text:
Passivation of metals in acids
It is commonly taught in chemistry textbooks that metals such as $\ce{Fe, Al, Cr}$ is "passivated" in cold concentrated $\ce{H2SO4}$ and $\ce{HNO3}$ due to them forming oxide films on the surface that prevents the bulk metal from reacting further. However, all those metals can dissolve in dilute solutions of the aformentioned acids, and dilute $\ce{HCl}$ as well.
However in metals such as $\ce{Ti}$ and $\ce{Sn}$ show somewhat reversed characteristics compared to iron or chromium. For example, according to Wikipedia:

Titanium is capable of withstanding attack by dilute sulfuric and
hydrochloric acids, chloride solutions, and most organic acids.
However, titanium is corroded by concentrated acids.

For tin, Greenwood and Earnshaw's Chemistry of the Elements state that:

Dilute $\ce{H2SO4}$ and $\ce{HCl}$ show little, if any, reaction but
dilute $\ce{HNO3}$ produces $\ce{Sn(NO3)2}$ and $\ce{NH4NO3}$. Hot
concentrated $\ce{HCl}$ yields $\ce{SnCl2}$ and $\ce{H2}$ whereas hot concentrated
$\ce{H2SO4}$ forms $\ce{SnSO4}$ and $\ce{SO2}$.

My question is that why some metals are corroded by dilute acids but not by concentrated acids; while other metals do the exact opposite but with the same mechanism?
My attempt to answer is that for the cases of $\ce{Fe, Al, Cr}$, concentrated acids oxidize them so quickly so that a thick layer of oxide is rapidly formed. The oxide is structured in a way so that the acid ions/molecules cannot get past it to attack the unreacted metal. In a diluted solution the oxidation does not occur as fast and the rate of formation of oxide layer is slower than rate of metal dissolving.
In $\ce{Ti, Sn}$'s case, the oxides $\ce{TiO2, SnO2}$ are more acidic but for some reason not as thick. Because they are more acidic, they require higher acid concentration to be dissolved, thus those metals only dissolve in concentrated acids rather than dilute.
Any help would be greatly appreciated, especially if you can give reliable journal papers or books.

related https://chemistry.stackexchange.com/questions/27954/what-causes-passivity-of-specific-metals-in-conc-nitric-acid

It is necessary to separate the case of $\ce{HCl}$ from the other acids like $\ce{H2SO4}$ and $\ce{HNO3}$ which may be oxidizing, because $\ce{HCl}$ is only an acid and does not or cannot act as an oxidizing agent. $\ce{HCl}$ cannot make an oxide layer. it cannot produce passivation. Nevertheless, the fact is that $\ce{HCl}$ is unable to react with some oxides like $\ce{Al2O3}$ and $\ce{Cr2O3)$ I am not sure there is a reason for this absence of reaction.

Hot concentrated H2SO4 may be an oxidizer, but whether hot or cold, Fe, Al and Cr will dissolve in it, not be passivated by it. HCl is also an active corrosive. HNO3 is the unique oxidizing passivator or corroder.


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Why does twistronic 'magic angle' graphene have only 180-degree symmetry?
From Graphene superconductors may be less exotic than physicists hoped[1]:

Excitement rose earlier this year with the discovery of
superconductivity in a similar
system[2]:
three layers of graphene twisted at their own special angle. Both
systems shared a rare, 180-degree rotational symmetry, which theorists
argued could support an especially exotic form of superconductivity
based on electron vortices known as skyrmions.

But the new incarnation of superconducting graphene appears strikingly plain.
Graphene has six-carbon rings like graphite, and is aromatic, so shouldn't it exhibit 60-degree symmetry or maybe 120?
What am I missing?
References:
[1]: Graphene superconductors may be less exotic than physicists hoped
https://www.quantamagazine.org/graphene-superconductors-may-be-less-exotic-than-physicists-hoped-20210614/
(accessed Jun 15, 2021).
[2]: A new twist reveals superconductivity’s secrets
https://www.quantamagazine.org/graphenes-new-twist-reveals-superconductivitys-secrets-20210316/
(accessed Jun 15, 2021).

$180^\circ$ rotational symmetry is, to put it mildly, about as rare as cow manure in the field. Yes, I know this is a direct quotation from Quanta, which I appreciate as a respectable source, but this particular statement quacks like a nonsense and walks like a nonsense, so I call it nonsense.

The publication lacks a clear illustration. Taking graphen's grid (this article points to this one), rotate it by 180 degrees (flat, staying in the plane of drawing) and move it a little, you probably may superimpose the position of the orange dots with those in turquoise.  Is it this?  Or the later one about Skyrmion pairs, one clock- and the other counterclockwise turning vortex?

@IvanNeretin I thoroughly disagree that the statement "quacks like a nonsense and walks like a nonsense".

Any further thoughts on this? I cant think of anything more to add to my answer, but I can add a bounty or you can post a link to this in one of the Matter Modeling SE chat rooms, there's some interest in twist there.

I will take a stab at this, tl;dr: graphene is flat, most 2D hexagonal lattice materials have a buckled structure so only have three-fold rotational symmetry even though their translational lattice has six-fold symmetry. But flat graphene's structure has sixfold rotational symmetry and sixfold includes two-fold whereas threefold doesn't.
And if that tl;dr is still too long so didn't read, translation and rotation are different animals, and hexagons are just plain weird.
From the link in the question:

Key to the skyrmion story is the 180-degree rotational symmetry that dictates electron transfers between the triangular sublattices. A rectangle has that same symmetry. A hexagon has it. A rectangular or hexagonal lattice has it. But stacking and twisting sheets of just about anything besides graphene breaks it. At last, Vishwanath and his colleagues were able to explain why the zoo of twisted lattices had failed to superconduct.

The question asks:

Graphene has six-carbon rings like graphite, and is aromatic, so shouldn't it exhibit 60-degree symmetry or maybe 120?

Yes, a single layer or stacked and aligned layers of a hexagonal lattice honeycomb Xene will have 120 or 60 degree symmetry, depending on if it's flat or buckled.

Are all stable Xenes (graphene-like 2D honeycomb sheets) buckled?

What am I missing?

Graphene sheets are flat whereas most Xenes are buckled; go around a ring and the atoms alternate up-down-up-down-up-down out of plane.

Why do we not see silicon in a structure like graphite?

The lattice (a mathematical thing indicating how you can translate the structure such that it looks the same) is hexagonal. A hexagonal lattice has 180 degree symmetry as well as 120 and 60 degree symmetry.
But the basis
doesn't have to have the same symmetry as the lattice!

The crystal structure consists of the same group of atoms, the basis, positioned around each and every lattice point. This group of atoms therefore repeats indefinitely in three dimensions according to the arrangement of one of the Bravais lattices. The characteristic rotation and mirror symmetries of the unit cell is described by its crystallographic point group.

In 2D we enumerate the symmetries of the basis in what's called the "wallpaper group". See the many links in my Math SE question Are planar symmetry groups and wallpaper groups the same things?
So...
while graphene looks the same rotated 0, 60, 120, 180, 240, 300 and 360 degrees, buckled honeycombs only look the same at 0, 120, 240 and 360 degrees, skipping 180 degrees.

Sigma and pi bonds in graphene. Sigma bonds result from an overlap of sp2 hybrid orbitals, whereas pi bonds emerge from tunneling between the protruding pz orbitals. image source

Example of a typical Xene that's buckled:

Balls and sticks model of a single layer buckled hexagonal lattice of silicene, the buckling is marked by buckling height δ. The in-plane unit cell composed of two silicon atoms is highlighted. source

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


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Does elimination occur when bromonapthalene is reacted with sodium amide in potassium hydroxide?
What should be the product of this reaction? Some books show that an elimination reaction would happen since $\ce{NH2-}$ is a strong enough base to take away a hydrogen on a benzene ring and create a triple bond. I've never seen that before and why would the stable ring react?

Are you trying to say that the ring would not react because of loss of aromatic stability?

@M.L regarding your latest edit, try to avoid using MathJax in title filed as the search engines can't read it and thus it makes a post harder to find.

Have you learnt about benzyne mechanism? If not, I would suggest you to read about that first. Books would generally have an overview of benzyne first before showing it on napthalene. Your reaction is similar to that.

Elimination does occur, but only to form an intermediate.
As TRC has mentioned above, this involves a benzyne mechanism.
$\ce{NH2-}$ is a very strong base; it is the conjugate base of ammonia, which itself is already a base.
Under such strong conditions, the aromaticity of the benzene ring can be broken with the $\ce{NH2-}$ eliminating a hydrogen from the aryl halide and forming ammonia. This leaves a negative charge on the arene, and naturally to resolve this negative charge, the halide, being a good leaving group, leaves the benzene ring as an anion.
This hydrogen is always eliminated alpha from the aryl halide, meaning to one of the carbons next to it.
Two of the carbons of the ring, one at the halide position and one alpha to it, now  have no substituents and form a triple bond between themselves. This is an aryne.
This Aryne is too reactive to remain for long, with a very strained triple bond and no aromaticity.
The aryne contains two locations, on each end of the triple bond, where the carbon has no substituents. Therefore, each end of the triple bond is now a prime target for an addition reaction.
Therefore, the ammonia attacks the aryne at one end of the triple bond, forming an brief intermediate before one of the hydrogens from ammonia undergoes a hydride shift to add to the other end of the triple bond. Net result is a substitution reaction. An amine group is substituted for a halide group.
However, because of how both ends of the triple bond can be attacked, the amine group may be at the same position as the original halide, or it may be at one of the alpha positions relative to it.
Libretexts has a good page on the mechanism which I shall link.
https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Benzyne_Mechanism

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How to calculate Kp without knowing the volume?
I have the following chemical equation:
$$\ce{Sb2S3 + 3H2 <=> 2 Sb + 3H2S}$$
I have $\pu{1000 grams}$ of $\ce{Sb2S3}$ that reacts with $\pu{10 grams}$ of $\ce{H2}$ in a reactor at $\pu{713 K}$.
I would like to calculate $K_p$ (equilibrium constant using partial pressures).
I know that $K_p$ is equal to the quotient of the partial pressure of $\ce{H2S}$ cubed divided by partial pressure of $\ce{H2}$ cubed, and that I need to use the law of perfect gases.
$$K_p = \frac{P_{\ce{H2S}}^3}{P_{\ce{H2}}^3}$$
I have calculated the amount of moles of $\ce{H2S}$ to be $\pu{2.264 mol}$ and the amount for $\ce{H2}$ is $\pu{2.02 mol}$.
However there is just one problem I don't have the volume in this exercise, since it wasn't given. Thus, I am a bit stumped.

Use the principle of partial pressures, Volume cancels out..

It does not seem this problem can be solved with the information given, please add information so that one can arrive at the number of moles stated here. Or, incorporate clarifications which you stated in comments into your post.

So you need $$K_p = \frac{p_{\ce{H2S}}^3}{p_{\ce{H2}}^3}$$ where $p_{\ce{H2S}}$ and $p_{\ce{H2}}$ are partial pressures of $\ce{H2S}$ and $\ce{H2}$ respectively.
Partial pressure is nothing but the mole fraction of the substance times the total pressure. You already have mole fractions, so that part is done.
Now notice that in the expression of $K_p$ , you have cube divided by cube, so the term of total pressure will get cancelled out.
So, you will end up having $$K_p = \frac{p_{\ce{H2S}}^3}{p_{\ce{H2}}^3}=\frac{(2.264 )^3}{(2.02 )^3}=1.407$$

Thank you very much for your answer. However, in my book the answer given is 0,426. Sadly I came to the same answer that you have given me but it appears to be wrong.
Edit: the amount of mols I gave you for H2 is actually wrong, my mistake, it's 2.685 mol.
the answer I get is 0.6, which still isn't right though :/

@StormCaster Initially I just put the values you had given without thinking much. Now when I think about it, how are you finding these values? Using the atomic mass of Sb as 122 g and sulphur as 32 g, we get the moles of $\ce{Sb_2S_3}$ as $\dfrac{1000}{122\times 2+ 32 \times 3}=2.941$ and moles of $\ce{H_2}$ as 10/2 =5. Now how will you find the equilibrium concentrations of these without knowing anything else? Are you sure you are not missing any other detail? May I know which book are you using?

Well, This is embarassing...
In the end it turns out there are 2,1302 moles of H2S and 2,8198 moles of H2.
so Kp is equal to (2,1302^3)/(2,8198)^3 which is equal to 0,4311 (the answer given in my book is 0,426 which is close enough I guess)
Thank you for your help, sorry for wasting your time!


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Why is chlorosulfite a good leaving group?
I came across the following website article[1]:

I understand the mechanism and also get the fact that this is nothing but an experimental result but why does the $\ce{ClSOO-}$ group departs spontaneously to form a carbocation? Also since carbocation is formed will there be any rearrangement(if possible)?
Reference:
[1]: SOCl2 mechanism for alcohols to alkyl halides: SN2 versus SNi
https://www.masterorganicchemistry.com/2014/02/10/socl2-and-the-sni-mechanism/
(accessed Jun 15, 2021).

Related: https://chemistry.stackexchange.com/questions/69063/if-socl2-reacts-with-alcohols-via-sni-why-doesnt-pocl3?r=Searchresults ... https://chemistry.stackexchange.com/questions/48489/substitution-at-thionyl-chloride-sulfur


text:
Anti-foaming agent for HCl + copper chloride solution
I am trying to etch my own circuit boards at home, and have whipped up some DIY cupric chloride etchant (just CuCl with a bunch of free HCl)
The problem I have found is that when I try to bubble air through it to regenerate it, the solution foams aggressively enough to come up over the top of my etching tank.
I'm more of an engineer than a chemist, so i'm unaware of what anti-foaming agents will work well in the strongly acidic solution.
I'm hoping for something I can buy in small quantites or create relatively cheaply.

If it was me, I would buy some “anti-gas” OTC stuff at the local pharmacy. Stomach acid is HCl, so it would be worth a quick try.

The circuit etch solution is made from CuCl$_2$ rather than CuCl. https://www.instructables.com/PCB-Etching-Solution-Cupric-Chloride/
After etching/dissolving some copper, you would get CuCl, and regeneration involves reoxidizing the Cu$^+$ to Cu$^{2+}$. When you use air, not all the O$_2$ is captured, and none of the N$_2$, so you have lots of gas bubbles forming and rising. One way to reduce the foaming would be to reduce the rate of aeration. A possible cause of foam is the presence of surfactant or a film-former from reaction with the circuit board.
One way to defeat the foam is to treat it with a low surface tension agent. Isopropanol (rubbing alcohol (70%) or 91%) sprayed onto the surface - just a little bit - can break the foam and let it settle down. Most of the isopropanol will evaporate easily so the treatment will have to be repeated. I have done this on a large scale.
Another additive would be a silicone lubricant. Spray cans of this type of product are used for nuts and bolts type of mechanics. A little goes a long way, and could interfere with your etch, but has potential.
Consider using hydrogen peroxide to reoxidize your Cu$^+$, as in the link above. This would significantly reduce the bubbling and might give you more control over the regeneration, although it does dilute the product.

Instead of oxygen bubbling, add some hydrogen peroxide $\ce{H2O2}$. It is not expensive. And it works perfectly well.
Here is a formula that does your job in less than 1 minute, without any foam.
Dissolve $\ce{20 g CuCl2·6 H2O}$ in $\pu{100 mL}$ water. Add $\ce{100 mL HCl}$ $35$% and $\ce{100 mL H2O2}$ $30$%. Here you are. Dip your circuit board in this solution. In less than one minute, the copper metal from the board is attacked and dissolved. The total equation is :
$$\ce{Cu + 4HCl + H2O2 -> [CuCl4]^{2-} + 2H2O + 2 H+}$$
To be true, this equation is the sum of a couple of equations taking place simultaneously, which are : $$\ce{Cu + CuCl2  -> 2 CuCl}$$ $$\ce{CuCl +  HCl -> HCuCl2}$$ $$\ce{2 HCuCl2 + H2O2 +  4 HCl -> 2 [CuCl4]^{2-} + 4 H+ + 2H2O}$$
And this reaction does not produce any foam !-


text:
Reason for GC peak splited
I ran GCMS and one of the peak of GC appears:

The peak shows some tailing, but there seems to be another peak at right-hand end of the signal. I have already checked the MS and confirmed that there is no other compound. What could this be due to?

Are you overloading the GC column by injecting too much? Try reducing the injection volume and/or increasing the split ratio and see what happens to the peak shape.


text:
How can the activity of ions be more than 1 if the activity of water is 1?
Now, I am quoting Chemlibretexts when they give the activity coefficient of $\ce{LiCl}$ to be $\pu{1.026}$ at the molality of $\pu{2.5 mol kg-1}$
For simplicity, assuming this is aqueous, I'll just convert the molality to concentration, since it shouldn't really affect it that much.
Given $y$ $\mathrm{=\frac{a}{[C]}}$
And here my question begins.
$\ce{LiCl}$ is a spectacularly soluble substance, $\pu{84.25g/100ml}$ or $\pu{19.8 molL-1}$, which indicates that at a concentration of about $\pu{2.5 molL-1}$ is nowhere close to saturation, and indeed, I do not believe there will be any repulsive effects between the water and the $\ce{Li+}$ and $\ce{Cl-}$ ions.
In aqueous solution, the activity of water is 1, because there is so much of it around that it cannot really affect direction of reaction by Le-Chatelier's principle.
Therefore, the equilibrium constant of a reaction even involving water in aqueous solution is written without water.
Activity is supposed to be a measure of effective concentration; i.e. the availability of a substance in a given volume.
So how is it that, $\ce{LiCl}$, can have an activity of $\pu{2.6}$ at a relatively low concentration of $\pu{2.5 molL-1}$, while water, which surrounds the entire reaction with roughly a concentration of $\pu{55 mol dm^-3}$ and is always in excess and available for reaction, can only achieve an activity of 1? Is there something special about a solute that allows it's activity to exceed 1?
This is because that $\ce{LiCl}$ having an activity above 1 somehow implies that $\ce{LiCl}$ is more available for reaction than water, which to me is bizarre.
I have seen other somewhat similar questions, but none of them touch on this paradox, or they only mention hydrophobic substances such as benzene where repulsion may be a cause.

Activity value depends on what activity is conventionally chosen for the compound as unit activity, by equation $\mu = \mu_0 + RT \ln{\frac{a}{a_0}}$. Activity coefficient of hydronium in concentrated HCl near 12 M is about 20, so hydronium activity is about 240. For water the activity is defined as 1 for pure water, for ions activity is equal to concentration as the limit value for infinitely diluted solution.

Hmm...it's still kind of odd to me. How is it that highly concentrated solutes can reach very high effective concentrations while much "concentrated" solvents are left with an activity of about 1? The way I see it, water doesn't affect equilibrium because changing it in any way does not affect the equilibrium position given that it is already so much in excess that it is always available for reaction. Should the same effect not apply to Hydronium at higher concentrations, lowering it's effective concentration to below that of water? In fact, what do conc/activity graphs look like?

You don't really compare the activities if different substances.

Still though, it seems very odd simply because the magnitude of the difference is a 240x magnitude in a situation where I would expect water to have the higher effective concentration, simply because it's concentration is 55.6 mol dm^-3 while that of hydronium is 12 mol dm^-3
Basically, I just fail to see how the effective concentration (in a way, availability for reaction) of a solute can exceed that for the solvent, given that the solvent is everywhere, especially by such a large magnitude.
I would expect the activity coefficient to decrease as concentration increases, like pH and activity

Activity can be considered as "effective concentration" just accidentally, if it is defined so for given compound state. Water activity = 1 has no relation to concentration.

Also the effective concentration is not linked to the spatial notion you have in mind. It goes the other way around,if I understand you correctly. In general the activity coefficient approaches its maximum with decreasing conc.

If it is not effective concentration, what exactly is activity and what makes the activity of a solute possibly much higher than a solvent? The wikipedia page mentions the idea of effective concentration.

Read the first comment few times more. See also https://en.wikipedia.org/wiki/Thermodynamic_activity

Aight. I understand what you mean to some extent, but I can't really formulate an explanation of all this.
Is there something fundamentally different between a solvent reacting and a solute reacting that leads to such wildly different values to come out? I think that is the core of my question.

The core point is that the unit activity levels are deliberately chosen, linked to well defined state and to the chemical potential chosen as the standard one. For solvent it is the pure solvent state, for ions it is infinite dilution state, ideally extrapolated to unit concentration, with activity numerically equal to concentration for c->0. Therefore, activities of different species are not directly comparable as concentrations are, because they are context dependent.

Related: Activity Coefficients of Concentrated Salt Solutions: A Monte Carlo Investigation

What makes you think that the activity of water remains constant?


text:
What would be the order of reactivity of these compounds in Sn1,Sn2,E1 and E2 reactions?
My problem is that these compounds seem as if their Sn1 and E1 reactivity are identical as the carbocation would of similar nature and stability. For the Sn2 part the first compound seems to have a hindered scope of a frontside attack (due to the tertbutyl group). However the answers given are completely the opposite. I am completely confused.

Well in both compounds 1 and 2 , the groups attached are the same but they have different stoichiometric alignments. So  in compound 1, the tery butyl and bromine are in different planes due to which there will be less steric hindrance there whereas in case of the second compound, bromine and the tert butyl are in the same plane hence it will provide a higher steric hindrance to the attcking nucleophile.

@Ritvish How about a backside nucleophilic attack on the compound 2

Maybe trying to draw the chair form of cyclohexane, then placing $\ce{C(CH3)3}$ group at equatorial position and then checking the positions of $Br$ atom can help...


text:
How measured content of Calcium Carbonate in eggshells change with the use of different acids?
I am investigating the calcium content of chicken eggshells. The only literature I can find on this consistently and solely depend on the use of HCl to conduct back titration. I conducted the experiment using Acetic, Ascorbic, Lactic, Oxalic and Hydrochloric acid and reached values of 66.8%, 37.8%, 58.3%, 95.5% and 56.5% respectively with a percentage uncertainty of 0.45% overall.
Since I have no literature to compare, I am not sure of my results (except HCl, which is acceptible when the difference in molarities is accounted). And I lack the theoretical background to explain why different acids could yield different values?
I would like to ask to provide theoretical background, or at least to direct me to it.

When downvoting, can you also state the reason please?

I down voted because this has almost no relation to science: it is essentially alchemy with eggshells. The results are poor, there is no reason stated for even picking the acids used, there were apparently no preliminary runs using carefully weighed samples of actual pure calcium carbonate (an extremely inexpensive substance), and so on. It is just a poorly conceived muddle.

I was actually curious about why HCl was always preferred in this experiment when it was demonstrated in one of our labs. I posted here as I was unable to find background information on this, but maybe it was not right to post without any strong methodology. Sorry for taking your time and thanks for your reply.

We use HCl because it is a strong monoprotic acid that is inexpensive, readily available in high purity and it is non-oxidizing. The other acids you used are weak acids, which can only add unnecessary complications.

Thank you very much, have a nice day!

I guess you did this experiment just as an academic exercise. All other acids are weak acids, and HCl is the strong acid. Surprisingly, only HCl should have provided correct results and oxalic acid should have the worst results. Since your HCl results are incorrect, and off by >50%, the results of others are somewhat dubious too.

Oxalic acid forms precipitates with Ca ions, so it should have the most incorrect results.


text:
Construction of the molecular orbital diagram for a species M4
I need to construct the molecular orbital diagram for the hypothetical species ${Li_4}$, which has the following geometrical arrangement:

The first step is to identify the point symmetry group. In this particular case, we consider that there is only one axis of rotation of order four (actually, other symmetry elements can be observed, but this is a previous consideration of the exercise):  $C_{4} $(Schöenflies notation).
Once the point group has been identified, we consult the literature for the table of characteristics. For this symmetry:

The associated reducible representation is then constructed. In this step, I have a doubt, because I do not understand the concept of "reducible representation" and its usefulness in this theory. According to what I have given in class, it is the number of atomic orbitals that remain unchanged when a symmetry element is applied on the solid. If we go by this definition, such representations would be:

Thus, the irreducible representation of this molecule would be: A + B + E
And, now, according to what has been taught in the course, it would be necessary to use the projection operator to determine the linear combinations adapted to the symmetry. But, here the truth is that I'm starting to make a mess.
Once here, how could I continue? Or, perhaps, they know of a simpler way of constructing orbital diagrams.

Could you explain why you think "It is clearly seen that the only symmetry element present is an axis of rotation of order 4"? Unless I am misunderstanding your diagram I would have thought there are more than just that. What about reflections for starters?

You are right. On closer inspection, you can see that there are more elements of symmetry, such as a mirror plane. I have commented that there is only one axis of rotation of order 4, because that is how it appears in the study slides. Perhaps this assumption is motivated by the fact that we want to study cyclic molecules ($C_{n}$ symmetry), and this exercise is presented as an introductory example. @IanBush

Try the book "Molecular Symmetry and Group Theory: A Programmed Introduction to Chemical Applications" by Alan Vincent (link). This book helped me a lot when I first started learning group theory for chemistry. Also, when using projection operator method, check whether you are using all of the symmetry operations. For example a $\ce{C_2}$ operation in the graph actually represents two symmetry operations, $\ce{C_2}$ and $\ce{C_2^{-1}}$, similarly $\ce{C_4}$ actually has 4 operations which you need to consider.

Thank you, but I checked and it is not available at my university library. @ShoubhikRMaiti

@aprendiendo-a-programar It's also available as an ebook. If you really really need it, there are some sites which would probably let you download it for free, but they are illegal, so I can't advise you to use them. Also, have you tried using all of the symmetry operations as I mentioned? Does that give you the right answer?

I didn't really understand what you meant. @ShoubhikRMaiti

Haha yeah it does sound a bit confusing now that I read it again. If I have time, I will try to write a more detailed answer tomorrow.


text:
What is the power density of graphite user the Wigner effect?
The energy that graphite released upon a cascade of defects by neutron radiation is 2.7 MJ/Kg, however there is no data on how quick this energy is released. I am interested in this as this not mentioned anywhere on the Windscale disaster

For those of us who had never heard of the Windscale disaster: https://en.m.wikipedia.org/wiki/Windscale_fire.

No actual facts, but relaxation of an interstitial atom to an available hole is rather instantaneous. Thus the reaction is limited by the diffusion of holes. (diffusion of interstitials would be orders of magnitude slower). Diffusions of either kind is slow in graphite, as the recrystallization temperature indicates. Combine these opinions into; rather swiftly, then slowing down. Milliseconds to reach a thousand degrees, would be my guess.

There are some papers in which Wigner effect was tested as e method of energy storage. A MORE interesting question is 'What OTHER materials can be used to store energy - how much and how fast can that energy be released?'.
The Russian patents on what is known as 'Red Mercury' are where the compound Hg2Sb2O7 is suggested and which I suspect MIGHT be based on theoretical work.
If you can produce a material that can store a LOT of Wigner energy and will also detonate would, I guess, allow all of the energy to be released VERY quickly indeed.
That said, I do not have the skillset to even estimate the veracity of the patents.


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Why is the only natural source of radium compounds uranium minerals?
A common mineral of barium is barytes, or barium  sulfate ($\ce{BaSO4}$). Because elements in the same periodic group have similar chemical properties, we might  expect to and some radium sulfate ($\ce{RaSO4}$) mixed  with barytes since radium is the last member of  Group 2A. However, the only source of radium compounds in nature is in uranium minerals. Why is it so?  I searched on the internet but couldn't find useful insights

The longest lived isotope of radium has a half life of 1600 years. So you look in uranium ores since radium is part of the decay chain from uranium to lead.

Very slightly adding to what Jon Custer commented, just look up radium in wikipedia: the decay chain for that longest lived radium isotope is right there.

My guess is because the radium ions are too large to fit into the crystal lattice of the mineral. So the radium is not incorporated into the baryte mineral.

@ĐỨcLêHồng No. The radium is formed in situ in uranium and thorium minerals. There is no “separate” radium available for mineral formation on typical geologic time scales.

Because radium is a part of uranium radioactive disintegration.

@DucLeHong. Even if some radium would have been incorporated into some barium substance, this phenomena must have happened more than one million years ago. And meanwhile, this radium would have disappeared by natural radioactive decay.

You can't find primordial radium because it's half life is too small compared to earths age.
Even the radium isotope with the longest half life, $\ce{^{226}Ra}$ has a half life of only 1600 years which is magnitudes smaller than the age of the earth, which is estimated to be around $4.54\times 10^9$ years[1].
This means that whatever radium we can find in today's age must be regenerating from a radioactive parent source. The most commonly known of those sources is as an intermediate in the radioactive decay of uranium and thorium to lead. From The Radiochemistry of Radium[2]:

Isotopes $\ce{^{223}Ra}$, $\ce{^{224}Ra}$, $\ce{^{226}Ra}$ and
$\ce{^{228}Ra}$ are part of the decay chains of natural thorium and
uranium isotopes; since thorium and uranium have very long half-lives,
these daughters are continually being regenerated by their decay.

References:
(1): Dalrymple, G. B. The Age of the Earth in the Twentieth Century: A Problem (Mostly) Solved. Geol. Soc. Spec. Publ. 2001, 190 (1), 205–221.
(2): Salutsky, M. L.; Kirby, H. W. The Radiochemistry of Radium; Mound Lab., Miamisburg, Ohio (Grace (W.R.) and Co., Clarksville, Md. Washington Research Center), 1964.


text:
What is an effective preservative for guar gum slime?
I am attempting to find an effective preservative for guar gum slime. I am using typical recipes:
A borax solution of 4%, a guar gum solution of 5% using distilled water, and a mix ratio of 20:1 (guar to borax). I am trying to find a preservative I can add to this homemade method to overcome guar gums' short shelf life.
I am going to try to use propylene glycol in the mixing of the guar solution next time. It's cited as a good way to disperse the guar powder to avoid clumping, but is also listed as a preservative on one of the many Mattel slime variants found on the patent, Guar Gum Based Flowing Gel Play Composition[1].
There are also some preservatives cited that are used as cosmetic preservatives but not having any knowledge of chemistry I don't know where to start trying those out in my recipes.
Primary questions:

What would be a good preservative for guar gum slime?
How and at what point in the process would I add the preservative?

The traditional web searches yield nothing but endless recipes for quickly expiring guar gum slime.
Reference:
[1]: Cordova, A. Guar Gum Based Flowing Gel Play Composition. 6387169, May 14, 2002.

When you say preservative, do you picture an antimicrobial agent or something to slow a purely chemical breakdown of the slime?


text:
Out of tropolone and tropone, which is more soluble in water?
According to me tropolone has intramolecular H-bonding, so extent of H-bonding with water will decrease and thus, will be less soluble.
Tropone (Cyclohepta-2,4,6-trien-1-one):

Tropolone (2-hydroxytropone):

The fact that trololone can do strong hydrogen bonding suggests to me it will be more soluble in water. tropone can't do strong hydrogen bonding (no -OH group). So what are you comparing tropolone to? 3-hyroxytropone might be more soluble than the 2-hydroxy isomer, but on what basis can you compare solubility to something which isn't an isomer at all and doesn't have any hydrogen bonding?

For the OSISIS property explorer, tropone yields a predicted value of cLogS of -1.39, tropolone of -1.32.  Thus, for the trained model implemented (theory, 2k compounds, pH 7.5, 25 C).  Of course, it is a model.

The presence of internal hydrogen bonding does not automatically mean lower water solubility. The water solubility difference between fumaric and maleic acid provides a good example.
Maleic acid (the Z-isomer) is capable of an internal hydrogen bond between the two carboxylic acid functional groups.

Fumaric acid (the E-isomer) is not capable of internal hydrogen bonding.

Maleic acid is about 100 times more soluble in water than fumaric acid.

Maleic acid: 480 g/L
Fumaric acid: 4.9 g/L

From these data, it is clear that the internal hydrogen bond is not an impediment to water solubility. Maleic acid is more soluble because it is more polar with the polar functional groups on the same side of the molecule. Fumaric acid is less soluble because it is less polar. The two polar functional groups are on opposite sides so the individual bond dipoles cancel.
This example may not extend to the comparison of tropone and tropolone, which are not isomers. Having reliable water solubility data for both compounds would provide useful information, but I was unable to find them.

Tropolone appears to have a solubility of about 5 g per L, but I couldn't find data for tropone, Thus, I made an analogy. Of course, data on the specific compounds will trump a generalization.

So in the end, which is more soluble?

I feel that this doesn't really answer the OP's question, it's more of an interesting tangent.

@BenNorris *“Maleic acid is ten times more soluble in water than fumaric acid.
Maleic acid: 480 g/L
Fumaric acid: 4.9 g/L“* Either the statement or the data has been given wrong by mistake, I suppose. Also, I agree with ShoubhikRMaiti that this really doesn’t answer the question OP has posed, though it is a important piece of information.

logP data is available from chemspider. It agrees with the trend described by the OP, if not proof that the suggested mechanism is correct. A better argument might be to compare structural isomers of tropolone (eg 2- versus 3-hydroxytropone).

@BuckThorn But isn't logP, regardless if computed / predicted, or based on an in virto experiment, about partition only?  It is not so evident (for me) how you relate a logP (octanol / water) of compound x into solubility of compound x in water if logP of x and molar mass of x are the only properties known in first hand.

@Buttonwood interestingly this post surfaced, which addresses your point: https://chemistry.stackexchange.com/questions/153155/looking-for-caffeine-partition-coefficients  In retrospect I think my previous comment was not entirely correct. Need to think a little more about this.

@BuckThorn Like there are logP and clogP, there are cLogS about predicted solubility, too (example theory / implementation).  Probably this is easier if the molecules don't contain groups easy to ionize.

@BuckThorn For curiosity, I used this tool equally on both isomers.  But the implemented model seems to posess some limitations, because the predicted data of cLogS are -0.25 (maleic acid), and -0.25 (fumaric acid).

@Buttonwood thanks for the links. Hmm, that is an impressively bad prediction of relative solubility, I wonder why. I think the approach of comparing solubility and partition coefficients might work better for different substituted tropones. Also, many predictive tools are based on ML algorithms, and the question was looking for a mechanistic explanation, so even a very accurate prediction is of little use if it doesn't provide insight (in this context).

@BuckThorn I agree. Like this work mentioned in one of the answers here, a simple click-and-use-the-results of programs without knowledge what algorithm and implementation may be dangerous; and chemistry is no exception to this (e.g.EuSpRIG).


text:
Linear equations for constructing molecules with covalent bonds
(By training I am a math-guy and my knowledge of chemistry is restricted to school level)
I have read a little bit about the electron spheres and sub-spheres, the Aufbau principle and covalent bonds from Wikipedia.
Is it possible to derive theoretically molecules based on covalent bonds using linear equations? I am asking this because I did some computations in Sagemath:
For instance:
$$
\begin{bmatrix}
1&3&6\\
\end{bmatrix}\\
\text{Free module of degree 3 and rank 2 over Integer Ring Echelon basis matrix:}\\
\begin{bmatrix}
3&1&-1\\
0&2&-1\\
\end{bmatrix}\\
\begin{array}{ccccc}
\ce{3H} & + & \ce N & -& \ce{Cr}\\
& + & \ce{2N} & - & \ce{Cr}\\
\end{array}\\
\begin{bmatrix}
1&3&7\\
\end{bmatrix}\\
\text{Free module of degree 3 and rank 2 over Integer Ring Echelon basis matrix:}\\
\begin{bmatrix}
1&2&-1\\
0&7&-3\\
\end{bmatrix}\\
\begin{array}{ccccc}
\ce{H} & + & \ce {2N} & -& \ce{V}\\
& + & \ce{7N} & - & \ce{3V}\\
\end{array}\\
\begin{bmatrix}
8&1&3\\
\end{bmatrix}\\
\text{Free module of degree 3 and rank 2 over Integer Ring Echelon basis matrix:}\\
\begin{bmatrix}
1&1&-3\\
0&3&-1\\
\end{bmatrix}\\
\begin{array}{ccccc}
\ce{Ti} & + & \ce H & -& \ce{3N}\\
& + & \ce{3H} & - & \ce{N}\\
\end{array}\\
$$
More such examples in Sagemath-Cell.
The example $\ce{3H} +  \ce N  - \ce{Cr}$ would correspond to a hypothetical molecule $\ce{CrH3N}$ (which I was able to find at pubchem).
My questions are:

How do I search online for such molecules if they exist or not? Is
there a standard way in specifying short molecule formulas?
Suppose I derive some equations for such hypothetical molecules
based on linear equations of covalent bonds. Is there any chemical
reason why these molecules might exist or not?


text:
What does the state of a substance at a specific T and P mean?
From the phase diagram of water, we see that water is a liquid at 20 °C, 1 atm. The state of the water at 20 °C, 1 atm is liquid. But there is actually also water vapor in equilibrium with liquid water.
What is the definition of “the state of a substance at a specific temperature and pressure?”

The state of a substance is probably understood as the most important phase describing this substance at a specific temperature and pressure. At $20°$C, and $1$ atm., the concentration of water in the liquid phase is greater than the concentration of water in the vapor phase. So water is considered as liquid under these conditions of $T$ and $p$.

The water phase diagram means the single component ( but its phases ) system. So at 20 deg C, the pressure 1 atm will finally convert all water vapour to liquid.

At 20C in a ONE component system the only head pressure being that of water applied pressure greater than the equilibrium vapor pressure will condense all the water vapor. If the applied pressure is less, vacuum chamber, liquid will evaporate until the vapor pressure is reached or all the water is vapor.

I will start by addressing the posted question: the "state of a substance at a specific temperature and pressure" refers to "the most stable phase of the homogeneous substance at the specific p and T". The phase is indicated in the phase diagram by looking up the appropriate T,p point. At $\pu{20 ^\circ C}$ and 1 atm pressure that would be pure liquid water.
Now, addressing the confusion, the system you are referring to (with vapour) is not at 1 atm, at least not 1 atm of water vapour. You are probably thinking of a closed container partly occupied by liquid water and a mixed gas phase with contents under a pressure of 1 atm. Yes, the liquid water in the container will be in equilibrium with vapour (if the liquid has a volume smaller than the container's, and if the temperature is not too high, causing all of the water to form vapour). But the pressure of the water vapour (that is, its partial pressure) will not be 1 atm. It will be much less, closer to the ideal equilibrium vapour pressure of water at that temperature.
Note the question is similar that posed here, from which I borrow the following phase diagram  (not of water):

That post states

However, this had me slightly confused because in the situation initially described (at a point somewhere within the liquid region - NOT on the phase boundary) some vapour existed above the liquid in the container ...

I quote the above because it is a related confusion. When water is in equilibrium with vapour the state is necessarily one described by a point on the liquid-gas coexistence line, and most decidedly not an interior point in the region labelled "liquid". An interior point is not one describing equilibrium with vapour (or, there is no coexisting vapour at such an interior point).

The statement of "But there is actually also a water vapor in equilibrium with liquid water" may have come from a confusion about the difference of a closed system of water and a beaker of water open to air, both under say NTP. When you talk about the state of water on the phase diagram, you are talking about the former case. As Buck answered above, the pressure must be the vapor pressure of water. In the latter case, the pressure comes from the air and it is applied to the liquid part only while the water vapor and air molecules including O2 and N2 etc form approximately ideal gas mixture, and there is a dynamic equilibrium between the vapor and the liquid phase at the vapor/liquid interface. That is the reason for humidity at a certain temperature. The effect of the atmospheric pressure on the liquid part can be seen by that the liquid can be brought into boiling state if the atmospheric pressure is reduced from 1 atm to the value at 25 °C shown on the curve for liquid/vapor equilibrium of the phase diagram. The reason the liquid boils under that condition is that the vapor within the bubble inside the liquid part is equal to external pressure, which is exerted on the liquid and passed to the vapor within the bubble. If you can understand this difference, then there would be no more confusion. Please see another relevant post here: What does the phase diagram mean by "liquid" at NTP? (The liquid and vapor coexist even when the situation is not on the curve).

The simple answer is that the phase diagram refers to pure water only—either in a single phase, or in some combination of phases.
So let's imagine we have a large piston that contains only water—nothing else—that is maintained at a constant temperature of $\pu{20 ^{\circ}C = 293.15 K}$ And suppose it starts at a pressure of $\pu{1 \times 10^-6 GPa \approx 0.010 atm}$.   Checking the phase diagram below, we can see that it will be in the gaseous state.  Now let's lower the piston to raise the pressure until we hit the liquid-gas coexistence line which, at $\pu{293.15 K}$, occurs at $\pu{2.34 \times 10^-6 GPa \approx 0.023 atm}$. That's the vapor pressure of water at that temperature.  At this point, liquid water will start to form.
As we continue to lower the piston, the pressure doesn't increase.  Instead, gaseous water converts to liquid water, but the pressure stays at the equilibrium vapor pressure of water.  This proceeds until the piston head reaches the surface of the liquid. With any infinitesimal increase in pressure beyond this, only liquid water will be present.
We can continue to increase the pressure to your desired pressure of $\pu{1 atm}$. At this point, we will of course still have only liquid water. Thus we can see that, when we have only pure water at ${\pu{T = 20 ^{\circ}C and p = 1 atm}}$, it will be entirely in the liquid state, consistent with the phase diagram.
Source of diagram:  Pierazzo, E. & Artemieva, Natalia & Ivanov, Boris. (2004). Starting Conditions for Hydrothermal Systems Underneath Martian Craters: Hydrocode Modeling. Special Paper of the Geological Society of America. 384. 10.1130/0-8137-2384-1.443.


text:
What unit and dimension for membrane permeability / permeation expression?
Context
When studying the diffusion of gases through thin membrane, one may use the diffusion coefficient $D$ used in Fick's laws as (one dimension form):
\begin{equation}
J = - D \cdot\frac{d\varphi}{dx}
\end{equation}
wherein:

$J$ is the diffusion flux (mol.m$^2$.s$^{-1}$)
$D$ is the diffusion coefficient (m$^2$.s$^{-1}$)
$\varphi$ is the concentration (mol.m$^{-3}$)
$x$ is the position

While this expression describes well what happens locally, it can be integrated over a membrane's thickness leading to:
\begin{equation}
J = \frac{D}{e} \cdot (\varphi(x=0) - \varphi(x=e))
\end{equation}
wherein $e$ is the membrane thickness and the membrane is located between $x=0$ and $x=e$. While $D$ is an intrinsic property of the membrane, if we only have access to the concentrations $\varphi$ before and after the membrane, it is the $D/e$ ratio that is of importance when studying the influence of the membrane on the diffusion flux $J$.
I am working on biological membrane samples of variable thicknesses. Thicknesses which I cannot—and do not want to$^a$—measure. Thus, from my external point of view, I can only measure the $D/e$ ratio to characterise the permeability of the membranes. Let:
\begin{equation}
X = \frac{D}{e}
\end{equation}
of unit m.s$^{-1}$.

Question
Hesitating between conductivity, permeability / permeation and diffusivity, I was wondering:
What would be the logical name for X?
Indeed, I came across several denominations for values expressing the same kind of phenomena:

Permeation and the definition of the barrer would lead me to think that $X$ is in fact analogous to the permeance of the membranes. Still, even at STP$^b$, the unit of permeance is m.s$^{-1}$.Pa$^{-1}$ while $X$ does not have the pressure dimension.
Maybe I could use the term lineic diffusion coefficient.
I also saw the hydraulic conductivity which has a unit of m.s$^{-1}$. However, If I understood well, it is linked to the behaviour of fluids flow in porous media, while I think it is diffusion which drives gas transport in thin membranes.

Does it make sense to you to use one of these $X$ definitions to characterise my membrane? Or should I better normalise it with ambient pressure, yielding a more conventional permeance?
Any advice would be welcome. If I do agree that any unit and name can fit as long as it is justified, I would also like to know what are the best practice when characterising thin membranes and what units are the most used to address such issues. I wouldn't want to use a misleading name that could loose the reader. Diffusion and permeation coefficients seems to be common[1-2] but I would like an external opinion on the matter, especially because I cannot access my membranes' thicknesses.

Footnotes and References
$^a$: even if I could measure the thickness of the membranes at some point, the idea is to build a general model that can be used without having to measure this thickness in real life condition. The only thing that matters for me is the actual gas diffusion rate through this membrane (in cm$^3$.cm$^{-2}$.s$^{-1}$).
$^b$: Standard Temperature and Pressure conditions.

Pasternak et al., Diffusion and Permeation of Oxygen, Nitrogen, Carbon Dioxide, and Nitrogen Dioxide through Polytetrafluoroethylene, 1970, DOI: 10.1021/ma60015a020
Neela et al., Permeability, diffusivity and solubility of carbon dioxide in fluoropolymers: an experimental and modeling study, 2014, DOI: 10.1007/s10965-014-0401-z

Why give it a name at all ? Only D is a property of the membrane's structure/ composition.

@porphyrin Because: (i) I cannot access D because I cannot access the thickness e, so D alone is not an option and (ii) it is convenient to use a name to refer to things in general. In particular if X is my figure of merit, I want to refer to it, and give it an appropriate name.


text:
Can dehalogenation reaction occur through E2 mechanism?
I have seen examples of dehydrohalogenation reaction that occur through E2 mechanism but never came across dehalogenation reaction that occur through E2 mechanism (for example, dehalogenation of 1,2-chloropropane).
Why is that all dehalogenation reaction occur through E1 reaction, if not can someone give a counterexample for it?

Isn't dehydrohalogenation same as dehalogenation?

@NisargBhavsar I thought dehalogenation was removal of 2 halogen atoms

How can 2 halogens be eliminated simultaneously? Is an halogen acidic enough to get this done?

@NisargBhavsar Reaction of 1,2-chloropropane with Zn dust

That's via a radical mechanism and not E1 or E2.

@NisargBhavsar I was taught that $2^{°}$Cl was removed from 1,2-chloropropane to form $ZnCl^-$ and carbocation.

You are suggesting a near-impossible mechanism. See this question's answer for the mechanism - https://chemistry.stackexchange.com/questions/57928/dehalogenation-of-vicinal-dihalides (Ignore the first few lines in the answer about Simmons-Smith reaction)

Please note that in 'E1' and 'E2' there is no need to use MathJax: the E should be upright and the number is not subscripted. If you want to use MathJax (again, not necessary) then you should use $\mathrm{E1}$ and $\mathrm{E2}$. There is also no need to enclose parentheses with math mode.

Dehalogenation of vicinal dihalides can occur through the E2 mechanism. After some digging, I found this article from 1955. The full text is behind a paywall for me, so I cannot cite the specific rate constants, but the general findings are that trans-1,2-dihalocylohexanes undergo iodide-mediated dehalogenation more rapidly than the cis-stereoisomer.

These findings suggest the anti conformation is necessary, which supports the E2 mechanism.

This answer suggests that the final step of the reductive dehalogenation of vicinal dihalides by zinc dust is E2-like.


text:
Why is cuprous oxide a p-type semiconductor?
Wikipedia — Copper(I) oxide says $\ce{Cu2O}$ has semiconducting properties without any explanation.
Nolan [1] talks about $\ce{Cu^+}$ vacancies making copper(I) oxide an extrinsic p-type semiconductor, but provides no explanation how these vacancies (I call them “atomic holes”) travel resulting in electrical conductivity.
I understand how silicone doped with boron results in an electron hole so it's a p-type semiconductor.
But for $\ce{Cu2O}$ the explanation I found is that one $\ce{Cu+}$ ion is oxidized to $\ce{Cu^2+}$ and to maintain electrical neutrality another $\ce{Cu+}$ ion is removed creating a hole which results in electrical conductivity and a p-type conductor is formed.
However, I don't find this explanation convincing.
For $\ce{Si}$ doped with $\ce{B}$ an electron hole was created and I can imagine that hole travelling.
But in $\ce{Cu2O}$ the hole is a vacancy as one $\ce{Cu^+}$ leaves the lattice. Such a big “atomic hole” can't appreciably travel for conducting electricity.
So, how is $\ce{Cu2O}$ an extrinsic p-type semiconductor?
What is really happening at the atomic level, namely:

What are the holes here responsible for p-type semiconducting?
Is $\ce{Cu2O}$ doped with something, since p-type or n-type semiconductors are usually made by introducing a dopant?

Reference

Nolan, M.; Elliott, S. D. The p-type conduction mechanism in $\ce{Cu2O}$: a first principles study. Phys. Chem. Chem. Phys. 2006, 8 (45), 5350. DOI: 10.1039/b611969g.

IMHO $\ce{Cu2O}$ as not fully stoichiometric oxide is doped by $\ce{Cu^2+}$, which acts as a  hole when the extra positive charge is effectively moving among copper atoms.

@Poutnik (a)So the reason i found(see 'My confusion' in my answer above) is wrong . But that same reason was implied in the research paper (though not clarified). (b)please elaborate what u mean by "which acts as a hole when the extra positive charge is effectively moving among copper atoms"...?(c)Is it just an opinion or you have  some sources...just asking...?

I do not claim I am an expert for semiconductors so I can be wrong. // I mean $\ce{Cu^{2+}-Cu^{+}-Cu^{+} ->  Cu^{+}-Cu^{2+}-Cu^{+} -> Cu^{+}-Cu^{+}-Cu^{2+}}$

Ok thanks, but let's  wait for some more inputs from other people....

$\ce{Cu^2+}$ and a $\ce{Cu+}$ vacancy take place of 2 $\ce{Cu+}$ to keep neutrality, but 2+ charge is movable among Cu  1+ charges.

Yes I get u but O2- ions are there also that may produce an hindrance to such charge exchanges.I mean it may be a less contributing reason. But, as wikipedia quotes about Cu2O , "In the history of semiconductor physics, Cu2O is one of the most studied materials, and many experimental semiconductor applications have been demonstrated first in this material" , there should be a more dominant factor at play.

Cu2O acts similar to a p type semiconductor due to the lack of an election, similar as in a standard p type semiconductor. In a standard p type semiconductor, the selection hole is present in the bond pair between B and Si. In this case, it is in the valence shell of Cu ( Cu2+: 3d9; Cu+: 3d10)

The $\ce{Cu^2+}$ ion you call "atomic hole" is responsible for the p-type conductivity. The way to understand is would be a $\ce{Cu+}$ ion with a h+ (hole) associated with it. Conductivity may occur by the hole jumping between neighbouring ions if this association energy can be overcome. Resulting in a temperature-activated hopping process. The hole is moving from copper to copper, not the entire $\ce{Cu^2+}$ ion.


text:
How many equivalents of Grignard reagent does a nitrile group consume?
I was looking at reactions of grignard reagent with certain kinds of compounds and came across it's reaction with nitrile group:

What I was wondering is what if we don't react the Imine formed after (1) and instead add excess of Grignard reagent? Would it consume the second equivalent as well?
I tried looking for something like this but couldn't find it anywhere. They added water after the Imine was formed to give the ketone.  I read somewhere that it might depend on the Grignard reagent, etc.
But if we consider a suitable one, would it be possible for the nitrile group to consume 2 equivalents of Grignard reagent?

A second eq of Grignard does not react with the imine-magnesium salt.

@Waylander how can we reason this out because there is still a pi bond between C and N right? Is it because of stability reasons? The negative formal charge on N is stabilised by (+)MgX ?

"Ketimines are, in general, less reactive than aldimines because of their low electrophilicity and the increased steric hindrance to nucleophilic attack on the C=N bond."  according to this reference  https://www.sciencedirect.com/topics/chemistry/ketimine

In my experience you have to activate ketimines to get organometallic reagents to add to them.


text:
what does Area(A) in Randles-sevcik equation refer to? geometrical area of electrode or electrochemical active surface area/
In the Randles-Sevcik equation,

$$i_p = 0.4463 \ nFAC \left(\frac{nFvD}{RT}\right)^{\frac{1}{2}}$$

we have an area $A$ in the formula. is this the electrochemically active area or geometrical area? I have seen in an article that they consider $A$ as both the geometrical area of the electrode and the electrochemical area. which one is true?

It is area in contact with the solution.

@EdV: I think it is also tacitly assumed that we are talking about one face of the electrode? In conductometry electrodes, one face is always buried or hidden so the area is one face of the electrode but what about CV?

@EdV Thanks. Looks great. I have seen the Hg drop ones but not the solid electrodes you have.

If I recall correctly, the area of electrode is determined with ferrocynanide standards, right? I have seen glassy carbon but I am sure carbon is bad boy in terms of reproducibility. The guy who invented carbon electrodes (Ralph Adams) had contaminated his lab air with carbon paritculates so badly that instrument would spark and die. Carbon was everywhere.


text:
Determining the leaving group for substituted benzene in Aromatic nucleophilic substitution

I could figure out that After the substitution of one of the halogens, with the $\ce{-OCH3}$ group(Electron donating group), the ring would be deactivated towards nucleophilic substitution hence the second halogen won't be substituted (option  (c)).
But I'm having problem deciding between (a) and (b). (a) seems right since Cl, is indeed a better leaving group but then (b) also seems right since $\ce{NO2}$ is para to the Fluorine.
How do we determine which Halogen would be lost?

Isn’t F a better LG then Cl in context of $\ce{S_NAr}$?

SNAr can take place only at para and ortho position of a strong electron withdrawing group

@Rishi oh yes that completely ran out of my mind. Thanks!

@PrajwalTiwari You can answer your own question. Go for it!


text:
Magnetic behaviour of potassium permanganate
I was going through the properties of potassium permanganate and found this-

It has two physical properties of considerable interest: its intense color and  its  diamagnetism  along  with  temperature-dependent  weak paramagnetism.

My teacher said that the intense color is due to the charge transfer from oxygen to manganese, but didn't said the reason for temperature dependent paramagnetism.
So, I thought for a while and thought about the Curie's Law.

Magnetization of a paramagnetic material is inversely proportional to the absolute temperature T
$M$ = $C$ $\frac{B_0}{T}$

Is this the correct reason for it? If not what is the reason?

Curie's law applies when the species has unpaired electrons - which there are none in KMnO4, so it can't be that. Rather the weak paramagnetism is normally explained by the Van Vleck mechanism, which requires a low lying excited state - and we know there is such a state due to the purple colour of the ion.

Related question for pot. dichromate: https://chemistry.stackexchange.com/questions/115841/magnetic-properties-of-potassium-dichromate


text:
Finding the molecule with the smallest bond angle in the following lists of elements

For each of the following lists of elements, find (and give a justification for your choice) the molecule with the smallest bond angle:
a) $\ce{OSF_2, OSCl_2, OSBr_2}$
b) $\ce{SbCl_3, SbBr_3, SbI_3}$
c) $\ce{PI_3, AsI_3, SbI_3}$

Here is what I think:
a) $\ce{OSF_2}$ will have the smallest bond angle because $\ce{F}$ is more electronegative and it will attract more pairs of electrons to it.
b) $\ce{SbCl_3}$ with the same reasoning.
c) $\ce{PI_3}$ has the smallest bond angle because it is more electronegative.
Is my reasoning correct?

Some of your answers are correct, but your reasoning isn't. Electronegative atoms don't attract lone electron pairs or the pairs that form bonds with other atoms.

@IvanNeretin Thank you! What is the correct reasoning then? How should I think about it?

All three sets of compounds can be explained using the Bent's Rule which states

Atomic s character concentrates in orbitals directed toward electropositive substituents.

Now, let's proceed case by case here.
Case 1: Substituent atom changes, Central atom remains same
In this case, as the electronegativity of the substituent increases, the p-character of the bond increases, which reduces the bond angle.
A case similar to your first case has already been discussed on the site. I would urge you to look at this question and its accepted answer. The only difference is that the thionyl group has been substituted for a carbonyl group.
Case 2: Central atom changes, the substituents remain the same
In this case, (your case 3), we are asked to compare between three molecules where the central atom becomes more electropositive as we go through the compounds.
Due to this increase in atomic s-character in the central atom, the p-character in the bond increases, which leads to a decrease in bond angle. Therefore $\ce{PI3}$ would have the largest bond angle in your third set of questions.
Also see:  Why does a larger bond angle correspond to greater s-character?


text:
Isomers of a Cycloalkane
What I know:

All alkanes are saturated hydrocarbons with the general formula, $C_nH_{2n+2}$
All alkenes are unsaturated hydrocarbons with the general formula, $C_nH_{2n}$

What's Confusing Me:

This is, if I am right, is called a "cyclobutane"; when I look back to the definitions I have, this compound seems to be bending them slightly. It's an alkane, which means that it is supposed to be saturated and be in the form $C_nH_{2n+2}$.

Saturated? Yes
Same general formula? No, this is $C_4H_8$, which means that it's an alkene?

Which homologous series does this belong to? Alkanes or alkenes? What is wrong with the "definitions" that I have?
If a were to draw its isomers, which ones would I draw?

Cycloalkanes do not follow the general formula for alkanes. They follow the general formula of alkenes because they have a single hydrogen deficiency like alkenes but it is not on the form of a double bond.

Why exactly would you think all hydrocarbons are either alkanes or alkenes? How about cycloalkanes, alkynes, dienes, arenes, etc. etc.

So cycloalkanes are in an entirely different group/homologous series?

Let me bring clarity to the matter.
All hydrocarbons are divided into saturated and unsaturated. In turn, each of them is divided into aliphatic and cyclic hydrocarbons.
So, alkanes and cycloalkanes refer to saturated hydrocarbons, BUT: alkanes are aliphatic and cycloalkanes are cyclic. Alkenes refer to unsaturated aliphatic hydrocarbons. They are unsaturated because they have double bond in their structure.
It is very important to understand that cycloalkanes and alkenes are just isomers due to having same general formula $\ce{C_{n}H_{2n}}$. But it doesn't mean that they are both unsaturated!
The general formula doesn't determine structure of compounds, but only the quantitative and qualitative compositions of their molecules.

I'm not so sure your definition of "unsaturation" is widely used.  See for examplle https://en.wikipedia.org/wiki/Degree_of_unsaturation , where it is clear that cyclohexane and 1-hexene would both have the same degree of "unsaturation", that is, that cyclohexane would be regarded as unsaturated.  I'd recommend revising this answer to make that clear.

According to Wikipedia, cycloalkanes are saturated hydrocarbons. For explaining this particular matter, they can be considered as saturated.
Link: https://en.wikipedia.org/wiki/Cycloalkane

What is wrong with the "definitions" that I have?
Yes, you are right. Employing common sense,
All alkenes are

unsaturated
hydrocarbons
the general formula, $\ce{C_nH_{2n}}$

Cycloalkane

is a hydrocarbon
is saturated
has the general formula, $\ce{C_{n}H_{2n}}$

All alkanes are

saturated
hydrocarbons
general formula, $\ce{C_nH_{2n+2}}$

There are three entities. If you will, the suffix -ane means saturation.

The formula $\ce{C_{n}H_{2n+2}}$ is useful only for saturated, aliphatic alkanes. Not for cycloalkanes.
Cycloalkanes belong to the homologous series with general formula $\ce{C_{n}H_{2n}}$.
For example, the following is a straight chain alkane:

Here, the two terminal alkanes have 3 bonded hydrogens and 1 bond with carbon, while the other intermediary carbons have 2 bonded hydrogens and 2 bonds with adjacent carbons.
On the other hand, here is cyclobutane:
]1
In this compound, each carbon has 2 bonded hydrogens and 2 bonds with adjacent carbons.
Thus, overall, any straight chain alkane (and even branched alkanes) has two hydrogens more than the corresponding cycloalkane with the same number of carbons. Both of them are saturated.
Hence they belong to different homologous series with different general formulae.


text:
What are the products when KCl and H2SO4 reacts?
I have been trying to find the right answer by googling and I just ended up finding multiple answers.
Ans #1: $\ce{2KCl + H2SO4 -> K2SO4 + 2HCl}$
Ans #2: $\ce{KCl + H2SO4 -> KHSO4 + HCl}$
Can someone help me?

Both of these are correct. It depends on how much KCl you add to the H2SO4.

The reaction between $\ce{KCl}$ and $\ce{H2SO4}$ depends from the temperature, and of the presence of water.
First both reactions need a total absence of water. In the presence of water, no $\ce{HCl}$ is liberated.
Second at room temperature, the reaction happens only at the surface of $\ce{KCl}$. So it is inefficient, because the product of the reaction prevents some further contact between $\ce{H2SO4}$ and the rest of  $\ce{KCl}$. If the temperature is near $210°C$ (melting point of $\ce{KHSO4}$), the reaction $(2)$ is efficient and produces $\ce{KHSO4 + HCl}$, as $\ce{KHSO4}$ is a liquid, and the reaction can proceed fast.
The first reaction must be made in two steps. First the mixture $\ce{KCl + H2SO4}$ must be heated at $210°C$, and then, when the first $\ce{HCl}$ is finished, $\ce{KHSO4}$ and $\ce{KCl}$ can be overheated at a higher temperature, near $1000°C$. And only then will $\ce{K2SO4 + HCl}$ be produced according to $(1)$.
In summary, the reaction $(1)$ is the sum of ($2)$ (occurring at ~$210°C$) plus the following reaction, happening at ~$1000°C$ :
$$\ce{KHSO4 + KCl -> K2SO4 + HCl}$$


text:
What is the rate constant k for the reaction between potassium manganate and oxalic acid?
For my school's chemistry classes, I am investigating the kinetics of the redox reaction between $\ce{K2MnO4}$ and $\ce{C2H2O4}$. I analyzed the data using a colorimeter set to $\pu{565nm}$. underneath are my results:

This picture shows the different results I obtained at different temperatures. Since absorption is directly related to concentration, I tried to take $\mathrm{ln(a)}$ over time and $\frac{1}{a}$ over time to determine the order of reaction and thus the rate constant k (by using the gradient of the straight line. however, none of the above resulted in a straight line as shown below.

As you can see, neither of these lines are straight and therefore I cannot find rate constant k. Is it because the reaction is auto-catalytic? If so, can someone please explain how to find the rate constant?

Are you sure that you have used potassium manganate $\ce{K2MnO4}$ and oxalic acid ? Because potassium manganate is a green substance that is stable in basic solution, and oxalic acid does not exist in basic solution : it is transformed in oxalate ion. So I suspect you have worked in acidic solution with potassium permanganate $\ce{KMnO4}$ which is purple in solution.

What is series $1$, series $2$, and series $3 $? What do you report in figure $2$ and $3$ ? Versus what ?

@Maurice I used manganate, a green substance which I then dissolved in water for a 0.001M purple solution. then I added 0.1M oxalic acid and placed it in a colorimeter.

@Maurice the series 1, 2 and 3 can be ignored, it's just excel placing it in. and for the second chart it is ln(absorbance) vs. time and the third is 1/absorbance vs. time.

Why did you choose potassium manganate ? it is not a good choice. Because, even without oxalic acid, potassium manganate gets slowly decomposed in water, due to the acidity of atmospheric $\ce{CO2}$ producing $\ce{MnO2 + KMnO4}$ $$\ce{3 K2MnO4 + 2 CO2  -> 2 KMnO4 + MnO2 + 2 K2CO3}$$

@Maurice Sounds about right, but I don't see how this is a problem. The reaction can also be carried out with permanganate just fine. My question is how to find the rate constant for the reaction shown in the first image. To clarify, I am studying the effects of autocatalysis on rate of reaction, and I want to use the Arrhenius equation for find the activation energy, but I have no clue how to find rate constant k.

Let us continue this discussion in chat.


text:
How many bonds does oxygen really make in minerals?
I'm studying some metallurgy things and interested in iron oxides.
The simple 2-D representation of a mineral like hematite shows $\ce{O=Fe-O-Fe=O}$.

But the crystal structure seems to show each oxygen making 4 bonds. Crystalline structure from Virtual museum of molecules and minerals[1]

So whats the deal here?
Reference:
[1]: Hematite – virtual museum of molecules and minerals
https://virtual-museum.soils.wisc.edu/display/hematite/
(accessed Jun 16, 2021).

This is a human.
Well, sort of.
All right, call it a simplified representation. But at least it shows the arms/legs ratio accurately, you can't deny that.

Now that was about as good as the molecular formula of $\ce{Fe2O3}$ (a compound which has no molecules in the first place):

As for the crystal structure image, that's what $\ce{Fe2O3}$ actually is. The sticks are drawn where the two atoms touch. Call them bonds or otherwise, the Nature does not care.
When you learn about the coordination compounds, the idea of atoms making far more bonds than their formal valence allows won't look that weird.
So it goes.

thanks for the answer. As im interested in the redox reactions taking place in these crystals with CO and H2 reduction, I'm really curious about the electrons being shared.

Reactions don't take place in the crystals. On the surface, maybe.

@NilayGhosh Still not as cursed as this $\ce{O=Fe-O-Fe=O}$ abomination.

There is no relation between the formula of a substance and the number of neighbors any given atom has in its vicinity. For example, each sodium cation $\ce{Na+}$ is surrounded by $6$ $\ce{Cl-}$ ions in the crystal of salt, although its formula is $\ce{NaCl}$. Each sodium atom has $6$ neighbors, but it stays at oxidation number $+1$.

So the crystal structure is not showing bonds, what is it showing?

It is showing the local nearest neighbours for each atom. Don't think of ionic bonding as occurring pairwise. As coulomb interactions are very long ranged an ion sits in the electric potential due to the whole of the lattice, and so is interacting with every other ion in the lattice. A bonds to B in such a situation is just not a good model.


text:
Why do we need salt bridge in a cell during potentiometric measurement as there is no or minimum redox activity at electrodes?
In electrochemical cell, we need salt bridge to maintain electrical neutrility in both halves cells during redox reaction?. But in during potentiometric measurement, no or minimum current is drawn and redix activity at electrode is negligible then why do still need salt bridge in this case?

Still you do not want change in potential due to resistance and you do need a full circiut. Recall Ohm's law V=IR.

Then how does salt bridge help to minimize this resistance as it's function is only to minimize junction potential and maintain electrical neutrility??


text:
What is 'absorbance power' in electrochemistry?
I have been asked

$\ce{250 mL}$ solution of unknown molarity of $\ce{CuSO4}$ in water was electrolysed using platinum electrodes. After the passage of a current $2 \times 10^{-3}$ ampere for $19.3$ minutes, the absorbance power of $\ce{CuSO4(\textrm{aq.})}$ was reduced to $50$% of its original value. What was the molarity of $\ce{CuSO4}$?

My questions are the following:

What is absorbance power?
What does "the absorbance power of $\ce{CuSO4(\textrm{aq.})}$ was reduced to $50$% of its original value" mean in the context of this problem?

Note: I am not asking for the solution to this problem.

Absorbance is directly proportional to concentration. To know what is absorbance read this: https://chem.libretexts.org/@go/page/3747

There is no such thing as “absorbance power”, except, perhaps, for diapers. Whomever set the problem meant absorbance.

What is absorbance power?

Sadly there is nothing called absorbance power. The question setter might have made an mistake by writing absorbance power when they meant absorbance.
Absorbance is the ability of an solution to absorb radiations of a particular wavelength. It is defined as the logarithm of the ratio of the initial and final flux(intensity if container is symmetrical) of the radiation.
$$A=\log \frac{I_0}{I}$$

What does "the absorbance power of $\ce{CuSO4_{(aq)}}$ was reduced to 50% of its original value" mean in the context of this problem?

Absorbance of an solution is directly proportional to it's concentration. Thus by saying that the absorbance of the solution was reduced to half, the question gives you information that the concentration of the copper sulphate solution became half of the original after the current was passed for 19.3 minutes.

Can the downvoter please mention the reason for the downvote? I am always open for improvements! :)


text:
what does it mean to say oxygen in a given context?
My textbook "chemistry by Julia" says the following

When it comes to expressing the molar mass of elements such as oxygen and hydrogen,
we have to be careful to specify what form of the element we mean. For instance, the element oxygen exists predominantly as diatomic molecules (O2). Thus, if we say 1 mole of oxygen and by oxygen we mean O2, the molecular mass is $\pu{32.00 amu}$ and the molar mass is $\pu{32.00 g}$. If on the other hand we mean a mole of atomic oxygen ($\ce{O}$), then the molar mass is only $\pu{16.00 g}$, which is numerically equal to the atomic mass of $\ce{O}$ ($\pu{16.00 amu}$). You should
be able to tell from the context which form of an element is intended.

and then it gives examples like the following:
How many moles of oxygen react with 2 moles of hydrogen to produce water?
here the book says it is $\ce{O2}$ which is true from the chemical equation.
another example says Air is approximately 21% oxygen.
and the meaning of oxygen here is again $\ce{O2}$ which is easy to deduce since oxygen found in nature is mostly $\ce{O2}$.
The last example I could not tell why and here it is:
Many organic compounds contain oxygen.
the meaning here is $\ce{O}$ a single atom, but why?
and how to tell from a certain context what oxygen means, whether $\ce{O}$ or $\ce{O2}$?

You can find useful  Notation basics and How can I format math/chemistry expressions on Chemistry SE.
See also upright vs italic and Math SE MathJax tutorial.

It can be ambiguous unless you use the explicit word molecular or atomic to qualify what type of oxygen you mean. Life can be full of ambiguities and you have to read between the lines sometimes and find what makes more sense within a given context. Such problems with usage crop up often in popular science. Chemists are usually careful with nomenclature to make sure it reflects how oxygen is bonded. Oxygen in molecules is usually present alone, that is, not bonded to another oxygen, except of course in molecular oxygen and peroxides.

Please note that the old atomic mass unit (amu), which referred to the relative atomic mass of oxygen, became obsolete in 1961 and is no longer used in chemistry.

Look at the formula of the compound containing oxygen. In some cases it's obvious, e.g. a mole of $\ce{O2}$ in the air. In other cases, you'd need to know how oxygen is liberated. For example, one mole of hydrogen peroxide, $\ce{H2O2}$, liberates only one half mole of gaseous oxygen: $\ce{H2O2 -> H2O + 1/2 O2}$ or better, $\ce{2H2O2 -> 2H2O + O2}$.
Er, umm... why write this as $\ce{1/2 O2}$, rather than one $\ce{O}$? The answer is because though the oxygen might be liberated in a very reactive atomic, rather than molecular, form, as in "piranha solution", it rapidly links together as molecular $\ce{O2}$.
Yes, it can be confusing, since there are a number oxygen allotropes. Aside from $\ce{O2}$, or dioxygen, which makes up about 1/5 of the air, there is $\ce{O3}$, ozone, and  solid oxygen may contain $\ce{O8}$. And, as Herr Lehrer states, "There may be many others, but they haven't been discavard".


text:
Why isn't canned food good forever?
So everything inside the can has been killed and nothing can get in. So why doesn't it last forever then?

https://www.today.com/food/when-do-canned-foods-really-expire-t119977

This is a good and interesting question! But this is probably the wrong site for it. It might be on-topic in Biology SE or Cooking SE and somewhat surprisingly in Space Exploration SE where there is a keen interest on keeping prepared food appetizing and delicious for months and in the future for years!

Not an answer, but food is full of all kinds of chemicals and they can interact with each other, and simply degrade over time. It doesn't become "gone bad" from bacteria, it does it all by itself.

Canned food slowly approaches its chemical equilibrium with itself and the walls, what state may not be as attractive,  tasty and healthy as the original state.

Why do you think nothing gets in? All things leak, the only question is how fast…

Consider that molecules can degrade themselves. You do not need food, as every chemist knows. It is just matter of time. If you are concerned with bacteriological danger, a well sealed beans can likely last as long as the can do. But that doesn't mean that the nutritional value or the palatability of the beans do not decay at all.

TL;DR Corrosion is the main culprit. USDA says low acidity canned foods may last up to 5 years, high acidity up to a year, because cans corrode with time, especially at higher temperatures.
An answer to this could range from the trivially philosophical ("nothing lasts forever") to challenging the premises of the question (who says everything has been killed/spores/nothing needs to get in because oxygen and water/does lasting mean being edible?).
It is however interesting that foods can be treated and packaged to retain near to original properties after extended time, and why this time can vary. The US Department of Agriculture -- in what would seem to be unnecessary exposure to potential lawsuits -- states on its official website:

Most shelf-stable foods are safe indefinitely. In fact, canned goods will last for years, as long as the can itself is in good condition (no rust, dents, or swelling). Packaged foods (cereal, pasta, cookies) will be safe past the ‘best by’ date, although they may eventually become stale or develop an off flavor. You’ll know when you open the package if the food has lost quality. Many dates on foods refer to quality, not safety. See FSIS’ Shelf-Stable Food Safety fact sheet for more information.

Avoiding the question of what is "shelf-stable", which sounds like a circular and not very useful definition (it is when it is), one can click on from the above website and come across further information from the USDA:

Will commercially canned foods last forever?
Commercial canning is done under tightly controlled conditions — careful sanitation and the necessary time and temperature under pressure, but there are still limits to how long it will preserve food. There are several factors that limit the shelf life of canned foods. First, cans can rust over time. Shipping accidents, where cans fall and dent or are crushed, also cause container problems.
Then there's can corrosion. In all foods, but especially in high-acid foods like canned tomatoes, natural chemicals in the food continually react with the container. Over several years, this can cause taste and texture changes, and eventually lower the nutritional value of the food.
High temperatures (over 100 °F) are harmful to canned goods too. The risk of spoilage jumps sharply as storage temperatures rise. In fact, canned goods designed for use in the tropics are specially manufactured.
Store canned foods and other shelf stable products in a cool, dry place. Never put them above or beside the stove, under the sink, in a damp garage or basement, or any place exposed to high or low temperature extremes. Temperatures below 85 °F are best. Check your pantry every few weeks and use canned goods you have had on hand for awhile. Don't purchase bulging, rusted, leaking, or deeply dented cans.

That still seems an incomplete answer. Assuming the can retains its original shape and does not rust, will the food inside last forever? What about the expiration dates, what are those for? On that the above site states:

Except for infant formula and some baby food, product dating — having a "use-by," "sell-by," or "best-if-used-by" date — is not required by Federal regulations. Dating is for quality, not safety. However, if a calendar date is used, it must express both the month and day of the month (and the year, in the case of shelf-stable and frozen products). If a calendar date is shown, immediately adjacent to the date must be a phrase explaining the meaning of that date, such as "sell by" or "use before." While there is no uniform or universally accepted system used for food dating in the United States, dating of some foods is required by more than 20 states. A shelf-stable product can be safely used after the "sell-by" date. Products displaying a "use-by" date, although still safe, may not be of acceptable quality after the "use-by" date.

Quality means that the contents have chemically degraded resulting in loss of textural properties, nutritional value, or flavor. It does not (necessarily) mean that it is unsafe.

I refer here to commercial canned food.


text:
Is "Sulfuric acid... makes a poor electrolyte... very little of it will dissociate into ions" really true? What does that actually mean?
While researching chromate conversion coating for edits to this answer in Space Exploration SE, I came upon the following passage in Corrosion Resistance of Stainless Steel to Sulfuric Acid

Sulfuric acid is quite corrosive in water although it makes a poor electrolyte due to the fact that very little of it will dissociate into ions, according to Chemical Land 21’s description of sulfuric acid. The concentration of the acid is what determines its corrosive effectiveness, as British Stainless Steel Association (BSSA) explains. Most types of stainless steel can resist low or high concentrations, but it will attack the metal at intermediate temperatures. The concentration is affected by temperature.

Wikipedia's Sulfuric_acid; Polarity and conductivity says:

In spite of the viscosity of the acid, the effective conductivities of the $\ce{H3SO4+}$ and $\ce{HSO4-}$  ions are high due to an intramolecular proton-switch mechanism (analogous to the Grotthuss mechanism in water), making sulfuric acid a good conductor of electricity. It is also an excellent solvent for many reactions.

Question: So one source explains that "Sulfuric acid is... a poor electrolyte due to the fact that very little of it will dissociate into ions", and the other seems to suggest that the "effective electrical conductivities" of the resulting ions are high "making sulfuric acid a good conductor of electricity". Is the fact that these appear to disagree to me due to my failure to understand what each means, or is one incorrect or at least incomplete?

What has the last image to do with any of the question? Given that the image source (please cite the actual source with words, because links can just die) has no obvious policy on reuse of their images, I'll simply remove it.  cc @MelanieShebel

Does this answer your question? Conductivity as a function of acid concentration

@Martin-マーチン thanks for your edit; yes I shouldn't have included the image, and in the past I've also addressed "[Source]" myself: Are meaningless link titles like “Source” sub-optimal, leaving the answer unrepairable if a link rots/breaks?

Besides the relative aspect treated by comments and answer.... The two sentences aren't in contrast. The second refers to a mechanism of conduction that isn't available in, just to give an example, Na+ and Cl- solution. Namely, you may think to freeze in place the ions, still current can flow between locations due to multiple proton exchange reactions.

The best known conducting aqueous solutions are that of strong acids in water because the hydronium ion (=protonated water) has the highest electrical conductivity known today. The infinite dilution conductivity of of hydronium ion are compared below from the Book Chapter "Proton Transfer Reactions and Kinetics in Water" by Stillingerenter. The units are cm$^2$/ohm equivalents. As you can see, the best conducting ions in water are the hydronium ion and the hydroxide ion. The "rest" of the ions are nowhere close.
$$
\begin{array}{rllll}
\hline t\left({ }^{\circ} \mathrm{C}\right) & \lambda_{\mathrm{H}^{+}}^{\circ} & \lambda_{\mathrm{OH}^{-}} & \lambda_{\mathrm{Na}^{+}}^{\circ} & \lambda_{\mathrm{Cl}^{-}}^{\circ} \\
\hline 0 & 225 & 105 & 26.5 & 41.0 \\
5 & 250.1 & & 30.3 & 47.5 \\
15 & 300.6 & 165.9 & 39.7 & 61.4 \\
18 & 315 & 175.8 & 42.8 & 66.0 \\
25 & 349.8 & 199.1 & 50.10 & 76.35 \\
35 & 397.0 & 233.0 & 61.5 & 92.2 \\
45 & 441.4 & 267.2 & 73.7 & 108.9 \\
55 & 483.1 & 301.4 & 86.8 & 126.4 \\
100 & 630 & 450 & 145 & 212 \\
\hline
\end{array}
$$
So, when people say good or poor aqueous electrolytes, these are all relative terms. You can choose these above values as your measuring bar. Dilute sulfuric acid in water therefore a very good conductor with respect to other salt solutions because it furnishes "protons" in water.
When it comes to the conductivity pure sulfuric acid, it will be an apple to orange comparison with its aqueous solution. There is no water (or very little), small amount of hydronium ions. The current carrying ions are now different. Look at the conductance (reciprocal of ohms, the old paper labels the y axis as conductivity) as a function of sulfuric acid concentration, it follows a very non-linear behavior and its resistance is relative high as compared to aqueous solution.
Thus the Wikipedia claim "sulfuric acid a good conductor of electricity." is a very relative comparision. Good conductor of electricity as compared to (??). The writer is silent after that! From this graph one can easily deduce that conc. sulfuric is not a very good conductor when compared with its aqueous solutions.
(Ref: Darling, Horace E. "Conductivity of sulfuric acid solutions." Journal of Chemical & Engineering Data 9.3 (1964): 421-426.)

Interesting - why the knee in the graph at 85-92% H2SO4?

No idea...these types of questions have kept chemistry or physics alive. Otherwise these sciences would have been dead by now.

@Vikki-formerlySean that's quite a compelling (and different and more specific and unique) question! Would you consider asking that as a separate question? I think it's worth asking if it's not answered in answers here or in any at the proposed dupe.

@uhoh: Aaaaand asked.

@Vikki just asked: Details of what actually happens in cold sulfuric acid between 80 and 90%; what molecular changes cause the viscosity to skyrocket?

The conductivity might be viewed relative to other electrolytes and by concentration.

By concentration: Up to about 25% $\ce{H2SO4:H2O}$, conductivity increases fairly linearly with concentration, reaches a maximum at ~35%, and then drops precipitately (no pun intended)! And sulfuric acid has such "affinity" for water that oleum, or fuming sulfuric acid, has more $\ce{SO3}$ than $\ce{H2O}$. Oleum is much less strongly ionized, and is a comparatively poor conductor.
In comparison to hydrochloric acid: At concentrations less than 25%,, $\ce{HCl}$ is the better conductor, but above that, $\ce{H2SO4}$ is better, up to ~50%. Above 50%, as mentioned, $\ce{H2SO4}$ conductivity decreases.

However, when you start a car, consider that a battery weighing perhaps 20 kg, with about seven liters of 35% $\ce{H2SO4}$, provides between 100 and 200 amps to turn over a four or six cylinder Otto-cycle engine... so $\ce{H2SO4}$ must be at least a fairly good conductor.

if you didn't mean the pun "precipitously" (like a precipice )

The apparent contradiction is due to the fact that none are absolute statements and to a misunderstanding of the term electrolyte.
The dissociation of sulphuric acid is concentration dependent, so more ions are in solutions at lower concentration.
Beside this, the second statement refers to a mechanism of conduction that is not general for ions in solution, namely proton exchange between anions. This intermolecular process, absent in, e.g. a NaCl solution, is an additional channel for the current flow. You can hypothetically think of not moving anions, still current can flow by the above protons exchange.
This nevertheless requires enough ions and a relative low distance between them.
As a result, the conductivity peaks at medium-low concentration, as in the graph you have attached.
In fact, a substance can be a relatively poor electrolyte in the sense that its dissociation is far then complete. However, as in the current case, its conductivity can be relatively high, at least higher than expected.


text:
Photobromination of cyclohexene
During bromination of cyclohexene in the presence of hv via free radical mechanism, there are two allylic carbons. Why does the major product have only one Br an not a Br on either side of double bond.

There are way more $\cdot$H than there are $\cdot$Br, This makes the major product the monobrominated one.

Depends on how much Bromine is used. Use an excess and you get di-bromo product

It is also worth noting that for $\ce{Br2}$, electrophilic addition to the pi bond is much faster than the allylic radical bromination. Dropwise addition of $\ce{Br2}$ to cyclohexene at room temperature in the near dark results in almost instantaneous decolorization of  bromine.

Thank you so much!!!. May i know why my post was downvoted ?


text:
How to model a molecule of benzene surrounded by 100 molecules of water with molecular dynamics?
I have experience calculating molecular properties (mostly DFT). However, I would like to spend some time on molecular dynamics methods. I would like to ask your advice to direct me towards a program with which to calculate systems such as the annealing of a benzene molecule surrounded by a hundred water molecules.

This question from matter modeling SE might be helpful: https://mattermodeling.stackexchange.com/q/6143

Do you mean simulated annealing? You can use Gromacs (https://www.gromacs.org/). Gromacs is quite easy to use. Also have a look at this post: https://mattermodeling.stackexchange.com/questions/4341/finding-parameters-for-a-molecule-in-various-forcefields/4374#4374 I have done MD simulations of small molecules in gromacs, so I can write an answer on that if you want. I haven't done annealing so I can't really write about that.

Nisarg Bhavsar, thanks, I will dive a little bit into NAMD and see if I can get something running, I see there are some tutorials on youtube, for example, https://www.youtube.com/watch?v=IArpsQsZ95U seems good! Shoubhik R Maiti thank you for pointing out Gromacs, also here I could dive into maybe looking at some tutorials will be useful: https://www.youtube.com/watch?v=rYZ1p5lXNyc&t=1s.

@PaulPerkele In my personal experience, learning gromacs was easier than learning NAMD (in fact right now I am trying to learn that). However, you may find the other one easier. Yes youtube tutorials are helpful. Do keep in mind though that for small molecules, you usually get parameters/topology from servers like https://www.swissparam.ch/ or from softwares (because you need general forcefield). As an aside, if you are responding to a comment on this site, make sure to tag them with an @