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Colligative properties and binary phase diagrams
I'm having troubles with colligative properties and how to relate them to binary liquid-vapour and solid-liquid phase diagrams.
First of all, in my book it is indicated that colligative properties are defined in the presence of a non-volatile solute, so much so that it simultaneously determines a lowering of the vapor pressure, an ebullioscopic raising and a cryoscopic lowering of the solution with respect to what happens in the ambit of the pure solvent. But this does not seem entirely correct.
In fact, even a water-ethanol mixture can undergo cryoscopic lowering, i.e. a eutectic is reached, although both constituents are volatile (therefore the solute is volatile, whether it is water or alcohol).
The same argument can be made, for the same volatile mixture, in the context of boiling point:

to the right of the azeotrope, an ebullioscopic rise is observed even though both compounds are volatile.
So, actually, when do these two colligative properties manifest themselves? Is it really necessary for the solute to be volatile?
The other doubt, as already mentioned, concerns about the correlation between state diagrams and the same colligative properties.

This state diagram refers to two substances that are miscible in the liquid state, but immiscible in the solid state, and which reach a eutectic, so the graph is based on the cryoscopic lowering. If freezing point lowering is defined only in the presence of non-volatile solutes, how does this graph hold true even in the presence of a volatile solute? (as in the case before, I mention the water+ethanol mixture, which reaches a eutectic)

This other graph, on the other hand, is a state diagram which refers to two miscible substances both in the liquid state and in the solid state. In this case a eutectic is not achieved and the graph, at least at first glance, is not based on the freezing point lowering, given that starting from A and progressively adding B, the melting/freezing temperature of the mixture increases.
Why is there this difference?
Thanks for every help, because i'm stuck.

When we say that nonvolatile solutes cause lowering of vapour pressure, boiling point rise or cryoscopic effect, it doesn't mean ONLY nonvolatile solutes cause it. It just happens that for nonvolatile solutes, there are simple equations to calculate those effects. For an arbitrary solute, opposite effects may occur like an increase in vapour pressure. Quite naturally, it depends on whether the solute is more volatile itself than the solvent.

@ViníciusGodim So what you mean is that, in the presence of a non-volatile solute, these colligative properties ALWAYS appear (freezing cryoscopic lowering, boiling point raising, vapor pressure lowering).
In the presence of a volatile solute, on the other hand, both these colligative properties or opposite events can occur and this always depends on the characteristics of the solute: for example, ethanol is more volatile than water, so by adding it progressively, an increase in vapor pressure will be observed and so we Will see a decrease in boiling point of the solution.

@ViníciusGodim At the same time, a cryoscopic lowering of the solution occurs, given that ethanol has a lower freezing point than that of water. However this does not explain why by progressively adding water to ethanol (so in this case the ethanol acts as a solvent), a cryoscopic lowering is still observed although water has a higher freezing point than that of the ethanol.

Nobody knows why a mixture of two compounds has or does not have a phase diagram with or without an eutectic or an azeotrope. Anyway these strange points may disappear when changing the pressure. Eutectic mixtures are not understood. They are simply here !

@Maurice oh, perfect. At this point I Guess that the colligative properties are defined in the presence of non-volatile solutes precisely for this reason: in the presence of a volatile solvent and a non-volatile solute, the cryoscopic lowering, the boiling point raising and the lowering of the vapor pressure ALWAYS occur.
In the presence of a volatile solute, on the other hand, it is not possible to exactly predict the behavior: a cryoscopic lowering can occur, as well as an elevation. Same for vapor pressure and boiling point.


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Why the change of the position of the salt bridge or metal piece leads to different voltage value?
I am doing an experiment where I use salt bridge to create galvanic cell. The goal of my experiment is to find if there is some correlation between the amount of solvent being used and the voltage being produced which means the accuracy of voltage is very important.
I am using zinc and copper metal pieces, zinc sulfate, and copper(II) sulfate with sodium chloride for salt bridge:

The experiment works, but the problem appears that when I modify the position of either the metal piece or the salt bridge, the voltage value will change, sometimes I will change a lot (e.g. $\pu{0.3 V}$ to $\pu{0.42 V}).$
Does the change of the position of the salt bridge or metal piece leads to different voltage value in galvanic cell?
If not, could it be the reason that my zinc sulfate powder does not fully dissolve in water? So my metal piece could touch the area where there are more Zinc powder which causes it to be more electron transfers happens. It is very hard to get it fully dissolved in distilled water. Any suggestions for improving solubility in water?
I am really looking for way to reduce the uncertainty of my experiment, but the different voltage measured by only changes the salt bridge or metal pieces makes me very concerned.

(+1) for including the photograph! Moving things around can change the cell’s internal resistance, thereby changing the potential you measure. This is a Daniell cell, so there is no need for high zinc ion concentration and the liquid in the anode reservoir (beaker) should not be cloudy.

@EdV, thanks for the comment. Do you have any suggestions for increasing  accuracy of the experiment? I am doing an experiment where I keep adding water to the solution to record the voltage, but the different position of the salt bridge or the metal piece seems to result in different voltage value after the voltameter measured stop changing.

I am not surprised at all. Even if you just add solution (or remove some), you change both the cell resistance and the electrode contact area with the solutions. Even worse for adding water because the solution concentrations change. So you have multiple things influenced by the one factor you are controlling. Hate to say it, but you might need to redesign a bit to get an unambiguous relationship between what you control experimentally (solution volumes) and the cell’s open circuit (or lightly loaded) response voltage. As well, the Nernst equation always is pertinent.

@EdV, thanks for that, so basically I am research if the amount of water has a relationship while making other condition be constant. My independent variable is the amount of water being used. I am wondering how do I have changed the cell resistance in the experiment? Do you mind explain it a little bit so that I could redesign the experiment. Sorry, I don't have too much knowledge with chemistry :(

Well, pure water is a poor electrical conductor because it has very low concentrations of ions. Your electrolyte solutions are much better electrical conductors: they have relatively high ionic content, so their electrical resistance is low. Diluting them lowers their ionic content, so their resistances increase. The cell output voltage is the net output after the voltage drop due to the cell resistance is deducted, as it were. So there are always trade offs and this is one such. Anyway, I will leave it to others to respond further. Best of success!

@EdV, sorry, may I ask another question, how does diluting my two solution (anode and cathode) diluting my electrolyte solutions (NaCl in salt bridge)?

So in this case will it be better to say my independent variable in concentation so that the change of cell resistance could be the dependent variable? Thank you sooo much

It may also improve your results if you polish your electrodes. That way you are ensuring the the surface available for reaction is proportional to the heigt of the solution. And as already state try using a more diluted solution of $ZnSO_4$. A solution close to 0,1 M shold work very well.

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How stable are paper book matches?
I have few hundred paper book match books that I collected about 25 to 45 years ago when many businesses had them custom printed.
They are from around the world although mostly US. They are in an open glass container in a closet. I am asking is there is any risk of them igniting with age

Today I learned aficionados of safety matches follow their hobby of phillumeny.  Some of them enjoy exchange on r/phillumeny -- maybe they can provide some insight from a practical perspective?  Worth to check Wikipedia's editions in other languages, too.

https://siarchives.si.edu/blog/their-heads-matchbooks-archives may prove useful.

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How did you convert primary alcohol using HCl (with ZnCl2)?
I kinda feel weird for this type of question. When primary OH (alcohol) is converted with HCl (with ZnCl2), how did the free Cl-ion come from? If, it is came from HCl, since when we get rid of the H from HCl?

Though the Lucas' reagent is a solution of concentrated $\ce{HCl}$ and $\ce{ZnCl2}$, it still is an aqueous solution containing $\ce{Cl^-}$, because

as a strong acid, practically all $\ce{HCl}$ in the aqueous phase dissociated *
$\ce{ZnCl2}$ is well water soluble

* One can purchase solutions of HCl in organic solvents which (depending of the solvent) may prohibit the dissociation.  An example of such a "dry solution"  is $\ce{HCl}$ in diethyl ether (entry at Sigma).


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Dependence of solubility on pH
I am studying the effect that pH has on solubility of metal salts. One example I came across is the idea that the solubility of $\ce{PbI2}$ is not strongly affected by the addition of strong acid to solution because $\ce{PbI2}$ contains the conjugate base of a strong acid ($\ce{HI}$). Online references explain that this makes sense, since the presence of additional $\ce{H+}$ ions won't successfully pull the $\ce{I-}$ ions out of solution (since $\ce{HI}$) is a strong acid that wants to dissociate).
However, is this line of reasoning actually correct? For example, what if the strong acid added to solution is $\ce{HI}$ itself? Wouldn't that lead to the presence of additional $\ce{I-}$ ions that increase the reaction quotient above $K_\text{sp}$ and thereby cause the solubility to decrease (more solid to precipitate out)?
I suppose the general question is: how should I think about the role of pH on solubility when the acid being added to tune the pH contains a compound that is being dissolved from the solid in the first place?

It's no pH, but ligands that change that are important.

pH and solubility of metal salts have very little connection. The rest is all extrapolation.

There can be impact of pH on PbI2 solubility, if Pb^2+(aq) are being turned to something else, like PbOH+(aq) or Pb(OH)2(s).


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Do "pseudo Van der Waals" gases exist?
In college, when deriving the Langmuir isotherm for gas-solid adsorption, the professor proposed a modified version of the Van der Waals state equation, what he called the "pseudo Van der Waals gas state equation". The derivation implied equaling the chemical potential for a certain compound in the gas and interphase, and when dealing with the volume for the interphase:
$$V\cdot (p+q) = RT$$
Where $V$ is the volume, $p$ the pressure, $R$ the gas constant, $T$ the temperature, and $q$ a correction term to account for the adsorbate cohesion. He argued that the interphase was something similar to a very condensed gas, so that equation of state was valid. Indeed, he obtained correctly the Langmuir isotherm, that later was derived using the kinetic approach.
However, I have been reading a lot of books and articles looking for that "pseudo Van der Waals gas", and I haven't found anything similar. I guess the approximation is just to neglect the excluded volume, but I am not sure whether my professor was right or he took some license by himself.

Neither van der Waals' gases neither pseudo van der Waals' gases exist. They are theoretical simplifications. It is similar as for ideal gases, just closer to real gases.

What I meant is if they are even defined. I know they both are just models.

Then it would be better to use other verb than exist. IMHO, it was just ad-hoc definition for the purpose of the lecture. Another variant would be with own volume, but without cohesion.


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Can one calculate the polarizability of Xe isotopes using Gaussian 09 and the 3-21G basis?
While trying to understand this paper, I read the following method for calculating the polarizability of different Xe isotopes:

Quantum chemical calculation of the exact polarizability of each
xenon isotope was performed by using Gaussian 09 software (version
D.01, Gaussian Inc., USA). The exact polarizability of each xenon
isotope was optimized by b31yp/3-21G and the density functional theory
method.

My understanding is that DFT can be used to assess polarizability of larger systems, but I'm a little befuddled by how it's applied here.
I couldn't identify what is meant by "b31yp" but it was fairly easy to read a description of the "3-21G" basis online, and it's apparent that the polarization of Xe atoms won't be included in that basis:

the 3-21G* basis set has polarization functions on second row atoms only.

So...can someone decipher what is meant here by "b31yp"?
EDIT: I think it must be a typo for the "B3LYP" DFT method, described here:

uses the non-local correlation provided by the LYP expression, and VWN functional III for local correlation (not functional V). Note that since LYP includes both local and non-local terms, the correlation functional used is actually:

C*ECLYP+(1-C)*ECVWN

In other words, VWN is used to provide the excess local correlation required, since LYP contains a local term essentially equivalent to VWN.

Does it seem plausible that the researchers correctly calculated that four different isotopes of Xe have the same polarizabilities to three significant figures?

I don't think it is of topic here, but it might be more on topic at [mattermodeling.se].

thanks! I didn't know about Matter Modeling! I think you're probably right, but I'm very happy by the quick and cogent responses below

In addition to the other answers, to do a proper calculation of the polarizability of Xe or any other "heavy" element, one needs to perform proper relativistic calculations in a larger basis set - ideally at the basis set limit since it's only a single atom. (The basis set used is entirely too small, as mentioned in other answers.)
So the number in the paper ($3.60 Å^3$) is simply wrong. Atomic polarizabilities, particularly of noble gases have been measured accurately and used for benchmark calculations:

Calculation: Sakurai, Sahoo, and B. P. Das 2018 - 27.508 a.u. = $4.076 Å^3$
Experiment: Olney, Cann, Cooper, Brion (1997) - 27.03 au = $4.005 Å^3$
Experiment: Hohm and Kerl 1990 - $4.1217 Å^3$

(Incidentally, the NIST reference cites $4.005 Å^3$ as the atoms polarizability of Xe.)
My guess is that one of the authors had used Gaussian, knew it handled isotopes (e.g., for vibrations), saw that it could calculate polarizabilities, and just ran the calculations without thinking too deeply.

As the other answer already says: it's entirely plausible and not surprising that these calculations came out this way. This maybe answered your question, but I'd like to offer some more context.
For all intents and purposes, B3LYP/3-21G is a terrible choice of computational method. By the way, you have (without me reading this paper) correctly identified this. I'm my personal opinion, using a phone book will produce more consistent results.
The basis set is way too small for any productive work, and in a computational setup of today's hardware, there's hardly any difference by using a better balanced one.
(If you'd really needed to cut down computations, go with a minimal basis, i.e. STO-3G. 6-31G is already a bad, but popular, standard, and only slightly more demanding than 3-21G. Most other split valence basis sets perform computationally better. With a dual core computer, a gigabyte RAM, for small molecules and on a human time scale, is not noticeable.)
Gaussian 09 is old and about 6 years past is prime. Revision D.01 isn't even the last one. However, with this level of theory, nothing much has changed, so it likely doesn't matter at all. However, if there were any bugs in the program, they would be in the final numbers.
Density Functional Theory is nothing without calibration.
The holy grail is still hunted for. And until we know about it with certainty, some functionals will be optimised for one scenario and others for something completely different. That's why there are quite a few benchmarks. Unless there's one that covers your field of study, you (them/they) should do those yourself (themselves/their selves).
N.B.: Most other quantum chemical programs use VWN5 for B3LYP.
Anyway, answering the titular question:

Can one calculate the polarizability of Xe isotopes using Gaussian 09 and the 3-21G basis?

You can calculate a number, but it's not reliable; so: no.
Answering a question you have not asked:

Can the exact polarizability of each xenon isotope be optimized with B31YP/3-21G?

Probably (not very likely) not. With DFA (density functional approximations) nothing is exact (yet, due to the limitations of the method.)

Different isotopes (of Xe but also any other element) only differ in the mass of the nucleus; the charge of the nucleus and the electron shells are exactly the same.  In the Born Oppenheimer approximation the electrons have their own wave function which includes point potentials from the nuclei (neglecting electron-nuclei cross terms).  Because of that, the mass of the nuclei (or kinetic energy of the nuclei) has no effect on the electrons at all.
From that perspective, it is not surprising that different isotopes have the same polarizability (or indeed any other properties) when calculated with DFT using the BO approximation.  What is maybe surprising is that the authors of the paper would think that is not the case.  I haven’t read the paper but this result would absolutely be expected in any formulation which doesn’t involve a wave function for the nuclei.

The BO approximation doesn't seem particularly relevant since they're just calculating the polarizability of individual atoms.

I was wondering this myself... whether BOA was applied here, because then yes, it's impossible for nuclear dynamics to impact electronic dynamics. Am I right that  DFT as applied here implicitly uses the BOA?

Yes, the calculation would imply the BO approximation.. but it's irrelevant for an atomic calculation.

@GeoffHutchison Not using BO is the only way nuclear mass can possibly affect the results here, it seems


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what is the meaning by 'species' of glycine?
The question ask for the species of the (i) glycine hydrochloride (ii) glycine titrated with NaOH. When they mentioned 'species', what did they mean by it?

Species here means the specific ionization state of glycine. Is it a neutral molecule or an ionized form? See the definition here: https://goldbook.iupac.org/terms/view/CT01038

The term species is used to describe a given type of atoms, molecules or ions. As a functional definition, a species is any atom, molecule or ion that could be part of a chemical reaction equation as reactant or product.
Here, it refers to the different protonation states of glycine, no matter whether anionic, neutral or cationic.


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Synthesis of a Substituted Indane
From the starting reactants given as the sole carbon source:

We need to synthesis the following compound (we can use any inorganic reagents).

Here is what I tried so far, but I don't think it looks "realistic"

This is yet another exmple of a poorly thought out question that bears no resemblence to real world chemistry. I presume the question setter wants to see F-C alkylation with the di-bromopropane to set up the indane ring. The reality is that this will mostly give the products of rearrangement.

see https://chemistry.stackexchange.com/questions/91606/propose-a-synthesis-of-indane-from-benzene

There is a difficulty with your last step. There are 3 possible benzylic sites for bromination and I think you will get a mixture. I would reduce the ketone to the benzylic alcohol and treat with PBr3

The reaction of the Grignard with bromobenzene is not going to do what you want and displace the Br. I would trap the Grignard with borate ester and couple that with the bromobenzene under Pd catalysis.

@Waylander "This is yet another example of..." are you talking about the OP's questions (I only see one at the moment) or are you lumping this question with those by other users and using your comment to rant against "them"? Your comment seems to me to be deliberately unwelcoming to a new user and to use the space under their first question to talk about things that belong in chat or meta. "Poorly thought out" may be true, but the "yet another example of" is gratuitous and unhelpful, and potentially discouraging. Let's use the space under a new user's first question to be more welcoming."

@uhoh my rant is directed at the setter of the question the OP is attempting to answer

@Waylander who it seems will not be a recipient; so perhaps, no longer needed?

@uhoh as that comment is quoted in the answer I'm leaving it up

@Waylander okay, one could argue that since the information is now captured in an answer it no longer needs to be there as well... It's no longer helping the page...

I'll be frank about it....you can't get there from here! In @Waylander's words " This is ...a poorly thought out question...". Allow me to pose the question as it should have been asked.  How does one synthesize 5-(1-bromo-2-methylpropyl)-2,3-dihydro-1H-indene 1 from benzene, propionic acid, and isobutyric acid?

Inspection of ketone 2 shows that a Friedel-Crafts acylation of indane 3 does not occur at $\ce{C4}$ but rather at $\ce{C5}$. The synthesis of bromide 1 from ketone 2 requires reduction to a benzylic alcohol (e. g.; $\ce{NaBH4/C2H5OH}$) followed by treatment with $\ce{HBr}$. Isobutyric acid and propionic acid are readily converted to their respective acyl chlorides with $\ce{SOCl2}$. As to the preparation of indane from benzene, this issue has been addressed previously on ChemSE.


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What is the highest temperature, ambient pressure type-1 superconductor as of 2022?
I was curious about the highest known Tc for type 1 superconductors.
This list suggests lead near 7K is as good as it gets for type-1 superconductors but I wonder if that is true given the current state of the art.

Just a warning, some aspects of that site are... contentious, especially the high Tc claims.

Somewhat confusingly some compounds exhibit both type 1 and type 2 behaviour (MgB2 for example). It has a Tc of ~39 K and has at least some type 1 character.


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When hydrogen and oxygen reacts, what defines if it will turn into hydrogen peroxide or water?
If you have a jar with oxygen and hydrogen inside, I'm interested in knowing what variables are most important to ensure that the mixture will become water (if that was the goal) or hydrogen peroxide, but not both. I understand that at room temperature those two gases will not react to form anything, so the question is more about the details of how the mixture could be manipulated to take a route to become X vs Y.

The idea of direct synthesis of H2O2 from elements is interesting. But with exception of reaction of residual molecules or atoms in high vacuum, I doubt it has any significance.

Here is an article claiming that fuel cells could produce either. In biology, there are catalases that degrade unwanted hydrogen peroxide.


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Are hydrocarbons compounds of hydrogen and carbon, or carbon and hydrogen?
Is it the same to say "Hydrocarbons are compounds of hydrogen and carbon." as saying "Hydrocarbons are compounds of carbon and hydrogen."
I got a B in my chemistry test just because of writing "hydrogen and carbon" instead of "carbon and hydrogen".
So i need to know if its the same and my teacher made the mistake or if it's me that made a mistake.

I’m voting to close this question because that not even chemistry, but how language works.

More precisely, it would be so called pragmatics here, it seems.

@Mithoron My original comment was it is not about chemistry, but English, logic and semantics ( or rather pragmatics, I have not known this meaning). But then I got a Xmas urge to be nice.

I think your teacher wishes to emphasize carbon, so carbon comes first. Organic chemistry is the study of carbon (in the limelight)

Kinda related: https://chemistry.stackexchange.com/questions/133655/are-there-any-organic-compounds-that-do-not-contain-hydrogen-and-if-so-what-a/133664 ... https://chemistry.stackexchange.com/questions/134101/is-c4o5-possible/134107

I'd vote to improve the heading to point out it is about the order of words. As of now it sounds too trivial to avoid downvotes.

Rather, in a way, you have been both right.
While there are subtle differences in context of pragmatics as below, it should not be the bases of rejecting the answer. It rathers looks like the teacher deliberately marked it as wrong, as it did not literally matched the given answer.
IMHO, instead of asking, you should rather stand up for yourself and confront your teacher. The fact you had to ask about it is evidence of your failure to be confident in your knowledge.
The right order A and B has two points of view(POV):
POV of logic: The order A and B is equivalent to B and A.
POV of pragmatics: A and B is usually used as "A is the main descriptor" while "B is the modifier". In similar sense as if A was a noun and B was a descriptive adjective.

Organic compounds are primarily compounds of carbon.
Not all carbon compounds are organic ones, but all organic compounds contain carbon.
Most of organic compounds contain hydrogen, but not always.
Typical exceptions are per-halogenated hydrocarbons, or mellitic anhydride $\ce{C12O9}$, containing just carbon and oxygen.
It is therefore natural to start with carbon, when specifying a subclass of organic compounds.

So, primarily, in context of organic compounds, "Hydrocarbons are compounds of carbon and ....."
Now, hydrogen comes as the modifier, specifying which carbon compounds are hydrocarbons.
Therefore, "Hydrocarbons are compounds of hydrogen and carbon." is logically right, pragmatically not exactly wrong, but suboptimal and less usual.

Nonsense! they are not called carbohydrens as in carbohydrates. The teacher was being pedantic. sort of like calling water oxyhydren because it is basically a compound of oxygen and insisting that acetic acid be ethanoic acid. To lower a grade because of that is suspicious of a hidden agenda. BTW water is an organic compound because its structure and properties are discussed in every organic text.

@jimchmst You confuse organic(1) in chemistry context and organic(2) in lifestyle context. // BTW, it has nothing to do with the order of word parts. // For OP: Generally, A should not ask B "Why did C say X?" . A should ask C.

I am not confusing anything. I am saying get over it. History is nice but lets stop reliving it.

Aggressive rhetorics does not make you right. Domain limited meaning of terms is widely used.

Frankly, I have to disagree, there really isn't any good reason for marking "hydrogen and carbon" as wrong. If the teacher says it's wrong because it emphasises hydrogen over carbon, that's their own subjective interpretation of it, which shouldn't come into play when assigning marks for a question which has an objectively correct answer.

@orthocresol While there are subtle differences in context of pragmatics, I do agree with you, and omitted to say so explicitly, that it should not be the bases of rejecting the answer.

That's fair, I don't know whether I was really disagreeing with you or with the teacher :)

I feel sorry the OP got stuck with that teacher. No good was served by deducting marks for such a pettifogging reason.

@EdV Some teachers hardly know the topic better than students and follow exact wording of given answers. If the answer deviates, they mark it as wrong, even if it may be and often is correct. The extra bad case are expectations of the particular numeric value with given valid digits to be filled in an input form of some webpage.


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Is an aqueous solution of lithium superoxide basic or neutral?
Is the aqueous solution of lithium superoxide basic or neutral?
I know that $\ce{Li}$ is a metal, thus its oxide ($\ce{Li2O}$) produces base in reaction with water.

pKa of $\ce{HOO^.}$ is 4.88.

It is hard to imagine lithium superoxide itself.

@NilayGhosh https://en.wikipedia.org/wiki/Lithium_superoxide // The better question would be "Is solution formed by dissolution of LiO2 in water basic or neutral?"

This assumes we can get the lithium superoxide into water at all. It seems to be stable only at very low temperatures or in nonpolar solvents.

@OscarLanzi Like that.

Lithium superoxide is only stable in a dry atmosphere and at temperatures below $\pu{70 K}$ $(\pu{-203 °C})$ according to Wikipedia. So it cannot be dissolved in water. And in the presence of water, it gets decomposed into $\ce{LiOH}$ and $\ce{O2}.$


text:
What is the determining factor between two potential bond candidates?
Say you have a vacuum (imaginary, of course) with three hydrogen atoms, how do we determine which two will bond and which one will be left out? If the same energy is present (equally) for the entire reaction, is the priority given strictly to the two that are closer to each other?

If the collision has too much energy. or if energy released by the bond formation is not dissipated by photon emission, collision, or vibration energy, then the bond is broken again. Similarly, $\ce{2 H}$ need it too to form $\ce{H2}$.
As a mechanical, everyday analogy, imagine a marble play with frictionless terrain, balls and a round enough hole. If you push a ball, it would not end in the (empty) hole, but will escape on the other side. It has to lose some energy somehow, not to be able to escape.

Thanks for answering, this is good information, but it doesn't answer my entire concern. I have rephrased my question — made it simpler, in case you have anything else to add.

As there are quantum effects, we cannot determine it, even if we knew exactly their state. // It is like this: 3 equally good wild west gunners randomly shoot to each other. Each shot has probability of kill  0.2. Which of those 3 survive ?


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Electrode potential and likeliness to reduce/oxidize (electrolysis)
I'm a little confused about situations where multiple chemical compounds can be reduced/oxidized and the likelihood of this happening to one compound over the others, depending on the electrode potential of that compound. Here are some exercises I'm trying to figure out on this topic and their answers as given to me:
First exercise: electrolysis with inert electrodes and aqueous solution $\ce{CuI_2}$, the question is: "why do we observe $\ce{I_2}$ gas release rather than $\ce{O_2}$".
The answer is: "If $\ce{H_2O}$ is oxidized at the anode (to form $\ce{O_2}$) rather than $\ce{I^-}$ (to form $\ce{I_2}$), we have $E^°_{cell}$ = $E^°_{Cu^{++}/Cu}$ - $E^°_{H_2O/O_2} = 0.337 - 1.23 = -0.893V$ which is lower than $E^°_{cell}$ = $E^°_{Cu^{++}/Cu}$ - $E^°_{I_2/I^-} = 0.337 - 0.5355 = -0.1985V$ and thus, since the potential is higher (in absolute value) for water, no $\ce{O_2}$ should be released as long as their is $\ce{I^-}$ ions available"
Second exercise: electrolysis with inert electrodes and aqueous solution at $\ce{pH = 0}$, $\ce{T=298,15K}$, $\ce{p=1bar}$ and containing $10^{-2}M$ of $\ce{Ag^+}$ ions and $10^{-2}M$ of $\ce{Cu^{++}}$ ions, the question is: "discuss the possibility of having a copper deposit without impurities from the silver".
The answer is: "[I'm skipping some calculations but it can be provided if required] We get the standard electrode potentials: $E^°_{Cu^{++}/Cu} = 0.337V$ and $E^°_{Ag^+/Ag}$ = $0.799V$ and from Nernst equation (so calculation of the actual -non standard- cell potential) we get for $\ce{H_2O/Ag}$ and $\ce{H_2O/Cu}$, respectively: $E_{cell}$ = $- 0.549V$ and $E_{cell}$ = $- 0.952V$ then we calculate the concentration of $\ce{Ag^+}$ at $- 0.952V$ (using Nernst law again) and we get: $[\ce{Ag^+}] = 1,7\times10^{-9}M$ , which is very low compared to [$\ce{Cu^{2+}}$] = $10^{-2}M$, so almost only $\ce{Cu}$ will deposit on the electrode."
So here is my question on all this: I have read everywhere that "more negative electrode potential means more likely to be oxidised", which actually correspond correctly to my first exercise ($E^°_{I_2/I^-} = 0.5355V$ is "more negative" than $E^°_{H_2O/O_2} = 1.23V$), is it correct to apply this reasoning here though?  And also that "less negative electrode potential means more likely to be reduced" and there I have a problem, in the second exercise $E^°_{Cu^{++}/Cu} = 0.337V$ is not "less negative" than $E^°_{Ag^+/Ag} = 0.799V$ but quite the contrary, still we calculated that almost no $\ce{Ag^+}$ was deposited (so reduced, to my understanding) compared to $\ce{Cu^{++}}$.
Could you help me to understand why we get this ? Simple and general guidelines for this kind of reasonings would really help.

EDIT (for those who would try to understand this topic): As Robert DiGiovanni pointed out (see the discussion below his message), the conclusion for the second exercise is incorrect. Since $E^°_{Ag^+/Ag}$ is more positive than $E^°_{Cu^{++}/Cu}$,  $\ce{Ag}$ will reduce first. Furthermore, the result from the Nernst equation actually confirms the fact that copper cannot be deposited pure. Since $\ce{[Ag^+]}$ is found really low from Nernst equation (giving the cell potential at equilibrium), it means that most $\ce{Ag^+}$ will be reduced when reaching equilibrium, and thus most of the Ag will be deposited on the electrode.

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

$\ce{2 Cu^2+(aq) + 4 I-(aq) -> 2 CuI(s) + I2(aq,s)}$

You should ask the author of the task to show you $\ce{CuI2}$ solution.

@Poutnik what copper gets reduced to is irrelevant.  Hydrolysis of water, written as suggested is H20 ---> H2 + O2  - 1.23 V.  Before this happens, 2 I- ----> I2 + 2e-   - 0.54 V.  Looking further, we find Cu++ + 2 e-  --->  Cu  + 0.34 V and Cu+ + 1 e- --->  Cu + 0.52 V. It seems metallic Cu would be formed at the cathode.

@RobertDiGiovanni It is relevant what iodide is oxidized to. It does does get chance to get electrooxidized in priority when it is already oxidized. Similarly for copper.  CuI precipitate is not easy to reduce on cathode.

Btw, the non standard redox potentials matter more in these cases than standard ones.

"Why do we observe I2 gas release rather than O2"  even if CuI precipitates, CuI is (very slightly soluble) available to continue to Cu metallic as long as current is applied.  The CuI precipitation reaction is interesting reading.

Copper in CuI will not be reduced before reducing iodine/triiodide. And even then non standard potential Cu/Cu+ will be probably below H2/H+  potential.

@Poutnik if you would kindly look at a redox chart, copper ions are are stronger electron acceptors than our beloved "H+".

@RobertDiGiovanni Redox potentials can swap the order easily, depending on activities of involved redox system components. It is similar as reactions with $\Delta_r G \lt 0$ do not always go forwards and vice versa.

@Poutnik correct, but we must remember the ratio in solution.  As long as (even minute) amounts of Cu + are in solution, they are preferentially reduced.  This is why it's really a voltage titration, with current flowing as metal cations are reduced.  But yes, with a precipitate and sufficient voltage, there will be other reactions.

Preferably is reduced iodine. Try to estimate Cu/CuI/I- potential.

Ah, but iodine leaves as a gas.

Part of it can, rest will dissolve or form I3^-. Iodine does not have zero solubility.

Let us continue this discussion in chat.

Iodide gives up its electron easier than water, silver accepts electrons before copper.
The best way is to draw the half reactions in the direction they occur.
Then the math is easy.  For example:
$$\ce{2Na metal + Cl2 ->  2 NaCl}$$
The oxidation half reaction is:
$$\ce{2Na -> 2Na+ + 2e- + 2.71 V}$$
You may find it on a Standard Reduction Potential Chart as:
$$\ce{ Na+ + e-  -> Na  - 2.71 V}$$
but that is not the direction the reaction is running.
Similarly, the reduction half reaction is:
$$\ce{Cl2 + 2e- -> 2 Cl-  + 1.36 V}$$
2.71 + 1.36  = + 4.07 V. Half reaction redox potentials are simply added.
Just remember to set your redox half reactions up this way and the comparisons will be easy.
Background reading on Potentiometric Titrations may be helpful.

So, if you want to plate out pure copper, plate out silver first, then replace the cathode.

So is the calculation I gave about $\ce{Cu/Ag}$ wrong ? (it's the correction from my uni, not saying they're never wrong though -everything I read on the internet was actually going in your direction-)

@c.leblanc "almost no Ag is deposited"??? One might simply plate out the silver first.  Apparently, they are saying, concentration wise, there is so little silver, don't worry about it.

To make sure I understood: does the in the direction they occur method you described also work for electrolysis (where we apply current, so it is the case of my exercises). If yes, when you write reactions in the direction they run: will all reactions that occur naturally (electrochemical cell) have positive electrode potential and will all reactions that occur when a potential is applied (electrolysis) have a negative electrode potential? Last thing: what is the rule of thumb to determine if is it more likely to be oxidised/reduced when you get that corrected value ?

not sure I understood your last answer. Do you mean it is possible what they say is correct or not ? I mean, at first both $\ce{Cu}$ and $\ce{Ag}$ have the same molarity. If $\ce{Ag}$ is more likely to reduce first, it doesn't make sense that it would be a very low concentration of $\ce{Ag}$ in the deposit compared to  $\ce{Cu}$ .. or am I missing something ?

Reactions that are not net positive can be run by applying potential from an outside source.  The rule of thumb is higher positive values for drawn half reactions happen first under all conditions.  This just in: Cu ++  ---> Cu  + 0.34 V.  Cu ++  ---> Cu +  + 0.15 V.

@c.leblanc copper will never plate out before silver, but Poutnik may be on to something.  Drop the silver out as AgCl (solid).  But (silver is worth more), I would simply plate out silver first. you can adjust your applied voltage to preferentially drop metals.  Silver first.

For the rull of thumb, does it mean (in the case of applied tension) that "less negative" (we could say "less negative" is "more positive") value for drawn half reactions will happen first ? I don't follow you in your last comment, does that make the statement in my exercise possible?

@c.leblanc. draw the half reactions in the order they proceed.  The more positive ones go first. Discuss this with your professor and others.  Incorrect test answers  can be very confusing.

I think I fully understand your logic about drawing the half reactions in the order they proceed (I just wanted to make sure that "more positive" also means "less negative" in the case of reaction with negative electrode potential happening thanks to applied tension, but it would actually make sense when I think about it). Thank you for that simple tip, it's actually much clearer and logical than anything I've read about it. You're right, those incorrect answers are my nightmare .. I will discuss that in class. Thanks for the clarifications !

@c.leblanc. from the Nernst equation, what we get is at that voltage most of the silver is already reduced. Think titration.  If you apply voltage to a copper and silver ion solution, the silver will ground the current is it reduces, showing a "plateau" until the silver ions are reduced.  Then voltage increases until copper grounds at the higher potential.  I had a great professor.  Very lucky!

I was actually thinking the same. Since we get that it is the silver ions that are really low in concentration, I guess it does make sense that it actually means they reduced almost fully (so silver deposit would be high) at equilibrium (since Nernst equation represent the potential of the reaction at equilibrium).

The main theme of the questions is a very good academic exercise but the question writer (whoever wrote the the two questions for students) made them vague and the answers are more convoluted than the question. It is bound to confuse students.
For the first question, $\ce{CuI2}$ does not exist and it cannot exist in solution. Copper(II) is a strong oxidizing agent that will oxidize iodide to free iodine. Let us correct the question and say, if we had a solution of $\ce{KI}$ i.e., potassium iodide in water and if we electrolyze it, what products do we expect at the anode?
The general theme for addressing such questions would be:

List all the possible ions and molecules in the solution: We have $\ce{H2O}$, free $\ce{K+}$ and $\ce{I-}$, $\ce{H+}$ and $\ce{OH-}$

Since we are interested in oxidation only, we will think about the species which can be oxidized at the anode, the only chemically realistic possibilities are that $\ce{H2O}$ and $\ce{I-}$. Forget the cathode for the time being.

Now check, which oxidation is thermodynamically favored by looking up the electrode potential tables. All modern electrode potential tables are written as reduction potentials. Note that all listed electrode potentials are provided under standard conditions, if they have $\ce{H+}$ or $\ce{OH-}$ in the equations, then under standard conditions, we mean that they have unit activity or say pH=0 or pH=14 respectively.

Please do not flip signs of electrode potentials. They are sign invariant. This action is frowned upon by modern electrochemists, although some general chemistry textbooks still do that.
$$\ce{I2 + 2e- ⇌ 2 I-     (+0.5355 V)}$$
Similarly,
$$\ce{O2(g) + 4H+(aq) + 4e-  ⇌  2H2O(l) (+1.229 V)}$$
If we interpret the above equations as follows given that we have $\ce{H2O}$ and $\ce{I-}$, then at the anode, it will require less energy to oxidize iodide ion to free iodine vs. water to free oxygen. Thus at the anode, thermodynamic oxidation of iodide will be favored instead of water.
As to the second question, the logic in the answer there is also convoluted. It will be far easier  to consider cathode reactions only. You have to consider reductions only this time. Follow the steps above list the reduction reactions under non-standard conditions. You may post the second question separately after attempting along similar lines of reasonings.

I personally found that flipping sign to make it match the actual direction of the reaction was really convenient, because you just have to find "which potential is more positive" to find which reaction is favored. Otherwise, you need different rules (more/less ... means more likely to reduce/oxidise), I find it more confusing and harder to see what's actually going on. Also, flipping sign if needed tells you some logical information: if sign is negative the reaction requires the application of a potential to occur, if sign is positive the reaction is spontaneous and produces a potential.

I agree, this is how it is taught, but I have personally discussed this with three world's leading electrochemists and they do not  agree with this sign flipping. Convenience does not mean scientific correctness. Hope the general guidelines helped.

@Achem I frequently compare it to analogy "Left river bank is on your left(naive approach)" versus "left river bank is on your left when looking downstream(geography)". And the latter ported to electrochemistry - it is related to Gibbs energy of reaction oxidating H2 to H+ in SHE (and not reversed).

@Poutnik, Good analogy but those electrrochemists have a different reasoning. Note that it is okay to reverse signs in thermodynamics but not in electrostatics. These electrochemists told me that it is an electrostatic sign associated with the electrode potential when written in reduction form- this is the true electrostatic sign of the electrode with reference to SHE, hence invariant. I cannot call a proton as a negatively charged particle, no matter how we look at it.

@AChem In a way, linking electrochemical potential to flipping Delta G of reversible reactions was not the best move.

@Poutnik, thanks to Gibbs and Latimer in America.


text:
Is delta S of universe always 0 when an ideal gas undergoes an isothermal irreversible process?
Since for an irreversible process,
$$dS_\mathrm{Surrounding} =-\frac{\text{dq}_\mathrm{irr,sys} }{T_{\text {surr }}}\tag{1}$$
where $\text{dq}_\mathrm{irr,sys}$ is heat exchange of system
and $-dq_\mathrm{irr}$ is heat absorbed by the surrounding.
Also $d u=d q+d w$.
Now in an irreversible isothermal process involving only ideal gas.
$$
\begin{aligned}
& du=d w+d q \quad . \quad(d v=C v \cdot d T) \\
& -d w=d q \\
& P_{\text {ext }} \cdot d v=d q
\end{aligned}
$$
Now Substituting in (i) Now $T_{\text {syst }}=T_{\text {surr }}$ As the process is Isothermal .  Clearly $dS_{\text {Surr}}=-dS_{\text {Sys}}$. Does that mean change in entropy of universe in irreversible isothermal process is always $0$?

Why call it irreversible if there is no change in entropy?

Can you please elaborate what you mean ?

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

Hey thanks I’m new here I will keep that in mind next time and will edit this question shortly

The amount of heat received by the system for the irreversible path is not the same as the amount of heat received by the system for the reversible path.

Hey I appreciate your answer but the case I asked is it’s obvious that the heat emitted by system is the heat absorbed by surrounding And if the sytem undergoes a irreversible isothermal process then work done is Pext.dv so the entropy of universe in such case should be always zero but that’s not the case why?

For the irreversible case, we are talking about a situation in which the ideal surroundings reservoir is maintained at the same temperature as the initial temperature T of the system, such that the final equilibrium temperature of the system is also T.  This is what we mean by an isothermal irreversible process, even though, within the bulk of the system during the process, the temperature may not be spatially uniform at T.
If, in determining the change in entropy of the system, you devise a reversible path between the same initial and final thermodynamic equilibrium states of the system as for the irreversible process, you will find that the final thermodynamic equilibrium state of the surroundings will not be th[e same as for the irreversible process.  So this approach gives you the correct change in entropy for the system in the irreversible process, but it will not give you the correct value for the surroundings.  Since the reservoir is ideal, the change in entropy of the surroundings for the irreversible process will be
$$\Delta S_{surroundings, irreversible}=\frac{q_{surroundings, irreversible}}{T}=-\frac{q_{system, irreversible}}{T}$$
The change in entropy of the system for both the irreversible and reversible path will be  $$\Delta S_{system,irreversible}=\frac{q_{system,reversible}}{T}$$
So, $$\Delta S_{total, irreversible}=\frac{(a_{system,reversible}-q_{system,irreversible})}{T}$$

Thanks a lot sir .  So such proceses should always have $ds_{total} =0$

No.  That's the opposite of what I said.  What I said was that for an irreversible process, $$\Delta S_{Total}>0 $$Only for a reversible process is $$\Delta S_{Total}=0$$


text:
Does wet fabric filter smoke better than dry fabric?
I have heard in various situations people use a wet fabric to breathe better in situations where a lot of fire-based smoke is present (like near wildfires). I wanted to know if there was any truth to this, specifically with regard to wet fabric vs dry fabric.
A quick Google search brought me to this website from FEMA where it says that dry material works better for aerosols and particulates, but it seems the primary focus of the article is chemical attacks, and I can't really find any other sources to back this up; almost ubiquitously across my searches, it seems that everyone is inclined to think that wet fabrics do better.
So my question is that: do wet or dry fabrics work better for filtering smoke? I would greatly appreciate sources when possible, seeing as this seems like an empirically determined fact.

Wet is good for cooling the air you breathe through the fabric. You don’t want to burn your lungs.

@JonCuster That is an excellent point; while I’m still interested in the answer here, it definitely seems there would be many instances where that benefit outweighs possible loss of filtering efficiency.

some possible information here though not sourced: https://physics.stackexchange.com/questions/685455/does-a-wet-paper-towel-absorb-gas-molecules-better-than-a-dry-one

The composition of smoke is not well-defined. On the one hand, some smokes could be considered aerosols of solid particles or liquid droplets only (no gaseous material worth considering). These smokes, as well as other aerosols (e.g., sneezes) can be filtered well enough by using fine weave fabrics (e.g., K95 medical masks). There does not seem to be any suggestion that wetting these masks improves their performance, perhaps because the fine weave would be blocked too much, even by a thin film of water or by swelling of the fabric material.
On the other hand, it might seem that a wet mask could trap liquid aerosol droplets (assuming they are aqueous, not oily) more easily, and this might prove advantageous if the mask is less protective or efficient than a K95 mask. Such a less effective mask probably wouldn't close down so much if it became wet.
On the other hand (if you have one), the gases in a fire combustion smoke may include NO2, SO2, HCN, H2S, Cl2 (from PVC), dioxins, methanol, and many others (Ref 1). These might well be trapped better by a wet mask because they won't be "trapped" at all by a dry one.
Smoke inhalation is the primary cause of death in victims of indoor fires. Particulate inhalation could certainly cause death, but I think the gases would cause incapacitation faster (choking, gasping, collapsing) than suspended particulates.
The graph below shows chemical compositions of volatile organic compounds in smokes from various fuels (Ref 1). The prevalence of oxygenates is obvious; these would likely be better absorbed by a wet mask.

In a reverse sense, wet masks seem to prevent transmission of large aerosol droplets outward better than dry masks (Ref 2). But for COVID protection, dry masks are preferable (Ref 3). But, wet or dry is better than nothing (Ref 4). And cotton exposed to humidity shows an improvement in filtering, whereas synthetic fabrics do not (Ref 5).
In general, specific testing, against the specific smoke (or its likely composition), of the exact mask or cloth/towel/etc., would be needed to determine what would "help you breathe better". In the short run, helping you get out of the smoke would be best; perfect elimination of contaminants with overly restricted airflow isn't the best solution. Perhaps every situation is unique.
Ref 1. https://en.wikipedia.org/wiki/Smoke
Ref 2. https://www.sciencedaily.com/releases/2021/11/211122135517.htm
Ref 3. https://www.wusa9.com/article/news/verify/does-a-wet-mask-make-a-difference-or-am-i-safe-wet-mask-efficiancy-vs-dry-masks/65-5d760229-9e3e-4649-b420-969c74f219cb
Ref 4. https://www.popsci.com/science/wet-masks-covid-protection/
Ref 5. https://www.nist.gov/news-events/news/2021/03/study-indicates-humidity-breath-makes-cotton-masks-more-effective-slowing

Exactly what I was looking for, thank you!

Good answer, the great problem is that the science of air purifiying respirators is complex. I know that in charcoals that while water adsorption normally reduces their ability to protect against gases / vapors in the case of phosgene the water can increases the ability of the charcoal to remove the toxic gas from air.


text:
Spin-preserving excitations
I'm currently working on quantum chemistry with quantum machine learning and one of the operations considered for the calculation is the excitation (and desexcitation) of an electron from an orbital to the other (which corresponds mathematically to a Givens rotation).
So far, these excitations have been taken as spin-preserving, meaning that : say I have initially two electrons in the ground state orbital, if I excite one of them to the excited state, its spin won't be change.
Apparently this is something common in quantum chemistry and I wanted to know why.
An argument I thought of was that the necessary energy to flip the spin of the electron is way higher than the typical electronic energy scale of the atom/molecule, hence making it impossible to flip.
Thank you for your response!

The electron (spin) angular momentum has to change to make spins 'parallel' and in a molecule this can be achieved via coupling of spin and orbital angular momentum, i.e. by spin-orbit coupling.  This is described in many quantum and spectroscopy text books. An example is that S.O. coupling allows triplet states to form from singlet excited states and hence phosphorescence to be observed as well as fluorescence.


text:
Why are these structural formulas not isomers of C5H12?
A Biology textbook question stated that there are three isomers of C5H12, the structural formula of the first two isomers pentane and 2-methylbutane were pictured, the reader was then asked to draw the third isomer of C5H12.
Probably a silly question, nonetheless, why are the following structural formulas I came up with not isomers of C5H12? If someone could elucidate my trouble, I and perhaps others, would be very appreciative.
1:
    H
    |
H — C — H
    |   H   H   H
    |   |   |   |
H — C — C — C — C — H
    |   |   |   |
    H   H   H   H

H atoms removed for simplicity:

C
|
C — C — C
|
C

three lines represent one single bond, so the hydrogen atoms can fit in the structural formula:

C — C — C
        |
        |
        |
    C — C

4:
C — C
|
|
|
C — C
    |
    C

5:
C — C — C
        |
        C
        |
        C

6.:
C — C — C
|       |
C       C

Edit: I know a similar question has been asked on this website, that bond rotation has to do with my confusion and I think I have just answered my own question  , so I am guessing all structures are pentane, except structure 2 which might be 2-methylbutane. Is this because single bonds allow the atoms they join to freely rotate about the bond axis? My trouble is visualising these structural formulas in 3D.
Finally that leaves us with 2,2-dimethylpropane:
    C
    |
C — C — C
    |
    C

Indeed, most of what you have drawn are conformers. They are essentially the same molecules but in different shapes. With time you will be able to visualise the molecules which are often written in 2D representations as actual 3D configurations. You can afford this along with a molecular building kit.

Write these compounds according to the rules for skeletal formula. Better yet look up some 3D representations of molecules.

There is n-pentane, there is 1 way how to side-attach C to n-butane and there is 1 way how to side-attach 2 C to n-propane.

According to IUPAC (in physical organic chemistry as well as sterochemistry), an isomer is one of several species (molecular entities) that have the same atomic composition (molecular formula) but different two-dimensional representations in which atoms are shown joined by lines representing single or multiple bonds, without any indication or implication concerning the spatial direction of bonds.
So ignoring the spatial direction of bonds, your examples 1, 3, 4, 5, and 6 are all the same; they are all simply pentane.
Only your example 2 is a different isomer; it shows 2-methylbutane.
The third isomer 2,2-dimethylpropane is missing.


text:
Does propene have resonance structures?
Propene is an alkene with a double bond between 2 carbon atoms. $\ce{C3H6}$ may have 2 resonance structures due to 2 possible placings of the double bond. E.g. it can be C=C-C or C-C=C (ignoring hydrogen atoms).
Delocalized electrons are a result of resonance structures. Can propene have delocalized electrons?

How would the proton move from one side of the molecule to the other? Your resonance structure doesn't take into account every atom in the molecule, just the double bond.

Do you want to say it should be like $\ce{CH2=CH-CH3 <-> CH2-CH=CH3}$  ?

@Poutnik Yes that's the structure I was suggesting

And why, exactly ? Isolated double bonds do not have resonance structures.

You can't really ignore hydrogen atoms.

@IvanNeretin Yes it cannot be ignored; I just meant it would be easier to recognise the molecular structures without hydrogen atoms around

Well, and I meant that hydrogens are the reason why you don't have the resonance that you anticipated.

Propene actually fulfil conditions for hyperconjugation (an alfa-CH adjacent to sp2 hybrid C atom) It is a delocalization of sigma bond electrons, also called no bond resonance. Mentioned C-H bond align in plane of the pi orbital. In such situation electrons of the sigma bond can be delocalized into the pi orbital. (Propene does not have resonance structures concerning only pi orbitals, as I suspect you think).
These are 3 "no bond resonance" structures of propene.

This site has diagrams concerning this problem, including mentioned contributing structures:
http://www.adichemistry.com/organic/basics/hyperconjugation/hyperconjugation-1.html


text:
Calculate the titer of a solution
I hope this is the right place to ask. We have this exercise without solution

One litre of diluted sodium hydroxide solution (c = 0.15mol/litre) is
to be prepared in the laboratory. The titer from the finished NaOH
solution is then determined. For this purpose, 0.62 g potassium
hydrogen phthalate (M = 204.2 g/mol, $pK_a = 5.4$) are dissolved in 0.05 litre
of water and titrated with the prepared sodium hydroxide solution
until turnover. The consumption is 0.022 litre. Calculate the titer of the
prepared sodium hydroxide solution.

On Wikipedia, I found "In titration, the titer is the ratio of actual to nominal concentration of a titrant" https://en.wikipedia.org/wiki/Titer
So if I understand correctly, the titer is just a fraction of the form $ t = \frac {c_{real}} {c_{ideal}} $, so a fraction of the real concentration over the ideal concentration, is this correct ?
In our case, we would simply need to use the titration equation at equivalence point: $$V_1 \cdot M_1 = V_2 \cdot M_2$$ where $V$ is the volume and $M$ the molarity
So we would solve
$$M_2 = \frac{V_1 \cdot M_1}{V_2} $$
and then the titer is $$t = \frac{c_{real}}{c_{ideal}} = \frac{M_2}{0.15} $$
because 0.15 mol/litre is given as ideal concentration in the exercise
$V_1$ and $V_2$ are given as 0.05 litre and 0.022 litre, and $M_1$ could be found by dividing 0.62 g by the molar mass (M = 204.2 g/mol), and then divide again by the volume of 0.05 litre

I am really unsure because there aren't a lot of results when searching on the Internet, so it seems to me that this isn't a frequently searched topic. Is my understanding above correct ? Why are there relatively few search results for the titer in Chemistry ? Even on this site, I did not find a similar question. Is this topic considered "not important" or am I missing something ?

Your approach is correct, although there is no information about which solution is considered as ideal, and which is real. It is also a pity you are not able to be more precise than $0.022$ liter as a final result. Two significant figures ($2$ and $2$) are not really reliable. Usually three or four significant figures are easy to obtain with ordinary burettes. The real volume should look like $22.02$ mL and not $22$ mL. Anyway the final titer should be equal to $0.935$.

Thanks a lot for your feedback ! Now that you say it, it's in fact strange to have two significant digits, this exercise comes actually from an old mock exam they gave us. I also wonder why there are few results for what a titer is, maybe I used the wrong terms for searching. And yes in fact now that you say it, there is no indication about which solution is real and which is ideal, I wonder why

@wengen. Usually a first solution is made to be "about 0.1 M" by dissolving an approximate amount of solute. Later on, a sample of this first solution is introduced in a conical flask and carefully titrated with a reference solution coming out of a burette. The volume so obtained can be known to ± 1 drop, so ±0.03 or ±0.05 mL. This titration allows you to calculate the concentration of the first solution with 3 - 4 significant figures. This concentration can be written down on the label (like a title = "titre" in French) on the bottle of the first solution. Titration comes from a French word.

Wikipedia is not the ultimate resource for classical volumetric titrations. This is why textbooks still exist. The definition of titer given in Wikipedia has nothing to do with the classical meaning of titer. There is no ideal or real solution in volumetric titrations. Titer in volumetric analysis means an expression of type:
1 mL (Given NaOH conc. X) =  Y (weight) of analyte.
In your case, it will be,
1 mL (Given NaOH conc. X) =  mg of KHP.
It is a quick way of analyzing solutions. Once an analytical chemist determines the titer of a titrant, all he/she need to know is the burette volume, and we directly get the weight of the analyte in one step. I will give you hints:
Write balanced equations of titration reaction. Calculate the actual molarity. This will give you X. From X, calculate the number moles of NaOH in 1 mL of that X NaOH. From the moles determine the moles of your target analyte based on the equation stoichiometry.

Interesting. That's almost how I did it in my lab courses, but never how I learned it. Do you have a source for this?

@Martin-マーチン, It would be good to start a new post on this topic with an example. I would be curious to learn the approach that was taught to you. I do not have a web reference but I think this was discussed in Vogel Quant. Analysis (1960s edition) if I recall correctly.


text:
Why don't serine and threonine have a 3rd pKa
Serine and threonine both have an extra OH functional group and hence they should be able to dissociate when the pH gets high enough. However multiple amino acid pH charts do not show the 3rd pKa of these 2 amino acids. Why is this so?

Serine and threonine have extremely high 3rd pKa values
Serine and threonine have a stable OH group, making it unlikely to dissociate except at extremely high pH values. To put things in perspective, physiological pH (the standard pH of the body) is about 7.4. The C-terminus is generally deprotonated at a pH of 1-3, so it's always deprotonated in physiological conditions. The N-terminus is generally deprotonated at basic conditions of 9-11, so it's unlikely to be deprotonated in the body. However, the pkR (side chain pka) for serine is 13, which makes it physiologically irrelevant. Thus, it is not found on amino acid tables together with other pka values.
For more on pKa values of amino acids, see here.
What makes the pKa of serine/threonine so high?
Other amino acids, glutamic acid and aspartic acid specifically, also have side-chain OH groups. These are stabilized by resonance from the carbonyl carbon. When deprotonated, the negative charge spreads out uniformly over both oxygen atoms. The same holds true for the C-terminus of every amino acid. They are stable in anionic form at low pH because of their strong resonance. However, serine and threonine, which have no resonance, don't have stable conjugate anions at low pH and therefore have high pKa values.


text:
what happens if benzene reacts with Cl2 in presence of FeBr3?
we know that halogenation of benzene adds a halogen to benzene in presence of a strong electrophile. but what if we use chlorine as a halogen but use an electrophile made out of a different halogen, like bromine?

Typically mixing a halogen electrophile with a catalyst using a different halide ion is not done in Friedel-Crafts halogenations. Here's why in the case described in the question.

The electrophile could be changed through halogen displacement. Here, that would be chlorine reacting with $\ce{FeBr3}$, independently of the benzene, so the electrophile in subsequent reaction with benzene becomes bromine instead of chlorine. Since the chlorine could alternatively act as an electrophile first, you get a mixture of halobenzene products.

When the proton is displaced, it will combine with a halide ion abstracted from the halide molecule/complex anion. But that could be bromide instead of chloride, giving a mixed hydrogen halide product and altering the catalyst (and perhaps its solubility and Lewis acidity).

So you risk mixing up the organic product, the displaced hydrogen halide and the catalyst. Don't go there. Match the halide ion in the catalyst with the halogen electrophile.


text:
I have plotted a Stern-Volmer Plot but cannot work out the fluorescence lifetime to be able to get the rate constant
I have plotted a stern Volmer plot and have my straight line equation. I understand that the gradient is kqtf but I need tf to work out kq but I don't have kd to do 1/kd.
How do I work it out?

Please give us more detail ! What are kq, kd and kqtf ? Which substance have you studied ? At which concentration ?

In a Stern-Volmer plot the ratio of fluorescence yield $\varphi$ is plotted vs the quencher concentration $[Q]$ as $\varphi_0/\varphi_Q =1 +k_{SV}[Q] $ where  the gradient is the quenching constant which is $k_2\tau$ where $k_2$ is the 2nd order rate constant and $\tau$ the fluorescence decay in the absence of quenching.
(If you search this site you will find examples https://applying-maths-book.com/intro.html )


text:
Why does FCC crystalline forms form hexagonal crystalline structures during CVD?
During the CVD process, Copper and Nickel which are both Face Centered Cubic crystalline structures are used as substrates for shaping hexagonal atomic structures such as graphene and hexagonal Boron-Nitride. Why is it that these form hexagonal atomic structures when they themselves aren't hexagonal?

But they are hexagonal, if you look at them sideways.

Sorry I dont quite grasp what you mean. Do you have a visual representation?

Fcc and hcp structures are similar stacking (fcc you need to look along the 111 body diagonal - a classic issue with the cubic conventional cell).

@AndiIacob for that check my answer to Silver (111) surface structure, and is bulk structure body-center or face-center cubic? Basically the "centered" lattices (fcc, bcc in 3D, centered rectangular in 2D) are contrived to be human friendly; the actual primitive unit cells are much smaller and oblique, and the 3D ones result in horizontal hexagonal planes when the "cubes" are stood up on their corners.

@uhoh so this means there are no modifications needed to the substrate correct? Om effect, any FCC could theoretically be used to make hexagonal shapes? And BCC as well?

@AndiIacob The internal fcc(111) planes for a simple metal will generally have a hexagonal arrangement, and a flat fcc(111) surface there certainly can be hexagonal as well. However sometimes clean, annealed fcc(111) (as well as other) surfaces have surface reconstructions, the most famous being Au(111) 22×√3  herringbone reconstruction, (amenable to googling). I don't know about BCC but I can find out in a day or so. Si isn't simple fcc, but the Si(111) surface has a very interesting 7×7 reconstruction.

@AndiIacob you can also try the hex face of a hexagonal crystal like hexagonal boron nitride (as you've mentioned) or hexagonal silicon carbide, or even cleaved HOPG.

[Ivan Neretin] But they are hexagonal, if you look at them sideways. –

[OP] Sorry I dont quite grasp what you mean. Do you have a visual representation?

Here is the visual.

[Jon Custer] Fcc and hcp structures are similar stacking (fcc you need to look along the 111 body diagonal - a classic issue with the cubic conventional cell).

Yes, both may be constructed by stacking hexagonal layers, either in an ABABA.. (hexagonal close packing hcp) or an ABCABCA.. (face-centered cubic fcc) fashion.

Nice visual there!

I've linked to your answer here +1


text:
Length of a 1D box in hexa-1,3,5-triene
Problem
From Hayward's Quantum Mechanics for Chemists [1, p. 36]

2.3. Calculate the wavelength of light that will be absorbed when
a it electron in hexa-1,3,5-triene, $\ce{CH2=CH—CH=CH—CH=CH2},$ is promoted from the highest occupied level to the lowest unoccupied level. The average $\ce{C—C}$ bond length in hexatriene can be taken to
be $\pu{144 pm}.$ Compare your answer with the experimentally observed
wavelength of $\pu{258 nm}.$

Answer

 2.3. Wavelength of light is $\pu{352 nm}$ (for box length of $\pu{864 nm})$

Question
Since $L = 5\times(\pu{144E-12 m}),$
$$E = \frac{7h^2}{8mL^2} = \pu{8.14E-19 J}.\tag{1}$$
After substituting the values I get the answer
$$\lambda = \frac{hc}{E} = \pu{244 nm}.\tag{2}$$
The textbook answer is $\pu{352 nm}.$ What confuses me is that they state that the length of the box is $\pu{864 nm}$ — but there are five bonds joining the six carbons together. So, would not we multiply the average bond length by $5$ and not $6?$
Is it acceptable to take the length of the box as the average bond length multiplied by the number of atoms?
Reference

Hayward, D. O. Quantum Mechanics for Chemists; Tutorial chemistry texts; Royal Society of Chemistry: Cambridge, UK, 2002. ISBN 978-0-85404-607-2.

Where is 7 coming from? The value of $\lambda$ is too high.

Please check the units, m, mass of electron in kg, c in m/s etc. then the longest transition wavelength is found as the difference from the top filled energy level  to the next one. The answer given is correct for the box length of 6 , perhaps someone else knows why 6 is used.

My apologies- I have edited this question and the proposed alternative answer using a box length of 5 should now be correct.

I remember reading that the pi electron cloud extends half a bond distance further than the last atom on the two ends of the molecule, which would add one more bond in total.

@porphyrin A length of 6 is typically used if you compare the particle on a ring model for benzene with the corresponding linear particle in a box model, see for example https://doi.org/10.1021/acs.jchemed.2c00523

@Loong,  Um, it is still an arbitrary fix; its the sort of thing that makes students sceptical about what they are taught, i.e. fix some parameter to get the result you want. I think that it would be better to say the agreement with expt. is expected to be poor, and it is,  as its such a simple model and start from there.

Related: https://chemistry.stackexchange.com/q/44676/4945

For this type of question, we not only consider the length of the bonds between the carbon atoms, but we also have to add on the atomic radii at both ends of the structure. I assume the atomic radii of carbon atoms taken in the text is $\pu{72 nm}$, which means doing $\pu{144 nm}\cdot5 + \pu{72 nm}\cdot2 = \pu{864 nm}$.
We can then solve it like you did doing $E = (4^2-3^2)\frac{h^2}{8mL^2}$ and the applying $\lambda = \frac{hc}{E}$. Ultimately, however, the addition of the two carbon radii to the length of the box would explain the textbook result (see page 6 of the link below).
Source: https://nanohub.org/courses/OED/01a/asset/5958#:~:text=Using%20the%20simple%20model%20shown,estimated%20to%20be%200.867%20nm.

But what is the reason for adding the atomic radii, it makes no sense and is just a fix, i.e. not rational or scientific, so as to get closer to the expected answer.

@porphyrin I mean, it is simply a model, but for it, I think that's what you have to include. From ChemLibreText: "A diagram of the particle-in-a-box potential energy superimposed on a somewhat more realistic potential. The bond length is given by β, the overshoot by δ, and the length of the box by L = bβ + 2δ, where b is the number of bonds."

yes of course its a only model but my point was that it the extension bit is not really based on anything other than to try to fit a number by fiddling the calculation by adding an arbitrary parameter, $\delta$ etc. The real point is that even though PIB is so simple it gets quite close in energy terms to something that be measured. That should be enough. Actually it is surprising it gets so close.


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My experiemental density of H2 does not match the published data. What causes this discrpenecy?
I placed 1 gram of NaBH4 in a balloon and placed it over a glass round bottle filled with water and acetic acid. While secured, I emptied the balloon into the bottle and made sure to wash the inside of the balloon. The reaction taking place is $\ce{NaBH4 + 2(H2O) -> NaBO2 + 4(H2)}$.
The yield in this case should be 213 grams of $\ce{H2}$. According to Wikipedia, "hydrogen" has a density of 0.08988 g/L. That should leave me with 2.369 L of $\ce{H2}$. But my little experiment showed a balloon with the about 13 cm in diameter, which a volume of sphere that wide would be 1.15 liters of H2. This would mean a yield of .103 grams or 48.5%. But that just doesn't make sense. My NaBH4 has not degraded by half.
If you were to double the given density of $\ce{H2}$ to 0.17976 g/L, the results are much more in line with what I'd expect. The yield of $\ce{H2}$ would be .206 grams or 97%, which would make sense since not everything is ideal and my $\ce{NaBH4}$  is a bit old. Obviously there will be minor errors in measurement, and its not a perfect sphere and I didn't control for temperature, but that cannot account for a 50% reduction in yield. Maybe the balloon is compressing the gas? But I don't think so.
So the only other thing I can think of is that the presented density of 0.08988 g/L for hydrogen is technically only for monoatomic Hydrogen, or $\ce{H}$, which doesn't make sense and no one ever uses. The density of $\ce{H2}$ at STP should be 0.17976 g/L then to match my observations. There must be something I'm missing here entirely that I'm not understanding.

It is always good to summarize and/or complement the verbal description by algebra.
Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger  permanent value.
You may find useful  formatting mathematical/chemical expressions/formulas.

Balloons leak hydrogen easily. The pressure inside an inflated balloon is greater than ambient pressure.

What is missing is that you have lept to assume the issue is with the density despite having chosen a way of measuring density that is a crude approximation and a way of collecting the gas that is unreliable for this purpose. Good technique would collect hydrogen by gas displacement (eg pushing water from an upturned measuring cylinder) which would give a reliable idea of the volume of hydrogen, a far better way of measuring the amount than estimating the approximate density.

The pressure inside your balloon is not approximately 1 atmosphere, even at ambient temperature. It is substantially greater, as it has to counteract atmospheric pressure plus the elastic force of the balloon latex trying to pull itself together. Therefore, your sample of hydrogen gas is not close to STP conditions, and its molar volume and density should not match. Also annoyingly, the pressure inside a balloon is sensitively dependent on how inflated the balloon is, and it varies non-monotonically - it starts at 1 bar, increases up to a maximum, then decreases again. This is not a very good way to measure gas volumes.
If you want a easy and fairly reliable way of measuring gas volumes on the order of a liter, a typical procedure is to fill a container (usually a graduated cylinder for extra precision) with water, invert it, and measure the water displaced by the evolved gas. For volumes on the order of milliliters, you can displace the plunger of a gas-tight syringe.
Edit: Actually when I wrote this answer, I trusted the calculations were overall correct, but Karsten's comment tipped me off to some errors, so I decided to double-check. So let's calculate the theoretical volume of hydrogen gas produced:
$$\ce{NaBH4 (s) + 4 H2O (l) -> NaB(OH)4 (aq) + 4 H2 (g)}$$
(Arguably your proposed sodium metaborate product is actually best described as sodium tetrahydroxyborate in aqueous solution, though that doesn't affect the proportionality between moles of borohydride used and moles of hydrogen gas evolved.)
$$\mathrm{1.00\ g\ NaBH_4 \times \frac{1\ mol\ NaBH_4}{37.83\ g\ NaBH_4 }\times \frac{4\ mol\ H_2}{1\ mol\ NaBH_4}\times \frac{22.4\ L\ H_2(@0\ °C, 1\ atm)}{1\ mol\ H_2}=2.37\ L\ H_2(@0\ °C, 1\ atm)}$$
(This volume can change by ~10% depending on the actual temperature of the gas inside your balloon. At 25 °C, you would get 2.60 L instead.)
So even though your calculation is a bit roundabout and contains some mistakes, it does seem the major error is experimental.

First, a minor consideration about gases:
In an experimental setup like yours, I would not really try to deal with a gas mass and density. The molar volume of any ordinary gas is ~22.4 liter. Easy to remember and one size fits all.
Second, the experiment results:
Armed with the above consideration about the molar volume, I would ask myself first not "what's wrong with my hydrogen?", but instead "why I am getting too little gas?" .
This alternative question has two important advantages:

It does not challenge a well-established fact
It is way easier to propose answers.

The possible answers may be:

The purity of the substances used (you state that "not everything is ideal with my..." . How can the possible impurities react in similar conditions?
Parasitic reaction paths that can yield less hydrogen (well, I cannot think of one in your particular case)
Did you collect all of the gas? Can some of it be dissolved in the solution? (there is solubility data available online - at least about hydrogen in water, you can get this as an estimate for your final solution)
What are my margins of error in both measurement and calculation? You start with 1.0g +/- 1% or 1.0g +/- 20% ?

For the point 1 question, you can stage a much easier experiment that actually involves measuring the mass of the hydrogen: weight your reacting substances, mix them and wait until the reaction finishes, weight again. Subtract the final mass from the initial mass.
You may, or may not get something like 213mg .


text:
Why does zircon hate lead? How do these tiny crystals so effectively exclude lead atoms during formation enabling accurate uranium-lead (U-Pb) dating?
Background
From the December 29, 2022 NPR article To peer into Earth's deep time, meet a hardy mineral known as the Time Lord:

"(Zircons) are really the best markers of Earth's time, or the history of the Earth," says Michael Ackerson, a geologist with the Smithsonian's National Museum of Natural History.)

From the podcast linked on the same page after 01:00:

He tells me that one of the crystals I'm looking at is 4.32 billion years old. Zircons are the oldest known pieces of Earth that still  exist on the surface today; the oldest go back as far as 4.37 billion years.
"They are the best markers of Earth's time, (for) the history of the Earth.
Ackerson says a zircon crystal originally forms in magma - hot molten rock. Other minerals do too; together they'll make up say, a granite mountain, that slowly weathers away.
"Most of the minerals don't survive, they slowly weather away. Quartz, feldspar, they're chemically or physically weathered or eroded, to a point where they're no longer quartz and feldspar. Zircon, and one of the reasons that zircons are so useful, is that zircon is very resilient.
The hearty crystals persist, and eventually get incorporated into another rock as it's forming. That means scientists can crush up the Earth's oldest rocks and pick through the debris, to find little grains of zircon, that are even older.
To know a zircon's age, they zap it with a laser.

Discussion of the use of lasers, argon plasma and detectors to count uranium and lead atoms.

The important atoms are uranium and lead. A growing crystal of zircon loves uranium and will take it in, but zircon hates lead, so if you find lead inside, it pretty surely came from the decay of uranium, which happens at a steady rate, like the ticking of a clock.

Question:
Why does zircon hate lead? How do these tiny crystals so effectively exclude lead atoms during formation enabling accurate uranium-lead (U-Pb) dating?
I'm interested in a basic understand of the chemistry that allows a growing crystal of zircon in magma to so effectively prevent uranium from being incorporated while allowing so many other metals in.
Besides uranium and obviously zirconium and silicon, Wikipedia's zircon gives the formula for zirconium silicate as $\ce{(Zr_{1–y}, REE_y)(SiO4)_{1–x}(OH)_{4x–y}}$ where $\ce{REE}$ stands for rare earth element:

...a set of 17 nearly-indistinguishable lustrous silvery-white soft heavy metals.

The Wikipedia zircon page also says:

Zircon precipitates from silicate melts and has relatively high concentrations of high field strength incompatible elements. For example, hafnium is almost always present in quantities ranging from 1 to 4%.

This seems potentially relevant: Chapter Six - Computer modeling of Zircon (ZrSiO4)—Coffinite (USiO4) solid solutions and lead incorporation: Geological implications

The structure of zircon, ZrSiO4 is modeled using interatomic potentials. The uranium end-member, coffinite (USiO4) and intermediate solid solutions of zircon and coffinite (UxZr1–xSiO4) are then modeled, allowing the prediction of lattice parameters as a function of uranium concentration. Finally, possible structures resulting from the radioactive decay of uranium to lead in coffinite are considered.

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

Zircon is rather picky because of its rather unusual composition. It has the composition $\ce{MSiO4}$ where $\ce{M}$ is predominantly zirconium (which is of course named after the mineral) but generally includes other elements in solid solution. Bevause the $\ce{SiO4}$ group with the normal oxidation states of silicon and oxygen takes four negative charges, the metal $\ce{M}$ must match the $+4$ oxidation state of the zirconium, at least on average. The metal must also have relatively large atoms in this oxidation state to substitute comfortably for the zirconium. Only lanthanides, actinides, and a few early-group transition elements in the fifth or later period fit these criteria. Wikipedia specifically mentions "high field strength incompatible elements" including $\ce{Zr, Nb, Hf, Ta, Th, U}$ and rare earth elements.
Uranium, which commonly adopts the $+4$ oxidation state, works well, but lead does not. Lead is too difficult to achieve the $+4$ oxidation state because of the relativistic impact that emerges in heavy late-group elements. Wikipedia discusses this relativistic effect:

Lead shows two main oxidation states: +4 and +2. The tetravalent state is common for the carbon group. The divalent state is rare for carbon and silicon, minor for germanium, important (but not prevailing) for tin, and is the more important of the two oxidation states for lead.[1] This is attributable to relativistic effects, specifically the inert pair effect, which manifests itself when there is a large difference in electronegativity between lead and oxide, halide, or nitride anions, leading to a significant partial positive charge on lead. The result is a stronger contraction of the lead 6s orbital than is the case for the 6p orbital, making it rather inert in ionic compounds.

Cited Reference

Greenwood, N. N.; Earnshaw, A. (1998), Chemistry of the Elements (2nd ed.), Butterworth-Heinemann, ISBN 978-0-7506-3365-9, p. 373.

This is fascinating! 1) Can you add some kind of supporting information for the assertion that Pb's difficulty achieving +4  is "because of the relativistic impact that emerges in heavy late-group elements"? 2) In "...substituents must match the +4 oxidation state of the zirconium, at least on average..." the "at least on average" seems to offer a back door for Pb to get in at a low concentration, as long as something else offsets it, whereas the underpinning of the U-Pb dating technique relies on essentially NO Pb. Anyway, can you add a supporting source for the "at least on average" also?

@uhoh Lead isotopic composition is significant too, as uranium-238 forms lead-206, uranium-235 forms lead-207, while all lead-204 is primordial. If there is no/trace lead-204, then all/near all lead-206/207 was formed by on situ decay of uranium. Isotopic composition can be conveniently determined by mass spectroscopy after zircon laser evaporation.

@Poutnik it may be a little more complicated; 1) if the source of the lead that I'm worried about is not a nice mixture of "natural abundances" but radioactive decay prior to zirconium crystallization, then one can't use natural abundances. 2) Also, one might call into question what "natural abundances" even means in the context of crystals that "go back as far as 4.37 billion years." I think there is a good opportunity for a follow-up question about all the assumptions involved, and precautions taken when using U-Pb dating in the first place.

Note that I have said nothing about natural abundances. There is no problem to determine lead isotopic abundances in adjacent rocks. // And there is remaining problem how to get lead to zircon.

@otoh From current uranium content in zircon, and its 235/238 isotope ratio, it can be determined the lead content and lead 207/206 isotope ratio, if all is product of in situ decay. So it can be also determined if it was not, if lead content and lead isotope ratio would not match.

@Poutnik I think you are converging on a great answer! Please consider posting one, thanks.

@uhoh  Lead-206/207 isotopes do not have different chemical properties than lead 204. If lead-204 does not get in zircons,  lead-206/207 do not either. Simple as that.

@Poutnik if you have an answer to the question as-asked, please go ahead and post it.


text:
Why do nuclei move considerably slower than electrons
I've been trying to learn quantum chemistry at an introductionary level. While reading I've found out that the Born-Oppenheimer approximation seems to be the reason for the basic and crucial model of molecules for chemists; molecular structure in space as nuclei occupying some positions with chemical bonds between them. Thus, I've tried to learn more about this approximation. However, I've gotten stuck at the reasoning for wanting to try to separate the wavefunction of the molecule into a product of two functions, one depending on the positions of the nuclei and one depending on the positions of the electrons as well as depending parametrically on the positions of the nuclei. All resources state that the reasoning is because nuclei move considerably slower than electrons, but I have not found one that could elaborate why this is the case.
Most books and other resources simply states that "Because the nuclei is over thousand times heavier than an electron, in the timescale of the movements of the electron the nuclei can be considered to be standing still.", or "Because of the great difference in mass, the electrons can respond almost instantaneously to the movements of the nuclei", and then quickly moves on. However, in this answer on "Born–Oppenheimer adiabaticity" it was said that "nuclei at average are moving considerably slower than electrons" because of the equipartition theorem.
In the book Introduction to Statistical Physics by S. R. A. Salinas, the theory of equipartition of energy is formulated by:

Consider a classical system of $n$ degrees of freedom given by the Hamiltonian $$H = H_0 + \phi p_j^2$$ where (i) $H_0$ and $\phi$ are functions that do not depend on the particular coordinate $p_j$; (ii) the function $\phi$ is always positive; and (iii) the coordinate $p_j$ is defined from $-\infty$ to $+\infty$.

It then shows how the expectation value of $\phi p_j^2$ equals $\frac{1}{2}k_BT$, where $k_B$ is the Boltzmann constant and $T$ temperature. Thus, we can conclude that for any system (in thermal equilibrium) that has energy contributions that are quadratic in such a variable, the average energy is $\frac{1}{2}k_BT$ for each of those degrees of freedom. Since kinetic energy is always quadratic in the momentum, it will have an average energy of $\frac{1}{2}k_BT$ in thermal equilibrium in each degree of freedom if the rest of the Hamiltonian does not depend on the momentum (which is the case for the Hamiltonian of a molecule). This elaborates the reasoning in the linked answer. However, this seems to "only" apply for macroscopic systems of molecules which also have to be in thermal equilibrium. I am not completely satisfied with this, especially considering when this is only about the average kinetic energy. I would assume that the distribution of kinetic energy for the different degrees of freedom in a general system at room temperature is such that a decent portion are thousands of times higher than others. I could not find any answer that rectified this.
One book I read mentioned something different than others; that the nuclei move considerably slower than electrons because of conservation of momentum. However, it gave no elaboration. One text I remember reading somewhere on a website said something along the lines of: "The electrostatic forces on nuclei and electrons from their pairwise interaction are the same, and so the changes in their momenta from these forces must be the same. Therefore, one might assume that the actual momenta of the nuclei and electrons are of similar magnitude." This is the explanation for the reasoning that makes the most sense to me, but I have not seen it anywhere else. Since it came from a source that wasn't necessarily trustworthy it made me unsure, until I saw the statement of conservation of momentum which felt related.
Does the statement of equipartition make sense as the explanation for the reasoning? If yes, why? Does the statement of changes in momenta make sense as the explanation for the reasoning? Or maybe there is something different that is the sensible explanation? Please do help me understand more clearly, I've been at this for weeks and I cannot seem to get any answer from books and other texts.

+1 just for "trying to learn quantum chemistry" :-) Welcome to Stack Exchange!

As kinetic energy is proportional to mass and velocity as $mv^2/2$ with the same energy the heavier particle moves more slowly.

If you consider the simplest atom, the mass of a proton is aproximately 1836 times the mass of an electron, so if both particles have the same kinetic energy, the ratio of their speeds is $v_e = \sqrt{1836} v_e$. Considering that in a stationary state both particles will have a constant energy and that you can partition it between as E = K + V,  the electron responds faster to a change in the fraction of potential energy.

Considering zero total momentum of the atom,  $\sum{m_i \vec v_i} = 0$

@uhoh Thank you. (:

@porphyrin I understand that part, but how do you know nuclei have the same kinetic energy as electrons?

If you hit a ping-pong ball with a paddle, and if you hit a billiard ball, just as hard with that paddle, imparting the same amount of energy, which ball will move faster?

@Atsjo, equipartition of energy


text:
When we take baths, how does the water in the bathtub absorb only the dirtiness on our skin?
I'm 16. I study chemistry in college along with biology, physics, and maths, and I have a general interest in all of them.
I always found this interesting: sometimes when I left the bathtub, even when I forgot to use soap, the bath would still have a somewhat noticeable, murky tint in the water, indicating that something had dissolved in it, even though I didn't take the time to rub it off manually. When we get in the bath, how does the dirtiness (small amounts of mud, bacteria, pen markings etc.) dissolve in the bath alone without any of our skin dissolving as well so that we can leave clean? Maybe dead skin might dissolve, that would make sense, is there an adaptive mechanism on the skin involved in this sort of process of hygiene? The same thing would happen if I put any other object in the water with something on its surface and the object would come out clean when the thing on the surface dissolved.
How does the water break the attraction? Some sort of cohesive force, between the atoms/molecules of the thing (dirtiness) on the surface of the object and the atoms/molecules of the object (our bodies) and separate them to make it dissolve? And how can it differentiate between the cohesion of the skin and the dirtiness (what determines solubility, if both polar and non polar molecules on our skin will dissolve in the water)?

Usually taking a bath involves the use of soap to speed up the process; in fact, it may be hard to answer because baths without any soap may be outside most people's everyday experience, and that might be because water alone does not work as well as you describe it. Yes the tub got dirty - probably some fraction of the top most layer of dead skin cells fell off and floated to the edge of the water, but I do not think you were able to get actually clean (yes, it's a subjective term) without soap.

You can "get clean" using water alone. The next time you experience a summer thunderstorm, modesty permitting, strip and wash thoroughly. You will be cleaner and more refreshed than any soapy shower.

Water is called a universal solvent, and perhaps it is rather a miraculous solvent with unusual properties. It is able to "wet" a lot a surfaces and thus able to disrupt the adhesive forces. (Cohesion refers to sticking of like molecules only.)
However, the murkiness in water is not due to dissolution of dead skin or dirt dissolving in water, but rather it is a suspension (hence murky). If you had a microscope, you would rather see particles of dust, mud, soil, oils, dead skin etc if someone has not taken a shower for a while. Just pouring water does not remove these things from objects, mechanical action is also needed to dislodge them from the surface (e.g., dishwashers and washing machines use mechanical forces to clean items with water).
Since you appear to like science, next time do a Tyndall light experiment. Take that murky tinted water (as you call it), and fill it in a transparent glass tumbler. Pass light through it, and view that water at ninety degrees (to the light beam). If you see a light beam, it indicates really small particles floating in water. It is not a true solution.

The outer layer of skin is water-insoluble. According to this paper, there is a structure specifically made in skin cells before they die and become part of the outermost layer of the skin:

The outermost, cornified, layers of the epidermis are composed of terminally differentiated keratinocytes known as corneocytes. Corneocytes lack a plasma membrane and instead are encased in a structure known as the cornified envelope (CE). CEs consist of highly insoluble, cross-linked proteins with covalently attached lipids and are essential for the mechanical integrity and water impermeability of the skin.

So not dissolving in water is one function of the skin. It also protects the inside of the body from dehydration by preventing excess evaporation of water on the skin (unless you are hot, when sweating helps to maintain a lower body temperature, also important for survival).

[OP] How does the water break the attraction?

That depends on what is on the skin. If you are muddy, the components of the mud are easy to wash off with water because the components are polar, and water is polar. If you are covered in motor oil (which is non-polar), water alone will not be efficient to wash it off, but soapy water will work (with some scrubbing).

[OP] When we get in the bath, how does the dirtiness (small amounts of mud, bacteria, pen markings etc.) dissolve in the bath alone without any of our skin dissolving as well so that we can leave clean?

While the layers of dead skin cells are not water-soluble (or soapy water soluble), we nevertheless continually shed these former cells, and continually replace them. In fact, much of household dust is made of material we shed (e.g. it contains quite a bit of cholesterol, as described here).

When we take baths, how does the water in the bathtub absorb only the dirtiness on our skin?

It doesn't. When you take long bubble-baths, your skin will also lose oils and other substances (giving you "dry skin"). Some people use oils or lotions ("moisturizer") to replenish these after a long soak.