text:
Why are the unsaturated fatty acids usually unconjugated?
My textbook gives the structures of the unsaturated fatty acids in food including oleic acid, linoleic acid, linolenic acid and docosatetraenoic acid. I figure that when two or more $\ce{C=C}$ double bonds appear in a fatty acid, they tend to arrange in this way:
$$\ce{-CH=CH-CH2-CH=CH\bond{-}}$$
with a methylene group blocking between them, rather than a conjugated form. Why is that?
And in the synthesis and degradation of fatty acids, a conjugated form seems to be more straightforward and more simple for both synthesis and degradation, needless to move the double bond after constructing the fatty acid chain or before β-oxidation to break down the chain. Furthermore, conjugated systems seem to be fairly popular in polyketones which share a similar synthesis mechanism with fatty acids.
So is it that the conjugated form is too easily oxidized to be stored for a long time, or is it that the rigid structure of conjugated double bonds would lead to difficulties in transportation and storage? Or, is there other explanation for that?

It goes to a question why life has evolved/has been created this way. A difficult answer, like asking an art paint producer why the painting has been painted the way it was.

Note that naturally-occurring conjguated fatty acids do exist, e.g., the conjugated linoleic acids, typically found in milk and beef (https://www.nature.com/articles/s41598-021-88870-9), and the conjugated linolenic acids, typically found in seed oils ( https://link.springer.com/referenceworkentry/10.1007%2F978-3-540-47648-1_1314 ). [See also https://en.wikipedia.org/wiki/Polyunsaturated_fatty_acid ] Nevertheless, your question (which is an interesting one) remains:  What is the evolutionary/functional explanation for why the predominant PUFAs are methylene-interrupted rather than conjugated?

@theorist Simple: they have different uses. Look up what the organisms make out of them (and how), and you  have the answer. ;)

@Karl.  Not so simple, I'd say. Sure, you can find epidemiological articles talking about their potential health effects, but those haven't been fully established and, besides, what we're talking about here is instead their actual role in metabolic pathways as regulators/precursors.  You can't get that kind of info. from summary articles (e.g., https://www.lipidmaps.org/resources/lipidweb/lipidweb_html/lipids/fa-eic/fa-conj+/index.htm or https://en.wikipedia.org/wiki/Conjugated_linoleic_acid)...and you can see the reason why by looking at the review articles in the primary literature, which...

indicate there continues to be significant uncertainty about just what role CLA plays (see, Fig. 4 in https://www.sciencedirect.com/science/article/pii/S175646461500167X , from 2015, and note all the ques. marks).  It seems it's just with the most recent articles (e.g., this, from 2021:  https://www.frontiersin.org/articles/10.3389/fphar.2020.587140/full ) that they seem to be demonstrating where they play specific precursor roles.  Bottom line: It appears this is not settled science--it's an active area of research.  Thus what we really need is a biochemist (which I'm not) to respond to this.

Anyways, if all this is simple for you, I'd guess you are a biochemist, and thus could provide the kind of judgement/overview/summary of the current state of the literature on this subject that I cannot.  So how about taking a crack at it? :).

@theorist You guessed wrong. ;) I just though feed 13C labeled fatty acids to a number of interesting organisms, and later check where the 13C ended up.

One factor is that the double bonds are generally all cis, and two consecutive cis double bonds make a rigid curl in the chain. More generally, avoiding conjugation preserves greater flexibility

Do the isoprene units have some point here?


text:
how to combine Raman spectrum data with different wavenumber
I have two set of spectrum data, the first record (in wavenumbers) looks like this:
...
967.0511723
968.9374611
970.8224407
972.7061127
974.5884787
976.4695403
...

And the second record (again, wavenumbers) umber looks like this:
...
955.9630179
957.8733753
959.7823965
961.6900828
963.596436
965.5014577
967.4051494
969.3075128
971.2085494
973.1082609
975.0066489
976.9037149
978.7994605
980.6938874
982.586997
...

How do i merge two datasets with the same wavenumber?

I am a little surprised. Your wavenumbers are not equally spaced. If they were, we could have done upsampling by interpolation methods.

could you recommend some python package or papers about spectrum interpolation? thank you very much !

I don't remember any, but you can start from MATLAB documentation on upsampling by interpolation.

Also upsampling to a higher value can be done by Fourier methods. The ke point is the sampling interval must be constant. Yours is not. This is the main issue.

Easiest is probably to combine your data in one vector for wavenumber and one for intensity and sort both vectors on the values of the wavenumber list. Then you could do a running average with $n=2$. Make sure to check that the peaks don't get wider due to the running average operator

Could you say a few words why you want to have equal wavenumbers? What is your following step?

@haidanli Re processing the data:, though written for XRD/X-ray (powder) diffraction data, fityk offers multiple options valuable to spectroscopy, too.  If specific to spectroscopy with Python, though, try spectrochempy, and addressing Raman in particular, rampy.

@haidanli Though not a programming code block, I borrowed the technique for an easier access of the data you present in your question (cf. the edit) for tentative work elsewhere.  Let me suggest that the addition of the second column (intensity in function of the wavelength) could be beneficial to advance further, toward an answer.

In the next step , i hope to merge two datasets with different wavernumber. so i will have a large samples of ranman spectrum and can do some regeression on this large samples.@Snijderfrey

source: Stack Exchange

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

sort: votes


text:
Overtreatment of vintage rubber parts with Wintergreen oil
I have read in many places that treating old rubber parts with wintergreen (methyl salicylate) will soften the rubber again. The method I chose was on this website . I have seen many ways to do the same thing, this one appeared to appeal to me, but I wish it had not. This method using heat, was one that kept the items being treated to only 4 to 6 minutes. Other methods were much longer.
Long and short of it, I followed the instructions and ended up swelling 3 rubber parts beyond use as they are now. 2 of them can be found online, so I am only concerned about 1 part that appears to only one left in existence, the current owner must know that because they are asking top dollar for it.
The question is, is there a solvent, fluid, compound, etc. etc., that will draw out what the methyl salicylate added to the rubber to make it pliable and swell so bad?

The parts in the picture are what was affected, the 2 that are not circled still have some firmness in their handling. The one circled, the one I cannot easily replace is supposed to fit into the ring on the right side of the circle. It is a push button on my 1973 Yamaha motorcycle. The rubber as gotten very soft compared to the other items, although the camera angle does not show it well, it is still reminiscent of the "derby hat" shape it needs to fit inside the metal ring that is its retainer.

Likely, the rubber is no longer as completely polymerized as before, and nothing can be done to salvage the parts. That said, the little "cap" part seems easy to fabricate from bicycle inner tube (it looks like a primer button for a carburetor), if you're not trying to restore an antique to appear as if it had OEM parts.

The part is now really super squishy, I know it is not a chemistry term, and its shape is not  very flat that could be made of an inner tube. On the inside, there is a shape, a protrusion,  in the center that keeps a spring in place to keep the switch button (the rubber piece in question) from grounding out on its own since it is a kill switch for my classic dirt bike.

I had a conversation with another rider that seemed very knowledgeable about things, at least about my plastic screw in gas cap that is for the same bike. It told him how it shrunk and the threads that tightened the cap in place would not engage anymore onto the tank. I had to get a new cap to replace the original. He explained to me that the hydrocarbons in the gas in the tank affected the hydrocarbons in the cap and the cap shrunk from the exposure. Or something like that. I was hoping that I could dunk the part in something that might do the same thing that happened to the original gas cap.

While I never rode a motor bike, I wonder why replacing the broken/outworn piece with a new one doesn't restore the function. By a brief search, suppliers like sumorubber, or Yahama parts house. Or, if there were communities to exchange elder, yet still functional spare pieces made back then, similar to brick link occasionally mentioned on bricks.se. Or maybe a visit at  mechanics.se?

Its simply a matter of keeping the originality of the bike, no more. The more original the bike the better. Some parts do wear out, like tires and drive chain. When those parts get replaced it is an accepted norm. A kill button/stop switch in my case, receives no real wear to speak of, so to keep it intact is desirable. But in all it is vanity.... And when all is said and done I may need to install a reproduction of the item. I have through another forum, they are available for a more reasonable price than the first one I found. But I need to try all avenues to restore this part if possible.

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text:
Why does Fluorine react differently than other halogens during Hydrohalogenation/Halogenation and also substitution with Alkanes?
From what I understand, during halogenation with an alkene or alkyne, diatomic gaseous halogen molecules can effectively 'break' the pi bond to produce a haloalkane/haloalkene. However I discovered that Fluorine cannot undergo this reaction, as it produces different products. I'm not so interested in what the other reaction is and products are produced, but more WHY the fluorine reaction is different to that of Chlorine, Bromine, and Iodine. I suspect it is due to Fluorine's extremely high electronegativity?
Furthermore, during a substitution reaction with alkanes and a halogen to produce a haloalkane and hydrogen halide as a byproduct, often catalysed by UV light, why does Fluorine react explosively in this case, whilst Iodine is generally unreactive? Is this also because of electronegativity?

Sulfur and selenium are also more similar than oxygen and sulfur, same for lithium, sodium and potassium.

The F-F bond is weak making the generation of F. radicals an easy process, but the C-F bond is very strong making the process highly energetically favourable and exothermic.


text:
Is there an error at page 138, 4.4 of Introduction of Computational Chemistry 3rd Edition?
Firstly, I found the RHF expansion of $\Phi_2$ and $\Phi_3$ is not symmetric at equation (4.29). Then I expand $\Phi_2$ by myself, the results showed it's
$
\Phi_2 = (\chi_A\chi_A-\chi_B\chi_B)(\alpha\beta-\beta\alpha) - (\chi_A\chi_B-\chi_B\chi_A)(\alpha\beta+\beta\alpha)
$
Moreover, even with this the equation (4.30)~(4.31) is also inconsistent, which the righthand side of the equation should be switched.
But this is a famous book in its third edition, so am I misunderstood or there is an error?

Reference:
(1) Jensen, F. Introduction to Computational Chemistry, 3rd ed.; John Wiley & Sons: Nashville, TN, 2017.

@Nisarg There may be people viewing this on mobile screens, for those it may be very, very inconvenient to read the smaller font that you are trying to fake with <sup> ... </sup> tags. There is also not really any point to it, since it actually increases line spacing and it does not save any screen real estate. And given that the source is a major point of this question, the choice to present it smaller than the rest is actually worse.

There is an updated list of errata* for the 3rd edition of Jensen's book at https://chem.au.dk/~frj/corrections_ICC3.html, which includes the mistake you pointed out (and more).
It reads:

page 138, (4.29), third line should read: [chiAchiA - chiBchiB](alpha.beta - beta.alpha) - [chiAchiB - chiBchiA](alpha.beta + beta.alpha)
page 138, (4.29), fourth line should read: -[chiAchiA - chiBchiB](alpha.beta - beta.alpha) - [chiAchiB - chiBchiA](alpha.beta + beta.alpha)
page 138, (4.31), a minus sign is missing, or change to (chiBchiA - chiAchiB)(alpha.beta + beta.alpha)

Or, put into maths:
$$\begin{align}
\Phi_2 &= [\chi_\mathrm{A} \chi_\mathrm{A} - \chi_\mathrm{B}\chi_\mathrm{B}](\alpha\beta - \beta\alpha) - [\chi_\mathrm{A} \chi_\mathrm{B} - \chi_\mathrm{B}\chi_\mathrm{A}](\alpha\beta + \beta\alpha) \tag{4.29} \\
\Phi_3 &= -[\chi_\mathrm{A} \chi_\mathrm{A} - \chi_\mathrm{B}\chi_\mathrm{B}](\alpha\beta - \beta\alpha) - [\chi_\mathrm{A} \chi_\mathrm{B} - \chi_\mathrm{B}\chi_\mathrm{A}](\alpha\beta + \beta\alpha) \tag{4.29} \\[10pt]
^3\Phi_+ = \Phi_2 + \Phi_3 &= - [\chi_\mathrm{A} \chi_\mathrm{B} - \chi_\mathrm{B}\chi_\mathrm{A}](\alpha\beta + \beta\alpha) \tag{4.31}
\end{align}$$

* Annoyingly, this was quite difficult to find, as Wiley only provides a link to a four-year-old list of errata. You need to go to Frank Jensen's homepage to get the updated link.


text:
What is the difference between heat of reaction (q) and enthalpy of reaction ∆H?
I am confused as I can’t understand the difference between the thermodynamic terms q and ∆H.
In my book,“general chemistry” by Ebbing and Gammon,it is stated that at constant pressure Qp=∆U+P∆V=∆H.
How come can q equal ∆H? The only occasion at which q=∆H is when the moles of the limiting reagent in the thermodynamic equation is 1, such that q=∆H/n.
What does constant pressure has to do with this equality? Because if we assume such equality of. Qp=∆U+P∆V=∆H at constant pressure, then it is also valid at constant volume when ∆V=0 such that Q=∆U=∆H.
Please help me understand this conundrum.
Many thanks in advance.

You may need to dive a little more into the definition of enthalpy in other sources if your textbook does not cover it in detail, for instance look through other posts on this site, which probably answer your question.

For instance: https://chemistry.stackexchange.com/questions/44129/what-is-the-qualitative-definition-of-enthalpy-in-beginner-terms?rq=1 https://chemistry.stackexchange.com/questions/10187/at-constant-volume-why-the-change-of-enthalpy-doesnt-equal-the-change-of-heat?rq=1 https://chemistry.stackexchange.com/questions/39988/does-enthalpy-equal-heat-when-pv-work-is-done?rq=1 and many many more....

I think the core of the question is about the confusion between enthalpy vs molar enthalpy (extensive or intensive quantity), and the question about constant pressure is secondary. Textbooks are notoriously sloppy about this, switching from enthalpy to molar enthalpy without using a different symbol (or bold vs not bold like in upperlevel texts). Heat is always an extensive property. I think there is a good question here, so I voted to reopen. @BuckThorn

@KarstenTheis I agree about the confusion that caused the post and agree that this is a common issue for a lot of students starting thermo. Maybe you are right that the molar quantity issue is worth being addressed in particular, so reopen as I single-handedly closed.

Related: https://chemistry.stackexchange.com/a/162243/72973 (which I just wrote to address both questions).

You are aware that, at constant volume, the number of moles present can change, right?  So $\Delta (PV)=RT\Delta n$.  So $\Delta U\neq\Delta H$

In my book,“general chemistry” by Ebbing and Gammon,it is stated that at constant pressure Qp=∆U+P∆V=∆H.

This is correct in the absence of electrical and mechanical work (e.g. the equality breaks down for a battery or a fuel cell doing electrical work, or shaft work heating up the reaction mixture).

How come can q equal ∆H?

You can express the change in enthalpy as an extensive quantity (changes with the size of the reaction) or, in the case of molar enthalpy, an intensive quantity (does not depend on the amounts reacting). The equation above refers to enthalpy, the extensive quantity (with units of e.g. kJ). On the other hand, enthalpy of formation usually refers to the intensive quantity and should be called molar enthalpy of formation (with units of e.g. kJ/mol).
For the enthalpy of reaction, things are a bit fuzzy in introductory textbooks, either interpreting it as extensive or intensive. If it is used as intensive quantity, a change in the coefficients of the reaction (i.e. doubling or halving all coefficients) will change the value of the molar enthalpy. This is a bit surprising.

The only occasion at which q=∆H is when the moles of the limiting reagent in the thermodynamic equation is 1, such that q=∆H/n.

If ∆H is understood to be the molar enthalpy of reaction, you would have to multiply by the amount that reacts, not divide, i.e.
$$q = \Delta_r H \cdot n$$
where $n$ is the amount that reacts (i.e. the change in the amount of any species divided by its stoichiometric coefficient).

What does constant pressure has to do with this equality? Because if we assume such equality of Qp=∆U+P∆V=∆H at constant pressure, then it is also valid at constant volume when ∆V=0 such that Q=∆U=∆H.

That is a different question, and there are good pointers in the comments to learn more. The generally valid equality is:
$$ \Delta H = \Delta U + \Delta (P V)$$
so $P V$ changes when the pressure is constant and the volume changes, or when the volume is constant and the pressure changes (or when both change). This equation refers to enthalpy, not molar enthalpy, as you can figure out from dimensional analysis ($P V$ has dimensions of an energy, not a molar energy).

Dear Karsten Theis,
I would like to offer you my sincere thanks from the bottom of my heart, I think I understood it know and my confusion has been addressed.
Your way of clarifying things in a yet not so complicating manner is peculiar, and I ask god to bless you and your knowledge, because people like you is what we need for constructive knowledge gaining in science. Thank you for using your experience with higher level textbooks to clarify the vagueness of the beginner textbooks when indeed clarity is key to understanding.
Once again, thank you so much.


text:
Preparation of m-nitroaniline from m-dinitrobenzene
Consulting various Organic Chemistry resources, I have found two reactions for the purpose stated in the subject:
First reaction:

Second  reaction:

But, I cannot find the reaction mechanism by which they take place. Which would be necessary for an exercise, the conversion of m-dinitrobenzene to m-chloronitrobenzene, in which I want to introduce as an intermediate step one of these two reactions. I suppose that the mechanism will be very similar, but I cannot find what it is.

This is known as Zinin reduction, a review is here (if you have institutional access) :   https://onlinelibrary.wiley.com/doi/abs/10.1002/0471264180.or020.04  otherwise look here https://chemistry.stackexchange.com/questions/34066/selective-nitro-reduction-of-poly-nitro-compounds  AFAIK the mechanism for this reaction remains obscure.

It does work. Have successfully reduced 4-methyl-3,5-dinitrobenzoic acid by the Zinin method.

I have also used this to selectively reduce dinitrobenzenes


text:
How should you report measurements and uncertainty for a measuring cylinder with 0.5 mL graduations?
I understand that a common convention is to report the uncertainty of an analogue instrument as plus or minus half of the smallest scale division.
By this logic, a measuring cylinder with $\pu{0.5 mL}$ graduations has an uncertainty of $\pu{\pm 0.25 mL}$.
Because the measurement should also be reported to the same number of decimal places as the uncertainty, however, this suggests that any measurements with this measuring cylinder should be reported to two decimal places (e.g., $\pu{(17.85 \pm 0.25) mL}$).
Would this not be an example of false precision? Reporting a volume measurement with 2 decimal places seems excessive for a measuring cylinder. Would it not make more sense to report an uncertainty of $\pu{0.3 mL}$ (0.25 rounded to 1 d.p.) and a measurement with only one decimal place (e.g. $\pu{(17.9 \pm 0.3) mL}$)?
Before you answer, I understand that the rules for uncertainty are not set in stone and depend on the experimenter's judgement to some extent. However, I'd love to know which approach would be more conventional in this situation. Thanks so much!

Related: https://chemistry.stackexchange.com/a/37788/103369

It depends on the quality of the graduate cylinder. It should say something like 25:0.5mL ±0.4 mL on the top of the cylinder (and a temperature). If it doesn't, all bets are off.

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers. Here especially: the \pu{} syntax, which allows you to type recordings without unwanted line breaks between the floating number, and the unit of the property.

common convention is to report the uncertainty of an analogue
instrument as plus or minus half of the smallest scale division

Yes, this is merely a poor convention propagated by general chemistry texts.
Actually, this does not reflect analytical measurement standard deviation, and this is not the way volume uncertainty should be expressed. The correct method to determine uncertainty is called volumetric calibration. Take for example a $\pu{100 mL}$ cylinder. You would measure, for example, $\pu{60 mL}$ of water or any other solvent of choice. The liquid would be transferred to a weighing vessel and the exact mass would be measured. The process would be repeated at least 10 times. The volume would be calculated based on the density of the liquid. The volume delivered by the graduated cylinder would be calculated as the average and standard deviation.
If we look up industry standards, such as ASTM (American Society of Testing and Materials, e.g., ASTM E1272-02(2019), reference), it says that a $\pu{100 mL}$ class A is supposed to have a tolerance (std. dev.) of $\pu{\pm 0.5 mL}$ but a different $\pu{100 mL}$ class B cylinder will have a tolerance of $\pu{\pm 1 mL}$. Same graduations, different prices! Class A cylinder is way more expensive. It is all about how precisely, the cylinder was manufactured under controlled conditions.
One can immediately see fallacies associated with this general chem rule.
A reader brought another useful point about TC (to contain) or TD (to deliver) glassware. ASTM mentions both TC and TD type cylinders and both types exist. Still in reality one might use a TC graduated cylinder to deliver. In order to determine the actual volume delivered by a TC/TD graduated cylinder, calibration is the way to go to determine how much actual volume is being delivered. To be pedantic, if a higher capacity balance is available, the cylinder could be measured directly on the balance (true TC measurement).

Measuring cylinders are calibrated ‘to contain’ (In); i.e. the contained (not the delivered) quantity of liquid corresponds to the capacity printed on the measuring cylinders. Thus, the suggested method of transferring the liquid to a weighing vessel and measuring the exact mass would not work as intended.

I added a paragraph on TC/TD glassware. This was a useful point.

@M.Farooq Hint: for chemistry.se, mhchem has a nifty \pu{} syntax.  An input like $\pu{(13 \pm 0.5) mL}$ coalesces into $\pu{(13 \pm 0.5) mL}$ with much less danger that a line feed will separate a floating number from its following units.  Have a look at the examples in the lower section of the documenting page.  (No, it is not -- and indeed does not intend to be -- a replacement for siunitx for LaTeX2e.)


text:
HCl labeled 31.45% concentration (20° Baume)
Just bought a standard concentration of HCl solution. The 31.45% is weight-by-weight & not weight-by-volume, correct?

Yes, this is mass concentration, reporting concentration by weight.
E.g., Oxy/OxyChem's documentation Hydrochloric Acid Handbook (link to pdf file), edition 08/2018 explicitly states how the two relate to each other on page 6 with:

«OxyChem produces and markets Technical Grade hydrochloric acid in two concentrations: 22° Baumé (35.21% HCl by wt.) and 20° Baumé (31.45% HCl by
wt.). Other dilute concentrations of hydrochloric acid may be available upon request.»


text:
Creating HCl acid
I have the following reaction:
$$\ce{2 AlCl3 + 3H2O -> 2 Al(OH)3 + 6 HCl}$$
My goal is to create hydrochloric acid, but for the reaction shown, hydrochloride gas is produced instead. There are two scenarios that I am facing where:

The aluminum chloride is an aqueous solution instead of a solid
The aluminum chloride is a solid after heating the aqueous solution and the result is water vapor and solid $\ce{AlCl3}$.

My confusion and questions are that:

If I had an aqueous solution that is $\ce{AlCl3}$ in water, then why does it not react in the first place?
Does the water in the product side need to be a liquid, an aqueous solution, or a gas?
How do I make sure that the reaction works nearly every time?
I am trying to create hydrochloric acid meaning that $\ce{HCl}$ needs to react with water to provide a concentration of $\ce{HCl}$ acid. Would the water need to be a gas in order for both $\ce{HCl}$ and water to condense to hydrochloric acid or is there something else that I missed?
Is there a way that if I added more water to $\ce{AlCl3}$ that I would get $\ce{HCl}$ acid instead of gas?

I know that there are other methods of producing $\ce{HCl}$ acid, but I want to use this method.

There's no such thing as "hydrochloric gas"; it's either "hydrochloric acid" for $\ce{HCl (aq)}$, or "hydrogen chloride [gas]" for $\ce{HCl (g)}$. On a side note, you can also check out FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange? to see how to write chemical equations here.

I meant to say hydrochloride gas.

Not really, it's hydrogen chloride, not hydrochloride.

By solid AlCl3, do you mean hexahydrate or anhydrous form? Their behavior in contact with water differ significantly, especially at low water/compound ratios, when anhydrous chloride undergoes dramatic hydrolysis with HCl fumes. The anhydrous one cannot be formed by dehydration of hexahydrate, hydrolysing during the process, producing HCl fumes. // Produced HCl would be either too diluted either contaminated by aluminium chloride.

Why produce a cheap substance ($\ce{HCl}$) from a more expensive substance like $\ce{AlCl3}$ ?

@Maurice That I wondered about too, like transmutation of copper to iron.

Per this source, a good account of the actual product of dissolving $\ce{AlCl3}$ in a large volume of water:

If aluminium chloride is dissolved in a large amount of water the solution is acidic, but this has nothing to do with formation of hydrochloric acid. The solution contains hydrated aluminium ions and chloride ions:

$\ce{AlCl3(s) + aq -> [Al(H2O)6]^3+(aq) + 3Cl^-(aq)}$

The hexaqua complex ion behaves exactly like ions of similar type formed from transition metals; the small, highly charged metal ion polarises (withdraws electron density from) the water molecules that are attached to the aluminium ion through dative covalent bonds. This makes the hydrogen atoms d+ and susceptible to attack from solvent water, which is acting as a base. The complex ion is deprotonated, causing the solution to be acidic from the formation of hydroxonium ions $\ce{H3O+}$:

$\ce{[Al(H2O)6]^3+ (aq) + H2O(l) ->  [Al(H2O)5OH]^2+(aq) + H3O+(aq)}$

So, no $\ce{HCl}$ per se, but adding a small amount of water to the dry salt will liberate fumes of hydrogen chloride that readily dissolve in water forming aqueous $\ce{HCl}$, but not likely a practical path either.
In my opinion, however, try adding $\ce{AlCl3}$ to a large volume of carbonated water as a possible path to very dilute $\ce{HCl}$. I expect the intermediate formation of an unstable aluminum carbonate followed by the deposition of $\ce{Al(OH)3}$ (interestingly, this reaction mirrors the wrong commonly cited hydrolysis reaction forming $\ce{Al(OH)3}$ and $\ce{HCl}$). Note, quickly separate the solid to limit the reverse reaction recreating the $\ce{AlCl3}$. This reaction, if successful, may also be viewed as paralleling the action of oxalic acid on various salts of mineral acids, successfully forming the mineral acid, along with a corresponding insoluble oxalate (there exists a thread on this).

Hint regarding mhchem, for future reference: On occasion, copy-paste from other sources into mhchem syntax drops «some of the (implicit) accidentals» to indicate subscript (mhchem describes with leading underscore) or superscript (by leading caret).  For a single plus sign (like in $\ce{Na+}$) this isn't problematic.  However, e.g., a dash with leading caret becomes an superscript minus, and without leading caret extends like a single bond to the next atom.  And arrows are more portable if composed with <, > and either - or =, too.

For those curious on my reference to a work on aluminum complex formation, it is because there is amazing effort to avoid any CO2/carbonate contamination. If this implies an issue with aluminum complex formation, and we are interested in other reactions, then it is possibly supportive.  Note: the alluded to CO2 > H2O > HCO3- > Aluminum carbonate > Al(OH)3 and CO2 release has a cyclic aspect so a small amount of CO2 impurity in time can apparently be problematic with aqueous AlCl3, for example.


text:
Can a vacuum on a closed system really “pull” things out of it?
So I’m learning gas discharge tube making (neon) and there’s a step where you cork one end of the tube and pull a vacuum on the other end to suck out all the impurities. The same thing was done when I was a mechanic working on an AC system that was opened up. Before you recharge it, you pull a vacuum for x minutes to remove the atmosphere in the system.
I’m having trouble comprehending that this is really possible. When you apply vacuum to a closed system, nothing seems to actually be moving. For instance, if you fill a drinking straw with coke, put one finger at the other end and suck, the coke goes nowhere. Remove your finger from the end and suck, and now you have movement; the coke goes right into your mouth. The negative pressure at the vacuum source (my mouth) works with the positive pressure at the other end of the straw, the air movement created pulls the air and the coke towards me. If vacuum on a closed system can’t pull liquid, how can it pull gas?

If you could suck hard enough with your mouth, then it would pull that liquid out too.

Only if the straw were to collapse…?

Well, I guess that's another instance where your examples aren't really comparable. The first point was that the pressure difference is far greater when using a vacuum pump or similar. The second point is that a glass vessel (hopefully) won't collapse under the pressure difference.

But my earlier comment stands, really. If you used a very robust glass straw, and you sucked hard enough, you would get the drink out. Fortunately (or unfortunately?), our mouths aren't designed to create strong vacuums.

Okay, I hear you. It’s just hard to comprehend or picture that because I’ve never been able to see that principle anywhere in life. Any time I see something sucking something to move it someplace else, it’s utilizing an open system. I can’t think of one example of being able to see the type of action you’re (we’re) talking about. Thanks. If you can think of something practical that does this that I can see a video of, that would probably help me wrap my mind around it.

It is a question from physics lessons like from the 8th grade ( when I was in the 8th grade long time ago ).

Note that water-based liquids are far less compressible (or expansible) than gasses so this isn't a good comparison. Consider a pump wine-saver - do you think that does not pull gas out of the gap between the wine and the top of the bottle? (You can actually smell it working)

That’s a great example, I never thought of that. Thanks abligh.

Wind is also a result of a pressure difference (among other causes). The underlying reason that is exists is the same like for what you describe, namely undirected movement of atoms/molecules that constitute the gas that leads to diffusion: https://en.m.wikipedia.org/wiki/Diffusion. You basically create a very strong wind by generating a large pressure difference.

Gasses don't get "pulled", they get "pushed".
When you empty a neon tube, you attach a device with almost no air in it to mouth of the tube.
The air that is in the tube then pushes itself into the device until the pressure is the same in both.
Meanwhile, the device expels its air to produce more low pressure, and more of the relatively higher pressure in the tube pushes more of its air into the device.
This continues, with the contained air pressure becoming less and less until you decide "good enough" or the limit of the device is reached.
It won't be a perfect vacuum, but it might have say a 99% vacuum, containing 1% of the original air content in the same volume.
For the drinking straw, again it isn't your mouth (the vacuum) that pulls, the liquid actually gets pushed up the straw.
What happens is that the weight of the atmosphere pushes down on the surface of the drink in the glass, and since that pressure is greater than the pressure of the partial vacuum you have created at the other end of the straw, the liquid is pushed up through the straw.
This is much like the way toothpaste is pushed out of its tube.
Similarly, a hand pump on a well works by producing a partial vacuum at the top of the pipe and allowing atmospheric pressure to push the water up the pipe.
There is a limit though.
The weight of the atmosphere won't push the water any higher than about 10 metres (11 yards).
In the case of a straw sealed at one end, suppose that it is made of steel or very strong glass so that it won't collapse, and that it is long and thin enough that bubbles of air won't flow from one end to the other.
The molecules in the liquid are attracted to each other, so the liquid will tend to act like a solid plug.
If a vacuum is applied to the other end, because there isn't any air pressure at the sealed end, no matter how good the vacuum you create at the mouth end, there won't be any pressure to push the liquid plug out.
Nothing happens.
However, if there is a small pocket of air at the sealed end, then the partial vacuum at the mouth end will cause some of the liquid to be pushed out.
The small pocket will become a larger pocket, but with lower pressure.
This will continue until the pocket's air pressure is equal to the pressure of the partial vacuum you are applying to the mouth, which could happen either before or after all the liquid has been removed.
Remember, gasses don't get "pulled", they get "pushed".
Even when you use a vacuum cleaner to clean a floor, the vacuum doesn't actually suck up the dust.
Instead, the machine creates a volume of low pressure in its canister, and the air in the room gets pushed into the pipe by atmospheric pressure in order to fill that void, dragging the dust along with it.
The best way to visualize the behaviour of gasses is to imagine a room containing thousands of ping pong balls that continue bouncing forever.

Temperature: the hotter the gas is, the faster the balls will move around.
Pressure: each ball bounces off a wall by pushing against it, faster moving balls pushing harder than slower moving balls.
Volume: the larger the room, the longer it will take for a ball to make the next bounce off a wall, and so fewer bounces will happen in a given time interval.

If a wall is moved outward to make a larger room, there will be a brief period when there are no balls near it (a vacuum), but the balls will quickly fill that space and the larger room will soon reach equilibrium again.
The wall didn't "pull" on the ping pong balls, it simply created a void where the balls were "pushed" into.
Meanwhile the larger room means less frequent bounces, which means less pressure.

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

"If vacuum on a closed system can’t pull liquid, how can it pull gas?"
It's back to that old thing: Gasses expand to fill the container, liquids don't. So the gasses will fill the entire container - reaching the vacuum, which will expel them, and then the remaining gas will do the same.
But the liquid (the coke) will stay put, unless the pressure is so low it boils.


text:
Simultaneous electrolysis on applying a suitable amount of potential difference
This question was asked in the JEE in 2020:

$\pu{250 mL}$ of a waste solution obtained from the workshop of a goldsmith contains $\pu{0.1 M}$ $\ce{AgNO3}$ and $\pu{0.1 M}$ $\ce{AuCl}.$ The solution was electrolyzed at $\pu{2 V}$ by passing a current of $\pu{1 A}$ for 15 minutes. Given $E^\circ(\ce{Ag+}/\ce{Ag}) = \pu{0.80 V}$ and $E^\circ(\ce{Au+}/\ce{Au}) = \pu{1.69 V},$ what metal/metals will be electrodeposited?

Why can't two ions be simultaneously deposited on the electrode provided that the applied potential difference is greater than both of their potentials?
I know that otherwise, electrolytes behave as if they are in parallel if a voltage is applied across them but why can't that happen here?

Unfortunately, it is impossible to make a solution containing $0.1$ M $\ce{AuCl}$. The Merck Index, under Gold monochloride, states first that this compound is insoluble in water. Second, a suspension of AuCl in water Is slowly decomposes according to : $\ce{3 AuCl -> AuCl3 + 2 Au}$.

If you replace $0.1$ M $\ce{AuCl}$ by $0.1$M $\ce{AuCl3}$, the problem makes sense, because the corresponding redox potential $\ce{E°(\mathrm{Au^{3+}/Au} )= 1.42 }$ V. So using  $\ce{Au^{3+}}$should lead to the same reasoning as using $\ce{Au^+}$. Unfortunately, $\ce{Au^{3+}}$ is usually obtained by dissolving $\ce{AuCl3}$ in water. And this solution will react with $\ce{Ag^+}$ ions, producing a precipitate of $\ce{AgCl}$. So, in order to avoid this reaction, $\ce{AuCl3}$ should be replaced by another compound àf gold(III). And unfortunately the gold(III) nitrate is not soluble in water.

Study materials in some regions frequently use made up tasks, ignoring well known experimental facts. It reminds me often the popular book "You are surely joking, Mr. Feynman!", specifically the chapters Richard Feynman described his physics education activities in Brasil in 1960s.

@Maurice, unfortunately, being a high school student, I am unaware of these indexes you mentioned :( , could you answer this whilst ignoring the solubility part, if you could explain it using two soluble salts and how they compete for electrolysis when a voltage is applied?

@Poutnik In India they conduct 32 shifts for the same exam and syllabus, they're ultimately compelled to do so you can't blame them too.

@Priyanshu Choubey. The only way of studying this problem of gold and silver ions is to dissolve gold(III) nitrate and silver nitrate into nitric acid, and perform the electrolysis in this solution. In this case you will deposit only gold, because if by a strange effect some silver would be deposited, it will be quickly redissolved by the reaction : $\ce{3 Ag + Au^{3+} -> Au + 3 Ag^+}$. The only trouble is that gold(III) nitrate is not commercially available...

@Maurice at high enough current density to saturate diffusion current , the cathode potential could be low enough to deposite silver too.

@PriyanshuChoubey You may find useful these links for text formatting (not to be applied to titles):
Notation basics ,  Formatting of math/chem expressions  and
upright vs italic

@Maurice India has ignored all of chemistry and physics in their NCERT or JEE stuff since the early 80s. And that's a realisation I unfortunately had to make here.


text:
Calculating internal energy of fragment from $D_0$, photon energy and interval energy + velocity of other fragment
I am trying to solve the following:
We consider a photodissociation of a molecule into two fragments. The energy available to the products can be calculated from the bond dissociation energy and the photon energy. If the internal energy and the center-of-mass recoil velocity of one fragment is known, how can you find the internal energy of the second fragment? (use conservation of momentum and energy)
So I know that [energy available to products]=[photon energy]-[bond dissociation energy]. But my problem is that I am not sure what [energy available to products] consists of. Is it the sum internal energies of the two fragments plus the sum of the kinetic energies of the two fragments?
Can someone shed some light one this problem?
Thanks!

I haven't thought about this closely enough to answer definitively, but I know it is possible to measure $D_0$ using velocity-map imaging (which manages to send fragments of the same velocity to the same point on a detector, hence determining product distribution of COM kinetic energy). One then combines this with pre-dissociation techniques where one excites a particular vibrational mode before dissociation so that fragment vibrational energy is known. The energy available to products is translational, rotational, and vibrational energy (ignoring electronic excitation).

See this paper about measuring $D_0$ of $\ce{(H2O)2}$ for more info


text:
What is the difference between milk and cream that causes disparate result when freezing and thawing?
If you freeze milk and then let it thaw/melt again you just need to shake and it is more or less indistinguishable from unfrozen milk. That is not true for cream - after letting frozen cream (40 % fat) melt you have something quite similar to whipped cream in consistence. Why is cream different from milk in this situation?

Milk and whipping cream are different in most situations, similarly as beer and whiskey.

So this question is not about ice cream?

I've observed that too, cream that accidentially froze turns into a funny solid form after thawing. Not exactly like whipped cream, more rigid.  Is it crystallisation of fat, or something with the proteins? Good question.

@Poutnik What are you talking about? They are very similar except for the fat content.

@Karl, perhaps it's like what happens when one freezes tofu. It forms a spongy material, I think from the water being squeezed out during the freezing, which is tough and chewy enough to use directly as a meat substitute.

So maybe beer and whiskey doesn't work as analogy but instead Virgin Bloody Mary and Bloody Mary.

@d-b Bingo! It is quite a difference.

@Poutnik No, you can make milk from 40 % cream and cream from 2-3 % milk, just let it rest and separate. They are interchangeable (although in practise you would of course not do that).

@d-b You can make 1% NaCl solution from the saturated NaCl solution and vice versa either, what does not change the fact their properties are very different. // I am sure the milk prepared from 40% whipping cream would be different. // Freezing 40% cream helps ice/fat separation ( for milk as well, in lesse extent - thawn milk tastes differently), as even without freezing it the cream is at the edge of spontaneous separation. Similarly salt solution freezing separates water and salt as well and they like each other much more than fat and water.


text:
Which theory could describe rate constant for photodissociation?
I am considering photodissociation reactions and want to find an appropriate theory to describe how one could determine the rate constant. I know Rice-Ramsperger-Kassel-Marcus (RRKM) theory is for unimolecular reactions but I have only seen the theory applied to isomerization reactions. Could one use RRKM theory to describe the kinetics of photodissociation? What other theories would make sense to apply?


text:
Burning wood with ammonium chloride
There is a method where you use a water solution of ammonium chloride to paint some pattern or write some text on a piece of wood and then blow on it with a heat gun. The result is that the pattern you painted becomes burnt into the wood.
I wonder what could be the mechanism behind how this works?
Here is a video where a guy does that: https://www.youtube.com/watch?v=zV0zC2qRFxE

Here is my guess as to what is going on: a solution of ammonium chloride is painted onto the surface of the wood leading to ammonium chloride being absorbed into the surface wood fibres. Ammonium chloride decomposes to ammonia + hydrogen chloride above 340C. Heating with the heat gun achieves this temperature (which is why it takes a while in the video). Either gas (or both) causes darkening of the wood revealing the design.
This seems to be backed up by the explanation here: https://hackaday.com/2011/05/16/chemical-wood-burning/

Heating surface with an application of $\ce{NH4Cl}$ results in the release of gaseous $\ce{HCl}$ and especially $\ce{NH3}$, which per Wikipedia on ammonia fuming, to quote:

Ammonia fuming is a wood finishing process that darkens wood and brings out the grain pattern. It consists of exposing the wood to fumes from a strong aqueous solution of ammonium hydroxide which reacts with the tannins in the wood. The process works best on white oak because of the high tannin content of this wood. Fumed oak is also called smoked oak.

I trust this helps.

Thanks. If that is the explanation then ammonium carbonate could work too. Here in Sweden the chloride is a bit more difficult to buy, whereas the carbonate you can buy in any food shop (ammonium carbonate is used as a raising agent when making cookies). I do have some doubts though regarding the ammonia fuming being the explanation, since it seems to take a long time to affect the wood so that it actually looks burnt. But, of course, in that method heat is not applied, whereas in the method we discuss the temperature is a few hundred degrees Celcius, so perhaps the process speeds up.


text:
Complete mechanism of benzotriazole synthesis from o-phenylenediamine
Does my suggested mechanism for benzotriazole synthesis look correct? Are the arrows for the cyclization in the right place?
Chemdoodle (which I drew this with) makes the curly arrows look a bit weird.

Too many steps, amino could substitute H2O, then deprotonation.

You mean the intermediate with the pendant hydronium would cyclize with expulsion of water?

Your solution is plausible but I am bothered by the use of structure 7 the way you have. Its too far  to reach! If you do an electrocyclization of 7 to form the benzene ring and a 4-membered ring bearing vicinal nitrogens, then you can have the remaining nitrogen as a neutral "nitrene" ready for 1,2-shift of nitrogen and ring expansion. OTOH, acid-catalyzed cyclization of structure 2 followed by dehydration gets you there.

When I was drawing it I figured the electrocyclization through the electronegative N wouldn't work but ended up drawing it anyway. So I was right to be concerned.

The amino-to-diazo electrocyclization and 1,2 shift is interesting, but I think I like Mithoron's answer where structure 5's nitrogen displaces water and then deprotonates better.

https://ibb.co/Lzthh3G

Jacob, you are free to like what you like. I'm not convinced. I'll post an answer.

Species 1 is your seventh structure. The C=N=N atoms are linear and a long reach for the remaining nitrogen. Electrocyclization of 1 to 2 restores aromaticity. Nitrene species 3 allows for a 1,2-shift to an electron-deficient center with concommitant loss of strain in the 4-membered ring to produce 1H-benzo[d][1,2,3]triazole 4.
Cyclization of 5 $\rightarrow$ 6 and dehydration via 7 gets the job done. Perhaps @Mithoron knows something that I am not aware of.


text:
Does a metal being "soft" equal being able to scrape onto paper?
When you scrape aluminum onto paper, it makes a faint mark. The question is, does softness of a metal correlate to how well a mark shows up when you scrape it onto a surface such as paper?

Are you aware of outliers? (soft not scraping or vice versa)

No, what are those?

Neither I am.. Only mercury, but that is a faux hit below a waist.

«How well a mark shows up» is an ambiguous description. Yet,

Does softness of a metal correlate to how easy material abrades? If you refer to Mohs' scale, yes.  Modern test kits contain pens with synthetic alloys to offer reproducible access to Mohs' levels of hardness

(credit, commercial source)
like it is mostly known in art for silver, the silverpoint.  Though similar to silver in Mohs' scale, I didn't hear (so far) about an aluminium point.  It equally is one of the techniques in geology, to compare the colour of the mineral as such with the colour of the streak of this mineral e.g., on a white tile you carry with you in the field; the two needn't be equal.

Is the trace of the material visually discernible from the paper?  This depends on the paper.  Paper often is sold bleached and containing fillers (e.g., $\ce{CaCO3}$, $\ce{TiO2}$) for a white appearance.  Again returning to art, your paper may be darker from the get-go, then allowing to use the technique known as trois crayons to introduce heights and lows with pencils used:

(example of a trois crayons, credit Wikipedia)

Does the trace adhere well to the paper? Like e.g., for charcoal, pen, chalk, in part this depends on the roughness of the paper if small particles enter grooves of the substrate.  Often, the adhesion to the substrate is enhanced after drawing by applying a coating called fixative.  (Hair-fixing spray sometimes is a cheaper alternative.)

In case of aluminum, you may bet that its metallic reflectance stays; on surface, it will oxidize to transparent, colourless $\ce{Al2O3}$.  In case of e.g., silver, this is going to get dark (think $\ce{Ag2S}$).


text:
Does red phosphorus with hydriodic acid reduce all alkyl halides?
$\ce{HI/P}$ is used for reduction of alcohols, aldehydes, ketones and acids to alkanes. As for alkyl halides, I have read that only alkyl iodides are reduced, whereas other alkyl halides do not:
$$\ce{CH3CH2I ->[Red P][\pu{1 eq.}~HI] C2H6 + I2}\tag{R1}$$
But I have also been taught that all alkyl halides are reduced to alkanes:
$$\ce{CH3CH2X ->[Red P][\pu{2 eq.}~HI] C2H6 + HX  + I2}\tag{R2}$$
Does red phosphorus with hydriodic acid reduces all alkyl halides or just alkyl iodides?

I don't think that in general, other alkyl halides react in this way.
When I searched Google Scholar for hydrodechlorination "hydroiodic acid", not much came up.  There was this article about the selective dechlorination of 2-chloro,5-bromo substituted pyridines.
I also tried searching in some proprietary databases for reactions where ethyl chloride is reduced to ethane.  There were plenty such reactions, but the for all of them, the reported reductants were zinc or hydrogen gas.


text:
Solid materials with dipolar molecules
I am trying to heat solids using controlled microwave radiation and I am therefore looking for materials that absorb microwaves well. I have understood that for radiation in the radio frequency and microwave frequency regimes the polarity of molecules and the conductivity are most important.
I am also reading this chapter about dry solids (A. C. Metaxas & R. J. Meredith. Industrial Microwave Heating. 1983.):

2.8 Dry, non-conducting dielectrics
Emphasis has been given so far to the conducting phase in dielectrics such as, for
example, water, because high frequency heating has been applied very successfully
to many drying applications. However, many applications require heating of dry,
non-conducting insulators such as natural rubber or other chemical substances,
in order to achieve specific reactions within the material. In this case, Maxwell—
Wagner type effects, in the absence of any trace impurities, play no part at all in
the heating process which is now governed solely by dipolar effects. In fact, if the
insulating material is non-polar, one may deliberately introduce a controlled proportion
of conducting particles, such as carbon black, to enhance the absorption
of the applied high frequency energy by the material to be heated. This is discussed
further in Section 3.9.

From this section I understand that it's possible to mix carbon into materials that are non-polar and non-conductive to make them absorb microwaves and I feel like this would be the case if I want to heat plastics for example. But my question is: are there (common) solids that are non-conductive but have polar molecules, so that they would also be suitable for microwave heating? And could they potentially have a higher absorption than conductive materials?
And does anyone know of any other absorbent solids?
I am thankful for any help!

Ice contains polar molecules, and is solid. One might quibble, not completely solid at home freezer temperatures, so you might test ice at liquid N2 temp, too.

Though not a material consisting of individual molecules like e.g., glucose, try silicon carbide ($\ce{SiC}$).
This material not only became popular because it is a good absorber of microwave reaction, but equally because of its high mechanical strength, high melting point, and low chemical reactivity.  These render $\ce{SiC}$ you might heard about earlier (abrasives, pads for breaks, heat filaments, etc.) suitable for (small) chemical reactors, too.

(taken from an overview by Anton Paar)

(Kappe, Acc. Chem. Res. 2013)
Since it is equally commercially available as shot and powder, you could blend it as a dispersion into a solid matrix, too.  Then, $\ce{SiC}$ were the antenna heating from inside of your sample.
Kappe, C. O. Unraveling the Mysteries of Microwave Chemistry Using Silicon Carbide Reactor Technology. Acc. Chem. Res. 2013, 46, 1579–1587; doi 10.1021/ar300318c.


text:
For Calcium and Potassium, as 4s is then lower energy than 3d,(and lower means or normally means nearer), would 4s thus be nearer the nucleus than 3d?
For Calcium and Potassium, as 4s is then lower energy than 3d, (and lower normally means nearer), would  4s thus be nearer the nucleus than 3d?
For example.
I read in the accepted answer to Why does the 3rd electron shell start filling up with scandium?
That  for Calcium and Potassium, the subshells in order of energy goes  4s then 3d i.e.
1s, 2s, 2p, 3s, 3p, 4s, 3d,..
4s is lower energy than 3d there.
I'm not talking about elements between Scandium and Zinc, where that link explains that for those 10 elements from Scandium to Zinc,  4s is higher energy than 3d but one or two electrons get repelled from 3d to 4s before 3d gets fully filled. Let's put aside the elements from Scandium to Zinc.
No question that for Calcium and Potassium, 4s gets filled before 3d, and 3d doesn't get any electrons there at all.
So for those two elements, Calcium and Potassium, is 4s nearer to the nucleus than 3d?

The question would have made sense in the case of a classical circular orbit.. An orbital being closer to or farther from a nucleus makes sense rather for the same orbital type, like for ns or np orbitals. And even then there is huge overlapping, so only a mean distance comparison makes sense.

Read the section on orbital energies for some context: https://pubs.acs.org/doi/pdf/10.1021/ed200673w


text:
Why are 2D block matrices considered irreducible?
I am currently reading Applications of Group Theory by M.S. Dresselhaus
On page 18 irreducible representations are defined and an example is given as follows:

The author claims that these representations are irreducible. However for E and A the $\Gamma_2$ 2D representation is in block form and it can be seen that it could be expressed in terms of the $\Gamma_1$ and $\Gamma_1'$ representations as follows
$\begin{pmatrix}
\Gamma_1 & 0\\
0 & \Gamma_1'
\end{pmatrix}$
, so why is $\Gamma_2$ considered irreducible? I am using the last line of the definition here.

See also this question on Math.SE about the same example in the same book (which I found while trying to search for the original text of the book for context).

The representation of the group (in this case) is given by a set of six matrices. Reducibility is a property of the representation as a whole, not the individual matrices that form it.
So, you can't say that the $E$ matrix on its own is (ir)reducible, or that $A$ on its own is (ir)reducible. You have to look at the set of six matrices as a whole, and whether they can all be simultaneously block-diagonalised. If they can, then the entire representation is reducible; if they can't, then the entire representation is irreducible.
In this case, $E$ and $A$ are block-diagonal, but the other matrices $B$, $C$, $D$, and $F$ are not. You could find some change of basis such that one or more of the latter become diagonal. However, it's not possible to find a change of basis which simultaneously diagonalises all six matrices. So, the representation $\Gamma_2$ is irreducible.
Another way of looking at it is that you are trying to decompose $\Gamma_2 = \Gamma_1 \oplus \Gamma_1'$. This is indeed satisfied for the $E$ and $A$ matrices, but not for the others. So, it's not possible to say that the representations add up, because the representation consists of all six matrices, and all six matrices must add up (technically, form the correct direct sum) for this to be the case.
The cited definition is actually very rigorous and does mention this (emphasis mine):

If by one and the same equivalence transformation, all the matrices in the representation of a group can be made to acquire the same block form, then the representation is said to be reducible [...]

Yet a bit more rigor might have been in order, as taking the quoted definition literally would seem to imply that $\Gamma_1$ would be reducible, all of its matrices being already in "the same block form" (and indeed identical).

In fact, compared to, say, Wikipedia's definition, the one given in the book seems pretty muddled. I can't even tell whether or not it's mixing up reducibility and decomposability, since the book never seems to define what it means by "same block form".


text:
How to make NaxPO4 buffer?
I am currently trying to make a buffer for protein expression. According to the article, I am to use a buffer with $\pu{50 mM}$ $\ce{Na_{x}PO4}$, $\pu{pH 8}$ (and $\ce{0.5 M NaCl}$). How do I make the $\ce{Na_{x}PO4}$ buffer?
I have previously made phosphate buffers from cold spring harbor protocols
(Mixing aliquots of $\ce{1 M NaH2PO4}$ (monobasic) and $\ce{1 M Na2HPO4}$ (dibasic) aliquots to get the desired pH and concentration. Is this the same buffer, just with another name? Or what to use to make $\ce{Na_{x}PO4}$?

The Cold Spring Harbor protocol is probably fine. There is another recipe that uses a mixture of potassium and sodium salts (for better solubility).

50 mM NaxPO4, pH 8 (and 0.5 M NaCl)

This is a weird notation, and the x should be a subscript:

Source: https://www.nature.com/articles/s41467-020-18907-6.pdf?proof=t
If you are concerned about correct stoichiometry, you have to include some hydrogen as well
$$\ce{Na_xH_{3-x}PO4}$$
At pH 8, you would expect some hydrogenphosphate ($\ce{HPO4^2-}$) and some dihydrogenphospate ($\ce{H2PO4-}$) ions. There is no simple way to write this combination as a formula, though.

Source: https://commons.wikimedia.org/wiki/File:PiSpeciation.svg

$\ce{Na_xPO4}$ is a non-official way of defining any mixture of $\ce{Na2HPO4}$ and $\ce{NaH2PO4}$ solutions. It can be used for describing all mixtures contained in the reference of Cold Spring Harbor.


text:
Relative Humidity inside a thermally insulated chamber for a NACL solution
I ran an experiment, where I had to measure the relative humidity of a thermally insulated chamber containing NaCl solution. I have read from a few papers, that the relative humidity of a NaCl solution remains at 75% over a wide range of temperatures (20C to 30C).
But when I can change the temperature inside the chamber from 20C to 30C the relative humidity changes from 80% to 60%.
I am having trouble understanding what exactly is happening here. I thought it might be the pressure change caused by temperature change, so I created a small hole (needle width) to compensate for it but the result remains the same.

You have ignored the kinetics. Water vapor needs some time to reach higher level of absolute humidity at equilibrium. // The initial high relative humidity was not well balanced either, unless the sensor was improperly calibrated.

Could you cite your sources? E.g. https://nvlpubs.nist.gov/nistpubs/jres/53/jresv53n1p19_A1b.pdf and https://www.teaforum.org/viewtopic.php?t=177. Also, how are you measuring relative humidity, and is your instrument calibrated just for room temperature or for the entire range?

@KarstenTheis I am using a calibrated SHT31 from Dracal which has both calibrated humidity and temperature. And here are my sources 1.) https://nvlpubs.nist.gov/nistpubs/jres/53/jresv53n1p19_A1b.pdf 2.) https://www.engineeringtoolbox.com/salt-humidity-d_1887.html 3.) https://nvlpubs.nist.gov/nistpubs/jres/81A/jresv81An1p89_A1b.pdf

Nicely done experiment... now just run it 10 or 100 times slower, to allow it to reach equilibrium.

@DrMoishePippik I try slowing down the process to see if it works. Thanks for the reply.

@Poutnik I try slowing the experiment. Thanks for the reply.

Note that NaCl solution is supposed to be saturated with excess of the solid salt. Geometry is adviced to have maximized liquid surface and minimal depth. Be aware that it is slow process, especially at cooling down, as condensed water has to dissolve the solid salt to keep solution saturated.


text:
Can chloralkali process be used with ocean water to produce sodium hydroxide?
Are there any modern-day Chloralkali plants that create $\ce{NaOH}$ using water from the ocean?
I'm trying to understand how modern-day plants that use the Chloralkali process to create $\ce{NaOH}$ (also chlorine, hydrogen, and $\ce{NaCl}$) obtained from brine solution.
Specifically, I was thinking of the increasing creation of desalinization plants. Are there any plants that intake the waste water from desalinization plants to produce Sodium Hydroxide ($\ce{NaOH}$)?

Intentional usage of dirty input ( not only NaCl ) would lead to dirty product. Seawater#Chemical_composition


text:
How to extract pure carbon dioxide from sodium bicarbonate?
What are the most commonly-used methods for extracting carbon dioxide gas from an aqueous solution of sodium bicarbonate?
Several industrial "Carbon Capture" plants today are able to extract carbon dioxide from unadulterated air by passing it through an aqueous solution of sodium hydroxide ($\ce{NaOH}$) to produce an aqueous solution also containing sodium bicarbonate. There's plenty of information on the internet about this step, and it's straight forward.
What's less straight forward is how the aqueous solution with $\ce{NaHCO3}$ is later reacted to produce a pure carbon dioxide.
How can carbon dioxide gas be purified from this Direct Carbon Capture (DCC) process using an aqueous solution of sodium hydroxide ($\ce{NaOH}$)?

Heating will do. You won't get your NaOH back, though.

Regrettably, though it is possible to capture CO2, energetically, it is impractical because the heat to regenerate the NaOH would expend huge amounts of fuel, generating far more CO2.

Sadly, I think I'm being downvoted because I used the word "Carbon Capture." To be clear, I'm not talking about DCC as a means to avert the climate crisis. I'm purely interested in the chemical process to collect pure Carbon Dioxide as environmentally friendly as possible for later use in a Sabatier Reactor. It's hard to imagine a more sustainable way to create CO2 than pulling it out of the air with a bit of NaCl & water.

@MichaelAltfield The question is being voted for closure due lack of details and clarity. That directly applies to the fact you have not clearly stated the background or context of your question in the question body. I suggest you move the esential content of your comment to the question, eventually more elaborated.

I could not find any verification of an industrial plant using this technology. The closest seems to be the plants in development by Carbon Engineering, which are planned to use a potassium hydroxide solution for the carbon dioxide capture step.
Their proposed process then precipitates the carbon from solution by addition of calcium hydroxide, yielding solid calcium carbonate. The $\ce{CO2}$ is liberated from the calcium carbonate by heating in a traditional calcination process such as is used in cement manufacture. This step is fueled by natural gas combustion.
In summary:

$\ce{CO2(g) + HO- (aq) + K+ (aq) -> HCO3- (aq) + K+ (aq)}$
$\ce{CaO (s) + HCO3- (aq) + K+ (aq) -> CaCO3(s) + K+ (aq) + HO- (aq)}$
$\ce{CaCO3 (s) + heat -> CaO(s) + CO2 (g)}$

This is described on the Carbon Engineering website here.

Yes, a technical answer to the question but with respect to the provided clarification:  "I'm purely interested in the chemical process to collect pure Carbon Dioxide as environmentally friendly as possible for later use in a Sabatier Reactor", the process  outlined in the provided answer involving "a traditional calcination process" is hardly "environmentally friendly", albeit, interesting info.

Further, since some are seeking a technically strict answer here, this provided answer does not address the question relating to either NaHCO3 or NaOH, or even a mention thereof. MODERATOR, where are you?

@AJKOER -  the difference between an aqueous solution of sodium bicarb and a solution of potassium bicarb does not qualitatively affect the described process, so this does provide an answer to a process that could be implemented. As for "environmentally friendly", without a definition of that term, I can only say that the main complaint against calcination as I understand it is the CO2 produced, which in this case is captured along with that from air. Some may consider that to be environmentally friendly, though some may not.

Hi all, I'm here - @AJKOER, please use a flag to get a mod.

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


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Degrees of freedom of a molecule of N atoms
Apparently a molecule of N atoms has 3N degrees of freedom. How?
Shouldn't it be dependent on the structure of the molecule? How exactly do we derive this?

Bonds in a molecule are flexible.  Thus, each atom can move in any possible direction in 3D space.
Note that they can't necessarily move any possible distance; indeed, if the molecule is stationary, the atoms may not be able to move much distance at all since, after all, they are bonded to other other atoms. In summmary, each atom can move a limited, non-zero distance in any possible direction.
Thus, to describe the motion of each atom, exactly 3 coordinates needed.  Further, since each atom can move in any possible direction regardless of what the other atoms are doing (again, possible distances are constrained, not possible directions), we need exactly 3N coordinates to describe the motions of all atoms in a molecule with N atoms.
Since describing the motions of all atoms in a molecule also completely describes the motion of the molecule as a whole, no additional coordinates are needed. Hence the number of degrees of freedom (DOF's) in a molecule is equal to 3N, independent of the molecule's structure.
Structure does matter when it comes to vibrational and rotational DOF's, however.   A linear molecule needs two angles to describe its orientation in space, while a non-linear molecule needs three.  Hence these have 2 and 3 rotational DOF's, respectivey.
From this, and from the fact that a molecule as a whole has 3N DOF's, we can calculate the number of vibrational DOF's by simple subtraction:
Linear molecule:
Total DOF's = 3 N
Translational DOF's = 3
Rotational DOF's = 2
Vibrational DOF's = 3N – 5
Non-linear molecule:
Total DOF's = 3 N
Translational DOF's = 3
Rotational DOF's = 3
Vibrational DOF's = 3N – 6

DOFs are defined as the number of ways a dynamic system can move.A linear diatomic molecule could translate along three axes, rotated about two, and vibrate about 3. So the number of DOFs I get is 8. Where am I going wrong?

Diatomic molecules have only two atoms.  Thus they have only one vibrational mode.  Hence DOF = 3 trans + 2 rot + 1 vib = 6 = 3*2 = 3N.  Note also that you wrote "linear diatomic".  Since diatomics have only two atoms, there was no need to add the "linear" qualifier, since they are all linear.

So can't the molecule vibrate along all the three axes?

Its vibrational motion consists only of the atoms moving apart and coming together.  So that's just one mode.  You're arguing that the molecule can do this in any orientation in space, which is true.  But that's already accounted for by its two rotational DOF's. I.e., those are not different vibrational modes, those are the same vibration taking place while the molecule is oriented in different directions in space. So when you get 8 DOF's you are double-counting the 2 rotational DOF's.

....Or you might be thinking of a movement in which the atoms move apart, but not along a straight line.  Within the reference frame of the diatomic molecule, that can't happen—they can only move back and forth on a straight line, corresponding to the molecule's one vibrational mode.  However, from the perspective of an external observer, the atoms can move apart along a curve.  But that's not a 2nd vibrational mode.  That's simply what happens, from the point of view of an external observer, when vibration and rotation occur at the same time.


text:
Why is this graph of vibration spectroscopy like this?

This graph shows the absorbance of different wavenumbers of infrared radiation causing different forms of vibrations of the water molecule. I would like to know why the graph turns out the way it is.
How do we decide which kind of vibration would take in the most energy? How do we know their respective absorbances?

Molecules' bonds vibrate at all temperatures, which manifests e.g., by change of bond lengths, or (intermolecular) bond angles.  At $T = \pu{0 K}$, molecules  still vibrate, too.
The energy in electromagnetic radiation is discontinous, discrete (you recall $\hbar$). For some energies in electromagnetic radiation, the electric field vector brings to the molecule in question a packet of energy which matches in frequency the frequency of an already existing vibration.  In an analogy, think of a pendulum; its swing is amplified if you tap with the matching frequency.  Similar with molecules (as if consisting of multiple swings) and IR radiation (the pusher); on certain energies, the molecule's vibrations get amplified.  The electric field vector gets attenuated.
A match in energy (here: in frequency) is not the only criterion to consider.  IR spectroscopy is selective to show only these excited vibrations which change the dipole moment of the bond/the molecule.  (Somewhat complementary to this is Raman spectroscopy, reporting about changes in the ease to polarize the electron cloud in a bond.)
Now for the shape of the spectrum you called graph.  On the ordinate, the energy is plot; wavenumbers is one representation of the different frequencies probed during the IR spectroscopy experiment.  The abscissa relates how efficient the transfer of energy from the electric field vector of the IR radiation to the molecule is: great absorbance equals to exciting these vibrations further is efficient, low absorption equals to there is not much change in intensity of the electric field vector passing the sample.  (Similar to the Beer-Lambert law typically introduced with UV-Vis spectroscopy, the number, the concentration of molecules in the optical path length contribute to these readings.)  Typically IR spectroscopy reports vibrations with high resonance frequency on the left, and vibrations with low resonance frequency on the right side.  Instead of absorbance, the ordinate may scale transmission.

(IR spectrum of ethanol as liquid film (neat), recorded by SDBS, entry 1300)
In your example, $\nu_1$ and $\nu_3$ are shown under one envelope. It is one of the characteristics of IR spectra for liquid/solid samples that the quantitized transitions are less easy to discern, because close molecules interact with each other.  Equally, the more complex the molecule, the greater the chance that signatures of similar vibrations overlap.  In addition, you may observe the vibrations as fundamentals, their harmonics, and combinations of them.  This one reason why often only a few characteristic absorption bands in the spectrum are assigned to vibrations in the molecule.

With the large body of recorded IR spectra, it is possible to assign typical bonds in molecules a force constant. Like in a spring, how strong their connection is.  For one, this relates the position of the associated vibrations in the spectrum in energy (the frequency scale, ordinate).  For two, it became possible to predict from scratch for still unknown molecules drawn where vibrations are likely to occur.
For the example of  water, paste the SMILES string O into MolCalc, hit once «optimize» to get bond lengths reasonable, then click «calculate properties».  On the results, select the tab «vibrational frequencies» to display the three vibrations of water:

(animation of the IR-active vibrations of water, molcalc)
Note: If you carefully compare the spectrum you got with the one with the one of MolCal's prediction, you recognize a energetic different sequence of the vibrations shown and different frequencies.  This is where the how the computation is performed becomes important.  Does it consider an ensemble of molecules in condensed state (like the one one records in the lab with a spectrometer), where the molecules of the matrix interact with each other?  A neat spectrum (only the compound itself, e.g. in ATR-IR spectroscopy) may very well show the bands at a different position than one recorded for the same compound in a suspension of nujol, or a pill of $\ce{KBr}$.  On the other hand, to obtain the result faster, computations often assume the molecule as isolated in vacuum.  And there are different levels of theory of computation; for teaching purpose (up to ten non-H atoms), the affordable one by MolCalc often suffices.


text:
Questions on Third Law of Thermodynamics
Third Law of Thermodynamics reads:
at absolute zero , the entropy of a perfectly crystalline substance is zero.
(https://simple.wikipedia.org/wiki/Third_law_of_thermodynamics)
And considering the Boltzmann equation:
$S=k\ln\Omega$
then we can infer that at absolute zero, a perfectly crystalline substance can be in merely one state. But is that true?
At absolute zero, a perfectly crystalline diamond containing a mixture of C-12 and C-13 is able to be in different states with different patterns of isotope. And a perfectly crystalline hydrogen containing a mixture of ortho-hydrogen and para-hydrogen is able to be in different states with different patterns of nuclear spin. Then can we conclude that the entropy of the diamond or hydrogen is greater than zero?
Since the dimension and realm of microcosmic structure has been expanded continually, there might be more isomerism forms other than isotope and nuclear spin isomerism to be revealed and we are not sure where is, or even whether there is, the ultimate of the microcosmic world. And how can we convince ourselves that all the isomerism of all dimension and in all forms is going to disappear when reaching absolute zero?
What's more, we don't need an absolute enthalpy but rather a relative one, while we indeed need an absolute entropy in physical chemistry as an essential part, so there should be an absolute origin for entropy and thus I shall not doubt the universally recognized Third Law, so what's wrong?
Does the word perfectly crystalline has in definition excluded all the forms of isomerism? Or are all the isomerism phenomena going to disappear when reaching absolute zero? Or is it that we simply omit the entropy produced by isomerism?(But, for example, the molar entropy produced by 50:50 isotope is $k\ln C ^{\frac{N_A}{2}}_{N_A}$ and thus not neglectable. )

https://physics.stackexchange.com/a/436054/160740; https://physics.stackexchange.com/a/150328/160740

Can a diamond crystal consisting of 2 isotopes be considered perfect, even if it has otherwise no flaws nor irregularities ?

https://en.wikipedia.org/wiki/Entropy_of_mixing

In fact, you don't need to speculate about other forms of, yet unknown, isomerism! The term you're looking for is residual entropy. (I notice this was mentioned in the links @andselisk posted.) (Also, the Wikipedia page on the Third Law seems slightly subpar...)


text:
Ionic radii of transition metal ions in the presence of weak field and strong field ligands

In the graph given in this image, I was told by my teacher that whenever we start filling the eg orbitals there is an increase in the radius because these orbitals face the ligand directly and hence the repulsion causes the radius to increase.
However, my reasoning says that due to ligand repulsion to electron cloud of that eg orbital, the size should rather decrease. I am confused
Is what Sir said true? help me in comprehending this.

Would this be for octahedral coordination of all the ions shown? We need to compare apples and apples when it comes to ionic radii, and crystal field or ligand field splitting changes with different coordination geometries.

Yes, it is for the octahedral geometry of the ligand around the metal ion.


text:
Calculation of hypochlorite concentration from active chlorine
So in the lab we got sodium hypochlorite solution. The concentration is in active chlorine (13%) on the label. How do I calculate it to concentration of hypochlorite? I need certain amount of hypochlorite (0,868g roughly) for the reaction I am trying to do and because it is not going as described in the article as they were using 5% hypochlorite solution I am checking for possible miscalculations on my part.
Thank you for the help and answers in advance

The active chlorine, in %, is the mass-percentage of $\ce{Cl2}$
dissolved in the solution, without taking into account the fact that this $\ce{Cl2}$ may be at least partially transformed into $\ce{ClO-}$ or $\ce{HClO}$, according to one of the two equations depending upon the pH value :   $$\ce{Cl2 + 2 OH^- -> ClO^- + Cl- + H2O}$$ $$\ce{Cl2 + H2O -> HClO + Cl- + H+}$$
One liter solution with $13$% active chlorine has been made with $130$ g $\ce{Cl2}$ being dissolved into enough $\ce{NaOH}$ solution. And $130$ g $\ce{Cl2}$ is $\frac{130 g}{71 g/mol} = 1.83$ mol $\ce{Cl2}$. This solution has a concentration [$\ce{ClO^-}] = 1.83$ M.
The origin of the word "active" is coming from the fact that, if such a solution is acidified with some strong acid, the previous equations will be reversed, and $1$ L solution with $13$% active chlorine will liberate $130$ g chlorine $\ce{Cl2}$.
Pleas take notice that sodium hypochlorite cannot be obtained easily as a dry substance by evaporating its solution. If such a solution is evaporated by heating it, $\ce{NaClO}$ will disproportionate according to $$\ce{3 ClO^- -> ClO3^- + 2 Cl^- }$$

From Wikipedia: "The anhydrous compound is unstable and may decompose explosively. It can be crystallized as a pentahydrate NaOCl·5H2O, a pale greenish-yellow solid which is not explosive and is stable if kept refrigerated."    I suppose that if you evaporate a solution of sodium hypochlorite, you might get the pentahydrate as a solid. I did this some years ago; it didn't explode till I added some sulfuric acid.

@James Gaidis. Heating a solution of $\ce{NaClO}$ is the industrial way of synthesizing sodium chlorate. So if you want to get $\ce{NaClO}$ crystals, you must evaporate it at low temperature, not far from $\pu{0°C}$. And this is not easy to do.

@JamesGaidis “it didn't explode till I added some sulfuric acid.” ;-)


text:
Why doesn't tetrachloromethane react with aqueous sodium hydroxide under reflux?
$\ce{CCl4}$ does not react with hydroxide nucleophile whereas $\ce{CH3Cl}$ does.
I had initially thought that it was due to symmetrical geometry of $\ce{CCl4}$ and so the dipoles cancel out and the molecule is non-polar, which means there is no partially positively charged carbon atom in $\ce{CCl4}$, and so there is nothing for the hydroxide ion to attack (I figured inducing a dipole on one of the $\ce{C-Cl}$ bonds by the hydroxide ion upon collision would not be possible and out of the question due to the large difference in electronegativity between the $\ce{C}$ and $\ce{Cl}$ atom, so any possible partial positive charge on $\ce{Cl}$ atom would go away almost instantly) but then I saw the mark scheme to this question and it said that it was to do with the large size of the $\ce{Cl}$ atom - preventing the hydroxide ion from colliding with the carbon atom. Ehich makes sense I guess, but does the argument I made answer this question in any way? (I also have a feeling I may have a misconception on whether a partial charge exists on any of the atoms for a non-polar molecule, so please do clear that up if that error is included in my statement).

Your explanation is correct. Chlorine is a big atom, and Carbon is rather small. This prevents the hydroxide ion from colliding with the carbon atom.

How, exactly, does CHCl3 react with sodium hydroxide anyway? By nucleophilic substitution or by deprotonation? If the latter, then CCl4 does not have a dissociable proton.

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

@OscarLanzi Deprotonation is the first step of two (if you want it).

Related: Why are DCM and chloroform so resistant towards nucleophilic substitution?

In chloroform, there are three electron withdrawing groups ($\ce{Cl}$) which are able to stabilize a negative charge.  Thus, the first step is the deprotonation of chloroform, typically with concentrated lye:
$$ \ce{CHCl3 + NaOH <=>[H2O] CCl^-_3 + H2O} $$
The second step is the loss of $\ce{Cl^-}$ to yield $\ce{NaCl}$ and dichlorocarben, $\ce{^{..}CCl2}$.  This may be dangerous if performed neatly (one reason you don't dry dichloromethane over sodium).  Under conditions of phase transfer catalysis ($\ce{NaCl}$ enters the aqueous phase, the carbene remains in the organic phase), it however may be tamed to be a useful procedure to install cyclopropyl groups:

An other base capable to deprotonate $\ce{CHCl3}$ is e.g., $\ce{KO^tBu}$ at $\pu{-30 ^\circ{}C}$ in diethyl ether (procedure in Org. Syn.).
This approach however has been superseded e.g., by the Simmons-Smith reaction (second line), attaching the cyclopropyl ring in one step with less safety concern, less drastic reaction conditions, and thus, wider scope of application.
Carbon tetrachloride, on the other hand, does not possess a proton $\ce{NaOH}$ could abstract.


text:
Can the expression be represented as a total differential?
I should determine whether the following expression represents a total differential:
$$\mathrm{d}f(x,y) = (\ln y) \,\mathrm{d}x + \frac{x}{y}\,\mathrm{d}y$$
But with the following method (physical chemistry):

Prove your statements assuming the existence of a "state function" $f$ and integration of the total differential. If necessary, give the function $f(x,y)$.
Note: Start with the definition of a total differential and do not use Schwarz's theorem.

So first I looked at the definition of a total differential.

In general, if the integral over an infinitesimal quantity is
independent of the path between the integration limits, this quantity
is called total differential. ... In general, if the integral over an
infinitesimal quantity depends on the path between the integration
limits, this quantity is not called a total differential. - Atkins,
Physical Chemistry, 6th edition

So if I now minintegrate $\mathrm{d}f(x,y)$ first as suggested, I arrive at:
$$f(x, y) = 2x\ln y$$
$f(x,y)$ could now be, for example, my internal energy $U$, $x$ my temperature and $y$ my volume $V$. Then the function would give me the change in $U$, since the following applies:
$$\Delta U = \int \mathrm{d}U$$
I have several more tasks of this kind here, could I not then arbitrarily say that is $V$, that is $T$, that is $U$ and therefore $U$ is always independent of the way, so all functions are total differentials? I somehow do not know what to do.

What makes you think that $U$, $V$, and $T$ are necessarily related by the function $U = 2V \ln T$? If this doesn't hold, then you have no right to replace $f$, $x$ and $y$ by anything else. As I understand it, you're basically asserting that $df$ is total because $dU$ is total (true, because $U$ is a state function) and $f = U$ (which is very sketchy).

If you want to show that $df$ is total, then all you need to do is to show that there exists some $f$ such that $(\partial f/\partial x)_y = \ln y$ and $(\partial f / \partial y)_x = x/y$ (thereby satisfying the definition $df = (\partial f / \partial x)_y,dx + (\partial f/\partial y)_x ,dy$, see https://en.wikipedia.org/wiki/Differential_of_a_function#Differentials_in_several_variables). Which you already did! So you could just have stopped there, and it would have been a perfect answer to the question (assuming that you had some working).

Don't you recognize the right hand side of the equation as $d(x\ln{y})$?


text:
How do compounds consisting of polar molecules dissolve?
In the case of compounds of polar molecules dissolved in water, I know it's to do with the formation of permanent dipole force of interactions between the water molecules and the polar molecules.  But how exactly does dissolution happen? I've thought about it, and the only reasoning that comes to mind is that the London forces between the polar molecules break, as it is weaker than the force between the water and polar molecule, although I'm not sure if this is the reason why..

Not only London forces (induced-induced), but Debye ( permanent-induced) and Keesom ( permanent-permanent) forces as well - being together van der Waals forces.

It might read like a nit pick, however (in standard lab chemistry), we dissolve compounds (which might consist of polar/apolar molecules) in solvents.  But dissolving e.g., glucose in water, does not dissolve the molecules into their parts.  In case of succrose, though, the split into glucose and fructose is a reaction (hydrolysis), a different process (break of a chemical bond).

Opposite charges attract each other (situation in the solid, undissolved state).  Water (high dielectric constant) is able to attenuate this attractive forces drastically (the state of solution).  Then it is thermodynamics' book keeping of enthalpy and entropy to favour one, or the other state, or an equilibrium.

Dissolving a substance in water means breaking hydrogen bonds. It requires energy. If enough new hydrogen bonds are made between solute molecules and water, this dissolution is energetically favorable. This happens when the solute is polar. If not enough new hydrogen bonds are made, the entropy effect is not sufficient to counterbalance the enthalpy effect : the substance will not be soluble in water.


text:
Why are higher-energy orbitals larger in size/radius?
I know the different atomic orbitals of atoms are derived from the wave function in the Schroedinger equation. Because I'm not a mathematician, though, I can't decipher the Schroedinger equation and therefore can't simply explain why higher-energy orbitals are larger than lower-energy orbitals (and therefore contain electrons that are, on average, further from the nucleus). For instance, the 2s orbital is larger than the 1s orbital for any atom, but why?
Is there a relatively simple answer to my question (that doesn't require complex mathematical understanding)? Or is the answer simply that the higher-energy atomic orbitals are larger because that's the solution given by the Schroedinger equation? I'm assuming this is the case, but I'm not entirely sure.

Addressing the title, applying the classical analogy of the central quadratic force (F = k/r^2).: It is like when a planet has higher mechanical ( potential + kinetic ) energy, if its orbit is farther from the Sun.

Roughly speaking, the potential energy is $-\frac1r$. Larger orbital means greater $r$ means smaller $\frac1r$ means greater $-\frac1r$ means higher energy.

The highest energy state ($n=\infty$) is a free electron, i.e. an electron at "infinite" distance. The excited states orbitals are merely stops on the way from the ground state ($n=1$) to infinity, both in terms of energy and average distance from the nucleus.
You might compare it to lifting a stone from the ground. It costs you energy to lift the stone, and therefore at any distance from the earth the stone has a higher (potential) energy. If you stop carrying the stone at any height, it will spontaneously fall back to the ground state; that's why we say the ground state has the lowest energy.
In an atom, the picture is more complex (for starters, it's quantized) but directionally the same.

In order to determine whether an orbital corresponds to a higher energy state you need to look at the combined potential+kinetic energy.
How to go about this is explained in this ChemLibreText (Ref. 1):

Although dot-density diagrams are very informative about the potential energy of an electron in an orbital, they tell us nothing at all about its kinetic energy. It is impossible, for example, to decide from Figure 5.6 whether the electron in a 1s orbital is moving faster on the whole than an electron in a 2s orbital, or even whether it is moving at all! Fortunately it turns out that this difficulty is unimportant. The total energy (kinetic + potential) of an electron in an atom or a molecule is always one-half its potential energy.

(A dot-density diagrams is a graphical representation of the electron density described by an orbital.)
As noted in the excerpt there is a simple relationship between the energies that allows us to relate the total energy to the potential energy (if you know the potential energy of an electron in a particular orbital, you know it's total energy). The relationship between kinetic, potential and total energy follows from the virial theorem. So how do you determine the potential energy?
The average potential energy of an electron can be computed as an average over all space of the Coulombic interaction energy between the electron in its orbital and the nucleus:
$$\left< PE \right> = \int |\psi|^2 \frac{-e^2Q}{r} dv$$
Here $\psi$ is the wavefunction of the electron derived by solving Schroedinger's equation, and  $|\psi|^2$ describes the electron density about the nucleus. These are functions of the distance from the nucleus.
For an orbital with larger principal quantum number, the electron density $|\psi|^2$ at greater distances from the nucleus is larger, so that regions of space where 1/r is smaller are weighted more, and the overall integral becomes smaller (meaning less negative).

I know the different atomic orbitals of atoms are derived from the wave function in the Schroedinger equation. Because I'm not a mathematician, though, I can't decipher the Schroedinger equation and therefore can't simply explain why higher-energy orbitals are larger than lower-energy orbitals (and therefore contain electrons that are, on average, further from the nucleus). For instance, the 2s orbital is larger than the 1s orbital for any atom, but why?

More commonly an orbital refers to an approximate representation of the probability density of an electron, which describes the likelihood of finding the electron somewhere around the nucleus. A larger orbital contains more of its density spread further from the nucleus. (The word "orbital" is also used as a synonym for a 1-electron wavefunction, which is an approximate representation of the probability amplitude of an electron. The wavefunctions are approximate solutions of the Schroedinger equation.)

Is there a relatively simple answer to my question (that doesn't require complex mathematical understanding)? Or is the answer simply that the higher-energy atomic orbitals are larger because that's the solution given by the Schroedinger equation?

The simple answer is that electrons are attracted to the nucleus and placing them further from the nucleus requires that the energy of the electrons be increased in order to overcome the nuclear attraction. Generally this means that the average kinetic energy also increases (which also follows from the virial theorem).
Reference

Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn (2020). Electron Density and Potential Energy, in: General Chemistry, ChemPRIME at Chemical Education Digital Library (ChemEd DL). Accessed online 2/2022.