text:
Isolating potassium chloride from sodium chloride and potassium hydroxide
Assume that you have an aqueous solution of sodium chloride and potassium hydroxide. Is it possible to isolate potassium chloride from this?
$$\ce{Na+(aq) + Cl-(aq) + K+(aq) + OH-(aq) -> Na+(aq) + OH-(aq) + KCl(s)}$$
Since all involved salts/salt combinations are highly water-soluble, I'm not sure how one could isolate the potassium chloride. Is there maybe a way to selectively precipitate the sodium via a non-water soluble salt?

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

I suggest to buy $\ce{KCl}$, or to prepare it by neutralization of $\ce{KOH}$ and $\ce{HCl}$.
If it happens you have mixed solution of $\ce{NaCl}$ and $\ce{KOH}$, neutralize it first by $\ce{HCl}$ (pH meter or pH indicator paper strips).
From Wikipedia solubility table one can conclude that while $\ce{NaCl}$ solubility almost does not depend on temperature, solubility of $\ce{KCl}$ is almost twice at 100 °C, compared 0 °C.
My suggestion is fractional crystallization:

Partially evaporate the solution until significant mass of salts crystallizes.

Heat the mixture to near boiling and mix well - part of $\ce{KCl}$ would dissolve.

Separate the precipitate enriched by $\ce{NaCl}$.

Cool down solution and let crystalyze salt enriched by $\ce{KCl}$.

With both precipitates and residual solution, you can recursively repeat the procedure.

You can set a system of N batches, where $\ce{NaCl}$ enriched precipitate would go to direction of the batch more enriched by $\ce{NaCl}$ and vice versa.

If it looks complicated to you, buy $\ce{KCl}$ or remember the Pierre and Marie Curie  struggles with the isolation of traces of $\ce{RaCl2}$ from $\ce{BaCl2}$, where their solubilities are much closer.

Another option is using about 3 times better solubility of $\ce{NaCl}$ in methanol, compared to $\ce{KCl}$. But as it is like $\pu{5 g/L}\ \ce{KCl}$ and $\pu{15 g/L}\ \ce{NaCl}$, there would be extensive solvent manipulation with price and safety issues.
As useful info, here is the question/answer about the solubility diagram of mixtures of both salts

The solubility of a salt like $\ce{NaCl}$ or $\ce{KCl}$ is not the same in pure water and in a solution of another salt.  Please look at solubility diagrams $\ce{NaCl - KCl}$ !

@Maurice I am well aware of it. :-) In fact, I have answered some question with such a diagram last year. The solubility in pure water is just a guidance, there is not totally messed up relation of their temperature dependencies.(seen at the diagram as well).

Ion exchange chromatography could be an option, too. A couple of (late) transition metals are separated from each other this way; the value attributed to them justifies the investment.  It very well may be a different story for NaCl, in economic perspective.

NaCl is about three times as soluble as KCl in methanol at room temp. That said, while it might barely be feasible to separate their salt that way, assuming nearly 100% solvent recovery, and near-zero energy cost (perhaps waste geothermal?), the most efficient commercial method to separate various salts is with evaporation ponds.
That said, this is not an answer for your question on separating a few liters of solution, but I've written it as one to add the links. If others think this should be placed as a comment, I'll try to do so.

I was thinking about methanol too, in the deleted comment, but decided against it.


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How does creatine increase DHT levels?
I'm interested in learning how creatine increases DHT levels in the context of 5α-reductase inhibitors (5ARIs).
A study (doi: 10.1016/j.urology.2013.02.045) found that 5ARI-induced DHT deficiency inhibited "autophagy and promoted apoptosis of smooth muscle cells in corpus cavernosum". The article concludes that DHT plays an important role in regulating erectile function. Creatine was shown NOT to increase testosterone levels in the same study that marked a sharp increase in DHT. I don't know if this means creatine produces testosterone which is all subsequently converted or if it facilitates 5ARI's enzymatic activity in some way. The study did not say, and this is what I'm trying to find out. If it's the latter, then in combined use with 5ARIs creatine should lead to less free testosterone (leftover from competitive inhibition) = less estrogen converted as an end result.

For putative connection to Lion's mane mushrooms, see e.g. https://www.reddit.com/r/Nootropics/comments/d8hdyx/i_challenge_anyone_one_this_sub_to_show_me_a/

@KarstenTheis in that same thread "I've posted this on another thread already:
The unsaturated fatty acids should be the cause of the 5α-reductase inhibiton of LM (and other mushrooms). Linoleic acid is found in LM and is a 5α-reductase inhibitor.
Vitamin B2 (riboflavin) is also found in LM and has 5α-reductase inhibiton effects.
Chart about the 5ari effects of different mushrooms. Source"

How does creatine increase dihydrotestosterone (DHT) levels?

To my knowledge, It is yet to know. I don't know anyone have clear path to explain this phenomenon. It has been found that three weeks of creatine monohydrate supplementation affects dihydrotestosterone (DHT) to Testosterone (T) ratio in college-aged $(\pu{18.7 \pm 0.53 years})$ rugby players (Ref.1).
Accordingly, these volunteering subjects loaded with creatine ($\pu{25 g}$ creatine plus $\pu{25 g}$ glucose per day) or placebo ($\pu{50 g}$ glucose per day) for 7 days followed by 14 days of maintenance (test subjects: $\pu{5 g}$ creatine plus $\pu{25 g}$ glucose per day or placebo: $\pu{30 g}$ glucose per day). The levels of DHT and T were measured after 7 days and 21 days (after 14 day meaintenance).

After 7 days of creatine loading, or a further 14 days of creatine maintenance dose, serum T levels did not change. However, levels of DHT increased by 56% after 7 days of creatine loading and remained 40% above baseline after 14 days maintenance $(P \lt 0.001)$. The ratio of DHT:T also increased by 36% after 7 days creatine supplementation and remained elevated by 22% after the maintenance dose $(P \lt 0.01)$.

However, these results have not been explained by the authors, and as recent reveiw pointed out that these results have not been replicated either (Ref.2). Further, considering creatine is endogenously formed from reactions involving the amino acids, arginine, glycine, and methionine in the kidneys and liver, and exogenously obtained from meat and/or as a dietary supplement one may come to conclude that it may have studied extensively. However, according to PubMed (archive of biomedical and life sciences journal literature at the U.S. National Institutes of Health’s National Library of Medicine) there are over 500 peer-refereed publications involving various aspects of creatine supplementation as of 2021 (Ref.2). Thus, creatine might not be extensively studied in all the aspects it might have created including elevation of DHT in the presence of $5\alpha$-reductase Inhibitors.
One posibility is creatine may act as $5\alpha$-reductase stimulant (opposit of inhibition) so that it may compete with $5\alpha$-reductase Inhibitor sites. Yet, that won't explain why serum levels of T is not changing but significant increase in levels of DHT as also obereved in rats (Ref.3).
The best explanation would have been creatine may alternate the fate of DHT (which includes several enzymes), increasing its time of existence:

References:

Johann van der Merwe, Naomi E. Brooks, Kathryn H. Myburgh, "Three Weeks of Creatine Monohydrate Supplementation Affects Dihydrotestosterone to Testosterone Ratio in College-Aged Rugby Players," Clin. J. Sport Med. 2009, 19(5), 399-404 (DOI: 10.1097/JSM.0b013e3181b8b52f).
Jose Antonio, Darren G. Candow, Scott C. Forbes, Bruno Gualano, Andrew R. Jagim, Richard B. Kreider, Eric S. Rawson, Abbie E. Smith-Ryan, Trisha A. VanDusseldorp, Darryn S. Willoughby, and Tim N. Ziegenfuss, "Common questions and misconceptions about creatine supplementation: what does the scientific evidence really show?," Journal of the International Society of Sports Nutrition 2021, 18(13), 17 pages (DOI: https://doi.org/10.1186/s12970-021-00412-w).
Min-Guan Zhang, Xian-Jin Wang, Zhou-Jun Shen, and Ping-Jin Gao, "Long-term Oral Administration of $5\alpha$-reductase Inhibitor Attenuates Erectile Function by Inhibiting Autophagy and Promoting Apoptosis of Smooth Muscle Cells in Corpus Cavernosum of Aged Rats," Urology 2013, 82(3), 743.e9-743.e15 (DOI: https://doi.org/10.1016/j.urology.2013.02.045).

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Why is there H^+ ions in Hcl?
I am new to chemistry, just learning the basics. So, my question is: Hydrogen and chlorine form a covalent bond, so shouldn't they both be negative as they both got an electron by sharing? Why is H positive even though it has formed a covalent bond? If Hydrogen had formed an ionic bond by giving an electron, then it would have been clear, but why covalent?

You mean HCl. // If you and your friend share 2 cookies, does it mean both of you can eat 2 cookies? No, it would need 4 cookies.

You might find answers also among the related posts listed next to the main post. For instance: https://chemistry.stackexchange.com/questions/2698/why-does-the-hydroxide-ion-have-a-negative-charge?rq=1 Please look through existing answers on the site, you are likely to find an answer to this basic question there.

Gaseous/liquid $\ce{HCl}$ has covalent bond, sharing 2 electrons. As they are shared, atoms do have them solely for themselves and therefore they cannot be both negative. 2 shared electrons means effectively 1 electron for each atom as before bonding.
The shared electrons are partially attracted toward the $\ce{Cl}$ atom, which obtains partial negative charge while $\ce{H}$ atom partial positive charge. This forms an electric dipole by displaced charges.
When HCl is dissolved in water, water molecules break H-Cl bond into ions, which get hydrated:
$$\ce{HCl(g) ->[H2O]H+(aq) + Cl-(aq)}$$

But in Na + H2O => NaOH is formed, which means Sodium gives up its electron to OH, right? And the left one H makes up H2 after we balance the equation, as you said, the electrons are more attracted to Cl, so in the way, same way electrons should also be more attracted to Oxygen in H2O. Then when does the one H ( Which forms H2) gets out of H2O with its electron, even though the electron is more attracted to Oxygen?

Sodium gives electron to water, forming highly unstable, but detectable hydrated electron. This attaches to partially positive H (H(+)-OH(-)), forming later H2, leaving OH- behind.

One more question, sir: why does only one H+ detaches from H-O-H?  Why not the 2nd one, too? Thank you!!

Because OH- is not H2O.

And because it is rather difficult to remove something positive from a negative target

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How to remove oil/grease contaminant from an aqueous solution?
I have a solution of water and calcium nitrate that is contaminated with some sort of heavy petroleum oil, semi-solid at room temperature a bit like petroleum jelly, and a strong odor of very crude refining. I wish to retain the salt solution and discard the oil.
The bulk of the oil has adhered to the walls of the glass beaker but a small percent remains in floating in the solution. What are some typical lab methods that could remove the remaining oil? Would adjusting pH assist the process? Would changing the temperature to solidify or melt the grease, or some cycling of temp?

You could attempt to wash it with a low boiling point organic solvent which is immiscible with water such as diethyl ether or hexane. Do you have a separating funnel?

Maybe a simple filtration will do the job. I would try.

I found that by transfering into a series of clean plastic bottles from the recycle bin the majority of grease has adhered to the bottles and the solution easy to decant, with minor rinse of distilled water for near zero loss. Solution is still a touch cloudy with very faint smell of the oil, it is now just barely "good enough" but I may try the non-polar solvent wash just to see if it improves, would certainly be handy for larger quantities. (Using as fertilizer supplement. Issue is caused by current supply chain for economical grades. Likely blind repack of untested lowest bidder import.)

Put it in the fridge. I'd be surprised if almost all of the oil/grease didn't fall out. Run the remaining solution through a coffee filter. Repeat the process until you're happy with the result.

I do not have a separatory funnel at this time, but I could rig something similar as this is not a high value or difficult to work with solution so small losses can be tolerated. I also intend to setup a solvent recovery distillation soon, both for lab and garage parts washing.


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How do computational programs decide how many virtual orbitals to compute?
How do computational programs decide how many virtual orbitals to compute? For example, a Gaussian job of benzene at the TD-DFT B3LYP/3-21G level of theory calculates from 42 electrons a total of 66 orbitals. Why not 65 or 67? Is there any significance to the number it to stop at?

The amount and kind of atoms in your molecule together with the selected basis set determine the number of orbitals and you can determine how many orbitals will be used by looking at the basis set definition. Larger basis sets can improve the accuracy of your calculation but it comes with an increased computational cost.

The basis functions you use describe a space within which your orbitals are vectors. If you have $N$ basis functions this space is $N$-dimensional, and in a $N$-dimensional space you can have at most $N$ vectors which are linearly independent, thus you can have at most $N$ orbitals. In the above example, the 3-21G basis set you are using has 66 basis functions, 9 on each carbon and 2 on each hydrogen, and $(6 \cdot 9) + (6 \cdot 2) = 66$, so you can describe at most 66 orbitals.

How do you know how many basis functions are on each atom?

That's what 3-21G implies - split valence with no polarisation functions

If larger basis sets are used the virtual space becomes a problem. There are (iirc) some methods that try to work around this by actually truncating it. Note that it's still the methods and not the program that accomplish that.

I did start writing in the answer that when a plane wave basis set you would never calculate all the virtuals, but thought that a complication too far. And even in a Gaussian basis set if you are just finding the ground state by the normal diagonalisation method it is easy to avoid calculating (all of) the virtuals, it's just as you get them more or less for free nobody bothers.


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Extent of Melting of Bonds or Intermolecular Forces
I'm currently having a revisit on the subject of periodicity of the periodic table, and I came across an issue that is kind of bugging me.
If we take a look at the melting and boiling point trends of just the second and third periods:

We can see that the difference in melting and boiling points for metals is much larger than that of molecular substances. I've heard two different versions of this:

The extent of breaking of metallic bond is much lower than the extent of breaking of other forces
The extent of breaking of non-directional forces is much lower than the extent of breaking of directional forces (This one doesn't seem plausible to me as hydrogen bonds are directional yet it plays an important role in determining the physical properties of liquids)

Now I want to clear one thing up first before I get to the point: I know that a lot of textbooks state that the stronger the bond/intermolecular forces, the more energy is needed to break them and the higher the melting/boiling point, and the data does support it. But is this kind of misleading? Since we know that it is the latent heat that breaks the attraction between atoms/molecules, and the only reason why temperature is related is because ∆S = L/T by the Clausius-Clapeyron Equation, where ∆S is more or less similar between substances at room pressure. (Please correct me if I'm not getting this right)
Therefore for the main question, I would be using enthalpy of fusion and vaporisation instead of melting/boiling temperatures to represent the extent of breaking of forces.
Now as I have said, I couldn't find any material online that shows how many bonds/intermolecular forces are broken (compared to when all of them are broken when boiling), so I tried to find out the trend from real data. I tried to quantify the extent of bond breaking with:
\begin{equation}
Extent of Melting = \frac{∆H_{f}}{∆H_{f}+∆H_{v}}
\end{equation}
And plotted the data in Desmos (using data from this website):

To me, there doesn't seem to be any pattern I can see other than the fact that metallic bond does seem to be pretty low in its extent of bond breaking.
So the main question I'm asking is: What kind(s) of atomic property(ies) can explain the above trend?
P.S. I'm pretty sure the method I used above is completely wrong, as the bonds are probably not broken only during the process of change in phase so as to satisfy the virial theorem (though I don't quite understand how the theorem is obeyed during phase change as KE seem to only depend on temperature only if we derive the KE term from the boltzmann factor, but that's a question for another day), and so if you believe what I did above is wrong, please also provide me the correct trend of "extent of melting" while answering the question. Thank you.

Latent heat alone  does not break the intermolecular bonds. The kinetic energy of molecules does, and measure of mean energy is temperature. The latent heat just supplies kinetic  energy lost by bond breaking, otherwise the net effect of bond breaking and creating would be zero.

I mean, at the end of the day, potential energy increases if we supply latent heat, even if it is by indirect means.

Sure, but melting and freezing is driven by temperature as the measure of the mean energy per degree of freedom, not by heat. Providing heat just manages disbalacing of melting and freezing rates. // As at m.p., there are always some molecules in solid phase to  have enough energy to go to liquid one and some molecules in liquid phase with low enough energy to stick on the solid phase.

I guess I do kind of see your point, but I think I still need some time to digest it. By the way, do you have any idea as to what the answer to my question is?

You compare incomparable, as various bonding properties kind of wipe off any sense to compare melting/boiling enthalpy ratios. It would make sense for similar structures of rather atomic or molecular substances, not for ones with large scale bonding. And metals are chapter of its own.


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What is the correct order of Relative Inductive Effect (-I) of different groups?
This confusion has put me in problem several times since last 2 years and I have been looking for a clarification but it seems that answer depends upon who you ask. I have several examples where I have searched and infact I changed my chemistry teacher to get this one right but again I am not sure which teacher is correct. We have two orders of Inductive effect from my two different teachers and backed by several websites but I don't know which one is correct.
Let's see the orders
For -I
From Teacher A

I am typing -I inductive effect from teacher A here for indexation as stated in comments
$$\ce{NF3+} >\ce{NR3+} >\ce{NH3+} >\ce{SR2+} >\ce{NO2}  >\ce{CN} >\\
>\ce{SO3H} >\ce{CHO} >\ce{CO} >\ce{COOH} >\ce{COCl} >\ce{COOR} >\\
>\ce{CONH2}  >\ce{F} >\ce{Cl} >\ce{Br} >\ce{I} >\ce{OH} >\ce{OR}$$
From Teacher B

I am typing -I inductive effect from teacher B here for indexation as stated in comments
$$\ce{NR3+} > \ce{NF3+} > \ce{SR2+} > \ce{NH3+}  > \ce{NO2} >\\
>\ce{SO2R}  >\ce{CN} > \ce{SO2Ar}  >\ce{COOH} >\ce{F} >\ce{Cl} >\\
>\ce{Br}  >\ce{I} >\ce{OAr} >\ce{COOR} >\ce{OR}  >\ce{COR} >\\
>\ce{SH} >\ce{SR} >\ce{OH} >\ce{CCR} (\mathrm{alkyne}) >\ce{Ar} >\ce{CHCR2} (\mathrm{alkene})$$
Where Ar = Aromatic
There are several websites which have different orders too
Please refer -I effect order on following websites and see how they are different from each other
https://www.vedantu.com/iit-jee/inductive-effect
https://en.m.wikipedia.org/wiki/Inductive_effect
https://www.quora.com/What-is-the-order-of-I-and-I-effect
Notice how some keep $ \ce{COOH} $ ahead of $\ce{CO}$ and vice versa, also notice that how some keep $ \ce{NR3+} $ahead of $\ce{NF3+}$ and vice versa although I know for this one that$\ce{ NR3+ }$has more -I effect. Please correct me if I am wrong.
And trust me, I have been trying for an answer for very long time but I didn't found an answer which is from an authentic source. Please consider my desperation for the answer before downvoting as I am going to appear for IIT JEE exam within few months. Thankyou to anyone who tries to answer this question.

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
Suitable formatting can be done according to formatting math/chem expressions/equations.

@Poutnik I have put the order in the text as expected, thank you

As resources, see also chem.libretexts.org - Bookshelves - Organic_Chemistry and CH SE: chemistry resources


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Determine reaction order of Cyclopentadiene

Reaction: $\ce{2 C5H6 -> C10H12}$

For this reaction it is stated that $\pu{0.609 M}$ cyclopentadiene was formed in an ethanol solution. When the solution was heated $(\pu{40 ^\circ C})$ and dicyclopentadiene was formed. The amount of dicyclopentadiene after time/h  was:

$$
\begin{array}{lr}
\hline
\text{time}/\pu{h} & c/\pu{M} \\
\hline
6 & \pu{0.017} \\
24 & \pu{0.059} \\
48 & \pu{0.099} \\
78 & \pu{0.135} \\
143 & \pu{0.18} \\
167 & \pu{0.191} \\
192 & \pu{0.201} \\
\hline
\end{array}
$$

When given data like this, i.e. the concentration for dicyclopentadiene and not cyclopentadiene, do I have to convert the concentration to the concentration of cyclopentadiene? And in that case, will the concentration of cyclopentadiene at each of given time become:
$$ [\ce{C5H6}] = 0.609 - 2x$$
where $x$ is the concentration of dicyclopentadiene at each of corresponding time?

Since the concentrations of product with time is given, you can find the order of the reaction by plotting time versus product concentrarion accordingly to find out it was 1st or second order.

Yes. You are right. The concentration of cyclopentadiene is equal to $c=0.609−2x$. And afterwards, you should plot the natural logarithm of this concentration c vs. time. If the points are well alined, the order is $1$. If the alinement is not good, plot the inverse $1/c$ versus time. If the points are well alined, the order is $2$. In both cases the slope of the line is the rate constant $k_1$ or $k_2$

Thank you! Perhaps you might help me with a follow up question. By doing this I got that the reaction order is $2$ and k = $0.0166$ M$^{-1}$h$^{-1}$. If I then want to calculate how long it will take for $10$% of cyclopentadiene to react at $40$°C if the initial concentration of cyclopentadiene is $0.609$ M and $1$ mM resp., do I simply use the equation $\frac{1}{a-x}-\frac{1}{a}$= kt, where the a-x is the $10$% of the initial value and a is the initial value? Thus, getting t${0.609} = 890$ h and t${0.001} = 542168$ h

Your $k$ value is OK. But your $t$ value is not. $a-x$ is not $0.1 a$. No ! It is $0.9 a$. Using $a - x = 0.9 a = 0.9·0.609 = 0.5481$, you must get $\pu{ t = 10.96 h}$

Oh yeah, I missed that! Thank you very much! :)


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Why did anyone invent the concepts of acids and bases?
I am home-schooling my son in first-year Chemistry, and I am struggling to teach Acids and Bases in a rewarding way.
I am not new to Chemistry.  I went to grad school in Physics and I even taught High School Chemistry for four years.  I admit, I am suspicious of the way Chemistry is taught in school today and I think it needs radical revamping.  It's either too much like stamp collecting (memorizing a bunch of terms and rules) or esoteric (obsessing about significant figures in 8th grade seems misplaced) or else just a bunch of unit conversions (Stoichiometry) without focusing on bigger glorious ideas.
I have read a dozen textbook treatments of acids and bases.  I can follow the expanding theories from Arrhenius through Lewis.  I can do all the calculations with no problem.
What mystifies me is:  Why do Chemists care so much about Acids and Bases?
Every treatment starts with Arrhenius studying dissolving ionic compounds in water and noticing that the resulting solution conducts electricity.  I appreciate that needs an explanation, but seeing charged particles floating in water seems good enough.  No need to invent "acids" and "bases" just for that.
Then, someone discovered that litmus paper and other indicator compounds will turn different colors in the presence of certain chemicals.
Maybe scientists realized that the same explanations for what makes water solutions conductors could also explain why these indicators change to different colors, and that was enough connection to warrant these new terms "acid" and "base"?   And this is important enough to spend three chapters of a Chemistry textbook explaining?
Do professional chemists use the concept of Acids and Bases in their daily work?  If so, how?
I know this is an unfocused question, but I hope I am conveying the type of answer that would be helpful to me.  "Acids and Bases are not just a historical artifact, like phlogiston.  They really are very important to understand in modern Chemistry because ...."

If I told you that a reaction yields one product at pH = 4 and another at pH = 8 - and that one of those products can kill you and the other can treat, say, high blood pressure - would the significance of acid/base chemistry be apparent?

@ToddMinehardt For that matter, our blood pH buffers are pretty helpful.

@EdV - True! Drinking water. Aquatic life. Precipitation/dissolution reactions everywhere. It's hard to argue it's NOT important!

Concept of acids and bases is one of most useful and most used concepts of everyday life of professional chemists.

And not only for chemists. It has direct application from geology  to biology. In a sense, it is like asking a physicist if quantum mechanics is anything useful, or if it was invented just to torture students. :-)

@Poutnik Fortunately, that question cannot be asked at physics SE because it would be closed as opinion-based. But between being asked and being quickly closed by a mod, it would get three or more upvoted answers. ;-)

I don't mean this to be rude, but I find it hard to believe that someone who claims to have taught chemistry for 4 years - and who claims a graduate-level education in physics - would ask this question. Perhaps a better question is why people bother with classical mechanics, since the concept is provably incorrect and only useful in cases where an approximate answer (as opposed to the exact, quantum mechanical one) can suffice.

Do professional chemists use the concept of Acids and Bases in their daily work?   Yes, absolutely. Consideration of solution pH at various stages of a reaction is crucial. At what pH does a compound ionise is hugely important for considering whether it will dissolve, interact with another compound etc

Check Newest 'ph' Questions - Homebrewing Stack Exchange for examples of why pH is important, even to non-chemists.

I am sympathetic to this question. Particularly at the beginning stages, the sciences are often introduced as a set of rules to be followed and terms to be memorized, with too little emphasis on the conceptual beauty that underlies it—getting students to understand why this is interesting. I think part of the problem is that it is challenging to teach both. The mechanics and terms do need to covered, and it is challegning to find underlying concepts that can be taught to beginning students in a meaningful and understandable way. Note also that intro physics has the same problem as intro....

...chemistry. High school phys. is mostly focused on the basic mechanics, and makes no significant mention of coarse-graining, or back-of-the envelope calculations, nor does it communicate ideas like this:  https://www.youtube.com/watch?v=o1dgrvlWML4  Focusing on your question:  The acid/base dichotomy is important to chemistry, because H+ and OH- are not merely electrolytes, they are oppositely-charged electrolytes; it's important in the same way the metals/non-metals dichotomy is. Indeed, acids are typically an H bonded to a non-metal, and bases are typically an OH bonded to a metal.

The words "acid" and "base" have both a historical origin.
The word "acid" is coming from the latin root "acies" which means "prickly". Its derivate "acidus" is an adjective used for describing a prickly, or hot, or sour liquid, because it pricks the tongue when drinking it. Another derivate is "acetum", latin name of vinegar, because vinegar is the first acidic liquid, the first known liquid pricking the tongue when drinking it.
The word "base" has another origin, which is more recent, but not so easy to explain. It needs a long explanation. In the Middle Age, it was known that acids can be destroyed (= neutralized) by adding soda, ashes or chalk. Soda ($\ce{Na2CO3}$) was obtained in Egyptian oasis. Ashes contains impure sodium and potassium carbonates, and chalk contains $\ce{CaCO3}$, but these data were unknown of course. Some alchemists called these strange substances "antacid". We will see that this name was never popular. The only fact that was accepted by all alchemists was that the reaction acid + antacid produces a new substance, a solid substance, which was called a "salt", because the sea salt was soon recognized to be also the result of an acid ($\ce{HCl}$) reacting on soda ($\ce{Na2CO3})$.
When heating a (solid) salt to a rather high temperature, it gets decomposed into two parts : nearly half of the stuff quits the solid phase as a gas (new word, coming from "ghost", created by van Helmont, b. 1577). Surprisingly enough, these vapors, when cooled, can be condensed and usually produce an acid !! And the remaining stuff, when cold, could have been described as "residue". But it was more informative to call it "base", because it stays in the lower part of the heated flask and refuses to pass into gas phase. Generally speaking, a "base" is the lower part of an object, or of a mountain. In these old times, a base is (or was) the part of the stuff that stay in the lower part of the hot container. Today, we know that this basic residue is made of oxides or carbonates. But even in the Middle Age, it was soon discovered that these bases were similar to the antacids that react with acids. So the word "antacid" was replaced by the word "base". The following "equations" could have been written .
        Acid + Base -> Solid salt (+ water + sometimes a gas)

        Salt (+ heat) -> Base(solid) + Acid(gas)

Although sea salt does not decompose this way, an important proportion of salts can be decomposed at high temperature according to this last equation, specially when the salt crystallizes with some crystallization water, like $\ce{ Na2SO3·7H2O}$, or $\ce{ Ca(NO3)2·4H2O}$, or $\ce{ Al2(SO4)3·12H2O}$. Examples : $$\ce{ Na2SO3·7H2O -> 2 NaOH(s) + SO2(g) + 6 H2O(g)}$$    followed at low temperature by : $\ce{SO2(g) + H2O(l) -> H2SO3(l)}$ $$\ce{ 2Ca(NO3)2·4H2O -> 2 CaO(s) + 4 NO2(g) + O2(g) + 4 H2O(g)}$$  followed at low temperature by : $\ce{4 NO2(g) + 2 H2O(l) + O2(g) -> 4 HNO3(l) }$ $$\ce{Al2(SO4)3·12H2O -> Al2O3(s) + 3 SO3(g) + 12 H2O(g)}$$followed at low temperature by : $\ce{SO3(g) + H2O(l) -> H2SO4(l)}$
The obtained residual base can be an oxide or a hydroxyde. It did not matter. And the acid may be partially decomposed. These facts were considered as secondary in these old times, where chemistry was not yet a science.
Today the words "acid" and "base" have a quite different meaning, of course. But it is interesting to know that these two names have a rather intuitive origin

" vinegar is the first acidic liquid, the first known liquid pricking the tongue when drinking it." Not at all, and that would be just a start...

So what is the first liquid pricking the tongue ?

Most under ripe fruit juice seems reasonable.

Acid and vinegar are nearly the same word in latin. So vinegar is the first known acidic solution. Furthermore, the English word "vinegar" is coming from the French word "vinaigre" which describes a wine (French "vin") which has become "aigre"; and "aigre" is a synonym of "acid" in French. The French word "aigre" is also coming from the latin origin "acies" which means "pricking".

Thanks SO much for all the effort spent here!
@Maurice I loved learning the historical origins of "acid" and "base".  It will certainly enrich my teaching.
If I understand correctly, it's not so much acids/bases that are crucial on their own, but acids/bases are a stepping stone to pH, and, for whatever reason, pH is a vital consideration in lots of biological processes.
Here's my attempt to coalesce the helpful comments into an ideal (to me, at least) intro textbook section on acids and bases:

Knowing whether a substance is a metal or a non-metal gives you lots of predictive power to guess how it will behave.  Is this substance a good conductor of electricity and heat?  Is it malleable?  What's the melting point?  All these questions can generally be answered simply by knowing whether it's a metal or non-metal.
When discussing solutions, another useful classification is "acid" or "base" or neither.  This lets you predict whether it conducts electricity, whether it tastes sour or bitter, and whether it feels slippery.  This list of predictions is clearly less important than those concerning metallic nature, but there is one consequence of the acid/base distinction that's, literally, a matter of life or death:
In Biology, there's a property called pH, which is intimately related to the acid/base character of a solution.  Biological processes are extremely sensitive to the pH of the surrounding solutions, and even small differences in the pH lead to wildly different outcomes in the underlying chemical reactions.
So, Chemists (and lots of other scientists) have learned to pay careful attention to pH, and to understand pH, you first need to understand acids and bases.

I think you have it backwards. Acid and base chemistry is a broad area, and pH is simply one way of characterizing acids and bases in liquid solutions. But there are also solid acids (https://en.wikipedia.org/wiki/Acid_catalysis) and gas-phase acids.  More broadly, you overly trivialize acids and bases.  They each have a rich chemistry associated with them, and are both critically important in many areas, including environmental chemistry, geochemistry, and modern industrial chemistry.   There's far more to them than sour vs. bitter, etc.

And pH itself is much more complicated when you look at the behavior of real acids and bases, e.g., if you look at a titration curve of a weak triprotic acid (an acid that can give off three H+'s):  https://chem.libretexts.org/Ancillary_Materials/Demos_Techniques_and_Experiments/General_Lab_Techniques/Titration/Titration_Of_A_Weak_Polyprotic_Acid
Separately, you might find this article useful:  https://pubs.rsc.org/en/content/articlelanding/2015/RP/C4RP00125G

I suggest you add this as an update to your question, reflecting your current understanding. There is a lot that is flawed in it btw. I recommend reading more Q&As on this site, and in particular the definitions of pH, acid, base etc. Also, you may find the history of science and mathematics SE site useful for this type of question, since you seek an explanation of how terms such as pH etc arose. Maurice's answer is also useful from that perspective.


text:
Can Positive ions satisfy primary valences according to Werner's theory for coordination compounds?
My book (NCERT- Chemistry Part for class 12) Coordination Compounds-NCERT (Page-2)
Mentions the following statements as a postulate of Werner's theory:

The primary valences are normally ionisable and are satisfied by
negative ions.

I get that the primary valences basically mean the Oxidation Number of the central metal atom in the coordination sphere. And they are ionisable as the ions that are outside the coordination sphere get dissociated in aqueous solutions.
But is it necessary that Primary Valencies are satisfied only by negative ions?
For example, in $\ce{[Co(NH3)6]Cl3}$ . It is clear that three Chloride ions satisfy the primary valence(+3 oxidation state) of $\ce{Co^{3+}}$ .
But if we consider $\ce{K4[Fe(CN)6]}$  there are no negative ions outside the coordination sphere.
So here are my doubts regarding this

Do the Potassium ions satisfy the primary valence/oxidation state of Iron here? Which means the book is wrong?
If not, then are the Cyanide ions satisfying the primary valence of Iron? If yes, then we can say Ligands can satisfy both Primary and Secondary valences at the same time? (As cyanide ions are already contributing to coordination number(secondary valence))
But I don't feel this could be true since that contradicts the fact that primary valences are ionisable.

Here the primary valence is not satisfied by the potassium ions, it is satisfied by the ligands. Werner's theory says that the primary valence of central atom is only satisfied by anions as the central atom in most complexes are transition metals which have a positive oxidation state.

Yes ligands can satisfy both Primary and Secondary valences at the same time.

According to the pictorial (geometric) representation of Werner's theory, species which satisfy the primary valence are joined by a dashed line, species which satisfy the secondary valence are joined by a solid line and those species which satisfy both valences are joined by a solid and dashed line (in parallel).
So yes ligands can satisfy both the primary and secondary valences.
(You can also refer this video if you understand Hindi.)

I believe secondary valencies are showed by solid lines as they are the ones which consist of actual bonds between Ligand and Central Metal Atom. Also if Ligands can satisfy both primary and secondary valences, How can we call primary valence as 'Ionisable'? As ligands exist inside Coordination sphere which is not ionisable.

Yes I have fixed that now. Note that all covalent bonds have some ionic character which is why these ligands can provide the primary valence. Werner's theory was not perfect and this ambiguity in the definition of "primary valence" is one of it's many drawbacks.

Also, other than anions, even water, which is a neutral ligand, acts as a counter ion in some cases through Hydrogen bonding (though an anion is always present), and is present as water of crystallization. Werner's definition of primary valence only being satisfied by anions is wrong and cannot be used in the current day. It would be better for you to learn the concepts of coordination compounds through CFT and VBT, though I am sure you already are given that you are preparing for JEE.


text:
Why do DFT calculations output molecular orbitals?
My understanding is that DFT finds the electron density which minimizes some energy functional. How does it make the connection from this optimized density to molecular orbitals?

Do you know how this works in time independent DFT? I'm worried you are confusing the two.

IIRC, most implementations of DFT and TD-DFT use Kohn-Sham orbitals, and the density is expressed in terms of those orbitals. So, the program always deals with orbitals, the density is calculated from the orbitals.

@IanBush No, I wasn’t aware there was a qualitative difference in how the connection was made in TD vs TI. Based on S R Maiti’s answer though I’m under the impression that there isn’t. What did you have in mind?

As long as you know there is no difference that is fine - I was just concerned you keep mentioning  the time dependent part when that has been totally irrelevant for your last two questions.

Oh I see what you mean. I was pretty sure that distinction wasn't necessary but included it just in case.

The short answer is they do not.
We can identify two distinct ways of doing DFT calculations, OF-DFT (orbital-free DFT) and KS-DFT (Kohn-Sham DFT).
Let us start with the OF-DFT formalism, as it is the formalism that is more in the 'spirit' of DFT.
In OF-DFT the energy is given as:
$$E\left[\rho\right]=\int_{\Omega}\phi_{\mathrm{ext}}\left(\boldsymbol{r}\right)\rho\left(\boldsymbol{r}\right)\mathrm{d}\boldsymbol{r}
+
\frac{1}{2}\int_{\Omega}\frac{\rho\left(\boldsymbol{r}_{1}\right)\rho\left(\boldsymbol{r}_{2}\right)}{\left|\boldsymbol{r}_{1}-\boldsymbol{r}_{2}\right|}\mathrm{d}\boldsymbol{r}_{2}\mathrm{d}\boldsymbol{r}_{1}
+
T\left[\rho\right]+E_{\mathrm{x}}\left[\rho\right]+E_{\mathrm{c}}\left[\rho\right]
$$
The first term is the external potential, the second term is the Coloumb interaction of the electronic density with the electronic density (electron-electron repulsion including self-repulsion), the third term is the kinetic energy functional, and the last two terms are the exchange- and correlation-contributions.
As can be seen in the above equation, there is no dependency on orbitals anywhere.
Now let us examine the KS-DFT formalism:
$$ E\left[\rho_{\Phi^{\mathrm{KS}}}\right] = \int_{\Omega}\phi_{\mathrm{ext}}\left(\boldsymbol{r}\right)\rho_{\Phi^{\mathrm{KS}}}\left(\boldsymbol{r}\right)\mathrm{d}\boldsymbol{r}
+
\frac{1}{2}\int_{\Omega}\frac{\rho_{\Phi^{\mathrm{KS}}}\left(\boldsymbol{r}_{1}\right)\rho_{\Phi^{\mathrm{KS}}}\left(\boldsymbol{r}_{2}\right)}{\left|\boldsymbol{r}_{1}-\boldsymbol{r}_{2}\right|}\mathrm{d}\boldsymbol{r}_{2}\mathrm{d}\boldsymbol{r}_{1}+
\left< {\Phi^{\mathrm{KS}}} \left| \hat{T} \right| {\Phi^{\mathrm{KS}}} \right> + E_\mathrm{x}\left[ \rho_{\Phi^{\mathrm{KS}}} \right] + E_\mathrm{c}\left[ \rho_{\Phi^{\mathrm{KS}}} \right] $$
We can see that in the KS-DFT formalism all of the terms, except for the kinetic energy, are the same as in OF-DFT formlism. Note, that the density here is contructed from the KS-orbitals.
In the KS-DFT formalism we get the kinetic energy contribution from the application of the kinetic energy operator on the KS-orbitals, whereas, in the OF-DFT formalism the kinetic energy is a functional of the density.
Now to address your question directly. In (KS)-DFT we do not construct the orbitals from the optimized density. Instead we construct the density from the orbitals. I.e. the density is:
$$ \rho_{\Phi^{\mathrm{KS}}} = \sum_{i=1}^N \left|\phi^\mathrm{KS}_i\right|^2 $$
However, this raises the question, why do we bother with KS-orbitals? Here, the unsatisfying answer is: Because we do not know any sufficiently accurate kinetic energy functionals.


text:
How to calculate rate of carbon dioxide from rate equation?
The concentration of carbon dioxide that changes with time:
$$\ce{CO2(g) + H2O(l) <=>[$k_\mathrm{fwd}$][$k_\mathrm{rev}$] HCO3-(aq) + H+(aq)}$$
How to calculate the integral for the following rate equation of carbon dioxide production?
$$\frac{\mathrm d[\ce{CO2}]}{\mathrm dt} = k_\mathrm{rev}[\ce{H+}][\ce{HCO3-}] - k_\mathrm{fwd}[\ce{CO2}]$$
Water is ignored in the rate equation due to its very large concentration that doesn't change over time.

Do you neglect the rate CO2(aq)+H2O <=> H2CO3(l)? CO2(aq) alone is not acidic.

@Poutnik  H2CO3 hardly exists. Even if hydration and deprotonation aren't concerted, this step could be well omitted. On the other hand, shouldn't the equation be with OH- instead of water?

@Mithoron It does exist. The equilibrium constant and both kinetic constants for CO2(aq)<>H2CO3 are known.

There is one reaction pathway from CO2(aq)<>H2CO3 however this is instantaneous compared to this reaction. I am really just wondering how to find the integral rate law for the rate equation above.

The key is to use stoichiometry to write the concentrations of your two products in terms of the concentration of CO2. Now you have a single-variable differential equation in [CO2]. Since it is first order and autonomous (coefficients are time-independent), the resulting ODE can be solved analytically using methods that you can look up.

Suppose we start with $n$ moles of $\ce{CO2}$ for simplicity. Then, at any point during the reaction, we will have $n-V_\text{gas}[\ce{CO2}]$ moles of each of the products. This means we could write their concentrations as
$$[\ce{H+}]=[\ce{HCO3-}]=\frac{n-V_\text{gas}[\ce{CO2}]}{V_\text{liquid}}$$
With all the concentrations expressed in terms of $[\ce{CO2}]$, we can rewrite your rate equation as
$$\frac{\mathrm d[\ce{CO2}]}{\mathrm dt} = k_\mathrm{rev}\bigg(\frac{n-V_\text{gas}[\ce{CO2}]}{V_\text{liquid}}\bigg)^2 - k_\mathrm{fwd}[\ce{CO2}]=k_\mathrm{rev}\bigg(V_\text{ratio}^2[\ce{CO2}]^2-2nV_\text{ratio}[\ce{CO2}]+A\bigg) - k_\mathrm{fwd}[\ce{CO2}]$$
Where $A=(\frac{n}{V_\text{liquid}})^2$. Since this now only depends on a single variable and a few constants (the starting amount of $\ce{CO2}$ and the volume of each phase), it can be readily integrated to solve for how $[\ce{CO2}]$ (and consequently $[\ce{HCO3-}]$ and $[\ce{H+}]$) vary over time.


text:
Systematic IUPAC name of buckminsterfullerane
Many of you have heard of buckminsterfullerEne, which is C60, also called the "buckyball".
Well, there's also buckminsterfullerAne, which is C60H60. This is a buckyball that's fully saturated (all carbon-carbon bonds are single bonds) and fully hydrogenated.
I'm aware that the nomenclature for polycyclic hydrocarbons can get pretty messy. For non-aromatic hydrocarbons, I've already determined why the systematic IUPAC names (the ones with numbers inside brackets that indicate the "backbones" and bridges) for tetrahedrane, cubane, and dodecahedrane are what they are:
Tetrahedrane: Tricyclo[1.1.0.02,4]butane
Cubane: Pentacyclo[4.2.0.02,5.03,8.04.7]octane
And from the Hamiltonian path of a regular dodecahedron, I was also able to verify the systematic name for dodecahdrane.
Dodecahedrane: Undecacyclo[9.9.0.02,9.03,7.04,20.05,18.06,16.08,15.010,14.012,19.013,17]icosane
As you can see, with a dodecahedron it starts getting pretty messy, but in the linked image above, if you number the rightmost point as #1, then whichever way you go the point that is connected to #1 via a dashed line would be #11. From then I was able to verify the locations of all the secondary bridges. From my intuition, it seems that you want to number the vertices such that at least one pair of vertices with a bridge has the vertex numbers differing by one-half the total number of vertices.
Now, that got me wondering, what would the same name for a truncated icosahedron be?
I started looking for Hamiltonian paths through a truncated icosahedron, and came across this paper. On Page 2, Figure 2, six possible Hamiltonian paths are shown. I numbered the first two of these and found out that the red figure has two bridges where the two vertices' numbers differ by 30 (10/40 and 25/55, the 32/52 in the circle is incorrect). I've numbered them as follows:
And then after locating the two bridges, I re-numbered the diagram with vertex #1 as one of the vertices in the black rectangle:
And from there, I tried to deduce the systematic IUPAC name for buckminsterfullerane. This is what I came up with:
Triacontakaihenacyclo[29.29.0.02,28.03,26.04,23.05,60.06,22.07,58.08,20.09,57.010,18.011,55.012,17.013,53.014,51.015,48.016,46.019,45.021,44.024,43.025,41.027,39.029,37.030,35.032,59.033,56.034,54.036,52.038,50.040,49.042,47]hexacontane
(man, what a mouthful)
where triacontahena- means 31 and hexaconta- means 60.
My question is whether this name is correct. And if it's not correct, where are the correct locations for the bridges?

What makes you think fully hydrogenated buckyballs exist?

https://www.science.org/doi/10.1126/science.253.5017.330 suggested that you'd need hydrogens inside the shell you not make it very strained. Such compound would have a lot of isomers you don't take into account.

@Mithoron Stereoisomers?


text:
Melting and boiling point of transition-metals, primarily for the groups of Cr and Mn
The NCERT Chemistry book for Grade 12 writes

The high melting points of these metals are attributed to the involvement of greater number of electrons from $(n-1)d$ in addition to the ns electrons in the interatomic metallic bonding. In any row the melting points of these metals rise to a maximum at $ d^5$ except for anomalous values of Mn and Tc and fall regularly as the atomic number increases.

However, it mentions no such reason as to why this exceptional/anomalous behaviour is observed. My initial thoughts towards approaching this issue were considering the electronic configurations that group members of Mn have an exceptionally stable half-filled configuration, but a similar half-filled configuration exists with group members of Cr as well. Secondly, I thought it could be due to the structures, the NCERT book also mentions Zn/Cd/Hg/Mn having certain exceptional behaviour when it comes to their structures with respect to the other members of the d-Block, but it doesn't mention the same for Tc which too shows a steady dip in it's melting point.
What could be the possible reason for such anomalous behaviour if not the electronic configuration or the solid structure? Here's the graph for the melting points for d-Block elements.

One of the factors affecting the melting point of transition metals is the number of unpaired electrons in $(n-1)d$  and in $ns$. These unpaired electrons take part in metallic bonding and the more there are, the more bonds and hence greater is the melting point. As we move from left to right in the 3d,4d, and 5d series, the number of unpaired electrons first increases (till $(n-1)d^{5}$) and then decreases as pairing starts.
Manganese shows an exceptionally lower melting point as due to the exceptionally stable half filled configuration, the electrons are not delocalised to a large extent. This occurs in manganese and not in chromium due to the fully filled $4s^{2}$ subshell in manganese, in which pairing has occurred.
This is also the reason why technetium shows a dip in its melting point in the 4d series as it has a $4d^{5}$ $5s^{2}$ configuration.


text:
Numbering order in IUPAC name of a trisubstituted alkatriene
During E/Z-nomenclature class, I found this molecule.

Using MarvinSketch 21.15.0 version I have (3E,5E)-6-bromo-3,4-dimethylhexa-1,3,5-triene for it. Why the name couldn't be (1E,3E)-1-bromo-3,4-dimethylhexa-1,3,5-triene? What I missing?

The name (1E,3E)-1-bromo-3,4-dimethylhexa-1,3,5-triene is correct. Therefore, it's hard to answer why MarvinSketch shows a different name; at least this is not because of any particular IUPAC rule.

Another question about problems with tie-breaking rules in MarvinSketch: What are the preferred IUPAC names of these two compounds?


text:
Hydroboration then oxidation with carbon monoxide and hydrogen peroxide

Thus is the question given in my textbook

This is my attempt at the question
This is the solution given in the textbook.
I don't understand how a ketone is formed instead of an acid. related question This is a related question but in this a tertiary alcohol is formed instead of a ketone. What is the mechanism of formation if ketone and why is my mechanism wrong?

There is a good explanation of the possible mechanisms of the ketone formation  here which I have reproduced below. No pathway leads to an acid. Your mechanism has multiple CO insertions and only one occurs.

What if excess CO is used, even then only 1 insertion will take place?

Only one insertion will occur. As the reaction is run under an atmosphere of CO there will always be an excess present.


text:
How do mechanisms of acidic cleavage of ethers differ for anhydrous, concentrated, and aqueous HI?
I have seen the usage of anhydrous, concentrated, and aqueous HI for the acidic cleavage of ethers. What is the difference in the mechanisms and which will follow either of SN1/SN2 mechanism?
Once the the oxygen is protonated, which cases would favor a carbocation formation (hence putting the iodide at a bulky site), and when the reaction follows SN2 mechanism as the leaving group is good (hence putting the iodide at a less bulky site)?

The stability of the carbocation in the different cases is going to make a significant contribution to which mechanism is followed. If it is a stable cation then SN1 can be followed more easily

Thank you.Actually the question that had come had an ether having ethyl and tert butyl groups on it . When treated with ANHYDROUS HI the shown product was ethyl iodide and tert butanol(SN2?) ,While I expected tert butyl iodide(SN1) as the product. The first part of my question actually asks about the reagent HI as well

I am unconvinced by the answer you have been given and a quick Google search reveals a number of sites that agree with you that tBu-I + EtOH are the outcome. There are also sites that give tBuOH + EtI as product. In real life I suspect you would get a mixture. There is also the possibility that if excess HI (as is likely) is present then it will convert the tBuOH to tBu-I

This looks like a a good summary https://www.masterorganicchemistry.com/2014/11/19/ether-cleavage/


text:
What kind of defects can I find in a knife made of high-carbon steel with a wooden handle?
I'm still studying the Japanese knives and must now prevent the different defects that could appear in this tool. I consider a high-carbon steel blade with a wooden handle, with the size of a Sakai Uchi Hamono knife (from 165 mm to 240 mm).
I must then find a non-destructive method to detect those defects before it's too late. Here is the list of methods I am allowed to use:

Penetrant Testing
Magnetoscopy
Eddy current
Radiography
Ultrasound
Infrared thermography
Shearography

My main problem is that those Japanese knives are famous to be extremely well manufactured, and they have a life-time guarantee, meaning that they basically can't break... Nevertheless, let's consider 3 possibilities:

The wooden handle gets a defect: highly improbable in my opinion since you only use it with your hand, without any particular stress on the wood;
There is a problem between the blade and the wood: as you can see it in my previous question here, the two components are not simply stuck together. Knowing the shape of the knife and the strength of the steel, I can't possibly imagine how I can separate the two objects without applying a disproportionate force;
The steel gets a defect. A "defect" would for instance mean that it doesn't cut properly anymore (maybe because of water when you need to wash it?) You can see how this kind of blade is forged here.

What kind of problem could I have, on the molecular scale for instance (fatigue failure ?) that would create such a defect? What kind of method in the list above can I use to find it?
Of course, if you find a potential problem linked to the handle, I'd be glad to hear about it.
PS: If you believe this is the wrong place to ask this question, could you please tell me where I should post it?

I am afraid this is very specialized knowledge area most chemists would have trable to give useful answers. It seems to me as a question for metallurgy or material science. Not sure if there is such site on SE network.

@blacksmith37 Could you provide an advice and/or reference guidance to the OP? I believe you know about the topic much more than me.

Steel is generally full of 'defects' from a material science point of view. Whether those impact their functionality under normal conditions is a different question. And a lifetime warranty just means they might last longer than some other products.

Wet mag would be best to examine for mechanical defects. Similar to dye pene but with fluorescent indicator and a magnetic field.. Usually done under UV light. Zyglo is one brand I recall. That will not find metallurgical flaws such as low hardness. And will not find non-metallic inclusions that may promote fatigue. These other flaws cold likely be found with eddy current (EMT) but it would likely be less costly to buy a hundred knives and throw any that fail away than to develop standards and do the testing. I suggest that you trust the vender and if there is a problem , buy another knife rather than test. PS; Same answer for traditional 13% chrome cutlery steels.

Thank you a lot ! Do yout agree that the potential problems will more likely appear in the blade than with the handle, thus ?

The tang portion of the blade will not not as much stress except at the transition to the main blade so relatively low risk.


text:
What happens during a displacement reaction?
When Zn metal is added to a solution of Cu(II)SO4, Zinc displaces Cu to form ZnSO4 as Zn is more reactive than Cu. What I don't understand is how.
1- What I'm assuming is: Zn metal collides with [SO4]2- ions and because it wants to be stable (by losing 2 electrons), it forces those 2 of its electrons onto [SO4]2- which then goes on to lose those 2 (of its previous) electrons as it also wants to be stable. Then, those 2 extra electrons collide with [Cu]2+ ions and are attracted to the ion because of the charge difference and then they form Cu metal.
2- But then, what stops Cu metal colliding with [SO4]2- ions (with Zn electrons) and making [Cu]2+ ions again.
3- Is it just the fact that the collisions are so frequent that Zn wins in the 'battle of displacement' because of its higher reactivity and forms some sort of equilibrium?
4- Surely, the electrons of Zn metal aren't different/stronger than the electrons of Cu metal to resist being moved out if the mechanism by which Zn metal reacts is by forcing its electrons to be accepted.
5- In the solution, the ions are separated due to the polarity of water molecules, so it's not like the Zn metal is next to the CuSO4 ionic compound, it would only be next to / 'collide' with the [SO4]2- ion. The only way for a displacement reaction I can think of then, is by the mechanism I'm describing in 1.
Please keep the answers relatively simple, I'm only in college. No quantum mechanics magic, ideally. I'm assuming the mechanism is completely wrong, but I can't find another way for it to happen (at least not in a non complex form).

You can benefit greatly by getting this and working through it: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95 (10) (2018) 1801-1810. The Zn and Cu Daniell cell is directly relevant to your example and it  is addressed in great detail. TL; DR Cohesive energy differences are the major factor in explaining the behavior of this famous galvanic cell. Look up the famous Daniell cell.

No. $\ce{Zn}$ does not force two of its electrons onto $\ce{SO4^{2-}}$ as you suggest.  It simply transfers them to the $\ce{Cu^{2+}}$ ions by collision. That's all. Copper metal will not get reoxidized, because its redox potentiel is high ($+0.34 V$

Useful links for text and formula formatting (not to be applied to titles):
Notation basics ,  Formatting of math/chem expressions  and
upright vs italic

Salts like $\ce{ZnSO4}$ or $\ce{CuSO4}$ do not exist in water solution as dissolved molecules. In fact, such molecules do not exist in the solid salts either (anhydrous salts or salt hydrates $\ce{ZnSO4 \cdot 7 H2O}$ or $\ce{CuSO4 \cdot 5 H2O}$). Those formulas just reflect the stoichiometric ratio of ions in a solid phase or solution. $\ce{ZnSO4}$ (or $\ce{CuSO4}$) exist in water solution as independent ions $\ce{Zn^2+(aq)}$ ( or $\ce{Cu^2+(aq)}$) and $\ce{SO4^2-(aq)}$. (aq) means hydrated state.
When zinc dissolves, there is ongoing reaction $\ce{Zn(s) -> Zn^2+(aq) + 2 e-}$, with electrons remaining in the metallic zinc, charging it slightly negatively. These electrons, when the potential of the zinc electrode drops low enough, start reacting with copper ions, that more easily accept electrons than zinc ions:  $\ce{Cu^2+(aq) + 2 e- -> Cu(s)}$, depositing metallic copper on the zinc surface.
The sulphate ions in the solution are just spectators/bystanders, "watching" indirect exchange of electrons between zinc atoms and hydrated copper ions.

The reason for this is the fact that Zinc is a more reactive metal.
(Note that, as Poutnik mentions (and I forgot to), $\ce{CuSO4}$ and $\ce{ZnSO4}$ do not exist in the liquid state but in the dissolved form and in this form they are ionised, i.e. they break up into their respective ions $\ce{Cu}^{2+}$,$\ce{Zn}^{2+}$, and $\ce{SO4}^{2-}$ )
What does more reactive mean? It means that zinc requires less energy to remove the 2 electrons than copper does.
So the difference between the energy released when zinc forms an ionic bond with $\ce{SO4}^{2-}$ and the energy required to remove electrons from zinc is larger than in the case of copper, i.e. more energy is released in the formation of $\ce{ZnSO4}$ than in the formation of $\ce{CuSO4}$.
As a rule of thumb, lesser energy means more stability. Hence $\ce{ZnSO4}$ is more stable and the reaction favours the formation of $\ce{ZnSO4}$.
You can also consider the electrochemical series (as Maurice has mentioned) in which zinc has a greater value of standard reduction potential than that of copper (which basically means that zinc can get oxidised more easily than copper) and hence zinc reduces copper and itself gets oxidised.

Thanks for the detailed answer regarding the reason for Zn being more reactive. Turned out I was making a super basic mistake. As Maurice pointed out, there was no reason for Zn to give electrons to SO42-, as it could just give them to Cu2+ instead.

Yep glad I could help and glad you understood.


text:
Gaussian type orbitals and p-type orbital
The Gaussian orbital for $p_x$ is defined as:
$$\psi_{p_x}=C(x-x_A)e^{-(\vec r - \vec R_A)^2}$$
Where $C$ is a constant.
Now if we multiply two such orbitals with two different centers, then
$$C_1(x-x_A)e^{-(\vec r - \vec R_A)^2} \cdot C_2(x-x_B)e^{-(\vec r - \vec R_B)^2}=C_3(x-x_A)(x-x_B)e^{-(\vec r - \vec R_A)^2} \cdot e^{-(\vec r - \vec R_B)^2}$$
I understand that the product of the exponential parts above can be transferred to a third center. But I am not sure about $(x-x_A)(x-x_B)$ part. This part, as far as I know, can't be transferred to $(x-x_C)=(x-x_A)(x-x_B)$. I don't think such $x_C$ exists unless $x=0$.
So my question is, usage of Gaussian orbitals only helps when the integral has all $s$ type orbitals. Whenever $p$, $d$ type orbitals get involved I can not transfer four center integrals to two center integrals.
Is my understanding above correct?

No, your understanding is not correct. Let the third centre at which the gaussian product is centred be $R_P$. Now we can write
$$(x-x_A)=((x-x_P)-(x_A-x_P))=(x-x_P)-x_{PA}$$
so defining $x_{PA}$. Similarly
$$(x-x_B)=((x-x_P)-(x_B-x_P))=(x-x_P)-x_{PB}$$
So
$$(x-x_A)(x-x_B)=(x-x_P)^2-(x_{PA}+x_{PB})(x-x_P)+x_{PA}x_{PB}$$
Thus
$$(x-x_A)e^{-(\vec r - \vec R_A)^2} * (x-x_B)e^{-(\vec r - \vec R_B)^2}=D(x-x_P)^2 e^{-(\vec r - \vec R_P)^2}-D(x_{PA}+x_{PB})(x-x_P) e^{-(\vec r - \vec R_P)^2}+Dx_{PA}x_{PB}e^{-(\vec r - \vec R_P)^2}$$
The first term in this is a d type function, the second a p type function, and the third an s type function, all centred at  $R_P$. Thus your product gets transformed to a linear combination of functions all centred at the same site, and thus the four centre integral can be reduced to a two centre one.
This trick of recentring all to the same site is common, and can be extended to any angular momentum.


text:
What does $n$ represent in "hydrated compound⋅nH2O"?
On Wikipedia it is written that in "hydrated compound⋅nH2O" $n$ is the number of water molecules per formula unit of the salt. I don't understand what that means and this is indicated in that I can't express it in other terms for being able to express a statement in other terms is a sign of understanding it. Can you express this in other terms? How do you imagine these n water molecules to be arranged in a formula unit?

It is the encyklopedia Wikipedia based  on wiki technology, with Wikiwand taking Wikipedia content and rewriting it for Wikiwand GUI. Wikiwand pages usually contain all sorts of 3rd party elements.

OrangeApple2 . n Plums means that for each orange and 2 apples, there is n plums, regardless of how oranges, apples and plums are mutually arranged in the sack.

@Poutnik I can now express it in other terms as well; $n$ represents the same thing that any other subscript number represents in the formula of any ionic compound ie the corresponding number obtained in finding the simplest whole-number ratio of the number of atoms in any of its crystal.

@Poutnik You may post it as answer so that I can close this post.

Does this answer your question? How to understand this form of writing the solution: (some salt • n H₂O)?

OrangeApple2 . n Plums means that for each orange and 2 apples, there is n plums, regardless of how oranges, apples and plums are mutually arranged in the sack.
Now more seriously:
When there is a chemical formula like $\ce{AB2. n H2O}$, then it means there is substance  $\ce{AB2}$ (ionic like $\ce{A^2+}$ + $\ce{2 B-}$, molecular like $\ce{B-A-B}$ or large scale bound with A:B ratio 1:2), which incorporates in its solid structure (crystals, microcrystalic, amorphous) molecules of water in ratio 1 $\ce{AB2}$ : $\ce{n H2O}$.
Usually, such $\ce{AB2}$ can exist as anhydrate ( being anhydrous) too, without water (like there is anhydrous citric acid and its monohydrate).
Some substances can exist in several variants with different values of $n$. E.g. calcium chloride $\ce{CaCl2}$ can be anhydrous (n=0),monohydrate(1), dihydrate(2), tetrahydrate(4), hexahydrate(6).
The formula gives generally nothing more nor less than the stoichiometric substance : water ratio, without any hint about the particular structure and molecular/atomic/ionic placement. In a case of salts, some water molecules wrap cations similarly as in solution, some may be independent or attached to anions.


text:
Stratification of gasses by mass
Suppose a homogenous mixture of mutually non-reactive gases is isolated in a container so that there are no external kinetic forces (vibration or rotation) that create turbulence or other mixing motion.  Will the gases always tend towards a perfect stratification by mass?  If not, what factors oppose stratification?
My guess is that temperature imposes a lower bound on stratification: i.e., even starting from a perfect stratification, higher temperatures cause more mixing across layers due to the random kinetic movement of gas molecules.
I'm also wondering if some non-reactive gases can nevertheless exhibit weak interactions that prevent or oppose stratification, similar to the way liquids can form solutions or emulsions that do not separate under gravity.

In 1905, Marcelin Berthelot filled a 1 liter bottle with Hydrogen gas, and then an identical  1 liter bottle with $\ce{CO2}$. He then joined the two bottles by their necks, with the hydrogen in an inverted position above the $\ce{CO2}$ bottle. He waited one night and then measured the amount of $\ce{CO2}$ in each bottle by adding a $\ce{Ca(OH)2}$ solution and measuring the mass of $\ce{CaCO3} $ produced. Surprisingly enough, he found the same amount of $\ce{CO2}$  in both bottles. Conclusion  : no stratification !

In theory, gravity imposes some stratification. But diffusion beats this effect everywhere except in the upper atmosphere even when you ignore turbulent mixing.

@matt_black I assume "diffusion" is the term for temperature-driven mixing?  So it would decrease as temperature decreases, and at some point not overcome gravitational stratification?

@feetwet - for a back-of-the-envelope exercise, consider comparing $kT$ (or $1/2 mv^{2}$ for thermal motion) vs $mgh$ (for some reasonable change of height, say 1 meter).

These questions are addressed well in this 2009 paper by Giovanni Badino: The Legend of Carbon Dioxide Heaviness.
Gravitational stratification does occur, but it is a result of the fact that change in pressure P with altitude z is a function of the molar mass of the molecules M (paper Equation 3):
$$\frac{dP}{P} = -\frac{M g}{R T}dz$$
The paper also notes the important but perhaps counter-intuitive independence of ideal gases: "Gases collaborate to create a final total pressure, but the partial pressure of each gas behaves as if the others did not exist."  Hence, the pressure (and therefore relative concentration) of heavy molecules decreases much faster as altitude increases.  So gravitational stratification only appears on scales of many meters (or kilometers when considering molecules of very similar mass).
Even if gases are initially stratified or otherwise separated, they will approach an equilibrium mixture through diffusion, the chemistry of which is complicated by many factors.  Nevertheless, w.r.t. the suggestion in the question: it is true that diffusion rates are proportional to $T^{3/2}$ (see paper Equation 7).

It would be worth including some real world evidence about where this is actual observable. There is little stratification at the top of Everest, for example.

Gases definitely will stratify. It is possible to collect CO2 by upward displacement of air. Chlorine gas spills are incredibly dangerous because they will fill low lying areas for hours or even days. I was familiar with this one: https://ebrary.net/131196/health/youngstown_february_1978_train_derailment_with_chlorine_cloud  Even a lower MW gas such as water vapor at a higher pressure will force all the air from a container such as a boiling teakettle. Once mixed, thermal motion and turbulence resist separation but lighter mass molecules more easily escape the Earth's gravitational field while they seem to have trouble getting away from Jupiter [we'll know more about stratification when we get there, I suspect that there is more H2 near the surface.]. The hexafluorides of U235 and U238 separate in a centrifuge.


text:
When would a R-O-R bond be linear?
While rummaging in my molecular model kit, I noticed that it contains a surprising number of $180^\circ$ oxygen atoms, i.e., oxygen that is meant to be covalently bonded to two partners in a linear geometry. Obviously this isn't a hard reference, but since the geometries contained in the kit usually represent common types of bonds, I started wondering when this actually occurs.
It's been a while since my organic chemistry course, but based on VSEPR, the first configuration I could imagine is a doubly positive oxygen with two double bonds on either side, that is, the isoelectronic configuration to carbon with a double bond on either side (as in $\ce{CO_2}$). But $\ce{O^{2+}}$ seems like quite an exotic configuration that I've never heard of. Alternatively, according to the Wikipedia article on linear configurations, the same might be possible in a linear (quasi-trigonal bipyramid) geometry given a central $\ce{O^{2-}}$ atom with two single bonds, isoelectronic to $\ce{Xe}$ in $\ce{XeF_2}$, which seems more in line with my intuition.
Are there other, more common linear oxygen configurations are there? And what are real-life examples of such bonds occurring?

Generally, the holes in that oxygen aren't for bonds.
My experience has been that aside from the tetrahedral looking oxygen, they also have an oxygen that has 5 holes. Two antipodally placed, and 3 in a perpendicular plane. This is supposed to represent a sp2 hybridized oxygen atom. I've seen model sets with a similar carbon atom. They also have fancy blobs that represent orbitals that you can put into the holes that are out of plane.

If you use the hole for a hydrogen bond, 180 degrees is possible (like in an alpha helix). https://www.flinnsci.com/globalassets/flinn-scientific/all-product-images-rgb-jpegs/ap10030.jpg?v=3d4a4dfbe03b41f6aa655229df2c785e

@KarstenTheis That seems like the answer I was looking for! Not perfectly 180 degrees, but close enough and the most organic/biochemical suggestion so far.

Oxygen can be bonded with 180° bond angles in perovskite structures, such as silicate minerals formed deep in Earth's mantle (the perovskite form of calcium silicate has been identified at Earth's surface by being trapped under internal pressure in diamonds). Such minerals have octahedrally coordinated silicon atoms connected by linearly two-coordinate oxygens, with the empirical formula $\ce{MSiO3}$. The calcium silicate compound referred to above has the structure given below (red = oxygen, black = silicon, blue = calcium)[Source]:

Since the silicon in this silicate-perovskite structure is a $p$-block element and yet octahedrally coordinated, the covalent bonds to it are shared over multiple atoms and thus each silicon-oxygen linkage is partially ionic (cf. $\ce{SF6}$).

Very ingenious answer! Not quite the sort of thing one looks to build with a molecular model though...

@orthocresol - maybe not for chemists, but materials science folk do odd things with ball and stick models…

@JonCuster I would love to see some of that...!

After posting the question, I also started thinking along these lines, good point. Since the kit is geared towards biochemistry, might there also be examples of such arrangements occurring in (organo-)metal coordination centers?

No, bond angles of virtually any $\ce{R-O-R}$ bond will differ from $180^\circ$, so there are no common linear oxygen configurations. Steric hindrance can cause a severe deviation from the ideal tetrahedral bond angle of $109.5^\circ$, but it will usually still be very far from $180^\circ$. For example, the paper below by Liedle et al. reports a bond angle for di-tert-butyl ether of $130.8^\circ$.

Reference:
J. Mol. Struct. 1989, 198, 1-15. https://doi.org/10.1016/0022-2860(89)80025-0


text:
Which vessel materials are compatable with magnetic stirrers?
I am shopping for some metal pans and beakers for use on my magnetic stirring hotplate. I do not want to block the magnetic coupling of the stir bar. The question is which common alloys are compatible? Is the compatibility essentially binary or are some materials weakly compatible?
I assume aluminum is fine because the hotplate is made of some type of cast aluminum, but I am uncertain of the effect of alloy or cold working. My ferro-magnetic stainless steel and carbon steel pans are not compatible. But how about austenitic stainless steels[non-magnetic, 304/316], copper, or brass? Unfortunately I don't have any known samples on hand to test.
Glass is ok in a pinch, or when chemical reactivity is an issue, but for simple heating baths it is a poor conductor, more fragile, and in larger sizes much more expensive than metal. (And ease of sourcing, I am near a commercial kitchen supplier and a scrap dealer with all shapes and material stock pots but I need to ship lab supplies.)

304 will be better than 316 because 316 can have fairly high ferromagnetism when cold worked. Cold worked 304 can also pick up some magnetism. Probably need to take magnet with you when you shop for a beaker

Yeah the empirical method is probably the practical solution. Still leaving the question open so people can add data, maybe I'll post what I find and make it a wiki.


text:
My book's claim about the shielding effect of s,p,d and f electrons
It's a relatively unknown Bangladeshi book. It's called "Chemistry-First Paper (Class XI-XII)" by Professor Haradhan Nag
According to my book, "Electrons of s and p orbitals produce more shielding than electrons of d and f orbitals. As a result, atoms of s and p-block elements experience lower effective nuclear charge. Consequently, their atomic sizes are greater than those of atoms of d and f-block elements".
My book's claim makes no sense to me. It's true that the electrons of s and p orbitals produce more shielding than electrons of d and f orbitals due to penetration, but why would that make atomic sizes of s & p block elements greater than d and f-block elements?
Let me take s-block as an example. In period 4, $\ce{K}(220 \pu{pm})$ is an s-block element and $\ce{Sc}(160\pu {pm})$ is a d-block element. It's true that the atomic size of K is greater than Sc; however, I don't think it's because s-orbitals have a stronger shielding effect than d-orbitals. If anything, we could ironically argue that Sc should have a greater atomic radius than K as s-orbitals have a stronger shielding effect than d-orbitals because the 4s-orbital is inner in Sc. In contrast, the 4s-orbital is situated outermost in K. So, why is my book making the claim that an outermost orbital's shielding power will determine the size of the atom? Outermost orbitals have no contribution to shielding!

Related

Edit:
It turns out that the 4s orbital is not actually inner in Sc. However, my point still stands. Why is my book making claims about shielding based on the outermost orbital (which has no contribution to shielding!)

See also Wikipedia: Slater's rules

Let's have a look at some illustrations:

Size of atoms by atomic number:

The size of Li, C and Ne with their orbitals (blue is completed and yellow is empty):

The shielding effect, illustrated with the density of probability of the electrons' presence (green):

Hydrogen probability densities of 2p and 2s orbitals:

We will need to use quantum physics to answer this question, so it gets further than the classical chemistry problems. I therefore consider electrons as matter waves, their position being non-deterministic, the probability of their presence at such and such a place determined by the Schrödinger equation.
On the illustration n°1, you notice that the atomic radius decreases with the number of electrons until reaching a new orbital, which necessarily increases the range of the electronic cloud (each orbital being more extended than the previous one).
The illustration n°2 shows you that the atomic radius decreases with the stability of the electron (number of completed orbitals).
The third and fourth ones just illustrate the consequences of the shielding effect.
Now, here is the answer to your question: within a set of neutral atoms that have the same number of orbitals, the charge of the nucleus and the electronic cloud increases with the atomic number Z. As the shielding effect of the first sublayers is limited to the maximum number that these orbitals can hold, the charge felt by the farthest electrons will be stronger and stronger, which will attract them more towards the center and decrease the atomic radius. Concerning the shielding effect of the higher orbitals, I have a little trouble understanding what your book means since the s and p orbitals always have a shielding effect (never replaced by the shielding of the other orbitals, but added to it). Generally speaking, a smaller shielding effect gives a higher charge of the nucleus and therefore a smaller radius. Atoms with d- and f-layers have a larger radius because the range of these orbitals is simply larger according to the Schrödinger equation (you can compare the range of 2s and 2p on picture n°4). I hope this helped you!


text:
Is collision theory applicable only for molecules?
$$\ce{Na (s) + Cl_2 (g)->NaCl(s)}$$
Is collision theory applicable for the above reaction, or is collision theory applicable only for molecules? In other words, do chlorine atoms collide with sodium atoms to produce sodium chloride , or does the chlorine atom get near the sodium atom, electron transfer takes place (as it is energetically favorable), and $\ce{Na+}$ and $\ce{Cl-}$ get stuck?
Edit after @BuckThorn's comment:

or does the chlorine atom get near the sodium atom, electron transfer takes place (as it is energetically favorable), and
$\ce{Na+}$ and $\ce{Cl-}$ get stuck?

In this process, the Na atom and the Cl atom don't touch, and electron transfer occurs at a distance (due to attractive/repulsive forces). Only after electron transfer takes place do $\ce{Na+}$ and $\ce{Cl-}$ touch and get stuck. That's why I don't think it is akin to collision.

You may want to explain why the two alternatives you present might or might not agree with collision theory (in particular the second). Collision theory combines a description of the rate at which reactants collide with a model of the fraction of those reactants that have enough kinetic energy to result in a productive collision. The details will change but, provided the assumption that a collision between reactants exceeding the required energy is appropriate, the theory should be able to describe reactions between gases and solids.

Note that $\ce{Na(g) + Cl(g) -> Na+(g) + Cl-(g)}$  is energetically unfavourable, unless ions end bound close enough ($\ce{Na+Cl-(g)}$) by coulombic force. Saying that, not speaking about energy needed for formal $\ce{Na(s)}$ atomization and $\ce{Cl2}$ dissociation.

@Poutnik On a related note: how do we determine if a reaction is energetically favorable or not? Do we add the ionization energy of Na (496kJ/mol) and the electron affinity of Cl (-349kJ/mol), and if the answer is positive(147kJ/mol) then we say that the reaction isn't favourable?

Yes it is like that. We would say such a reaction is endothermic, consuming thermal energy. Strictly speaking, I should use thermodynamically favourable via negative Gibbs free energy change, not enthalpy change. ( $\Delta G = \Delta H - T \Delta S$ ), but at these big $H$ changes the entropy term plays just minor role. When energy is released, it is usually dissipated elsewhere, so it cannot be reused to go back. OTOH, if energy is needed, you have to gain it somewhere.

@Poutnik "unless ions end bound close enough (Na+Cl−(g)) by coulombic force."- sir, I couldn't understand exactly what you mean by this. Do you mean that large amounts of energy are released (energetically favourable) when Na+ ion and Cl- get closer and get stuck together?

@TomHardy Energy outcome of producing ions depends on how far from each other the are  If far enough, it is endothermic. If close enough, it is exothermic. The best if many opposite ions are close enough, forming solid ionic lattice.

Your example is one  of many that have been studied by molecular beam scattering techniques so this is a rather general answer.
Collision theory calculates the rate constant averaged over all geometries and energies at at given temperature. It is the product of cross section, average collision velocity and the Arrhenius factor. Thus it depends on the cross section. The simplest model assumes hard spheres so will work for atom-atom, atom -molecule etc. but less well as the species become more complex. But actually the cross section is very complicated especially as the potential between molecules varies with distance and orientation perhaps as coulomb potential or Lennard Jones or something more complicated. In this case The trajectories of the interaction between species has to be calculated and these averaged to get the collision theory rate constant.
In fact the rate constant is not that important what is important is the potential energy profile between species as this reflects the electronic properties of the molecules and can be calculated from Quantum Theory. The rate constant can then, in principle, be calculated from the potential although experiment is always essential.
Experiments of this sort are studied using atom/molecular beams and numerous reactive scattering reactions have been studied such as H+H2, D+H2, N+O2, F+H2,H+F2, O+CH, O+Cs and many more.
You should look at two excellent books on this topic Levine & Bernstein 'Molecular Reaction Dynamics, and Chemical Reactivity' publ OUP 1987, and Steinfeld, Francisco & Hase, 'Chemical Kinetics and Dynamics', publ Prentice Hall 1999.
The picture below shows how complicated the trajectories become when there is a Lennard jones potential between two atoms as one approaches the other from the left but at different position vertically. You can see that the species get pulled together at a larger separation than hard sphere (the inner circle) but can orbit and escape if there is enough energy. In this picture no reaction occurs but you can appreciate if the species get close enough with enough energy a condition could be made to make them react, for example being within the wide grey ring.

The picture below shows the Cl+H2 $\to$ HCl+H reaction. The species approach with the same energy in each case but collide at different point in the H$_2$ vibration and this leads to different outcomes, i.e no reaction or more of less vibrational excitation in HCL. The blue colour is low energy and the transition state can be seen at the bend. To calculate the rate constant the number of successful transition state crossings vs. total number at a given energy must be calculated. This could then be compared with the collision theory rate calculation and an effective cross section derived.

"The species approach with the same energy in each case but collide at different point in the Cl2"-You probably mean Cl instead of Cl2.

"The picture below shows the Cl+H2 → HCl+H reaction"-how does Cl remain as Cl? Isn't it a diatomic molecule?

"The simplest model assumes hard spheres so will work for atom-atom, atom -molecule etc."-Just to be clear, you agree with me that collision theory should be applicable for the reaction Na(s)+Cl2 (g)⟶NaCl(s), right?

you seem to be considering a solid /gas reaction which is different.

BuckThorn made a comment about this matter. Do you agree with him, sir?

@user119245 I just now noticed, it is a typo it should be H$_2$  vibration not CI$_2 I will edit it shortly$


text:
Huge variation of the atomic size of Uranium
I've already posted this question in Physics Stack Exchange, but the answer that I received (actinide contraction similar to lanthanide contraction) is not convincing for me, or at least is not detailed enough to explain the huge difference.
According to Villars and Daams [Journal of Alloys and Compounds, 197 (1993) 177] the atomic volume of U is 2.073×10−2nm3, whereas that of thorium is much larger, 3.295×10−2nm3.
This is confirmed by comparing bond distances in pure elements. Even if Th and U crystallize with different structures, they have the same coordination number (12). In U the U-U bond distance is 0.275 nm, whereas in Th the Th-Th bond distance is much larger 0.360 nm. Why this difference? This huge difference is not observed for Ce and Pr (where 0.333 and 0.363 bond distances are measured).
The external electronic configurations of Th and U atoms are:
Th: [Rn]6d27s2
U: [Rn]5f36d17s2
What's the origin of this strong difference?
Why, then, their ionic sizes (for example the sizes of Th4+ and U4+) are quite similar?

Due different relations of 4f vs 5d  and 5f vs 6d orbital energies, the first actinoids, in contrary to the first lanthanoids, behaving more like d than f elements. So, in a sense, you are like comparing $\ce{Hf}$ and $\ce{W}$. OTOH, when ionized enough, this deviation more or less disappears.

Are you sure the coordination number of U in the crystalline metal is 12? Greenwood and Earnshaw state "For U, Np and Pu are rather irregular so that the coordination number is not a precise concept". https://www.webelements.com/uranium/crystal_structure.html gives the structure as orthorhomobic with space group Cmcm, and references C. S. Barrett, M. H. Mueller, and R. L. Hitterman, Phys. Rev., 1963, 29, 625

12 (anti-cuboctahedron) is the coordination number reported in the Pearson's crystal structure database for the Cmcm structure of U.

Uranium has more valence electrons capable of populating bonding molecular orbitals (or bonding electronic bands) when compared to thorium - basically, uranium is closer to the middle of its block, compared to thorium. More bonding interactions will pack the atoms tighter. Indeed, uranium has an atomic weight ~2.5% greater than thorium, but metallic uranium is 63% denser than metallic thorium (I am assuming that these densities were measured for metallic phases with the same packing for both elements).
Now, if the way you define "atomic size" is by using distance measurements performed on the solid metallic elements (e.g. from x-ray diffraction data), you will pick up on this extra bonding effect, and it will look like uranium atoms are much smaller than thorium. Indeed, the ratio of the atomic volumes you quote is 1.59, which is suspiciously close to the (inverse) ratio of 1.63 for the densities of the pure metals, so I'm guessing this is how these values were obtained.
However, you could in principle make a measurement of atomic size for isolated atoms in the gas phase. In this situation, there are no bonding interactions, so uranium and thorium are likely to have much more similar sizes. So as it almost always happens in these discussions, it comes down to what do you mean by size?
The reason there is no such difference between cerium and praseodymium is that the 4f orbitals lack radial nodes and are subject to a considerable effective nuclear charge, so they behave almost as core electrons in the lanthanides, where they should be valence electrons. Therefore, placing or removing electrons in the lanthanide 4f orbitals as you move along the block has comparatively little effect on the bonding interactions for these elements.

I have since realized the atomic volume ratio of 1.59 is reproduced exactly by the ratio of the densities of the pure metals (1.63) multiplied by the ratio of the atomic masses (232/238 = 0.975), so the atomic volume data quote in the question indeed almost certainly is based on the pure metals.


text:
What does the heat capacity really represent?
According to Wikipedia, heat capacity or thermal capacity is a physical property of matter, defined as the amount of heat to be supplied to an object to produce a unit change in its temperature:$$C = \lim_{\Delta T \rightarrow 0} \frac{\Delta Q}{\Delta T}$$
According to another source (which is PDF only, and not in English), in the context of thermodynamic properties of gases due to the rotations
of diatomic molecules, at very high temperatures:$$C_{V,\mathrm{rot}} \approx k_\mathrm{B} = \pu{1.38E-23 J/K}$$
Does this imply that adding $\pu{1 J}$ to this system will raise its temperature by $10^{23}$ kelvin?
I believe there is something in this definition I didn't get. Could you explain the sense of $C$ in a more intuitive way please?

Heat capacity [J/K] ]is property of a system/object, being an extensive property.. The property of matter/substance are the intensive parameters specific or molar heat capacities [J/K/kg] resp. [J/K/mol]

Are you aware that energy  1 J is for rotational energy of a single molecule an astronomically huge energy, available only at astronomicaly high temperature? (ignoring the fact any chemical bond would break at temperature many many orders lower). // The above C_V,rot is the heat capacity due rotational motion of a single molecule.

Oh, I see, I should have stayed at the $1 \ eV$ scale ^^ Thanks !

Typical thermal energy at 20 °C is $\frac{kT}{e} \approx \pu{25.2 meV}$

It is easier to think of the heat capacity at constant volume as the slope of the internal energy with temperature. This is effectively your first definition. The correct interpretation of the heat capacity is only possible via quantum theory, that is by realising that atoms and molecules have discrete energy levels.
Thermodynamics implicitly assumes that all energy levels are so closely spaced compared to temperature (at all temperatures) that they are effectively continuous.  This is only true for translational energies thus the translational energy is the sum $\sum mv^2/2$ for $x,$ $y$ and $z$ directions giving $E=3k_\mathrm BT/2$ and so heat capacity $C_\mathrm{tr}=3k_\mathrm B/2$ for atoms and molecules. Doing the same calculation for rotation and vibrations assuming classical behaviour gives for a diatomic,  $C_\mathrm{rot}=k_\mathrm B$  and $k_\mathrm B$ for the vibration, the total is $C_V=7k_\mathrm B/2$.  These values overestimate the true experimental because by using classical arguments we have assumed all levels are fully occupied whereas we know that because levels are discrete, at any given temperature only some levels are populated according to the Boltzmann distribution.
This then returns to the calculation of the internal energy, which is the sum of every level's energy weighted according to its Boltzmann population. At very low temperatures only the zero point energies are populated and so the heat capacity tends to zero, a result impossible to obtain from classical thermodynamics. At very high temperatures the classical values are obtained for all molecules, but in many cases this temperature is far, far above room temperature.

You wrote: "Thermodynamics implicitly assumes that all energy levels are so closely spaced compared to temperature (at all temperatures) that they are effectively continuous."  That's not quite right.  Thermodynamics makes no assumptions about the microscopic character of the substance.  For that reason, when we say, e.g., that Cv for a diatomic gas is 7/2 nR generally, we aren't using thermodynamics at all.  Rather, we are using statistical mechanics uninformed by quantum mechanics.  I.e., we are using classical statistical mechanics. It's the latter that also gives us the Dulong-Petit law.

In sum, values for heat capacity, like Cv=7/2 nR for a diatomic gas, are quantitites derived outside of thermodynamics that can be used as an input into thermodynamic expressions.  But thermo is just as happy taking, as input, an expression for heat capacity as a function of T derived using quantum mechanics.  Thermo has no opinion on the matter; it is a big tent that welcomes both. Thermo only complains if you attempt to use a negative heat capacity, since that causes a 2nd law violation.  Take-home message: Don't blame thermodynamics for the limitations of classical statistical mechanics! ;)

@Theorist Thanks for you comments. I was trying to explain simply without being too technical. Of course thermo was developed before atomic theory was understood and treats matter as continuous undefined stuff but I think that it is easier nowadays to explain this in terms of energy levels as this is more familiar.  Its great generality is its strength, but this I think,  makes thermodynamics hard for students.

As @Poutnik wrote in the comments, the heat capacity $C$ quantifies the influence of energy on the temperature of the system. There is no contradiction to write $C = k_\mathrm{B}$ since at the scale of a single molecule $\pu{1 J}$ is huge (energies are of the order of the electronvolt $\pu{eV}$ with $\pu{1 eV} \approx \pu{0.6E-19 J}$. Typical thermal energy at $\pu{20 ^\circ C}$ is $k_\mathrm{B}T/e \approx \pu{25.2 meV}$.

Useful links for text and formula formatting (not to be applied to titles):
Notation basics ,  Formatting of math/chem expressions  and
upright vs italic // For more, see Math SE MathJax tutorial.

Consider $\pu{1 eV} \approx \pu{1.6E-19 eV}$ as  $\pu{1 eV} \approx \pu{1.6E-19 eV}$. //  BTW $\pu{1 J} \approx \pu{1.6E19 eV}$ is wrong as $\pu{1.6E19} \ne \frac{1}{\pu{1.6E-19}}$.

Thanks for the \pu{} command, that changes a lot of things, didn't know it. Sorry for this mistake

P.S.: Be aware of e like \cdot  versus E like \times: $\pu{6e-6}$ as $\pu{6e-6}$ versus $\pu{6.4E-6}$ as $\pu{6.4E-6}$. 10^6 in context of \pu{} can be written also as $\pu{e6 mg}$ seen as $\pu{e6 mg}$.

And would you know how to properly write "$C_{V, rot}$" such as "$rot$" appears like "rot" ?

$C_{V\mathrm{,rot}}$ as $C_{V\mathrm{,rot}}$, a variant  $C_{V\text{,rot}}$, with \text{} taking the string literally.  $\mathrm{a b}$ displayed as $\mathrm{a b}$ ,   $\text{a b}$ as $\text{a b}$

It is not about quantum vs classical physics, but simply the size of the system. 1 J, 1 mol, 1 g etc. -- these are scales used for macroscopic quantities, i.e. amounts of chemicals that we can see. 1 eV, k_B, etc. - these are things suitable for single molecules.

Oh, sorry, it's corrected


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Can I scrape off compounds from normal TLC like in prep TLC for MS analysis?
As far as I know, the limit of detection of mass spectrometry is very low compared to other analysis. In principle, so I think I can scrape off some compounds from normal TLC plate after development like in prep TLC, filter with appropriate solvent, and get the filtrate. But I have never seen this application. I think there are some reasons for which one can't carry out it, but can't occur to definite reason.

Yes, you can do this.

@Waylander Oh it is possible. Thank you.

I might well be tempted to run a line of product mixture up the plate instead of a single spot so that a greater amount of material is harvested


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Melting stainless steel (or high carbon steel in general) by artisanal techniques (forge welding)?
Here is a new question about the making of Japanese knives (you can find the first two here and here).
According to many sources, the best Japanese knives are made in a traditional way with forge welding. However, I learned that forge welding is only for iron and low carbon steel, while the best knives are made of stainless steel (high carbon).
Would there be a way to heat stainless steel blades to melting so that they can be hammered using this forge welding, or should more advanced welding techniques be used? If so, could you give me an example?

From https://en.wikipedia.org/wiki/Forge_welding : "Since the critical temperatures are affected by alloying agents like carbon, steel welds at a lower temperature-range than iron. As the carbon content in the steel increases, the welding temperature-range decreases in a linear fashion."

So their must be another reason why the welding can't be done... seems like I mixed up 2 elements

There is no melting in forge welding ; melting would be a large error. Two solid  surfaces are pressed firmly together and reheated; at high temperature, diffusion improves the bond. Damascus and samurai blades are similar, layers of high carbon and low carbon steels folded several times with reheating for diffusion bonding producing effectively a medium carbon steel. Long ago iron came 2 ways; strong ,brittle cast iron, or soft ductile very low carbon wrought iron. The technology did not exist to produce  strong tough medium carbon steel. The forge welding of high and low carbon was a good step but the relative high cost of man power mostly limited it to high value products like blades. I am unaware that the process was ever used for chrome stainless cutlery steels. I expect it could be done as a hobby, but certainly not a practical process to make a blade. It is a bit of catch 22, with 13% chrome in steel very little carbon will produce some hardening. So ordinary 410 with 0.15% carbon will harden enough to make a fair blade. To get 410 that won't harden ,you need to pay a premium for 410 S or, 409, or 405. So it will be an expensive hobby.

Seems another person wants to join the secret society of metallurgy.

I see, so when producers say "my japanese knives are 100% traditionals" they can't say the truth since there is 1,7% of carbon in their blade !

I should add you in my report as co-author of my work ^^