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Help understanding how "steric effects" are distinct from "electronic effects"?
@jakebeal's excellent answer to Why do animal cells “mistake” rubidium ions for potassium ions? includes the following passage:

In the case of potassium versus sodium, which are both very important in biochemistry, a recently discovered mechanism in sodium-potassium pumps uses binding properties to grab both, then atomic size (via steric hindrance) to distinguish sodium versus potassium as described nicely in the answers to this question.

Wikipedia's Steric effects says:

Steric effects are nonbonding interactions that influence the shape (conformation) and reactivity of ions and molecules. Steric effects complement electronic effects, which dictate the shape and reactivity of molecules. Steric repulsive forces between overlapping electron clouds result in structured groupings of molecules stabilized by the way that opposites attract and like charges repel.

Since "the way that opposites attract and like charges repel" sounds "electronic" to me, why does one say that "Steric effects complement electronic effects" instead of simply being electronic effects?

I’ve not the time for a proper answer any time soon. But yes, the waters are slightly murky here: steric effects are because of electrons so fundamentally are also “electronic” effects. One way of phrasing the difference is that steric effects can be fully explained simply by the position of each atom and the repulsion between their associated electron clouds (hence can be explained with molecular models), but electronic effects are more subtle in that they not only depend on atomic positions but also the exact distribution of electron density between different atoms (i.e. orbitals).

@orthocresol that actually explains it (to me at least) quite well. Something like the difference between classical Coulomb force analysis and a proper quantum mechanical treatment perhaps?

I wanted to say what I see what orthocresol already said. You may find the same situation in many case when a general principle or cause is underlying. The same happens when energy of a molecule is decompose in terms of bond length, angle, substituents, etc. This leads to a function that is certainly not analytic, as for all terms are in reality variously interconnected. It makes much sense anyway. Take conformers of CH2X-CH2X, as a simple example. It is another case in which we can consider a term as a function of the angle.

If you wish stearic effects are electronic effects themselves, this also somewhat answers you question.

@uhoh Yes, in that one could be said to represent a "deeper" layer of analysis than the other, but I'm hesitant to draw any more parallels beyond that. Resonance (an electronic effect) is definitely a QM thing, but the inductive effect (also electronic) doesn't necessarily need QM to be explained, it just needs the basic concept of electronegativity, which can kind of be boiled down to just Coulombic attraction. Though it might be getting a bit philosophical here. :-) I think Matt's answer is clearer than my original comment.

The normal distinction between "steric" and "electronic" is based on whether the effect is transmitted through space or through bonds
All the normal physical interactions we experience are arguably electronic. When you touch your desk, you feel force because of interactions between the molecules of the desk and the molecules of your hand because the molecules of one interact "electrostatically" with the molecules of the other.
But your hand and your desk are not bonded together.
When a noble gas condenses into a liquid, it has a finite volume. The atoms become liquid because there are weak attractive forces (not usually described as "bonds") but have a fixed volume because the strong forces between their filled electron shells repelling each other prevent the atoms packing together any more closely. This second effect is very similar to the force stopping you hand penetrating your desk.
Arguably this is all electronic interactions.
But, if you are a chemist, it is worth drawing a line because it makes it much easier to talk about the properties of molecules. Electronic interactions to a chemist are usually interactions transmitted via bonds in a molecule.
So a molecule like fluorescein (shown below) has an extended pi-system so the oxygens at either end of the three fused rings can interact via bonds (and the overall molecule has electronic transitions that make it a useful fluorescent dye).

But in a molecule like this substituted biphenyl the electronic interactions might drive the molecule to be planar (so the two pi-systems can communicate) but the steric size of the 4 methyl groups prevent that because the methyl groups would come too close in 3D space to allow a planar configuration.

So the molecule will not have two aromatic rings in the same plane but will have significant angles between the rings. A "space filling" view (below) shows the methyl groups take up too much space to allow the rings to be coplanar.

Chemists think of the forces trying to make this molecule planar as "electronic" as they are transmitted via the bonds but the forces preventing this as being "steric" interactions. They are more like the forces preventing your hand from penetrating your desk than they are alike the forces holding the molecule together.
It is a useful distinction even when the forces are internal to a molecule.

+1 immediately because even though I will read through more carefully later, I know an excellent answer when I see it ;-)

That's better than my comment.

It is easy to find papers that distinguish "steric" and "electronic" this way (for example, https://doi.org/10.1021/cs501813v and https://doi.org/10.1016/j.poly.2011.09.035 ). So, that's what people do. Personally, I feel like "steric" is a subset of "electronic" and find this restricted usage of "electronic" objectionable.

I observe that "steric" effects would still hold if we think of the space filling model as made of neutral pieces.  It's that atoms take up space that matters.  Sure, Pauli exclusion (of charged particles) prevents interpenetration, and the charges soften the potential barriers we might imagine from the space filling model, but those are details -- the core idea is that parts of the configuration space are excluded because atoms have volume.

@WaterMolecule you might find it objectionable, but it is highly practical. Unless you have some magic way to completely solve the wavefunction for all the electrons in every complex molecule. The practical difference is clear and easy to understand.

tl;dr– Effects are:

steric (or non-steric) if they're related (or not) to physical shape/configuration;

electronic (or non-electronic) if they're related (or not) to electronic charge.

I would suggest seeing these as distinct concepts; effects can be qualified with respect to sterics or/and electronics, as appropriate (or just called an "effect" if no qualification is needed).

Basically:

"Steric" basically just means "related to shape".  Steric effects are those related to shape/configuration.

"Electronic" basically just means "related to electric charge".  Electronic effects are those related to electric charges.

Chemistry students often learn about chemical reactions in the context of simple molecules.  For example, a substitution reaction:
$$
\text{R-Br} + \text{OH}^{−}
~\Rightarrow~
\text{R-OH} + \text{Br}^{−}
\,,
$$
where $`` \text{R} "$ is a placeholder for a generic substrate.  Reactions like this are understood through theories about reaction mechanisms, which focus on how the electrons move around and just generally why the reaction happens as it does.
The situation can get more involved when we're talking about reactions with larger molecules, such as proteins in Biochemistry, where the substrate, $`` \text{R} " ,$ can be massive and wrapped around the reaction site, significantly obstructing it.  Or another chemical component (another functional group) could be maintained in a certain configuration in close proximity, significantly affecting reactivity.  These are discussed as steric effects.

Note: "Steric" and "electronic" aren't alternatives.
With respect to electronics:

"Electronic" clarifies that an effect is related to electronics.

"Non-electronic" clarifies that an effect isn't related to electronics.

With respect to sterics:

"Steric" clarifies that an effect is related to sterics.

"Non-steric" clarifies that an effect isn't related to sterics.

An effect can be qualified with respect to electronics, sterics, both, or neither, as appropriate.

Mostly just trying to clarify some confusion I perceived as possibly being reflected in the Wikipedia articles, where "electronic effect" seemed to be presented as a synonym for "non-steric effect".

@uhoh: Generally we just say "electronic" when we're referring to electronic effects as opposed to non-electronic effects, e.g. a kinetic-isotope effect.  If we don't need to draw such a distinction, we merely don't say either "electronic" or "non-electric".  The Wikipedia articles attempt to make a case for a system in which sterics and electronics are dichotomous, rather than just using the "non-" prefix, which is just.. well, honestly it's just really silly.

"electronic... if they're related... to electronic charge." So are you saying that steric effects are not related to charge? If not by the Coulomb force, by what effect can one molecule respond to the shape of another?

@uhoh: I'm saying that sterics and electronics should be spoken of separately rather than assumed related in specific ways.  For example, yes, there are non-electronic steric effects, e.g. steric isotope effects.

+1 okay thanks, this has given me much more to think about than I'd planned, I'll take some time think about it :-)

Arguably, if we had the computational power to fully solve the complete Schrodinger equation for molecules, there would be no distinction between "electronic" and "steric" effects as the solution would consider all interactions and produce the full configuration space of the molecule. But we can't. So chemists need to make the pragmatic distinction between electronic and steric effects to make any sense of their data.

@matt_black: If the distinction's between [steric, electronic effects] and [non-steric, electronic effects], then if we assume "electronic" is provided by context, then they'd be "steric effects" and "non-steric effects".


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What is the IUPAC name of this organic compound?
Variety is the spice of life. I have made an organic compound myself. Can you please tell me the name?
I thought that the correct answer is -
3,3-dibromo-2,4-dichloro-1,4-difluoro-2,4-diiodo-1-methylbutan-1-ol, because the haloalkane group is to be given in the alphabetical order and the priority of the alcohol group is higher than the haloalkanes.
Is it correct? What will be the derivation of the name if the alcohol group is replaced with hydrogen?

Find the longest chain, principal functional group gets lowest number.

You are nearly right, as it is : 4,4-dibromo-3,5-dichloro-2,5-difluoro-3.5-diiodo-pentan-2-ol

and for the change alcohol -> haloalkane, it suffices to change the suffix ol to donkey because there is no particular suffix for them

user213305, you've enlightened me.

Please don't answer in comments; especially since all three answers were not correct (however, @Maurice was very close, just a wrong hyphen).

As of Loong's suggestion, I'd put the answer removing it from comment section: The name should be 4,4-dibromo-3,5-dichloro-2,5-difluoro-3,5-diiodo-pentan-2-ol as Maurice pointed out. However, keep in mind that there are 3 chiral centers within the molecule. But, the indicated name is okay since stereochemistry of the molecule is not shown.
If $\ce{OH}$ group is replaced by $\ce{H}$ atom, the name of new compound is 2,2-dibromo-1,3-dichloro-1,4-difluoro-1,3-diiodopentane. Again, there are still 3 chiral centers within the molecule.

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Rutherford's Alpha Ray Scattering Experiment & Plum-Pudding Model
My teacher said, "Rutherford thought that all the Alpha particles would directly go through the gold foil without any collision if the plum pudding model was right. Because, as said in plum pudding model, positive charges are arranged in a balanced way in the atom with electrons like plum in pudding. As there is no densely accumulated charge in the center, so the Alpha particle will not deviate by anything."
But two question raised on my mind.

Is my teacher right?
If not, then why?
If yes, then why evenly accumulated positive charge will not deviate the Alpha rays? I mean, when the alpha ray will come near to the pudding like positive charges, it should deviate more sooner nah? Moreover, as there are evenly arranged positive charges, so, no empty space is in the atom. So, no alpha particles would come across the foil. Is my assumption true if thought from plum pudding model?

A plum pudding model prediction would be like results of shooting bullets through a wall formed from dust bound by polyurethane foam. But the experiment results were like if there were occational stones and not all dust in that shooted at wall.

@Poutnik as a high school student, don't have much idea about polyurethane foam but understood the main theme! Thanks

https://chemistry.stackexchange.com/questions/45115/rutherford-gold-foil-experiment?r=SearchResults And other questions are pretty similar.

It is a very common foam, used e.g. for sponges or padding of products in shipping boxes. We use in Czech a generic name molitan, but I was not sure about what is used in English. See wikipedia polyurethane

Details, details. It isn't that there would be no repulsion in the plumb pudding model, it is that the repulsion would be very different. A large, diffuse mass of positive charge might bend the beam slightly, but would not cause large deviations. But a tiny, hard nucleus could occasionally cause big deviations.

First of all, yes, your teacher is right for most of the part but it is not that they will go without any collision. Instead, they will penetrate through the foil like a bullet shot at paper.
It is not that the alpha rays will not be deviated but that if the charge was evenly spread throughout the atom, it would not be able to provide sufficient repulsive forces to wear out the kinetic energy already given to the alpha particles and slow them down. As a result, the alpha particles should have penetrated the atom like a bullet penetrates a sheet of paper.

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Angle made by orbital with y-axis
What is the hybridization and angle (in degrees) with respect to the y-axis for the following hybrid orbital $ψ=\frac{1}{√3}c×(s-\frac{1}{√3}×p_x+\frac{√3}{√2}×p_y)$
The answer given is 30 degrees but I couldn't figure out how.I was able to figure out that it is an $sp^2$ hybrid orbital wavefunction.Any help is appreciated.

It is difficult to read, because  there is an open parenthesis before s - 1, and it is not closed

This orbital isn't normalised, even though you seem to have included what appear to be intended as normalisation factors. Also, $x/y\times z$ is pretty ambiguous: do you mean $x/(yz)$ or $(x/y)z$?, and even though one can guess which it is (since it makes little sense to divide by an orbital), you shouldn't write that way. My suggestion: use $$\frac{a}{b}c$$, it's far clearer (and prettier): $$\frac{a}{b}c$$

the orbital is normalized. if we add the squares of the coefficients of s,px and py orbitals, it adds up to 1

Well, now that you've added some constant $c$, then yes it is obviously normalised as long as you choose the right $c$: but if that $c$ weren't there, then I count $$\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{3} + \frac{1}{9} + \frac{1}{2} = \frac{17}{18}.$$

It seems like someone was trying hard to get 30 degrees.  1/sqrt(3) is the tangent of 30 degrees, and sqrt(3)/2 is the cosine...


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In dative bond S→ S, why are the Oxidation state of donor S = +2 and recipient S= -2 although both of them have the same electronegativity?
We know that if the dative bond is directed towards less electronegative element, it is ignored while calculating the oxidations state and it is considered as +2 for donor and -2 for the recipient when it is directed towards the more electronegative species. But in the case of the same electronegative species, Why the second rule (same as directed towards more electronegativity) is used? What is the significance? How is it different from above two condition?

Oxidation states are mere human convention. You may count them 0 and 0 instead, it doesn't matter.

https://chemistry.stackexchange.com/questions/24928/oxidation-state-of-the-sulfur-atoms-in-the-thiosulfate-ion


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Classically permitted position values in a quantum harmonic oscillator
In  a quantum harmonic oscillator that is in the state with quantum number $n$, what range of the position $x$ is allowed classically? Does it have something to do with Bohr's correspondence principle?

When you diet "classically" does that mean usually or according to classical mechanics?

Assume for a minute that instead of having a quantum oscillator with total energy $E$ you had a classical one with the same amount of energy. That energy is the sum of potential and kinetic contributions:
$$E = E_\textrm{p}+E_\textrm{k}.\tag{1}$$
$E_\textrm{k}$ is zero and the total energy equal to the potential energy when the extension (amplitude) is at a maximum, when
$$E = E_{\textrm{p,max}} = \frac12 kx_\mathrm{max}^2\tag{2}$$
which, solving for $x_\mathrm{max},$ leads to
$$x_\mathrm{max} = \sqrt{\frac{2E}{k}}.\tag{3}$$
All that is left to do is to substitute in the right value for $E.$ That is dictated by the expression for the quantized system,
$$E_n = h\nu\left(n + \frac12\right),\tag{4}$$
so that the corresponding classical maximum amplitude is
$$x_\mathrm{max} = \sqrt{\frac{h\nu(2n + 1)}{k}}.\tag{5}$$

Thanks @Buck Thorn.  To others, am a first time user of Stack Exchange, without any prior experience. Oindly understand my situatiom


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Choice of lower bound of curve for PV work
I am told that the area under the PV curve gives the magnitude of work done by a gas.
Let $V_i$ be the initial volume, $V_f$ be the final volume, and the pressure as a function of volume be given by $P=f(V)$. To my understanding, the work done is given by the area bounded by $x=V_i$, $x=V_f$, $P=F(v)$ and $y=0$.

Proceeding in the above stated manner, the work done comes out to be the sum of the areas of green and yellow region, but the answer stated neglects the yellow region.
Origin of Question : JEE MAIN
Here's the answer according to the official answer key released by the testing agency:

The error comes from reading the graph: the crossing of the axes corresponds to the point (2,2) and not (0,0)

If that is correct, then we must be dealing with a thermodynamic cycle here, with the lower boundary of the cycle p = 2.  Otherwise, where we placed the lower boundary on the graph would not matter.

@ChetMiller but the question states that the inital state of gas is A and final state is C, which does not form a thermodynamic cycle.

I still think that implicitly, it is necessary to add a point D (2,2) and that the gas describes the cycle ABCDA otherwise we cannot effectively neglect the yellow region in the added image

I agree with Nicolases'.   And, if the yellow area is included, then the work is 64.

@ChetMiller, please check the recent edits.

@SuchetSaxena even the biggest can make mistakes! Nothing justifies the given value of $ \pu{48J} $ other than that of a cycle such as the one shown in the figure!

@Nicolas, seems like a huge error went unnoticed. Odd that nobody has picked it up even after a year.

I did not quite receive the closure I was looking for. If the answer truly is 64, then a test that has huge stakes got it wrong and nobody even rectified it.

I understand your disappointment but I do not see how they can find the advertised value


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calculate cell potential for the voltaic cell where its involving the two cobalt species with Copper
Calculate the cell potential for the voltaic cell that results when the following two half-cells are connected at $\pu{25 ^\circ C}$:
(1) A platinum electrode inserted into a solution of $\pu{0.10 M}$ $\ce{Co^3+}$ and $\pu{0.0010 M}$ $\ce{Co^2+}$
(2) A copper electrode inserted into a solution of $\pu{0.010 M}$ $\ce{Cu^2+}$ ions

Answer. $\pu{1.66 V}$

how it's calculated?
I know how to calculate the second cell half but what about the first half?
Edit 1
will I need more than this equation?
Cathode
$\ce{{Co^{3+}} + e− ⇌ {Co^{2+}}  ->   1.92V}$
Edit 2 from Nicolas Answer
$\ce{E=E^0 -0.059*log \frac{[Co^{2 +}]}{[Co^{3 +}]}}$
$\ce{E=1.92 -0.059*log \frac{0.001}{0.1}}$
$\ce{E=2.038}$
Now how to use the Nernst equation again, which concentration I will use for Co?

I Thought I will use it for the cell, not just half cell which requires two concentrations

Edit 3
Applying Nernst to the second half
anode:
$\ce{Cu^{2 +}(aq) + 2e- → Cu(s) ->  +0.34}$
$\ce{E=E^0 - \frac{0.059}{2}*log \frac{1}{[Cu^{2 +}]}}$
$\ce{E=0.34 -\frac{0.059}{2}*log \frac{1}{0.01}}$
$\ce{E=0.281}$
E cell = E cathode - E anode = 2.038 - 0.281 = 1.757
The answer is 1.66 did I make everything right and That's a book's mistake?
Edit 4
trying to solve it in one step
Cathode
$\ce{{2Co^{3+}} + 2e− ⇌ {2Co^{2+}}  ->  1.92V}$
Anode :
$\ce{Cu ⇌  Cu^{2 +} + 2e- ->    0.34V}$
Cell:
$\ce{{2Co^{3+}} + Cu ⇌  Cu^{2 +} + {2Co^{2+}} ->    1.58V}$
$\ce{E=E^0 - \frac{0.059}{2}*log \frac{[Co^{2 +}]^2[Cu^{2 +}]}{[Co^{3 +}]^2}}$
$\ce{E=1.58 -\frac{0.059}{2}*log \frac{0.001^2 * 0.01}{0.1^2}}$
E = 1.76
So I think it's a book's mistake written 1.66 instead of 1.76

Hint: Find a half-reaction involving the two cobalt species.

In both cases (in all cases besides), the principle is the same: one writes the corresponding half-equation of the couple concerned then one establishes its potential using the relation of Nernst

Is there any link for a question like this?

Do you know about Nernst's relationship? Otherwise, it might direct you to seek it out and apply it to your half cell.

Yes I know Nernst's relationship but it requires the concentration of the two half cells

Start by writing the equation corresponding to the first half-cell then the associated Nernst relation

I am trying after more searching, does these steps seem right?

in the statement, you indicate that the 1st half-cell corresponds to a mixture of $ \ce{Co^{2 +}} $ ions and $\ce {Co^{3 +}} $ ions but in your equation guess, you used copper!

omg sorry, I will start searching again

will I need more than this equation?

Reading you, I still have the impression that Nernst's relationship is not very familiar to you: a little tour in https://en.wikipedia.org/wiki/Nernst_equation should help you understand how to use it.

ok I will read about it again

by the way, you ask how to do for $ \ce {Co} $ except that in your statement it is question of copper for the second half-cell

The equation noted corresponds well to that of the first half-cell: on the other hand, the potential indicated is not the good one but just that of the standard potential.
The Nernst relation is written $\ce{E=E^0 +0.06*log \frac{[Co^{3 +}]}{[Co^{2 +}]}}$: the potential therefore depends on the concentrations
You can therefore calculate the potential of this half-cell.
You do the same thing with the second to arrive at its potential: from the 2, you can calculate the potential of the cell

did I solve it right?

@A.Hany, Your approach is correct. The book may have round off errors.


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Activity of Polonium in half-time in Becquerels
I got an exercise but I don't know the exact answer.
In the catalog note of the reference source of Polonium-210 radiation with a half-life of 138 days, it was stated that one day the activity was $\pu{2 \times 10^6 MBq}$. Calculate activity after 558 days.
After my calculations the final answer was ${126468.14}$, but my professor has told me that my answer should be about 1,7 times higher.

The number of significant figures in your answer should be cut in half. (I meant all figures, not just those after the dot.) As for the answer itself, well, now that's strange: mine is slightly different from yours.

Actually, I got only specified data from my professor, I used the half-life formula to calculate the final answer.

126468.14 +/- 30000 would not look so great, would it ?

Why would there be as much as 30000 units approximation?

Well, your answer and would be the answer are both wrong by the fact that after 552 days, the activity should be $\frac{2 \times \pu{10^6 MBq}}{16} = \pu{1.25 \times 10^5 MBq}$. Thus, answer should be lower than this number (disregard sig numbers).

@zajkevv You do not know as input data the activity with 8 significant fugures, do you ? There is just one. Otherwise it would be $\pu{2.0000000e6 MBq}$

Well, first of all, I should point out that your answer and the would be the answer ($1.7$ times your answer) are both wrong by the fact that after 552 days (4 half-lives), the activity should be $\displaystyle\frac{\pu{2E6 MBq}}{16} = \pu{1.25E5 MBq}.$ Since this is smaller than both your answer and the would be answer, they both would be incorrect. Accordingly, correct answer would be smaller than $\pu{1.25E5 MBq},$ because the sample would be further decays for another six days. Let’s do the proper calculations:
The half-life $\left(t_\frac{1}{2}\right)$ of a radioactive element is defined as
$$t_\frac{1}{2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}.\label{eqn:1}\tag{1}$$
The $t_\frac{1}{2} = \pu{138 d}$ in the equation \eqref{eqn:1} for your case. Thus from the equation \eqref{eqn:1}, you can calculate for $\lambda:$
$$\lambda = \frac{\ln 2}{t_\frac{1}{2}} = \frac{0.693}{\pu{138 d}} = \pu{5.02E-3 d-1}. \tag{2}$$
Also, the radioactive decay is according to the equation:
$$A_t = A_0\mathrm e^{-\lambda t}.\label{eqn:3}\tag{3}$$
where $A_0$ is initial activity and $A_t$ is the activity of the same same sample after time $t$ (Note: $t_\frac{1}{2}$ and $t$ should have same units). Thus, applying the $\lambda$ value on the equation \eqref{eqn:3}:
$$A_t = \pu{2E6 MBq}\times\exp\left(-\pu{5.02E-3 d-1}\times\pu{558 d}\right) = \pu{1.21E5 MBq}.$$
Note: I disregard the correct significant figures because I believe the initial activity should be given in three significant figures.
For fun, you can also checked the validity of the answer using my initial remark. You can use the activity of the sample after four consecutive $t_\frac{1}{2}$s (after $138 \times 4 = \pu{552 d})$ as its initial activity and find the activity after another six days using the equation \eqref{eqn:3}:
$$A_t = \pu{1.25E5 MBq}\times\exp\left(-\pu{5.02E-3 d-1}\times\pu{6 d}\right) = \pu{1.21E5 MBq}.$$


text:
Total Chloride Vs Organic/Inorganic Chloride
I am currently running a project to detect chloride level in crude palm oil. On most lab service websites, some offer total chloride detection and others offer organic chloride detection. What are the differences between total chloride and organic chloride?

How long have you been thinking and searching to get an answer to this  question, before asking ? What is the difference between ( fruits and vegetables ) and ( just fruits ) ?  Is chlorine in DDT organic or inorganic ?  Is chlorine in table salt organic or inorganic ? Note that there is not expected inorganic chlorine in vegetable oils as lipophilic liquid, unless contained in hydrophilic part of eventual emulsion.

BTW, you may confuse chlorine and chloride, unless it is meant the total/organic/inorganic chlorine content is expressed as chloride.


text:
Why does having equivalent resonance structures give more stability (Related)?
This question is based on @Heisenberg's Question with the same title.
I recently offered a bounty for a more detailed answer for the same question.
These are the things I would like to be in the answer:

Can molecules with 'n' canonical structures be taken as an n-state  system with coupling(At least as an approximation) ?

Suppose if it is possible.
Consider a molecule with 2 canonical structures represented by $|A\rangle$ and $|B\rangle$
Without the coupling let the energies of the two states be $E_1$ and $E_2$ respectively.
So the Hamiltonian is represented by:
$H=\pmatrix{E_1& 0\\0&E_2}$

Now with coupling the Hamiltonian in the most general case will be:
$H=\pmatrix{H_{11}& H_{12}\\H_{12}^*&H_{22}}$
The energy eigen-value with the lower energy in this case is:
$E_1 = \frac{H_{11}+H_{12}}{2} - \frac{\sqrt{(H_{11}-H_{22})^2 + 4H_{12}H_{12}^*}}{2}$
So my doubt is if the 2 states are symmetrical then $H_{11}$ and $H_{22}$ are equal right?
Then the term inside the square-root $(H_{11}-H_{22})$ becomes zero, and it seems to me that when this term becomes zero, the energy splitting becomes the least.
And yet it is resonance with equivalent canonical structures that are more stable. Why?

Are molecules with equivalent resonance structures actually more stable? Please include references or experimental evidence if the it is true.

If molecules with equivalent resonance structures are in fact more stable than others, then does it mean that resonance in molecules cannot be approximated as a discrete n-state system even for ones with 2 canonical structures (Since there might be a contradiction with arguments in 2)) Can resonance be understood only using MOT or Huckel Theory ?

Note: Please add your answer in Heisenberg's question so that I can award you the bounty.
You might understand why I'm having these doubts and the doubts itself if you read these chapter of the Feynman lectures on Physics Chapter 8 , Chapter 9 , Chapter 10

Comments are not for extended discussion; this conversation has been moved to chat.

2) is answered here


text:
Why does allyl anion have only two resonance structures?
There are only two resonance structures of allyl anion with negative charge distributed over positions 1 and 3:
$$\ce{\overset{-}{C}H2-CH=CH2 <-> CH_2=CH-\overset{-}{C}H2}.$$
What's the criteria for electrons forming a double bond? Why isn't there a third resonance structure
$$\ce{\overset{+}{C}H2-\overset{-}{C}H-\overset{-}{C}H2}?$$

Without even checking the correctness of the formula, it is always the same question.... We can perhaps draw other limiting forms, but their weight would be negligible as those hypothetical structure would be highly energetic (for instance you have placed two adjacent opposite charges. This is unfavourable already in terms of coulombic repulsion, and so on. Look for standard rules, they work in most cases.

Wouldn't the adjacent negatives be stabilized by the adjacent positive and negative charges?

Well, yes in a way they do. Forming a double bond. But even coulombic treatment suffices. The outermost right charge would anyway feel the adjacent more than the outermost left.

Right, got it. It's all a matter of knowing it, and you can't always make your own predictions, right?

sometimes is more tricky, and the rules of thumb don't work.

Don't put too much trust in ephemeral things. Resonance structures only exist in your head and vanish when you stop thinking about them.

Right, thanks. Do you want to post it as an answer so I can mark it?

To which comment do you refers? My one doesn't deserve to be an answer. The @Ivan Neretin's one is a bit too much of disenchantment :)

Yours, but his comment also helps.


text:
Why does higher temperature affect the rate of electrolysis?
I understand that more heat energy= higher rate of electrolysis, but can someone explain using higher-level terms why this occurs and if there are any theories or rules that explain this?

Temperature affects the ion mobilities and therefore electrolyte conductivity as the ratio of local current density and local potential gradient.

Rate of electrolysis is only dependent on the current (Faraday's law) passing through the solution. Remember that in an electrochemical experiment you cannot independently vary voltage or current. One factor has to be fixed. These two modes are called constant current electrolysis or constant potential electrolysis.
Once you have fixed voltage, you can ask you yourself, what factors affect current rather than electrolysis. The key factor is solution's resistance or better conductance (reciprocal of resistance). Conductance increases with temperature.


text:
Writing a chemical equation with a reactant of unspecified chain length
Would appreciate it if I got some general tips and hints and not the solution cause I'm trying to learn from this question :)
The question is as follows:

FA1 is a solution containing $\pu{5.00 g dm-3}$ of a dibasic acid, $\ce{HOOC-(CH_2)_n -COOH}$.
FA2 is a solution containing $\pu{5.00 g dm-3}$ of $\ce{NaOH}$.
$\pu{25.0 cm3}$ of the acid solution FA1 needed $\pu{17.0 cm3}$ of the sodium hydroxide solution, FA2, for complete neutralization.

Determine the value of $n$ in the formula of the acid.

The issue I have is in starting the question. I genuinely have no clue how to start a chemical equation with something like $\ce{HOOC-(CH2)n -COOH}$. Tips and hints would be greatly appreciated thanks.

Start by writing the equation of the reaction
from the volume added to achieve neutralization, you can write a relation between the quantities of matter so the unknown will be the molar mass of FA1
you will be able to determine FA1 since you know its structure

Tips and hints would be greatly appreciated.

My tips and hints for you (I apologize for if my tips and hints direct to me treating OP as a novice):

Your dibasic acid is $\ce{HO2C-(CH2)n-CO2H}$. Thus, $\ce{A^2-}$ in the equation $(2)$ of Karsten Theis's answer is $\ce{^-OOC-(CH2)n-COO^-}$.
Molar mass of this dibasic acid is: $2 \times (12+32+1) + 14n = 90 + 14n$
Find the concentration of FA2 solution in $\pu{mol dm-3}$ (in other words, in $\pu{M}$. Hence, you can find the the amount of $\ce{NaOH}$ (in $\pu{mol}$) in FA2 solution ($\pu{17.0 cm3}$).
Determine the amount of dibasic acid (in $m_\ce{H2A} \ \pu{ mol}$) in FA1 solution ($\pu{25.0 cm3}$) using the equation $(2)$ of Karsten Theis's answer.
Now, you can find $n$ using following equation:
$$m_\ce{H2A} \ \pu{mol} \times \pu{(90 + 14n) g mol-1} = \pu{5.00 g dm-3} \times \pu{0.025 dm-3}$$

Since you have already calculated $m_\ce{H2A}$ in step 4, you can calculate the $n$ $(\pu{1.0 dm-3} = \pu{1000 cm-3})$.
Note that $n$ is an integer so final answer should be rounded to closest integer.

I genuinely have no clue how to start a chemical equation with something like $\ce{HOOC(CH2)_nCOOH}$. Tips and hints would be greatly appreciated thanks.

The only thing that an acid does in an acid/base reaction is to lose hydrogen ions. You can write the neutralization of a monoprotic acid with $\ce{NaOH}$ like this:
$$\ce{HA(aq) + NaOH(aq) -> A-(aq) + Na+(aq) + H2O(l)}\tag{1}$$
This is already balanced. For a diprotic acid, the unbalanced equation would be
$$\ce{H2A(aq) + NaOH(aq) -> A^2-(aq) + Na+(aq) + H2O(aq)}\tag{2}$$
It is possible to balance this even though we did not specify what $\ce{A}$ is.
$\ce{HOOC(CH2)_nCOOH}$ is a diprotic acid as well, so its neutralization is described by a similar chemical equation. The bothersome $\ce{-(CH2)_n -}$ group prevents you from counting atoms on both sides of the equation; however, as it does not partake in the reaction, you can ignore it - it will appear on the reactant and product side in the same form.

Tips and hints would be greatly appreciated thanks.

You got some tips and hints in the comments how to proceed once you figured out the stoichiometry. Another way of stating the steps is:

Determine the mass of FA1
Determine the amount of FA1 (how many moles)
Determine the molar mass of FA1
Figure out what $n$ is

Thanks! This was helpful. Am I right to think that in this case A is OOC(CH2)COOH?

almost ! Here you assay a diacid $ \ce {H_2A} $ -> the group A would therefore be $ \ce {OOC-(CH_2)_n-COO} $

Ahhhh. So the salt that will be formed is Na2(OOC−(CH2)−COO) am I correct?

Yes, if you want to write it as salt. I chose to write it in the dissociated form in equation (2)


text:
Are phenyl and allyl groups in prop-2-enylbenzene coplanar?
Wouldn't C1 and C2 in prop-2-enylbenzene be on the same plane as that of the phenyl ring, right? If the double bond connected to C2 was between it and C1, they would be, but is it correct that now they're both planar, but on different planes?

That's right, there is no conjugation, hence no reason for the whole thing to be planar.

@Ivan Neretin; wait, ring planarity is due to conjugation? I thought planarity was a requirement for the latter.

It is a mutual thing. Planarity is a requirement for conjugation, and conjugation enforces planarity.

@HarryHolmes with reference to your last comment - to which Ivan Neretin already replied - I suggest that you read my answer to https://chemistry.stackexchange.com/questions/146178/what-is-the-difference-between-resonance-and-%cf%80-conjugation as it is strongly related as well as it clarifies some uses of terms.

Yes sp2 carbon is planar and sp3 isn't It depends if that is a transition state, a byproduct or the main product (probably not). To find the molecule with the lowest energy you have to do a theoretical calculation or measure the XRD in certain properties.

Probably not
Part of the problem with the question is geometry. Any two points always form a plane. What I suspect you mean is "is the allyl double bond in the same plane as the benzene ring?" in which case the answer is probably not.
To make the situation clearer here is a different numbering of the atoms in the allyl benzene molecule:

The question I think you intended is whether the 2* and 3* bond is in the same plane as the benzene ring.
There is no chemical or geometric reason why it would be. the 1-1* bond and the 1-2 bonds will freely rotate so there is no bond-related force driving the orientation of the 2-3 double bond. to illustrate here is a 3D view of one possible configuration:

I say one possible configuration as it is likely that the barrier to rotation of those two bonds is small and, at room temperature in the liquid, the molecule will explore all the possible configurations (it might be different in a solid crystal where packing forces between units might lead to one being favoured).
In short: there is no reason to suppose the double bond and the benzene ring will be in the same plane because the two single bonds connecting it to the ring will freely rotate.

I get this; it's just that I've heard of some sort of rule that states that a carbon attached to an $sp^2$ hybridised one is in the same plane as it. In that case(using your numbering), 1 and 1*, and 1* and 2*, would be coplanar. Could you clarify why this rule doesn't work?

@HarryHolmes When sp2 carbons are attached to other sp2 carbons, then there are electronic reasons that might drive them to be coplanar (if your example where vinyl benzene then this might matter). The reason is that pi-bond systems will interact. But the allyl group isn't directly connected to the benzene so its pi-bonds can't interact with the bonds in the ring: there is a tetrahedral sp3 carbon in between.

I think you mean conjugation. Yeah, that makes carbons coplanar, but this is something separate.

@HarryHolmes To be even clearer, the 1* carbon is in the same plane as the ring (it has to be because the 1 carbon is sp2). But this does not constrain rotation about the 1-1* bond or the rotation about the 1*-2* bond. Perhaps that explains the confusion?

If its being sp2, as you said just now, is why 1-carbon makes 1* carbon stay on the same plane as it, wouldn't the same apy for the sp2 carbon, 2*? Wouldn't that effectively make them all planar?

@HarryHolmes The sp2 nature of the 2* carbon constrains it to be in the same plane as the 2*-3* double bond. But the 1* to 2* bond freely rotates so this will not, in general, be the same plane as the benzene ring. Consider the similar molecule diphenyl benzene: two flat benzene rings connected by an sp3 carbon. The sp3 carbon between the rings has two freely rotating bonds. So Most configurations will not have the benzene rings in the same plane and there is no force driving them to be in one.


text:
Why is U the symbol for lattice energy?
I often see $U$ or even $\Delta U$ being the symbol for lattice energy. Why is that? Does it refer to internal energy?

Yes, it corresponds to the internal energy of the crystal. IUCr uses a more well-thought symbol $U_\mathrm{L}.$

Why do we use $\Delta U$ and not $\Delta H$? Is the measurement of lattice energy done at constant volume, or something?

@Quantonium I suppose because the lattice energy it is the internal energy of the lattice.


text:
Calculation of concentration at equilibrium for extraction

The equation for equilibrium of solvent-solvent extraction is:
$$\ce{PhOH(toluene) <=> PhOH (aq)}$$
An aqueous solution contain $\pu{1.5 \times 10^{-2} M}$ phenol and is shaken with the same volume of toluene. Determine the equilibrium concentration of $\ce{PhOH(toluene)}$. $K_\mathrm{D} = 14$

I have done lots of similar questions and each time I think I've gotten the method of how to do these types of questions it seems like the method doesn't apply anymore. In this question, I am really confused because I initially solved it and got the right answer. But when taking a closer look, I noticed that my calculations were wrong, and did it again in the correct way (or at least what I think is the correct way), but I got the wrong answer.
My attempt at solving the problem
Given that the aqueous solution had the concentration $1.5 \times 10^{-2}$, I made the following table, where $n_a$ is what I want to calculate.
$$\begin{array}{c|c|c|}
 & \text{PhOH(toluene} & ⇌ & \text{PhOH (aq)} \\ \hline
\text{t=0} & - & &\ \pu{1.5 \times 10^{-2} M}\\ \hline
\text{n equilibrium} & n_a & &n_0-n_a \\ \hline
\text{c equilibrium} & \frac{n_a}{V} & & \frac{n_0-n_a}{V}
\end{array}$$
My question is, when writing the formula for $K_\mathrm{D}$, is the organic phase always in the numerator? and the aqueous phase in the denominator? Meaning that if my reaction is written as above, do I still have the organic phase divided by the aqueous phase or do I follow the rule of products over reactants. If I follow the rule of products over reactants, I get the correct answer, but if I take org/aq which is the formula our professor refers to, then I get the wrong answer.
Which way is the correct way to solve a problem like this? All help is truly appreciated!

Why not to just take $$\frac{c}{c_0-c}=K_\mathrm{D}$$ ?

I'd like first to answer your question:

My question is, when writing the formula for $K_\mathrm{D}$, is the organic phase always in the numerator and the aqueous phase in the denominator?

The answer is yes. The IUPAC Recommendations 1993 (Ref.1) defines Partition Ratio ($K_\mathrm{D}$) as follows (also see the Goldbook):

Partition Ratio ($K_\mathrm{D}$): The ratio of the concentration of a substance in a single definite form, $\mathrm{A}$, in the extract to its concentration in the same form in the other phase at equilibrium, e.g. for an aqueous/organic system.
$$(K_\mathrm{D})_\mathrm{A} = \frac{[\mathrm{A}]_\mathrm{org}}{[\mathrm{A}]_\mathrm{aq}}$$

Thus, if you have used equal volumes of organic and aqueous phases, the equation you can set up to solve is:
$$(K_\mathrm{D})_\mathrm{A} = \frac{[\mathrm{A}]_\mathrm{org}}{[\mathrm{A}]_\mathrm{aq}}= \frac{[\mathrm{PhOH}]_\mathrm{org,eq}}{[\mathrm{PhOH}]_\mathrm{aq, initial}-\mathrm{PhOH}]_\mathrm{org,eq}} \\ =\frac{[\mathrm{PhOH}]_\mathrm{org,eq}}{\pu{1.5 \times 10^{-2} M}-[\mathrm{PhOH}]_\mathrm{org,eq}}= 14 \tag{1}$$
When you solve the equation $(1)$ for $[\mathrm{PhOH}]_\mathrm{org,eq}$, you get your answer:
$$[\mathrm{PhOH}]_\mathrm{org,eq} =\frac{\pu{1.5 \times 10^{-2} M} \times 14}{15}= \pu{1.4 \times 10^{-2} M} $$
Hence, $[\mathrm{PhOH}]_\mathrm{aq,eq} = \pu{1.5 \times 10^{-2} M} - \pu{1.4 \times 10^{-2} M} =  \pu{1.0 \times 10^{-3} M}$
Note: If you perform extraction in different volumes of aqueous and organic phases, you should read this Wikipedia article to get help for solution.
References:

N. M. Rice, H. M. N. H. Irving, M. A. Leonard, "Nomenclature for liquid-liquid distribution (solvent extraction) (IUPAC Recommendations 1993)," Pure & App. Chem. 1993, 65(11), 2373-2396 (DOI: https://doi.org/10.1351/pac199365112373).

But I want to calculate $[PhOH]_{toluene,eq}$ which is why I am confused. Because $1 *10^{-3}$ is the correct answer but I get that answer when solving for the aqueous phase and not the organic phase..

That "correct answer" answer is incorrect, according to definition of $K_\mathrm{D}$, which is 14, indicating $\ce{PhOH}$ is more organic like. You should consult your teacher about it.


text:
How to carry out cyclization of 1‐(2‐methylbutyl)‐2‐nitrocyclohexane?
How the cyclization can take place as there is no other functional group than $\ce{NO2}?$

One way that I can think of is photochemical bromination which occurs selectively at tertiary center which then can be promoted to give cation in polar medium. Nitro group on being reduced to amino group will attack on carbocation, thus closing the ring.
The problem here is that there are three tertiary centers and radical close to functional group will be more stable than the one at the far tertiary center on side chain.
How would you handle this conversion?

You might want to read up on the Hofmann-Loeffler-Freytag reaction. https://en.wikipedia.org/wiki/Hofmann%E2%80%93L%C3%B6ffler_reaction

The Hofmann-Loeffler-Freytag reaction as suggested by @user55119 in the comments above is the way to go.
First you need to reduce your nitro group. The reduction of alkyl nitro compounds is less simple than aromatic nitro groups and lower yielding in general. I would suggest the use of Al/Hg amalgam.
Acetylate the resulting amine.
N-chlorinate the amide - trichloroisocyanuric acid is good for this conditions
Irradiate the chlorinated amide to form the [5] ring similar to this example Baldwin gelsimicine synthesis
Finally hydrolyse off the acetyl group with cHCl.


text:
What determines solubility (or miscibility) in organic solvents?
I'm actually quite surprised that this isn't really mentioned anywhere on the internet (or I'm just bad at searching)! Solubility in water is known to be directly related by multiple factors, and for organic molecules it's usually ability to form hydrogen bonds, polarity, bulk etc. I would imagine this trend would continue for most of the smaller alcohols (methanol, ethanol, isopropanol etc.) to some extent. However, I can't find even a brief mention on how solubility in common solvents is determined.
Let me share my (admittedly very elementary) logic:
For a solvent like cyclohexane, which is non-polar, I imagine other non-polar solutes would be able to dissolve. However, if the non-polar compound is solid, it will require heating to break the van der Waals forces first. I assume little to no enthalpy change occurs in this process. Other more polar solutes would prefer to continue associating with themselves, and therefore not dissolve.
For a solvent like diethyl ether, it gets a tad more complicated. The oxygen atom on the ether is a hydrogen bond acceptor, and the molecule is polar. So, I imagine that non-polar solutes are poorly soluble. The same should go for compounds like amines, alcohols and carboxylic acids because they rather hydrogen bond with themselves. However, for meekly polar compounds with functional groups like aldehydes, ketones and halogenoalkanes, I imagine it depends on its polarity? It's polarity should be near that of the solvent; if it's more polar it will rather be with itself, if it's less polar the solvent would rather be with itself. So for example, if I mix ethyl chloride with diethyl ether, will they be miscible?
If anyone can support or debunk my logic, it will be greatly appreciated.

There was a plethora of textbooks published from 1970s to early 1980s about chemistry happening in organic solvents. These books are good at delivering fundamental principles of intramolecular interactions and simple enough models for understanding basic principles of miscibility and selection of solvent for a particular synthesis.

From the top of my head, I'd suggest 1. Nonaqueous Chemistry; Topics in Current Chemistry; Springer-Verlag: Berlin/Heidelberg, 1972; Vol. 27, and 2. Popovych, O.; Tomkins, R. P. T. Nonaqueous Solution Chemistry; Wiley: New York, 1981. Also, get to know Hansen solubility parameters.

@andselisk Thanks! Will check it out.

Not really clear what we shall confirm or debunk. Yes solubility is a huge topic with a simple principle (like likes like) and a lot of details. That is why polarity of a solvent is not, or is not necessarily its electrical dipole moment. .

@Alchimista I was just referring to the two examples I gave above (cyclohexane and diethyl ether), which can help me understand solubility as a whole ;)

@chematwork I see it better now. More or less it should be like that, plus shape. You basically means that without reactions or particular reaction everything boils down to electrical polarity. It makes sense, considering the  surfaces effectively exposed to the counterpart by each member of the couple.


text:
Does 99.8% acetic acid cause severe skin burns like formic acid?
I've so far handled only concentrated fuming nitric, sulfuric & hydrochloric acids from the inorganic side. Even during an accidental spill on the hand I did say only the $\ce{HNO3}$ affects the skin by turning it yellow and the top layer sometimes peels off.
But in the organic acid side I've had once experience in the past long long ago with formic acid and it was pretty nasty. A really pungent smell and if it gets on the hand then immediately the top layer of skin will peel off. Worse than there inorganic ones. Is 99.8% acetic acid this strong?
I'm about to prepare some homemade stop bath solution for Black and White film development. I'm going to prepare a 10% acetic acid solution. Just wanted to know though I will wear nitrile gloves, in case a few drops come on the skin how bad will it be? I can't seem to find any video or much info of it's effect on YouTube or anywhere either? Just want to take all precautions.

It's not so bad as formic, afaik. Pure acetic acid is glacial, though.

Glacial acetic acid will cause a nasty burn. It might not be as bad as formic acid, but it took most of the skin off my right hand.

@MaxW Yikes! Thanks for the info. I never though acetic acid would be that bad. Well again the oner I've is glacial. I read some people mention it's flammable too. Does it catch fire when near a flame? IS that what people mean or is it because it reacts with metals to produce H2 gas?

One can find the flammability of a substance  by looking up its lash point, e.g. https://webwiser.nlm.nih.gov/substance?substanceId=402&identifier=Acetic%20Acid&identifierType=name&menuItemId=32&catId=58 . That means it would not easily catch fire until heated to just above body temperature.

Safety data sheets for chemicals exist for a reason. Read what the sheet says about glacial acetic acid (and worry about the vapour as well as the liquid: it is pungent and dangerous if you breathe it).

@The_Vintage_Collector If you're going to be playing with chemicals you really need to educate yourself on basic lab safety. One critical part of that is learn about safety data sheets (SDSs) - every chemical that can be bought or sold must have one and you must get into the habit of reading the SDS before you even think about opening a chemical you are not familiar with. It tells you everything that is known and not known about a chemical's properties, hazards, and safe handling requirements. Don't trust internet people - read the SDS.

@J... I did read the msds. The reason I asked for is because all acids and types of bases that even mildly will irritate skin are classified as ones that will cause burn. But this is the first time I’m handling highly concentrated acetic acids. 97% of the time I’ve had to handle only inorganic acids and mainly HNO3, H2SO4 & HCl. And a splash of few drops on hands that were not covered or torn nitrile gloves should cause immediate damage. I’ve had only my first experience with organic acids with formic like I mentioned above.

@J... That incident is what made me to respect all acids and bases with extreme caution. Like I said in my reply tot he answer below. This time I did use special acid protective gloves along with safety glasses and a respirator. I just wanted to get an idea what I’m getting into.

@The_Vintage_Collector If you read the SDS and still had questions then my point is that it's worth learning how to read an SDS.  If it's just a mild irritant, it will say it's a mild irritant.  If it will cause severe skin burns, it will say "causes severe skin burns".  SDSs don't use hyperbolic language for flair - if it says "severe", it means it.

Yes it is corrosive to skin.  The Safety Data Sheet identifies it as Category 1A for skin corrosion, which means:

A substance that, according to animal data acquired from a scientifically validated method, produces irreversible damage to the skin after an exposure of three minutes or less, and within one hour of observation, in at least one of three animals.

Don't let the "organic" part fool you.  In terms of handling hazards glacial acetic acid rivals concentrated hydrochloric acid, and actually beats it in skin corrosion classification.

I hurt my hand about 1971. Long before any sort of protective gloves were common. What I was doing was setting up the reagents for the experiment where acetic acid and ethanol form an ester. Being an idiot I thought glacial acetic acid was just like vinegar.

Oscar Lanzi & @MaxW This is what happened with me when I was like 15 years old when I handled Formic acid. I managed to prepare the solution and used some special gloves used to handle acids and other corrosive liquids when handling them.


text:
iupac name please?

I am kind of confuse here whether to  number $\ce{NO_2}$ or double bond in the ring.

The double bond and the nitro group are the two last things to number in this compound. What do you have so far?

2-(6-nitrocyclohex-3-en-1-yl)ethanoyl chloride should be the correct name.
According to Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a lower locant should be assigned to the substituent only written as a prefix (i.e. the nitro group) than a double bond '-ene' ending, since the unsaturation is considered to have a higher seniority or priority.

Is using "acetyl" instead of "ethanoyl" wrong? Chemdraw says acetyl.

Acetyl is the preferred IUPAC name while Ethanoyl is the systematic IUPAC name. You can use either.

@VijayBharadwaj "2-(6-nitrocyclohex-3-en-1-yl)ethanoyl chloride" is not the preferred IUPAC name. You might want to read the Blue book again about the preferred name for acetyl chloride and the omission of locants.

I think the OP's query was regarding the correct way to assign locants to the substituents. Other than the ethanoyl-acetyl issue, I think the locants I've assigned are correct. Please correct me if I am wrong though.


text:
Could Carbon Monoxide be reduced by oxidising a more reactive metal?
After studying displacement at school, I wondered if it is possible to reduce the toxic gas of carbon monoxide to less harmful and more useful substances, like so:
Carbon monoxide + metal -> Carbon + metal oxide
Where the metal is of course more reactive than carbon, higher in the reactivity series, and so would displace it.
Is this chemically possible, or does it require too much energy? Forgive me if I appear foolish, but I just wonder if it could actually happen.
After researching extensively online, the main things which come up in searches are reduction of carbon dioxide, rather than carbon monoxide.

This must be possible with at least some metals, in that it won't require too much energy. But it will require too much money.

Oh right, that's interesting. Why would it be expensive? And would using something like aluminium work? As it's more reactive than carbon.

You don't have pure CO. You have a ton of CO2 and a little CO, and your metal is going to reduce both (at best).

Oh right. I was thinking more electrochemical reduction of something like smoke, which has a really high CO percentage in comparison to CO2. But would this fail too?

Smoke hardly ever has all that much CO, not to mention having a lot of O2. If you had pure CO some pretty interesting stuff can be made from it via reduction with alkali metals.

Magnesium is known to continue burning in CO2, producing carbon. Just guessing it could burn in CO as well, but the spontaneity should be calculated.

@Poutnik Well, $\ce{CO_2}$ is a lot more oxidizing (carbon at higher temperatures will burn exposed to it) than $\ce{CO}$. I have ruined many a yearly salary's worth of precision machined graphite parts to $\ce{CO_2}$ exposure, believe me.

@StianYttervik I am aware of CO2 being much better oxidant than CO. But as Mg reduce CO2 to C and not just  CO says something. As I say, it has to be calculated, or deduced from wikipedia: Ellingham diagram. Implying from the Mg burning temperature and the chart, Mg is able of reaction $\ce{Mg + CO -> MgO + C}$ above some 1850 deg C.

While possible, this way of reducing CO is not very practical. We already have good ways to reduce CO emissions in some circumstances. Vehicles use catalytic converters, for example, to oxidise CO to CO2 to reduce emissions, for example.


text:
Why is oxygen more stable than oxygen dication?
According to molecular orbital theory (MOT), $\ce{O2^2+}$ has a greater bond order than $\ce{O2}$ and two less antibonding electrons. So why is molecular oxygen $\ce{O2}$ more stable than the molecular ion $\ce{O2^2+}?$
One possible reason that comes to mind is that the antibonding (AB) orbitals, although higher in energy than the constituent atomic orbitals, are still negative in energy, so adding electrons to AB orbitals still lowers the energy. But then, by that same argument, $\ce{O2-}$ would increase stability further, which it doesn't. Am I overlooking anything?
Note: I know that from an electrostatics point of view, $\ce{O2}$ is obviously more stable as bring two negative electrons close to positive $\ce{O2^2+}$ decreases the electric potential energy of the system. I am interested in how MOT explains relative stability (or if it actually even does).

You have to distinguish reactivity (with commonly found substances such as water and other components in air) and bond strength. Just because something has a strong bond doesn't mean it does not react - just compare alkanes and alkenes and alkynes.

I am strictly talking about thermodynamic stability. O2 is more thermodynamically stable than O2+2 by elementary electrostatics, but MOT predicts otherwise. I am asking for an explanation for this

You can only compare substances when they have the same set of atoms and the same charge. How would you assign thermodynamic values to a naked electron?

Well, I think it is not more stable. For cation enthalpy is lower, bond stronger, but... getting those electrons still gives off energy, by aforementioned electrostatics. So, why try to get answer when you already have it?

@Mithoron I'm afraid I don't understand. O2 is NΟΤ more (energetically) stable?

How can it not be more energetically stable if adding electrons to the dication releases energy?

If you compare just the two things - the cation is just better, but if you add electrons to the picture, ah then pesky charge sucks them up and spits out surplus energy. It's like Earth is perfectly OK without the Moon, but if the satellite didn't have enough energy to stay in orbit, it would crash, big boom lots of energy comes off, but result has still more internal energy then Earth in the beginning, being bigger.

You have to consider the system as a whole - you can't directly compare $\ce{O_2}$ and $\ce{O_2^2+}$ because they have different numbers of particles. To put it another way when you consider the relative stability of two interconvertible specifies you really have to write down a chemical reaction that connects them, and then consider which direction is thermodynamically favoured, and then consider whether it is kinetically feasible. Stability is a really tricky and more often than not misused word!
So here we have to look at $$
\ce{O_2 -> O_2^2+ +2e^-}
$$
As you note $\ce{O_2^2+}$ has a stronger bond than $\ce{O_2}$. But that is not all that is going on. We are also binding two electrons to the Oxygens - how does that affect the energetics? Well the reaction above is simply the sum of the following processes
$$
\ce{O_2 -> O_2^+ +e^-}
$$
$$
\ce{O_2^+ -> O_2^2+ +e^-}
$$
In other words the energy required is the sum of the first two ionisation energies of molecular oxygen. Do we know these? Well after a good 10 seconds looking I can't find the second energy, but the first is very well known, it is after all related to the discovery of compounds of Xenon and the start of noble gas chemistry. It is 12.0697 eV (from https://webbook.nist.gov/cgi/cbook.cgi?ID=C7782447&Mask=20) which in more familiar units is 1165 kJ/mol - a huge energy, bigger than any bond energy I am aware of, and note there is at least this much to come as the second ionisation energy will be even larger, as it is pulling a negative electron away from an already positively charge centre. Thus the energy gained by attaching the two electrons will massively out weigh any increase in bonding in the molecule.
Can we relate this to MO theory? Well as Koopman's theorem tells us the ionisation energy is an approximation of the orbital energy, this is telling us that relative to having a free electron it is more energetically favourable to put the electron in the lowest energy unoccupied orbital in the molecule. That this orbital is antibonding is irrelevant. Antibonding simply means it weakens any bonding in the molecule; it doesn't say anything about the energy of that orbital relative to the free electron. And you know this must be true - many molecules have electrons in antibonding orbitals and those electrons don't just rush away to infinity, hence occupying the orbital must be a lower energy proposition for the electron compared to being unbound. Thus $\ce{O_2}$ is thermodynamically more stable than $\ce{O_2^2+}$ and two free electrons because the decrease in energy caused by the electrons occupying the highest unoccupied molecular orbitals is (much) greater than any possible decrease in the bond energy.
That's the thermodynamics. Kinetics is difficult in general, but very unlikely to be an issue here. At this point I chicken out.

Would it be correct to say that the bond order in MOT and the 'stability" it describes is relative to the unbonded state, NOT infinite separation of each electric particle, and O2 & O2+2 have unbonded state with different energies (infinitely separated atoms vs infinitely separated cations)? If this is correct, wouldn't that nullify the use of MOT comparing relative stability (which is pretty much all we use it for at this level)

Secondly, what do you exactly mean by the extent of "bonding"/bond energy ? Wouldn't having more electrons in the system further lowering the electric p.e make it more difficult to pull the nuclei apart and thus increase bond energy too?

Not totally sure I understand what you are saying, but I'm taking it as saying that you have to be careful in making sure you have a consistent zero of energy. If so yes, obviously. If you do that you can make meaningful energetic comparisons. That this is poorly taught or even omitted is a different matter - and that observation may very well be true but I have been out of the education system at least for below graduate level for so long that I don't nowadays feel qualifies to comment.

To your second comment - they go into antibonding orbitals. What do antibonding orbitals do?

Oh ok. I was picturing the 2 extra electrons between the nuclei for some reason.  A quick refresher on antibonding electron densities explains this. So basically, while it is more difficult to pull apart each electron from the nucleus in the molecule, it is more difficult to (symmetrically?) pull apart the 2 nuclei in the dication?

All in all, would it be more accurate to describe the bond order s a measure of bon energy as opposed to thermodynamic stability?(extremely sorry for continuously bothering you )

Careful about thermodynamic stability - you have to consider that in the context of the WHOLE system. But "would it be more accurate to describe the bond order as a measure of bond energy", yes, precisely. Bond order applies to the isolated molecule, and is independent of the environment molecule is in (more or less). Thermodynamic stability requires the whole system to be considered.


text:
Chemistry Precise
What would be more precise to find the volume of a cube of metal and why?
1: Measuring cube with a ruler and calculating volume
2: Calculating the water displacement of the cube with a graduated cylinder
(1-2 Sentences)

Depends on how accurately performed is each measurement. One would have to take into account precision of ruler and asymmetry of cube, for example, in the first case, and adhesion of water to cube and cylinder, causing meniscus and other errors.
However, in some cases, ruler measurement would be far more accurate than water displacement: Archimedes would have shouted something other than "εὕρηκα" if he were measuring a cube of sodium (or congeners) in his bath. Doubt he'd want to lye in that!


text:
In a heating curve, when it is plateau, where does all the energy go since the temperature does not get higher?
So between the change of phases within the matter(solid liquid gas), the temperature reaches a plateau, no matter how much heat you add to it, it just stays the same. Where did all those energy go?

You may have learnt at some point that a change of phase is endothermic (i.e. needs heat to happen). So, when you transition from a phase to another and keep adding heat, the temperature will remain the same, as all the heat you'll add will be used for the change of phase. Hence, the temperature will remain the same until the transition is over (i.e. you gave your system enough heat for the transition to be complete), then will vary again.


text:
Steam distillation vs fractional distillation
I am a high school student and I am a little confused in the techniques used to purify organic compounds. For example, in steam distillation we reduce the external pressure on the compound to be purified by passing steam over it, which reduces its boiling point.
First of all, how is that possible? At first I thought that the external pressure on that compound should be increased because now not only air, but also steam will also exert pressure over it.
OK, if I assume that it will decrease then also ,why do we use this? If we have to reduce the external pressure like this, I think  we can do this in fractional distillation — also by taking the mixture of water and a water-immiscible compound. Shouldn't the boiling point also decrease?

The point is that saturated vapour pressure is—at least ideally—proportional to the molar fraction of volatile compounds. Steam distillation is used for substances insoluble in water, so its molar fraction in is phase is ( or near to ) 1. the consequence is, vapour pressures add each to other and when the sum equal to external pressure, the heterogenous mixture is boiling. Steam distillation is typically used for preparation of essential oils from herbs and plants, e.g. for aromatherapy.

Your question arises from a wrong premise. Steam distillation is not at reduced pressure, and depending on the apparatus the pressure can be even a bit higher. For how it works see above.

Steam distillation simply adds the vapor pressures of mutually insoluble substances. It happens at any equilibrium T and P and can be effected by just boiling the mix. The normal boiling point of water is a convenient T and works for compounds with a reasonable vapor pressure Perhaps you have noticed while gently cooking meat with water in a covered skillet the inside of the lid accumulates a thin layer of fat. That is mostly steam distillation of the lower MW fats. It also works in reverse the azeotropic drying of solvents such as benzene, toluene removes water. Injected steam agitates.

Let me start from the basics, Distillation is used to purify a mixture that has volatile liquid and non-volatile impurities.
We use Steam distillation in which the oil has a significantly high boiling point than water atleast and it is also immiscible with water.
1)Steam does not increase its external pressure rather it increases the mixture's vapor pressure as well as partial pressure due to water vapor. What happens is that the steam which we pass into the oil makes a mixture with the oil and creates an oil-water mixture which is an azeotrope, so now the water in the oil-water mixture has its own vapor pressure which is now added to the vapor pressure of oil, which increases the mixture's internal pressure.
2)As I have said earlier, the oil-water mixture is an azeotrope, so the boiling point of the mixture becomes significantly less than the boiling point of the oil.
For example, by data for Pehnylaniline and water mixture, saturation pressure of phenylaniline is 7.07 kPa and for water, it is 94.30 kPa and hence for the mixture, it is 7.07 + 94.30 = 101.37kPa and hence the mixture will boil just below 100-degrees Celcius but if we take Phenylanline only then its boiling point is 184-degree Celcius which makes it easier to carry out distillation.
Also, we use steam to provide heat to the oil-water mixture. The steam provides heat to the oil by condensing which also reduces its loss to the flask, environment, etc.
Some links for sources,
Steam distillation- how it actually works?
Internal vs external steam distillation
http://www.chemguide.co.uk/physical/phaseeqia/immiscible.html

Steam distillation can also be used in reverse to dry a solvent, such as benzene, pentane, hexane, immiscible with water.  Gentle distillation will remove the water as an azeotrope. It seems to work also for chloroform and methylene chloride even tho there is some solubility.  This is used to remove water from esterification type reactions.


text:
Equilibrium for the synthesis of methanol

Liquid methanol is obtained with carbon monoxide gas and hydrogen gas in a reactor at $\pu{300 °C}$ and $\pu{250 atm}.$ $K_p = \pu{9.28E-3}.$ Find $K_c.$

With only these data, do I have to use the following equilibrium for the calculations
$$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)},$$
or do I have to suppose first methanol is obtained as a gas and then liquified:
$$\ce{CO(g) + 2 H2(g) <=> CH3OH(g)}?$$

To find out, it suffices to know the state of methanol under the reaction conditions: have you looked for it?

Yes, I saw in a phase diagram of methanol that under these conditions it is a supercritic fluid (hybrid between gas an liquid...)

Carbon monoxide is CO. Further more, equations are not balanced, unless you managed conversion O -> 2 H

Yes, sorry for the mistake

This question will be completely impossible to answer unless the text matches the chemical equations. are you talking about the reaction of carbon monoxide or dioxide?

methanol critical temperature is 240.2 deg C, so methanol cannot be liquid at 300 deg C at any pressure.

Note that under the given conditions, methanol cannot exist in the liquid state, however since you are given in the problem that methanol is obtained in liquid state, you will only consider the equilibrium (for problem solving purpose)
$$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)}$$
And since $K_p$ is related to $K_c$ by the relation
$$K_p = K_c(RT)^{\Delta n_\mathrm{g}},$$ where $\Delta n_\mathrm{g}$ represents the change in the number of moles of gaseous species, so here $\Delta n_\mathrm{g} = -3,$ Therefore $K_\mathrm{c}=K_\mathrm{p}(RT)^3=9.28\times 10^{-3}\times (300 \times 0.0821)^3 \approx \pu{138.65 atm^3 L^3 mol^-3}$.

Comments are not for extended discussion; this conversation has been moved to chat. (Since I have this comment here, pardon me for adding a link to the newly redesigned FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?, as it teaches you how to use \ce{...} and \pu{...} correctly.)

Hm, OTOH, not all comments are discussion.


text:
How do I extract anthocyanin from a blueberry?
I am doing a school project to extract anthocyanin from a fruit. I chose blueberry. What is a step by step method to extract the anthocyanin? I have to do it in school so hopefully its not a very complex procedure. Also, im not allowed to use methonal as it is flammable. From what i have researched, i should use ethanol and hydrochloric acid. But i do not know what else to add. Thank you.

Huh?!? Ethanol is flammable too...

In what form do you want them to be obtained? To do what ? Because depending on the purpose, we can proceed through more or less complex experiments ...

Ethanol/water, HCl procedure here - though whether you can do it as described depends on how well equipped you school lab is. https://www.researchgate.net/publication/276264844_Optimal_Extraction_of_Pure_Anthocyanins_from_Fruits_of_Sambucus_nigra

@Nicolas Im obtaining the anthocyanin to conduct a kinetic study to explore the antioxidant properties of the extracted anthocyanin

@MaxW actually im not too sure why we are not allowed to use methanol. Maybe the school just does not have it

Methanol is a lot more toxic than ethanol, which is probably the reason.

@Waylander so i crush/blend the blueberries and add 70% ethanol-water 1%citric/hydrochloric acid. After that what is the extraction process? I filtrate the solution with filter paper and obtain the filtrate, mix it with silica gel, filtrate again and then heat up the filtrate to get rid of the ethanol. Lastly i purify it by adding silica gel, acetic acid and ethyl acetate. I dont have to filtrate the solution again or heat it up to get rid of the silica gel, acetic acid and ethyl acetate? Also it says that i add acetic acid to remove water soluble compounds, but isnt anthocyanin watee soluble?

Anthocyanins are hot water souble but note the specification of the silica quoted in the procedure - it is reverse phase silica.


text:
Why are 13C CP MAS NMR signals sometimes split by N, and absent in dipolar dephasing?
A past paper question on solid state NMR I was looking at asks why, in $^{13}$C CP MAS NMR, carbons directly bonded to nitrogen are "occasionally" split, and why they're usually absent in the dipolar dephased spectra.
I assume this is probably due to quadrupolar interactions, since $^{14}$N has spin 1; the size of the quadrupolar interaction depends on the electric field gradient at the quadrupolar nucleus, and hence the symmetry of the environment, so the "occasionally" presumably occurs when the nitrogen is in a particularly asymmetric electronic environment and so has a very large quadrupolar coupling? I'm not sure why we don't see this splitting all the time.
Presumably the absence of these carbons from dipolarly dephased spectra is due to a short T$_2$ time - just like CH and CH$_2$ carbons have a short T$_2$ time due to dipolar coupling, I assume the quadrupolar interaction reduces T$_2$ time.
I'd be very grateful for help understanding this.


text:
How does hydraulic equilibrium work on the particulate level?
I recently learned that osmosis occurs when there is a semipermeable membrane between pure water and a solution of water. The pure water then fills up the solution's side until, as my textbook put it,

"The pressure exerted by this column of solution counterbalances the pressure exerted by the water moving through the membrane from the pure water side."

How does water naturally want to move from one gradient to the other? Doesn't water move through the membrane both ways? Also, why does water "want" to go to the more concentrated side when the particles move at random?
In other words, why doesn't the flow of water molecules depend on the number of water molecules in the membrane instead of the concentration gradient?

There is water concentration, or rather activity, difference. More water molecules tries to get to the other side from the less concentrated solution.

The membrane has plenty of holes that are a little bit bigger than small molecules like water. Bigger molecules or bigger ions like usual solutes cannot go through these holes. Of course water has the tendency to cross the membrane both ways. When a water molecule hits a hole in the membrane from the pure water side, it can cross it. Nothing prevents it. On the other side, when a solute molecule hits a hole of the same membrane, it blocks the passage for a couple of microseconds. No water molecules can go through this hole from the solution side of the membrane during this time. So on average, more water will cross the holes of the membrane in the direction pure water to solution than backwards.
Is it what you wanted to know ?

Water diffuses across the membrane in both directions. The higher concentration of water [AKA Activity] in the water with less solute means more collisions with the membrane more penetration into the membrane and more transport across the membrane. The process equilibrates when the water activities are the same on each side of the membrane. The solute does not "block the passage".  Membranes, especially in reverse osmosis where pressure is used to reverse the natural osmosis, will coat, foul,  and the membrane becomes nonpermeable. When this happens the holes are plugged.

Instead of saying "blocks the passage", you could just say that the rate of water getting to the holes is lower because the concentration of water is lower. If you make a molecular simulation, it does look like blocking a bit. Because two molecules can't be in the same place at the same time, one molecule always blocks the other in that sense.


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What is a possible conventional interpretation of Wilfred Hagen's puzzling Electron Paramagnetic Resonance spectrum?
In searching for a conventional interpretation, Wilfred Hagen of Delft Technical University is stumped by an Electron Paramagnetic Resonance spectrum (aka Electron Spin Resonance spectrum) in "Distinguishing electron paramagnetic resonance signature of molecular hydrino":

A sharp, complex, multi-line EPR spectrum is found, whose detailed
properties prove to be semi-quantitatively consistent with
predictions20 from hydrino theory with an average error less than 0.09
G (0.2%) over a 39 G span of 37 lines. We have sought but failed to
find reasonable alternative, ‘conventional’ interpretations for the
detected paramagnetism.

Figure 2. EPR of postulated molecular hydrino H2(1/4) caged in solid
Ga(O)OH polymer. Trace a: wide-scan overview spectrum showing a single
feature only close to the free electron g value. Trace b: zoom-in of
the single feature in trace a shows two main lines of equal intensity,
separated by circa 4 Gauss, and whose center is distinctly shifted
from the free electron value to g = 2.0045. Trace c: further zoom-in
on one of the lines now recorded with a very small modulation
amplitude reveals a fine structure of multiple lines with apparent
peak-to-peak derivative line width of 0.17 Gauss and separated by
circa 0.32 Gauss. Trace d: extensively averaged spectrum taken under
conditions optimized for maximal signal-to-noise ratio at the expense
of minor over-modulation and power saturation, exhibits a complex
pattern of triplet satellite lines. Data collection times for traces
a-d were 10, 16, 375, and 2400 min, respectively. All spectra were
taken at ambient temperature. Trace e is a simulation using field
positions predicted by hydrino theory. Other experimental conditions
and simulation parameters are given in the METHODS section.

He provides the most plausible conventional interpretation:

Alternative interpretation. Alternative to the hydrino analysis in Fig
2e the spectrum in Fig 2d can also be approximately reproduced under a
conventional phenomenological spin Hamiltonian assuming an unusual
combination of two isotropic radicals of unequal intensity each with a
g value of 2.0046. This model would require the two main lines to be
due to an isotropic S = 1/2 system split by an I= 1/2 nucleus with
$A_{iso}\approx$ 3.9 Gauss with additional hyperfine structure form a
combination of some 5 nuclei the majority of which has also I = 1/2. A
second S = 1/2 system should give rise to the satellite lines due to a
different combination of some 5 nuclei, one of which should have I = 1
(e.g. $^{14}N$) to account for the repeating triplet pattern (see Extended
Data Fig. 7 for a detailed analysis).

However, he rejects it:

We consider this alternative explanation of the EPR highly unlikely on
the following grounds. The reaction mixture only contains H2, O2, H2O,
and Ga. Even in the presence of trace contaminants of, e.g., C, N, we
cannot envision how the high-temperature plasma reaction conditions
and sample formation in strong aqueous base could lead to the
formation of stable radical structures of considerable complexity. The
ToF-SIMs, EDS, and XRD analyses also eliminate alternatives.
Furthermore, since the sample is a solid, for complex radicals one
would expect to see anisotropy in the spectra. In particular
absorption-shaped peaks that come with axial or rhombic symmetry of
the spin Hamiltonian are not observed.


text:
Why does the specific conductivity against concentration curve for NaCl curve more than KCl?
I understand why KCl has a higher specific conductivity than NaCl.. but I dont understand why the curvature is more significant for NaCl (KCl is a straight line). Ive provided a picture of this here, and this is also what I get from my experiment. Does anyone mind explaining why?

Could you provide units for the abscissa?

Also, mS is not a unit of specific conductance (conductivity).

@BuckThorn this is a diagram i found online, i think it is mass %, im not sure. is specific conductance (conductivity) different than specific conductivity? i also got similar curves in my own experiment.. KCl was a straight line and NaCl was curving, where my y axis was micro siemen/cm and the x axis was mol/dm-3

Why don't they all start from the same data point at 0%?

@ManRow hmmm good point, i have no idea? any theories? do you think the greater curvature of NaCl is plausible? tbh im kinda unsure about the unit of the. y axis here and if it is the same as specific conductivity (which is siemen/m)

I remember very similar queation was asked within last months.

@Poutnik I also remember one. Can you find the link?

Could you provide a link to where you obtained the data?

@BuckThorn i found may pictures that show this type of curvature.. it's also what i got from my own experiment

Ok, the reason I asked is because the non-zero intercept is highly suspicious (actually, it doesn't make sense) unless there is some background electrolyte of course.

I think something is wrong. At 0% point, all lines must start from the same point. So an error has occurred.

For equally charged ions a smaller radius implies a higher charge density, and therefore stronger Coulombic interactions with other charges in solution, including stronger ion-dipole interactions with water molecules, resulting in a more stable hydration sphere, and stronger interactions with counterions (here chloride). Such interactions reduce the mobility of the ion by opposing the effect of the applied external electric field, by either increasing solvent drag or generating an opposing electric field (see for instance this chem libretext for an explanation). This slowing down can be interpreted as an increase in the hydrodynamic or Stokes's radius, and also results in a reduced diffusion coefficient, as shown in the following table (source CRC, $\pu{ 25^\circ C}$):

Ion
$\Lambda_+ \pu{(10^-4 m2Smol^-1)}$
$D \pu{(10^-5 cm^2s^-1)}$

$\ce{Cs+}$
77 .2
2 .056

$\ce{Rb+}$
77 .8
2 .072

$\ce{K+    }$
73 .48
1 .957

$\ce{Na+}$
50 .08
1 .334

$\ce{Li+}$
38 .66
1 .029

$\frac12$ $\ce{Sr^{2+}}$
59 .4
0 .791

$\frac12$ $\ce{Ca^{2+}}$
59 .47
0 .792

$\frac12$ $\ce{Mg^{2+}}$
53 .0
0 .706

$\frac12$ $\ce{Be^{2+}}$
45
0 .599

Therefore the smaller monovalent cation, $\ce{Na^+}$, and divalent cation, $\ce{Mg^{2+}}$, have lower molar conductivity than the species lower in the periodic table ($\ce{K^+}$ and $\ce{Ca^{2+}}$).

Additional comments:
(1) Instead of discussing NaCl or KCl, start discussing individual cations and anions, recognizing that these salts completely dissociate into ions in solution. Notice that the different salts share a common anion (chloride, Cl-). Therefore any differences are due to the choice of cation. Sounds obvious, but then you can look at everything from the perspective of the cation's interactions (even if those of the anion will also change between salts, and affect conductivity).
(2) Distinguish between the size of an ion before and after hydration. Bare (unhydrated) $\ce{Na^+}$ is smaller than $\ce{K^+}$, but hydrated $\ce{Na^+}$ is bigger than $\ce{K^+}$.
(3) Bigger bare (unhydrated) cation means weaker interactions and therefore faster mobility.
(4) What drives the formation of a larger hydrated ionic radius is deep down the same thing that drives interactions with other ions. The hydration shell is not that tight. Interaction with other ions will readily override them.
(5) Stronger interactions with other ions means more deviation from ideality. Ideality here means that individual ions behave the same (their mobility is the same) at all concentrations.

I thought KCl has a smaller hydrated ionic radii than NaCl though? I am using the molar conductivity values but Im just not sure why the slowing down affect is greater for KCl given it has a smaller hydrated ionic radii

@Riqueza $\ce{Na^+}$ interacts more strongly with other ions and this causes it to respond more strongly to an increase in concentration.

why does Na+ interact more strongly than K+ when it has a greater hydrated ionic radii? if I were to plot  specific conductivity  against concentration, should I expect a greater curvature for NaCl or KCl?

@Riqueza $\ce{Na+}$ interacts more strongly with other charges period. This includes charges on other molecules or charges on other ions such as chloride ion. As you increase concentration of chloride this means $\ce{Na+}$ has more ions to slow it down.

what concept could I use to substantiate or justify this? ive been researching into activity and ionic strength but tbh im kinda lost... i thought NaCl would interact the least tbh cause it has the largest hydrated ionic radii

@Riqueza Please see my additional edits to the answer.

Ion association should not be neglected.  For example, magnesium sulphate exists as both contact and solvent-shared ion-pairs in seawater

@Poutnik Thanks, you are probably right, although we are discussing chloride salts, and I am just trying to clarify some fundamentals. You are discussing a higher term in this expansion.

@buckthorn While I agree 2+2- salts have higher association tendency,  compare also Q concentrations versus Mg^2+/SO4^2- seawater concentration. Sure effects due ionic neighborhood comes first, but when concentration is high enough, solution gets quite crowded.


text:
What spectrum of light does penetrate fog?
I am wondering about what spectrum of light does penetrate fog the best. I would like to know if IR light penetrate fog better than UV light and why is that. Thank you for answers.

A good guide is that the scattering intensity is $\sim 1/\lambda^4$, ($\lambda$ is wavelength) but there are different regimes depending in the size of the scattering particles vs the wavelength. On this basis ir has lower scattering than UV.

And continuing to longer wavelengths, there is terahertz radiation, for body scanning in airports, etc., and, of course, radar! Without radar, we would know very little of the surface of Venus, which, to put it mildly, is heavily fogged in!

This is more of a physics question than a chemistry one. So, you should try physics.SE.


text:
How can a detergent "designed for sensitive skin" warrant the "corrosive" hazard symbol?
The dish-washing detergent, which you can find an image of here, has the following hazard symbol on its back:

On the back, it additionally says that if it comes in contact with the skin, you should immediately wash with lots of water, however the detergent also says "specifically developed for sensitive skin" on the front.
How can these two statements make any sense together? Are the chemical regulations that it has to follow just too on-the-safe-side, or what is going on? If I google up its safety data sheet in my local language, it contains the following phrasing:

When handling chemical products, chemical resistant, impervious gloves complying with an approved standard should be used at all times if a risk assessment indicates the need for. (translated from Danish)

It also contains:

Personal protective equipment for the body should be selected on the basis of the task to be performed and the risks involved and should be approved by a specialist before handling this product. (translated from Danish)

So my question is, why does a product like dish-washing detergent that is developed for sensitive skin need these kinds of warnings?
Edit: Image of the back can be found here.

Can you provide an image of the back of the bottle similar to the front?

I added an image. The one I have myself is not in English, but this picture appears to be the same product.

It has different properties at operation concentration and as a concentrate E.g. few % acetic acid is used as natural food additive as vinegar, while concentrated acetic acid would peel your skin away, as MaxW can testify.

The answer, putting on my legal hat (albeit, as disclosure, I have little to no professional legal credentials or training but have, at times, perform some legal research) is that the term 'sensitive skin' is simply not legally defined or has been adjudicated in the courts.  So, from a legal perspective, no contradiction here.

If you post your answers as a comment, I can't accept them.

Even in the light of @Poutnik comment - which must be the answer, this is hilarious :) However, azulene (in chamomile) is a soothing and decongestant principle entering eyedrops formulation. Working with it I've discovered that it is also, perhaps not antiriots police level, a terrible really terrible lacrimator! Paracelsus at work, again. 5 ml to a l'iter, it probably means that is specifically less aggressive than other products. Pity is that for assessment of bullshit like this animals are used.


text:
Does population of vibrational states depend on population of electronic states?
We know that molecules have different energy states (vibrational, rotational, electronic). Using Boltzmann distribution we can find the popoulation of each energy state. If we have only electronic states (e.g. atoms) then we can find the population of the ground, first excited state etc. If we have a molecule (neglecting rotational levels) will the population depend on the population of the electronic state? For example in this diagram Vibrational levels distribution:

will the population ($\%$) also depend in which electronic state the distribution is evaluated?
I think that the percentages would not change. For example the ground electronic state in $υ=0$ vibrational state would have the same population with the excited electronic state in $υ=0$ vibrational state. The percentages will be the same while the number of molecules in each state will be different, e.g. in ground state with $υ=0$ we will have $100$ molecules while in the excited state with $υ=0$ we will have $30$ (numbers are just random). Otherwise it wouldn't make sense to say that at low temperatures the most populated electronic state is the ground state.

Well, to be precise one should consider combined electronic-vibrational states over which to perform the Boltzmann distribution calculation. (These are basically products of the electronic and vibrational states, although note that the different electronic states will also have different vibrational spacings, since the bond strength / force constant will be changed by electronic excitation.) But at room temperature, the population of excited electronic states is essentially zero.

The vibrational frequency and hence energy spacings will generally be different in the ground state compared to bound electronically excited states due to different bonding after electronic excitation. Hence the populations of these levels will be different to the ground state at the same temperature. Of course you have to produce the excited state first and it has to live long enough for thermal equilibrium via collisions with inert species to be produced.

Until "Otherwise it wouldn't make sense to say that at low temperatures the most populated electronic state is the ground state." I follow your reasoning. It does not really matter as we do not normally deal with absorption from higher vib states in the el ground states, thus even less with those within an excited one. Porphyrin/orthocresol (note the heterogeneous phases :)) more or less answered to your question. However, the last sentence that seems to pose trouble, is right the answer. At low T there are no molecules in the excited state, thus no need for looking at the corresponding vib

In the case we want to include translational energy should we treat it as rotational energy with smaller seperation between the energy levels?


text:
How was pH measured back in the day if you had nothing to calibrate to?
We learn about indicators but these seem awfully useless if you have no clue what pH the color change occurs at. How did early chemists figure this out? I guess more broadly, how was the hydronium content even measured in the first place?

Voltage of electrochemical cells whose reactions involved protons

The wikipedia article on S. P. L. Sørensen is a good starting point and references the original literature, https://babel.hathitrust.org/cgi/pt?id=inu.30000091476832&view=1up&seq=141

For future reference: Questions about historical aspects of chemistry can also be a good fit for History of Science and Mathematics Stackexchange.

"Hydronium" content can be easily determined by simple acid base titration. Keep in mind that there is nothing fundamental about pH. It is just a convenient scale for expressing hydrogen ion concentration. For example, a kilogram is a recent invention for expressing mass measurement, it does not mean that ancient people did not weigh anything before kilogram was known.
In the same way, acid-base titration is more than 300 years old, with litmus among the earliest indicators. pH is a relatively new concept. Of course, the titrations were not analytically precise and accurate 300 years ago.
Historically, the simplest way to do acid base titration was to use potash (naturally occurring mineral $\ce{K2CO3}$). One could continue titration until the evolution of bubbles stopped.* No indicator needed. This is the way vinegars were assessed. Note that in those time moles, molarity, did not exist. There was the concept of equivalent weights, i.e., this weight of X will react this that weight of Y completely. The concept of equivalents is still taught in some countries, although it has vanished from Western curricula.
Kolthoff (father of modern analysis) has written books on acid base indicators. Check internet archive. Your question is worth an entire essay.
*From Bishop, Indicators: International Series of Monographs in Analytical Chemistry

fizzy titration! :-)

@uhoh, carbonate titration is still considered among the standard ones. Amazing that some processes are still too good.


text:
Calculate EAN for [Fe(diars)2Cl2]Cl?
I know EFFECTIVE ATOMIC NUMBER (EAN) as the following
• The sum of the number of electrons, donated by all ligands and those present on the central metal ion or atom in the complex is called an effective atomic number (EAN).
EAN = [(atomic number of central metal) – (the oxidation state of the metal) + (the number of electrons gained by the metal from the ligands through co-ordination)]
I need for this component $\ce{{[Fe(diars)}2{Cl}2]Cl}$
I saw (diars) & couldn't calculate it
Edit 1 Thanks to Mathew Mahindaratne comment
EAN = [(atomic number of central metal) – (the oxidation state of the metal) + (the number of electrons gained by the metal from the ligands through co-ordination)]
EAN = $\ce{26 – 3 + (2*4 + 2*2) = 35}$
The question is do I take the 2 electrons that came from Cl2?

The metal complex you need to consider here is $\ce{[Fe(diars)2Cl2]+}$. each $\ce{diars}$ donate $\ce{4e-}$. Now you can calculate EAN ($\ce{diars}$ is 1,2-bis(dimethylarsino)benzene).

is it correct now?

It is correct now. Yes, you have to count all electrons from ligands in the complex ion so should count electrons from $\ce{2Cl-}$  ions as well.

In your oxidation state calculation you render chlorine in the complex as $\ce{Cl^-}$, which then donates two electrons to the central atom.  Hence, the term for the chlorine ligands is $+(2×2)$ as given by Matthew.
This part of a broader rule in transition-element-complex electron counting.  Halogens are to be rendered as halide ion ligands that usually donate two electrons, but add one to the oxidation state of the metal.