text:
Out of the given reagents, find the ones to be used to convert benzene to p-aminophenol

$\ce{Br2/FeBr3}$
$\ce{CH3Cl/AlCl3}$
$\ce{KMnO4}$
$\ce{HNO3/H2SO4}$
$\ce{NBS}$/hv
$\ce{NaOH}$
$\ce{Sn/HCl}$
$\ce{NH4SH}$
$\ce{NaBH4}$

I'm able to get the amino group onto the ring by using 4 and 7.
But I have no idea how to get $\ce{-OH}$ in para position (also specify the order).

I am not happy with the idea of trying to convert a bromobenzene into a phenol using sodium hydroxide. Normally an aryl halide is very inert to nucelophiles, the nitro group will increase the rate but it will not make much difference.
What I would do is to do the following

Nitrate benzene with a mixture of nitric and sulfuric acids to form nitrobenzene
Reduce the nitrobenzene with tin in acid.
To do the synthesis I would want to use nitrous acid and allow the diazonium salt to decompose forming the phenol.
I would nitrate under very gentle conditions to form a mixture of nitrophenols. After separation I would then go to the next step.
I would reduce again with tin in hydrochloric acid to form the target molecule.

I think that a fault exists in the question, I suspect it is an exam question and that the person who wrote the question did a bad job of the question writing.

I haven't actually seen SNar with hydroxide but I would have no doubt at all that an alkoxide would do it, this is a key step in the synthesis of fluoxetine.

Which step in the synthesis of prozac are you thinking of, I have seen some methods for the synthesis of prozac but I can not recall the route used.

The SNAr addition of a secondary alkoxide to 4-F-trifluoromethylbenzene. I no longer have Scifinder access so I can't look up an example of hydroxide doing SNAr on 4-Br-nitrobenzene

Waylander's answer is also correct.. You can replace Br with OH in brominitrobenzene using NaOH via SNAr mechanism

Please give a reference to the reaction of hydroxide with an aryl halide to give the phenol. I have checked the literature and  Sekar and  Thakur in Chem. Commun .,2011,47,6692–6694 were able to get KOH to react with aryl bromides but they used copper and glucose. The fact that they needed to use a transition metal cataylst suggests that the reaction is not a simple nucelophilic attack. Also see International Edition, 2009, vol. 48, # 5, p. 918 - 921 where Pd was used to do the same reaction, it is clearly not a simple nucleophilic attack.

Also the aryl flouride is a special case, the high electronegativity of F increases the partial positive charge on the carbon bearing the F atom. This then increases the rate of the reaction of the aryl halide with nucleophiles. So I think that the comparison of the reaction of para fluoro-trifluoromethylbenzene is not a helpful thing

@nuclear-chemist chloronitrobenzene to 4-Cl-phenol http://chem-guide.blogspot.com/2010/04/chemical-properties-of-haloarenes.html

Steps:

Use $\ce{Br2/FeBr3}$ to make bromobenzene.
Nitrate with $\ce{HNO3/H2SO4}$ to make 4-bromo-nitrobenzene.
Treat with $\ce{NaOH}$ to displace the $\ce{Br}$ making 4-nitrophenol.
Reduce with $\ce{Sn/HCl}$ to make the required p-aminophenol.


text:
Forging steel inside a zero oxygen environment
Intro: I have watched one too many blacksmithing videos and now I have a question: is it possible to forge steel (either just reshape or "forge-weld") inside a no-oxygen environment?
Scenario: considering the amount of material that turns into iron oxide during the forging process, would it be possible (or plausible) to fill a room with some non reactive noble gas (maybe helium or neon) and proceed to forge inside this room? Or would it be possible to forge in a complete vacuum?
If any of these two options are possible, what are the hindrances and requirements?
Extra info: I am not from the chemistry area and I haven't  studied chemistry in years. I took a look at some old textbooks of mine and also did some research on on Google but did not find anything specific to this question. This is purely out of curiosity, and since I don't know how reddit works (therefore can't do a more relaxed inquiry there) , I thought it was not a bad idea to ask here.

Look at it this way: why bother about iron turning into oxide? Because it means lost money? Well, helium costs money too, and a good deal more than iron. Other than that, sure, why not.

Absolutely true. I just wanted to know if it was possible, or if there are any other possible limitations. Not so much for the loss of material and costs indeed.

Yes it is possible, and in fact it is done when you weld Al or Ti. But with iron it is just not worth the effort.

Removing O2 will yield a steel that is different from one made via traditional methods - see below. Also, you can't have a combustion process delivering the heat for the forging if you do not use O2 at all.
https://www.britannica.com/technology/steel/Basic-oxygen-steelmaking

Yes, but would it be possible to use electric induction to heat up the steel? Would that have any impact on the surrounding gas? (I believe oxigen would ionizer, but I'm not sure either how would that impact the result, nor if it would happen to a noble gas)

Since they sell such devices, it should be possible. http://www.mtixtl.com/15KWSmallVacuumInductionMeltingSystem-EQ-SP-15VIM.aspx

@J.Ari I think that heat could be delivered also by conductive heating.

You can forge Fe2 metals in a low O2 environment. Its going to cost you a vacuum chamber, and an induction forge.

It is possible to forge iron in a low atmosphere environment, but it isn't worth it because there are few if any benefits to doing so.
This isn't true for some other metals, though, or even for some specialist steels. In many other metals and alloys the presence of oxygen or water vapour has a very bad effect on the quality of welds. But the common way to exclude air from these systems is to use an apparatus that flows some inert gas around the spot being welded rather than to fill the room with an inert gas. Some examples are given here.
This is mainly worth doing for many specialist aluminium, copper and titanium alloys where the metals are oxidised more easily than typical irons or steels and effect of impurities in the welds is much larger. It isn't as simple as just having an inert gas blanket though, as this brochure for commonly used gas mixtures shows.

"Stick" welding of steel is still very common; no gas except what is generated by the flux coating . MIG is  common for steel ;  Two methods - solid wire with gas and flux cored without gas.  And for heavy steel weldments - subarc uses flux with no addition of gas.

A forge weld contains oxides, mostly iron, which reduce its mechanical properties. The extent of property loss depends on the amount and orientation of the oxides. Modern steel welding techniques shield the weld puddle from oxygen with fluxes and/or gas ( mostly CO2 ). One could make some expensive equipment to shield a forge weld from oxygen but it would be much more costly than other welding processes. Likewise , forging could be done in a protective atmosphere ; it would be very expensive and give no advantage other than saving a little surface steel. I think some forging operations for titanium use protective atmospheres because it is so expensive and will absorb O , N , and H from air ( at forging temperatures). And the various "blacksmith" programs are entertainment, not science or technology.

I'm going off the top of my head but Im pretty sure that is how CPM (crucible particle metallurgy) steels are made. Basically powdered iron and alloying elements are sprayed into a nitrogen filled cylinder and then put under high heat and pressure.
Edit. Just checked and I wasn't 100% wrong.
The steel is actually pre melted and then made into particles, not just sprayed in as iron and alloys, the cpm process is to distribute carbides and alloys evenly and with a fine grain.

I've wondered about this too. Not from a practical sense, but more of a "that would be kinda cool" perspective.
For iron I'd go with nitrogen. All you'd be doing is flushing out the oxygen. (I'm not sure about other metals) I guess you'd have to use an induction forge to heat your  material. You could just have some kind of basic breathing mask or tube supplying you with air, maybe even one just going outside of your enclosure. You'd just need to get enough oxygenated air to you to breathe comfortably. I don't think you'd need to seal out the nitrogen because it's 78% of our atmosphere anyway and not harmful to us. I would definitely make sure you had a quick and easy method of exit, or a way to just open the enclosure to outside air. Maybe some kind of dead-man switch so that if you became unconscious or incapacitated for any reason, the enclosure would automatically open to outside air. I'd also have that same switch controlling the induction forge so that it would shut off if you tripped the switch. I mean, like others have said, it's probably not worth the cost and effort, but I think it would be interesting to see if it meant you didn't need flux, and if it improved bonds when people forge weld billets from things like bicycle chains.
disclaimer: I'm not a blacksmith or a metallurgist. I'm just commenting because I like the question and I enjoy working it out in my head.

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text:
How do you stop a Pharoah's Serpent reaction?
I am a novelist and working on a book. I have been searching google for an answer to the above question. How does one stop a Pharoah's Serpent reaction? Thanks!

Pharoah's serpent is the exothermic decomposition of Mercury(II) thiocyanate https://en.wikipedia.org/wiki/Mercury%28II%29_thiocyanate.   As it is strongly exothermic, I think it will be pretty hard to stop once initiated.

You don't. $!$

Just put water on it.  A temperature of $\pu{165^\circ C}$ is required to sustain the decomposition. Adding water will remove heat from the reaction and will dissolve a small amount of the mercury thiocyanate. The heat removed by heating and boiling the water will cool the contents below the decomposition temperature.  Adding soap will help the water penetrate into the powder and more quickly stop the reaction.

The water will rapidly vaporise, possibly creating a mercury-bearing aerosol. Bad idea.

I somehow think that a small amount of aerosol is likely not as bad as the additional fumes from a continued reaction.

I'm not sure it's small. Compare it with throwing water on hot oil: bad idea.

Water is denser than oil which causes the splattering of oil.  that mechanism is not available to a dense solid powder.

Can you give a reference that addresses the safety aspect?

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text:
Is alkyne hydration possible without a catalyst?
As the title says, is alkyne hydration possible without a catalyst at any temperature? i.e. will it ever occur within a reasonable amount of time? (excluding acetylene)

Generating an unstabilized vinyl cation doesn't seem that great.

Any temperature? Not pressure?

Does hydroboration count?


text:
Presence of precipitates
Does a cloudy solution indicate presence of precipitates or not?The solution isn't completely opaque neither it is completely clear.

Yes it does. $,$

No it does not always always, if you have very fine gas bubbles in a liquid it can appear to be cloudy.

Cloudiness indicates light scattering by small particles, with size on the order of the wavelength of light and larger. If those small particles are denser than the liquid, they may eventually sink and become the settled precipitate.
However, some small particles are so small and so nearly equivalent in density to the liquid that they do not sink, but remain suspended. Brownian motion helps. Some clays can remain suspended indefinitely; in water treatment plants, FeCl3 is used as a flocculant to bring such turbidity down.
Sometimes intermolecular interactions keep particles suspended. For instance, many surfactants form micelles comparable in size to the wavelength of light and may, in pure form, be considerably more, or less, dense than the liquid, and yet remain suspended indefinitely because they are at least partially hydrophillic.
Centrifugation may help bring down suspended particles (or float them if they are lighter than water).


text:
CAS numbers for salts of molecules
Often times Pharma molecules come as different salts where the Cation changes. Do these have typically the same or different CAS numbers?
e.g. Take Glucosamine Hydrochloride.
https://en.wikipedia.org/wiki/Glucosamine
It is available as as both the Na and K salt. But I cannot seem to find if they all have different CAS numbers.
What I have now is: 3416-24-8
Further complication seems that there is a Glucosamine Hydrochloride as well as a Sulfate. The wikipedia page describes both but I cannot see different CAS numbers. The sulfate itself seems available as either a Sodium or Potassium salt.
Any thoughts?

They are different compounds; why would they have the same CAS number?

@IvanNeretin: Makes sense!

Every different compound has a different CAS number, even if they are enanthiomers when isolated and published (their racemic mixture has another CAS number). No exception! See following examples:
L-(−)-Glyceraldehyde: CAS Number [497-09-6].
D-(+)-Glyceraldehyde: CAS Number [453-17-8].
DL-(+/-)-Glyceraldehyde: CAS Number [56-82-6].
For your question, I found the CAS Number of D-(+)-glucosamine hydrochloride, which is [66-84-2]. The CAS Number [3416-24-8] is for glucosamine (free base).
The CAS Number of D-glucosamine sulfate is [14999-43-0]. Each molecule contains two molecules of glucosamines and one molecule of $\ce{H2SO4}$, the two acedic hydrogens on each of amine groups. Its molecular formula is $\ce{[C6H14NO5]2.SO4}$, giving the formula weight: $\pu{456.42 g/mol}$.
The CAS Number of D-glucosamine sulfate potassium chloride is [216699-44-4]. Each molecule contains two molecules of glucosamines, one molecule of $\ce{H2SO4}$, and two molecules of $\ce{KCl}$. Its molecular formula is $\ce{[C6H14NO5]2.SO4.2KCl}$, giving the formula weight: $\pu{605.52 g/mol}$. However, NIH website listed [1296149-08-0] as the CAS Number of glucosamine sulfate Potassium chloride, which must be issued a fairly recent time. Spectrum Chemical Co. sells the compound as a dietary supplement under the same CAS number.


text:
Balancing the reaction of potassium permanganate and sulfuric acid with hydrogen peroxide
A problem on redox reactions invites the reader to balance the actual reaction from which the redox simplification arose. After some hunting I found the balanced equation and attempted to re-derive the coefficients using:
$$\ce{a H_2SO_4 + b KMnO_4 + cH_2O_2 -> dK_2SO_4 + eMnSO_4 + fH_2O + gO}$$
We can quickly derive some valid relations by looking at each element:
$$
\begin{array}{cll}\\
\ce{H}&\quad a+c &=& f\\
\ce{O}&\quad 4a+4b+2c&=& 4d+4e+f+g\\
\ce{S}&\quad a&=& d+e\\
\ce{K}&\quad b&=& 2d\\
\ce{Mn}&\quad b&=& e
\end{array}
$$
This can be massaged a little but because there are 5 equations and 7 unknowns it seems that the system is underdetermined.
If we let $a=3,~b=2,~c=5,~d=1,~e=2,~f=8,~g=10$ the equations all work.
My question is:

Could we arrive at these values without guessing or actually doing an experiment to measure (say) the oxygen evolved?

FWIW, I notice an old paper - J. Am. Chem. Soc. 1889, 11 (6), 94-98 - of which I can only see the first page seems to take a similar approach and gets two possible equations, maybe for the reason above.

Shouldn't it be $g\ce{O2}$? Nascent oxygen doesn't exist, so is that an error in the reaction product?

@GaurangTandon: I don't think it matters for purposes of balancing, if you look at the paper, the authors put it this way. Just divide by two to get the amount of  oxygen gas.

We could arrive at these values just as well as any other possible values, of which there are plenty. This is a sum of two reactions, and it could be balanced to anything. Related: https://chemistry.stackexchange.com/questions/64202/more-than-one-way-of-balancing-a-chemical-equation/64233#64233

@IvanNeretin:  This seems like it may be an answer to the question, thanks.

https://chemistry.stackexchange.com/questions/63369/multiple-solutions-in-the-balancing-of-redox-reactions was marked as duplicate of that ^

It seems to me that this is simply a redox reaction in the presence of acid.
If you write the 2 half reactions, and balance in the presence of acid (sulfuric acid), you end up with the same answer (with 5 O2 instead of 10 O on the product side).
This seems to be more of a mathematical manipulation rather than a chemistry problem, IMHO.


text:
Whats the symmetric, anti-symmetric and non-symmetric part of the direct product of irreducible representations?
I am trying to understand how to see if a vibrational mode is Jahn-Teller active or not. According to the group theoretical description of the Jahn-Teller effect one needs to check if the symmetric part of the direct product the irreducible representation (=irrep) of the electronic state with itself contains the irrep of the distortional mode in its symmetric part. And this is where I have trouble to understand. I can understand that one can decompose tensors (tensor products) into a symmetric and an anti-symmetric part, but I fail to understand how to apply that to the the direct product of two irreps, as they are in most cases one-dimensional. Can someone explain to me what exactly means "symmetric, anti-symmetric and non-symmetric part" in this context and how to see that in specific cases?

When they are one-dimensional, there is nothing to decompose.

@IvanNeretin: OK, so lets take the example E$g$ in D${4h}$.

You should probably find some 'direct product' tables to speed up multiplying symmetry species. The symmetric stretch, say in CO2, is when CO bonds both stretch in phase, anti-symmetric when one stretches as the other compresses, $\pi/2$ out of phase. Anti-symmetric is somewhat subjective in a complex molecule and so this and anything else can be called asymmetric.  (Asymmetric means 'no symmetry' which in not the case as all symmetry species are determined by the point group.)

In $D_{4h}$ product of E with anything but E is E, and $E \times E \to A_1+[A_2]+B_1+B_2$ where [] mean anti-symmetrized product  of a degenerate representation with itself.

My tables tell me: $\mathrm{E_g}\otimes\mathrm{E_g}=\mathrm{A_{1g}}\oplus\mathrm{A_{2g}}\oplus\mathrm{B_{1g}}\oplus\mathrm{B_{2g}}$, while they charcterize $\mathrm{A_{2g}}$ as being "antisymmetric with respect to particle permutation". So whats then symmetric, non-symmetric and whats asymmetric. I am also specifically interested in the "anti-symmetric" part since I want to analyze the system also relativistically (in double group $D_{4h}^*$)

@porphyrin: Yes, and now I want to know, whats the "symmetric part" and whats the "anti-symmetric" part and whats the "non-symmetric" part, how does it relate to this peculiar "antisymmetrized product", how does it relate to the (possibly inversion-anti) symmetry of the stretch you mention and how this is defined exactly? I do also not see how a non-symmetric part can be left over, since decompostion into symmetric and anti-symmetric part of a tensor should leave no rest.

@porphyrin: So what you describe in your CO$_2$ example seems to mean totally symmetric and not totally symmetric, like A$_1$, or similar, but that does not make sense in this context.

There seems to be a description of what you want, re symmetrized and anti - symmetrized direct products in section 5.14 of the 3rd edition of Atkins & Friedman 'Molecular Quantum Mechanics' publ OUP.

@porphyrin: In 5.36 there, shouldn't it rather be $\chi^+(R^2)$, instead of  $\chi^+(R)$?

5.36 is some other equation in my copy. In general the symmetric $\chi^+$ and antisymmetric $\chi^-$ components of a direct product have characters given by the formula character $\chi^+=([\chi(R)]^2+\chi(R^2))/2$ and $\chi^-=([\chi(R)]^2-\chi(R^2))/2$ where $\chi(R)$ is the character for operation $R$ and $\chi(R^2)$ the character for the square of the operation.

@porphyrin: So what would be the "direct product" part here, then?

In  $D_{4h}$ in the product $E\times E$ using $\chi^+$ should produce the $[A_2]$ term given above.

Sure, but where is it in the equation?

I don't think that I can help anymore. To see how it works you will have to work out the direct product by multiplying the characters and then form the irreps from that result.

The symmetric and antisymmetric part of the direct product here refers to the fact that certain irreducible representations in the direct product are symmetric while others are antisymmetric (w.r.t some operations like $\sigma_v$, $C'_2$). So, for instance, when we write:
$$\Pi \times \Pi = \Sigma^+ + [\Sigma^-] + \Delta$$
the $[ \ ]$ simply means that the irreducible representation $\Sigma^-$ is the antisymmetric part of the direct product. It doesn't mean that you are somehow decomposing $\Sigma^-$ into a symmetric and antisymmetric part, and then selecting the antisymmetric one. This won't be possible since $\Sigma^-$ is a one-dimensional representation.

So what are the "defining" group elements then, that have to be checked, is that an arbitrary definition? One might suppose something like the highest order generator element....


text:
Can a π bond form between the py and d(x2-y2) orbitals along the x-axis?

π bond does not form due to overlap between $\ce{d_{x^2-y^2}}$ and $\ce{p_y}$ along x axis? True or false?

Here is a diagram of the orbitals.

This pair of orbitials should be distingished from $\ce{d_{xy}}$ and $\ce{p_y}$, which are shown below.

I think it is true.

may i know why?

Because there is very little chance of overlap. In $d_{x^2 - y^2}$ the lobe oriented along x axis will overall make no overlap with $p_y$ as those are orthogonal, and also the lobe oriented along y axis is much far from $p_y$. If you draw the diagrams it will be more clear. Then you can also see $p_y$ and $d_{xy}$ can form stong $\pi$ -bonds.

When $y$ changes sign, $p_y$ changes sign, but $d_{x^2-y^2}$ doesn't, hence their overlap equals $0$ by parity.

Perhaps you may (or may not) be familiar with a quantity known as the overlap integral, defined as:
$$ S_{AB}= \int_{\text{all space}} \psi_A(r)^*\psi_B(r) d^3r$$
for 2 atoms, A and B.
In simple terms, it is the amount of "overlap" of 2 orbitals of an atom. Its value ranges between $0$ and $1$, implying no overlap and complete overlap respectively.
Now, to see if two orbitals overlap, in your case $\ce{p_y}$ and $\ce{d_{x^2-y^2}}$, one simply needs to compute the quantity defined above. Now, of course this would be very difficult if you don't know the functional form of the two orbitals, and even if you did know the functional form, evaluating the integral would still be messy.
So we perform some mathematically chicanery. If you are familiar with "even" and "odd" functions. An even function times an odd function is odd, and the product of two odd functions is even.
The orbital $p_y$ is odd as it changes sign if $y \rightarrow -y$. However, $\ce{d_{x^2-y^2}}$ doesn't change signs under the same transformation and thus is even. This is evident if you look at the shape/diagram of the orbital.
Thus their product is an odd function, and when evaluating the integral of an odd function over a symmetric interval (and doing so over all space is necessarily a symmetric interval because 3D space is isometric) is necessarily zero.
Simple proof of this mathematical fact is given below:
Consider an integral of the form (this is the same form as the 3D integral given in the first line, wherein your boudns are $-\infty$ and $+\infty$ for each of the three directions $x,y,z$
$$I=\int_{-a}^a f(x)\,\mathrm{d}x$$
$$I=\int_{-a}^a f(x)\,\mathrm{d}x$$
If $\,f$ is odd or even, then sometimes you can make this simpler. We can rewrite that integral in the following way:
\begin{align*}
I=\int_{-a}^a f(x)\,\mathrm{d}x
&= \int_{-a}^0 f(x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \\
&= \int_0^a f(-x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x
\end{align*}
For an odd function, we have $f(-x)=-f(x)$, whence
$$I = \int_0^a (-f(x)+f(x))\,\mathrm{d}x = 0$$

The p(y) and the d(x2-y2) orbitials have no net oribitial overlap when the centre of the p(y) orbitial is on the x axis. Where the origin is the centre of the d orbitial.
On the otherhand the p(y) and d(xy) orbitals will have net overlap if the centre of the p(y) orbitial is on the x axis. Again the origin is the centre of the d orbitial.


text:
What exactly is entropy?
I have looked at other questions, videos and articles but  I kept reading disorganization, messiness or energy distribution (probability) or a thermodynamical equilibrium and stuff. What do we mean by disorganization? And some say that it is a wrong definition. Also, there is this spontaneity and what is its relation to entropy? I am very confused and I would be very pleased if anyone explained it.

By disorganization we mean a huge number of possible states ($\Omega$), and entropy equals $k\ln\Omega$.

To add a little to what @Ivan Neretin has written, words such as 'disorganisation, randomness' etc are not helpful and are best forgotten; from memory they arise from a mis-interpretation of a throwaway comment Boltzmann made.  If a system at equilibrium  has an energy $E$ to $E+\delta E$ and $\Omega$ accessible states then entropy, as first  defined by Boltzmann, is uniquely defined by $S\equiv k_B\ln(\Omega)$. In a quantum system the number of states is completely determined.

One devotes the entirety of a course on thermodynamics to understand what entropy is and does. I'm afraid there is not space here to answer this question, although it is an important one.

Possible duplicate of Entropy - "Wiggle"?

Entropy has two general precise definitions.  That which is most useful here is from Boltzmann, S(E) = k ln W(E) where S is the entropy, E is the energy of an isolated system (which is the easiest system to consider), k is Boltzmann's constant, and W(E) is the number of states of the system that have energy E.
For example, if the system has exactly one state with energy E, then S = 0.  If there are 2 states with energy E, then S = k ln 2.  And so on.
Now "organization" has no precise definition, but is rather an descriptive label we might apply to categorize states of the system.  Some of those states look "organized" to us, some look "disorganized."  For example, consider a model of atoms adsorbed on a crystal surface.  Let's model the surface as a square 3x3 array of sites and put 5 atoms on it.  A priori, there are 9!/(9-5)!4! = 126 ways of putting 5 atoms on 9 sites.  Now which of these states are "organized?"  That's to some extent a matter of opinion, but a common definition is "even" or "symmetrical."  So, for example, the state where there's an atom on every other site, in a checkerboard pattern, looks "organized" -- as if the sites formed a 2D crystal.  There's only one such state.
So now let us supposed we observe  this system, and we have some way of knowing, either from measurement or method of preparation, that it is in an "organized" (symmetric, checkerboard-like) or "unorganized" state.  There is only one "organized" state, so if we know the system is in an organized state then its entropy is k ln 1 = 0.  On the other hand, if we know the system is in an "unorganized" state there are 125 possible states it could be in, so its entropy is k ln 125, higher.
This is generally the case.  States we are willing to call "organized" generally have severe restrictions on the degrees of freedom: their values are distributed symmetrically in some way, e.g. the atoms are distributed uniformly, or in a regular array, the velocities are all the same, and so forth.  Because of these restrictions, it is almost always the case that there are fewer states of the system that satisfy the restrictions than that don't.  So the entropy of the system when it's known to be "organized" is less than when it is known to be not.
In short, it might help if when you hear "organized" you mentally translate that to "has some symmetry restrictions on how the degrees of freedom (e.g. atoms) are arranged."  Hopefully it is intuitive that there are far fewer ways of arranging atoms symmetrically than not, and therefore that the entropy of systems known to be symmetric is lower than the entropy of systems that are not.

Your discussion about 'organised' and 'symmetry' can be misleading. Entropy is only about counting the number of accessible states at a given energy. If the system has atoms on alternate sites only (as you describe) then this state has the lowest energy and no other sites can be reached. Entropy zero. If it is 'disorganised' as you call it then the energy must be higher and more states are occupied. This is a different situation as the energy is greater. As the energy is greater  more states are occupied (including alternate sites) and so  the entropy is greater. Both cases cannot co-exist.

I made the normal and physically realistic assumption that the adsorption energy is the same for every site.  Hence, an "organized" state with 5 atoms on a certain set of 5 sites has the exact same energy (let us say -5e, with e some per-site adsorption energy) as a "disorganized" state with 5 atoms on any one set of a large collection of sets of 5 sites.

Your energy is in effect the enthalpy, the free energy G is what is important in determining what the state of the system will be and this depends on the entropy. Your regular pattern will be at higher free energy than a disorganised one and so will not be at equilibrium. Only by lowering the temperature can the regular pattern state become the most stable.

Remember we're working in the microcanonical ensemble here, because I explicitly said the system is isolated.  So working out the thermodynamic potentials is less useful than it usually is.  Nevertheless, since there's no meaningful pV terms, you're right that H is E in this case, and for both kinds of states H is the same (since E is the same).  S when the system is known to be "disorganized" is higher, as I said, so G for that situation is indeed lower, just what you'd expect from the general rule that G should be minimized for the macroscopic state of highest entropy for a fixed H.


text:
Why is hexacyanoferrate(II) colored and at the same time diamagnetic?
I am wondering why $\ce{[Fe(CN)6]^4-}$ is colored and at the same time diamagnetic. For a transition metal to be diamagnetic all the electrons should be paired up in the d orbitals and at the same time a transition metal is colored because certain wavelength of a white light is absorbed during the electron movement to a higher energy d orbital.
With regard to this, I see a paradox. Since $\ce{[Fe(CN)6]^4-}$ is colored, it means that electron movement happened. If there has been an electron movement, not all the electrons of $\ce{[Fe(CN)6]^4-}$ are paired up
However, why is $\ce{[Fe(CN)6]^4-}$ still considered diamagnetic?

Have you heard about charge transfer phenomena (Ligand-Metal, Metal-ligand, Inter Valence etc.) ?

Color does not indicate the presence of unpaired electrons by itself.

The ground state ion is $d^6$ and low spin which means that there are 3 pairs of electrons in the $t_{2g}$ orbitals. On absorbing a photon an electron gets promoted to the $e_g$ orbital: this gives the compound its colour. This excited state species has unpaired electrons but like all excited states is relatively short lived and will decay back to the ground state. In any sample the number of excited state is tiny compared to the ground state so we can simply ignore the number of excited states.


text:
Why isn't ammonium chromate called diammonium chromate?
I have to name $\ce{(NH_4)_2CrO_4}$. I know the answer is ammonium chromate. But I don't understand why it isn't called diammonium chromate.

https://chemistry.stackexchange.com/questions/14141/why-is-cacl2-called-calcium-chloride

https://chemistry.stackexchange.com/questions/69400/why-is-alcl3-is-known-as-aluminium-chloride-while-pcl3-is-known-as-phosphrus-tri?noredirect=1&lq=1 is also a parallel question. The answer in there is valid here as well.

You don't need the prefix because you know (or rather are assumed to know) the charges on the constituent ions of an ionic compound, which this is. The ammonium ($\ce{NH4^+}$) cation has a charge of +1 and the chromate ($\ce{CrO4^{2-}}$) anion has a charge of -2. Since the formula unit has to have a net charge of zero, it follows that "ammonium chromate" must have the empirical formula of ($\ce{{(NH_4^+)}_2CrO4^{2-}}$). Similarly "ammonium nitrate" must have an empirical formula of $\ce{NH4NO3}$ because the nitrate anion has a charge of -1.  In both cases, you can deduce the formula from the name and your pre-existing knowledge of the properties of the ions.
The prefixes are used much more often in simple molecular compounds because the elements in periods 2 and 3 can often be in multiple oxidation states, and you need the prefixes to figure out the actual molecular formula, the names of the elements are not enough. The most famous example is probably nitrogen and oxygen, which form at least half a dozen neutral compounds. You need prefixes so you can distinguish between $\ce{NO2}$ (nitrogen dioxide) and $\ce{N2O}$ (dinitrogen monoxide).  "Nitrogen oxide" wouldn't cut it, because you wouldn't know the oxidation state of $\ce{N}$ and $\ce{O}$ and could not deduce the formula from the name.
Even when the constituents of ionic compounds can form more than one ion, e.g. for transition metals, we usually choose to write the oxidation state of the varying ion instead of using prefixes, e.g. $\ce{FeO}$ is "iron(II) oxide" instead of "iron monoxide" and $\ce{Fe2O3}$ is "iron(III) oxide" instead of "diiron trioxide."
You may observe that this is all shot through with logical inconsistency and redundancy. That is a reflection of the chaotic history of chemical nomenclature and the need to balance systematic naming conventions to make it easier for the newcomer with the ability to use the vast existing knowledge base that uses older and more ad-hoc or individually varying schemes.

"You may observe that this is all shot through with logical inconsistency and redundancy."
How so?

Well, the first that comes to mind is the fact that when naming ionic compounds we give the oxidation states explicitly in the name and ask the reader to deduce the subscripts ("iron(III) oxide"
=> ox states of +3,-2 => subscripts 2,3) whereas when we name molecular compounds we do the reverse, we indicate the subscripts  in the name and ask the reader to deduce the oxidation states ("phosphorus trichloride" => subscripts 1,3 => oxidation states +3,-1).  I can't think of any good reason for this, aside from historical accident.

As for the second, just consider the madness that is the naming patterns of hydrides.  The systematic names of the Period 2 hydrides are lithium hydride, beryllium hydride, borane, methane, azane, hydrogen oxide and hydrogen fluoride.  Whoa.

Prefixes such as mono-, di-, tri- etc. are typically used in naming covalent/molecular compounds, and not for ionic compounds.
Ammonium chromate is an ionic compound, so the prefix is not used. The ammonium cation has only a 1+ charge and so the Stock system of Roman numerals is not required. Thus, the only correct name would be ammonium chromate.


text:
Is didehydrobenzene valid IUPAC nomenclature?

I don't know if this compound exists in real life or not. If it did, what would be the correct IUPAC nomenclature for it? According to me, it could be:

1,2-didehydrobenzene
cyclohexa-1,3-dien-5-yne

For reference, the general class of compounds like this is called benzyne.

Pretty sure that the specific compound is also called benzyne, much like is the case for phenol. Not exactly a nomenclature book, however there is direct usage of both 'benzyne' and '1,2-didehydrobenzene' in the IUPAC GoldBook (see here and there). The third option also looks OK to me. I would guess the PIN is benzyne but could be off base.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) for the compound that is given in the question is 1,2-didehydrobenzene. The name cyclohexa-1,3-dien-5-yne is acceptable for general nomenclature. The formerly used name benzyne is no longer recommended.

According to this site 1 your second option Cyclohexa-1,3-dien-5-yne is IUPAC acceptable

I think it would be good if you answer the questions on nomenclature along with the reasoning on how to derive the name. Most students are not allowed to access ChemSpider (and other such tools) during exams.


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quantum number - question about the ms-value
I've got the quantum number $\ce{(3,2,2,-1/2)}$, and the question is which electron belongs to this quantity. The answer is orbital $\ce{3d^1}$. But I don't understand why. I would have understood it, if the ms-value was $\ce{+1/2}$. But how can you have $\ce{-1/2}$?

There is no difference between 1/2 and -1/2 in the problem you are solving. Independently of it there is one electron on d orbital

Okay, thank you. But how do I then know the answer is $\ce{3d^1}$ instead of $\ce{3d^2}$

The quantum numbers are $n,;l,;m_l$ for the orbital and $S$ and $m_s$ for the electron spin. The ranges are $l=0,;1,\cdots n-1$, $m_l=0,\pm 1,\pm 2,\cdots \pm l$ and $S=1/2$ and $m_s=\pm 1/2$. You have $n=3,\ ; l=2,; m_l=2$ making one of the $2l+1$ or five $3d$ orbitals. One electron makes $3d^1$. The $-1/2$ can only refer to the spin quantum number $m_s$ and is not usually included.

As porphyrin clearly explained, the first three quantum numbers usually are enough to determine the orbital of a given electron.
You probably know this but I'd like to let it here for everyone else.
Each different value of the principal quantum number $n$ yields a different wave function and a different state of the electron. In atoms, $n$ defines the order of all states according to energy.
The second and third quantum numbers $l$ and $m_{l}$, arise from the quantization of rotation where $l$ gives the shape of an orbital and its angular distribution and $m_{l}$ determines the number of orbitals and their orientations in a subshell.
The spin quantum number of an electron is denoted $s$ and it has the value of $s=1/2$. The spin magnetic quantum number is denoted $m_{s}$ gives the projection of the spin angular momentum on the z-axis. Therefore, one electron can have only one orientation: either $m_{s}=+1/2$ which we commonly denote up-spin or $\alpha$ and $m_{s}=-1/2$ denoted as down-spin or $\beta$.
What I see in the comments are partial answers because to associate an electron (not an orbital) to these numbers you need to use $m_{s}$. Then, you have a spin-down or $\beta$ electron in a $3d$ orbital.
Answering your question in the comments. To have a $3d^{2}$ you'd need two pairs of four quantum numbers: $(3,2,2,+1/2)$ and $(3,2,2,-1/2)$. Here, two electrons occupy the same region in space but they have different orientations with respect to their spin angular momenta.


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Determine minimum C-H bond length in ethane derivatives

Which of the following compound has minimum C-H bond length?

ethane
ethene
1,2-dibromoethene
1,2-dibromoethane

How can we know the comparison of $\ce{C-H}$ bond length (not $\ce{C-C}$)? Also, there is not any reagent given, so that we would do dehydrohalogenation reaction at it. How to compare it simply?

One way to compare $\ce{C-H}$ bond lengths is to compare their stretching frequencies using IR spectroscopy. However, since ethane and ethylene are gases at ambient conditions, we are going to have some difficulties in accuracy. Yet, it is worth comparing. Since we are comparing $\ce{C-H}$ bond in each case, we can suggest that the stronger the stretching frequency of $\ce{C-H}$ bond is, the shorter the bond should be.
1,2-Dibromoethane is a liquid at room temperature and $\pu{760 torr}$ with boiling point in the range of $\pu{131-132 ^\circ C}$. 1,2-Dibromoethene is from Aldrich is a mixture of cis- and trans-isomers, which boils at $\pu{110 ^\circ C}$ under $\pu{754 torr}$ pressure. IR spectra of both are extracted from NIST.gov webbook and illustrated as follows:
1,2-Dibromoethene: Range of $\ce{C-H}$ stretching is around $\pu{3150-3050 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3100 cm^{-1}}$):

1,2-Dibromoethane: Range of $\ce{C-H}$ stretching is around $\pu{3010-2950 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3000 cm^{-1}}$):

Accordingly, we can easily suggest that $\ce{C-H}$ bond in 1,2-dibromoethane is longer than that in 1,2-dibromoethene. However, this is hardly a surprise since we are comparing a $\mathrm{sp^3-s}$ bond versus a $\mathrm{sp^2-s}$ bond. Nonetheless, this is a good exercise because it confirm the validity of the test, in a way. Yet, the $\ce{C-H}$ stretching at around $\pu{3000 cm^{-1}}$ is undoubtedly suggested that  $\ce{C-H}$ bond in 1,2-dibromoethane is shorter than a $\ce{C-H}$ bond in a regular alkane.
Now, let's see two good liquid alkane and alkene examples to represent gasious ethane and ethene, respectively.
2-Methylpent-1-ene (a terminal alkene) is a liquid at room temperature and $\pu{760 torr}$ with boiling point of $\pu{62 ^\circ C}$. n-Pentane (the shortest n-alkene at ambient condition) is a liquid at room temperature and $\pu{760 torr}$ with its boiling point around the range of $\pu{35-36 ^\circ C}$.
2-Methylpent-1-ene: Range of $\ce{C-H}$ stretching is around $\pu{3100-2800 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3075 cm^{-1}}$):

2-Methylpent-1-ene: Range of $\ce{C-H}$ stretching is around $\pu{2990-2800 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{2900 cm^{-1}}$):

Thus, we can confidently suggest that the $\ce{C-H}$ bond lengths of given compounds are in following order:
ethane > 1,2-dibromoethane >ethene > 1,2-dibromoethene

Unfortunately vibrational spectroscopy is not a technique with which to obtain structure. There is only a general trend in bond length of, say, single bonds, vs. frequency so any answers based on spectra are just guesses. Rotational spectroscopy in the gas phase can and is used to obtain bond length for small molecules.

Your comments are well taken. I'm giving a relative comparison to OP's request for simple explanation. Otherwise, we are all speculating a lot of things, which can't explain otherwise.

Got it, Thanks for the explanation sir :) !..

According to me, answer should be (c) less than (b) less than (d) less than (a).
Reason- Due to lone pairs present on bromine, in (c) the Carbon would acquire a partial sp character.
But in (d), due to higher electronegativity of $\ce{Br}$ w.r.t $\ce{H}$, more p character would be in $\ce{C-Br}$ than $\ce{C-H}$ due to which $\ce{C-H}$ bond length would reduce.

@Mathew Mahindaratne, is my ans correct ??

Your argument is logical. :-)


text:
No.of isomeric ethers of formula C3H6O
I managed to get 3 ethers:-
1.4-member ring ether(Oxetane)
2.Vinyl ether
3. Epoxy ether
This question was asked in my school exam and the teacher had given its answer as 4..Could anyone please tell what would be the 4th isomer ??

The methyl epoxy ether is chiral, with the chiral center being the middle carbon in the chain, so it has two enantiomers.


text:
Is the electrical conductivity only measurable in aqueous solutions?
I have been looking for other ways to measure it, but all I've found is through aqueous solutions. So I want to know if there's another way to do it, or why it isn't possible through other means.

Ultra pure water has very low conductivity, it is the dissolved ions that causes most of the conductivity. A high dielectric solvent is necessary to solvate ions thus any other high dielectric solvent can, in principle, also be used, eg. acetonitrile, n-methyl-formamide.

One can measure conductivity of everything that... conducts. If physical property and you seem to check too much about general chemistry instead.

Did you look into ionic liquids? Also in ionic complexes like poly (ethylene oxide)? For instance, look into these references:

P. V. Wright, Electrical conductivity in ionic complexes of poly(ethylene oxide), Polymer International, 1975, 7(5), 319-327 (https://doi.org/10.1002/pi.4980070505).
Q.-G. Zhang, S.-S. Sun, S. Pitula, Q.-S. Liu, U. Welz-Biermann, J.-J. Zhang, Electrical conductivity of solutions of ionic liquids with methanol, ethanol, acetonitrile, and propylene carbonate, J. Chem. Eng. Data, 2011, 56(12), 4659–4664 (https://pubs.acs.org/doi/abs/10.1021/je200616t).


text:
Why enol content of ethyl acetoacetate is more in toluene than water?

Ethyl Acetoacetate has enol content of 0.4% in water and 19.8% in toluene. What is the reason behind this? Does this anything have to do with the intramolecular hydrogen bonding in the enol form?

The extent of enolization is affected by the solvent, concentration and temperature. Ethyl acetoacetate has enol content of 0.4% in water and 19.8% in toluene. This is because water reduces the enol content by hydrogen bonding with the carbonyl group making this group less available for intramolecular hydrogen bonding. The effectiveness of intramolecular hydrogen bonding in stabilizing the enol, with respect to the keto form is thus seen on varying the solvent and particularly on transfer to a hydroxylic solvent.

In water, due to more number of electropositive H present, the O will tend to H-bond with water molecules due to which enolisation hardly occurs although active methylene carbon. But in toluene, since it is a non-polar solvent, intramolecular H-bonding is more favorable due to which % enolisation increases in toluene. But since ester group has some resonance in ethylacetoacetate, the % of enolisation does not increase too much in toluene.

In non -polar aprotic solvents intra H.B is present in enol form hence, it is more stable but in polar protic solvents H.B is formed with polar solvent molecules hence, intra H.B is absent in the enol form


text:
What is the product of the reaction of diol with tosyl chloride followed by lithium aluminum hydride?

This question is about the structure of products X and Y.
According to me, the answer should be only A (as OTs will substitute on benzylic carbon).
But in the answer key, they have given B and D.
Please help.

I'd suggest you that, in your future questions and answers, not to abbreviate word for the benefit of most of the readers. Thank you.

Done @MathewMahindaratne. Will keep this in mind for future questions...

The picture is blurry. Please upload a better quality one/ consider sketching through chemsketch.

There are two replaceable $\ce{H}$s in the starting material, but only 1 equiv of $\ce{TsCl}$ is given. Thus, primary $\ce{-OH}$ will win the battle due to steric reasons, and therefore, B represent the intermediate product X. The second reaction is firstly acid-base reaction between secondary $\ce{-OH}$ and $\ce{H-}$, and then, nucleophilic substitution ($\mathrm{S_N}$2) reaction to replace $\ce{-OTs}$ group by $\ce{H-}$. That leaves D as the product Y, after acid workup.

So u mean that since the benzylic carbon is sterically hindered, its  -OH won't be substituted??

The faster reaction is acid-base reaction. Once, acetic $\ce{H}$ removed from $\ce{-OH}$, it is basically not removable.


text:
How is more than one pinacolone is possible?
This question came in one of my school tests. They had given us an unsymmetrical pinacol i.e 1,1-diphenyl-2-methylpropane-1,2-diol and asked us whether it will give only one pinacolone or not.
According to me, I had assumed that carbocation will be formed on the diphenyl-substituted carbon due to which a methyl shift will occur and we will get only one product. But my teacher marked it wrong and said that more than one pinacolone is possible here.
How is it possible?

Then most probably he is also considering the possibility( minor but can also occur in a very small fraction) of forming the tertiary carbocation and a phenyl shift to give the other Pinnacolone.

The problem is what is a reaction which gives "only one product", the problem is that very few reactions give a 100 % yield of a substance. For almost all real life reactions there are a small fraction of molecules which react he wrong way.
I think that when more than 99 % of molecules react the "right" way that those which react the "wrong" way can be ignored normally. Unless the wrong product has some very harmful effect on the product when it is used for a purpose.
What I think will happen is that the carbocation formed by protonation of the hydroxyl bonded to the carbon bearing the two phenyl groups, followed by the loss of a water molecule will have a much lower energy than the other carbocation. Thus it will be more easy to form.
But there is a problem, for a rearrangement of a carbocation to occur normally we need to rearrange to a lower energy carbocation. By lowering the energy of the carbocation we can lose the thermodynamic drive for rearrangement. In the ideal world we would consider the rate of rearrangement of the two different carbocations and their formation. I think that the reactions forming the carbocation will be reversable while the pinacol rearrangement will be irreversable.
But in some cases such as many homogenious asymmetric rhodium hydrogenations of prochiral alkenes the lower energy alkene rhodium complex which is present at a far higher concentration than the higher energy isomer, the lower energy isomer has far too little reactivity. As a result while the majority of the complex might be in the R form the tiny trace of the complex in the S form is responsible for the vast majority of the hydrogenation reactivity.


text:
Are there contradictions about intermolecular forces in the postulates of kinetic theory?
Regarding the postulates of the kinetic theory, in one place I read:

There are no intermolecular forces except during the collision between molecules.

And yet in another place this appears:

Collisions between two molecules do not exert unattractive or repulsive forces.

Aren't these statements contradictory? Are there or are there not intermolecular forces during the collision, according to the postulates of kinetic theory?

Welcome to Chemistry! Take the tour to get familiar with this site. Regular text can be formatted with Markdown, and mathematical expressions and chemical equations can be formatted using $\LaTeX$ syntax. If you receive useful answers, consider accepting one.

Both seem to me to be ok. The collisions are between hard spheres, so only at contact is this effective, which is what statement 1 says, and 2 effectively says the same: only contact at hard sphere collision is important, i.e. the particles do not feel the effect of one another (due to any intermolecular) potential until collision. (Should 'unattractive' really be attractive ? It is true either way but attractive is more general.)

The kinetic theory of gases is, empirically, a good approximation to how most gases behave most of the time. Its key assumptions are:

there are no attractive forces between the molecules or atoms that make up the gas
the molecules or atoms interact only by elastic collision
the molecules or atoms have negligible size

The two statements you have are (especially the second) are poorly worded versions of these assumptions. For the collisions to be elastic there have to be repulsive forces between the entities that make up the gas.
But don't forget that the assumptions are not strictly true, they are just good approximations under most conditions. Molecules are not infinitely small and attractive forces do exist. But, at least when most gases are far from the point where they liquefy, the approximation is good and it explains the way gases behave. For example, a given quantity of a gas (specifically the number of molecules or moles) always takes up the same volume at the same temperature and external pressure however big or heavy the component molecule or atom is. We observe this to be true (approximately).
The apparent contradiction you worry about is due to a careless way of stating the key assumptions and is not a problem with the (well validated) theory.

I tried to find the place where you found the two statements, and I could not.
Absence of long-range forces
The first statement makes sense to me:

There are no intermolecular forces except during the collision between molecules.

The OpenStax Chemistry textbook, for example, has this statement:

Gases are composed of molecules that are in continuous motion, traveling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.

Both statements say the same thing: there are no long-range forces acting on the particles.
Elastic collisions
Your second statement does not make sense to me, and I think it contains a typo.

Collisions between two molecules do not exert unattractive or repulsive forces.

The corresponding statement in the OpenStax Chemistry textbook is:

Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).

This statement also does not make sense to me. If there are collisions, there are repulsive forces. The second part is a known component of kinetic theory. The particles can exchange kinetic energy, but the total kinetic energy is conserved (or more stringently, momentum is conserved).
What were they trying to say?
Two billiard balls hitting each other are a good model of an elastic collision. Two balls covered in velcro are not - they are sticky. So in that sense, you want an absence of attractive interactions so that particles don't stick together (and form a condensed phase). I am puzzled why they think repulsive short-range interactions are incompatible with elastic collisions. Even attractive interaction are compatible with elastic collisions when they are sufficiently weak that they break when the repulsive interactions push the particles apart after colliding.
How do molecules actually behave?
Molecules have attractive interactions at short range that turn repulsive when they come too close together, for example modeled by a Lennard-Jones potential for noble gas particles. A substance like water make hydrogen bonds at short range. Using water as a example, these interactions are not magically turn off when liquid water turns into a gas. Instead, they become negligible because the molecules have high kinetic energy (or for low pressure, the number of collisions decreases compared to normal pressure). So for a gas much hotter than its boiling point, we can neglect those interactions and still get a good explanation of the temperature vs. pressure behavior.


text:
Is it possible to prevent agglomeration and sedimentation of microparticles just by increasing the viscosity of the medium?
I am considering a suspension of a fine powder in a high-viscosity liquid.  The suspension shall be stable over long time scales, i.e. the particles only move a few mm per year, and I have the following boundary conditions:

The particles sizes shall be between 0.1 and 1.0 microns, so the suspension is colloidal
The concentration of dispersed particles in the medium is high (the phases have approximately equal volume).
The viscosity of the medium must be high. I am considering >10000 cSt, which is comparable to thick honey or shower gels
The medium will also be free of water; possibly long hydrocarbons

In a first naïve treatment with Stokes' law I thought that such a suspension should be sufficiently stable. Then I read about interparticle effects, which can lead to agglomeration of the particles and sedimentation (mainly on wikipedia). I don't know yet which material will be used as the solid powder, so I have to consider the worst case of attracting forces between the particles, so that there will be agglomeration and sedimentation in my case. This has to be prevented, but for several reasons I would like to avoid steric stabilization.
However, all examples of agglomeration considered a "thin" liquid like water, while polymeric stabilization technique is based on long polymer chains in the medium. Since in my case the viscosity of the medium is very high in my case, I wondered if this might be enough to prevent agglomeration. The high viscosity is due to long molecular chains, so I'm thinking that this might count as a case of polymeric stabilization. Is it possible to prevent agglomeration and sedimentation of microparticles just by increasing the viscosity of the medium?

If you can make this, it will be stable. At equal volume of the phases, this will already be solid-like (0.5 is above percolation threshold, which is in the range of 0.25-0.3, depending on the geometry of your particles.).
See https://en.wikipedia.org/wiki/Percolation_threshold#Thresholds_for_3D_continuum_models
The absolute particle size is irrelevant, only the size distribution matters. When you reach the percolation threshold, the only thing that matters about the continous phase is that it's incompressible, which is the case for practically all condensed matter. If it's viscosity is very low (and the solid particles large, > ~100 µm), it could flow out of the compound, but surface tension would usually preclude that.
Also a very lowly viscous continuous phase could mean that the compound yields already to gravity. https://en.wikipedia.org/wiki/Yield_(engineering)
However it is not so simple to make such a suspension. If you try to mix powder and polymer, your stirrer (of whatever kind) will stop or break long before you get a homogeneous mixture, because your sample becomes solid, and there is no way of melting it or anything. A helper solvent can lower the solid fraction, below the percolation threshold, but is hard to get rid of later.
So you'll have to find out where the percolation threshold actually is, and just use a bit more. It varies a bit with temperature (your polymer phase will have a larger expansion coefficient, not sure if that is enough to make the difference). Anyway the yield stress is still tolerable just above the threshold, and processing is possible.

This is a very interesting aspect I didn't think of before. But I cannot really make much sense of the link. I can't quite relate the mathematical concept to the physical suspension. In particular, there is no dependence on the particle size or on the viscosity of the medium, which seems counterintuitive to me.

I updated the answer.

Thank you, these were valuable additions. I think I'll have to dig a bit deeper into the matter.


text:
Boltzmann Transport equation: Is df/dt=o for collisionless only?
$$\frac{\mathrm df}{\mathrm dt} = \left(\frac{\partial f}{\partial t}\right)_\text{force}+\left(\frac{\partial f}{\partial t}\right)_\text{diff}+\left(\frac{\partial f}{\partial t}\right)_\text{coll}$$
where the "force" term is the external force, the "diff" term represents the diffusion of particles, and "coll" is the collision term.
Source: https://en.wikipedia.org/wiki/Boltzmann_equation
So shouldn't $\frac{\mathrm df}{\mathrm dt}$ be 0 when it is not just collisionless but also force and diffusion must go to zero too?


text:
Quantization and Bohr's model
According to quantization it's said that emitted or absorbed energy is quantized.
Then, when it's said in bohr's model an electron changes its orbit (Let's say it goes to a higher energy shell from $n=2$ to $n=3$) The energy difference between the orbits is $100~\mathrm{kJ}$.  Now, according to quantization is the $100~\mathrm{kJ}$ itself a quantum. Or the $100~\mathrm{kJ}$ will be absorbed in quantized way?

No not the energy you give as this is so large it has to be that for a  mole of atoms therefore the quantum of light (radiation) used to excite one atom is this amount divided by Avogadro's number.

The energy difference between $n=2$ and $n=3$ in Bohr's model is
\begin{equation}
\Delta E=13.6\ \mathrm{eV} \times \left(\frac{1}{4}-\frac{1}{9}\right) = 1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ} = 181.4\ \mathrm{kJ/mol}
\end{equation}
Thus, for this excitation to occur, a single energy quantum (photon) with an energy of $1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ}$ is required.
The excitation does not happen if:

the photon has more or less energy than $1.88\ \mathrm{eV}$ (The energy difference needs to be matched exactly!)
the same amount energy is distributed over multiple photons (The energy needs to be provided by a single photon!)

As already has been pointed out, the $181.4\ \mathrm{kJ/mol}$ itself is not a quantum, but one mole of quanta, where each quantum has $1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ}$ of energy. As each atom can only absorb this energy once (assuming it does not relax back to $n=2$ by emitting the energy), to absorb so much energy will also require one mole of atoms.


text:
Sulphur trioxide problem
I'm having minor difficulties solving this problem:

How many grams of sulfur (VI) oxide can be derived from 8 g of sulfur and 6 g of oxygen? (S-32)

I've tried to solve it this way:
$\ce{2S + 3O2 ->2SO3}$
$\ce{S}$ = 8g : 32 g/mol = 0.25 mol
$\ce{O2}$ = 2 x (6 g : 32 g/mol) = 0.375 mol
Neither oxygen nor sulfur is in surplus, therefore:
$\ce{S}$ = $\ce{SO3}$
$\ce{SO3}$ = 80 g/mol
$\ce{SO3}$ = 80 g/mol x 0.25 mol = 20 g
But the correct answer is 10 g of $\ce{SO3}$. What did I do wrong?

You write an equation with O2, and then use n(O). That's a contradiction.

I've corrected it. I thought that only one atom of oxygen is relevant.

What would you do with the other atom? Of course both are equally relevant. Now your n(O2) is wrong.

Is it correct now?

No. Why the "$2\times$"?

I think that the "correct answer" is wrong, either that or the question is missing out some fact we need to know about

The way to approach these limiting reagent stoichiometry reactions, as they are referred to, is to first determine the molar quantities of each of the reagents.
You have correctly done so for sulfur, but not for diatomic oxygen. There is no reason, as Ivan Neterin has pointed out, to multiply by 2 the number of moles of oxygen. Doing this will in effect, cancel out the 32 g/mol of $\ce{O2}$ and you will solve this problem like there is 6 g monatomic oxygen [$\ce{O}$], rather than what there really is [diatomic oxygen [$\ce{O2}$]].
The next step is to set up a sort of "need v. have" proportion in order to determine which reagent is the limiting reagent.
Since there are .25 moles of $\ce{S}$ and .1875 moles of $\ce{O2}$, we can determine the limiting reagent is $\ce{O2}$
This is because, if sulfur was the limiting reagent, we would need greater than 1.5 times the current quantity of $\ce{O2}$
In essence, if $\ce{S}$ was the limiting reagent, the reaction would be
$\ce{.25S + .375 O2->.25SO3}$,
while if $\ce{O2}$ was the limiting reagent, the reaction would be
$\ce{.125S + .1875 O2 ->.125 SO3}$
As you can see, when $\ce{S}$ is the limiting reagent, we write too much $\ce{O2}$ than we really have [our reaction says we have .375 moles $\ce{O2}$ when we really have .1875 moles$\ce{O2}$]. In the second reaction, we have more than enough $\ce{S}$ than the reaction calls for.
Therefore, $\ce{.125 moles or ~10 g SO3}$ is produced.
I believe your problem was converting the $\ce{O2}$ to moles; you must remember that the molar mass of the compound is 32, and you have 6 g of that compound, not double 6, or 12 g.


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Molecular kinetic interpretation of pressure
Regarding the molecular kinetic interpretation of pressure, we arrive at a point in the $F=\frac{\Delta P}{\Delta t}=\frac{2mvx}{\Delta t}$, that is to say, the force exerted in the shock on the wall is calculated by dividing the change in the amount of movement of the wall by that time, which I understand should be the time of duration of the shock, but nevertheless the time used is the time between two shocks, the time in which the molecule travels the distance of the edge of the vessel for one side and returns...this does not contradict the definition of impulse that refers to the duration of the shock??

Essentially, we are talking about the time-averaged force.

The theory works as well for hard-sphere molecules, where the contact time is zero. Force is mass times acceleration, and the latter is ($2 v$ per (time interval that in average contains exactly one wall collision))

The derivation can be obtained without having to explicitly consider the time of impact. Suppose the molecules have kinetic energy $E=mc^2/2$ where velocity is $c$ and mass $m$. The molecules are in a fixed box of volume $v=xyz$. The molecule can move in any direction and so has components $u, \;v,\; w$ along each axis meaning that $c^2=u^2+v^2+w^2$ but as no direction is special (the gas fills the volume uniformly) then we can let $c^2=3u^2$.
The pressure on the vessel's walls is due to the change in momentum and for a wall collision after moving along the $\pm x$ direction is $2mu$. A molecule will on average hit opposite walls $u/2x$ times per second when moving in the $\pm x$ direction. By Newtons second law the rate of change in momentum equals force, thus a molecule exerts a force $2mu \cdot u/2x=mu^2/x$ on the face of the container perpendicular to $x$ which has area $yz$. As pressure is force /area and as the collision area is $yz$ then $p=mu^2/(xyz)\equiv mu^2/V$. Finally, substituting for $u$ and allowing for $N$ molecules the pressure is $p=Nmc^2/3V$ i.e pressure is twice the kinetic energy divided by three times the volume, $\displaystyle p=\frac{2E}{3V}$


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Lactic acid - Elementary analysis
Lactic acid is a hydroxy carboxylic acid that can form in the muscles and lead to soreness. The molecular weight of the lactic acid is $\ce{90.1 g/mol}$. Elemental analysis shows that the substance contains $\ce{40.00 \% w/w C}$,  $\ce{6.71 \% w/w H}$, and $\ce{53.3\% w/w O}$. Determine the molecular formula.
My method:
$\ce{C}$ has an atomic mass of $\ce{12.0107}$. $\ce{H}$ has an atomic mass of $\ce{1.008}$. $\ce{O}$ has an atomic mass of $\ce{15.994}$. We start by determining the moles:
$\ce{C=40/{12.0107}=3.33}$
$\ce{H=6.71/{1.008}=6.66 mol}$
$\ce{O=53.3/{15.5994}=3.33}$
We determine the molar ratio:
$\ce{C=3.33/{3.33}=1}$
$\ce{H=6.66/{3.33}=2}$
$\ce{O=3.33/{3.33}=1}$
But the answer is $\ce{C_3H_6O_3}$. What am I doing wrong?

The molar ratio is correct but you need to consider that the molecular formula can be some multiple of C1:H2:O1, not just CH2O

You can check if the calculated molecular weight corresponds to lactic acid if not try next multiple like 2, then 3 etc.
You get $\pu{30g/mol}$ for $\ce{C1H2O1}$ and
$(\pu{90g/mol})/(\pu{30g/mol})=3$
So the multiple to be applied is 3:
$\ce{C3H6O3}$


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Why can't negative charge on a carbon have resonance with a neighboring alcohol?

In these resonance forms of a conjugate base of ascorbic acid, why can't the electrons on the negatively charged carbon in the third form be used to form a double bond with the alcohol and thus delocalize the negative charge over the oxygen?

Because octet rule.

Hypothetically, if the electrons on the negatively charged carbon in the third form be used to form a double bond with $\ce{O}$ of the $\ce{-OH}$ group, it would have made ten $\ce{e-}$s around that $\ce{O}$. That would violate the octet rule because $\ce{O}$ is a second raw element, and hence, no $3d$ orbitals to accommodate extra pair of $\ce{e-}$s.
That's been said, one would question whether these resonance forms are even possible because it is impossible to form a negative charge on $\ce{2C-OH}$ group (the second form) without forming a negative charge on $\ce{3C-OH}$ group at first. This is because, $K_\mathrm{a}$ of $\ce{3C-OH}$ in ascorbic acid is $6.7\times 10^{-5}$ ($\mathrm{p}K_\mathrm{a}=4.17$), due to the conjugation through $\alpha,\beta$-unsaturated carbonyl function, giving ascorbic acid an organic acid status. Note also that $\ce{2C-OH}$ is also fairly acitic with $K_\mathrm{a}$ of $2.5\times 10^{-12}$ ($\mathrm{p}K_\mathrm{a}=11.6$) because it is also an enol function. Nonetheless, in free ascorbic acid under basic conditions, $\ce{3C-OH}$ ionizes first before $\ce{2C-OH}$.
Interesting fact: Ascorbic acid is one of unique organic acids, which have no $\ce{-COOH}$ group. Two other notable examples are picric acid ($\mathrm{p}K_\mathrm{a}=0.3$) and squaric acid ($\mathrm{p}K_\mathrm{a1}=1.5$ and $\mathrm{p}K_\mathrm{a2}=3.5$). All of them are more acetic than benzoic acid ($\mathrm{p}K_\mathrm{a}=4.20$).


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Predicting the temperature of an object using the heat of its surroundings after a certain amount of time?
how would you predict what the temperature of something would be if it spends a certain amount of time exposed to heat from its surroundings? is this possible?
ex. what temperature would a hot bottle of water at initial temp 48 C be after being put in a 4 C refrigerator for 2 hours?

The calculations are possible, but pretty complicated.

As @Ivan Neretin comments you will need to use Fick's laws of diffusion, and know all the thermal diffusion coefficients etc. for the materials used. This will be a calculation that will have to be done numerically and would be standard-ish if you were a chemical engineer. In fact I suspect that it will be quicker and more accurate to do an experiment.

Sure. Start here. You can make quick and moderately accurate estimates with some simplifying assumptions about your system.

You're basically looking at radiation, conduction, and convention as methods to dissipate heat.
Heat loss by radiation can be computed from Wein's law or something similar.
Conduction requires knowing the thermal properties and dimensions of the materials in contact.
Convection is the trickiest one since it requires some model for how the air or other fluid around the object is moving and coupling this to conduction.
All three are subject to significant errors in computation if you're not careful though if you're trying to estimate rough order of magnitude, you might not care.
All in all, this is not an easy problem to compute from first principles.
Note also that your example is not nearly well-defined enough to solve this problem. What is the surface area of the bottle and the size of the refrigerator. How well does the refrigerator maintain its internal temperature? Is there a fan? Is the water inside a thermos bottle?


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Half-Life of Radioactive Isotopes: Why? How?
Why do radioactive isotopes have a half-life? I know that they decay in order to become stable but why would it take out enough subatomic particles to be half? Or am I approaching this question the wrong way: the remaining radioactive substance is going to be half in x amount of time and the other half will be stable atoms.

Half life is a half life because it is defined this way. Wait less than that, and the decay will take out less than half atoms; wait more, and it will take out more than half.

The half life is used as a convenient measure because a sample of isotopes decays  exponentially with time. Each atom has the same chance of decaying in a fixed time interval and this leads to an exponential decay. All atoms behave the same way and continue to decay until all have been changed, whether this takes millennia or seconds.

You can just as easily define a "one-third" life, a "quarter-life". Radioactive isotopes decay exponentially; half-life is just convenient measure that captures the kinetics of the decay.

https://chemistry.stackexchange.com/questions/69536/in-a-sample-of-radioactive-substance-all-the-atoms-dont-disintegrate-simultaneo

@Bluedragon01313 We generally discourage crossposting without at least mentioning that you have put the question in a different location.

Duplicate: How does a half-life work?

It would be better if we answer the question and clear the OP's misconception instead of voting to close as unclear.

As commenters have pointed out, there is nothing about half-life that makes it special; we could as easily measure and tabulate one-tenth lives or one-fifth-lives.
Half-life has two marginal advantages: the half that has decomposed and the half that hasn't are equal, so whether we are referring to remaining or already decomposed nuclei is unambiguous, while for a third-life we could be confused as to whether we are talking about the time it takes for a third of the nuclei to decompose or for all but a third of the nuclei to decompose; and experimentally speaking, half is convenient because it's distinctly smaller from the initial amount of nuclei but large enough to be easily measured compared to the initial amount. However, these two advantages are fairly unimportant, and if we had decided to measure and tabulate one-tenth lives for all radioactive isotopes, we would be fine.
What's conceptually more important is that we can use half-lives as a proxy for isotopic disintegration rates, because radioactive decay follows first-order kinetics. Nothing prevents us from using half-life to describe any kinetic process, such as a chemical reaction rate. However, most of the time that's of limited value, because in general, half-lives are not constant, but change depending on the amount of reagents. It is only for first-order kinetics (i.e. when the amount of nuclei decays exponentially with time) that half-lives (or any fractional life) are constant.
For a general kinetic equation with only one reagent $\ce{A}$,
$$ -\frac{\mathrm{d} [\ce{A}]}{\mathrm{d} t} = k [\ce{A}]^n$$
where $n$ is the kinetic order of the process, we can always define the half-life of the process by integrating from an initial amount of $\ce{A}$, $N_\ce{A}$, to half that amount:
$$ \int\limits_{N_\ce{A}}^{N_\ce{A}/2} \frac{\mathrm{d} [\ce{A}]}{[\ce{A}]^n} = \int\limits_{0}^{t_{1/2}} - k\,\mathrm{d} t $$
which means that the half-life can be expressed in terms of the integral on the left side:
$$ t_{1/2} = - \frac{1}{k} \int\limits_{N_\ce{A}}^{N_\ce{A}/2} \frac{\mathrm{d} [\ce{A}]}{[\ce{A}]^n}  $$
This integral will in turn depend on the kinetic order $n$. While in general the half-life will take the form
$$ t_{1/2} = \frac{1}{k} \frac{1}{n-1} \left( (N_\ce{A}/2)^{1-n} - N_\ce{A}^{1-n} \right) $$
and therefore will be dependent on the starting conditions $N_\ce{A}$, and will change as $\ce{A}$ is consumed, for first-order kintetics ($n=1$)
$$ t_{1/2} = - \frac{1}{k} \left( \ln{|N_\ce{A}/2|} - \ln{|N_\ce{A}|} \right) = \frac{\ln{2}}{k} $$
so half-life does not change with the amount of $\ce{A}$.
As long as the decay is first-order (or exponential, which is the same), half-life is such a constant that we can measure and tabulate, as it will be the same under all possible circumstances; no matter if we are talking about tonnes of fissile material or a handful of nuclei, the value for half-life will be the same.
If we start with $N_\ce{A}$ radioactive nuclei, after a time $t_{1/2}$ you'll have half that amount, $N_\ce{A}/2$; since now you have $N_\ce{A}/2$, after another $t_{1/2}$ you'll have $N_\ce{A}/4$; after a third interval equal to $t_{1/2}$ you'll have $N_\ce{A}/8$, and so on.
This makes tabulating half-lives equivalent to tabulating kinetic coefficients $k$, as these can be transformed into each other quickly and unambiguously, while for other kintetic orders we'd need additional informations to perform that transformation.
Note as well that the last step also shows why we could arbitrarily decide to use any other fractional lives in the same way: if, for instance, we used one-tenth lives, we'd get
$$ t_{1/10} = \frac{\ln{10}}{k} $$

I think we should have measured decay based on one-tenth lives, measured as the time it takes for 1/10th of the material.  Then we could say we "decimated" the material using the original definition of that word.  A chemistry measurement and an English history lesson in one!

@CortAmmon Decimate means to destroy one-temth, and it's not clear if a tenth-life would mean one-tenth survives or has decayed. With half-life either definition works just fine.

@aeismail Which is precisely why it would become an English history lesson!

We should really use 1/e-life, which is the mean life, i.e. mean lifetime of an atom before decay. It's more consistent with other measures like attenuation length or electrical time constant.

@user71659 I agree, and "pi is wrong" but we still use it ;-)

After the half-life time, half of the atoms you initially had are still there good as new, the others have split off an alpha particle, undergone beta decay, and might even have decayed further.
Sit down 10000 people with three dices each, which they throw every five seconds, and tell them they can leave when they throw a triple of sixes. After a certain time $t_{1/2}$, which I could calculate if I weren't too lazy now, about half of your people will be gone. After another $t_{1/2}$, approximately half of the rest will have left, and so on. With a hundred people, the prediction might be off, but of atoms, you usually have at least billions.
Problem with atoms is I don't know their game. But they always play the same one, and I always measure the same half-life time for the same kind of atom (isotope).

I don't think your dice example explains what it needs to explain. Any gradual process reaches a point where it's half-complete. The question is why is this a useful measure. The key point with the dice (and radioactive decay) is that, if it takes an hour for half of the people to leave, it will take an hour for half of the remaining people to leave, and so on. This is in contrast to, e.g., the time taken for half of those people to die: if you take 100 people in their 20s, it will probably take 50 years for half of them to die, but it won't take 50 years for half of the remainder to die!

@DavidRicherby good point


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Help with Carbon Fibre
I'm having trouble finding articles about carbon fibre that don't contradict other articles.  I just have a few questions about the atomic structure and the intermolecular and intramolecular forces.
So, if we compare graphite's molecular structure to carbon fibre, is carbon fibres structure 3-dimensional compared to graphites linear sheets of carbon???

Also, I would like to know what forces exist inside carbon fibre.
Thanks in advance for your help.

There is more than one type of carbon, unfortunately. Some do transform to graphite under reasonable conditions, some do not. Thus, properties and structure of carbon fiber depend on  its structure and preparation, and even minor changes in the process can lead to drastic changes in the result.

my primer into carbons was this paper https://onlinelibrary.wiley.com/doi/abs/10.1002/9780470403709.ch1 , though it might be hard to obtain legally.

Ideally carbon fiber is mostly graphitic, but it does contain defects where the carbon has a 3-d non-aromatic structure and allows for cross-linking of the sheets.  We can know that fiber is mostly graphitic because it would not be high temperature stable if it were not.

You also have to consider scale. It is possible that on a small scale the structure is graphitic, but each "flat" graphite sheet bends on a larger scale creating tubes which on a small scale look like graphite but on a large scale look like fibres.


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What are the similarities and differences of adiabatic and diabatic states?
The idea of an adiabatic potential energy surface or adiabatic state is closely related to the Born-Oppenheimer approximation. This has been discussed on this site before in some detail. People frequently talk about diabatic states as well. It seems based on the reading I have done that a diabatic state is not really counter to an adiabatic state as one might expect from the names, but is sort of a different representation when looking at processes involving multiple potential energy surfaces.
So, my question is, how are adiabatic and diabatic states/surfaces/processes (what word should I be using here??) similar and how are they different? Can both pictures be used to describe the same processes all of the time or only some of the time (or perhaps I'm mistaken and they are never the same)?
Also, I think this is just terminology, but where does a nonadiabatic process fit into all this? I assume this just means there is a crossing of surfaces and the Born-Oppenheimer approximation breaks down, but that isn't quite the same as what I'm asking above I think.

If I understand correctly it is about at what point you add the interaction between the two states. In the diabatic case one forms two PES then the off -diagonals are the couplings. In the adiabatic the coupling is added when forming the PES. Non-adiabatic is about how to treat the crossing point, as in Landau-Zener. Does not arise in the adiabatic case as the surfaces are already separated and so do not cross. I stand to be corrected on this, however.

I'd be interested in seeing a clear explanation to this as well. If your bounty doesn't work out, I might add a post to the Canonical Q&A on meta.

A potential energy surface (PES) is generally the electronic energy written as a function of the nuclear coordinates. This is obtained by solving the Hamiltonian, considering explicit dependence on the electronic coordinates $r$, and parametric dependence on the nuclear coordinates $R$.
$\hat{H}\psi(r;R)=E\psi(r;R)$
Solving this, gives us the electronic energy as the function of the nuclear coordinates, $E(R)$. A plot of this function is the PES. This PES is obtained within the Born-Oppenheimer (BO) approximation, which allows us to assume that the kinetic energy of the nucleus does not matter. The physical interpretation is that the nucleus is very heavy compared to the electrons and, therefore, so much more slower that it can be considered almost stationary. This PES is generally written in the adiabatic basis, where the diagonal terms represent the electronic energy and the off-diagonal terms are the nuclear kinetic energy, $T_n$.
The off-diagonal terms can be neglected, as we already saw that $T_n$ is nearly zero. But in cases where two PESs come very close together (conical intersections and avoided crossings), there is coupling of the nuclear and electronic coordinates. This means that the nuclear kinetic energy can no longer be neglected, and there is a breakdown of the BO approximation. Further, the numbers become very high when the two states (PESs) cross as in a conical intersection and become too hard to handle. In such cases, a unitary transformation is done to change the basis from an adiabatic to a diabatic basis. By doing this, the diagonal term, $T_n$, gets converted to a potential energy term and is easy to handle.
Therefore, adiabatic and diabatic PESs differ by the basis they are written in. They can both be used to describe the same process, but there is no real need to use the diabatic representation when the BO approximations holds.
They can be referred to as adiabatic/diabatic PESs, or states, or surfaces, or representations. The term 'process' here is incorrect as it refers to thermodynamics terminology.

Since posting this question I have done some reading and what you have written is my understanding as well. The difference is whether you choose a basis that diagonalizes the kinetic energy operator or potential energy operator. Thanks for the clear answer.


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What determines the geometry of Nitrogen to be planar or trigonal pyramidal?
In the following molecule (1-(4-fluorophenyl)-4-[[1-(2-methoxyethyl)tetrazol-5-yl]-phenyl-methyl]-piperazine)

I have seen its 3D structure using Jmol, and I see that both nitrogens on the middle ring have either tetrahedral or trigonal pyramidal geometry. I get this, no problem. However, the ring that is made of four nitrogens is entirely planar. Shouldn't it also have some trigonal pyramidal nitrogens too? As for instance, the N on the point connecting to the -OH chain, it has 2 bonds and one unshared pair of electrons, yet it is still planar!

Lone pair determines that. The piperazine nitrogens have it, while the tetrazole N with side chain doesn't.

The simplest example for $\mathrm{sp^3} ~\ce{N}$ showing planner geometry is N-methylpurrole.

@MathewMahindaratne Just why would that N be $sp^3$?

@Ivan Neretin: Then, what you'd like to call it as an answer for a question from a college student who is still going for their first degree?

@MathewMahindaratne I'd say a college student must realize that the pyrrole N doesn't have a lone pair and hence is planar.

You can predict the geometry around the nitrogen atom whether it is Pyramidal or trigonal planar by noting the hybridization of $\ce{N}$. If the lone pair on $\ce{N}$ atom doesn't delocalise at all i.e there are all $\sigma$ bonds and no $\pi$ bond occurrence,the hybridization remains $sp^3$ and the geometry remains tetrahedral( leaving the lone pair it is Pyramidal), but if the lone pair on $\ce{N}$ is in continuous delocalisation and there is a case of formation of $\pi$-bond, the geometry has to be planar (hybridization close to $sp^2$).  In middle ring if you observe carefully the lone pair on lower $\ce{N}$ is in continuous delocalisation with the benzene ring so its geometry will be planar, but there are all $\sigma$-bonds associated with the upper $\ce{N}$, so that has a tetrahedral geometry.  Similarly observe the tetrazole ring (ring made of four nitrogens). Draw the resonance structures of that. You will find that all the $\ce{N}$s are equivalent (forming 2 $\sigma$-bonds and $\pi$-electron cloud throughout the ring) and the ring is actually aromatic. Thus all the nitrogen atoms have to be planar for effective delocalisation.

I follow your explanation, but I have been finding counterexamples (not on purpose) to your rule. Here in Sulfadiazine: https://pubchem.ncbi.nlm.nih.gov/compound/5215#section=Top How do you explain the N in contact with the nitro ring is not planar, while the terminal N is?

Nitrogen becomes planar when its lone pair becomes involved in pi-bonding. The five-membered ring has significant delocalisation of electrons to produce a cloud system similar to that in benzene.


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Mathematical definition of a symmetry operator
I had a hard time understanding what it means to apply a symmetry operator to a function, so I wondered if there is a formal way to define this? As far as I understand, applying a symmetry operation to a function means the following:
$$ \hat Rf(x) = f(\hat Rx) $$
The result has then to be the same function with a coefficient in front of it. Otherwise it would not be symmetric in regard to the symmetry operator $\hat{R}$.
Is this correct and can it be shown that this is how to apply symmetry operators to functions ?

Well, yes, that's pretty much the size of it.

Do you now any source that formulates it explicitly like this ?

Try to work  out the functions on the rhs of any point group. Only if the operation say C2(x)  on function f, results in 1 then this gives symmetry species, say B1. For squared functions such  $(x^2 \pm y^2)$ repeat the operation on x twice and y twice.

Would love to see an answer to this, because the implication in the OP looks odd to me and I haven't seen this notation before. Bishop's discussion of, say, d orbitals seems clear and there is no mention of this property--it seems we could have vectors of functions operating on a matrix operator and the dimensionality could be wrong?

Symmetry operators produce coordinates transformations, so yes. I have seen it written as you propose in some documents. And as far as I understand that's a property of symmetry operators, not of every operator. You wouldn't write R=partial derivative.

That's not entirely correct. You're making at least an extra assumption here. You're assuming that $\hat{R}$ is a function from $D\mapsto D$ where $D$ is the domain of $f$, but there's no reason why $f$ must operate as a function on $D\mapsto D$. It might just as well be $f: D\mapsto \mathbb{R}$.
I think what you're really asking for is that a symmetry element $\hat{R}$ such that $\hat{R}: D\mapsto D$, and $\forall x\in D$, $f(x) = f(\hat{R}x) = (f\circ \hat{R})x$.
Then for example, for $f(x) = (\cos x, \cos x)$, the transformation $\hat{R}$ mapping $x \rightarrow -x$ would would a valid symmetry element.


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Can I use pure zirconium as an anode in electrolysis?
I read that zirconium is not very reactive even when supplied with a current in an electrolyte. If I had a sizable piece of it, can I use it to perform electrolysis for rust removal of items? Would it have the same or similar results to graphite which results in a less messy process, namely the black film left on parts? Is it toxic?

No, you cannot use zirconium as an anode for electrolytic rust removal.
Zirconium is a member of the family of 'valve metals' (an archaic term), which form a strongly passivating oxide film when exposed to air and/or water.  From here:

Owing to their low electrochemical potential the group IVB and VB valve metals Ti, Zr, Hf, V, Nb and Ta readily react with water or oxygen to form a dense, protecting passive layer.

You're right that these metals, including Zr, tend to be exceptionally chemically resistant (same source):

Because of these protecting oxide films, the valve metals show an exceptional resistance towards corrosion in many aggressive environments. This explains why valve metals are widely used in the construction of chemical apparatus.

However, a feature of these metals is that the resistance of the passivating oxide film increases significantly when an anodic bias is applied to them (same source):

Valve metal oxides are formed conventionally electrochemically ... .

The Wikipedia article on anodizing concurs (emphasis added):

Anodic films are most commonly applied to protect aluminium alloys, although processes also exist for titanium, zinc, magnesium, niobium, zirconium, hafnium, and tantalum.

Thus, were you to construct an electrochemical cell with a zirconium anode and a rusty steel cathode, I expect you would get a brief current transient as the Zr-oxide passive film grew a bit, but then it would tail off to essentially zero.  While I haven't worked specifically with zirconium, the above is the observed behavior of all of the valve metals I have experience with: tantalum, niobium, titanium, and aluminum. I would be quite surprised if zirconium behaved any differently.


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Equation to find specific gravity (or density) from a solution of sugar and water
Is there an equation or simple way to calculate the final density of a solution made from x grams of sugar and y liters of water?
And would it work in reverse? if we have y liters of water at z density or specific gravity, can we know how much sugar is in the solution?
I'd like to run an experiment for my nephews, something like This.
However, lacking any background in chemistry, the precise math is tough to nail down. It would be great to have some deeper understanding of how this works...

You might use something like this graph. A rule of thumb used in brewing is that 1% (w/v) increases 0.004 g/mL above the density of water (~1 g/ml), i.e., 12% (w/v) sucrose would be around 1.048 g/mL. This works fine for low concentrations.

To do the experiment for your nephews, you don't really need to know the density, nor do any math. The more sucrose you have per unit volume of water, the more dense that solution will be. This is the same a making a density gradient in biochemistry lab. Just take your time and be careful while layering one solution atop the other. If you can find a pipet, that will make it a lot easier to do.

Yeah, but it could be a nice opportunity to learn some nice things myself Dr J, so why not? Also Felipe, that was a lot to unpack and waaaaaay above my understanding of chemistry. Any chance to get the 'For dummies' version?

@MikeMeyers - The link that Felipe posted has curves for a lot of chemicals. You want the one for sucrose (table sugar). You can ignore the others.

Here an empirical approach:
1) A simple way to calculate density: mass/volume
mass = x g of sugar + (y L of water*1000 mL/L*1g/mL)
1 g/mL = water density
Then measure the total volume (total mL) of the mix:
density = (x g of sugar + (y L of water * 1000 mL/L * 1g/mL))/(total mL)
2) For the reverse calculation, you can determine density as a function of sugar content by preparing solutions with different sugar concentrations and use this to interpolate your specific density value to a sugar concentration.
You can also use preexisting tables or graphs as Felipe suggested


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Mixing alcohol with aluminum sulphate
I am trying to make a mixture of the following ingredients: distilled water, alcohol, aluminum sulphate and glycerin.
The problem is that when I add alcohol to the water and aluminum sulphate mixture (the aluminum sulphate is completely dissolved), the alcohol solidifies the aluminum sulphate again, so I am left with even more solid aluminum sulphate then in the beginning.
What I need is for the mixture to be completely mixed, clear and without solids.
Thank you everyone for your time!

What's the solubility of aluminum sulfate in the solvent system you want? It may not be fully soluble.

Do you have access to an ultrasound bath?

An ultrasonic bath can help a soluble solid dissolve more quickly. In this case however, the OP has demonstrated that the solid is insoluble in the water alcohol mixture and nothing is going to help dissolve the solid in this solvent other than possibly changing the temperature.

I don't have  access to an ultrasound bath unfortunately.
I did however try mixing the ingredients at high temperatures, but as soon as the mixture cools down the same thing happens...

It looks like you added the alcohol before the glycerin.  Possibly the aluminum sulfate would be more soluble with the glycerin in there, so try adding that before the alcohol.
Wikipedia calls aluminum sulfate only "slightly soluble" in alcohol, so the alcohol addition would be expected to lower the solubility of the salt.  Nothing is reported there about the impact of glycerin.
Upon further review
Nope, we now find this did not work.  Hence as other answers indicate, we need a more careful approach.

I tried this, it didn't work. The glycerin changes nothing, no matter how when or how much I put in.

The solution occurs extremely slowly, but the ACS has an article on dissolving it by putting ethanol and aluminum sulfate (or aluminium sulphate, if you prefer, but aluminum sulphate would be a joint UK/USA contrivance) in a blender for five hours (!), then adding water. This procedure produced between 1% and 2.9% by weight $\ce{Al2(SO4)3}$ in $\ce{C2H5OH}$.
N.B. The friction of stirring and heat from the motor might volatilize enough $\ce{C2H5OH}$ to catch fire, particularly with sparking from brushes in the motor. Perhaps a hood with automatic fire suppression might be a safe place to try this... but I would be concerned even then. And this should be watched, despite the incessant noise of the blender!

Thanks for the input, but blending for 5 hours does not seem as a very practical way of getting what I neeed.


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Photodegradation vs photooxidation
I have been writing a paper about polymers being "aged" by light. Most of this aging happens because light hits atmospheric oxygen and either excites it to its singlet state or causes atmospheric water to produce hydroxyl radicals. Either way, the $\ce{^1O2}$ or the $\ce{OH^*}$ causes some sort of radical mechanism which usually ends up with the polymer chain being cleaved or some $\ce{-OOH}$ group being attached on.
Anyway, that's not the important part for this post. What I am wondering about is: Should I be describing this as photo-oxidation or photo-degradation, or is either appropriate? In reviewing my paper I've noticed I constantly flip between the two. I know what I mean, but I'm worried somebody reading this will think I mean two different processes, or complain about it.

"Oxidation" is a neutral description, and would often have a defined product. "Degradation" is unwanted, or the resulting structure is at least unimportant.

You it should add a sentence in your paper to explain that it could be one or the other or either etc depending on what you know about what is going on.


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Why do gnocchi float when they are cooked?
Gnocchi is a kind of pasta made with flour, potatoes and eggs. They are cooked in boiling water and when they are ready to be eaten they float.
Why do they float when they are cooked?
I found two quite different answers on the web:
1) "The buoyancy is responsible for the water vapor generated from boiling water as it creates bubbles that stick to the gnocchi surface bringing them to the waterline." (first answer)
2) "when you cook or boil it, the water inside of the gnocchi will evaporate. As the water evaporate, the water will turn into air and the air will be trapped inside of the gnocchi. It is indicated by the size expansion of the gnocchi.  Those trapped air will make the gnocchi relative lighter to the water which resulting it to float above the water."(second answer)

Just turn off the heat for a few seconds. If they don't go back to the bottom, it means the second answer is correct.

„As the water evaporates, the water will turn into air …“ – This is not exactly what happens during evaporation of water.

Related: https://cooking.stackexchange.com/questions/62303/why-do-dumplings-float-when-they-are-ready

@Loong maybe "air" has an alternate meaning of "gas". But yes, "steam" or "vapor" would have been a better choice of words.