text:
How does Entropy influence the racemization of an optically active compound?
We were doing the topic "Isomerism" in our Organic Chem class and there's a question in our worksheet whose answer I haven't really understood, the question is as follows:

The racemization of optically active compounds is driven by? (a): Entropy (b): Enthalpy (c): Entropy and Enthalpy (d): Element of Symmetry

The worksheet answer key says the answer is (a): Entropy.
But I wasn't so sure of this answer, although I could rule out that it isn't (d) for sure since, for an optically active compound, there shouldn't be an element of symmetry (or it would be Inactive). But I can't quite understand why the answer should be Entropy, I always thought entropy just means the degree of randomness like particles just vibrating.
It would be of great help if you could tell me if the answer is in fact Entropy; and if so, could you perhaps explain how Entropy would influence this optical activity.
Thank you for your time!

The two species are related only by mirror symmetry so have the same enthalpy. Starting with just one enantiomer racemization produces two types of molecules so has increased the entropy

See also:  https://chemistry.stackexchange.com/questions/103915/is-there-a-thermodynamic-driving-force-for-racemisation

Where there is a mechanism to interconvert enantiomers, entropy will drive racemisation
There is some context that is helpful but omitted from the question. Not all chiral compounds racemise under normal conditions and you need an easily accessible mechanism for racemisation to occur.
If there is such a mechanism, then a pure chiral compound will racemise over time (as the thalidomide molecule does in typical biological conditions with tragic results). If there is no mechanism, it will not (there is no easy way to convert the two chiral forms of camphor). In the first case the driving process is entropy. The two isomers are indistinguishable thermodynamically so the racemisation is not driven by differences in the energy involved. The point is that the chemical process is random and can have no memory of the original optical isomer, so the result of the interconversion reaction is random and gives both isomers equally. This is one way of saying that the apparent direction of the reaction (from optical purity to racemic mixture) is purely entropy driven.


text:
How do you rewrite molarity to weight percentage or volume percentage?
Suppose I have a molar concentration $[C]$ in $mol/m^3$ in a continuous reactor, from which the differential equation is as follows:
$ \frac{\partial [C]}{\partial t} =  k[A][B] + \phi C_{,in} - \phi C_{,out}$
Where the first term is the rate law with rate constant $k$, and with concentrations of reactants $A$ and $B$, as $A+B \longrightarrow C$.
The second and third terms are inflows and outflows of substance $C$ in a continuous reactor.
The outflow of $C$ is measured, and therefore I need to have each term correlate with each other by having the correct dimensions. I have the following question:
Suppose $C$ is a gaseous substance, and the measured output is in $vol\%$ (volume percentage), how would I convert $\phi C_{,out}$ or $\phi C_{,in}$, which is in $[Nm^3/s]$ (normal cubic meter per second), to $vol\%C$? I found the following:
$\hspace{30pt} \phi_{C_{,out}} = \dfrac{p\dfrac{vol\% C}{100\%}}{RT}\phi_{out}\hspace{10pt}$ but this leads to a wrong dimension $[m^3/mol]$ for the $vol\%$

Try to derive both percentages from molar concentration from the scratch, using the respective definitions and known ideal gas laws.

Could you maybe give a hint? I'm a bit stuck

Describe being stuck. // Write down for yourself definitions of considered quantities (molarity, weight percentage, volume percentage) and relations between them. Then you are done.

Relation of molarity and molar fraction ( = volume fraction for ideal gas approximation), using ideal gas state equation:
\begin{align}
c &= \frac{n_1 }{ V} \tag{1}\\
pV &= nRT\tag{2}\\
c = \frac{n_1 }{ \frac{(n1+n2)RT}{p}}&=\frac{p}{RT}\frac{n_1}{n_1+n_2}\tag{3}
\end{align}
Computing volume fraction and percentage from molar concentration:
\begin{align}
\varphi&=c \cdot \frac{RT}{p}\tag{4}\\
vol\%&=\varphi \cdot 100\tag{5}
\end{align}
Dimension of $\frac{cRT}{p}$ is
$$\frac{\pu{mol m-3}\pu{J K-1 mol-1}\pu{K}}{\pu{N m-2}}= \frac{\pu{ m-3 J }}{\pu{(N m)(m-1 m-2)}}= \frac{\pu{ m-3 J }}{\pu{J m-3}}=1\tag{6}$$
Computing mass fraction and percentage from volume fraction:
\begin{align}
w_1&=\frac{\varphi_1 M_1}{\varphi_1 M_1 + (1-\varphi_1)M_2}\tag{7}\\
mass\%&=w \cdot 100\tag{8}
\end{align}
Legend:

$c$ - molar concentration $\pu{mol m-3}$
$n_1$ - analyte molar amount [$\pu{mol}$]
$n_2$ - molar amount of the mixture but the analyte[$\pu{mol}$]
$V$ - volume of mixture [$\pu{m3}$]
$R$ - gas constant [$\pu{8.314 J K-1 mol-1}$]
$p$ - gas pressure [$\ce{Pa}$]
$\varphi$ - volume fraction
$\varphi_1$ - the volume fraction of the analyte
$w$ - mass fraction
$w_1$ - the mass fraction of the analyte
$M_1$ - molar mass of the analyte [$\pu{g/mol}$]
$M_2$ - molar mass of the major gas [$\pu{g/mol}$]
( or the effective mean molar mass of used gas mixture, e.g. $\pu{28.8 g mol-1}$ for air)

Thank you so much! What would be the difference between $\varphi_1$ and $\varphi$?

Aside of general and particular context, it was just a typo of missing index in the equation.

That seems logical, thanks. If I understand correctly, this is only useful for gaseous analyte $A$ in a mixture or major gas $B$ right? What would be different if it would be a solid mass for the mass fraction?

For volume/mass fraction/percentage, you would need density of the components and the mixture at given conditions. Rest is trivial high school knowledge. // Do not expect receiving answers  to your questions on StackExchange network  without explicit a priori effort to answer it yourself.

source: Stack Exchange

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text:
Why sodium and calcium swapped their positions in electrochemistry series and metal reactivity series?
According to metal reactivity series, Na is more reactive than Ca. But according to the electrochemical series, Ca is a stronger reducing agent than Na. Why is that?

What is the definition and criterion of reactivity? It is undefined.

The electrochemistry series is based on the thermodynamic parameter - standard redox potential, directly linked to standard molar Gibbs energy of a formal reaction (See e.g. this libretexts doc):
\begin{align}
\ce{Ox^{\mathit{x}} + \frac {n}{2} H2(g) &-> Red^{x-n} + n H+(aq)}\\
\Delta G^°_\mathrm{r}&=-nFE^°
\end{align}
The latter equation is reformulation of the equation $\ce{W=q.U}$(non-V work, charge, voltage), known from electronics or electrostatics.
The reactivity series is vaguely defined and there must be specified scenario and criteria. The reactivity is primarily a kinetic parameter, that is loosely related to thermodynamics. But can be defined in thermodynamic context as well. Reactivity order may be different for different reaction scenarios, including also dry and hot scenarios with melted reagents.
Comparing metals from the same group, lithium and sodium, the former has lower standard redox potential = stronger reducing agent in water context. But, lithium is less reactive in contact with water, with slower reaction rate than sodium.
In an outdoorsy analogy, the difference between the trip start and destination altitudes (thermodynamics) and the altitude difference from the start to the saddle (=activation energy = kinetics) are mutually independent.
The "calcium trip" goes to a  lower-altitude valley.
The "sodium trip" goes across a lower-altitude saddle.

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text:
Calculating conductivity from Nyquist plot, plot has no semi-circles?
I am familiar with the following process for calculating conductivity from a Nyquist plot:

Run impedance on potentiostat
Plot - Imaginary Z vs Z
Generate equivalent circuit
Fit data
Calculate conductivity by entering the resistance value, thickness, and diameter of sample

My confusion is that, in the past, my spectra always had a semicircle. Now I am running samples which are giving basically a 45 degree line that starts to the right of 0 on the X axis. I believe I have an equivalent circuit (Resistor + Constant Phase Element/Resistor + Warburg element). My question is, are you able to calculate conductivity as long as you can extrapolate the resistance? In other words, if you have an equivalent circuit which contains a resistor, can you always calculate the conductivity? Or do you need a semi-circle? Thanks.

Interesting question. I guess most of chemists do not consider the complex impedance or admittance.


text:
Troubleshooting synthesis of halazone from dichloramine-T
I was trying to oxidize some dichloramine-T to halazone. It should be easy enough since dichloramine-T has a free methyl group para to the para-sulfomyldichloro group. I followed a published procedure [1], but got very different results.

By oxidation of easily available dichloramine-T (3) with $\ce{KMnO4}$ in mild alkaline medium, we have been able to obtain as high as $95\,\%$ halazone containing $18-20\,\%$ chlorine. In this procedure, the sodium salt of halazone (4) is formed in the first step which, after hydrolysis with dilute $\ce{AcOH},$ gives a white crystalline compound with a chlorine odor, melting at $\pu{196 °C}$ with decomposition.

Experimental
In a $\pu{250 ml}$ flask, $\ce{KMnO4}$ $(\pu{2.9 g})$ was placed and dissolved in water $(\pu{150 ml}).$ Dichloramine-T $(\pu{2.5 g})$ and $\ce{Na2CO3}$ $(\pu{5 g})$ were mixed and added to the flask and
stirred for $\pu{5 min}.$ The reaction mixture was refluxed on a steam bath until the dark violet color disappeared $(\pu{2 h}).$ The mixture was then cooled at $\pu{25 °C}$ and filtered. A dark brown precipitate was separated $(\ce{MnO2}).$ The filtrate was yellowish with a chlorine odor which was concentrated and later neutralized with $\ce{AcOH}$ $(50\,\%).$
While neutralizing the solution, a vigorous reaction occurred and a white substance precipitated which was filtered and dried. Yield $95\,\%,$ m.p. $\pu{195-196 °C}.$ This compound was identified as N,N-dichlorosulfamoylbenzoic acid by direct comparison (ir, mass, m.m.p.) with an authentic sample.

My two main problems are:

I can't precipitate the product out of the solution as mentioned. My mother liquor stays colorless and no precipitation occurs upon acidification.
I can't separate the $\ce{MnO2}$ particles from the solution using normal methods like filter papers/cloth and have to use Hyflo® Super Cel® filter aid each and every time.

I've used chromic acid, chromyl chloride, $\ce{KMnO4},$ $\ce{H2O2}$ all to try and oxidize the compound with no avail. Really need some help with this.
Reference

Saljoughian, M.; Sadeghi, M. T. An improved procedure for the synthesis of p-(dichlorosulfamoyl)benzoic acid (halazone). Monatsh Chem 1986, 117 (4), 553–555. DOI: 10.1007/BF00810903.

Does your mother liquor have the chlorine odour?

@Waylander Yes and it smells bloody awful

Have you checked the pH of the ML that you are trying to precipitate from?

@Waylander Yep. Its about 11.85-12.05. Everything checks out. I did the same reaction but I oxidized para toluene sulfonamide instead of Dichloramine T everything else (amount of reagents etc) being the same and I isolated the oxidized para sulfamoyl benzoic acid with about 70% yield. I just can't seem to oxidize Dichloramine T

Can you concentrate down an aliquot and get an NMR of the residue?

I don't have an NMR available at hand so I'll just boil down the aliquot to get an FTIR and if I detect C=O and C-O peaks, I'll atleast know the way I am proceeding is right

@Waylander So I detected COOH group in the FTIR and re ran the experiment taking the Raw Material in excess and isolated the product with about 10 % yield.

Whenever a procedure says "as high as 95%" you know they usually get a lot lower, but I'd still expect better than 10%.

Can you help me @Andrew ? I have tried nothing and I am all out of ideas. Also, Dichloramine T is water insoluble. Nothing else can be oxidized on that ring. I just don't understand where is all the Dichloramine T going. I mean atleast it should precipitate out of the system right? Why did I get only 0.6g ppts out of a 6g batch


text:
Modelling barrier-less reactions
I am currently exploring a few dissociative reactions whose PES is barrierless (no first order saddle point). I have found a few approaches in literature like flexible nudged elastic band theory, Newton string method, URVA etc to identify the Transition state structure. Say I use one of these methods to find the structure and obtain its vibrational frequencies, can I use it in standard RRKM calculations to find the rate constants ? Or should I follow theories like VRC-VTST which finds the rate constant directly ? What is the preferred approach in problems like this.

This is slightly related, but from a different time, where the methods you mention were not as wide spread. https://chemistry.stackexchange.com/q/9957/4945 I would recommend explaining the abbreviations you have used. This enhances not only searchability but it'll help other people get familiar with the terms. I personally found NEB a good tool to look in theses things. Iirc it is available with orca, so you might be able to do that for your level of theory already. In total, I don't really think that there is a preferred method (yet).

Another note: [Mattermodeling.se] would also have been a good home for this question. It is too old to migrate now though. Maybe you find some more help there. Just to avoid confusion: I'm not saying it is off-topic here. If you would want to ask it there again, you may do so, but please delete the question here first.

I would suggest you to use VRC-VTST for rate constant calculations especially if you want to calculate it for a large temperature range, and the reason is that in barrierless reactions the position of the Transition state along the reaction path change dramatically by changing the temperature (canonical) or the energy (microcanonical), thus using one structure for TS would may be lead to a good result only for very small temperature range.


text:
Why is diequatorial trans-1,2-dimethylcyclohexane more stable than diaxial trans-1,2-dimethylcyclohexane?
Consider diequatorial trans-1,2-dimethylcyclohexane and diaxial trans-1,2-dimethylcyclohexane as shown in the figure below [1, p. 178]. In both compounds, the ring is free of angle strain.

trans-1,2-Dimethylcyclohexane has a conformation in which both methyl groups are equatorial and one in which both methyl groups are axial. As we would expect, the conformation with both methyl groups equatorial is the more stable one.

It can be clearly seen from the figure that in the diaxial, the methyl groups are much farther away than they are in the diequatorial. Hence, the diaxial conformer should be more stable due to less torsional strain or less repulsive dispersion forces. But my book states the opposite. Why?
Reference

Solomons, T. W. G.; Fryhle, C. B.; Snyder, S. A. Organic Chemistry, 11th ed.; Wiley: Hoboken, NJ, 2013. ISBN 978-1-118-13357-6.

This might be of help. And the word is "equatorial", just as in equator.

@user55119 I think this is a common mistake. For e.g. the source of this answer wrote equitorial instead of equatorial.

The key concept is 1,3-diaxial interaction. See the figure 1 below using methylcyclohexane as an example.

Figure 1: Equatorial and axial conformations of methylcyclohexane with 1,3-diaxial strain highlighted.
If a methyl group is in axial position of a cyclohexane ring, the angle strain-free chair conformation means that it ends up rather close to the axial hydrogen atom two carbons along the ring. This steric strain is depicted in the axial case with the inverted brackets. In trans-1,2-dimethylcyclohexane, the di-axial conformation would introduce this strain into the equation twice, destabilising the molecule.
Now you might want to argue that including a second methyl group into the depictions below would cause other types of strain to appear and you would be correct. In the di-axial conformation the two methyl groups are anti-to one another while the di-equatorial configuration has them gauche. However, realise that the cyclohexane ring also exists. Each methyl group in the di-equatorial conformation is gauche with respect to the other methyl but anti with respect to the cyclohexane ring. In the di-axial conformation, each methyl group is anti with respect to the other but gauche with respect to the cyclohexane ring. (See figure 2 below.)

Figure 2: di-equatorial and di-axial conformations of trans-1,2-dimethylcyclohexane with 1,3-diaxial strain highlighted for both methyl groups.
Comparing it in that way shows that the strain between the methyl groups and the ring remains more or less the same in both cases. Probably having the two methyl groups anti would be slightly preferred as the ring is conformationally less flexible and would thus induce marginally less strain. However, the clear driving force here is 1,3-diaxial strain which only occurs in the diaxial conformation.

Thanks, I understand that here the steric repulsion between 1-3 positions was the cause. But, I can't understand this point
"" Each methyl group in the di-equatorial configuration is gauche with respect to the other methyl but anti with respect to the cyclohexane ring"" . How?

Shiva: The diequatorial configuration [sic; conformation] is anti with respect to the cyclohexane ring. Consider methylcyclohexane in the equatorial chair conformation. The dihedral angles Me-C1-C2-C3 and Me-C1-C6-C5 are both 180 degrees. This "interaction" is taken as 0 Kcal/mol. To see how heats of formation play a role in this type of analysis, look here.

@user55119 But please reconsider the figure of di-equatorial 1,2-dimethyl cyclohexane . How can both methyl groups in them are anti? Clearly their dihedral angle is less than 90° and 180° is greater than 90° . How are they anti with respect to ring?

Shiva: There are two types of interaction. Substituents with the cyclohexane ring and substituents with each other. In the diequatoral-1,2-dimethylcyclohexane the interactions of the first kind are taken as non-existent because of the anti relationship of each methyl group with the ring. However, there is a gauche interaction of the two methyl groups with each other (~0.9 Kcal/mol). In the diaxial conformation there are two gauche interactions of each methyl group with the ring (total: 4 x 0.9 Kcal/mol = 3.6 Kcal/mol). continued

... See this link besides the one in the Comment I made in your original post.

@user55119 I keep forgetting which is configuration and which is conformation; if I used the wrong words, feel free to suggest a correcting edit!

@Jan: Cis vs. trans is a configurational change. Bonds must be broken. Equatorial vs. axial is a change in shape. No bonds broken. But you knew that already. Gave you an upvote (#13) for an otherwise solid explanation. Why not make the changes yourself?

@user55119 I mean, I could but I know myself and I run the risk of getting it wrong again. I haven't found a suitable mnemonic that my brain can use to keep them apart. It seems, you can, and since anyone can improve is a core feature of the Stack Exchange model why not go for it? ;)

@Jan: You want a mnemonic...I'll give you a mnemonic! Are you familiar with the architect Louis Sullivan's dictum, “form follows function”? Change that to “form follows function(al groups)”, which translates to “conformation follows configuration”.  To wit, the functional groups must be properly placed before one can worry about the shape of the molecule. Moreover, it follows that configuration precedes conformation in the dictionary. Now, if I only had a mnemonic to remember how to spell “nemonic”! ;-) PS: check the edit I made.

Basically, groups attached at the axial position are under some strain. There is angle strain between 2 axial groups and some repulsion between them.
The more stable form usually contains bulkier ps at equatorial position, which can relieve some strain in them and the lighter groups at axial position. It is a fact than equatorial positions are more stabilising, due to molecules being oriented in an ‘equator like’ manner.
You may want refer to:
https://www.researchgate.net/post/Why_is_a_molecule_most_stable_when_its_heavier_substituent_is_at_the_equatorial_position

Generally , equatorial conformers are more stable due to less torsional strain . However in this case , where the ring is di-subsituted at vicinal carbons , the repulsion between electron clouds is a dominating factor than the repulsion between the methyl groups . The repulsion between electron clouds is more in axial position ( though their bonds seem somewhat to be antiparallel ) than it is between electron clouds of bonded pair of equatorial one .


text:
Why is the Conductivity trend same in fused and in dissolved or aqueous state of ionic compounds?
For an example if we take $\ce{KCl}$ and $\ce{NaCl}$ then I have understood the part where in aqueous state $\ce{Na+}$ becomes bigger than $\ce{K+}$ hence the conductivity reverses but in fused state ions exist on their own hence the conductivity order must remain $\ce{Na+} \gt \ce{K+}$

Consider proper capitalization of chemical element symbols to be taken seriously. k is the Boltzmann constant.

"Naked" $\ce{Na+}$ ions in melted salt, being smaller and more polarizing, may have bigger tendency to form ionic pairs with $\ce{Cl-}$, similarly as $\ce{Mg^2+(aq)}$ and SO4^2-(aq) have tendency to form ionic pairs is sea water as $\ce{MgSO4(aq)}$

So basically in fused state as they are not solvated the ions having more effective nuclear charge(polarizing power) are able to interact more effectively with each other for an instance say Na+ and Cl- hence they reduce each others movement??

Ionic pairs do not electromigrate.

could you please elaborate on that ?

Does Na+Cl- pair migrate to the cathode or the anode ?

As per my knowledge they move individually Na+ to cathode and Cl- to anode

While they form a bound ion pair, then no, they do not, similsrly as ion pair MgSO4(aq) is sea water. There is formation equilibrium and the ion pair fraction causes effective ion mobility slowing down.

Thank You now I think I have understood


text:
How to get the total number of numbered positions in a molecule/compound
I'm working on a metabolite database where "labeled" compounds can contain isotopically labeled atoms in specific positions.  When a new labeled compound is entered into the database, we'd like to be able to perform a simple validation on the submitted labeled isotope positions.  E.g. there's a Carbon-13 at position 2 in a glucose compound.
The first suggestion by another developer was to check that the positions submitted are less than or equal to the number of atoms of the given type in the compound based on the formula.  It's been a long time since my BS in Biochemistry, but I recalled that positions are not (necessarily?) relative to the given atom.  The positions (as I recall) are based on multiple types of atoms and are numbered in order based on its weight and the weight of atoms it's bonded to, though I don't remeber the specific rules.
I've been googling for awhile now and flipping through my old chemistry books, but I have yet to turn up those rules that govern:

What atoms receive a position (e.g. carbon and nitrogen atoms are numbered 1-9 in purine but oxygen doesn't get a position in glucose)
What atom gets position 1, or any other position for that matter

For now, all I want to do is determine the total number of numbered positions so that if an isotope position is submitted outside of the range, we can throw an error and say that position is invalid.  Aside from having more C13 positions than the number of carbons in the compound, that's all our requirements currently call for.  (While it would be nice to know, we don't have a requirement for ensuring the given position actually is supposed to be a carbon.)
And while I'm at it, are there different atom position/numbering schemes in use currently?

Could it be as simple as summing all the atoms from the formula that excludes oxygen and hydrogen for example - could that be thwarted by some edge cases, like diethyl-ether, where there's an oxygen in the middle, or other un-numbered "side-groups"?

I'm not sure I get you, why not just use the number of atoms of the element? Using the number of magnetically equivalent atoms would be better, but I don't see a point, if you just want to cut down errors.

@mithoron. The point is we don't want erroneous data in our database. We want validated/correct data. Not sure I'm following you, though I could take the total and subtract all the hydrogen atoms as the most basic heuristic, but the point is, I would like to know the rules enough to be able to make an intelligent check. Like, aside from hydrogen, are there any atoms that don't get a position? Do hydrogen atoms not get a position or do they get positions, but their positions are always last, numerically? Basically, what are the numbering rules? I'd like to do better than to low bar of all atoms

Wait, is there a miscommunication? I'm not counting protons/neutrons of an "element". I'm counting occurrences of elements (I.e. atoms) in a compound.


text:
Why is CaO a lewis base
We know that $\ce{CaO}$ is a base. But when I wanted to know which theory proves this fact, I got to know that Lewis theory explains this through a reaction with $\ce{SO_3}$. But the definition of Lewis base are those substances which can donate lone pair electrons. We all know $\ce{CaO}$ is an ionic compound. So how can it possibly have lone pair electrons? Moreover, there is no central atom in $\ce{CaO}$. So whose lone pair electrons will we be counting if there were one?

Why does calcium oxide react with sulfur dioxide?

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.

I think your conceptual problem is with the man-made classification of acid-base reactions and other associated labels. Chemical elements and compounds do not care what labels humans are applying onto them...they will combine the way Nature and its laws of physics have allowed them. In the link suggested by andselisk, the answer hinting at the Lux–Flood theory is a better answer.

O^2- ion has 4 pairs ready to react, making it a base much stronger than OH-, which is its conjugate acid.


text:
Activating alumina before column chromatography
Recently I synthesized some amine compounds, and because it showed better separation in the neutral alumina TLC plates than silica plates, I am planning to purify the reaction mixture using neutral alumina column. In some texts, people often activate, that is, remove the water from silica or alumina by heating it at around 130-160 ℃ for 30-60 minutes in the oven before the column chromatography. However, in the catalog of Sigma Aldrich, I found this sentence.

For chromatogrpahy, alumina is usually used with organic solvents, and is treated with water to achieve different affinities, or activites, for specific uses. Sigma products are all activity #1, and water can be added to convert alumina to a lower activity grade

As far as I know, the Brockmann activity is determined by the water content. Then, if I remove water from it, is the Brockmann acitivity of the alumina changed? Or does it have nothing to do with column chromatography and I don't need to care about it? Is the removal of water of silica or alumina before column chromatography essential procedure?

If it is sold as "Alumina for chromatography" you do not need to activate it.

@Waylander Thank you. Then in that case occasional reactivation because of long-term storage would be enough?

It depends on how it has been stored. I would try using it first to see whether it give an adequate separation without activation.


text:
How can brass still be made even though the crystal structures of zinc and copper are not the same?
Based on the Hume-Rothery rules, to dissolve an element into a metal, the crystal structures for both must be the same. But the structure for zinc is HCP, whereas copper has FCC structure. It does not follow the rule. But why can brass still be made?

The $\ce{Cu-Zn}$ phase diagram is, to put it mildly, complicated.

(source)
See that wide area denoted $\alpha\rm (Cu)$? Now, when we add a little amount of one metal into the crystal lattice of another metal, typically the minority metal just "agrees" to coexist in that structure, even though different from the structure that it would form on its own. And it just so happens that "a little" can sometimes be not that little.
Now, there are different kinds of brass with different composition, and not all of them fit into the said area. What happens to those which don't? Well, then we have a whole bunch of other Greek letters, and each of them means a particular phase which has a structure of its own, different from both metals. Sometimes cooling will produce grains of different phases, which is of course very different from having just one phase. This explains widely different mechanical and corrosion properties of different brasses.
As for Hume-Rothery, it is good for simple cases. This one, like I said, is not simple. Neither is $\ce{Cu-Sn}$ (bronze), or $\ce{Fe-C}$ (steel), or pretty much any other binary phase diagram of any importance, for that matter. Actually, simple cases are hard to come by. If you still want to see one, look for $\ce{Pb-Sn}$.

And two different crystal structures coexist; such as 60 : 40 yellow brass is a mix of alpha and beta grains. Cartridge brass ,70 : 30 is a single type crystal , alpha.

Without "so it goes" Ivan's answer can be potentially flagged as NAA:)

I'd hardly call that a 'complicated' phase diagram, there are certainly worse ones out there. But, yes, various brass compositions will have various microstructures/phases and thus have different properties.

@andselisk without that line, it is not considered an Ivan answer ;)

Can I award points for answering a question about a yellow alloy with a phase diagram that looks like a giraffe?

While molten, copper and zinc (and tin) are miscible. As the alloy cools and crystalizes, the metals do indeed separate, forming grain boundaries. These inclusions "pin" slippage between planes of copper atoms, making the alloy harder and more brittle (or less malleable, if you prefer). From Wikipedia: "Precipitation hardening relies on changes in solid solubility with temperature to produce fine particles of an impurity phase, which impede the movement of dislocations, or defects in a crystal's lattice." Perhaps a lattice-and-precipitatoe sandwich?
BTW, that is why brass does not survive aging as well as pure copper: corrosion occurs at grain boundaries, where dissimilar metals meet.


text:
Why is it wrong to draw the Lewis structure of a molecule's atoms and then figure out the Lewis structure of that molecule
When drawing for example the Lewis structure of nitrate ion (NO3)^-1 whould it be wrong to draw nitrogen and oxygen separately and then try to figure out the structure of the ion? In that case does the nitrogen give one of its electrons to the oxygen? cause a nitrogen atom does have a lone pair of electons but in the nitrate ion they are gone.

Each Oxygen atom has 8 electrons surrounding it and so does Nitrogen. 2 electrons form a bond pair and hence a bond, thus they are no more a ‘lone pair available for donation.
In each N-O bond 2 electrons are used. The point is that nitrogen uses all its 8 surrounding electrons (which constitute an octet) in forming a total of 4 bonds, 3 single bonds and 1 double bond with Oxygen atoms.
So it has no electrons availed freely to become lone pairs, it thus has a formal positive charge on it and no lone pairs available for donation. Essentially what you did in a way, is absolutely correct, just the last reasoning you missed.
Note: N, when forms 3 bonds usually has 1 lone pair for donation, when 4 bonds are formed, it has none and +1 charge (generally).
Hope it helps.

Yes nitrogen doesn't have any lone pairs in the nitrate ion and i am asking why, since it has a lone pair in its atomic form. Is this because it gave one electron, of its lone pair  to the oxygen beside and made a bond using the other electron of that lone pair.  Thank you for trying to explain this

Nitrogen atom has a lone pair and three free electrons around it.  If you surround the nitrogen atom with three oxygen atoms, three electrons of the nitrogen atom can make a bond with an electron belonging to the oxygen atoms. This would give to the nitrogen atom a structure respecting the octet rule. But not the oxygen atoms. They do not respect this octet rule, even with adding a supplementary electron to make $\ce{NO3^-}$. The only way to give oxygen an octet is to break the lone pair, as you did previously..


text:
Why is the carbocation on the left more stable?

I assume resonance is coming into play here due to the hydroxy group on the carbon adjacent to the positive charge. What about the electron donating effect of the alkoxy groups?

Inductive effect no where comes near to resonance, both in electron withdrawing/giving.

What resonance are you entertaining?

Both hardly make sense, so it's rather pointless exercise.

@Mithoron what do you mean by both hardly making sense?

You can draw anything, but it doesn't mean something exists.

Remember that molecular structures of organic compounds do not always show all four valence to each carbon atom and you have to fill in the missing valence with hydrogen. When you do this you find that the double bond is isolated in both carbocations. There is no resonance.
So, we fall back to the inductive effects. Functional groups with oxygen are relatively electronegative and tend to be electron-withdrawing, and the carbocation center in the right ion is closer to two such groups rather than one.


text:
What determines whether a mechanism is nucleo/electrophillic addition/substitution?
So I recognise what an electrophile and a nucleophile is, but not all mechanisms (that I know of at least) have one species substituting or adding. For example, propanone added to NaBH$_4$ has an initial step of a hydride (nucleophilic addition), and then a
H$^+$ being added to an O$^-$ (electrophilic addition).
I'm a bit wary of this example as the second step doesn't involve a carbon. So, another example would be the addition of HBr to ethene. The first step has the addition of H$^{\delta+}$ (electrophile) and then Br$^{-}$ being added as a nucleophile.
This website offers a solution by specifying that the rate determining step takes precedence, but I can't find anything to corroborate this

so what exactly is your doubt, is it like identifying which step occurs first or whether it is entirely an SN or EN mechanism reaction ?

The latter; (The site suggest it's to do with the rate determining step)

in the end it is the rate determining step and stability of intermediates formed which decides whether an SN or EN reaction occurs. Eg : in a tertiary halide E1 is most preferred, followed by SN1 and SN2 and E2 are almost negligible.

@PSR_123. Thanks. This answers my question

You know the difference between addition and substitution. Nucleophilic/electrophilic reactions are similar to redox reactions with reapect to reagents involve. There is always a reduction reaction when oxidation invoved. Similarly, there is always nucleophilic reaction when electrophilic reaction involved.

@PSR_123 your throwing around these mechanism schemes like these are supposed to be a universal truth. If anything, they are idealised model reactions and they don't happen in nature. The rate determining step approximation should also carefully viewed as a crude approximation. It only produces reasonable (not good) results of one (and only one) step is magnitudes faster than all (including side reactions) other steps. I find your comment counterproductive.

Any reaction that is either a nucleophilic or an electrophilic addition/substitution is in fact both a nucleophilic and an electrophilic addition/substitution. Whatever a nucleophile combines with can also be called an electrophile that attacks the nucleophile; textbooks just adopt a single perspective.
We also have examples where the same reagent is both an electrophile and a nucleophile to different parts of the "substrate". Grignard reactions, in which the Grignard reagent has both a nucleophilic carbon atom and an electrophilic magnesium atom, are one such example.


text:
"No heat, light or electrical energy may be evolved or absorbed during the formation of a mixture." What about heat from friction?
I was reading about mixtures in my chemistry textbook. One of the characteristics of mixtures that was stated was "No heat, light or electrical energy may be evolved or absorbed during the formation of a mixture." Doesn't the heat evolved from the friction between the particles of a mixture count in this context?

Eventual friction heat is not released because of forming mixture, but because of mixing. If the same work were done by mixing the components separately, the same heat would be released. Note that some mixtures do release mixing heat, when different intermolecular bonding is involved.

Please cite the source of the quote.

No heat, light or electrical energy may be evolved or absorbed during
the formation of a mixture

By including the word "may," the book may be saved from being criticized too harshly. The formation of many mixtures involves the evolution or absorption of heat Heat of Mixing
In your mind, you are picturing a mixture that is being agitated by a mechanical body (a mixer), and as a result, the mixture becomes warm. Mechanical heat is the result of an agitating motion. In mechanics there is a concept of "Mechanical Equivalent of Heat". It was a historical experiment of its time. See Mechanical Equivalent of Heat

Textbooks often go overboard in trying to differentiate between mixtures and compounds, physical and chemical changes and similar subtle happenings by setting up rigorous rules that have to be broken if one thinks more deeply about them. However, if one does think more deeply about it don't completely discard the original intent. ALL interactions with matter involve chemical forces; there are some general groupings such as intermolecular, intramolecular bonds, nuclear bonds; there are always instances that are not rigorous. As you learn chemistry you will grow to appreciate and look for and perhaps study these instances or at least not be too surprised if they are encountered. Mixtures are not only mixtures They involve interactions between the components these interactions involve energy, entropy, bond breaking and formation [usually weak] sometimes maybe not. I was just sitting on my porch during a thunderstorm. These are caused by a rapid rising of humid air, condensation of water vapor releasing huge amounts of energy and somehow forming ions. These were initially all physical changes but a nearby lightning bolt and tremendous thunder was far from simply physical. So we got heat, light, sound and even fear from physical changes[there were also chemical changes N2 + O2 reacted to give NO, Interesting because NO is an endothermic compound a chemical change that required energy. The book over simplified! Chemistry is far too interesting to simply learn rules and not think about them. [don't rely on friction as a solution to thinking about things either, it is also a chemical process to be thought about.]


text:
Are molecules exchanged between a solid and liquid in equlibrium?
In Vapor-Liquid Equilibrium, it is my understanding that molecules exchange between the gas and liquid continuously; even when no difference in temperature or partial pressure exists.

Regarding Solid-Liquid Equilibrium; is there a similar continuous exchange of molecules? For example on a single crystal of ice submerged in pure water at perfect and constant 0°C, will molecules on the surface of this crystal exchange with the liquid? In the past I have assumed this to be the case, but now realize I have not seen a reference that explicitly details this like is common for vapor-liquid equilibrium. Are there any references that calculate the rate at which this occurs? Side question, is there also a continuous exchange for solid-gas equilibrium?

Yes, and yes. All solids have some vapor pressure, and (often negligible) solubility.

Generally, molecules/atoms/ions have no means for awareness the system is in equilibrium, being "instructed to do nothing". Additionally, they do not have temperature. A single kinetic energy value belongs to a wide range of possible system temperatures, so they are changing a phase at will.

Your question was already answered https://chemistry.stackexchange.com/questions/73344/proof-of-dynamic-nature-of-equilibrium

Does this answer your question? Proof of Dynamic Nature of Equilibrium

Even two solids in contact may exchange atoms, such as gold in contact with lead. Accidental cold-welding is a serious problem in space technology, in particular.
You can prove to yourself that ice/water, in equilibrium, exchanges molecules: immerse an ice-cube on a string in water at 0°C, and the cube will slowly change shape. "No chemist ever observes the same ice cube twice..." as Heraclitus or Heisenberg stated (I'm not certain).


text:
Red P+HI with some specific functional groups
I've seen the related question (Does RedP+HI reduce all carbon functional groups to alkane?). Although the question is admittedly similar, neither the answer provided nor the document linked in it answer my question, since they don't include the functional groups I am most interested in(since my sources have been mostly in consensus regarding the others), it's basically just alcohols, aldehydes, and ketones(with one ester, that too of the sort that gives unexpected reactions sometimes due to strange mechanisms).
Does Red P+HI reduce esters, acid anhydrides, acid halides, cyanides, and oximes? I've checked Solomon as well as some exam-prep-type books, and just searched the web, finding a couple of pages here itself, a few sites meant for exam prep, which I know to not be 100% reliable. My teacher has basically said it reduces everything, but my generally more reliable HOD has said it would only reduce alcohols, aldehydes, ketones, and carboxylic acids to alkanes; all of this just resulting in me being more confused than before.


text:
Thermodynamics derivatives: Does $(\partial A/\partial B) = (\partial B/\partial A)^{-1}$?
If I need to find, for instance:
$$\mathcal{T} := \left(\frac{\partial U}{\partial S}\right)_{V, N}$$
knowing:
$$\left(\frac{\partial S}{\partial U}\right)_{V, N} = \frac{1}{T}$$
Can I say that $\mathcal{T} = (1/T)^{-1} = T$ ? In general terms, let $A$ and $B$ are two thermodynamics values, is it true that
$(\partial A/\partial B) = (\partial B/\partial A)^{-1}$ ?

The question is better suited for Math.SE, where it already exists.

are thermodynamics variables "usually" strictly monotome ?

There are singularities like near 0 K or the liquid critical point, where some partial derivatives converge to 0 or $\pm \infty$

ok, and between those singularities ?

Usually monotonous , but there are exceptions, like local minima for water heat capacity near 36 °C, or for molar volume near 4 °C. Or maxima of electric conductivity of concentrated electrolytes.

Additionally, such partial derivatives may be undefined at some points, typically at phase changes.

Ok thanks, I'll avoid using this formula then

@andselisk  This question is about thermodynamics, and belongs here.  The Math.SE answer says it's not generally the case that, for a function of more than one variable (e.g., z(x,y)), that (∂z/∂x)_y = 1/(∂x/∂z)_y.  However, in thermodynamics, we're generally dealing with equilibrium systems, which means we are operating under the constraint that dz = 0. And under that constraint, we can derive the cyclic rule and the relationship that (∂z/∂x)_y = 1/(∂x/∂z)_y.  So yes, this formula should be used (and often needs to be used) in thermodynamic derivations.  Just watch out for singularities.

As a mathematics student- although dB/dA is not always equal to (dA/dB)^(-1), it IS equal for "nice" functions A, B. Most functions appearing in applied mathematics are "nice", so in your case dB/dA can be considered to be equal to (dA/dB)^(-1).


text:
Rate order when the slow step only contains intermediates?

I feel a bit sheepish asking this, but I was told that the stoichiometry of the slow step of a mechanism indicates the order of a reaction with respect to a reactant. However, in the slow step of this reaction, there is no reactant (of the overall equation) reacting. So how does one determine the rate order? Would it be in terms of the intermediate.
I don't see how the rate could be $k[C_4H_{11}O^+]$ as that is not a reactant in the overall equation
Image source: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(McMurry)/10%3A_Organohalides/10.05%3A_Preparing_Alkyl_Halides_from_Alcohols

Who tells you that the rate could be $\ce{k[C4H_{11}O^+]}$ ?

@Maurice. No-one said anything about this specific example. But in all examples that I have been given, slow steps always are made up of, at least in part, the original reactants.  The only other idea I have is that the overall rate order is 0

In this case the $\ce{tBuOH2+}$ is in rapid equilibria so you can rewrite it in terms of $\ce{[tBuOH]}$ easily. In more general cases you can read up on the steady state approximation and pre-equilibria approximation

@ yolo. So what is your question?

How to tell the rate equation of a reaction when the slow step only contains intermediates (i.e what the rate order is wrt each reactant)

The rate equations only truly work for elementary reactions, for everything else they are mere approximations. At any point where there are equilibria involved they start to break down. This is an empirical business and the rate determining step approximation is terribly crude.

@Martin-マーチンah I see. What should I do from an SE standpoint?

Step 1 is fast and reversible so that in effect an equilibrium amount of $\mathrm{C_4H_9OHH^+}$ is formed which reacts slowly in step 2. Once the $\mathrm{C_4H_9^+}$ is formed it reacts rapidly in step 3 so the reversible part of step 2 never occurs to any great extent. Thus the limiting step is step 2 with first order reaction rate $-k_2[\mathrm{C_4H_9OHH^+}]$ and the product forms at minus this rate.

I highly doubt that this reaction actually follows the SN1 pathway or is best described with that.


text:
Why is desalination waste usually disposed back into the sea instead of converting it into chemical products?
I was reading about the desalination processes such as reverse osmosis and capacitive deionization and their alternatives. They all work on removing ions from water and disposing them back to sea/ocean in the form of brine.
This got me thinking why do they have to move the previously dissolved ions to another liquid, forming brine? Why don't they dispose these ions in their “normal” form without dissolving them again? Does it has anything to do with their reactivity/danger?

@BelalBahaa - Do you mean something like why isn't the ionic aqueous sodium removed as sodium metal, and the chloride removed as gas?

It costs far more to completely separate ions from seawater than it does to partition it into less-concentrated (not perfectly pure, but drinkable and usable in agriculture and industry), and more-concentrated waste (i.e., brine, containing the unwanted ions). Water is used in massive amounts, and the most-economical process wins.

@ToddMinehardt I mean why aren’t they removed as a sodium ion and chlorine ions separately without dissolving them in another electrolyte for disposal?

@DrMoishePippik Is cost the only reason or are there other reasons for it?

@BelalBahaa, Sodium and Chloride ion cannot exist separately in different solutions.

They could try evaporating the brine in evaporation fields, and then extracting valuable minerals and/or metals from them.  It's really just a matter of whether that's the most cost-efficient approach.

There is no dissolving them again. Ions remain dissolved all the time.
It is like taking

$\pu{11 m3}$ of sea water with salinity e.g. $\ce{30 g/L}$,

producing

$\pu{1 m3}$ of (about) pure water (with residual salt)
$\pu{10 m3}$ of sea water with salinity $\pu{33 g/L}$

by active redistribution of water (like in reverse osmosis) or ions(like in electrodialysis) between 2 separated volumes. All numbers are just illustrative.
It is technologically/economically advantageous in desalination to use big excess of input sea water to keep the salinity difference between the input water and the waste water small. With big difference, the process would be less efficient and more demanding.
This is kind of analogical to efficiency of heat pumps,  used for building heating. Maximal efficiency is reached at the minimal difference between interior and exterior temperature.
There would be just minor difference between processing of the waste water from desalination and raw sea water, if we wanted to get minerals from the waste water. Additionally, there would be large water volumes, questioning capacity to process it and demand for its output.

For curiosity, here is the Wikipedia table of energy consuption of chosen methods:

Desalination Method
Multi-stage flash  distillation-MSF
Multiple-effect  distillation-MED
Mechanical Vapor  Compression MVC
Reverse  osmosis-RO

Electrical energy (kWh/m3)
4–6
1.5–2.5
7–12
3–5.5

Thermal energy (kWh/m3)
50–110
60–110
None
None

Electrical equivalent of thermal energy (kWh/m3)
9.5–19.5
5–8.5
None
None

Total equivalent electrical energy (kWh/m3)
13.5–25.5
6.5–11
7–12
3–5.5

See Desalination Technologies for the extensive list.

Also, by not making waste brine too brackish, it washes away waste which might precipitate and clog filters and plumbing.

You seem to know your water separation so can you explain why they do electrodialysis or RO when in hot environments, they can use vacuum distillation for it and then reroute that steam using a closed system in some form of vapor recirculation. The incessant insolation will supply enough energy I'd guess and then its just vacuum pump's operating cost and you can use water to cool its internals so that heat isn't wasted either. Route the facilities cooling system's heat to the brine as well. Total heat capture and then reuse.

Distillation is energetically less favourable.


text:
Reason of being 2nd ionization energy
Why is the second ionization energy always greater than the first? By shielding effect, it shouldn't have been true. Let's take $\ce{Mg}$ and $\ce{Mg^+}$ for example. Effective nuclear charge(ENC) for $\ce{Mg},Z_{\mathrm{eff}}=Z-$ no. of core electrons $=12-10=2$. For $\ce{Mg^+}$, it's also $12-10=2$, since there are still $2$ shells before valence shell like in $\ce{Mg^+}$. The result is giving equal number of attraction in this case. So energy needed to extrapolate one electron should be the same in both cases. Moreover, according to ENC, the size of the cation and anion are equal here, but we know cation<atom. I will be very grateful if someone sheds light on this atomic radius and attraction property in light of shielding and screening effect since that's how atomic and ionic radii are compared.

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.

The electron energy is not determined by the effective nuclear charge. which is just a contributing factor. BTW, the ENC is better computed by Slater's rules. The integer counting you have used would have been applicable if orbitals had been non-overlaping regions with spherical symmetry of probability density.

Stealing one electron from a magnesium atom is hard and results in a positively charged magnesium ion. Then stealing another electron, which is negatively charged, from the positively charged magnesium ion, is harder still. Each subsequent theft gets harder.

@ed v you may be aware that magnesium can actually be stopped at one ionization, though it usually must dimerize to do so. See https://en.wikipedia.org/wiki/Low_valent_magnesium_compounds.

@OscarLanzi I have only heard very little about low valent magnesium species, so thank you for the link! I was just trying a simple rationalization of the typical trend. I used to tell my students in first year chemistry that if a rich guy had multiple wallets and a mugger stole one, it would be harder to mug the guy for another wallet. So, just the usual trend, not pesky details about energetics, etc.!

A good mugger renders the victim at least senseless and takes both wallets. It is basic high school physics the first electron is removed from an effective nuclear charge of +2 each successive electron Has one more charge hence needs more energy. Compound formation is the effective mugger It returns the energy in much higher lattice and or solvation energies

One way to think about this is to consider the integral of the force required to bring the electron from its radius to infinity.
When you remove the first electron from a neutral species like $\ce{Mg}$, the electron leaves behind a positive ion of charge $1$. When you remove the second electron, it leaves behind a positive ion of charge $2$. Therefore, the second ionization energy will generally be about twice as large as the first ionization energy.
In addition, there can be a significant radius effect if the electrons belong to different subshells, but that is not the case for $\ce{Mg}$, as you are removing two electrons from the $\ce{3s}$ shell. As a matter of fact, for $\ce{Mg}$ the ratio of 1st and 2nd ionization energies is very close to 1:2.

Exchange energy effects, which favor half-filled subshell configurations, can also enter. They cause phosphorus and chlorine to each have IE2 < 2 × IE1.


text:
Notation for acid-base reactions of differing strengths - equilibrium?
I'm having trouble finding anything meaningful in textbooks or searches. My question is what is the correct notation for writing acid-base reactions when their strengths are varying? I know that theoretically all reactions are reversible but when is it appropriate to use equilibrium arrows for certain acid-base reactions?
e.g. would a weak acid - strong base reaction be regarded reversible?
for example would $\ce{NaOH + CH3COOH $\leftrightharpoons$ NaCH3COO + H2O}$ be suitable (disregarding the lack of states and my ineptitude in writing MathJax)
or would $\ce{NaOH + CH3COOH -> NaCH3COO + H2O}$ be more suitable?
Do we just simplify it into only considering the forward reaction for the sake of stoichiometric calculation?

Consider rather $\ce{CH3COOH(aq) + H2O(aq) <=> CH3COO-(aq) + H3O+(aq)}$ and $\ce{CH3COOH(aq) + OH-(aq) <=> CH3COO-(aq) + H2O(aq)}$

I understand this, so when the task is to write an equation for the above neutralisation would an equilibrium be more suitable, with sodium ions as spectators? Because most likely the equation is meant to be used for titration calculations, so it isn't appropriate to have the net ionic equation here, and having sodium spectators on either side seems somewhat clumsy. I suppose I should just be using the first equation: $\ce{NaOH + CH3COOH ⇋ NaCH3COO + H2O}$

I have thought it through again @Poutnik and the following approach seems more appropriate:
$\ce{NaOH(aq) + CH3COOH(aq) -> NaCH3COO(aq) + H2O(l)}$
$\ce{CH3COO-(aq) + H2O(l) \leftrightharpoons CH3COOH(aq) + OH-(aq)}$
as the first equation cannot be equilibrium as $\ce{NaOH}$ will not form again in any measurable quantity

In fact, NaOH(aq) is just a shortcut for independent Na+(aq) + OH-(aq). The same for sodium acetate. But if one insists on molecular reaction notation, one must accept the equilibrium for molecular notations of the acid and the salt.

I'm aware - I was just a little conflicted as to whether the whole reaction should be shown in equilibrium because that seems to insinuate that $\ce{NaOH}$ may be formed, which would neglect Sodium Hydroxide being a strong base. Thanks for your help.

In that formal reaction, in reality no NaOH is either consumed either created, as it is OH- that is both consumed and created by the opposite ongoing reactions at the equilibrium. So if NaOH can be formally consumed, it can be formally created as well. One can use both uni-/bi-directional notations, depending on the context and aspect to be point out.

Interesting - so in the context of titration (where one compound is gradually added) would the unidirectional notation be more correct for the neutralisation, and then bi-directional acetate-water reaction to explain an alkaline equivalence point? (As for neutralisation it doesn't make much sense to use bidirectional)

Let us continue this discussion in chat.


text:
Is there a material that ions can pass through but water can’t?
I was wondering whether there is a material that can pass ions (such as the ions existing in seawater) but not water. I am aware that water molecules are smaller, so they can pass through some materials but ions don’t. However, is there a material that is the opposite? Or is there a hydrophobic material that can be used as a coating for a porous material, and doing the same job by preventing water from passing but allow ions to pass through it?
Thanks in advance

Some potassium channels are water free, DOI: 10.1126/sciadv.aaw6756

@KarstenTheis, I think he is asking about non-living materials as used in water purification.

@AChem It's all atoms, whether the context is non-living or living, so you can take one domain and learn from it to understand the other. For water purification, the opposite is used (let water molecules pass while retaining ions) in reverse osmosis.

Your H3O+ and OH- ions might form a problem.

There are so many materials that allow the passage of ions but not water. They are called ion-exchange membranes. I believe you are interested in water purification based on your previous posts. The trick is to attract the ions on one side. Student always tend to forget that a solution must remain electrically neutral by all means. If you want a certain ion to leave water, you must provide another ion of the same charge that has left the solution.
An example is given here Image below

Why do other ions need to pass in the opposite direction? Couldn't electrons pass between the two? For the sake of argument, an electrolysis cell with an external battery and an ion-exchange membrane would appear to send ions in one direction across the membrane?

@abligh, Good question: There are no free electrons in ordinary aqueous solutions. The charge in solution is carried by ions. If you are interested, ask a separate question with a diagram in your mind.

done: https://chemistry.stackexchange.com/questions/165688/do-ion-exchange-membranes-always-require-ions-to-pass-in-both-directions


text:
Can primary amine be produced using Grignard reagent?
I found a way to produce Secondary Alkyl Amines using Grignard Reagent.
(source: wiley.com)

Ref: Z. Huang, J. Lv, Y. Jia, Chemistry Select 2016, 1, 5892. DOI: 10.1002/slct.201601464
But I want to know if I can use $\ce{HNO2}$ instead of $\ce{R-NO2}$ here to obtain $\ce{R^1-NH2}$ [say, $\ce{H-NH-R^1}$] or is there any other way to produce Primary Amines ($\ce{R-NH2}$) using Grignard Reagent ($\ce{R-Mg-X}$)?

Yes indeed. Primary amines can be made using Grignard reagent. The first reported instances of such reaction was found during the 1930s where they used mono/dihaloamines and reacted with Grignard reagent to form primary amine, secondary amine and other side products like ammonia and mixed magnesium halide reaction (Ref.1a).
$$\ce{RMgX + NH2Cl ->  RNH2 + MgXCl + (other side products)}$$
$$\ce{RMgX + NH2Cl ->  RCl + MgXNH2 (side reaction)}$$
When dihaloamines is used, the yields are not that promising (Ref.1b):

Dibromoamine reacts with Grignard reagents to produce primary amines, secondary amines, ammonia, and nitrogen. The percentage yields of these products if Grignard reagent used is n-butyl magnesium chloride are n-butylamine 7.8 per cent, di-n-butylamine 2.2 per cent, ammonia 79.0 per cent, nitrogen 5.9 per cent;

There are tons of other electrophilic aminating agent beside one mentioned in Waylander's answer. Some examples being  4,4′-Bis(trifluoromethyl)benzophenone O-sulfonyloximes (Ref.2a), 2-Imidazolidinone O-sulfonyloxime (Ref. 2b).  You can find tons of other reaction which uses Grignard agent to produce amines. (Ref. 3)
References

The formation of primary amines from Grignard reagent by George H. Coleman and Charles B. Yager:

From Monochloroamine (1929 paper) DOI: 10.1021/ja01377a029
From Dibromoanime (1933 paper) Link: https://scholarworks.uni.edu/pias/vol40/iss1/53/

Preparation of Primary Amines from Grignard reagents by Koichi et.al.

by the Alkylation of O-Sulfonyloximes of Benzophenone Derivatives(1999 paper) DOI: 10.1246/bcsj.72.1869
with 2-Imidazolidinone O-Sulfonyloxime (2003 paper) DOI: 10.1246/bcsj.76.1063

Synthesis of amines by the electrophilic amination of organomagnesium, -zinc, -copper, and -lithium reagents by Tahir Daşkapan, DOI: 10.3998/ark.5550190.0012.520

Grignard reagents, in addition to being nucleophiles, are strong bases. Attempting to react a Grignard $\ce{RMgX}$ with nitrous acid is simply going to result in protonation of the Grignard to give $\ce{R-H}$ and magnesium nitrite.
Primary amines can be prepared from Grignard by reaction with 1,3-dioxolan-2-one O-sulfonyloxime.

Ref: Synthesis of Primary Amines by the Electrophilic Amination of Grignard Reagents with 1,3-Dioxolan-2-one O-Sulfonyloxime
Mitsuru Kitamura, Takahiro Suga, Shunsuke Chiba, and Koichi Narasaka
Organic Letters 2004 6 (24), 4619-4621 DOI: 10.1021/ol0479951


text:
Can a hydroxyl group be attached to a triple bonded (sp hybridized) carbon?
Alcohols are defined as organic compounds containing one or more than one hydroxyl group attached to a sp3 hybridized carbon atom.
And it is mentioned in the book I am referring to [errorless chemistry] that compounds in which hydroxyl group is attached directly to sp2 hybridized carbon are known as enols or phenols (aromatic compounds).
Does this mean that -OH group is never attached to triple bonded carbon?

Look up “ynol”. . .

It seems an OH directly attached to a triple bond is unstable with respect to a ketene H2C=C=O.  I think I can see why, The OH would be relatively acidic and the carbonyl double bond more stable. This is an interesting question something that I never thought of before. Too bad your text didn't even hint about this. [or maybe they did by omission congrats on picking up on it.]

The keto-enol tautomerism is in most cases favoured for the ketone. It does actually show that both forms are stable. The same is likely to be true for the ynol and ketene equilibrium, just even more favoured.

There is a reasonable mechanism for keto-enol isomerization the alpha protons are acidic. The vinyl protons in the ketene are much less acidic so in spite of the law of Microscopic reversibility this looks like a one way street.  Some reactions such as some radical combinations just don't reverse unless a lot of energy is involved. Also the mechanism must be very elementary.

@jimchmst Thanks,I get it now. Its just that the definition does not make sense to me. I'd like to classify ynol as alcohol too


text:
pH calculation of a mixure containing 2 strong acids and a buffer mixture
Calculate the pH of a mixtureof 10 mL 0,1 M HCl ; 5 mL 0,1 M HNO3 ; 10 mL 0,2 M HAc ; 15 mL 0,4 M NaAc.
My understanding was that I have to find the [H+] by:

finding the concentration of the products by:
a) finding the amount of moles (n) = c*V
n(HCl) = 0.1M * 0.010L = 0.001 moles
n(HNO3)= 0.1M * 0.005L = 0.0005 moles
n(HAc) = 0.2M * 0.010L = 0.002 moles
n(NaAc) = 0.4M * 0.015L = 0.006 moles

b) Total volume = (0.01 + 0.005 + 0.01 +0.015)L = 0.04L
c) the concentration of the products are: c = n/V(tot)
c(HCl) = 0.001n/0.04L = 0.025M
c(HNO3) = 0.0005n/0.04L = 0.0125M
c(HAc) = 0.002n/0.04L = 0.05M
c(HANa) = 0.006n/0.04L = 0.15M

Because strong acids wil react first I added the c (= [H+]) of the acids = 0.025M + 0.0125M = 0.0375M

pH = -log (cA) = 1.4259...

Because HAc and HANa are a weak acid and it's conjugate base I used the buffer formula to find the pH:
pH = pKa + log (cb/ca)
pKa of CH3COOH = 4.75
ca = 0.05M
cb = 0.15M
pH = 4.75 + log (0.15/0.05) = 4.75 + 0.47712 = 5.22
[H+] = 10^-pH = 5.928*10^-6

[H+] = 0.0375M + 5.928*10^-6 = 0.037505928
pH = -log(0.037505928) = 1.4

The correction indicated that the answer should have been a pH = 4.88.
I don't know where it went wrong, but I have the feeling I terribly misunderstood a concept.

Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. Advantage is: Focus on principles, Easier spotting of mistakes, Improved orientation, Q/A reusability and the permanent value.
You may find useful  formatting mathematical/chemical expressions/formulas.

First, you have to allow the strong acids to react with the weak base.
$$\ce{HCl + NaOAc -> HOAc + Na+ + Cl-}$$
$$\ce{HNO3 + NaOAc -> HOAc + Na+ + NO3-}$$
This goes to completion. As a result, you make some acetic acid and use up some, but not all of the acetate.
Now you should be able to figure out the final concentrations of the major species (acetic acid, acetate) and plug it into the buffer equation.
To turn this into a rule: you will never have a strong acid and a weak base in the same solution (same goes for a strong base and a weak acid). They always react with each other until at least one of them is gone.


text:
Does mayer's equation change for two mole of ideal gases or is it remains same for any number of moles
Mayer's Formula is given by: Cp – Cv = R
But is the relation remains same for two mole of ideal gas, or is it
Cp - Cv = 2R
As, far as i know the number of moles are cancelled out in the derivation.
∆H = ∆U + nR∆T
nCp∆T = nCv∆T +nR∆T
Cp - Cv = R   (remains same)
(∆H is enthalpy change, ∆U is internal energy change, n is no of moles, R is gas constant and ∆T is change in temperature, Cp is molar specific heat capacity of an ideal gas at constant pressure, Cv is molar specific heat capacity of an ideal gas at constant volume )

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Note that dimension of $R, C_p, C_V$ is $\pu{J K-1 mol-1}$ and not  $\pu{J K-1}$.
Molar ($\pu{J K-1 mol-1}$) and specific ($\pu{J K-1 kg-1}$) heat capacities are intensive properties, independent on scaling (amount of gas).
Heat capacities ($\pu{J K-1}$)  are extensive properties, proportional to scaling (amount of gas).


text:
Do ion-exchange membranes *always* require ions to pass in both directions?
In an answer to this question, it was asserted that:

Student always tend to forget that a solution must remain electrically neutral by all means. If you want a certain ion to leave water, you must provide another ion of the same charge that has left the solution.

As per discussion in the comments to that answer, is the second quoted sentence strictly true?
Can electrical neutrality not be achieved by electrons (rather than ions) migrating one way or another to provide balance? For instance, perhaps imagine an aqueous solution under electrolysis which is divided by an appropriate membrane. A quick search brings up AEM (Nature paper) which appears to depict unidirectional migration of hydroxyl ions.

The requirement for neutrality is not as strict as to a single ion excess. It just means forming local excess of one kind of ions rapidly changes local potential, which is acting against such excess.

@Poutnik yes indeed, but I don't think that's what the quoted text meant to exclude - I suspect the author meant "in order for ions to continue to pass in one direction you must provide ions going in the other" (or similar). Perhaps they will reply here.

Can electrical neutrality not be achieved by electrons (rather than ions) migrating one way Electrons react with water in fractions of milisecond, what can be seen as short flashes of blue color of hydrated electrons during reaction of alkali metals with water. It was observed using an ultrafast camera with thousands FPS.

@Poutnik what can I search to see this please.

@htmlcoderexe  Search for articles (co)authored by Czech chemist Pavel Jungwirth. Some of them, like the one in Nature, may be behind a paywall. the main novelty is the explosion of alkali metals in contact with water is primarily physical coulombic explosion, not explosion of hydrogen. See e.g. Google: "Pavel+Jungwirth" alkali metal explosion electrons blue

@htmlcoderexe  See also Jungwirth group publications, with possible search filtering on site and Nature:  Coulomb explosion during the early stages of the reaction of alkali metals with water

@Poutnik thank you so much, I will look through this as soon I have the time!

I've done my best to distill these comments into an answer

Yes-but.
An inert cathode (e.g. the standard hydrogen electrode used as a cathode) does release electrons into water, but those electrons almost immediately react with the water, according to what we normally think of as a half-reaction,
$$\ce{2e- + 2H_2O -> H_2 + 2OH-}$$
so the effect is to produce hydroxide ions in solution (and hydrogen gas).
A series of experiments involving high-speed cameras and alkali metals have demonstrated that free electrons are very briefly in solution in water before reacting, e.g.

Coulomb explosion during the early stages of the reaction of alkali metals with water
Mason PE, Uhlig F, Vaněk V, Buttersack T, Bauerecker S, Jungwirth P.
Nature Chemistry 7(3):250-4 (2015)
A Non-Exploding Alkali Metal Drop on Water: From Blue Solvated Electrons to Bursting Molten Hydroxide
Mason PE, Buttersack T, Bauerecker S, Jungwirth P.
Angewandte Chemie 55(42):13019-13022 (2016)

(Free electrons can be a stable solute for much longer in other solvents.)

... but in the case of an aqueous solution under electrolysis, don't the electrons go through the wire (and power supply)? Which is why we see electron flow in the wire and (I presume) no free electrons / blue flashes of light.

@abligh The electrons go through the wire and power supply, yes, but those are in between the anode and the cathode.  Once electrons reach the cathode, they go into solution and then react with the water as I described.  I think the reason we don't observe blue flashes with the standard hydrogen electrode is just that the electrons enter solution much more slowly than they do when being shed by alkali metals, so the concentration never gets high enough to be observable.


text:
Where is the LITHIUM in NMC Cathode?
This maybe a noob question, but regarding NMC Cathodes, it is said that 33% each of Ni, Mn and Cobalt. So where is the LITHIUM supposed to be if all Ni Mn and Co already make up 100%?

The lithium is present as ions that intercalate into the cathode and anode.

@Poutnik So from a Chemistry point of view, can you please give me an example of NMC Cathode. There what would be the weight in % of Lithium, when we are already saying Ni Mn and Co mixed oxides make up 100%?

This info is easy to be found, such a question should be rather closed due lack of own a priori explicit effort.

This info maybe easy to find, but it may not be very easy to understand for people who are not from a background of chemistry (or have been out of touch with school chemistry for long). Given that you have answered correctly, and I will of course close the question, it wouldn't hurt to be mindful of the fact that Stack overflow is a forum for asking questions, easy the answer for it may be. The added deprecative comment from your side just leaves a bit of a sour aftertaste for others who may not even ask questions seeing knowledgeable people like you for whom the answer maybe in the face.

Your question in its current formulation is better suited for sites like Quora, which, in contrary of SE sites, has practically no question policy, aside of inappropriate questions.  Questions containing (more or less) just a question(s) are usually closed on the SE network site for lack of own explicit a priori effort. How do I ask a good question?. // As you originally accepted the answer, you understood it. So you would, if you found it by yourself, eventually asking for final clarification. At least trying is essential on SE,

(Reusing my deleted comments:)
$\ce{Ni + Mn + Co}$ do not make 100%, as they are there in form of mixed oxide. Each 33% rather means they are in 1:1:1 ratio. E.g. for purely $\ce{Co}$ cathodes, the composition is $\ce{Li_xCoO2}$, with 0<x<1. The key is not to make x=0 nor 1, what would lead to irreversible changes.
In fact, it is formally mixed $\ce{Li + Ni + Mn + Co}$ oxide, where $\ce{Ni + Mn + Co + O}$ form an anionic lattice, while Li+ ions are intercalated within it, in variable amounts according to the charge state ( More charge = less Li, as it is moved to the graphite anode)
See also Wikipedia: Lithium nickel manganese cobalt oxides

Lithium nickel manganese cobalt oxides (abbreviated Li-NMC, LNMC, NMC or NCM) are mixed metal oxides of lithium, nickel, manganese and cobalt. They have the general formula $\ce{LiNi_xMn_yCo_zO2}$. The most important representatives have a composition with $x + y + z$ that is near 1, with a small amount of lithium on the transition metal site. In commercial NMC samples, the composition typically has < 5% excess lithium.

From neicorporation.com: Lithium Nickel Manganese Cobalt Oxide (NMC) Electrode Sheets:

Layered Lithium Nickel Manganese Cobalt Oxide ($\ce{LiNi_xMn_yCo_zO2}$) is a widely used class of cathode material with NMC111 ($\ce{LiNi_{1/3}Mn_{1/3}Co_{1/3}O2}$), also known as NMC333) being the most common representative. However, Ni-rich NMCs (NMC811, 622, 532) are becoming more and more in focus now in current research due to their higher specific capacity and energy. Among these materials, NMC811 ($\ce{LiNi_{0.8}Mn_{0.1}Co_{0.1}O2}$) can deliver a high capacity of ~$\pu{200 mAh/g}$ with an average discharge potential of ~$\pu{3.8 V}$ (vs. $\ce{Li+}$/$\ce{Li}$), making it a promising positive electrode material for high energy density lithium-ion batteries.


text:
How do I show that a transition is electric dipole allowed with group theory/symmetry?
This is actually a follow up of this question
The follow-up is not because of the electric instead of magnetic dipole (this is trivial).
It is because I'm interested in extra info.
Suppose I have a system with $O_\mathrm{h}$ symmetry. The electric dipole allowed transitions from the ground state belong to the $\mathrm{T_{1u}}$ irrep. Following the argument in the answer to the previous question, the ground state (or initial state) is $\mathrm{A_{1g}}$, the electric dipole $\mathrm{T_{1u}}$ and states which belong to $\mathrm{T_{1u}}$ satisfy the requirement that $\mathrm{A_{1g}} \in (\mathrm{A_{1g}} \otimes \mathrm{T_{1u}} \otimes \mathrm{T_{1u}})$. Nice.
Now suppose that I create an excited state with an electric field. Then I will end up in a state which belong to the $\mathrm{T_{1u}}$ irrep. To know which transitions are electric dipole allowed from such states (i.e. which can be probed via an electric field), I need to find the irreps $Y$ for which $\mathrm{A_{1g}} \in (\mathrm{T_{1u}} \otimes \mathrm{T_{1u}} \otimes Y)$.
Since $(\mathrm{T_{1u}}\otimes \mathrm{T_{1u}})  = (\mathrm{A_{1g}} \oplus \mathrm{E_g} \oplus \mathrm{T_{1g}} \oplus \mathrm{T_{2g}})$ and $A_{1g} \in (Y \otimes Y)$ is always verified, any state belonging to any of $\mathrm{A_{1g}}$, $\mathrm{E_g}$, $\mathrm{T_{1g}}$, $\mathrm{T_{2g}}$ can be bright.
Here the new part. In a real experiment, which transitions are possible also depends on the relative polarization of the pump and the probe electric field. Say that the pump selects the $x$-polarization (let me call it $\mathrm{T}^x_\mathrm{1u}$ state) and the probe can be polarized along $x$ or along $y$. Can I say something about
$(\mathrm{T}^x_\mathrm{1u}\otimes \mathrm{T}^x_\mathrm{1u}\otimes Y)$ vs $(\mathrm{T}^x_\mathrm{1u}\otimes \mathrm{T}^y_\mathrm{1u}\otimes Y)$?

The xyz in the point group also describes the photon's polarisation as it were, the probability of absorbing a photon goes as polarisation  direction of photon as a dot product with  the dipole direction in the molecule  $q_{photon} \cdot q_{dipole}$  where $q$ is any of $xyz$ in same frame of reference

@porphyrin. Sure, what you writeis completly general. However it does not reply to the question above, i.e. how this can be translated in terms of the irreps.

The T${1u}$ symmetry species is the same in x, y or z, in Oh point group so there is no difference, i.e. any polarisation of your radiation will be absorbed depending in what Y is, e.g. ground state A${1g}$. You will need to look up the direct product of T${1u}$ (or work it out from irreps) and multiply each of products by your choice of Y. So if you want to probe from the state T${1u}$ you have produced it will depend on what the final state symmetry is and on dipole direction you choose.

Yeah. This is the point. My question can be maybe changed like that. I know that $T_{1u}\times T_{1u} = (A_{1g}+E_g+T_{1g}+T_{2g})$ where $T_{1u}$, as you say, does not distinguish between x, y ,a z. Now, suppose I define a way to distinguish the direction, notation $T_{1u}^i$ with $i=x,y,z$ (not sure if this is possible). Can I say for example something like $T^x_{1u} \times T^x_{1u} = A_{1g}+E_g $ while  $T^x_{1u} \times T^y_{1u} = T_{1g}+T_{2g} $ ?

No: I don't see how this would work. The irreps are the irreps, thats the point about group theory there are only so many ways of arranging things.

Can't I define, for each irrep, something like an ortho-normal basis set which spans the irrep. $T_{1u}$ has dimension 3, and $T^i_{1u}$ would be the there elements of the basis set. Then fixing the polarization of the filed, would be equivalent to choose a basis set with specific properties. Finally I would be able to define to which irrep the product of two basis function belong to. Looking to the fact that $T_{1u}$ transforms like (x,y,z), $A_{1g}$ like $x^2+y^2+z^2$, $E_g$ like combinations of $x^2$, $y^2$, and $z^2$, and $T_{2g}$ like $xy$, $xz$, $yz$ ...

@porphyrin . I've finally found the answer. In the book on gruop theory by Dresselhaus the partners $T^i_{1u}$ and similar are defined. See updated answer below

I found the reply in the book by Dresselhaus.
See also discussion here: https://arxiv.org/abs/2211.12241
$T_{1u}$ transforms like $(x,y,z)$
I assume that selecting the pump polarization along $x$, I select the state which transforms like $x$.
Sending a probe polarized along $x$ I obtain something that transforms like $x^2$, i.e. which belong to
$A_{1g}$ ($A_{1g}$ transforms like $x^2+y^2+z^2$), or $E_g$ ($E_g$ transforms like combinations of $x^2$, $y^2$, and $z^2$).
As a result, for pump and probe with same polarization, these are the allowed transitions:
$T_{1u}\rightarrow A_{1g},E_g$.
For orthogonal pump and prove I obtain something which belongs to $T_{2g}$ ($T_{2g}$ transforms like $xy$, $xz$, $yz$) or $T_{1g}$ ($T_{1g}$ transforms like $R_x$, $R_y$, $R_z$, since i.e. like $x p_y - y p_x$).
As a result, for pump and probe with orthogonal polarization, these are the allowed transitions:
$T_{1u}\rightarrow T_{1g}, T_{2g}$.

If you pump with any polarisation from the ground state you will get $T_{1u}$ and so probing with any polarisation will measure ground state bleaching, if in region of gs absorption, or gain if in emission region, or absorption from excited state if elsewhere and the final state must be $A_{1g}$ in each case as molecule has such high symmetry, i.e.  only $A_{1g}\mu_{x,y,z}T_{1u}$ contains $A_{1g}$.  Operators $x^2,xy\cdots$ are  for Raman transitions not dipole as Raman measured polarisability which is a measure of volume and so cross section is a squared coordinate term.

Here I'm discussing absorption from excited state. As explained in the answer (And I've also verified numerically), the final state is either $A_{1g}$ or $E_{g}$ for probe polarization parallel to the pump polarization, or $T_{1g}$ or $T_{2g}$  for probe polarization perpendicular to the pump polarization. I'm going to upload soon an arxiv with the numerical results on LiF and hBN. I'll add the link to the reference here.


text:
Open shell TDDFT excitaion contribution
If I have a open shell system (10 alpha electrons and 9 beta electrons) and running TDDFT calculations, is it possible to have a 9a→10b transition? Or does it have to be 9b→10b? Because both exciation won't change the multiplicity of the system, ending with one un paired electrons.

If by alpha and beta you mean spin up and spin down, you would need spin flip TDDFT.
Standard TDDFT is for the density - density response function $\chi_{\rho\rho}$. It describes the response of the system to a scalar potential $V$ which couples with the density. This captures the effect of an external longitudinal electric field. The density operator is composed by a pair of operators with the same spin, and moreover the spin index is summed. $\rho=\sum_\sigma \rho_\sigma=\sum_\sigma \psi^\dagger_\sigma\psi_\sigma$. This is why you cannot flip spin.
First you need to move to TDSpinDFT. A theory in terms of $\rho_\sigma$ and $V_\sigma$. This captures both a longitudinal electric field and the z component of a magnetic field.
Then you need to generalize this to include the spin flip channel, using $\rho_{\alpha\beta}= \psi^\dagger_\alpha\psi_\beta$. With that you also capture the response to the x and y components of an external magnetic field. This is what can flip the spin.

Welcome! It is not necessary to post your answer again. For future reference, if you cannot undelete your post, raise a custom flag on it, so moderators can do it for you.


text:
What is the burette solution called in a titration, when its concentration is unknown?
Generally in titration, the unknown solution is in the conical flask, called the analyte; the standard solution is in the burette, called the titrant. How do the terminologies change when one swaps the contents in the burette and the conical flask?
Is the solution in the burette always called the titrant? Or is titrant the solution with known concentration?

There is no fundamental requirement in analytical chemistry that the unknown or the known solution is in the burette. Traditionally, the standard solution is filled in the burette. All we need is the correct volume of unknown and the known (called a standard) to perform calculations.
The reagent (known or unknown concentration) being added to a receiving solution is called the titrant. The receiving solution is called the titrand. The latter term is commonly used as a jargon but it is not an accepted word in leading English dictionaries (Websters, Oxford etc.).


text:
Proof for equation of radial probability distribution curve
So my textbook says the equation for this is
$$P(r) = 4\pi r^2\Psi^2 $$
It also gives the volume of the shell formula
$$\mathrm{d}V = \frac 43 \pi(r+\mathrm{d}r)^3 - \frac 43 \pi r^3$$
which I understood
But what is the derivation method for the radial probability distribution curve equation mentioned above?

@Poutnik thank you, can you please elaborate on the proof of the latter?

The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.
Effort not shown can be considered as effort not done and such a question may be  closed.  How do I ask a good question?.

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https://chemistry.stackexchange.com/q/107097/16683

There is no experimental proof about the radial distribution function. But there is a proof about the value of the wave functions. If the wave function is twice derived and the result is added to the potential energy multiplying  the wave function, like in the Schrödinger equation, the result gives the exact values of the measured transition energies of the hydrogen atom. And this is a great success.

More generally, physical equations can never be proven in the rigorous mathematical sense. All we can say is that they are consistent with all known experimental data and with other similarly supported models.

I expect that OP means derivation, not proof. But yeah.

Yes I want derivation, sorry. Just edited the question

You are not much clear what exactly you want to derive.Should the derivation include why Psi^2 is the probability density, or, have you already accepted that? As for the latter, it is really trivial for case Psi^2 is spherically symmetric. The equation is not true if not symmetric, where integral over fi and omega angles must be present.

Let take for the question context as the axiom of the quantum atomic theory this:

$|\Psi|^2$ is the probability density an electron occurs at the given point. $\Psi$ itself is a wave function as a particular solution of Schroedinger wave equation.

Then the probability differential for the given volume differential is:
$$\mathrm{d}p=|\Psi|^2\mathrm{d}V=|\Psi|^2\mathrm{d}x\mathrm{d}y\mathrm{d}z$$
The wave equation and functions could be formally formulated for Cartesian coordinates x,y,z. But as a general rule, equation solving and the solution results are more simple and elegant, if the symmetry of the problem and the symmetry of the used coordinate system match each other. For that reason, both the equation and functions are formulated for spherical coordinates $r$, $\varphi$, $\theta$, as the electrostatic force with the central charge is spherically symmetric.
$$\mathrm{d}p=|\Psi|^2\mathrm{d}V=|\Psi(x,y,z)|^2\mathrm{d}x\mathrm{d}y\mathrm{d}z=|\Psi(r,\varphi,\theta)|^2r^2\sin{\theta}\mathrm{d}r\mathrm{d}\varphi\mathrm{d}\theta$$
For a spherically symmetric case of s-orbitals, $\Psi$ is a function just of $r$. The infinitesimal volume of a spherical shell is
$$\mathrm{d}V=4\pi r^2 \mathrm{d}r$$
(where $4\pi r^2$ is the surface of a sphere with the radius $\mathrm{r}$), therefore:
$$\mathrm{d}p=|\Psi|^2\mathrm{d}V=|\Psi(r)|^2 4\pi r^2 \mathrm{d}r$$
Then, $P(r)=|\Psi(r)|^2 4\pi r^2$ is the radial probability distribution function in the equation $\mathrm{d}p=P(r)\mathrm{d}r$
for spherically symmetric wave functions of s orbitals.

For general wave function $\Psi(r,\varphi,\theta)$:
$$P(r)=\int_0^{2\pi}{\left(\int_0^{\pi}{|\Psi(r,\varphi,\theta)|^2r^2\sin{\theta}\mathrm{d}\theta  }\right)\mathrm{d}\varphi},$$,
as the radial probability distribution function $P(r)$ is $|\Psi(r,\varphi,\theta)|^2$ integrated over the spherical surface of the radius $r$.
As a check for the symmetrical case above, $\Psi(r,\varphi,\theta) = \Psi(r)$
$$P(r)=\int_0^{2\pi}{\left(\int_0^{\pi}{|\Psi(r,\varphi,\theta)|^2r^2\sin{\theta}\mathrm{d}\theta  }\right)\mathrm{d}\varphi} = |\Psi(r)|^2r^2 \int_0^{2\pi}{\left(\int_0^{\pi}{\sin{\theta}\mathrm{d}\theta  }\right)\mathrm{d}\varphi}= \\|\Psi(r)|^2r^2 \int_0^{2\pi}{\left(2\right)\mathrm{d}\varphi}=|\Psi(r)|^2 4 \pi r^2$$

Just to expand a bit on the detailed answer already given: the problem can be solved in arbitrary coordinate system, and it all comes down to finding the volume element in different coordinate systems, which has a relatively simple general algorithm.
So instead of just stating on how to write a volume integral in spherical coordinates, you can start by writing it in an arbitrary coordinate system as
$$ I = \int_{\pmb{V}} f(u_1, u_2, u_3) \left| \frac{\partial (x,y,z)}{\partial(u_1, u_2, u_3)} \right| du_1 du_2 du_3 \, ,$$
with $\left| \frac{\partial (x,y,z)}{\partial(u_1, u_2, u_3)} \right|$  is the determinant of the Jacobian matrix, measuring the change of a given volume element when going from Cartesian to the arbitrary $(u_1,u_2,u_3)$ system (we don't even need to be restricted to 3D by the way).
Then you write the definition of the spherical coordinates
$$ x=\rho\cos\theta\sin\phi\\
y=\rho\sin\theta\sin\phi\\
z=\rho\cos\phi $$
Then you take all the partial derivatives to build the $3x3$ matrix, take the determinant, and you end up with the correct formula.

Good idea and applicable generally for any coordinate transform. But, especially if we keep in mind quite simple way how spherical coordinates are defined, it may be easier to write it directly in them. // Also, integration is easier there.


text:
What happens to undissolved salt during electrolysis of brine?
Say we have a container of brine with excess NaCl in the anode, and a seperate container with just water at the cathode. Both are connected by a salt bridge. If we perform electrolysis, Sodium ions will migrate away to the cathode. Does this mean that some of the undissolved NaCl will now be able to dissolve?

Try it :-) Or simulate it with an electric circuit with very big serial resistor.

Does that mean no?

It means you should think about it long enough before asking. // The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.
Effort not shown can be considered as effort not done and such a question may be  closed.  How do I ask a good question?.

If the cathodic region contains only pure water, there will be no electrolysis. Pure water is not conducting the current. After some time, some ions will pass from the salt bridge to the cathodic region, which becomes slightly conducting. But the current will be extremely small


text:
Violation of pauli's exclusion principle? ( During formation of neutron star or black holes)
Sorry it's a bit long
so i was watching Action labs video on how neutron stars are formed and in that he says " if there's enough mass, then the mass will keep on collapsing in on itself in the star and will collapse more until actual electrons are so close that the only thing that is stopping them from overlapping each other is Pauli's exclusion principle ,but if there is too much mass then even the Pauli's exclusion principle cannot keep the electrons from getting closer together ,so electrons get closer to protons leading to electron capture " so i wanted to ask if formation of neutron star, or formation of Black hole leads to violation of Pauli's exclusion principle?
I tried searching this on quora and physics stack but i couldn't get a simple explanation, there were some contradictory answers. some people said in case of formation of Black holes, degenerate mass is formed and that Pauli's exclusion principle cannot be violated on earth but in space during formation of Black holes or neutron stars, while some said it can't be violated ever.

I would not underestimate physicists  for this topic. Chemists are renown of refusing to put a piece of black hole into a test tube.. (joking, but seriously)

To be clear, it's not that Pauli's exclusion principle suddenly "stops applying" to electrons, it's just that new states of matter with fewer electrons become more energetically favorable, exactly because PEP still applies to electrons - it gets to a point where the overall energy of a system is lowered by destroying the electrons so PEP doesn't force there to be super-high energy electrons. A similar thing happens to neutrons in a neutron star at the brink of collapsing into a black hole.

I’m voting to close this question because this is completely off-topic.

The key point about Neutron stars is that they don't have any electrons, so Pauli is irrelevant. Nuclear processes lead to the consumption of electrons by protons to give neutrons, leaving no electrons.

Imagine the Pauli exclusion principle (PEP) in white dwarf context like side shrinking of a layer of identical balls laying on the floor. When shrinked enough and no free place for all balls, the extra balls would start another, upper layer. And another layer. And another... And you would need progressively higher side pressure to continue.
When matter is being squeezed in white dwarfs, PEP forces electrons with overlapping wavefunctions to fill progressively higher and higher energy levels, not to be in the same quantum state. That leads to progressively higher needed pressure which stops dwarf contraction.
If matter is squeezed enough, electrons have high enough energy and are tightly localized near protons. That leads at some tipping point to the nuclear reaction capturing electrons:
$\ce{p + e -> n + \nu_e}$
and gravitation squeezing can continue.
For white dwarfs it is interrupted by runaway nucleus fusion and supernova 1A event.
Massive stars skip the white dwarf state, but the matter degeneration, driven by PEP, happens for the iron non-fusable core.
Such stars end in a supernova II event due runaway fusion due shock wave bouncing fusing layers and form a neutron star. More massive ones form a black hole.

It's simple. Pauli's princinple, for the electron at least, is about the fundamental mathematical formulation of wavefunctions in chemistry- which do not matter to black holes as they deal with the nuclei. Just like how chemical properties do not matter to radioactivity.


text:
Comparison of boiling points of n-propyl chlorine and ethyl iodide
I know that boiling point of alkyl halide is proportional to the length of the alkyl group, and that the order of BP is RI > RBr > RCl > RF for an alkyl group R. But which of the factors is dominant?

It depends on occasion, in this case iodide has higher boiling point.

Does it depend on the occasion? Perhaps some investigation of simultaneously removing a CH2 and converting a chlorine to an iodide could be investigated as a function of alkyl group. First step is to generalize the question, then determine analysis approaches.

In the online resource I found, it is clear that the boiling point of $\ce{C2H5I}$ is more than that of $\ce{C3H7Cl}$.

I wouldn't draw any general conclusions from this since the values are comparable and so the trend may change when comparing branched/ring hydrocarbon moieties. This trend also works out when comparing fluorides and chlorides. However, this trend is reversed when comparing chlorides and bromides, probably, because of the addition of d-subshell.
Edit:
Although I have not referenced the CRC handbook (since I do not have the handbook), I did find more evidence regarding the reported values here and here. You will also be able to cross-verify the data at ChemLibreTexts.
I also checked some values with Wikipedia and the corresponding first search engine result and the values seem to match.
Edit: While doing further reading from Wikipedia, I found that, besides dipole moment, boiling points are also related to molecular masses. I drew this graph showing the variation of boiling point (on $y$-axis) with the mass of the haloalkanes (on $x$-axis)

An expanded trend could also count as semi-answer, but the data at least needs to be presented 1. in a readable form as a properly formatted table with data sources and units and not a low-quality screenshot, and 2. from an authoritative source, which a random PPT file from the internet is not. Also, both chemical formulas and symbols for orbitals are upright.

@andselisk okay noted

That's great. So, when are you going to fix the answer? We can help with the formatting, but the data source is something you need to sort out yourself.

@andselisk i'll fix the data source and formatting by tomorrow, quick question: do we accept wikipedia?

Strictly speaking, Wikipedia is not a reliable source, but it the absence of other sources it is at least trackable in most cases where the data comes from. I'd probably use CRC Handbook of chemistry and physics for data.

I'm not sure what's going on here, but citing two other obscure websites doesn't help. Why not search for the source of the table, reformat it as text so that it is searchable (not as an image), and cite the textbook it has been taken from? These websites ripped the table 6-2 from Vollhardt's Organic Chemistry 2014, so cite the origin and break the circle of copypasting stuff without proper attribution.

@andselisk I don't have Vollhardt's book either. Though I wanted to ask, what are your thoughts on the boiling point vs. molecular mass of the haloalkanes graph?

Two types of intermolecular forces influence the boiling points of alkyl halides. London forces and dipole dipole interaction. In case of alkyl halides the dipole dipole interaction has a smaller influence when compared with the influence of London forces. Now, London forces are surface attractions, arising from coordinated temporary dipoles and molecules with larger surface area have larger London attractions so resulting in higher boiling points. So that's why the boiling point of Ethyl iodide is about 72°C and that of Propyl chloride is about 47°C. And this is also reason that for halides boiling point order is
R-F < R-Cl < R-Br < R-I.