, the overlapping of atomic orbitals is not possible due to out of phase approach. I feel the formation of $\pi$ bond between the sulphur atom and the second oxygen atom must take place between $2p_x$ orbital of the second oxygen atom and $3d_{xy}$ orbital of the sulphur atom. Kindly tell whether I am correct or not.
Kindly clarify my doubt.

This won't help solving your problem, but it is important: Sulfur does not use d orbitals for bonding. The statement in the book is old, it is no longer considered correct.


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Why are bridge bonds weaker than terminal bonds and, H-B-H terminal bond angles larger than H-B-H bridging bond angles in diborane?
In Concise Inorganic Chemistry by J.D.Lee (Adapted by Sudarsan Guha, Fourth Edition), on page 83, under the topic "Bridge Bonding" it is given for diborane $(\ce{B2H6}):$

The $d_\ce{B-H}(\text{terminal bonds}) < d_\ce{B-H}(\text{bridge bonds})$
The energy required to replace hydrogen atoms from the bridged position is more than that needed for the terminal position. This is supported by the fact that on reaction with methyl chloride, terminal hydrogen atoms are replaced by methyl group in preference to the bridged hydrogen atoms.
The reason for this could be the fact that to break the bridge bond, the overlapping zone should be broken from both sides and the overall energy required is more as compared to do so for terminal bonds.

Structure of Diborane:

Image Credit: Wikipedia
Based on the second paragraph it can be concluded that terminal bonds are weaker than bridge bonds. But some sources on the internet including Wikipedia state the bridge bonds are weaker than the terminal bonds without any satisfactory reason. I wish to know the reason for this.
Further, kindly explain why the terminal H-B-H bond angle is larger than bridging H-B-H bond angle as evident from the molecular structure as given above. I have searched my book as well as google to my best, but I didn't get any reason for this.
Kindly note, I read the unaccepted answer to this question. But still my doubt persists and makes me type this question by adding relevant details from my book and internet.
Kindly clarify this doubt.

Well, thanks for linking, I'd need to look for it otherwise. In future do not ask about things that are already asked and answered and perhaps get a better book...

Even though I agree that this question and sub-questions have been more or less covered previously, this shouldn't discourage you from asking new questions. From what I noticed, so far your questions were nicely formulated and well referenced (especially for a new user), and if you don't manage to immediately find the answer by searching the Chemistry.SE (BTW see this Meta post), it's better to ask rather than keep staying in the dark.

@Mithoron, I read the answer and the comments following it. One of my doubts is clarified. I have highlighted the other doubt and added some details about a related problem. Now, I feel this question is different from the one mentioned as a possible duplicate. To be honest, the book I am currently reading is considered to be the best Inorganic Chemistry book for JEE Advanced. But I came to know (after asking questions on Chem.SE) many concepts are outdated. But we are supposed to learn these!

@Mithoron, The answer to that question is good, but I feel the statement in my book is equally good proving bridge bonds are stronger contrary to what most reliable sources say. I need to know why this kind of reasoning is incorrect. That's why I posted it.

@andselisk, Thank you for your comment :). I read that Meta post. First of all, I thought of editing that post and starting a bounty for that. I felt that would be compressing another question into the existing question, and hence started off with a new one.

I have read this post. But it didn't clear my doubt.

It depends on what you define as a bond.  A three-center two electron bond is in general stronger than an ordinary two-center bond because delocalizing the bond over three atoms instead of two makes the bonding MO more stable.  But, the bridge bond is also shared between two linkages so each individual linkage has less bonding than a single two-center bond linkage.
The question of bond angles is a bit more complex because it hides a couple other issues.  Superficially, the bond angles in the bridge region are forcibly reduced by forming a four-membered ring.  But that begs the following questions:

Why is the dimer formed instead of a trimer or tetramer, which would enable a larger ring and more natural bond anges?

Why is the bond angle through boron actually greater than the bond angle through hydrogen?  Since hydrogen forms smaller atoms steric factors would favor the reverse.

An answer to both questions may lie in the tendency of 3c-2e bonds to form closed structures.  The bonding orbital described above gets an extra bonding linkage by bringing the end atoms of a chain together.  Such a tendency is seen in the formation of a triangular trihydrogen cation in interstellar space, and in the intermediate stage of a carbocation rearrangement.
In the case of diborane this tendency shows up as boron-boron bonding. The dimer has a favorable geometry for forming such a boron-boron bond, whereas an oligomer with larger bond angles in the center would tend to hold the boron atoms too far apart.
The nature of this boron-boron bonding is revealed in molecular-orbital calculations reported by Prof. Bruce Tattershall at Newcastle Universitywinnoplus=2&outputype=HTML5.0_JSmol&basis=MO&orbital=B(p_x)_H_bridge*&plane=xz&orbno=10).  The diagram taken from this source is shown below to reveal six occupied valence MOs and defines the $x$ and $z$ axes in the molecule for later reference.

Below are the quantitative data behind this diagram for the occupied molecular orbitals.  Entries marked with a blue dot indicate boron-boron bonding, while those with a gold dot are boron-boron antibonding.

The boron $s$ and $p_x$ interactions along the sigma-bonding axes show a mixture of bonding and antibonding overlaps; but the $p_z$ interaction, using orbitals that overlap the hydrogen atoms, shows a single bonding interaction with large boron components. This indicates that the closure of the 3c-2e bonds in the dimer occurs through the $p_z$ lobes rather than along the actual boron-boron axis, forming a "$\pi$ bond over a gap" analogous to what an organic chemist might find in a homoaromatic molecule.
Note also that the 3c-2e bond closure occurs around a hydrogen atom. A larger group in the bridging position could leave the boron atoms too far apart to overlap well and thus inhibit formation of the b9ron-boron bond in a dimer. This would explain the greater tendency of trialkylboranes to exist as monomers compared with the parent (di)borane.

Thank you for your answer. I have made some corrections to the question. I learnt from other sources the bridge bond - a 3c-2e bond is intact weaker than ordinary bonds. I agree with that fact since even Wikipedia says the bridge bonds are weaker. But I feel the statement (2nd paragraph in the quoted text) is also equally true. I am confused. Kindly help me clarify my doubts. Sorry for irritating you.

Slight correction.  The 3c-2e bond is weaker in each of its individual linkages, but adding up all the linkages gives a stronger bond overall.  Otherwise the 3c-2e bond would just break up to make an ordinary bond.


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What solvents can solve aromatic urethane resins?
Do you have any information about chemical resistance of hexa-functional aromatic urethane acrylate?
I want to know what solvents can solve aromatic urethane resins.

You are talking about EBECRYL 220, right? Hexafunctional means that this is going to form a rather dense network. Its likely not going to be soluble. But why dont you read the application notes from the manufacturer?  --> homework

@Karl thank you for your answer .I want to remove UV INKS from plastic.these inks have been formed from some resins such as aromatic urethane .I can remove some color that have no aromatic urethane but I dont have what should do to clean aromatic urethane .


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Interference between 2 electrons
If 2 electrons undergo destructive interference (as they show wave nature)  will they disappear or will they have no wave nature left?

Charge can't disappear; neither can mass. All in all, the wave nature of electrons should not be taken that literally.

So what should happen after interference

There is no interference.

So will they follow laws of motion then..

Everything follows the laws of motion; why would two electrons make an exception?

But laws of motion dont work in quantum world, so how can they follow laws of motion.

Why, they do, only in a quantum way.

So they must undergo an elastic collision(ideal conditions)

This question also arises with photons, so please check this out: https://physics.stackexchange.com/q/23930

The mathematics of quantum mechanics predicts an unevenness of the observation, similar to wave equations in that there are points where high values of electron density/reaction and low (~zero) values occur. If you have places (~50% of the places) where "interference" has yielded zero electron density/reaction, then the other 50% of the places will have twice the expected (average) electron density/reaction.
The question should include the entire observation, the whole experiment. For example, it is easy to imagine a photon experiment where diffraction causes interference effects upon a photographic film: photons abandon their even spread by being diffracted, and don't go to certain places on the film, but do go (in twice the expected density) to other places. Two photons don't just cancel out in some places - they "cancel out" and also "add up" in other places. Oh, heck, just think of a sine wave (square its amplitude) plopped down on the film and where the high peaks are, the film turns dark and where the wave bottom is, nothing happens. Two photons, two sine waves; there has to be a slight difference in path so that when you superimpose them, they add up more in one place and subtract more in another.
That thought experiment was reasonably easy because we think of photons as sort of fuzzy, wavy, intangible things. We usually consider electrons to be touchable, therefore firm on some sort of basis, so how can they interfere and cancel out of existence? Well, the wave nature of electrons means that they have a spread in space which has some of the characteristics of a wave, and they can, under the right conditions, exhibit this effect strongly. Getting back to the photographic film experiment, electrons can be made to exhibit the same effects as diffraction does to photons. Where it becomes more difficult to understand is that while the photon (say, ~500 millimicrons wavelength) is much fuzzier than a silver ion (~0.25 millimicron diameter in AgCl), you would expect an electron to be, if anything, a bit smaller than an atom, since, after all, electrons (many of them!) will fit into an atom or ion.
So you can see that a photon with a spread of 500 millimicrons will smash down over close to a million Ag+ ions, but can only activate one, but you might expect a little electron to be like a bullet and just hit one silver ion. How could you miss? Consider the bullet to be wobbly (wavy) enough to hit either one Ag+ or its neighbor. The faster it goes, the wobblier (wavy) it gets. And these waves interfere so that some paths just don't happen, and others are travelled twice as frequently - this doesn't mean that two electrons cancel out of existence or form a bi-electron.
At least that's the way I look at it without the comfort and brevity of mathematical equations. But  words only poorly describe the situation and it takes forever to try to explain.

1 millimicron = 1 nm, I assume. While I understand that you are trying to make a comparison between dimensions easier, Is it really so difficult confusing to use nm?

Otherwise good way to introduce the concept!

Thanks, but I'm old. I would have liked to use Angstroms. I still have to get used to nm - and pm.


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Entropy for a closed system
Case -1
In a free expansion of an ideal gas in a region or container, we cannot trace a pressure-volume path since the pressure, temperature and volume of the gas fluctuate unpredictably when the gas fills the region (container).
But since entropy is a state property, the difference in entropy must depend only on initial and final state. So we can replace the irreversible free expansion with a reversible process with same initial and final condition.
Consider an insulated container along with a thermal reservoir attached to it whose temperature can be varied by a knob. It's current temperature is T. A piston is also attached to it. Above the piston is kept a small container containing iron balls (to balance the pressure due to atoms below it). The initial state is the same as the initial state of a free expansion.
Removing the iron balls one by one, we increase the volume with an infinitesimally amount and also by keeping the temperature constant. Since it follows an isothermal path, the heat absorbed by gas from reservoir is equal to the work done by the system (positive).
W=Q
After the final condition by the reversible isothermal process is achieved,
Change in entropy = Q/T
And so, Entropy increases
To sum it up:- we can write :-
To find the entropy change for an reversible process occuring in a closed system, we replace that process with any reversible process that connects the same initial and final states.
Case - 2
Since the process is reversible we can proceed in a reversible manner, i.e we can add more and more iron balls to decrease the volume and consequently the energy must be extracted from the gas and be given to the thermal reservoir, i.e the heat Q is negative. Thus entropy must decrease.
But this does not happen since the thermal reservoir's entropy increases and the system consisting of the gas and reservoir, the total entropy change is 0.
Note - By taking the gas as the system, we do not meet the requirement of the entropy's definition of having a Closed System.
To sum up this :-
If a process occurs in a closed system, the entropy of the system increases for irreversible processes and remains constant for reversible process. It never decreases.
I have explained here in short about what has been given in my textbook for entropy.
My Question - In case 2, we take the system to be the gas as well as the reservoir and so the total entropy change is 0 while in case 1, we take the system to be only the gas. If I take the system, in case 1, to be gas and reservoir, the entropy should be zero. Since heat is being removed from the reservoir thus Heat Q in the entropy formula for reservoir should be negative. And thus the total change should be zero in case 1 as well.
Why doesn't this happen?

Some people define a "closed system" as one that can exchange neither heat nor work not mass with the broader surroundings.  Others, like us engineers, define a "closed system" as on that can exchange both heat and work with its surroundings, but not mass.  We engineers call a system that can exchange neither heat nor work with the broader surroundings an "isolated system."  Which definition of a closed system are you subscribing to?

Isolated system

In case 1, for the irreversible path, the system is isolated.  But, when calculating the change in entropy using an alternative reversible path, the system no longer needs to be isolated.  The alternative reversible path shows that, as expected, for the actual irreversible path in the isolated system, the entropy increases.  In case 2, you carry out a reversible path in an isolated system and surroundings, and, as expected, the combined entropy change is zero.


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what is the structure of bamboo kun?
There are claims that the bamboo fiber has inherent antibacterial properties due to "bamboo kun".
However I could not find anywhere what is the chemical composition of this "bamboo kun" and also no evidence to substantiate the antibacterial claims.
Does anybody know if bamboo kun is a myth or it really has a known chemical structure?
Edit1:

check out wikipedia: https://en.wikipedia.org/wiki/Bamboo_textile

Bamboo also contains a substance called bamboo-kun – an antimicrobial agent that gives the plant a natural resistance to pest and fungi infestation

google books: https://www.google.com/search?tbm=bks&q=bamboo+kun

Edit2:
Also I could not find this Japan Textile Inspection Association which lots of sites making the antibacterial claim above are citing as the source of antibacterial validation (although there is a Japan Textile Federation that has nothing to do with the above claims):

http://www.bambrotex.com/second/bc_nab.htm
https://www.textilemates.com/bamboo-fibre/
https://highlineorganix.com/pages/why-bamboo
https://www.weber-textilien.ch/bademantel/50-210-bambou-bademantel-mit-schallkragen.html
https://web.archive.org/web/20111005122758/http://www.fashionandearth.com/us/bamboo-clothing-facts.html

It is not implausible that bamboo should have antibacterial properties, after all plants generate lots of protective chemicals. However, the use of this particular word (kun or kunh) by a number of commercial textile websites seems like a promotional gimmick without a strong scientific basis.
Claims of an intrinsic antimicrobial property of processed bamboo fibers are likely exaggerated, as the excerpt below (1) suggests:

3.3. Antibacterial Properties
The antibacterial properties of the control sample (OB) and the
  composites (ZB, GB, ZGB) were qualitatively investigated by the
  bacteriostatic circle method against one Gram-negative (E. coli) and
  one Gram-positive (B. subtilis), and the results were shown in Figure
  6. Figure 6A displayed the antibacterial activities of bamboo samples against E. coli. It can be observed that OB had little antibacterial
  activity without an inhibition zone.

This is not to say that bamboo lacks antimicrobial properties. Plants employ a myriad of chemical substances as sources of physical and chemical protection. Extracts of bamboo (2) apparently contain bactericidal compounds, although the concentration in processed fibers appears insufficient to grant these significant antibacterial properties.
It's also worth adding that the above refer to specific species of bamboo and there may be significant variety in antibacterial properties across species.
In conclusion it is difficult if not impossible to answer the OPs question as it is unclear what "kun" or "kunh" refers to, it may just be a catchphrase to describe unknown substances granting antibacterial properties to raw bamboo.
References

Junyi Zhang, Bo Zhang, Xiufang Chen, Bingbing Mi, Penglian Wei, Benhua Fei, and Xindong Mu. Antimicrobial Bamboo Materials Functionalized with ZnO and Graphene Oxide Nanocomposites. Materials (Basel). 2017 Mar; 10(3): 239.
doi: 10.3390/ma10030239
The origin of the antibacterial property of bamboo. T. Afrin , T. Tsuzuki , R.K. Kanwar & X. Wang.  The Journal of The Textile Institute. 2012; 103(8): 844-8491. doi: 10.1080/00405000.2011.614742