text:
How can I synthesize 2‐amino‐3‐(chlorosulfonyl)benzoic acid from toluene or benzene?
I am trying to design a proper synthesis route for 2‐amino‐3‐(chlorosulfonyl)benzoic acid, though methyl 2‐amino‐3‐(chlorosulfonyl)benzoate—using an ester group in lieu of a carboxylic acid—would also be acceptable.

I am thinking a good starting material would be toluene or benzene. Here is my proposed route.

I determined the ortho-para ratio for toluene nitration from p. 816 of "Organic Chemistry, 6e" by Loudon & Parise.

Typically, para substitution predominates over ortho substitution, but not always. For example, nitration of toluene gives twice as much o-nitrotoluene as p-nitrotoluene. This result occurs because the nitration of toluene at either the ortho or para position is so fast that it occurs on every encounter of the reagents; that is, the energy barrier for the reaction is insignificant. Hence, the product distribution corresponds simply to the relative probability of the reactions. Because the ratio of ortho and para positions is 2 : 1, the product distribution is 2 : 1.

Also, according to the same page,

The boiling points of o- and p-nitrotoluene, 220 °C and 238 °C, respectively, are sufficiently different that these isomers can be separated by careful fractional distillation.

Is there a better method? Please identify any mistakes I may have made in my synthesis and suggest improvements. Note that I install t-butyl as a blocking group in order to direct the sulfonyl chloride group ortho to the amino group.

I would be concerned about how you are going to isolate the final product from the Aluminium residues

There's no way to isolate such compound. What magic do you have to prevent reaction between sulfonyl chloride and amine groups?

I suspect that with 2 strong electron-withdrawing groups on either side the amine will not be very nucleophilic. That said the material looks horrendously difficult to handle.

@JonG do you need the free carboxylic acid? An ester would make this a great deal simpler.

I'm curious as to how you get these numbers 67% and 33% for nitration of toluene. I would be pretty surprised if those were the real numbers. There are two ortho positions and one para position which naively suggests that 2:1 is the ratio, but these positions are not the same, and should not have the same weighting.

This is a cmpd that is likely to self-condense with an amino and a sulfonyl chloride group. How do you avoid protonation of the amino group during sulfonation?

@Waylander an ester substituent would be acceptable.

@orthocresol I have updated the body of my question to address your concern.

@Waylander also, what about the material would make it so difficult to handle?

The sulfonyl chloride group is very reactive and water sensitive so solvent/aqueous extractions are out.

@Waylander regarding the use of an ester, would that make it easier to separate the aluminum residues? I would be curious to know why you think the ester group would make the synthesis pathway (or some individual element of it) simpler.

I was thinking primarily of alternative ways of installing the sulfonyl chloride - making from the sulfonic acid, which would also give the acyl chloride if you had the carboxylic acid present.

Here is a proposed route starting from aniline.
First protect the -NH2 with t-BOC
Directed lithiation of t-BOC-aniline (sBuLi/Et2O, -75C) and quench with MeSSMe to install -SMe in the 2 position.
Directed lithiation under the same conditions quenching with Ethyl Chloroformate to install  -CO2Et in the 6 position.
Oxidise the -SMe group to sulfoxide (MCPBA, DCM)
Pummerer rearrangement of the sulfoxide (Ac2O, aq. base work up) gives the free thiol.
Oxidation of the thiol to the sulfonic acid, one example of possible conditions hydrogen peroxide with Rh catalysis
Form the sulfonyl chloride (conditions using trichlorotriazene here. Depending on your choice of conditions you can remove the -BOC group or preserve it. I think preserving it will give a flexible late stage intermediate that is reasonably easily handled. It is possible to directly form the sulfonyl chloride from the -SMe with chlorine in aq acetic acid conditions here but the conditions are harsh and risk removing the -BOC group

Would it be acceptable to convert the thiol-containing intermediate directly to the sulfonyl chloride using H2O2 and SOCl2 in acetonitrile at room temperature?

I think that would remove the tBOC group from the aniline as HCl is generated. Is this a paper exercise or are you actually doing it?

I am working it out on paper before I actually implement/"do" it.

What is your next step with this product?

Wikipedia (https://en.wikipedia.org/wiki/Pummerer_rearrangement) illustrates an extra carbon, with two H-atoms attached, in-between the two 'R' groups. If I treat the 'Me' (attached to the S-atom) as one group, and the benzene ring with its other substituents as the other 'R' group, that does not leave me any space for a CH2 in-between. I have seen a couple of examples on the internet with a CH in-between, but none with no H's in-between. I am curious if you have a source, given that the Pummerer rearrangement   seems to usually involve an alpha-carbon.

Pummerer on PhSOMe creates PhSCH2OAc initially which hydrolyses on basic workup to give PhSH + CH2O + AcOH. I do not have database access to give you a reference but I have done this reaction

Let us continue this discussion in chat.


text:
Work done in expanding a gas reversibly and irreversibly
So, my chemistry teacher gave the class following $P_{external}$ versus $Volume$ diagrams for reversible and irreversible expansion of a gas which are as follows.

(Reversible expansion)

(Irreversible expansion)
Then he gave us following results to memorize:
$$\mid \left(W_{reversible}\right)_{expansion}\mid \gt\mid \left(W_{irreversible}\right)_{expansion}\mid$$
Where W is for work.
According to him, these results are always true.
But consider the following graph

If I'm not wrong, this can also be considered an irreversible expansion.
But as we can see clearly,
$$\mid \left(W_{reversible}\right)_{expansion}\mid \lt\mid \left(W_{irreversible}\right)_{expansion}\mid$$
This thought is really bothering me. Where did I go wrong?

The upper case looks like reversible isothermic expansion, the middle case like irreversible adiabatic expansion, the bottom case like reversible isobaric expansion and reversible isochoric compression.( All implies the system contains a constant amount of an ideal gas ).

@Poutnik - Is it right to call it a compression while already considering it "isochoric"? And why not call the middle case  isochoric then isobaric expansion instead of irreversible adiabatic expansion?

Hehe, you are right, I revoke it. It is an isochoric cooling down.

The middle case can have 2 possible explanations. It is either irreversible adiabatic expension after sudden decrease of external pressure, either it is reversible isochoric cooling down and then reversible isobaric expansion.

Is the eternal force per unit area exerted by the inside face of the piston on the gas equal to the internal force per unit area exerted by the gas on the inside face of the piston, or is Newton's 3rd law of action-reaction no longer valid?  Do you think that, in an irreversible rapid expansion, the "pressure" of the gas (force per unit area) is still determined by the ideal gas law (or other equation of state), or does the equation of state only apply to thermodynamic equilibrium states and reversible processes (comprised of a continuous sequence of thermodynamic equilibrium states)?

@ChetMiller - As far I can understand, in case of irreversible rapid expansion, the external force per unit area applied on the face of piston would not be equal to the pressure applied by internal gas. The difference would infact act as a driving force for the expansion/compression, whatever takes place. I don't think pressure of the gas could still be determined by the ideal gas equation in irreversible expansion as the system would be erratically expanding that too in a very small time period. It can be applicable once the equilibrium is restored.

In other words, you don't believe that Newton's third law applies at the interface between the piston and the gas?  You are correct that the pressure at this interface cannot be determined by the ideal gas equation for an irreversible expansion.  So you are forced to rely on the value of this pressure imposed externally, say by applying an external atmosphere or controlling the motion of the piston manually to achieve a desired pressure history.

Would you like to know way the force per unit area exerted by the gas on the piston face is equal to the force per unit area exerted by the piston face on the gas, even in an irreversible expansion?  Or, in other words, would you like to understand what's actually happening physically?

@ChetMiller Is the eternal force per unit area exerted by the inside face of the piston on the gas equal to the internal force per unit area exerted by the gas on the inside face of the piston... I think I previously could not understand what you meant by "external force" here. My thinking goes like there would be action-reaction pair established between both piston and internal gas and piston and surrounding, as described by Newton's 3rd law. But the forces would be different in magnitude acting as an unbalanced pair on the piston causing it to move in or out (or compressing or expanding).

@ChetMiller - and yes, ofcourse I'd like to know what actually is happening in the physical realm.

Are you familiar with the physical property of a gas known as viscosity?

@ChetMiller - (Apologies for late reply) well yes, literally, I'm aware of the viscous nature of the gas, but formally speaking, I have very less knowledge about that aspect. More or less, I've been dealing with ideal gases.

I will call your processes (1), (2), and (3). I assume (1) is reversible isothermal expansion, (2) is irreversible isothermal expansion, and (3) is unspecified.
The key idea is that the inequality $w_\text{rev} \geq w_\text{irrev}$ can only be applied to reversible and irreversible analogues of the same process. (1) and (2) satisfy this requirement because (1) is the reversible analogue of (2), but (1) is not the reversible analogue of (3).
Physically, (1) is like having a pile of sand on top of a container (containing an ideal gas) whose lid is a piston, and then slowly blowing away the sand. (2) is the same scenario, except you sweep off the sand in one grand gesture. But (3) is a different scenario entirely.
If we refer to the first law, $\Delta U = q + w$, it's clear that, given an initial and a final state, we can make $w$ as large as we want as long as we have a commensurate value of $q$. Just because processes (1), (2) and (3) all have the same initial and final states does not mean that it makes sense to apply the inequality $w_\text{rev} \geq w_\text{irrev}$ to any two of them.

I think you're right. I didn't think about the situation this way.

The ideal gas model represents the limit of the behavior of real gases in the limit of low pressures.  Real gases exhibit a non-zero viscosity, even in the ideal gas limit.  So ideal gases must be considered to exhibit viscous behavior.
When a fluid (liquid or gas) is deformed very slowly, the state of stress within the fluid is isotropic (acting equally at all directions at each given location) and the force per unit area on any surface is equal to the pressure determined from the equation of state (e.g., the ideal gas law).  However, as the rate of deformation becomes more rapid, the state of stress is no longer isotropic, and the viscous contributions to the stresses are proportional to the rate of deformation.  For a gas in a cylinder experiencing a rapid (irreversible) deformation, this translates into the force per unit area exerted by the gas on the piston face varying not only with the volume of the gas, but also with the rate of change of volume.  A crude  approximation to the compressive stress on the piston face in a rapidly deforming gas can be expressed as:  $$\sigma=\frac{nRT}{V}-\eta\frac{1}{V}\frac{dV}{dt}$$The first term represents the ideal gas behavior, and the second term represents the viscous contribution; the parameter $\eta$ is proportional to the gas viscosity.
In a rapid irreversible expansion of compression at constant externally applied pressure, we can set $$P_{ext}=\sigma=\frac{nRT}{V}-\eta\frac{1}{V}\frac{dV}{dt}$$  Note that, even if this is different from the original pressure and from the pressure calculated from the ideal gas law, the force per unit area exerted by the gas on the piston face can still match the external pressure if the gas deforms rapidly enough, such that the viscous contribution makes up the difference.  In fact, this is what actually takes place as the gas response automatically adjusts to the externally applied pressure in a rapid (irreversible) process.

You are misunderstanding how pressure is changing.
Work is given by $$\delta W = -p_{ext}dV$$
Work is performed when the external pressure on the system is smaller than the internal pressure of your gas.
$p(A)$ and $p(B)$ are the final and initial external pressures of your system. If you are expanding isobarically (as in your second image), $$W = -p(B) \Delta V$$
in the reversible and non-reversible case. Therefore, work done in an isobaric expansion is the same for reversible and non-reversible process. It may happen than once the pressure drops, the system might expand again, but we are not interested in that regime right now. No work will be done in an infinitesimal amount of time after a sudden instantaneous drop in pressure.
If the pressure change is adiabatic from $p(B)$ to $p(A)$, you will have the curve in image 1.
If the pressure change is irreversible i.e. you have a sudden change in pressure (in image 2, the pressure drops suddenly, not in a smooth, differentiable manner), you can apply
$$W_{irrev} = -\int _{V_A} ^{V_B} p_{ext} dV = p(A)\Delta V$$
If it were reversible and adiabatic,
$$W_{rev} = -\int p_{ext}dV = -nRT\ln \left( \frac{V(B)}{V(A)}\right)$$
assuming an ideal gas. You can check that
$$W_{rev} \geq W_{irrev}$$
Note the $\geq$ rather than $>$ in the above expression. That was a slight error on the part of your teacher.

Could you please explain these lines. Sorry, but I think, I don't clear understand it.- "Work is performed when the external pressure on the system is smaller than the external pressure of your gas"

Oops, sorry I made a mistake. It should be internal pressure of your gas. When your gas is pushing the boundary with a certain pressure $P$, the system will only expand if the outside pressure is less than $P$.

One more thing please. Should the third diagram be considered irreversible or not?

Since there is a discontinuity at $V=V(B)$, it is irreversible. Generally, any smooth curve on a $P-V$ plot is a reversible process, unless explicitly stated otherwise.

If you have a massless frictionless piston,, how can you say that the external pressure is different from the internal pressure.  Or, if you regard the boundary of your system as the inside face of a piston, irrespective of whether the piston is massless or frictionless, how can you say that the external pressure is different from the internal pressure.

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Did JJ Thomson know about Eugen Goldstein’s experiment discovering canal rays?
We learn that JJ Thomson discovered the electron in 1897. Several years EARLIER in 1886, Eugen Goldstein performs the same experiment but with the anode and cathode switched to produce positively charged canal rays. My question is, did JJ know about Goldstein’s experiment? I’m surmising he did not, as it wouldn’t have taken such a leap to conclude that Goldstein’s  was a positive particle as well as the negative electron JJ discovered. Does anyone know the history who can confirm this is true or prove it is false?

Stack Exchange History of Science and Mathematics will get you the response. Secondly have you tried looking at the original paper of JJ and does he cite Goldstein using Google Scholar? Most likely he does.

See the original paper. http://web.lemoyne.edu/~giunta/thomson1897.html#footnote. The charge of cathode "rays" was determined by Perrin not by JJ.

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CaCl2 and Citric Acid react /w metal pot
I'm building a dehumidifier based on $\ce{CaCl2}$-solution as descant, which is to be regenerated by heating and evaporation.
I thought it would be a good idea to add some citric acid to combat chalk-buildup due to rather hard tap water.
To test for possible reactions of $\ce{CaCl2}$ and citric acid, I mixed a 30% $\ce{CaCl2}$ solution by weight, added different amounts of acid (10%, "rather more", "very much") and boiled them. (My main fear being that some Ca might react with Citric acid to form a precipitate, which did NOT happen.)
While heating, the solutions with more citric acid turned dark turquoise, but remained clear. Using vinegar did not cause a color-change. Heating $\ce{CaCl2}$ or citric acid alone does not cause color change. No other reactants (I know of) were present. It seems that cooking for longer turns the solution darker, but I may be mistaken on that.
The reactions I could think of should all be colorless/white, not blue/green/turquoise: Ca-citrate, decomosed citric acid, I didn't find sources for citric $\ce{Ca}$ complexes. Maybe a $\ce{Cl}$-complex is formed, but $\ce{Ca}$  is the only available metal, and again I couldn't find resources for chlorine $\ce{Ca}$ complexes. Impurities are probably not at fault, since the color is very strong/dark.
What I'm asking is specifically what reactions might occur between $\ce{CaCl2}$ and citric acid. I know of enough ways to get around my hard-water problem.
Edit:
The pot I used for heating was at fault. Heating in an enamel pot showed no color change. Thanks @poutnik and @m-farooq
I would still be very curious as to what reaction might cause a green/blue color change. From what I can tell, the metal pot is made of stainless steel.

Can you add a picture? These blue/green salts indicate transition metals. What was the container made of in which you were heating?

Calcium forms a citrate complex, but the real issue may be thermal stability of it. There can occur a bunch of complicated organic reactions. As the acid alone does not get such colour, transition metal interference is excluded, unless present in CaCl2.  It may be the complex thermal decomposition along an organic chemistry path.

Thank you for your input. My kettle indeed had a metal bottom which I thought to be stainless steel. Heating in an email-pot had no color-changing effect.

Is email pot enamel?


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What is the electron configuration of platinum?
Some textbooks write the electron configuration for platinum (Pt) as $\ce{[Xe] 4f^14 5d^9 6s^1}$. However, some also write it as $\ce{[Xe] 4f^14 5d^8 6s^2}$. Which is correct?
I hope someone can help me with an authoritative source.

Please, learn to use better grammar when writing your text. Start sentences with a capital letter. You don't need to put so many spaces between each word.

See Madelung rule exception: https://chemistry.stackexchange.com/questions/35086/exceptions-to-the-madelung-rule-for-electronic-configurations

https://chemistry.stackexchange.com/questions/2660/anomalous-electronic-configuration-of-platinum

The NIST Atomic Spectra Database,[1] which is as authoritative as it gets, shows that $\ce{5d^9 6s^1}$ is the correct ground-state electronic configuration for platinum.
Various inorganic textbooks also say the same. Wikipedia cites Meissler and Tarr for this table: https://en.wikipedia.org/wiki/Electron_configuration#Other_exceptions_to_Madelung's_rule

Kramida, A., Ralchenko, Yu., Reader, J., and NIST ASD Team (2020). NIST Atomic Spectra Database (ver. 5.8), [Online]. Available: https://physics.nist.gov/asd [2021, January 6]. National Institute of Standards and Technology, Gaithersburg, MD. DOI: https://doi.org/10.18434/T4W30F


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why do we see characteristic colors of alkali metals in oxidizing flame?
I am a high school student and I am a little confused in a topic, My confusion is that:
We know alkali metals and their salts imparts characteristic color in oxidizing flame, for example: Lithium gives crimson red color, sodium gives yellow , potassium gives violet and Caesium gives blue, My school textbook says that this is because the heat from the flame excites the outermost electron to a higher energy level and when it returns back it will radiate in visible range as I mentioned above, but if this is the correct reason then we know the outermost electron  of lithium is the most attracted  among all alkali metals .So, it means that it would take high energy to excite its outer electron so a short wavelength of light is needed i.e probably in violet region, so when it returns back it should emit the same wavelength. Then why do we see these weird pattern of colors? I think this is not the complete reason for why do we see those colors? please explain this in brief only so that I can easily understand.

https://chemistry.stackexchange.com/questions/50467/why-lithium-gives-flame-coloration

Does this answer your question? Why is the photon energy of lithium flame coloring lower than that of cesium?

There was no need to close or downvote it. Sorry to say but those answers are quite incorrect, especially the "Accepted" answer is hopelessly handwaving. What a fallacy with accepted answers in SE...the person who is asking it is accepting it as a correct answer!

@M.Farooq Then CV https://chemistry.stackexchange.com/questions/141382/wavelengths-of-colors-emitted-by-alkali-metals-during-flame-test - is your own answer also incorrect?

I don't see any factual inconsistency there. If there is one, will be happy to correct it.

I don't think it is a good idea to connect ionization energy with the color of flame emission, especially in a Bunsen burner.
Ionization energy means that you have separated the electron out of the nucleus attractive field. However, electronic transitions, corresponding to visible wavelengths, take place while the electron is still bound to the nucleus. Thus the flame color of alkali metals might correlated with the IP  but not the causation.
Let us look at the resonance wavelengths of emission of Li, Na, K, Rb, and Cs. Resonance wavelengths means the transition from the $\mathrm{1s}$ to $\mathrm{2p}$ orbital or $n\mathrm s$ to $(n+1)\mathrm p$ orbital in the case of alkali metals.
$$
\begin{array}{r|ccccc}
\hline
\text{Element} & \ce{Li} & \ce{Na} & \ce{K} & \ce{Rb} & \ce{Cs} \\
\hline
\lambda/\pu{nm} & 670 & 589 & 769 & 794 & 894 \\
\hline
\end{array}
$$
You cannot even see these wavelengths with your eyes after 700 nm which are in the deep deep red. Some people can see the red lines of K 769 nm. However, you can see the energy gap is indeed decreasing as we go down the group. The ionization energy also goes down the group. However, you can see, the colors which you see in the Bunsen burner have nothing to do with ionization energy.
The wavelengths have been extracted from a old book titled Graphische Darstellung der Spektren von Atomen und Ionen mit ein, zwei und drei Valenzelektronen by Dr. W. Grotrian, 1928. I think NIST atomic spectra database will be better today but I am not sure if they list resonance lines clearly. This book has graphically labelled transitions for almost all elements known in 1928.

Yes, this is the answer. First ionization means total expulsion of the outermost electron. Meanwhile, electronic transitions occur when the outermost electron is energetically excited and gets promoted to a superior level of energy and back, emitting the remaining energy under the form of photons.

@M.Farooq It would be nice to see the source for the wavelength values. Your another answer which I suppose you've copy-pasted a good chunk of data from states "very old book by Grotrian". That's kinda cryptic. Could you please specify the title, year, edition, page number and IDs, if any?

I extracted data from the Grotrian diagrams, it is not listed in tables as one would expect. It is indeed from 1928, really old book by Dr. Grotrian. Will add the reference shortly. Most likely readers will benefit from the NIST database rather than this ancient inaccessible book. I am not sure if they have Grotrian diagrams.

The relationship between ionisation energy and emission spectra is complex
You make the assumption (implicitly) that the colour you see in emission spectra is from the single complete ionisation of an electron from the highest energy electron orbital in lithium. But the colour you see is not that simple for two reasons.
One is that you only see visible light (roughly the octave from about 400 nm to about 750 nm of wavelength). Many emission lines occur outside that range in the infra red or ultra violet, so the colour you see will rarely be a product of the complete range of possible emission lines.
Another is that there are far more transitions that the one that is related to complete ionisation of the outermost electron. All atoms have many possible electron orbitals  higher than the lowest occupied one. Transitions between most pairs of energy levels are possible (there are complicated but irrelevant reasons why some can't happen related to quantum mechanics). And flames often have enough energy to put some electrons in many of those orbitals leading to a large number of possible emission lines when electrons can fall from any energy level to any other energy level. (see this introduction to the concept that explains the complexity for the simplest atom, hydrogen, which shows the calculated orbital energies and the possible transitions that result).
So colours result from filtering of the possible emissions to visible light only and the complexity of the transitions that are possible. So there is no meaningful relationship between colour and the ionisation energy.

"Transitions between every pair of energy levels are possible." This ignores the atomic spectroscopy selection rules. You are somehow implying that there is a relationship between observed flame color and ionization energy but it is complex. The point is, does the relationship exist? You might wish to reconsider these points.

@M.Farooq I'm answering at a basic level appropriate to the question. Transition rules are one level of complexity too far (and they explain brightness not existence of transitions). And I imply nothing at all about the relationship with ionisation energy: I explain why you should not even expect one when trying to explain colour.

We should not teach something to students which has to be unlearnt later. One can say that certain transitions are allowed rather than absolutes "Transitions between every pair of energy levels are possible." Selection rules in atomic spectroscopy have nothing to with the brightness of the line, it is either yes or no (see for example Laporte selection rule). Brightness of a line depends on the population of the atoms in particular excited state, and this is governed by Boltzmann's distribution and temperature only (as long as it is allowed).

At this level of explanation this is not a problem. Besides selection rules are not that absolute in many systems otherwise we wouldn't have phenomena like phosphorescence. My point was not not to state an absolute rule but merely a simple explanation avoiding concepts unnecessary to the core explanation–like triplet and singlet states. Explaining those would add nothing of value to the idea that there are many energy levels and many possible transitions and the ones you that contribute to colour are unlikely to be all of the possible ones.

There is no atomic phosphorescence phenomenon in the flames or otherwise in gas phase. Some ions may show it in a solid state.

I am really sorry, I didn't mean that, I have gone through each similar question but the answer there are either very complicated for me to understand or not up to the point,{ I am again saying its my fault that I am in high school and don't have much knowledge its not the fault of answerer , I just want to learn more in a simple way that's why I asked it here} so I just asked what I think should happen but something unexpected is happening, I just wanted to know what's wrong in my thinking? where I am going wrong and what is the reality:)

@M.Farooq I didn't claim phosphorescence in flames, read carefully. But, more importantly, you again focus on an aspect of the explanation that is completely irrelevant to providing a suitable answer at this level to this question.

Matt_black, It is not about a completely irrelevant point but a blatantly incorrect assertion. It is about editing one small sentence "Transitions between every pair of energy levels are possible" which is scientifically, factually, wrong. In public, please don't assume that this grade-12 is your only audience, there will be hundreds of other students /experts who will read it in future. The Sun does not rise from the West, regardless of what grade you are talking to.


text:
Oily residue on old toy rubbery tyres
I have some sets made by well-known Danish plastic toy manufacturer from my childhood which I've been reviewing recently. Most of the elements was in a condition I was expecting (some yellowing, a little bit damage), but not with the smaller tyres. Originally, I found them quite oily/greasy by touch with some even leaving a bit of, what seemed to look like, liquid residue when removed from plastic wheels. Oh, and they were also full of domestic dust.
Didn't pay too much attention as I was going to clean them with the rest of bricks in washing machine using mild detergent. Upon drying I've check affected elements again and found that plastic elements exposed to rubber tyres were clean. These small tyres however, while their conditions was much better, they still had a bit of oily feeling by touch.
Tyres having this issues are up to 30 years old and rather small. Bigger ones are from the same time period but seem to be made of different material and did not show any signs of being oily even prior cleaning.
Affected tyres look like that if it matters (pictures just show the form, not the condition):

So I'd like to know few this around this issue:

What is going on here?
Is it possible to clean rubber tyres of this residue and get them back to normal condition? How long would it last?
Is it safe for plastic elements made of ABS and poly-carbonate (two most common plastic types used by this manufacturer) to have contact with rubbery tyres exposing issue explained above? Any other possible hazards?
Is there anything I could do to prevent from such issue happening to similar tyres in the future?

Disclaimer. I have not included a brand name for couple of reasons. First, my interest is purely from science/practical point of view. Next, I am not looking to accuse any one of some issues in case there are any. And last reason - idea was to hint a company, as I thought some may have better knowledge of exact materials used in above elements.

Not entirely sure of the reason for downvote. I have read the guide for this section of SE and my understanding this question fit here best. In case this is related to the way I've mentioned brand/company - I have included a disclaimer.

The "rubber", or elastomer, likely has plasticizers to soften the material and make it more elastic and less brittle. These often migrate to the surface as parts age. Some of these plasticizers for elastomers include:

Sebacates, similar to animal fats
Adipates, based on adipic acid, which is found in beets
Phthalates, esters of phthalic anhydride
Some are innocuous, but others, such as bisphenol A (BPA) may have long-term toxic effects. Washing the parts, as you have done, should remove them from the surface. You might need to repeat washing, using different detergents.

Another possibility is depolymerization, actual breakdown of the plastic. At room temperature, this process is slow, though it could contribute to that sticky surface layer.
As for safety, you might check on rules at the time and place the toys were made. Modern toys are made with safety in mind, but plastics have had a bad rap in the past, such as causing liver damage to workers in vinyl production. Caveat emptor.


text:
Comparing basic strengths between pyridine and 1,2-dihydropyrazine

Compare the basic strengths of above two compounds.

In the first compound (1,2-dihydropyrazine), the lone pair of one N will participate in resonance but the lone pair of other N will not, similarly in pyridine, lone pair will not participate in resonance.
I think that since the first compound atleast has $2$ nitrogens (and only one in pyridine), it should be slightly more basic than pyridine, but I came to know that it's the other way round. So how do we compare their basic strengths and what are their $\mathrm{p}K_\mathrm{b}$ values?

The additional nitrogen in 1,2-dihydropyrazine rather pulls the electrons towards itself since it is a electronegative atom causing decrease in basic nature.

pKb(pyridine) = 8.77 at 25°C [CRC HANDBOOK for Physics & Chemistry; 2004 - Edition 8]

Related : https://www.alfa.com/en/pyrazines/?page=2

What I would do is I would compare their acid strengths. Due to inductive forces, deprotonating the 1,2-dihydropyrazine would result in a more stable compound than that in pyridine--the nearby nitrogen (electronegative) will share the burden of the negative charge. Therefore, 1,2-dihydropyrazine is more acidic and thus is less basic.

is more acidic and thus is less basic. This linking is valid for conjugate acid/base pair, not for eventually amphoteric components.

First of all the first compound has two nitrogen atom so we need to know which one is more basic . One of the nitrogen atom has its lone pair involved in resonance due to which they are not fully available for donation so it is less basic nitrogen. Now since we know the more basic nitrogen in first compound we can compare the basicity  between the two compounds in first compound there is a -I effect off nitrogen on the basic nitrogen atom and in the second compound there is only + I effect on the basic nitrogen atom due to which the nitrogen of first compound is slightly electron deficient for donation and second one is electron rich so the second compound is a better base.


text:
Variation of Boiling point with pressure from Clausius-Clapeyron equation
For a first-order phase transition such as water to steam , we write Clausius-Clapeyron equation as:
$\frac{dP}{dT}= \frac{L}{T \Delta V}$
Now  what I infer from this equation is that if the phase-transition takes place at a lower T (that is boiling point is lower) then the rate on the Left Hand Side increases which basically means that at that temperature any small change in T will cause high change in P. This inference is unsatisfactory to me and as such I want to understand what information this equation essentially gives.
More specifically, I wanted to understand a statement I had come across which said that this equation explains why boiling point of a liquid increases with increases in pressure. Although, physically I understand why this is true but how is this fact deduced from looking at the Clausius-Clapeyron equation?
I tried varying "stuff" in this equation to check whether I could get some sort of understanding but I just couldn't deduce that from this equation.

Is the right side of that equation positive or negative? If it’s positive a positive change in P must be accompanied by a positive change in T. If the right hand side is negative, the reverse would be true.

If the right side is positive, change in P accompanies a change in T. Right, but then is this T the boiling point? What happens to the right hand side T then?

The correct name is the Clausius-Clapeyron equation.

What exactly is LHS ? Left Hand Side ( of the equation ) ?  Remember that saved words are often figuratively expensively paid by clarifications. LHS is not a common abbreviation.

@Poutnik Ok got it. Also, do you have something to contribute towards answering my question?

No resources for now.


text:
How to recycle Lithium-Polymer battery?
I can't manage to find any information about the recycling process and the chemistry behind it. The only thing I see everywhere on the internet - is a guide like this:

Step 1 - Discharge the battery.
Step 2 - Store at room temperature.
End of the article.

I have a DELL XPS15 battery, slightly swollen. And when I asked local repair service about recycling - they said: "nobody recycling Li-Pol here" (in my city).
So I decided to investigate this topic. Any idea where I can start?

I'm not too sure about lithium-polymer batteries unfortunately but I've been dong a lot of reading on lithium ion battery recycling and one of the reasons they have such a poor recycling rate is because they have a lot of different components and it's not usually standardised between manufacturers. Meanwhile lead-acid batteries have a recycling rate of I think 99% in the US since the components are more simple. There's a lot of research on it at the moment though! I'd be cautious about discharging the battery yourself since depending on the electrolyte it can cause the release of HF. If you li

I'll be careful. The Battery itself works (was) fine. I noticed the problem in the early stage, so I guess I can discharge it with my laptop. Anyway, seems the process is not very easy for "make it home", so I most likely will send this battery to the factory in Romania. (closest one) But still very curious about the recycling process.

Lithium ion batteries are pretty poorly recycled as is and so when I was looking through literature reviews I found only one mention of lithium ion polymer batteries being recycled. The paper that was using this method was also recycling lithium ion batteries the same way though so it wasn't clear if anything different had to be done.
The main thing that seems to be different with lithium ion polymer batteries is just the electrolyte. The electrolyte is often not recycled anyway since the main focus is typically on recycling the cathode because it contains cobalt and/or nickel and is profitable to recycle.
If you want a more detailed overview of the different processes lithium ion batteries are recycled this review is open access and explains the ways different companies currently recycle lithium ion batteries:
https://www.mdpi.com/2313-0105/5/4/68
There's no mention of lithium polymer batteries in the review but this patent mentions recycling lithium ion and lithium polymer batteries together so I assume some of the same methods can be used:
https://patents.google.com/patent/US8979006B2/en
There's no standard method for recycling lithium ion batteries but usually methods involve mechanical, pyrometallurgical and/or hydrometallurgical processing. Usually the batteries are first discharged and then shredded or dismantled.
If mechanical processing is carried out it usually relies on different physical properties in the materials that allow it to separate out components. For example, the electrodes are a fine powder so if you put the shredded battery through a sieve you can separate the electrodes from the other materials. The cathode and anode can also be separated using froth filtration since they have different hydrophobicities. This can allow the cathode to be isolated and further processed to obtain metals like cobalt if the battery has a lithium cobalt oxide cathode. The review goes into more detail about the other mechanical steps that can be carried out but there's quite a lot of them.
Pyrometallurgical processes involve putting the battery in a furnace and you end up with an alloy of metals (containing cobalt, nickel and iron usually) and a slag (containing aluminium, lithium, manganese and some other metals). The metals in the alloy can then be extracted.
Hydrometallurgical processes involve using solvents to leach metals. This is generally combined with other methods like mechanical processing so the battery components you have are slightly more homogenous and easier to process. Sulfuric acid is the most common acid used to leach cobalt from lithium cobalt oxide. The leached metals can be precipitated out and recovered.
It's a bit difficult to clearly explain how lithium ion batteries are recycled since it does seem like every company does it differently. This just mentions some of the processes that are being used at the moment but there's lots more as well that are being researched. I know this doesn't clearly mention how lithium polymer batteries are recycled but I hope it's of some help anyway!

Thank you for your detailed answer, @Liv! I got the main point - it's much more complicated than it might seem and not a good idea attempting to do it home. I hope the industry will go forward and batterie's production will become more waste-free soon. I'll go through the links - it looks very interesting topic. Thank you very much!


text:
I need to reduce the Nitrile group without affecting the boronic acid. How to do that?
I need to Reduce the Nitrile group (CN) in the (3-cyanophenyl)boronic acid in to a primary amine without disturbing the Boronic acid group.Using LAH can trigger reduction of Boronic group. what are my options?

Options : https://en.m.wikipedia.org/wiki/Nitrile_reduction

The answer and  comments to this question are relevent https://chemistry.stackexchange.com/questions/74580/what-is-an-easy-way-to-reduce-a-boronic-acid-to-borane

SmI2 looks a possibility  Org. Lett., 2014, 16, 1092-1095.

Also could protect the boronate, perhaps with MIDA?


text:
are the exact and kohn-sham electron densities totally symmetric?
For any molecule with open or closed shells considering the electronic state to be the ground state:

Is the exact electron density totally symmetric?
is the Kohn-Sham electron density totally symmetric?

(spatially upon symmetry operations of the molecular point group).
I refer to "exact" as the density built using the wave function of the molecular Hamiltonian in the Born-Oppenheimer approximation. And I refer as the KS electron density as the one built by a single Slater determinant solution of the KS equations with the exact exchange-correlation functional or approximations. In principle due to the Hogenberg-Kohn theorems these two electron densities should be the same, nevertheless it seems that they cannot be in certain cases. This is why I distinguish the two densities.

Symmetric with respect to what (i.e. symmetric with respect to permuting electrons or spatially symmetric with respect to a group operation)? Or do you mean the density matrix itself?

I mean, totally symmetric with respect to symmetry operations of the point group of the molecule

Yes, if the molecular geometry is in a point group, the electron density will transform as one of the irreducible representations of the point group,  with a very few exceptions.  For the ground state, it will almost always be the simplest irreducible representation (the A1 representation for C2V, for instance.)  This representation is all 1's, so it retains the symmetry under group operations without multiplication by -1.


text:
What is difference between density of the equilibrium and vapour density of the mixture
While solving question on calculating degree of dissociation,  I came across this question:

At 627 degrees Celsius and 1 atm $\ce{SO_3}$  is partially dissociated into $\ce{SO2}$ and $\ce{O2}$ by the reaction;
$$\ce{SO_3 <=>  SO2 + 1/2O2}$$
the density of the equilibrium mixture is 0.925 gram per liter what is the degree of dissociation?"

In  answer, they calculate the the molecular mass of mixture at equilibrium(Mmix) by using density of equilibrium which comes 63.348 and we know molecular mass of $\ce{SO3}$ is 80 , vapor density of $\ce{SO3}$ will be 80 by 2 is equal to 40 ,vapour density of mixture is equal to molecular mass of mixture by 2 is equal to 68.348 by 2 is equal to 34.174 and from here we can calculate degree of dissociation by using the formula(x=D-d/(n-1)d ) but my doubt is what is the difference between vapor density of mixture and density of the equilibrium mixture?

I read your explanation of the answer a half dozen times and I can't follow it. However the gist of the problem is that the various gas and gas mixtures follow ideal gas law PV=nRT. Thus $\ce{SO3}$ is one molecule but $\ce{SO2 + 1/2O2}$ is 1.5 molecules. So let's invoke magic. At 627 C I have pure $\ce{SO3}$ at 1.00 atm. I now cast another spell and convert all of the $\ce{SO3}$ to $\ce{SO2 + 1/2O2}$. The volume, mass and density are all the same. But the number of molecules has gone up by 1.5 times. So the new pressure is 1.5 atm.

Please rewrite your calculations ! The value 63.348 suddenly becomes 68.348, as if 3 has been changed into 8 !

Prefer using symbolic algebra to literal numbers. You can easily derive the average molar mass from density via the ideal gas state equation. And then the equilibrium composition.

The two answers below can't both be right, but I can't figure out which is wrong and where the mistake was made.

Obviously we have to assume ideal gas behavior.
The gas constant is $\pu{0.082057 L⋅atm⋅K−1⋅mol−1}$. The molar volume of any gas at 900 K and 1 atm is:
$$V = \dfrac{nRT}{P} = \dfrac{\pu{1 mol}\times \pu{0.082057 L⋅atm⋅K−1⋅mol−1}\times \pu{900 K}}{\pu{1 atm}}= \pu{73.851 L}$$
The molecular mass of $\ce{SO3}$ is 80.066 g/mol, at 900 K and 1 atm so for pure $\ce{SO3}$, which is 0% decomposed, we'd expect a density $\mathrm{D_0}$ of:
$$\mathrm{D_0} = \dfrac{\pu{80.066}}{73.851} = \pu{1.0842 g/L}$$
Now let's consider a mixture of 2/3 $\ce{SO2}$ and 1/3 $\ce{O2}$ where the $\ce{SO3}$ is 100% decomposed. The volume of a mixture of $6.023\times10^{23}$ molecules is still the same, but the density of the mixture, $\mathrm{D_{100}}$ , is different. First we need to calculate the average molecular mass.
$$\mathrm{M_{100}}= \dfrac{2}{3}\times 64.066 + \dfrac{1}{3}\times 15.999 = 48.044$$
$$\pu{D_{100}} = \dfrac{\pu{48.044}}{73.851} = \pu{0.6506 g/L}$$
So at 0% dissociation the density is $\pu{1.0842 g/L}$ and at 100% dissociation the density is $\pu{0.6506 g/L}$. It is linear in between. Let $x$ be the % dissociation. Then:
$$\mathrm{D_x} = 1.0842 - \dfrac{1.0842 - 0.6506}{100}\times x = 1.0842 - 0.004336x$$
For a density of 0.925
$$ 0.925 = 1.0842 - 0.004336x$$
$$\therefore x = 36.7\%$$

Let's start from $m = 0.925 g˙\ce{SO3}$ that will be heated at $900 K$. This is the amount of $\ce{SO3}$ that will occupy a volume of $1.00 L$ after heating it at $900 K$ at $1 atm$. and after partial decomposition. In moles it is also  $n_o$ moles $\ce{SO3}$,  $$\ce{ $n_o = \frac{m}{80 g/mol} = \frac{0.925 g}{80 g/mol} = 0.01156 mol$}$$ At $900 K$,  a fraction $\alpha$ of these moles are decomposed into $\alpha$ moles $\ce{SO2}$ plus $\alpha/2 $ moles $\ce{O2}$. Then the total number of moles at equilibrium will be : $$\ce{$n$ = $n_o$˙($1$ - $\alpha$ + $\alpha$ +    \alpha/2}) = n_o(1 + \alpha/2) = 0.01156 mol (1 + \alpha /2)$$ The total volume is : $$\ce{$V = nRT/p= 0.01156 mol$(1 + \alpha/2)$RT/p = 0.001 m^3$}$$  Simple rearrangement of this formula gives : $$\ce{1 + \alpha /2  = \frac{0.001 $m^3·p$}{0.01156 $mol·RT$} = \frac{0.001 $m^3$·101325 $Pa$}{0.01156 $mol$·8.314 $J mol^{-1}K^{-1}·900 K$} = 1.172 }$$
The degree of dissociation $\alpha$ is equal to $\alpha = 2 ·0.172 = 34.4$ %

0.925 is density of equilibrium mixture , not mass of SO3..

$0.925 g$ is the mass of the $\ce{SO3}$ that will occupy $V = 1 L$  at $900 K$ and $1 atm$. after partial decomposition.

Ok , got your logic and answer is also close to given answer 34% ,so density of equilibrium mixture means density of SO3 after partial decomposition?, can we like this..

@Rover. OK. So please remove the down point you gave me one hour ago !

that wasn't me.. and can we conclude that  eq mix means density of products after partial decomposition ?

@Rover. Yes. Density of equilibrium mixture is identical to density of products after partial decomposition. Equilibrium is obtained after partial decomposition.

M_avg = rho . RT/p, and from that is easily calculated equilibrium composition.


text:
Is "Hydrogen Bonding" coordinate covalent bond?
My textbook says:

The Oxygen atom of other molecules links to form coordinate covalent
bond with H-atom using lone pair.

Now I am confused. Is H-bonding coordinate covalent bond?

Please understand that just stating random paragraphs from your textbook will confuse us as well. We need to understand more than you've mentioned in your question. Consider adding a picture of the text that has led you to ask this question. Also give examples of the 'other molecules'. You could be knowing the context of the question but we here on the other side feel like we are searching for a needle in the grass!

H-bonding happens when a hydrogen attached to a electronegative atom like F,O,N gets electrostatically attracted to ( in this case I assume ) oxygen with lone pair.

Hydrogen-bond is not a coordinate covalent bond. it is an electrostatic attraction. The sentence from your textbook describes the third bond happening around the Oxygen atom in ions like H3O+, which is not a Hydrogen-bond.

The electrostatically attracted might not be the most clarifying description.

"In a way yes". However, rather than starting from nomenclature to get facts, do the vice versa. Otherwise and again "in a way", they could be seen as permanent dipole-dipole interactions, too.
Hydrogen bonds have energy of about a tenth as compared to covalent bonds. So they are considerably weaker but still stronger than other intermolecular interactions.
Thus, they have a special, energetically intermediate, status that was given a name.
How much one goes into quantum mechanical description (the partially donated lone pair is in a deformed orbital between the X and the H)* or if one stops at considering a pure electrostatical attraction between partial and essentially point-like charges as done in not too advanced courses I do not know in details. I guess it depends on how deep one want to go with the description.
But the experimental facts are as I have described above.
*https://www.tandfonline.com/doi/full/10.1080/23746149.2018.1428915 is an example.


text:
Lack of elaboration in IR spectra of compounds
I've recently started learning about IR spectroscopy, and a common trend I'm seeing in my textbook and other sources is that not all the features in the IR spectrum are explained in detail. The focus is primarily on features whose results are known, and there is quite a bit of backward reasoning in deriving the structures of compounds from their IR spectra.
Consider the IR spectrum for Decane:

Here, what are the features causing absorption in the $\pu{720 cm^{-1}}$ range? Morrison Boyd states that C-C stretching causes absorption in the $\pu{1500-1600 cm^{-1}}$ range, and the absorption in the $\pu{3000 cm^{-1}}$ is explained in the figure, but I am unable to explain the other absorption.
Another example is the IR spectrum for Butanamine:

Here, what is causing the large absorption in the sub-$\pu{1000 cm^{-1}}$ region? Wouldn't such a large absorption at this frequency indicate unsaturation? How are we deducing then that it is an alkane?
The lack of explanation is a common feature for a lot of IR spectra presented in the text. Is this normal for someone learning IR spectra at this level? Also, any suggestions for other references for learning about IR/NMR spectra in more detail are appreciated.

Hesse, Meier, Zeeh is the standard OC spectroscopy textbook.

https://chemistry.stackexchange.com/questions/60019/why-is-the-segment-of-the-ir-spectrum-below-1500-wavenumbers-called-the-fingerpr

Spectroscopic Methods in Organic Chemistry by Williams & Fleming is a good textbook for this https://www.springer.com/gp/book/9783030182519

Please remember that IR spectroscopy is just a tool. It is mostly used for identifying important functional groups. Hence, instead of analysing the fingerprint region, which is tricky, we could simply use other methods, such as HNMR (which could be explained in depth).

In IR, wavenumbers below 1500/cm are called the "fingerprint" region. It's very tricky to assign anything there, because the transitions don't relate to specific bonds, but come from larger structures.
If you look deeper into the literature, you will surely find an explanation for that big hump at 800something inverse centimeters.
Generally, two substances with matching fingerprint region in the IR can safely be regarded as identical, but you cannot draw conclusions about the molecular structure. Hence the name.


text:
What would be the identification reactions of selenate(IV) ion?
I have not been able to find any $\ce{SeO3^2-}$ identification reactions anywhere. Are there any?

Can you be more specific? Classic German AC handbook Lehrbuch der analytischen und präparativen anorganischen Chemie by Jander and Blasius lists six different methods (ten if counting various reductants). Where are you determining selenate(IV) content, exactly?

You can use reducing agents to reduce the $\ce{SeO3^2-}$ salt to elemental selenium. This paper1 discusses the use of iron(II) salts in acidic medium (phosphoric - hydrochloric acid) as reducing agent. The reaction proceeds at r.t.
You can use other reducing agents like hydrochloric acid, sulfur dioxide, hydroxylamine hydrochloride, hydrazine hydrochloride. The reduced selenium has a red color which turns greyish-warm on warming, but be careful on overheating (boiling or evaporating turns to serious losses of selenium by forming volatile $\ce{SeCl4}$).
References

Rajkumar Kalaparthi , Srija Korapu , Padmarao Chekuri, 2020, New Spectrophotometric Determination of Sodium Selenate [Selenium (IV)] and Sodium Selenite[Selenium(VI)] with Iron(II), and Their Quantitative Analysis of Selenium (IV) – Selenium (VI) Present in A Binary Mixtures, INTERNATIONAL JOURNAL OF ENGINEERING RESEARCH & TECHNOLOGY (IJERT) Volume 09, Issue 05 (May 2020), DOI: 10.17577/IJERTV9IS050094
Vogel, A. I., & Svehla, G. (1996). Vogel's qualitative inorganic analysis


text:
Equilibrium with an E2 step
E2 mechanisms are known to be irreversible in general, because the forward step takes 2 species and produces three species, and a trimolecular reverse step is unlikely, purely on entropic grounds. (As mentioned in this post.)
Now does that mean that any reaction that is reversible cannot have an E2 step, on the grounds of microscopic reversibility?
Particularly, I want to ask about this hypothetical mechanism, of an isomerisation:

I am not sure how to draw its mechanism. Clearly, some elimination reaction is taking place: it can be E1, E2 or E1cB. Because this is an alkene compound, E1 seems unlikely, only E2 and E1cB are possible. $\ce{MeS-}$ seems like a bad base for both E1cB and E2, so I can't say which one is better. Can I remove E2 from the options purely because the reaction is reversible?
[Note: This is an exercise question which asks me to draw the best possible mechanism if the intermediate detected was the one in the picture. It does not matter if the actual reaction happens in that way or not, the question is hypothetical]

Note also that the entropy argument might not apply if you were, say, in methanethiol as a solvent. Well, maybe not methanethiol exactly since it has a very low boiling point.

I think your thought process is correct, but you can push it further.
The way I would solve this problem is to think about how you could go from the alkyne to either alkene.
Then, via microscopic reversibility, you would argue that the reverse mechanism--the elimination--is very similar.

 The opposite of conjugate addition + protonation would be most similar to E1cb.


text:
Epoxide opening mechanism
Consider the following reaction:

Our interest is the second step. Notice that after protonation of oxygen, the water molecule attacks in an S$_{\text{N}}2$ fashion.
My question is, why does this reaction follow S$_{\text{N}}2$ mechanism? A protonated oxygen is an excellent leaving group. Therefore it must result in a carbocation formation. In other words, the reaction must follow S$_{\text{N}}1$ mechanism.
What am I missing here?

Source: Chapter-9, Organic chemistry by David Klein

"Must" is a very strong word. Why can't a good leaving group take part in an SN2 reaction? Surely you are aware that iodides react faster than chlorides in SN2 reactions, precisely because iodide is a better leaving group than chloride. So leaving group ability clearly affects both SN1 and SN2 reactions.

@orthocresol I understand. My intuition here was that $\ce{R-O+}$ doesnt require a water molecule to initiate the opening, thereby forming a carbocation on its own

So I guess this is a purely experimental determination?

The main difference here is that because the leaving group is tied to the electrophile, even if it does leave, it snaps right back. Also, in general, you frequently require coaxing to get an SN1 reaction to take place. In other words, you need to drive that reaction forward. The electrophile doesn't just pop off that easily.

@Zhe Im not sure why thats the case. In my opinion the oxygen wants to get rid of the $+$ charge ASAP. Why would it join back again? Am I missing something?

Would you rather have that positive charge on an oxygen or a carbocation?

The facts dictate that the mechanism is SN2. The product is a trans-diol.


text:
Why doesn't fluorine harm the ozone layer the way chlorine and bromine do?
I don't understand the chemistry here: fluorine is a halogen too.
I read that fluorine in the atmosphere readily forms HF, which is (somewhat) stable and doesn't catalytically break down ozone.
But, shouldn't loose chlorine in the atmosphere form HCl similarly?

Have a look at the well known chemistry of ozone depletion. If you replace the chlorine atoms in there with fluorine, you get compounds which are either more or far less stable. And HF is a very stable molecule in the gas phase. There is no liquid water in the upper stratosphere.

https://chemistry.stackexchange.com/q/14230/102629

Karl (+1): Your brief comment is actually quite accurate upon my further research. While F2 and F- do not readily lend themselves to radicalization (apparently the need for VUV light and .H presence), once activated highly active radicals are formed leading to stable compounds.

TL;DR: The mechanism is different for fluorine and chlorine atoms. Fluorine atoms react and forms stable HF molecules while chlorine atoms turns into a radical by the action of UV which helps in the destruction of ozone.
Long answer:
Ozone depleting substance(ODS) are gases that take part in ozone depletion process. Most of the ODS are primarily chlorofluorocarbons(CFC) and halons that contain chlorine and bromine atoms respectively which reach the stratosphere and reacts with ozone and lead to ozone depletion. In this context, the chemical component responsible are chlorine radical (Cl·) and bromine radical (Br·). The ODS travel to the stratosphere without being destroyed in the troposphere due to their low reactivity and once in the stratosphere, the Cl and Br atoms are released from the parent compounds by the action of ultraviolet light, e.g.
$$\ce{CFCl3 ->[UV] Cl· + ·CFCl2}$$
Ozone is a highly reactive molecule that easily reduces to the more stable oxygen form by the catalytic action from the halogen radicals. Cl and Br atoms helps in destruction of ozone molecules through a variety of complex cycles. A simple cycle is as follows:
$$\ce{Cl· + O3 → ClO + O2}$$
$$\ce{ClO + O3 → Cl· + 2 O2}$$
Chlorine radical is released and the process is continued and the overall effect result in ozone depletion. Halon gases contains bromine and the process is similar to that of what CFCs does but the action is more potent and is more efficient in this ozone depletion process.

What about fluorine and iodine?

CFCs also contains fluorine atoms which gets released along with chlorine and bromine on photodissociation in the stratosphere. But this fluorine by itself does not contribute to ozone depletion. Fluorine atoms released are quickly sequestered into carbonyl compounds and subsequently into hydrogen fluoride by reacting with water and methane molecules which is very stable in the stratosphere. As such, CFCs are now being replaced with Hydrofluorocarbons (HFCs) according to Montreal protocol (but they do contribute to global warming).
Iodine containing organic molecules react so rapidly in the lower atmosphere that they do not reach the stratosphere in significant quantities and does not play role in ozone depletion.
References

https://en.wikipedia.org/wiki/Ozone_depletion
https://csl.noaa.gov/assessments/ozone/2010/twentyquestions/Q7.pdf
https://hal.archives-ouvertes.fr/hal-00256296/document

Nice writeup, however, it technically despite its length (as compared to my answer, which was 1st presented with virtually the same argument) my reading notes the implicit citation of only UV light as the agent radicalizing chlorine, which technically does not impact fluorine (only VUV as my answer and sources notes in further conjunction with the hydrogen atom radical). The question relates to halogen differences.

Also, you imply that F2 does not form similarly an active atomic fluorine radical, that is wrong. Read my source which, in a reaction chain, clearly cites the $\ce{.F}$ radical. As having presented a competing (actually correct) answer, I am, per professional courtesy only, not downgrading your response.

Here is a depiction of how 'chlorine' destroys ozone per an education source:

CFC molecules are made up of chlorine, fluorine and carbon atoms and are extremely stable. This extreme stability allows CFC's to slowly make their way into the stratosphere (most molecules decompose before they can cross into the stratosphere from the troposphere). This prolonged life in the atmosphere allows them to reach great altitudes where photons are more energetic. When the CFC's come into contact with these high energy photons, their individual components are freed from the whole. The following reaction displays how Cl atoms have an ozone destroying cycle:

$\ce{.Cl + O3 → .ClO + O2 (step 1)}$

$\ce{.ClO +O. → .Cl + O2(step 2) }$

$\ce{O3 + O. → 2 O2(Overall reaction)}$

Chlorine is able to destroy so much of the ozone because it acts as a catalyst.

Note, the reaction cycle depends on the presence of chlorine as the chlorine atom radical.
However, with fluorine an apparent difference in the degree of relative stability as to quote a source work: Effect of VUV radiation and reactive hydrogen atoms on depletion of fluorine from polytetrafluoroethylene surface:

• Already only VUV radiation itself can cause depletion of fluorine.

• The strongest F depletion observed after a synergistic exposure to VUV and H atoms.

• Only H atoms without the help of VUV were not effective enough.

So, unlike chlorine and bromine, where UV radiation, itself, is apparently sufficient to induce $\ce{.Cl}$ and $\ce{.Br}$ formation, the stability of fluorine and its tendency to create stable $\ce{HF}$, for example, is likely the explanation as to why it is not usually cited in a corresponding ozone depletion cycle.


text:
Why are halogens base sensitive groups (Wolff-Kishner Reduction)?
So while studying about Wolff-Kishner reduction reaction to form alkanes from carbonyl groups. I read that it is not suitable for base sensitive groups as the environment is highly basic and as an example the following reaction was given:

Now the desired product that should've formed is

But this doesn't happen as -Cl is a base sensitive group and so reacts with KOH. But how can a halogen be base sensitive when it already  has 7 e- and is highly electronegative? How is it possible for it to react with OH-?
Also, I don't understand the reaction mechanism of this reaction so can someone please explain how this takes place?

Have you studied the E2 mechanism? https://en.wikipedia.org/wiki/Elimination_reaction

I only have a vague idea about it and haven't yet studied it. So can you please elaborate how it is relevant here and what is the reaction mechanism of this reaction, perhaps in an answer instead of a comment?

The chloride can also displace directly with hydrazine

You've probably realised already that a possible side reaction is the SN2 nucleophilic substitution of Cl by HO-, which would look like this:

Evidently, this is not the product we actually get. Hydroxide ions are considered 'hard' nucleophiles, as they have a high negative charge density, and they tend to react best with 'hard' electrophiles with high positive charge density. The electrophilic carbon here is reasonably 'soft' whilst the proton adjacent to it is considered 'hard' due to its small size (you can learn more about this here). As a result, hydroxide ions like to act as bases instead of nucleophiles, and the mechanism favoured is not the SN2 shown above, but in fact the competing E2 elimination shown below.

This proton may not look particularly acidic, but formation of a new alkene alongside loss of a reasonably stable chloride leaving group drives the deprotonation.
Another contributing factor is that whilst the substitution has a reasonably small change in entropy, the elimination results in a greater number of species than in the reactants, so has a large positive entropy. This is relevant because Wolff-Kishner reduction is usually done at a reasonably high temperature, and entropy becomes a greater influence on the spontaneity of a mechanism as the temperature increases. This further favours E2 over SN2.
The chloride is considered a base-sensitive group for the above reasons, and this second pathway is inevitable alongside a Wolff-Kishner reduction in a highly basic solution.

I think the reaction as you told will be very inefficient. To explain why, I quote the paragraph from page-905 of David R Klein's Organic Chemistry (*):

"The pH of the solution is an important consideration during imine formation, with the reaction being greatest when the pH is around (4.5). If the pH is too high (i.e: no acid catalyst is used) , the carbinolamine is not protonated, so the reaction occurs more slowly. If the pH is too low, most of the amine molecules will be protonated to give ammonium ions, which are not nucleophilic. Under these conditions, step 2 of the mechanism occurs too slowly. As a result, care must be taken to ensure optimal pH of the solution during imine formation"

Though the above paragraph concerns imine formation, it applys equally well to hydrazone formation which occurs as an intermediate in the standard scheme of wolf Kishner reduction of ketones.
The mechanism of the reduction can be found here

Formation of hydrazones are irreversible, unlike imines. That is one reason why they are used as derivatizing agents (2,4-DNP's, semicarbazones, etc.) The Wolff-Kishner is conducted under basic conditions, not pH 4.5.

Hmm but you had written in this answer that it's irreversible. Correct me if I misread anything.


text:
Where can I find literature values for standard molar enthalpies of neutralization?
I'm looking for enthalpy of neutralization literature values for inorganic weak acids. NIST, PubChem, etc. don't list them. Is there somewhere else I should look?

I suspect this data might be somewhere in JANAF thermochemical tables.

Doesn't look like it. Everything there is formation.

Try to discover the molar enthalpies of formation of 1) the ion H+ (or H3O+), 2) the ion OH-, and 3)  the liquid water. Take this last value, subtract first the enthalpy of formation of H+ and then the enthalpy of formation of OH-. And here you are. This gives you the molar enthalpiy of neutralization acid + base in solution. OK ?


text:
Criterion for gravimetric analysis of silver

If the only criterion for gravimetric analysis of silver is solubility of  precipitate, which one of the following two compound are we supposed to choose? $K_\mathrm{s}(\ce{AgCl}) = \pu{1.78E-10}$ and $K_\mathrm{s}(\ce{Ag2CO3}) = \pu{8.13E-12}.$
a) $\ce{AgCl}$
b) $\ce{Ag2CO3}$
c) both of them are suitable
d) none of them

As I learned from textbook, one of the criteria for choosing the right one is: $K_\mathrm{s} < 10^{-7}$—thus I eliminate option d.
I also read that it is meaningless to compare the solubilities of two salts having different formulas on the basis of their $K_\mathrm{s}$ values. Therefore, I would choose option c—both have $K_\mathrm{s} < 10^{-7}.$ But from experience I found out that $\ce{AgCl}$ is used way more often—but that is probability because of other criteria—which I do not have to bear in mind in this question.
Is my reasoning correct?

The solubility of $\ce{AgCl}$ is equal to $\sqrt{K_\mathrm{s}} = \pu{1.3E-5 M}.$
The solubility $s$ of $\ce{Ag2CO3}$ is such that  $K_\mathrm{s} = 4s^3.$ So that its solubility $s$ is equal to $s = \pu{1.2E-4 M}.$ This is ten times more than the solubility of $\ce{AgCl}.$
For gravimetric purposes, $\ce{AgCl}$ is a better choice.

Be aware that in the solubility comparison context, solubility product constants can be directly compared for compounds with the same number of ions created from the formula, where a greater solubility product means a greater solubility.
For compounds with different ion counts, one has to compare ( molar ) solubilities in $\pu{[mol/L]}$ , calculated from solubility products as $K_\mathrm{sp}^{\frac 1n},$ where $n$ is number of ions. $n=2$ for $\ce{AgCl}$, $n=3$ for $\ce{Ag2CO3}$.
So $K_\mathrm{sp,\ce{AgCl}}^{\frac 12}$ versus $K_\mathrm{sp, \ce{Ag2CO3}}^{\frac 13}$

That is a nice short cut for comparing solubilities.


text:
Safe to wash weighing paper?
I accidently washed the Fisherbrand weighing paper (cat. 0989812A) with acetonitrile : water (1:8) to remove some sample. The weighing paper appeared to deform a little bit after the wash. I wonder if ACN:H2O mixture dissolved anything from the paper or the paper deformed just because it absorbed some water?

"Rinse" might be a more appropiate word than "wash" here.

Why on earth do you care about the paper? Do you want to re-use it?

@Karl, Rinsing or so-called washing the weighing boat, or weighing funnel is a standard procedure for quantitative transfer to the receiving vessel. He must have applied the same logic on the weighing paper.

The response to your concern depends on the application. If it were just an academic exercise, forget about it, since you may have already done the experiments. However, the point to keep in mind is that paper based weighing materials should not be rinsed or washed for standard preparation until and until unless it is specifically allowed to do so. Paper can have leachable materials which can contaminate the standard- this applies very well to trace analysis. If you note, Fisherbrand specifically mentions "low nitrogen" content.
The second reason why the paper based weighing paper should not be rinsed is that the bottom of the paper has been in contact with the balance pan (which are often horribly dirty in academic labs due to misuse). Since that paper is thin and permeable to solvents, rinsing the paper is akin to washing the dirty pan into your standard solution rather indirectly. Of course, balances in a professional or research settings are better maintained.
Sometimes it appears that some solids particles are still sticking to the weighing paper after transfer process. In such cases, one uses the "weighing by difference" method. You know the weight of the paper+solid, after transferring, you would weigh the paper again, and record the actual weight transferred to the flask or other container. No washing is required.
Analytical work, where extreme care is required, paper is not used at all for standard preparation. One should use "weighing funnels" which are commercially available and there is no concern in washing or leaching or soluble materials.


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Calculation of Gibbs free energy of a chemical reaction
So, I'm trying to derive the equation: $$\Delta G = \Delta G^o + RT \ln Q\tag{1}$$
And trying to follow the steps detailed on this website here. I am having trouble deriving equation 7, which tries to sum chemical potentials. Previously, it derives the change in chemical potential, $\Delta \mu _i$ when the reaction is poised away from equilibrium, and for the reaction $A + B \rightarrow C + D$, four equations, each for $A, B, C$ and $D$ respectively.
$$\Delta \mu _A = \mu_A' -\mu_A = RT \ln([A']/[A])\tag{2}$$
Where $[A']$ is the activity away from equilibrium. Afterwards, it proceeds by:
$$\Sigma\Delta\mu_i = RT \ln\left(\frac{[C'][D'][A][B]}{[C][D][A'][B']}\right)\tag{3}$$
My confusion is regarding the arrangement of the activities in the $\ln()$ function. I assumed that:
$$\Sigma\Delta\mu_i = \Delta\mu_A + \Delta\mu_B + \Delta\mu_C + \Delta\mu_D = RT \ln\left(\frac{[A'][B'][C'][D']}{[A][B][C][D]}\right)\tag{4}$$
It says that we assume the concentration of the products increases by convention, but I don't get it. Even if the concentration of the reactants decrease and the products increase, the change in sign will be due to the $\ln()$ function. If you flip the activities of the reactants around, it doesn't give the net change in Gibb's free energy anymore.
If anyone can help explain this, I will be very grateful.
P.S. Sorry if the question seemed low effort, but the context is mostly given in the website, and I can't find an explanation anywhere :(

Equation (4) lacks the stoichiometric coefficients, which are taken as negative for reactants (because they are used up, not made). Equation (9) of the source cited shows the correct relationship between Gibbs energy of reaction and chemical potentials.

It seems to me, it goes like this:
$$\mu_A=\mu_A^0+RT\ln{[A]}$$
$$\mu'_{A}=\mu_{A}^0+RT\ln{[A']}$$
So, $$\Delta G=\mu'_C+\mu'_D-\mu'_A-\mu'_B=\mu_{C}^0+RT\ln{[C']}+\mu_{D}^0+RT\ln{[D']}-\mu_{A}^0-RT\ln{[A']}-\mu_{B}^0-RT\ln{[B']}
=\Delta G^0+RT\ln{Q}$$where $$Q=\frac{[C'][D']}{[A'][B']}$$
But, $$\Delta G^0=-RT\ln{K}$$where
$$K=\frac{[C][D]}{[A][B]}$$


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How to model CRISPR/Cas9 binding and cleavage using a generalized Kittel's zipper model?
I am thinking about how to model the binding of a CRISPR/Cas9+guide-RNA complex to a double-stranded DNA that cause its cleavage.
This post describes four steps at which a CRISPR/Cas9 system binds and cuts a target DNA strand:

PAM Binding:
Phosphate Lock Loop
Guide RNA binding
Cleavage

I believe it is possible to develop a fairly simple model based on Kittel's zipper model or its generalization to estimate the propensity of a Cas9 and gRNA to cleave the DNA.
Thus, I would appreciate some advice how to develop such a zipper-like model (e.g., maybe by having a double zipper).
Ultimately, I would like to compute the free energy of the different states based on the sgRNA sequence: $\Delta G_\text{bind}$, $\Delta G_\text{melt}$, $\Delta G_\text{cut}$.
Here is another schematic illustration by Addgene (in this illustration, the binding step includes the DNA melting and the binding of the RNA with the DNA):


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SMART string for alkyl group excluding alcohols
I'm quite new to the usage of SMARTS strings. I would like to find all alkyl groups (but no $\ce{CH3})$ in a component, here e.g. cyclohexanol (C1CCC(CC1)O).
My SMARTS string is: [CX4;H0,H1,H2]
However, this matches every carbon. I would like to write my smart string, so that C:3 is not matched, because it is bound to something else than a carbon (here: oxygen). How is this possible with SMARTS?

I think the question is on topic here, but I guess it would get better attention and answers at [mattermodeling.se]. I suggest migrating it.

How to find carbon

with four connections
not -CH3
and only connected to carbon

As in your SMILES there are only implicit hydrogen, you do not need to count Hs.
Count the explicit bonds and exclude the carbon with one explicit bond to avoid -CH3.
[CX4;D2,D3,D4]
To find the carbons that are connected to others than carbon
C[!C]
Now you can use recursive SMARTS, and set the second pattern to NOT.
[CX4;D2,D3,D4;!$(C[!C])]
Now only carbons would be found that match both patterns.
CX4   C with total connections of 4
D<n>  number of explicit bonds
$()   recursive SMARTS
!     NOT
;     AND
,     OR
In cyclohexanol it will find

and here is a dummy molecule for testing it.
CCC(F)C=CCC1CCC(O)C(C(C)(C)C)C1

I presume you mean cyclohexanol, not cyclohexanone.

@z1273 Changed -one to -ol. Thank you.

Thank you very much for the detailed answer!


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What factors promote honey's crystallisation?
Honey is a complex mixture containing, probably, more than hundred compounds.
I understand that the % of those components will affect the crystallisation.
However, I am wondering what factors may speed up the crystallisation, apart from heat?

different question but related: Is honey really a supersaturated solution? Does heating to un-crystalize redissolve it or melt it?

The decisive factor is the plant origin of honey. The major components ( reportedly by some beekeeper books of my passed away father ) have ratio glucose : fructose about 1:1 to 1:1.7. Glucose reportedly crystallizes more readily, triggering fructose crystallization as well. So honey closer to 1:1 ratio has stronger tendency to crystalize.(**)
Aside of already mentioned factor of temperature, another factor is obviously the ripeness of honey. Lower water content in more matured honey means higher crystallization probability.
Seeding by already crystalline honey helps to trigger or speed up crystalization.
If crystallization is already happening, or if it is intended and desired, then stirring speeds up crystallization.
Seeding and stirring leads to many small crystals, what is applied during production of the pasted honey. This form is favourite one among people. It often dissolves faster then liquid honey during drink sweatening. It is also very convenient for bread spreading, as its flow is almost frozen.
Slow spontaneous static crystallization leads to big crystals, forming mechanically tough product, requiring big force to separate a chunk of honey(*). The crystallization rate, the size of crystalsfraction of residual liquid phase and physical properties of the product seem to be composition dependent as well.
If such honey is wanted to be liquid again, it is recommended to carefully heat it up in $\pu{40 ^{\circ}C}$ warm water bath. High temperatures affect the present probiotics and enzymes. Note that honey is along fermented milk the major source of natural probiotics.

(*) My father stored honey in 4 L flasks for home-made pickles. When honey was crystallized, we often digged it up by table spoons or forks. But unless they were very strong, like heavy stainless steel ones from Soviet Union, they had tendency to bend.
(**) Infamous one is larch tree honey. It has so strong crystallization tendency it often crystalizes in the honeycomb. It is not being able to be directly centrifuged out without honeycomb destruction.


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Is mixing silicone oil with other types of oil safe?
I'm looking to lubricate some nylon gears in a servo motor, and online sources tell me silicone lubricants are one of the best ways to go.
However, when I opened up the servo, I found that it already has lubrication in place, most probably a mineral oil-based one. It looks dry though, and I'm inclined to re-lubricate it anyway.
Would having the two types of lubricant in there be a cause for concern? Should I exert effort to clean it out first?
Conversely, would there be any issues if I apply a mineral oil-based lubricant to gears that are known to have silicone oil?

So, either (1) clean the motor and regrease, or (2) determine from the data sheet what kind of lubricant they specify.

Different types of oil might not mix well. They might not be unsafe to mix, but the result might be worse lubrication.

I've erred on the side of caution and used multi-purpose synthetic lubricant. Their compatibility chart says that basically all plastics in my house are compatible. I've verified that the existing lube isn't silicone. I still wonder about the question though; please feel free to answer.


text:
What are the compounds in oxygen and carbon dioxide indicator solutions that gives their color?
My brother had some leftover oxygen and carbon dioxide indicator solution for one of his take-home experiments from school, so he let me have them. I want to use them for an experiment and write out some reactions, but I don't know what compound is in the indicators that gives them their color.
The oxygen indicator is green and comes with a color chart which says the following in regards to concentration of oxygen: 2ppm is red, 4ppm is purple, 6ppm is yellow, 8ppm is light green, 10ppm is dark green/teal, and 12ppm is either very dark green/blue or black.
The carbon dioxide indicator is orange and comes with a color chart saying low CO2 concentration gives a red color, medium is orange, and high is yellow.
A few websites said that you could make a bicarbonate indicator with thymol blue and cresol red, but I don't know how that makes sense if the color chart only ranges from red to yellow. As for the oxygen indicator, I have no idea what the compound(s) can be.
Thanks in advance!


text:
Why do electrons jump back after absorbing energy and moving to a higher energy level?
Electrons in a shell absorb energy and move to higher energy levels, but they release their energy and jump back to the shell they originally were in. Why do they jump back? Why can they not keep revolving around the nucleus?

If you throw a ball up in the air, why does it come back down?

I think we can explain this by saying that, the electron has its allocated place based on Aufbau principle and in any higher energy level the electron would be in an excited state and will lose its energy to jump to the lowest energy level possible.

This is a very difficult question to answer because it asks "why" for a fundamental process.

Similar question on Physics.SE: Why do electrons in an atom 'fall' back to the ground state?

Note that despite common misconception, electrons do not "revolve" around the nucleus.

It touches the fundamental question Why systems or objects have tendency to reach the lowest available energy level ?

@chrylis-cautiouslyoptimistic- what you say is about as much a misconception as saying the opposite. We simple don't know how to interpret quantum mechanics, therefore don't really understand the behaviour of quantum entities like electrons.

It makes sense that excited atoms in bulk de-excite: They are perturbed by those other atoms. So maybe we can refocus the question to ask: Why do isolated atoms de-excite? Do they also need to be perturbed (e.g., by vacuum perturbations)? And are there models that that do a good job of calculating the excited state lifetimes of isolated atoms? If so, then understanding those models will get you on the road towards understanding why isolated atoms de-excite. Related?:  https://mattermodeling.stackexchange.com/questions/4602/can-you-estimate-the-electronic-excited-state-lifetime-of-a-molecule

This is a very fundamental question and for really understanding the "why" some advanced physics is involved. I will describe the process rather superficially.
As you might know, the level energies of atoms and molecules can be calculated (in principle) using quantum mechanics. The simplest system is the hydrogen atom as it consists of a single proton and a single electron. Ignoring higher order effects (such as interactions of electron and nuclear spins and QED effects), the quantum mechanical calculation gives the same result as the Bohr model, that is, the level energies of hydrogen are given by the Balmer formula, which you probably know.
The calculation does not predict that the excited levels fall back to the ground state. An electron in an excited orbital will, according to this calculation, always stay in this orbital if nothing happens to the system.
Because we know that excited states decay, something must happen to the system to induce the decay. It turns out that in our calculation we have ignored the interaction of the atom with the photon field. Atoms can absorb light and emit light and we have completely neglected this. If you do take this coupling into account, you will find that excited states have a limited lifetime and that an excited electron will fall back to a lower level eventually.

Many things are only stable in their lowest energy state: electrons are no different
Hold a ball in your hand. It is, in effect, in an excited state. Open your hand and the ball falls to the floor, without much effort or any push. Set the ball on the floor and it doesn't move. It is in its lowest energy state and won't move around unless given a push.
Many things are like that in the world. A cone is stable resting on its base. But resting on its tip, the very slightest fluctuation will cause it to fall over. This is a little like an electron being in an excited state, as is the case with a ball.
An electron in a high energy state is (to leave out a whole bunch of complicated quantum stuff) like the cone or the ball. Mostly, the slightest nudge it gets will cause it to fall to the state where it can exist at the lowest possible energy level.

what provides the required nudge to an electron or what leaves the electron like we did with balls ?

@Ankit The important thing is that there are nudges, knowing exactly what they are is probably very complex. One (obvious) source is interactions with other atoms or photons, which are common and very very frequent. Another could be random quantum fluctuations (compare the question to "what causes a radioactive atom to decay" there really isn't a simple explanation.

There is a general heuristic in quantum physics, often referred to jokingly as the totalitarian principle, that everything not forbidden is compulsory. That is, any process that can occur will occur, with some rate, probability, or cross-section, provided that it doesn't violate any conservation laws.
An atom in an excited state can in most cases emit a photon without violating any conservation laws. (There are states like the 2s state of hydrogen that can't decay by emission of a single E1 photon because of conservation of parity.) Therefore we expect that this process will occur at some rate.

Matter can gain or lose energy in small, specific amounts called quanta. A quantum is the minimum amount of energy that can be gained or lost by an atom. For example, it might seem that you can add any amount of heat to water, but what is actually happening is that the water molecules absorb quanta of energy which increases the water's temperature in infinitesimal steps. Because these steps are so small, the temperature seems to rise in a continuous, rather than a stepwise manner.
Now, what does this have to do with energy levels and electrons? Well, the reason why matter can gain or lose energy only is small, specific amounts is because of the energy levels in the atoms. Presented by the equation here created by Planck: Quantum E = hv, (h is Plank's constant and v is the frequency) for a given frequency, matter can emit or absorb energy in whole number multiples, like legos. You can only take or add a whole lego, not a part of it. Therefore, quantities of energy between the whole number multiples do not exist, meaning there are fixed energy levels in an atom with quantized energy (because energy is absorbed through electrons in energy levels) and electrons can only exist within the energy levels, not between them. Each energy level has a higher quantized energy than the previous. So, to answer your question, when an electron absorbs energy (a specific amount of energy), the electrons moves up to higher energy levels because the electron has a higher quantized energy than its own energy level. Similarly, when the electron releases energy (a specific amount of energy), the electrons moves down to its original energy level because it has lost the absorbed energy and has the same quantized energy as its own energy level.
Hope I answered your question!

Hi @Jake, this is not an answer to the question, which is about the stability of excited states. You give a description of quantization of energy levels.


text:
Rate of change and kinetics order of reaction question
The rate of reaction is given by the following equation, where [Ker-S-S-Ker] is just the molecule being broken down and [RSH] is an abbreviation for thioglycolic acid
The thioglycolate is in large excess and so essentially remains unchanged.

Can somebody explain why the answer is C?
I thought the answer was E (because the rate of change, i.e. the derivative, is equal to the concentration of the reactant itself and so, I thought this signifies a decaying curve?)
Many thanks.

Indeed, the only thing you'll need is solving the first order differential equation (but before that, let me define $R_0 := \ce{[RSH]_0}$ and $C :=\ce{[Ker-S-S-Ker]}$.
Now, let's solve it assuming (that's what is given) that $R_0$ can be considered a constant.
Then, we have : $$\frac{\operatorname{d}C}{\operatorname{d}t} = -kR_0C \implies \frac{\operatorname{d}C}{\operatorname{d}t} + kR_0C = 0 \implies C(t) = \exp\left(-\frac{t}{kR_0}\right) + C_0$$ (integration constants can be determined using initial values, I won't be giving further details here on how to solve this...)
Hence, since the concentration of your molecule is an exponential function of time, it is inevitably a convex function (so only curves C and E can work).
Finally, a simple thought on the values of $k$ and $R_0$ show that curve E cannot be the answer as the decay is to "fast".
Hope this helps, if needed I can also provide a "variables separation" method of solving the differential equation.

Thanks @Thomas Prévost ! You lost me at some parts but I think I get it up to the point where you narrow it down to C and E. Why, however, is E 'too fast' to be logical? I don't understand the reasoning for that...Many thanks!

Well, as I don't have the whole exercise giving the precise context of the reaction, Isuppose that some information are given, or at least some clues explaining why the decay should be that slow. In fact, the "speed" of decrease is given initially by the value of $-\frac{1}{kR_0}$ (as it is the derivative of $C(t)$ for $t = 0$).


text:
Sensitivity vs. Limit of Detection of rapid antigen tests
I'm comparing a bunch of SARS-CoV2 rapid antigen tests:

Columns 4 and 6 list the values for sensitivity and limit of detection (LOD). How come that a test with a several times lower limit of detection can have a worse sensitivity?
As an example, consider tests #2 and #4:
#2 - Sensitivity: 97.7% - Limit of detection: 2.0*10^2.4 = 502 TCID50/ml
#4 - Sensitivity: 91.4% - Limit of detection: 2.5*10^1.8 = 158 TCID50/ml
Shouldn't these values correlate, i.e. the more sensitive a test is the smaller the amount it is still able to detect?

Detection limits depend on both sensitivity and the relevant noise levels. A short list of detection limit publications is here.

You have to ask the authors how did they determine the limit of detection. In the  German Standard DIN 32645, Nachweisgrenze has a specific protocol.

In this situation, "sensitivity" refers to how often the test is able to correctly detect the presence of analyte. A test with high sensitivity means that there are few false negatives.
Limit of detection refers to how much of the analyte must be present before it is considered positive.
Imagine two different tests: test A uses an antibody with very high affinity for a specific spike protein sequence. Because of the high affinity, it can detect the spike protein even when there's not much around (good limit of detection). However, the antibody is less robust against other sequences, so it sometimes reports a false negative in the presence of spike protein variants (poor sensitivity).
Conversely, test B has an antibody with medium affinity for several different spike protein variants. It relies on high levels of the protein being present, but it works with many different variants and rarely gives false negatives. This test would have a high (aka poor) limit of detection and high (aka good) sensitivity.

The usual definition of sensitivity in chemical analysis is that it is the slope of the response function or calibration curve, as relevant. It is also as you have given it, especially for ROC methodology. And, unfortunately, it is sometimes mistakenly used as detection “power”, whatever that is supposed to mean. Of course, the standards folks also weigh in. ;-)


text:
MgO as food supplement - solubility in water
A number of magnesium food / sports supplements contain it as MgO - which is practically unsoluble in water. The recommended quantity is somewhere around .3g / day, so not an insignificant amount.
How is that supposed to work? Does it rely on the pH shifts in the human gut? I've tried to check MgO solubility in water depending on pH, but the publication I've found is behind a paywall.
Edit: IvanNeretin has pointed out that MgO will react with stomach acid, not dissolve in it. I suppose that's what the manufacturers go for.

Are you absolutely sure it is magnesium oxide and not magnesium(II) content calculated/normalized for MgO?

@andselisk I'd like to link to some without advertising for them, so here's one that I just googled, no endorsement whatsoever: https://images-na.ssl-images-amazon.com/images/I/81c97Tz0qmL.AC_SL1500.jpg OR https://images-na.ssl-images-amazon.com/images/I/81tcFDPbR8L.AC_SL1500.jpg

Then again, it might be genuine MgO, because why not. Of course it will dissolve in the stomach.

@IvanNeretin Can you link to any data about MgO solubility depending on pH? Or otherwise whatever makes you sure about it?

Any entry-level textbook on chemistry will do. Basic oxides react with acids. MgO is a basic oxide. HCl is an acid. That's it.

@Zubo Well, it might as well be indeed magnesium oxide then. I was a bit skeptical since I happen to take Maxler's Magnesium B6, which lists magnesium lactate dihydrate as the source of magnesium as well as magnesium stearate as a filler component, I presume. Photo: https://i.stack.imgur.com/wlKwH.jpg (please ignore the scraps, I had to rip off the label with translation glued on top).

@IvanNeretin Yeah, that slipped my mind - I was only looking for solubility. I guess that would still depend whether or not there's actually enough of HCl, but I suppose that's what they were going for.

Using MgO and relying on gastric juice to do its job is not a reliable strategy. Oftentimes people might suffer from low acidity, and if I were manufacturing food supplements I would not take the risk of getting complaints about digestive problems caused by undissolved magnesia.

Then again, some people suffer from high acidity, and they may use MgO to lower it.

@IvanNeretin That's not how pharma works. They want maximum coverage for maximum profit with lowest possible risk of side effects. A soluble organic salt is a much cheaper solution (sorry for the tautology) for the business.

@andselisk If you want Mg inside, yes. If you want to lower acidity, no.

@IvanNeretin Sure, if you want a short-term laxative with a high chance of developing diarrhea:)

@andselisk Well, Maalox is a thing.

@IvanNeretin Maalox is not a thing since 2012.

@andselisk In Russia, it still is.

@IvanNeretin Ah, a good old Russian custom of treating patients with drugs prohibited in the West:)

@andselisk Yeah... In Soviet Russia, drugs test you.

I actually tried to find the approximate amount of HCl in a given human stomach, but all I can find is the acidity of gastric juice or its production rate, not actually how many equivalents are available at any time.

As a matter of fact, the tablet might actually skip the gastric acid (if, say, taken on empty stomach or whatever reason) and find itself in the intestine with high pH - I wonder if there would be any absorption after that.

MgO will react with stomach acid, not dissolve in it. I suppose that's what the manufacturers go for.

This is not factually right at all. Stating the metal content as an oxide is an century old tradition of quoting the metal content as oxides. Magnesium percentage quoted as percent MgO does not mean that the tablet actually contains magnesium oxide. The links OP provided in the comments shows that the tablet actually contains magnesium stearate. This is done to avoid upsetting the stomach because a very water soluble magnesium salt can induce diarrhea in a few hours.
Magnesium steareate is a salt of a fatty acid, which is rather water insoluble as expected but it has some fat solubility. So when you take it with food which has some fats along with this fatty magnesium salt, there must a be slow absorption of the fatty salt by the body hence the tummy is happy.
Basically, you are eating a magnesium soap!

The second link shows no stearate (which is precisely why I added it). I agree, there are supplements that basically mix every which salt possible - citrate, maleate, stearate, etc. But some contain only the oxide, hence my question.

If you can read the first sentence carefully, once again, you will learn that your question negates all the basic principles of chemistry. MgO is a basic oxide, it will dissolve in the stomach acid. It is simple acid-base chemistry.


text:
Does the electron in a neutral hydrogen atom ever exist in the p, d or f sublevels?
I learned that the electron in a neutral hydrogen atom can be excited to higher levels. But there are different sublevels in the levels. For example s, p, d and f.
Does the electron only jump from s sublevel to s sublevel or can it jump to a p sublevel ?
I also learned that an electron always fills the sublevels in order. Does this mean that an electron in a neutral hydrogen atom can only exists in s sublevels ?

An electron in H always sits in 1s, unless excited. When excited, it can go anywhere.

Can it go to $2p_y$ ?

$2p_y$ is not really a state.

@IvanNeretin Does carbon not have one electron in the $2p_y$ orbital ?

Orbitals are not quite the same as states.

You should read up about selection rules which govern what spectroscopic transitions are allowed and which not. As a rule of thumb, s-p, p-d, d-f are allowed but not s-s, p-p, s-d etc.

@IvanNeretin Can a hydrogen atom that has only one electron have it's electron in the $2p_x$ orbital ?

An excited hydrogen atom can have an electron in the state with $n=2,;l=1,;m=1$. That would be $2p$. But you can't tell whether it is $2p_x$, $2p_y$, or $2p_z$.

@IvanNeretin Great , that answers my question.