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Sublimation as a separation technique
During a lecture, my professor said that sublimation at atmospheric pressure (1 atm), as a purification/separation technique, can be performed when the sublimating substance has a triple point falling at pressures above 100 mmHg. Why is this assumption made? Shouldn't sublimation at atmospheric pressure only occur when the atmospheric pressure (1 atm) falls below the triple point? In fact, only under this condition can be found a temperature at which the vapor pressure of the solid equals atmospheric pressure.
I don't understand why we consider 100 mmHg as the limit.
Furthermore, it has been said that the sublimation point of vanillin, at atmospheric pressure, falls at 48°C, so at these pressure and temperature values we actually fall on the sublimation curve. Right?

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@Poutnik so, what you mean is that sublimation can happen at any point (of pressure and temperature) that falls in the area of the solid in the phase diagram. As one moves away from the sublimation curve, the extent of the same decreases. Therefore, if we are above 100 mmHg, the sublimation will take place with an acceptable speed at 760 mmHg, even if it is not possible to find a temperature at which the vapor pressure of the solid equals the atmospheric pressure (760 mmHg), because we would trespass the liquid zone.

@Poutnik The most we can do is raise the temperature slightly, to raise the vapor pressure. Right?

Water triple point is $\pu{+0.01 ^\circ C}$ at about $\pu{610 Pa} \ll \pu{100 kPa}$ and water does sublime at freezing temperature. But slowly. I assume $\pu{100 mmHg}$ is a practical threshold for sublimation to have a possibly sufficient rate. There is no need to fit atmospheric pressure. Liquid evaporation occurs with vapor pressure below external pressure too.
Phase diagrams are about equilibrium.
At temperatures below their melting point, solids sublimate until the vapor pressure reaches the solid-gas line. Higher temperature means faster sublimation with higher vapor pressure. If the triple point has too low a pressure, we can use only a low maximum vapor pressure which may not be practical for larger quantities.

What do you mean with 'which may not be practical for larger quantities'? And so, is what I wrote in the comments above right?

Try sublimation of 100 g of ice at -5 deg C ice to another place, cooled to- 24 deg C.

More or less correct. Sublimation and evaporation occur at any temperature of the solid or liquid, it is just a question of vapor tension at given temperature which is proportional to rate of phase conversion.

Great! Thanks! So, obviously, ice at -5 °C will sublimate greater. But, we are always at a pressure (760 mmHg) really far from pressures that individuate the sublimation curve, so It Is Always a really slow phenomenon

The external pressure is not much relevant.


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Differentiating identical and enantiomers

Why is this example not considered enantiomer but identical?
And as it is identical what would be the enantiomer pair?

Have you tried R/S configuration?

Yes i have . But my question here is why are these identical

Almost a duplicate: https://chemistry.stackexchange.com/q/13730/72973

Regardless if you just started to study organic chemistry, or not; there is nothing wrong to go to the shelf and to pick a real model kit (an example).

Try 3D pin and ball visual model from home available items, creating respective 3D models according to the Fisher projections of your image. Then try to superimpose them, rotating the whole models or the bonds. If you succeed, you will know why. If not, you will know they are not identical.
You can formally rotate the bottom part of molecules around the central $\ce{C-C}$ bond to have the side methyl on the bottom. Then it is clear it is the case of the same symmetry as for optically inactive meso tartaric acid.  Both stereogenic centers mutually cancel each other. The molecule has a plane of symmetry and therefore is not optically active, as it's mirror image can be superimposed with the original.

The suggested turning does not change configuration. Mirror images of your case and of meso tartaric acid can be turned around the bonds to become original image, what disqualifies them from being enantiomers.
With bonds properly rotated, the upper molecule half is the mirror image of the lower half. If you mirror the whole molecule, the upper half becomes the lower half and vice versa, leading to the identical molecule. That is revealed if you turn the mirror image by half turn clockwise.

With methyls on top and the bottom, the enantiomers would be those isomers with H and Br atoms alternating sides. They form mirror twins that cannot be superimposed. Again, they have the same symmetry as optically active isomers of tartaric acid.

If I consider flipping the molecule that would result in every pair of enantiomers being identical.

I have suspicion you take the charts as 2D flat structures, what is wrong. Fischer projection on your image is convention to display 3D bond structure, with the vertical bonds arranged in a way of bending top and bottom away from you, while horizontal bonds are stretching toward you. Like if there was a horizontal wooden log behind and you tried to wrap the molecule around it. Remember four C bonds are not rectangular in a single plane, but are pointing to the corners of a tetraedr.

The answer of the question is that these are identical. My answer was that these are enantiomers but it is wrong, so what I am not understanding is that how are these identical. If I consider flipping the molecule that would result in every pair of enantiomers being identical.

Sorry, I still don’t get it. Just as you said these are 3D arrangements , if I turn this into a wedge and dash structure to visualise it in 3D space I am still getting two different molecules which are mirror images. How are they identical. I am sorry I really don’t understand.

Imagine a ball, laid on a table. Imaging joining 4 other balls by 4 pins, pointing up, down, left, right. that is how it is drawn on your image above. But the reality is like if you lift this flat model by hands and force up and down pins to point little down and left / right pins little up.  Follow also the Fischer projection link.

Properly rotated, the upper molecule half is the mirror image of the lower half. If you mirror the whole molecule, the upper half becomes the lower half and vice versa, leading to the identical molecule.

I see it now, Thank you. What I was doing wrong was trying to match upper half to upper and not considering rotation of the molecule.

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At what temperature does Brass blacken in air?
A machine has overheated in an air filled enclosure, and a major brass component of mass around 1kg turned black. It was probably not in the high temperature regime for more than 2 hours at a guess and perhaps much less.
So, can anyone estimate what the approximate temperature might have been?

It might be useful to first make sure what the black surface really is. Is it the assumed CuO or is it the residue of some other decomposed or evaporated material from around? Could you show an image of the situation?

The black color is due black copper(II) oxide:

It can be formed by heating copper in air at around 300–800°C: $\ce{2 Cu + O2 -> 2 CuO}$

Possible range of temperature seems quite wide.
Appearing and intensity of brass blackening is combination of temperature and duration. It can be formed faster with growing temperature due faster kinetics of formation, but getting unstable at very high temperature due thermodynamic decomposition.
As organic substances are not reportedly present (above natural trace background), carbon from pyrolyzed  organic matter is not considered.

And any hydrocarbons in the air also contribute to blackening.

@JonCuster Yes, but it was not explicitly mentioned to be present.

sure, but ‘machine’ implies lubricants which get everywhere really easily.

@JonCuster True. But OP is not very explicit generally. // OP: Asking for answers and not providing enough relevant details, purpose, context or background of questions are contradictory decisions.

@Poutnik Commercially sensitive with legal implications. No lubricants or other organics present. Air was dry

It should be noted that air on Earth invariably contains some organic molecules, which come from a variety of materials, plants, and animals: https://doi.org/10.1111/j.1600-0668.1994.t01-2-00007.x

@WaterMolecule It does, but scientific zero is not the same as mathematical zero. In context of the scenario, the natural background is negligible, not distinguishable from not present.

@Poutnik Yes, but these molecules have a tendency to collect on surfaces, especially hydrophobic ones, resulting in measurable effects. https://doi.org/10.1038/s41467-022-34641-7 Whether there's enough to form a black layer on heated brass probably depends on the local concentration.

@WaterMolecule I agree. But the measureabililty is very context dependent. If one considers thermal evaporation and oxidation of traces + major component forming black oxide.....  // Turning black would not be limited to brass, even if it would be more noticeable there.

A metallurgist would guess at the temperature with a metallographic examination (grain size, twins, etc.) and hardness tests. (It would anneal to some degree depending on temperature.)
The first step would be to determine the brass/bronze alloy. Common wrought brass products are 70:30 or 60:40, and interestingly the 70:30 is most yellow. The 60:40 picks up a pinkish cast.

The blackening of copper alloys (e.g., basic brass is 33% zinc with 67% copper) is due to the formation of black oxide of copper (the process is known by the trade name Ebonol C: Accordingly, if the copper alloys is containing 65% or more of copper, a black oxide treatment can be applied to alloy surface to blacken by converting the copper to cupric oxide $(\ce{CuO})$. According to Wikipedia, the temperature required to blacken brass alloy is about $\pu{400 ^\circ F} \ (\pu{204 ^\circ C})$.
It is also noteworthy to mentioned here this somewhat cool educational demonstration. According to this Illini website:

[...] penny coins, starting in 1983, were made of zinc with a thin layer of copper plated on the surface. If these coins are heated, the zinc will diffuse into the copper layer, producing a surface alloy of zinc and copper. These alloys are brasses. Not only does the zinc change the properties of copper, but also the color of the brasses changes with zinc content - reaching a golden yellow color at around 20% zinc and golden at 35-40% zinc. Copper also oxidizes when heated in air, producing a black layer of copper oxide $(\ce{CuO})$. Thus when heated, there is a competition between the rate of oxidation (making the surface black) and the rate of diffusion (making the surface a golden-yellow color).

You may use this demonstration to find out the exact temperature when the coin starts blackening (hopefully)!

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looking for the absorption spectrum of polyene
I am looking for the absorption spectrum of linear polyenes (1,3,5-hexatriene, 1,3,5,7-Octatetraene, 1,3,5,7,9-decapentaene, etc..) in the range of UV and Visible light wavelength.
I see the data at the 2nd page of the PDF (*1). In the table, I see the max absorption wave length for hexatriene is 268nm.

name
$\lambda_{max}$

ethylene
163nm

butadiene
217nm

hexatriene
268nm

octatetraene
304nm

But I see another webpage(*2) says hxatriene's wave length is 258nm.
The absorbance due to the π – π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm.

Where do you look for linear polyenes absorption spectrum data?
(*1) https://www.chem.purdue.edu/courses/chm424/Handouts/14.2%20Alkenes%20and%20Aromatics.pdf
(*2)
https://courses.lumenlearning.com/suny-mcc-organicchemistry/chapter/interpreting-uv-spectra/

Hexatriene can exists as cis- and trans-isomers.

These are the wavelengths of the maximum of an absorption band so are probably only accurate to a few nm but I found 268 nm for the hexatriene, so 258 could be a typo.

@ porphyrin: According to this article, both isomers of hexatriene shows absorbance at $245, 255,$ and $\pu{265 nm}$.


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When we boil water, does the generated steam remain at the same temperature of boiling point until all water is boiled?
Suppose we are boiling water in such conditions so that the boiling point of water be $\pu{100 ^\circ C}$.

Does the generated steam have the same temperature of 100 C until all water is boiled?
And only after complete boiling does the temperature of steam start to rise?
Does it make a difference if we are boiling with a lid on or not?
Does only water have a temperature of 100 C and the temperature of
steam increases even if all water is not boiled?
In slightly  technical terms, is the heat provided to boil the water used only in changing the phase of water or is it also used in raising the temperature of the vapor formed?
If the temperature of steam does remain constant until all water is boiled the question arises why, why does it remain constant?
What is stopping the heat to not raise its temperature?
What forces it to be only used do drive phase change?

Please read this webpage and re-formulate your question: http://www.sites.hps.cam.ac.uk/boiling/steampoint.htm

Also, the myth of the boiling point: http://www.sites.hps.cam.ac.uk/boiling/index.htm

Note that posted StackExchange(SE) questions containing merely just questions are usually closed on SE network. What particular things did not you understand while reviewing offline and online resources about this topic? Asking should be the last thing to do, but the questions seem to be the opposite. // The steam temperature can be at, below or above boiling point, depending on particular scenario and warming or cooling effects acting on the steam.

@AChem boiling is surprisingly very complicated. I could not understand the first linked page so I can not re-formulate the question,

@Osmium, As you can see, boiling process is indeed very complicated. Don't rely on  and digest web "answers" without thinking until and unless someone comes up with solid data and experimental measurements.

As the heating device has routinly surface $\pu{t \gt 100 ^\circ C}$, steam can have easily $\pu{t \gt 100 ^\circ C}$ as well. OTOH, if the surface is at steam section colder, or if is being mixed with air, steam can have easily $\pu{t \lt 100 ^\circ C}$.

Steam (actually, water vapor) above a rapidly boiling can be hotter than the average temperature of the liquid because the situation is not in equilibrium.
Macroscopically, consider that in nucleate boiling the greater part of water in the container is not boiling; only in certain parts of the vessel has the water been superheated, and then flashes into vapor. "When the temperature difference is [sufficient], isolated bubbles form at nucleation sites and separate from the surface... [at a greater temperature difference] the vapor escapes as jets or columns which subsequently merge into slugs of vapor."
Those regions that formed bubbles were above the boiling point of water. Though the temperature of the bubble drops due to rapid expansion, it must still be above 100 °C or it would condense before reaching the surface (That oscillation of bubbles forming and collapsing can be observed at an incipient boil , i.e., liquid entrainment stage of boiling.). Therefore, the vapor immediately released from the bubble can be hotter than the body of the liquid, which can be verified by putting a thermometer in the spout of a steaming kettle and in the liquid.
On a microscopic level, consider the fastest moving particles, i.e, the hottest, are the ones that escape. Of course, that is on average, since there can be areas in the liquid that are superheated and have not yet boiled.
Note, though, that in film boiling, the Leidenfrost effect maintains a layer of exceedingly hot water vapor against the surface of the container, below the level of the water! Of course, that is not liquid water, but it is below the surface of the water... so the answer to your question depends on the rate of boiling and where in the container you measure the temperature.

The key question before we speculate too much is where do we measure the temperature. A thermometer dipped in boiling water will read 100 Celcius under ordinary conditions. Where do measure the steam temperature? Right above the water, 12 inches away ? Let us think experimentally.

In addition to the nonequilibrium factors described well by DeMi8she Pippik, we may also consider the behavior when the water is not pure. Assuming we have set up nucle-ation sites and other factors so that the temperature no longer differs measurably from equilibrium, that equilibrium temperature will, in most practical cases, tend to rise as boiling proceeds and the composition of the liquid inevitably changes.

With nonvolatile solutes, the frugality and therefore vapor pressure of the water will drop with the composition shifting to more solute. In the case of sugar solutions in water, this effect should be familiar to anyone who cooks candies or jams.

With volatile solutes such as ethanol, the boiling starts off with predominantly the volatile solute at lower than the boiling point os the water alone, and the temperature rises as the liquid composition shifts towards more of the less volatile water component. Here the manufacturing of distilled liquor comes to mind.

I think the OP assumes standard, normal, usual and expected behavior of boiling water and wants to focus on behavior and properties of steam/vapor during the boiling process at various scenarios.

I think you meant "nucleation sites" rather than "nucleotide sites"? Nucleotides are biochemical compounds.

@andrew the largest number used in mathematics is not Skewes' number of Graham's number, but the number of times I have been burned by Autocorrect. I'm sure of it! Now to defeat it, I added a hyphen.


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Determining the number of stereoisomers of a compound
I am confused about:

Does stereoisomers include both cis-trans isomers and enantiomers (those with chiral carbons)

If a cyclic compound exhibits cis trans isomerism, is it able to exhibit optical isomerism on the same carbons

How to determine the total number of stereoisomers if a compound has both cis-trans and optical isomerism

The question is as follows:

The prostaglandin molecule you have given has four stereogenic centers: Three on ring substitutions and one in a one of long side chains $(\ce{-CH(OH)(CH2)4CH3})$. Since none have other symmetric factors such as plain of symmetry, usual numbers of stereo isomers of the molecule are $2^4 = 16$. However, both long chain substitutions contain disubstituted double bonds, each of which can exist as $E$- and $Z$-isomers ($cis/trans$). Thus, each of the original 16 isomers can exist as four other srereoisomers: $EE$-, $EZ$-, $ZE$-, and $ZZ$-isomers. For example: $(2S,3S,4S,3''R)$- isomer can exist as $(2S,3S,4S,3''R)$-$(2'E,1''E)$-, $(2S,3S,4S,3''R)$-$(2'E,1''Z)$-, $(2S,3S,4S,3''R)$-$(2'Z,1''E)$- and $(2S,3S,4S,3''R)$-$(2'Z,1''Z)$-isomers, all four of which are stereoisomers with optical activity. Thus, you may find total of 64 optical active isomers in given prostaglandins.

$1)$ Cis-trans isomers are also stereoisomers. They are called as diastereomers.
$2)$ Consider, 1-bromo-3-methylcyclohexane. It shows cis-trans isomerism and each of the isomers(that is cis and trans) have enantiomers.
$3)$ Again, consider the above example. 1-bromo-3-methylcyclohexane has $4$ stereoisomers, each of them is chiral.
The compound you have given have 16 stereoisomers excluding cis-trans isomers.

Hi for point 3), if 1-bromo-3-methylcyclohexane has one site for cis-trans isomerism and two chiral carbons (on carbon 1 and 3) does that mean that the total number of stereoisomers is 2^(1+2) = 8?

Also, for the compound in my question, how many sites for cis-trans isomerism are there and what would be the total number of stereoisomers possible?

Yes. In general the maximum numbers of stereoisomers a molecule can have is $2^{m+n}$, where $m=$ number of chiral centres and $n=$ number of double bonds that can give cis-trans isomerism . For the compound you gave , there are $2$ sites of double bond that give cis-trans isomerism.

Remember that cyclohexane exists in chair conformation , so cis corresponds to both functional group facing upwards and trans corresponds to one group facing upwards and the other downwards. So your first comment, there are no sites for cis-trans due to double bonds, so total number of stereoisomers is $4$.

I see! Thank you very much!


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Hannay smith equation derivation
Our chemistry teacher taught us about Hannay smith equation for calculating the percent ionic character. He did not told how that equation was derived. I could not find any satisfactory result from google search either. But, I am curious to know how it was derived.
$$ \% \hspace{2mm} ionic \hspace{2mm} character=[16(\Delta E.N.)+3.5(\Delta E.N.)^2]\%$$

I assume it is the result of empirical interpolation of data, rather than derivation.


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Why does cold welding not work with carbon steels?
Cold welding is the process of joining metals using pristine surfaces without contaminents(oxide layers, oil, dust...) using pressure through atomic diffusion between the metal joints. Why does this process not work with carbon alloys such as stainless steel? I've only seen 1 article mention it worked with low carbon steel but it cannot be done with higher carbon steels but no explanation as to why.

I guess the metal has to be at least somewhat soft and pliable, which ordinary steel just isn't.

While it wouldnt exactly be cold welding anymore, would a modified process which heats the higher carbon steel to be more ductile work with pressure? Or is the carbon the real problem?

The key properties of many steels are based on careful creation of the domain structure of the bulk material (eg grain size and composition) and surface effects (eg strong oxide layers on the surface to protect against corrosion). They are often products of the specific production method which involves the heating history. Cold welding will alter the key properties in ways that don't preserve the key features of the original alloy. This matters a lot less in more uniform alloys, and might explain the differences.

Its good for welding aluminum in the 7xxxx series though, which isnt exactly uniform. it just seems like carbon diffusion is the problem, probably due to ductility being decreased. But it would be great to have a research paper dedicated as to why atomic diffusion doesn't take place and in what circumstances it does.

The carbon inhibits flow at the grain boundaries. That said, cold welding still takes place, if you wait long enough. Clamp the cleaned parts under high vacuum and wait a few years.

@DrMoishePippik could you clarify as to why it inhibits the flow? Does it have something to do with its electron configuration? Or atomic size/weight? What makes carbon inhibit the flow at the grain boundaries under pressure but not other atoms?

I guess (from having heard a few lectures about solid state inorganics, no real experience on the subject) that complex crystalline structures that form a polycrystalline surface just don't fit together sufficiently, crystallographically, with a second surface. Even if both are polished flat down to the atomic scale.
That is especially true if they are not just metallic, but have some covalent character bonds in them too. The chance that even a tiny fraction of domains on both sides would by chance fit together is practically zero.
For the two pieces to form a continuous phase, a lot of surface atoms have to move around, and they're not going to do that if the jump to the next energetically favourable position is more than one or two atomic diameters.

So why domains of carbon steel are worse fit together?

They never fit perfectly together, also with plain iron. Unless you take two single crystals and polish and align them to atomic precision. The opposing domain/grain surfaces have to rearrange a bit to form a single, continous grain boundary between them.

Ok, so why carbon steel rearranges/aligns/fits/etc worse than plain iron? This was the question you are trying to answer.

@sa7 Because plain iron has a simple structure, where all lattice positions are more or less equivalent.


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Why must the beta term in the rigid-rotator PDE take the form J(J+1)?
Here is the partial differential equation for which the solutions Y are are the wave functions for the rigid-rotator (Equation 5.5 in McQuarrie and Simon's Physical Chemistry).
$$\sin\theta\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial Y}{\partial \theta}\right) + \frac{\partial^2 Y}{\partial \phi^2} + \left(\beta \sin^2\theta\right)Y = 0$$
In McQuarrie and Simon's Physical Chemistry, it is said that

When we solve Equation 5.5, it turns out naturally that $\beta$ ... must obey the condition: $$\beta = J(J+1)$$

My question is why must $\boldsymbol{\beta}$ take the form $\boldsymbol{J(J+1)}$ (where $\boldsymbol{J}$ is a natural number))? Since $\beta$ is essentially a constant, this implies that using any value of $\beta$ (e.g., 1.5) would lead to a partial differential equation with no solutions. (1) is this assumption correct and (2) if so, what is the mathematical proof that $\beta$ in any other form would lead to no solutions?

$\beta$ must take this form because any other value of $\beta$ would lead to a partial differential equation with no nice solutions. As for the mathematical proof, it can be neither told nor heard without doing quite a bit of math.

I guess, in the linked question I only explicitly show that the term must be of the form J(J+1), and I don't actually prove that J must be a non-negative integer (assuming that we're talking only about orbital angular momentum). As Ivan's comment notes, the proof of that is pretty ugly (you must explicitly solve the differential equations), and you are best served by looking it up in a textbook of your choice. I think the solutions are called associated Legendre polynomials, so that will give you a search term. Btw, if you look it up, skip the chemistry books—go straight to physics or maths.


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AlCl3 stuck to Pyrex
I made $\ce{AlCl3 . 6 H2O}$ the easy way, with $\ce{HCl}$ and foil. After filtering and reducing on the stove, I get a nice yellow crystal of contaminated $\ce{AlCl3 . 6 H2O}$. The yellow should be from $\ce{Fe}$ contamination.
The problem: it is stuck insanely hard to my old Pyrex pan and I’m having trouble getting it out. I can re-dissolve it in water, but that’s a step back. Would nonstick work? Any thoughts?
Thanks

What's "reducing on the stove"?

Heating it up to remove the water.

Then you don't have any AlCl3 anymore.

Aluminum oxide is insoluble in water. These crystals re-dissolve. I think I still have AlCl3 my only problem is scraping it off the bottom. I specifically used very gentle heat to prevent the oxide from forming.

To clarify I am not trying to remove the hexahydrate. Only the excess water.

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

Heating $\ce{AlCl3·xH2O}$ will never produce anhydrous $\ce{AlCl3}$. Never ! Hydrolysis will always occur, yielding $\ce{AlCl_{3-x}OH_x}$ plus some $\ce{HCl}$

The «easy way» is $\ce{2Al + 3Cl2 -> 2AlCl3}$ in a flow reactor, but not in a beaker.  An aqueous solution of $\ce{AlCl3}$, stabilized by hydrochloric acid, still will not  provide you an access to water free $\ce{AlCl3}$. You likely got something in the range between $\ce{AlOH3}$ (one can dissolve in either acid, or lye) and $\ce{Al2O3}$ because $\ce{Cl^-}$ and $\ce{H+}$ of water became $\ce{HCl}$ and left the container (you mention heating, the escape is more likely then).  Not necessarily corundum (one form of $\ce{Al2O3}$) though.

I would use polymer Petri dish (easier to scrape off) and vacuum evaporation instead of stove to reduce hydrolysis.


text:
Volume of a solution in terms of molality, molar mass and density
I found this equation in "Experiments in Physical Chemistry" by Garland, Nibler, Shoemaker. This is on page 174 (Experiment 9).
$$V=\frac{1000+mM_2}{d}$$
where $V$ is the volume of the solution, $m$ is the molality, $M_2$ is the molar mass of the solute and $d$ is the density of the solution.
The equation was in the middle of a derivation and is shown without any background information. I suppose it should be something pretty straightforward but I am not able to obtain this equation from the definition of molality, density and molar mass.

Please typeset the equations using mathjax: https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange Copy-pasted images are not searchable.

Recall the definitions of molality and molarity, it is mol solute / kg of solvent and mol solute/ Vol of solution in (L). It is not kg of solution in molality. Instead of memorizing a plug and chug formula, would not it be far better to understand the conversion and do it yourself?
I quote an example from General Chemistry by Ebbing and Gammon (2016). Read the strategy:

Molality to Molarity: An aqueous solution is 0.273 m
$\ce{KCl}$. What is the molar concentration of potassium chloride,
$\ce{KCl}$ ? The density of the solution is 1.011 x 10$^3$ g/ L.
Problem Strategy You are asked to determine the molarity of the
solution, which is the moles of solute $(\ce{KCl})$ per liter of
solution (volume of $\ce{H2O}$ and $\ce{KCl}$. The molality tells you that the solution
contains 0.273 mol $\ce{KCl}$ in $1 \mathrm{~kg}$ of
water. If we determine the volume of solution that contains $0.273
 \mathrm{~mol} \ce{KCl}$, we should be able to calculate the
molarity. Because the solution is made up of only $\ce{KCl}$ and
water, the solution mass is the mass of 0.273  mol
$\ce {KCl}$ plus the mass of 1.000 kg of water. Knowing
the total mass of solution, we then can use the solution density to
convert from the mass of solution to the volume of solution. The
molarity is obtained by dividing the moles of solute by the volume of
solution in liters.

EDIT: Now that the OP posted the text, we can discuss the hints.
The equation $$V=\frac{1000+mM_2}{d}$$
is derived from mass balance principle. Keeping the above discussion in mind,
$$Total\hspace{0.15cm}mass\hspace{0.15cm}of\hspace{0.15cm}solution, M=mass _{solvent} + mass_{solute}$$
This is in grams. Now recall the mass volume relation from density. Here $V$ is the volume of solution not the solvent.
$$d=\frac{M}{V}$$ or $$M=dV$$
These hints should be enough to derive the desired equation.

I don't want to memorize the formula. I want to derive it. It is more general that a simple example from Gen. Chem. The equation is just part of a longer derivation.

I edited my answer in case you want to see where the equation appears.

Thanks, I will have a look. Please add the full book reference.

"Experiments in Physical Chemistry" Garland, Nibler, Shoemaker. This is experimental 9, page 174

@user130775, Check if you can derive it now.

I believe I got to the point of Msolution = 1000+Msolute.
What I can't get is how Msolute = molality * (Molar mass) solute

@user130775, do a dimensional analysis. molality x formula weight gives the mass of solute in grams added to 1000 g of solvent.

Got it. For the mass of the solvent = 1000 g = 1Kg, the molality = mols. Then mols * Molar mass = mass(g). Thanks!

The equation given is
$$V = \frac{1000 + m \cdot M}{d} \pu{cm3}$$
It only works if you know to enter the quantities' numerical values after expressing them in the conventional units (g for mass, mol/kg for molality, g/mol for molar mass and e.g. g/mL for density). With these specifications, the expression gives the volume of a solution containing 1000 g of solvent.
Another way of writing this, using the modern abbreviation for molality, $b$, and mass, $m$:
$$V_\mathrm{solution} = \frac{m_\mathrm{solvent}  + m_\mathrm{solvent} \cdot b_\mathrm{solute} \cdot M_\mathrm{solute}}{d_\mathrm{solution}}$$
The numerator is the mass of the solution as sum of the mass of solvent and solute. In the cited equation, the 1000 reflects that molarity has units of $\pu{mol kg-1}$ but the molar mass conventionally has units of $\pu{g mol-1}$.

Good grief! IUPAC has come up with a new symbol for molarity as ($b$)? It is a joke that IUPAC mentions under amount concentration "Also called ...in older literature molarity". Molarity is as current terminology as possible. It is not going anywhere for several decades. What is the need to come up with new terms when there is no problem with molarity and or its symbols?

$$m = \frac{n}{m}$$ is confusing as a definition.

As is $$M = 3 M$$

One would write [NaOH] = 3 M to imply NaOH concentration is 3 Molar. I wish IUPAC also justified their choices with some context rather than impose it. Do they make their meeting minutes public? On the other hand, $m=\frac{n}{m}$ has symbolic and algebraic problems.

Sorry typo, I fixed it.

@Poutnik, Earlier it was molarity.

@Poutnik, "As you let it there..." There is a time difference in various continents :-). Yes, it is shocking how blatantly IUPAC denigrates molarity, a unit which has worked for centuries and come up with amount concentration.

@AChem Do you mean molarity as quantity, sharing the unit mol/L with its synonym "amount concentration", or the molarity context specific unit M = mol/L ?

@Poutnik, Yes, I was wondering about the need to invent amount concentration when molarity does a good job. What is the need to coin a synonym?

@Achem Molarity is too short, cannot be confused with its vaque generic meaning (amount) in everyday life and perhaps IUPAC guys are paid by the length of the terms. :-P  (deleting older or OT comments)

@Poutnik as much as I appreciate your sarcasm, there are parts of the world, where the sounds for r and l are hard to distinguish. Therefore molarity and molality are also hard to keep apart, and the typo potential is high. I personally dislike the term molarity very, very much. And why are you guys hating so much on IUPAC?

@Martin-マーチン If one of terms molarity vs. molality should go to avoid confusion, I would vote for molality, as it is in my experience used much less (never in my case, but during studying). I do not hate IUPAC, just sometimes wonder over their decisions. In next few decades, I do not give much chances to "amount concentration" against molarity among practical chemists. If it must be, I would prefer molar amount, molar concentration, like we already have e.g. molar conductivity, molar TD state function values, or molar heat capacity.

@Poutnik One criticism of the term molarity (or molar concentration) is that molar quantities have the chemical amount in the denominator (molar volume, molar mass), but here it merely says to give the amount of substance rather than the mass or the volume of the solute as part of the measure of concentration. It is also a matter of languages. "Stoffmengenkonzentration" is not considered a overly long word in German, and at least it is a single word.

@Karsten well, molar conductivity is not conductivity per 1 mol either. :-P


text:
Equilibrium concentrations of CO and Cl2 by dissociation of phosgene
I am working through a problem set from MIT's open courseware course Principles of Chemical Science.
The question is:

Phosgene $\ce{(COCl2)}$ is a chemical warfare agent that decomposes by the reaction:
$$\ce{COCl2(g) <=> CO(g) + Cl2(g)}$$
$K= 8.3 \times 10^{-4}$ (at $\pu{360 ^\circ C}$)
Calculate the $\ce{[CO], [Cl2],}$ and $\ce{[COCl2]}$ when $\pu{10.0 mol}$ of phosgene decomposes at $\pu{360 ^\circ C}$ and reach equilibrium in a 5.00-L flask.

I am a bit unsure whether equilibrium constant given is the ratio of partial pressures, or concentration of gases. ($K$ or $K_\mathrm{c}$)
My first approach is assuming that $K$ is given (not $K_\mathrm{c}$):
Given: $T =\pu{360 ^\circ C} = \pu{633.15 K}$

Convert $K$ to $K_\mathrm{c}$
$K = (RT)^{\Delta n_\mathrm{r}} \cdot K_\mathrm{c}$
$K_\mathrm{c} = \dfrac{K}{(RT)^{\Delta n_\mathrm{r}}}$
$K_\mathrm{c} = \dfrac{8.3 \times10^{-4}} {(8.3145 \times10^{-2} \cdot 633.15)^1}$
$K_\mathrm{c} = 1.577 \times10^{-5}$
Calculate starting concentration of $\ce{COCl2}$
$[\ce{COCl2}] = \dfrac{\pu{10.0 mol}}{\pu{5.00 L}} = \pu{2.00 mol L-1} $
Calculate equilibrium concentration of each species:
$\begin{align}
& [\ce{COCl2}] = 2.00 - x \\
& [\ce{CO}] = x \\
& [\ce{Cl2}] = x \\
& K_\mathrm{c} = \dfrac{[\ce{CO}][\ce{Cl2}]}{[\ce{COCl2}]} = \dfrac{x^2}{2.00 - x}
\end{align}$

Using ($K_\mathrm{c} = 1.577 \times10^{-5}$ above): (ignoring negative solution for $x$)

$x = 5.608 \times10^{-3}$

$\begin{align}
& [\ce{COCl2}] = 2.00 - 5.608 \times10^{-3} = \pu{1.99 M} \\
& [\ce{CO}] = \pu{5.61 \times10^{-3} M }\\
& [\ce{Cl2}] = \pu{5.61 \times10^{-3} M} \\
\end{align}$

The solutions given here are:

$$\begin{align}
& [\ce{COCl2}] = \pu{1.96 M }\\
& [\ce{CO}] = \pu{0.041 M} \\
& [\ce{Cl2}] = \pu{0.041 M} \\
\end{align}$$

I'm not quite understanding why my answer is different.
My second approach is to determine the partial pressures of each species at equilibrium:

Calculate initial partial pressure using
$\begin{align}
& P_\mathrm{\ce{COCl2}} = \dfrac{n_\mathrm{\ce{COCl2}}RT}{V} \\
& P_\mathrm{\ce{COCl2}} = \dfrac{10.0}{5.00}RT \\
& P_\mathrm{\ce{COCl2}} = \pu{105.2 bar}
\end{align}$
Calculate equilibrium pressures of each species:
$\begin{align}
& P_\mathrm{\ce{COCl2}} = 105.2 - x \\
& P_\mathrm{\ce{CO}} = x \\
& P_\mathrm{\ce{Cl2}} = x \\
& K = \dfrac{P_\mathrm{\ce{CO}}P_\mathrm{\ce{Cl2}}}{P_\mathrm{\ce{COCl2}}} = \dfrac{x^2}{105.2 - x}
\end{align}$

Using ($K = 8.3 \times10^{-4}$ above): (ignoring negative solution for $x$)

$x = 0.294 $

$\begin{align}
& [\ce{COCl2}] = \dfrac{P_\mathrm{\ce{COCl2}}}{RT} = \dfrac{105.2 - 0.294}{RT} = \pu{1.99 M}  \\
& [\ce{CO}] = \dfrac{0.294}{RT} = \pu{5.59 \times10^{-3} M} \\
& [\ce{Cl2}] = \dfrac{0.294}{RT} = \pu{5.59 \times10^{-3} M} \\
\end{align}$

My last approach is to assume that the problem set has given me $K_\mathrm{c}$
I get much closer to the given answer.

Calculate equilibrium concentration of each species:
$\begin{align}
& [\ce{COCl2}] = 2.00 - x \\
& [\ce{CO}] = x \\
& [\ce{Cl2}] = x \\
& K = \dfrac{[\ce{CO}][\ce{Cl2}]}{[\ce{COCl2}]} = \dfrac{x^2}{2.00 - x}
\end{align}$

Using ($K = 8.3 \times10^{-4}$): (ignoring negative solution for $x$)

$x = 0.0403$

$\begin{align}
& [\ce{COCl2}] = 2.00 - 0.0403  = \pu{1.96 M} \\
& [\ce{CO}] = \pu{0.040 M }\\
& [\ce{Cl2}] = \pu{0.040 M }\\
\end{align}$

I'm still off by one significant figure.
Thanks for your help. Sorry in advance for any inconsistent formatting.

First off, you're only off by 0.0002 because that's what it would take to round up. Second, sig figs are a terrible way to do this in general, even though we teach this all the time... Also, I suspect they took the first root (-0.041) and didn't look clearly at the negative sign.

Your last approach is the correct one. Given answer might have approximation problem (see Zhe's comment).


text:
Which reaction is most likely to take place in a voltaic cell?
I was confused by this MCQ:

The answer is A. I thought that the spontaneous reaction was usually between the most negative and least negative. Could someone please explain why this is the answer.

The important thing is to remember that for a formal redox half-reaction, oxidizing H2 to H+,
there is relation of the standard reaction Gibbs energy and  the standard redox potential
$$\Delta_r G^\circ =  -nFE^\circ .$$
The lower is the potential, the stronger(thermodynamically, not necessarily kinetically) is the respective reductant and weaker the oxidant. And vice versa.
So $\ce{Mn(s)}$ is the strongest reductant,  $\ce{Pb(s)}$ the weakest one.
So $\ce{Mn^2+(aq)}$ is the weakest oxidant, $\ce{Pb^2+(aq)}$ the strongest one.
From above must be clear that the only spontaneous reaction is :
$$\ce{Mn(s) + Fe^2+(aq) -> Mn^2+(aq) + Fe(s)}$$

Thank you! However, it says that the answer is A, That's why I'm confused. Is it because this reaction is.not in the options?

Sorry, I have confused metals, but the result is the same.

You can also state that, when comparing two redox couples, the metal belonging to the redox couple with the most negative values will react. Here, comparing the three metals $\ce{Mn, Fe, Pb}$, the metallic $\ce{Mn}$ will always react and produce $\ce{Mn^{2+}}$, because of its highest negative value ($E° = - 1.18$ V) and $\ce{Pb}$ will never react: its redox potential is not negative enough.

One of the easiest ways is to draw the reactions in the direction they occur.  For example, considering:

$$\ce{Mn + Fe++ ->  Mn++  + Fe}$$

$$\ce{Mn++ + 2e-  -> Mn    - 1.18 V Does not want 2 electrons}$$
Can be viewed as:
$$\ce{Mn  -> Mn++ + 2e-                + 1.18 V glad to give them away}$$
$$\ce{Fe++ + 2e- -> Fe.           - 0.45 V happy without electrons, but not as much as Mn}$$
Then just add:     1.18 + (-0.45) = + 0.73 V positive, it will occur spontaneously.  Electrons flow toward Fe++.
Do this with all the others, only A is positive.  The sign (positive or negative) of the sum of the 2 half reactions shows whether or not it is spontaneous.  This is how batteries work.
Most importantly, direction of electron flow is relative to the to metals involved, with most positive to most negative (as written!) producing the highest voltage.

Sign changing of electrode potentials is something which is disliked by electrochemists. Please search previous discussions on this topic. The sign of the electrode potential is invariant.

@AChem I have heard this, but it is a very valuable tool for this specific purpose.  + favorable ... - unfavorable.  Writing out the half reactions in the direction they run and simply summing makes it easy.  Try it!

I often compare it to marking left/right river bank naively (on my left/right) or cartografically (on my left/right when looking downstream.) Similarly, marking cathodes/anodes where reduction/oxidation occurs(Phys chem.) vs  where reduction/oxidation occur when in active galvanic mode (engineers at design of rechargable cells.) // For potentials, it is unlucky they did not established different formal electrochemical quantities. Perhaps using dual-sign redox potential versus single sign electrode potential. Having flipped signs may confuse many beginners, when confusing both.

@Poutnik points very well taken.  Downhill and uphill?  But I am more into mechanism than math, which is why half reactions are done first.  Rote and rhyme failing, Al metal gives, atomic chlorine takes.

Potentials (not limited to electrochemistry) should have IMHO a single sign, as property of a state, not direction of a process. The flipping nature of redox potentials is abusing of general terminology, unfortunately already burned into foundations of electrochemistry.

Use what works, but yes, certainly be aware.  I suppose the convention could be to always write it from the 0 charge state.

Thank you. It's a lot clearer now.


text:
Is milkshake a heterogeneous mixture?
In my textbook, one of the examples of a heterogeneous mixture is milkshake, I don't see how ?
I think the answer depends on the fact that How many phases are there in a milkshake, which I think should be only 1 (if we don't consider any "added stuffs":)

It depends on the scale. Milk is homogeneous when observed with our naked eyes. It becomes heterogeneous under the microscope.

Rule of thumb: If a solution is not transparent, it is most probably a heterogeneous solution such as milk (colloidal). Thus, milkshake is definitely a heterogeneous mixture.

For a consumer, the milkshake is heterogeneous, as there are seen air bubbles and eventually small pieces of fruit. Fresh milk would be homogeneous for him. For a scientist with a microscope, both would be heterogeneous. For across the street observer with a binocular, both would be homogeneous.

Yes, a milkshake is heterogeneous.

milk is heterogeneous all on its own; fat globules in the milk are a different phase than water solutions in the milk.  (Curds vs. whey).

ice cream adds two additional heterogeneities: it contains microscopic ice crystals, and ice is a different phase than both water and fat globules.  It also contains tiny air bubbles, and this is a fourth phase.

Homogenity/heterogenity is dependent on evaluation conditions.
A mixtures is heterogenous if both these conditions are true:

Local variability of mixture properties are above the threshold of

property value resolution AND
mixture application requirement.

Spacial granularity of the above variability is above the threshold of

spacial resolution AND
mixture application requirement.

Otherwise, a mixture is homogenous.
If values vary less than we notice or care
OR if values vary within smaller distance than we notice or care
THEN the mixture is homogenous.
Every mixture can be considered homogenous or heterogenous at certain ( sometime extreme) conditions.

If there are floating evenly distributed potatoes in water, observed by a telescope with resolution $\pu{1 m}$, it is homogenous mixture.
If there is an old favourite red T-shirt, a "16-Windows(3.x)-colors man" may say it is still homogenously colored, even if his wife disagrees.

Milk is a colloid dispersion, i.e. a

"substance consisting of microscopically dispersed insoluble particles is suspended throughout another substance."

(source: English Wikipedia)
As such, visible light is scattered when it passes this medium; equally known as Tyndall effect.  In case you get your hands on a bright light source, you then can monitor the optical path across this medium.  It needn't be a laser as in this video recording about a very dilute milk:

(screen photo of the mentioned video, 1:28 min into the recording)
Addition after a comment:
Replication with «true milk» requires a brilliant light source passing an aperture narrow enough.  On the other hand, because of the high concentration of particles with a dimension about light's wavelength, scattering occurs so frequently that the beam fuzzes out rapidly:  (For these two transmission mode observations, equally note color of the milk and glare (and apparent color) of the glass in the second run with milk which is less prominent for the experiment with tap water in the experiment.)

Setup: daylight-white light of a LED torch (focus to infinity) passes a flat bottom glass, perpendicular to the surface (aperture: $\pu{3 mm}$ diameter, thickness glass wall: $\approx \pu{2 mm}$).  Path length $\approx \pu{10 mm}$ for either water (first experiment), or fresh milk (second experiment; 1.5% fat, purchased less than $\pu{12 h}$ earlier, which passed both Pasteurization and homogenization.  Homogeneous hereby is not about a discern between solution (e.g., $\ce{NaCl}$ dissolved in water), or collide (e.g., fat dispersed in milk), but refers to a narrowed (i.e., adjusted) particle size distribution of the fat globules suspended in milk after passing homogenization for which Wikipedia reports

an average diameter of two to four micrometers [of unhomogenized cow's milk] and with homogenization, [an] average [of] around 0.4 micrometers.

($\pu{1 µm} = 1 \times \pu{10^-6 m}$, and $\pu{1 nm} = 1 \times \pu{10^-9 m}$; so $\pu{0.4 µm} \approx \pu{400 nm}$. For comparison: average diameter of a human hair $\approx \pu{100 µm}$ (ref), wavelength of visible light $\approx 250\ldots\pu{700 nm}$.)

Hmm, but if the Tyndall effect occurs in the diluted milk the same observed way in every place, with liquid not manifesting spatial variations of it ... we have homogeneous liquid in Tyndall effect measurement context. If it varied e.g. due fat separation, it would be in the given context heterogenous.

@Poutnik If there is no agreement / convention / definition about homogenous / heterogenous (which may vary over time, perspective / community) the same word may refer to different qualities.  This is why your comment prompted an experiment to extend the answer.

There are procedural conventions in some particular cases, but generally, homogenity as having uniform properties depends on how thoroughly and zoomed in/out we look. Evaluating dough, a baker and a food scientist would have very different criteria for homogenity.


text:
Calculating rate constant for reaction between potassium permanganate and oxalic acid
I'm writing a lab report investigating the effects of temperature on the rate constant in the reaction between potassium permanganate and oxalic acid. For my practical I've measured the time of colour change (purple to colourless).
I've confirmed that the rate expression is: $$\mathrm{rate}=k[\ce{KMnO4}][\ce{H2C2O4}]$$
However, I am unsure about how to calculate the rate constant for the given reaction. I have rate as:
$$\mathrm{rate}=\frac{[\ce{MnO4^-}]}{t}$$
where the $\ce{MnO4^-}$ ion is disappearing. However, calculating this and using the rate expression only gives me an average value. I understand that I would use the integrated rate laws if the reaction was second-order with one reactant, but here there are two reactants so I am unsure what to do.
Or am I just overthinking this?

Note that measuring the rate constant of this reaction is very tricky as the reaction is autocatalyzed by the forming Mn^2+ ions.

If a reaction is second order in $\ce{A}$ and $\ce{B}$, with $\ce{[A]_0 ≠ [B]_0}$, and with a rate law given by $\pu{d[A]/dt = d[B]/dt = k[A][B]}$, the integrated law is : $$\ce{\frac{1}{[B]_0 - [A]_0}ln{ {\frac{[A]_0 [B]}{[A][B]_0}}}} = kt$$
As it is pretty long to establish this law, I recommend that you open a book on chemical kinetics like A. A. Frost and R. G. Pearson, Kinetics and Mechanism, John Wiley & Sons, New York, 4th ed. 1965, p. 16.


text:
Separating alkane/alkene isomers on the GC but the isomers have the same retention time
I'm trying to differentiate between n-octane, 3-octene, and 4-octene from a mixed sample using gas chromatography, but n-octane and 3-octene seem to have the same signal and 4-octene is shouldering the n/3 peak. Does anyone have suggestions for a way to separate the peaks? I've currently been trying to change the temperature method but the ones I'm trying seem to not be working. Thanks!

You may need different column, e.g. with more of aromatic character, better distinguishing existence or location of double bonds.

You can vary the GC column (capillary/packed; material of coating/packing; length of the column); the flow rate ($\ce{N2}$ vs. $\ce{H2}$ as carrier), the temperature of operation as fix (and the level as such can be altered), or as temperature gradient.  Similar to liquid chromatography, programs may offer some guide of selection (e.g., EZGC chromatogram modeler).

For our organic chemistry lab program we have developed GC program for one of the separation labs and we use alkane homologues as the analyte mixture to be separated. The chromatogram give a baseline separation on shorter (3-m column) Carbowax-20M packed column. Thus, for curiosity, I checked for separation of alkene isomers on similar column. Sure enough, I found one separating $\ce{n-C16H32}$ alkene isomers:

The column is $\pu{300 m}$ Garbowax-2OM capillary column operating at $\pu{96 ^\circ C}$ with $\pu{0.15 MPa}$ $\ce{N2}$ mobile phase (Ref.1). As indicated in the diagram, the method has even separated most of $cis/trans$ isomers at same position.

References:

Ladislav Soják, Ján Krupčík, and Jaroslav Janák, "Gas chromatography of all $\ce{C15}$-$\ce{C18}$ linear alkenes on capillary columns with very high resolution power," Journal of Chromatography A 1980, 195(1), 43-64 (ODI: https://doi.org/10.1016/S0021-9673(00)81542-3).


text:
As per Lewis structure, can oxygen molecules in ozone be interconnected with one another through covalent single bonds?
Ozone has three oxygen molecules. Since all three are the same atom, there is no difference in electronegativity. As per theory, formation of a polar covalent bond is not possible but there are two resonance equivalent structures with one single and one covalent bond.

As I searched in the net, those are the only two structures repeated often. Also whenever there are more than two Lewis structures, the structure which is valid as per VSEPR theory will be taken are my understanding. So, technically even though the three interconnected single bonds are valid as in this diagram,, why doesn't the structure hold good? Is it because there are more number of close lone pair of electrons which will cause repulsion as per VSEPR theory? But, still how does the polar bond form when there is no difference in electronegativity between the oxygen atoms? I am asking this because, I am not able to get validate this question with third structure in the google search.
I am adding one more detail which I found regarding this, called 'formal charge':

The formal charge of an atom in a molecule is the hypothetical charge
the atom would have if we could redistribute the electrons in the
bonds evenly between the atoms. Another way of saying this is that
formal charge results when we take the number of valence electrons of
a neutral atom, subtract the nonbonding electrons, and then subtract
the number of bonds connected to that atom in the Lewis structure.

Thus, we calculate formal charge as follows: formal charge = # valence shell electrons (free atom) − lone pair electrons −1/2*(  bonding electrons)
I calculated the formal charge for O1, O2, O3 and all three resulted in +1. Sum of the formal charges didn't give net charge of zero. I guess, that could be one of the explanations. If anyone can confirm, it will be helpful. I will meanwhile also search. Thanks for the help!

For options of formatting of the posted plain text, see as inspiration SE - help - formatting.

The cyclic structure with three single bonds is good, as far as Lewis structures go. It is just that the real molecule does not follow it.

https://chemistry.stackexchange.com/questions/22290/can-ozone-have-a-triangular-structure

also https://chemistry.stackexchange.com/questions/128554/lewis-structure-of-o3 https://chemistry.stackexchange.com/questions/51719/do-cyclic-oxygen-molecules-exist-and-their-theoretical-stability

I have seen that the question is closed and a link has been provided. However, I found recently VSEPR theory fails for isolelectronic species, which are elements, ions and molecules that share same number of electrons. Also VSEPR theory fails where central atom is a transition metal. It does not apply to ionic substances which are bonded by charge instead of electron pair sharing. This could be useful to someone.

Because reality, not theory, dominates chemistry
Ozone, by the way, has three oxygen atoms not molecules.
But while many theoretical structures are possible and look OK in different bonding theories, that isn't how chemists work out the real structure. Most bonding theories are too weak to make accurate structural predictions for many "difficult" structures.
When it was realised that benzene, for example, was a ring some proposed it was essentially the same as cyclohexatriene, with alternating double and single bonds. But observations of the crystal structure showed that, despite simple bonding theories ideas, the bond lengths were all equal. Better theories were developed to explain the actual structure.
Ozone, in some theories, could be a 3 membered ring or, indeed, a variety of other structures. But it isn't. Microwave spectroscopy show it to be a bent molecule as does the fact is has a dipole moment. That's what is is, theory be damned. The central angle is about 116° and the central oxygen has a net positive charge withe the terminal oxygens having net negative charges but bond lengths between those of a single and double bond.
There are theories that can account for this, but the predictions of simpler theories are useless. What we observe the structure to be is far more important than theory.

Thanks, then Lewis structure is a simple theory ? Are there more complicated theories that could account for all molecule formation with better accuracy than Lewis structure? Should there be any one grand theory in place, as I read there are more elements that are not in the periodic table ? Please correct me, if I am wrong


text:
Cause and Prevention of Palladium Mirror in Stille Coupling
I am conducting some monomer and polymer syntheses using Stille coupling or other palladium-based coupling. I usually use the combination of $\ce{Pd_2(dba)_3}$ and $\ce{P(o-tol)_3}$ in Stille coupling, but sometimes use other ligand when I use other palladium-based coupling such as Buchwald-Hartwig coupling. I usually observe palladium mirror precipitated on the wall of the reaction flask and this led to unsatisfying yield or molecular weight of polymer. I have thought that this palladium mirror precipitation is due to the oxidation by oxygen, so I attributed it to the insufficient inert atmosphere. But recently, I tried some other polymerization using Stille coupling, and there was no palladium mirror at all after 48 hours, even I didn't flow Ar gas! (We don't have Schlenk line, so we alternatively flow Ar gas continuously into the reaction flask during the reaction) The reaction system only had the adaptor packed with silica gel to prevent moisture. So I rethink: Can the palladium mirror be formed because the reagents are not favourable to the Stille coupling, not the oxidation by oxygen? And in this case, does adding extra Pd catalyst and ligand help when I observe the palladium mirror during reaction?


text:
Why is the local coupling constant (vibronic coupling) given in eV?
Should the local coupling constant g not be without a unit, as the Holstein Hamiltonian suggests?
$
H=\underbrace{\sum_{P, Q} h_{P Q}^{(0)} a_P^{\dagger} a_Q}_{\text{Fixed geometry}}+\underbrace{\sum_{P \in \Omega} \sum_\lambda \hbar \omega_\lambda g_{P P}^\lambda\left[b_\lambda^{\dagger}+b_\lambda\right] a_P^{\dagger} a_P}_{\text{Atom displacement}}+\underbrace{\sum_\lambda \hbar \omega_\lambda\left(b_\lambda^{\dagger} b_\lambda+\frac{1}{2}\right)}_{\text{Vibrations}}$
$H$  = Holstein-Peierls Hamiltonian  .$P$ or $Q$ = electronic levels
λ = normal mode   g = coupling constant   h(0) = transfer integral
a† and a =  creation and annihilation operator for fermions
b† and b = creation and annihilation operator for bosons   ℏω = energy of phonon
To give more context, in Ref. 1 for example, the local coupling constant is given in eV.
References
(1) Shizu, K.; Sato, T.; Tanaka, K. Vibronic Coupling Density Analysis for α-Oligothiophene Cations: A New Insight for Polaronic Defects. Chemical Physics 2010, 369 (2–3), 108–121. https://doi.org/10.1016/j.chemphys.2010.03.014.

The local coupling constant is typically given in units of energy, such as electronvolts (eV), because it represents the strength of the coupling between the electronic and vibrational degrees of freedom in a system (which is thus an 'energy-related' quantity). The Holstein Hamiltonian, which is used to describe the interaction between electrons and phonons in a system, includes the atom displacement term, which represents the coupling between the electronic state $P$ and the vibrational state $\lambda$, with the coupling constant $g^\lambda$ being the coefficient in front of this term for unit equality in the equation.

Yes, the coupling should have units of energy, but a dimensional analysis of the Hamiltonian presented above tells me that g lambda itself should be dimensionless - the units of energy  are provided by hbar omega in front of it. Is the language just lax here, and the quantity in eV really the product hbar * omega * g ?

I believe g lambda should be dimensionless as it's a ratio between the strength of the coupling between the electronic and vibrational degrees of freedom to the energy of the vibrational mode. It's often reported in units of energy as well as the product, for consistency with the vibrational energy scale.

OK - but could you say mathematically exactly what is being reported in units of energy?


text:
Electrolysis of Solutions
So far I've only encountered formulas to solve for solutions of single molecules while learning about electrolysis, i.e. H2O or FeCl3. This generally has involved determining the resultants, identifying whether they are attracted to the anode or cathode, and performing some multiplications.
What approach is used for analyzing complex solutions? For example, an everyday soda solution may contain water (H20), fructose (C6H12O6), phosphoric acid (H3PO4), and carbon dioxide (CO2). I'm not sure how to approach the resultants. It would seem the decomposition potential should for each molecule should first be considered to identify which molecules are likely to decompose, but given so many elements how can we determine what resultants will remain? It would seem C6H12O6 alone could result in 6(C) + 6(H2O) or in 6(C) + 6(H2) + 3(O2), for example.

I'm not quite sure what you mean by "formulas". To know what the result of performing electrolysis on any solution would result to one would need to consider all the different electrochemical reactions and their different standard reduction potentials, their kinetics etc... It's quite complex. Performing electrolysis does not simply lead to a separation in the different elements. Consider for example nitric acid in water, given a high enough applied potential one would form ammonia (most reduced nitrogen form) and oxygen.

@Noah perhaps "formula" is poor word choice on my part - the lectures I've watched thus far only deal with single molecules which have very basic decompositions. I'd like to learn how to approach more complex problems. Is there a standard set of steps to apply, perhaps iteratively?

Let's say you have a mixture with a certain set of chemicals, a first step is to look up any electrochemical reactions these chemicals might undergo. Next comes the hard part, the rate of an electrochemical reaction is very much dependent on the electrodes used, the concentration and the potential at the electrodes, you'd need to know all the kinetic data, the mass transport in your system and the conductivity. Let's say you use copper wire, you might make a whole range of stuff from CO2; methane, CO, ethene, alcohols... you might make all kinds of oxidized things from fructose.

In general, there's no "straightforward" solution. To understand that it's not trivial you might need to learn a bit more about electrochemistry, but definitely look up these things if you are interested!

Fructose and $\ce{CO2}$ are covalent solute molecules. They are not modified by electrolysis when dissolved in a solution of phosphoric acid. Fructose, for example, does not produce ions in solution : it will never be decomposed (by electrolysis) partially into fragments, or totally into $\ce{C, H2}$ and $\ce{O2}$. Water is a solvent, made of covalent molecules. It can be electrolyzed.

I'd like to learn how to approach more complex problems. Is there a
standard set of steps to apply, perhaps iteratively?

Every electrolyte has its own decomposition potential and a stability window. One can carefully design mixtures and study their electrolytic behaviour, for example, what happens if we electrolyze an aqueos solution of transition metal salts. There are no standard set of steps or iterations, although for simple ionic cases, people look up the electrode potential table to predict to products of electrolysis. For example, in higher classes one might study the electrolysis of KCl and KI in water. What products do we expect under standard conditions? These can be answered systematically. Electrolysis products of random mixtures like soda are difficult to predict, because once you have organic compounds, they start reacting at the electrodes via radical reactions.
For the conceptualization of electrolysis, take binary mixtures which represent soda such as a very dilute phosphoric acid and citric acid solution. If you electrolyze this mixture at the decomposition potential of water, you will the see expected products, hydrogen and oxygen.
Let us say you have water and tons of fructose in it (this is why sodas are bad!), it will almost be a non-conductor of electricity. Simple underivatized sugars are not good electrolytes. On the other hand, carefully controlled (pH adjusted) solutions of sugars can be oxidized. Search electrolysis of carbohydrates.
Again think about the electrolysis carbon dioxide saturated water; this solution will also be a poor conductor of electricity. Still since we will have plenty of water molecules, and using the right decomposition potential, hydrogen and oxygen will be the main electrolysis products.
If you are still curious, search voltammetry of mixtures.


text:
How can I measure the energy contained within water vapor?
Let’s say I have a kilogram of water vapor at $\pu{423 K}$. Can I directly measure the energy, contained as heat, within the vapor? All the laws I know depend on knowing the changes that happened that led to forming this vapor at that temperature.

The only option I see is having a device, measuring vapor temperature and pressure (
or assuming a pressure), having a given volume, computing vapor energy on computer or dedicated smart device.

The temperature is not enough. You may need pressure and the volume of container, at least.

If I have the pressure and volume of the container, how can I go on and find the energy within the vapor?

All the laws I know depend on knowing the changes that happened that led to forming this vapor at that temperature. - by applying them. Heating up water + boiling up water + isobarically heating vapor + isothermally compressing vapor.

Oh. I thought you meant measuring instantaneous values. So, it all comes back to taking the process step-by-step. Thanks a lot

Just for fun, how would you imagine measuring instantaneous value? You could let absorb controlled amount of vapor in water, measure the warming up and calculate the passed heat by the way of calorimetry. Impractical and potentially dangerous. OTOH, One can create an empirical a formula or a chart, where reading vapor state variables would point to vapor thermal energy.

You should describe the scenario, purpose, context and background in more details. Is it theoretical or practical task? For the latter, what is available?

The absolute values of internal energy $U$ or enthalpy $H$ (which could be more useful for water vapour) at a single state point are meaningless. It is only the difference between two different state points that matter. Thus, the value for a single state point can be set to any arbitrary value. Therefore, this arbitrary meaningless value cannot be directly measured. Many handbooks set the arbitrary state point so that the values of these properties are positive for most liquid or gas states. For water, the triple point is selected as the reference state, where the internal energy $U$ of saturated liquid is assigned a zero value.
That's why all the laws you know depend on knowing the changes that happened that led to forming this vapour at that temperature. Heating a mass of $m=1\ \mathrm{kg}$ of water from the triple point  with $U=0$ to a new state with $T=423\ \mathrm K$ and $p=10^5\ \mathrm{Pa}$ would give you $U=2582.7\ \mathrm{kJ}$ and $H=2776.3\ \mathrm{kJ}$.
Furthermore, you want to directly measure the energy that is contained as heat. You should remember that internal energy $U$ is a state quantity or state function because it describes an equilibrium state of a thermodynamic system, regardless of how the system has arrived in that state. For a closed thermodynamic system,
$$\Delta U=Q+W$$ where is $Q$ amount of heat transferred to the system and $W$ is work done on the system (according to the sign convention used by ISO and IUPAC) provided that no chemical reactions occur. Thus, simply by looking at the value of $U$, you cannot know how much was supplied by heat. You also cannot know how much heat can be extracted from it. Again all the laws depend on knowing the changes.
However, since the internal energy $U$ describes an equilibrium state of the thermodynamic system, regardless of how the system has arrived in that state, you do not need to know the exact path taken from the reference state to the observed state; you just need to assume a possible path. So instead of measuring $U$ directly, you can measure the values for a set of thermodynamic parameters (like mass $m$, temperature $T$, and pressure $p$ in the example above) that describe the state of the thermodynamic system. Then you can calculate the change in internal energy $\Delta U$ for a system that changes from the reference state to the new state with the observed parameter values assuming any possible path.
There are precalculated values (so-called steam tables) that already did this job for you. So you can simply measure the values for a set of thermodynamic parameters and then look up or interpolate the corresponding value for the internal energy $U$.


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Can lipases cleave fatty acids linked with amides instead of the usual esters?
The typical substrate for lipases are esters, such as triglycerides where fatty acids are linked to the glyerol through ester linkages. I'm interested in whether lipases (such as pancreatic lipase) can cleave fatty acids linked to proteins by amide linkages. In lipopeptides, fatty acids are often conjugated to peptides at the N-terminal amine or lysine side chain, yielding an amide (also known as a peptide) linkage.
I have found some examples where some lipases can be used to catalyze the opposite reaction (forming an amide between a fatty acid and an amine of a polypeptide). So it seems like such cleavage might be possible. Does any one know of work demonstrating cleavage of amide bonds between fatty acids and amines?

Yes they can, but they have higher specificity for esters. Here is part of the abstract of a paper using directed evolution to increase the activity:

A double mutant F207S/A213D gave the highest molecular activity of 1.1 min−1 for the amide, corresponding to a 2-fold increase compared with that of the wild-type lipase. A structural model of the lipase indicated that the mutations occurred at the sites near the surface and remote from the catalytic triad, but close to the calcium binding site. This study is a first step towards understanding why lipases do not hydrolyze amides despite the similarities to serine proteases in the active site structure and the reaction mechanism and towards the preparation of a general acyl transfer catalyst for the biotransformation of amides.

Note that the abstract is a bit self-contradictory, first saying that they increased the activity two-fold, then saying lipases have no amide hydrolysis activity. It is possible that a turnover of $\pu{1.1 min-1}$ is considered not active, considering the much faster turnover for the lipase reaction. The introduction, which has lots of references, clarifies this.
I found this in a search for "lipase specificity amide" on google scholar. I did not look for the specific substrate you had in mind because that feels like giving out fish instead of hints on fishing.


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Which one is more Bronsted basic?

pyrrole anion vs imidazole anion
I at first thought it was the right one because it has 2 lone pairs so I would assume it would have more affinity towards hydrogen protons. However, the resonance forms of the imidazole anions place a negative charge on both nitrogens, so the resonance hybrid would have a partial negative charge on both nitrogens. While for the pyrrole anion, the negative charge is localized on one nitrogen, making me think that it is less stable than the imidazole anion and more basic. Any thoughts?

Yours thoughts seem to be correct.  Imidazole pKa is 14.5 and 6.8.  Pyrrole pKa is 16.5.  You do have a resonance form for the imidazole ring.  Additionally, nitrogen is electron withdrawing.  The localized negative charge on deprotonated pyrrole will be more basic.
Both nitrogens in imidazole can be protonated, hence 2 pKa values.


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Why don’t Grignard reagants react with their alcohol products?
I’ve read that Grignard reagants react pretty quickly with alcohols and carboxylic acids to form their respective conjugate bases. I’ve also learned that Grignards can be used for the prepararion of alcohols and carboxylic acids too. How is it possible to perform these reactions if the reagant would also react with the end product?

The alcohol is the last product after acidifing. There is no acidic hydrogen is present before acidifing.

The addition of a Grignard to a carbonyl doesn't form an alcohol—not directly at least. It forms an alkoxide ion (i.e. the conjugate base of an alcohol).
Only when that reaction is complete, is water (or some other acid, ammonium chloride is used quite regularly) added to the solution. The water protonates the alkoxide ion to form the alcohol, and also destroys any remaining Grignard. This step is usually referred to as 'protic workup' or just 'workup'.
So the Grignard never sees any alcohol.
The same is true of the carboxylic acid formation (assume you're referring to Grignard addition to carbon dioxide). It doesn't form a carboxylic acid straight away, it forms a carboxylate ion.

Ah, but then why is it that alcohols on the Grignard’s reagant themselves DON’T exist in this alkoxide form instead? I was told that Grignard’s reagant will, under normal conditions, react with an alcohol and this is the reason we use protection with TMS.

I'm not entirely sure what you mean. You mean why you can't have a hydroxyl group on a Grignard reagent? You can, it would just be very wasteful, as you'd be using up half of your Grignard reagent to deprotonate the hydroxyl group on the other half first.


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Can an oxidising/reducing agent oxidise/reduce itself?
I am new to the study of this divine science. So Just a query: Can an oxidising/reducing agent oxidise/reduce itself? If it can can anyone give an example and explain it?

You were told already in your prior question.

A simple example can be the disproportionation of chlorine when bubbling a current of chlorine gas $\ce{Cl2}$  into a solution of $\ce{NaOH}$ (containing $\ce{OH^-}$ ions):
$$\ce{Cl2 + 2 OH^- -> ClO^- + Cl^- + H2O}$$
Here the chlorine atoms have the oxidation number equal to zero in the reactants, and get the final oxidation number $-1$ in $\ce{Cl^-}$ and $+1$ in $\ce{ClO^-}$.  Here one of the $2$ original $\ce{Cl}$ atoms oxidises the other one from $0$ to $+1$. In this operation it is reduced fro $0$ to $-1$.
The final solution is the famous bleach, used for cleaning dirty surfaces.

My answer has been downvoted. Why ? It is a perfect answer to the question as written in the title.


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If we say that neon , krypton, aryon and oganesson are stable beacuse they follow octate rule, then why are xenon and radon are stable?
Neon is $\ce{K^2 L^8}$,
Argon is $\ce{K^2 L^8 M^8}$,
Krypton is $\ce{K^2 L^8 M^{18} N^8}$,
But Xenon is $\ce{K^2 L^8 M^{18} N^{26}}$ and radon $\ce{K^2 L^8 M^18 N^{32} O^{26}}$
They don't follow octate rule but they are stable. How? And....
Erbium is $\ce{K^2 L^8 M^{18} N^{32} O^8}$
It follows Octate rule but not stable. Why?

Becaust the octet rule is a simple rule of thumb, that works for many common elements and compounds, but not (strictly) for all of them. Why are you excluding Helium from your list of noble gases? Exactly.

Please check your orthography and grammar.

Strictly speaking, the octet rule works well for the ten first elements. In the series Na - Cl, there are already exceptions to the rule, like sulfur in covalent $\ce{SF6}$

Some of your electron configurations are wrong.
Xenon has 18 $N$ electrons ($s+p+d$ sublevels full) and eight $O$ electrons ($s+p$ sublevels full). Radon has 18 $O$ and 8 $P$. Erbium does not have all the $f$ subleblvels full and ends up with $N^{30}O^8P^2$.
You may want to look up the Aufbau Principle. It's imperfect but more accurate than the shell-by-shell ordering you assume. Basically, when you have more than one electron the repulsions between electrons raises the energy of sublevels that have higher angular momentum and don't get close to the nucleus, and then they overlap with lower angular momentum orbitals of what you thought were higher shells.


text:
Chromatography and Height Equivalent to a Theoretical Plate (HETP)
HETP can be expressed in two ways:

HETP = L/N
HETP = A + B/u + Cu

where:
L: lenght of the coloumn
N: Number of theoretical plates
A: Eddy-diffusion parameter, related to channeling through a non-ideal packing
B: diffusion coefficient of the eluting particles in the longitudinal direction, resulting in dispersion
C: Resistance to mass transfer coefficient of the analyte between mobile and stationary phase
u: flux speed
regarding equation 1), N is not constant for a column, but depends on the peak. Consequently, HETP should vary depending on the chromatographic peak considered and not constant for a given column.
Concerning equation 2), instead, the parameters present are column-fixed, therefore also HETP should be a column-fixed constant.
So, is HETP constant, given a column, or does It vary with the peak being considered, as N?

You are correct, $N$ is dependent on a peak, early eluting peaks are narrow and later eluting peaks are wider in an isocratic separation. The equation $H=L/N$ is derived from distillation theory, so the height equivalent for a particular component is a constant for a fixed flow rate, at a given temperature, pressure, mobile phase, stationary phase, analyte, column packing, and particle diameter.
The second equation is called the van Deemter equation (from 1951!!). As one can readily see, it shows the flow rate dependence of $H$, because $H$ is a function of linear velocity $u$. It is highly dependent on the type of analyte being investigated.
In short, neither efficiency nor HETP are fixed for a given HPLC or GC column. You have to specify all the conditions when quoting these values.
P.S., the concept of efficiency, HETP equation, and the Van Deemter equation are not valid for gradient separations.

so H is a parameter that is valid only for a specific analyte, set specific conditions in the column?
That is, is it not possible to associate the SAME value of H with ALL the peaks that are obtained or with the whole column?

Yes, you are right. All peaks will have different H values dependeing on their time of elution.

Great! Thank you.


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What determines strength of nucleophile--Polarisability or Electron density
My understanding is that nucleophiles are lewis bases. Apparently nucleophile and base strength are not the same though. I thought Nu strength/Base strength is determined by how well it can donate an e- pair which is thus affected by the electron charge density on the e- donating atom.
However I've come across weak bases but strong nucleophiles Eg: Br-. It's a weak base cause it is large and has low electron density. However Br- is a strong nucleophile and this was attributed to its polarisability.
However I thought the issue of polarisability was more of deciding the hardness of the lewis acids/bases and am thus very confused as to what really determines the strength of a nucleophile/base and how polarisability comes into play.

The most correct answer would be that the bromide ion is both a strong nucleophile (since the electrophiles in reactions involving bromide are usually soft) and both a weak base in aqueous solution(since now the electrophile is the proton, which is too hard to be friends with the soft bromide).
(For the definition of "soft" and "hard" see e.g. this Wikipedia page)

Nucleophilicity is a kinetic concept, and differs from basicity which is a thermodynamic concept.
There are some molecules which are both strong bases, like the ethoxide anion, and hence some factors, like having a negative or delta negative charge will contribute to a molecule being a strong base and nucleophile.
However, due to the mechanism of nucleophilic subsitiution via an SN2 reaction, sterically hindered molecules will not be strong nucleophiles. A good example is the text-butoxide anion- strong sterically hindered base, but poor nucleophile.
Likewise, some factors, such as conjugate acid stability will be more important for promoting basicity than nucleophilicity.
Back to your example, Bromide ions are a weak base, because although they have a negative charge, they are fairly stable, but they are strong nucleophiles due to the lack of steric hindrance.
A chloride ion, for example would be both a stronger base and nucleophile. A stronger base because the Kpa of HCl is lower than that of HBr and so the Cl- ion is stable than the Br- ion, and a stronger nucleophile as the Cl- ion is slightly smaller and so less sterically hindered.


text:
How it was decided that the freezing point of water in the Kelvin scale of temperature should be 273.15 K?
While discussing about the different temperature scales, our teacher told us about the freezing point and boiling point of water in different scales. I asked my teacher that how these numbers were decided, like why it was decided that 273.15 should be the freezing point of water in Kelvin scale and not some other number, but could not get a satisfactory answer. So, here I am requesting if anyone could tell me the reasons for giving the different specific numbers in different scales for the same temperature, in the simplest way possible.

Not a decision. A measurement.

Have a closer look at the  Wikipedia page: https://en.wikipedia.org/wiki/Kelvin

Try to learn the habit to make at least the basic research about the topic before asking about it. As asking should be rather the last than first thing coming to your mind.

@Martin-マーチン I have already gone through it, but still posted the question so that someone could give a simple to-the-point answer; but thanks! I guess, I should go through it again!

Freezing point of water is at 273.15 K, because absolute zero is at -273.15 °C and absolute scale is just Celsius scale shifted by 273.15 degrees.

History: the Celsius scale was defined first; the zero on the Kelvin scale is dependent on that definition
It may seem awkward and annoying that the zero point on the Kelvin scale (and therefore the freezing point of water) isn't an exact number, but it is a consequence of history.
There used to be a variety of different temperature scale in the early history of chemistry. The Fahrenheit scale annoyingly used by a small number of countries to confuse science students, is one legacy of that era. And this matters because of how the scales were defined.
The original definition of the celsius scale was based on the freezing point and boiling point of pure water under very carefully defined conditions. The difference between the two temperatures was divided into 100 for convenience (it is certainly more convenient than the Fahrenheit scale which divides that interval into 180 parts though the original definition of the endpoints was more complicated and convoluted). The Kelvin scale came later but uses exactly the same interval as the celsius scale. The bp was set at 100 celsius and the freezing point at 0 celsius.
But this definition occured long before we knew there was an absolute lower limit for temperature. And it was based on empirical observations. When chemists and physicists got to grips with thermodynamics and started to explore really low temperatures, it was too late to define a new scale with different definitions. So when experiments revealed that there was a lower limit, absolute zero, they could only measure what that number was on the celsius scale as it made little sense to devise a new scale and mess with all the existing definitions.
Hence the awkward value of absolute zero and the melting point of water. The Kelvin scale, for convenience, simply adds the empirical value of absolute zero to the celsius temperature, which minimises any need to redefine anything else and therefore all normal temperature measuring devices.
So the freezing point of water was not decided to be 273.15, it was decided that it was 0 in celsius and, when absolute zero was measured and the new Kelvin scale set, that demanded that the freezing point of water had that value in the new scale.


text:
Direction of electron flow in X-ray Tube in X-ray Crystallography
In X-ray crystallography, X-ray tube may be used to generate X-ray. The set-up is the following:

Here it shows the electron beam goes from cathode to anode. If anode is the negative electrode, why will electron travel towards it? Why don't electrons travel from anode to cathode?
Thanks!

Why do you think the anode is the negative electrode?

A negative particle goes to the anode.

@AChem isn't by definition the anode is the negative electrode?

@YushiLi, No. You might be confusing the electrochemical terminology. For a DC source, in evacuated glass tubes, anode is the electrode which is connected to the positive terminal of the DC source. So, here in the X-ray tube, anode is the positive electrode.

In context of electrochemistry, anodes are electrodes where oxidation occurs, positive in the passive, electrolytic mode, negative in active, galvanic mode. In context of physics, circuits and vacuum bulbs, anodes are positive electrodes.

Ah I see. Thank you both!

For one, the question potentially displays confusion about electric current as either flow of positive charge (and by convention, the direction of electric current among e.g., electricians from $+ \rightarrow{} -$), or flow of electrons.  See e.g., Wikipedia's article about this:

For two, the anode of X-ray tubes not only is where the electrons travel to, but often literally is called the target they hit, because its composition has a great influence on the brightness, brilliance, the position of the characteristic emissions (in first approximation e.g., $\ce{Mo} K\alpha \approx \pu{0.71 Å}$, $\ce{Cu} K \alpha \approx \pu{1.54 Å}$), and underlying bremsstrahlung of such a source.  As they carry much energy to a small spot, largely released as heat, you need good cooling (water on the backside of the anode, the Coolidge X-ray tube), use a rotating disc as target (a frequently used design in protein crystallography), or switch to a liquid jet altogether (one form of a  microfocus source).

(image credit English Wikipedia, X-ray tube).

In case the next question is where the X-ray photons travel to: the spatial distribution of their paths is not uniform (an earlier question).

Thanks for the details. What I don't understand is why electrons flow towards the anode. Like your image from Wikipedia shows, the electron is flowing from negative electrode (anode) towards the positive electrode (cathode). Why here in the X-ray tube, it can be the opposite? Don't like charge repel? Thank you.

With the battery/alimenting direct current in their back, a cathode is the negative electrode and gives the electrons away (like a catapult). All while an anode accepts incoming electrons and is positive (relative to the cathode).


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Effect of catalytic NaOH on atom economy
I conducted a base catalyzed aldol condensation using NaOH. Now I'm supposed to calculate the atom economy, AE:
$$\text{AE} = \frac{{\text{MW of desired product}}}
{\sum{\text{MW of reactants}}} \times 100\%$$
but I'm unsure on how to interpret the role of NaOH: It reacts with the reagent by deprotonation (base), but it also gets regenerated (catalyst) and is not reused (waste).
Since the AE is also a way of depicting waste generated during a reaction my approach would be to include it as a reactant although it has a catalytic role which as I learned is not considered in AE?
What would be the correct way?

You may consult e.g., the Wikipedia article about the topic, and use the literature references it cites as an entry.
Though the very first example of Sheldon's paper is about an acid catalyzed reaction, you can draw a conclusion by analogy:

Both reactions intend to oxidize the alcohol to yield the ketone. In the first instance, the the acid is used in stoichiometric amounts and if you are not able to recycle / reuse the acid, it is waste; low atom efficiency. In the second case, you have a solid catalyst which after the reaction is separated, filtered-off and can be reused for the next batch of reaction; this improves atom efficiency.
So, how do you run your aldolization?  Do you run it e.g., by phase transfer catalysis, can recover the base for the next batch of synthesis?  Because you say no, so you count it fully into MW of reactants.
Sheldon, R. A. Atom efficiency and catalysis in organic synthesis. Pure and Applied Chemistry 72, 2000, 1233–1246; https://doi.org/10.1351/pac200072071233 (open access).


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Which gas at 15 °C and 1 atmosphere of pressure has the highest refractive index in visible light?
For my own curiosity's sake, I am attempting to find the gas that at 15 °C and 1 atmosphere of pressure in pure form has the highest refractive index on average and overall, for visible light. Assume visible light means having a wavelength in the range of 380 to 750 nanometers.
Please keep in mind that I am not looking for the densest gas whether active or inert. Density does not necessarily correlate with a materials refractive index.
If for some reason volume needs to be specified, please assume a 1x1x1 kilometer region.
I am being very specific in what I am seeking and have specified wavelengths, pressure, temperature, purity, and even volume.
If I must for whatever reason, I must be any more specific then please assume that the gravity is the same as on Earth at sea level on the equator though I do not actually require that level of precision.
Sea level conditions are commonly used as a common standard for predicting the behaviors of different materials here on Earth and a common standard of that nature was exactly what I needed.
The reasons for the described conditions specificity is to provide a decent set of common standard conditions for measurement and avoid the mistakes of a similar previous question, see:
What gas has the highest refractive index?
https://en.wikipedia.org/wiki/Standard_sea-level_conditions

Get hold of Landolt-Bornstein handbook. Small alkanes like n-propane and n-butane will have reasonable refractive index in the visible region.

The full reference is Landoldt-Bernstein (1962) Zahlenwerte und Funktionen. Eig-
enschaften der Materie in ihren Aggregatzustanden. 8. Teil (edited by Hellwege K. H. und Hellwege A. M.), pp. 871-877.

The Engineering Toolbox gives indices of refraction for various gases, presumably at one atmosphere pressure(and thus requiring temperatures above 15°C in many cases). Among substances that are actually gases at 15°C and one atmosphere, (di)methyl ether at 1.000891 tops this list. This makes dimethyl ether about three times as strongly refractive against a vacuum as air.
But that list is not exhaustive. The refractive index of uranium hexafluoride as a function of temperature and pressure is given in Ref. 1 yields 1.001455 at 632.8 nm and 1.001496 at 457.9 nm, both at 15°C and 760 torr (1 atmosphere). Uranium hexafluoride is thus over 60% more refractive against a vacuum than dimethyl ether, and more than five times as refractive as air.
Reference

Wright, S.P. (Aug 1982). Index-of-refraction measurement of uranium hexafluoride vapor at 6328 and 4579 nanometers (K/AIS--5003). United States


text:
How to electroplate vias and THT holes on PCB with copper?
How can I electroplate copper into PCB's vias and THT holes?
I found several instructions, but they all use palladium (which is expensive), or calcium hypophosphite (which I can't get in my country), or hydrogen peroxide (which I can only get with a concentration of 8%).
Are there any alternative ways using chemicals which are easier to get?

https://chemistry.stackexchange.com/questions/81714

The connection of vias in multilayer PCB's used to be done via (no pun intended) electroless plating, but that process is a bit fiddly, demanding extra precleaning and postcleaning, and use and disposal of hazardous chemicals, such as chromium compounds. You can still buy an electroless plating kit, for US$150 and up, but the solutions have limited shelf life and contain various hazardous compounds.
Commercially, electroless deposition has largely been replaced by the black hole process, so-called because a thin layer of black conductive carbon is deposited in the via, then plated with copper.
An individual might find conductive ink an alternative method for connecting vias. This can be automated, as in this video. Conductive ink can be found for ~US$120 for 50 grams. Also, see this paper on filling vias with conductive copper paste.


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Decarboxylation of THCA in cannabis after plant material is mixed with ethanol
When making a tincture of cannabis, an early step is decarboxylation by roasting cannabis flowers at low temperatures (such as 230F for 30 minutes). My question is, how can one decarboxylate cannabis that is already in a solution of ethanol?
There are many questions about this online from people who have started making a cannabis tincture but did not decarboxylate the plant matter first. I am asking the question here to learn about this from a chemistry perspective, rather than anecdotes.
Of the anecdotes about this, here are a few ideas which seem credible based on my research on decarboxylation. Which of these processes are safe and effective for decarboxylating cannabis tincture, after the plant matter is mixed with ethanol?
A. "Do nothing": Curing/drying the plant matter first does some decarboxylation, so tinctures made with dried cannabis will likely have psychoactive potency even if not decarboxylated by heat before dissolution.
B. Heat the solution: Given ethanol's boiling point is about 173F, it seems unsafe to heat-by-flame a solution of cannabis and ethanol to decarboxylation temperatures (200-250F). Some suggest using a water bath to apply heat, evaporating the ethanol from an unsealed container in the process, resulting in a concentrated and decarboxylated solution.
C. Dry then heat the plant matter: Cannabis can be filtered out of the solution then left to dry (ethanol evaporating within a day, remaining water taking longer but being of less concern to heat). That relatively-dry and ethanol-free plant matter can then be roasted at low temperatures, as it would be prior to being mixed in a solution for a cannabis tincture. It's unclear to me how well this would work, given cannabinoids and terpenes would already be extracted and remain in the ethanol solution; I expect that after alcohol extraction has begun (e.g. in solution >1 day), decarboxylation of filtered plant matter would be less effective.

Why close this question? There is nothing about personal medical advise in here, it is a question about chemistry methods for decarboxylation and it received a valid and helpful answer.

Efficient production of tetrahydrocannabinol $(\Delta^9-\mathrm{THC})$ from cannabis is important for its medical application. If that's what your intention is (indicated by studieng the chemistry), you should follow the methods described in two recent patents (Ref.1 & Ref.2):

$\Delta^9$ THC acid is obtained from plant material and extracted into an aqueous solvent under conditions of $\mathrm{pH}$ control. The acid is converted to a salt and the salt extracted into a polar solvent, yielding acid of high purity. The $\Delta^9$ THC acid is then converted to $\Delta^9$ THC which is further purified and combined with a carrier for pharmaceutical use.

I put only the abstract of Ref.1 but recommend to read the whole patent to get better understanding. Also, I'd recommended the reading of Ref.3 for some studieds on decarboxylation and its proposed mechanism.
References:

Neil J Goodwin, Nicolas J. Archer, Christopher Murray, Alan K. Greenwood, and Derek McHattie, "Production of $\Delta^9$ tetrahydrocannabinol," United States Patent 2009, US 7,592,468 B2.
Frederick R. Ferri, "Method and apparatus for processing herbaceous plant materials including the cannabis plant," United States Patent 2015, US 9,199,960 B2.
Helene Perrotin-Brunel, Wim Buijs, Jaap van Spronsen, Maaike J. E. van Roosmalen, Cor J. Peters, Rob Verpoorte, Geert-Jan Witkamp, "Decarboxylation of $\Delta^9$-tetrahydrocannabinol: Kinetics and molecular modeling," Journal of Molecular Structure 2011, 987(1-3), 67–73 (DOI: https://doi.org/10.1016/j.molstruc.2010.11.061)(PDF).