text:
Sonication to purify solid product- the crude product just disappeared
I am forming Fmoc-valine-citrulline dipeptide, and below is the literature protocol I am following (from Bioconjugate Chem 2002, 13 (4), 855-869):

Formation of Fmoc-Val-Cit

Fmoc-Val-OSuc (succinimidyl ester) ($\pu{14.91 mmol}$) in DME ($\pu{40 mL}$) was added to a solution of Cit ($\pu{2.743 g}$, $\pu{1.05 equiv.}$) and NaHCO3 ($\pu{1.315 g}$, $\pu{1.05 equiv.}$) in water ($\pu{40 mL}$). THF ($\pu{20 mL}$) was added to aid solubility, and the mixture was stirred at room temperature for 16 h. Aqueous citric acid (15%, $\pu{75 mL}$) was added, and the mixture was extracted with 10% 2-propanol/ethyl acetate ($\pu{2 \times 100 mL}$). The solid product began to precipitate but remained in the organic layer. The suspension was washed with water ($\pu{2 \times 150 mL}$), and the solvents were evaporated. The resulting white solid was dried in vacuo for 5 h and then treated with ether ($\pu{80 mL}$). After brief sonication and trituration, the white solid product was collected by filtration

I confirmed the formation of the crude product through HRMS and NMR. I then proceeded to carry out sonication to purify. I thought the purpose of sonication was to initiate crystallization. However, when I sonicated the crude product (~20 mins), I observed that the product was well-dissolved in ether, rather than forming a crystal. I used a Büchner funnel to collect the product but there was nothing left on the filter paper. Would pouring more slowly over the filter paper solve the problem? And more importantly, I want to understand the purpose of sonication step.

Did you do the trituration?

@Waylander No...do you think trituration is critical?

@chemrese I think so. The product is clearly soluble in neat ether, so trituration with something less polar (hexane?) will be needed to crystallise it.

I would also note that the procedure calls for "brief" sonication, 20 mins is way too much

@Waylander @ Alchimista I will take the crude, put a minimum amount of ether and hexane (perhaps 1:1), and triturate with a glass rod.  I will sonicate for 5 mins only.

@Waylander Would there be a way to remove residual water or isopropyl alcohol in the crude? I think it also caused a problem when sonicated

That's what the drying in vacuo in the original procedure does

I think you got it: the most likely hypothesis from what is described here is that the solid you treated with ether was not soluble in ether (it's not even very soluble in EtOAc!), but it still contained too much iPrOH (and maybe water?), and that caused its dissolution. I mean, it would be silly to advise to triturate and sonicate a product in a whopping 80 mL ether if it were so soluble in it. You would not normally dry a solid for 5 h before trituration, so drying was meant to remove a solvent; one must check that it has actually happened, before continuing.

So I think we have established the problem (thank you @user6376297) . The material still contained enough iPrOH to cause it to dissolve in ether rather than give a filtrable solid because it was not adequately dried. What you need to do with your ethereal solution of product is remove the solvent under reduced pressure then dry the resulting solid under vacuum for at least 5hrs and preferably overnight. Then briefly (~1 min) sonicate the dried solid with ether and collect by filtration.
There is a lesson here: if you have a literature procedure that delivers your desired product in good yield then you follow it - without shortcuts.

I think the problem arose also because I ignored the scale of work-up solvents. I did 0.2 mmol scale to test out the reaction, but didn't properly calculate the amount of work-up solvents needed relative to the scale I was using.  I only cared about the reaction solvent amount (DME, THF..). I realised I shouldn't have done like that

I did hard vac it overnight, actually, but maybe there was too much of the IPA and water.

I followed everything in the second time but still, the product all disappeared... maybe my hand is just bad.

Then your product is in the ether solution. Concentrate that down, add toluene and vac that off whic hwill remove residual iPrOH. You should get solid product at that point that you can recrystallise.

There is some confusion on the purpose of sonication here, which is understandable, as to quote, for example, from Wikipedia on Sonication:

Sonication is the act of applying sound energy to agitate particles in a sample, for various purposes such as the extraction of multiple compounds from plants, microalgae and seaweeds.[1]

which relates to physical separation techniques. However, a key further comment:

Sonication has numerous effects, both chemical and physical. The chemical effects of ultrasound are concerned with understanding the effect of sonic waves on chemical systems, this is called sonochemistry.

Related to this question, assuming possible sonochemistry (also referred to as sonolysis) occurring, to quote a source: Free radical formation from sonolysis of water in the presence of different gases:

Free radical formation by sonolysis of water has been considered to be generated by cavitation especially when ultrasound was irradiated with low frequency (from a few dozens to several hundred kHz).(8–10) Cavitation is a phenomenon which refers to the formation, growth, and collapse of small bubbles formed in liquids. During the process of cavitation, the extremely high temperature (several thousand degrees K) and high pressure (hundreds of atmospheres) of imploding cavitation bubbles lead to the thermal dissociation of water molecule, which in turn results in generation of •OH and •H.(11–13)

And importantly, further comments:

Only •OH was detected in the O2-dissolved water sample, and the amount of the radical was much greater than that determined in any of other gas-dissolved water samples. One of the possible reasons to explain why the •H radical was not detected in the O2-dissolved water is that the •H reacts with O2 to form •OOH. In the H2-dissolved water, only •H was detected, suggesting that H2 reduces or neutralizes •OH. In the N2-dissolved water, both •OH and •H were detected at comparable level. In the water samples dissolved with rare gases (Ar, Ne, and He), the amount of •H was almost double as compared with that of •OH, and both •OH and •H yields increased in the order Ar > Ne > He.

The above implies that application of a sonication treatment other than 'brief' in the presence of water could have crossed a boundary into sonochemistry.
However, as to what particular impact a radical presence could have produced, it is not entirely clear, as product chemistry varies distinctly, for example, between the •OH and •H radical.
However, expecting some possible active radical chemistry products also is likely reasonable with a '20-minute' session.

Here US seems to replace some trituration. It has to do with removing impurities, it is basically trituration. This is what I get from the posted procedure. Anyway informative...


text:
what happens if you add the concentration of BOTH product and reactant at equilibrium
It is seems obvious that if you increase the REACTANT ONLY, the position of Equilibrium will move to the product side. Thus making more product if you measured it again in the second equilibrium. HOWEVER, what will happened if you increase BOTH of the product and reactant?

The result depends on on which
side of equilibrium the reaction quoting would be. https://en.wikipedia.org/wiki/Reaction_quotient

If you increase the amount of product and reactant in the proportions that they exist in, at equilibrium, then nothing will happen. It would be as if you had two flasks of reactants at equilibrium and connected them together.
If the product and reactant were not added in their equilibrium proportions then the reaction would convert one into the other until they were at their equilibrium proportions.
This assumes that it is a closed system etc.

Your answer assumes that the reactant and the product being discussed have same stoichiometric coefficient, which isn't necessarily true.

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Can iron oxide in well-point water be removed by electrolysis?
Can iron oxide dissolved in well point water stored in a 1000l plastic tank be removed by using a 12V battery charger and simple electrodes? If so what material should the electrodes consist of?

No ! Iron oxide cannot be dissolved by electrolysis. It can only be dissolved by concentrated acids, like concentrated hydrochloric acid.

If the plastic tank has access to air ,you do not need to do anything . Ferrous (
+2 ) is soluble in water . Ferric (+ 3) is not . So any ferric particles will settle out as oxides/hydroxides ,etc . Soluble ferrous iron will be oxidized by dissolved oxygen from air in hours to ferric , and settle out as oxides / hydroxides , etc. So ,with residence time to settle ,no agitation , and possible filtering , the iron in the water will be at a very low level. I accidently saw this chemistry in action once. I filled a swimming pool with private well water ( which I knew had some iron but did not treat for it) . This reqiured 12 or more hours . I turned off the pump at night , the next morning there was a couple inches of reddish jel-like layer on the bottom - the precipitated ferric iron. The standard swimming pool filter cleaned it out in about a day.

A jug with no cap does not constitute aeration. My experience with a ferrous salt with application of an air pump produced no results until the solution was acidic. This answer is also not in accord with any health agency recommendation, fails to provide any sources (one would think on matters of public health at least one reference, especially on a forum that enforces and disapproves of even promulgating accurate personal health advice ) and should be deleted. By the way, is there even an answer on the question of "electrodes" in this anecdotal story?  Otherwise, enjoyed reading this reply.

I recommend following a refined version of this remedy suggested by a state health department, addressing iron in well water, to quote:

Aeration (injecting air) or chemical oxidation (usually adding chlorine in the form of calcium or sodium hypochlorite) followed by filtration are options if iron levels are more than 10 mg/L. Sediment filter, carbon filter, or water softeners can remove small amounts of iron, but the iron will quickly plug the system.

My refinement on the aeration technique is to couple it with water electrolysis (liberating hydrogen and oxygen) employing carbon electrodes. And yes, the carbon electrode will breakdown, but it can be filtered out and the elemental carbon may further absorb toxins in the water and even improve taste.
As to why both at once, it is to foster the transient presence of active radical oxygen species (including H2O2, which will not introduce or interact with any organics present to possibly create any persistent problematic side products).
A comment on the chemistry without the application of electricity (just aeration), if iron is presence as Fe(HCO3)2, the action of oxygen could proceed as follows:
$\ce{ 2 Fe^{2+} (aq) + 3 O2 (d) + 6 H^{+} (aq) -> 2 Fe(OH)3 (s)}$
creating red insoluble ferric hydroxide (more commonly known as rust) through the action of aeration (or electrolysis) formed dissolved oxygen.
[EDIT] Relating to aerated electrolysis, a patent citing an application of caption gases from electrolysis that is further fed back into the system to effect a more prolonged and beneficial aeration effect. Also, a paper: Aerated Electrolysis For Reducing Impacts of Shale Gas Production Wastewater on Water Sources regarding Disinfectant Byproducts Formation where the aeration process is of aid in reducing more problematic water contaminants that can be created via the electrolysis.

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About alternative magnetic effect on porcelain (for induction furnace purposes)
I am trying to use the Induction furnace mechanism (like described via this video), for heating part of the Muffle furnace, as one dental porcelain furnace prototype.
So, my plan is to put copper coil with frequency around 10 ~ 4000 HZ, and placing some iron bar inside the ceramic container (one sample shown below!):

.
I have googled the induction muffle furnace, and find only one patent by this topic (you can see below):

So, I am in doubt that some part of this project must be wrong. So, I asked here about the effect of changing the magnetic field on the characteristics of dental porcelain, if that effect were too much and create some deformation via dental porcelain coating, this prototype would not be proper for this purpose.
As I am not familiar with chemical terms and ..., If possible, suggest some proper tag to faster finding the effect of the alternative magnetic on the dental porcelain coating.
Update:
For clarifying question it must be said:

induction furnace mechanism are working fine, and for creating high
temperature the plan have had some additional metal bar to become hot
(based of induction mechanism), and create the desired heating for
melting the porcelain coating. But my question is about the effects of
alternative magnetic fields on desired characteristics of porcelain
(target).

Thanks for your consideration.

On reason you don't see much information is that your plan will not work. The copper wire will get hot by resistance and conduction and at some temperature it will react with the refractory = fail. Copper has essentially no use as a heating element because the oxide spalls off after each heat cycle. Use Nichrome wire ( this alloy family has a few different names ). However , as I understand induction heating it needs an conductive target. I am guessing your dental porcelain has poor electric conductivity , so will not heat by induction.

thanks for you answer, the induction furnace mechanism are working fine, and for creating high temperature the plan have had some additional metal bar to become hot (based of induction mechanism), and create the desired heating for melting the porcelain coating. But my question is about the effects of alternative magnetic fields on desired characteristics of porcelain (target).

So you are heating the porcelain by conduction of heat from a metal bar heated by induction . Difficult for me to see any advantage.

Thanks, but i asked about chemical engineering to be more aware about the special changes made by magnetic alternative fields on porcelain materials.


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"Encyclopedic" / semi-exhaustive reference work on Chemistry
I'm buying my brother a present—It's the holiday season after all. He's about to start his academic career in chemistry. I would like to give him a book on chemistry, but not something that is considered "chem 101", "popular", or too specialized and probably off-topic. I am a mathematician myself, and in my field we have the beautiful "Princeton companion to mathematics". It is edited by a Field's medallist, has contributions of many high-profile mathematicians, and is both incredibly varied and in-depth. It contains many, many beautiful phenomena, but is by no means a "popular book" or an easy read. I was wondering if anything like this exists in Chemistry. So to clarify: the book would ideally introduce an advanced reader to several concepts in other branches of chemistry than the person's speciality, in a way that is appealing to a serious academic. Many thanks!

Greenwood-Earnshaw could work - it's about inorganic compounds, not concepts, though. BTW this topic is rather too broad :/

Thanks! In what sense is it too broad? Perhaps such a thing simply does not exist within the field of chemistry...

I am definitely not after a university textbook...

G. E. isn't a textbook, but reference book, so it's OK in this way, but covers only inorganic, not whole of chem. It's still a huge and interesting thing. Dunno about more general book. Such not exactly specific questions with multiple more or less fitting possible answers are considered too broad.

Not specifically what you're looking for, but consider the CRC Handbook of Chemistry and Physics.

Although not the encyclopedic work you asked for as as gifts go, consider a 1st edition [1939] of Linus Pauling's, "The Nature of the Chemical Bond."  Truly a classic.

Or consider the Merck Index, which lists the properties and practical applications of about $10'000$ chemicals by alphabetic order, covering more than $2000$ pages. Afterwards, there is a list of about $500$ different types of chemical reactions, And it is not expensive (about $100$ €) as it has been printed and reprinted so many times since its first edition (1899). It is the Bible for chemists. And it is not commercially related to the chemical factory Merck, apart from the name.

Nitpick: It is now (15th edition) licensed by Merck (the Merck known as MSD) to the RSC.  Not sure about now (because of RSC's online edition) but in the past, earlier editions were available as a less heavy, easier to (structure)search CD, too (https://pubs.acs.org/doi/pdf/10.1021/ci0400462).

One of the most comprehensive descriptive chemistry books is Holleman and Wiberg's Inorganic Chemistry. It fits your encyclopedic criterion with about 2000 pages with all major atomic structure concepts plus descriptive chemistry of the elements. It is being published for more than 100 years and originally it was in German.
Another classic, again of about >1000 pages was Glasstone's "A Textbook of Physical Chemistry". It is slightly dated (1940s) but very exhaustive in terms of covering all the classical concepts to be ever encountered in physical chemistry.


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How can x-rays show how many protons are in a nucleus?
My textbook states that: Henry Moseley found using x-rays that the atomic number of each element has a characteristic positive charge.
I don't know much about x-rays and can't find an answer online or elsewhere. How does this technique work ?
My thoughts are that maybe all protons emit x-ray photons at the same frequency and rate. If this is true then some kind of x-ray detector could be used to detect the energy at that x-ray frequency that a nucleus emits. The number of protons could then be calculated.
I could be way off here.
Can someone explain how x-rays can be used to count the number of protons in a nucleus?

See Wikipedia article Moseley's law

X-rays are not emitted by protons. They are emitted by atomic electrons when they fall down from an upper orbit (or level) to another one near the nucleus. Of course the electrons have first to be excited (= sent to this highly excited level). But whatever the way of exciting the electron, the energy of the transition corresponding to the return to the lowest level is proportional to the number of protons in the nucleus. If you are able to measure the energy of the emitted X-ray produced during this transition (or fall), you can calculate the number of protons of the nucleus. I am sorry. I have used a rather simple language. Pure theorists will prefer using a better vocabulary, which may not have been understood by Kantura. I may be wrong.


text:
Why is the carbon fluorine bond stronger than the carbon oxygen bond?
From this wiredchemist.com, I obtained values of the bond dissociation energies of the $\ce{C-O}$ and $\ce{C-F}$ bonds:
\begin{align}
D_0(\ce{C-F}) &=  \pu{485 KJ mol-1}\\
D_0(\ce{C-O}) &=  \pu{358 KJ mol-1}
\end{align}
I have read an explanation saying that is because of fluorine being more electronegative than oxygen, consequently the dipole moment across the bond is greater. As a result the delta negative on $\ce{F}$ and delta positive on $\ce{C}$ attracts each other more compared to the situation in $\ce{C-O}$.
There is another explanation of better overlap resulting in stronger bond. However in this case the orbital overlap seems quite similar and if not $\ce{C-O}$ has better overlap than $\ce{C-F}$, so this explanation doesn't apply.
However, if this is the case, then why in many organic reaction mechanisms, the polar bonds are more prone to break when the atom with delta positive on is attacked by a nucleophile? Would you give examples for nucleophilic addition-elimination and for nucleophilic substitution?

Bond Dissociation Energy (BDE) is a homolytic process. More difficult to break a polar bond homolytically than a less polar bond.

You may want to read this. https://chemistry.stackexchange.com/questions/124896/c%e2%88%92h-bond-length-in-organic-compounds/124909#124909

Thank you for this, @user55119. If so, then e.g. why reaction of propan-1-ol with HI has higher rate than with HCl? Textbooks use bond enthalpy to explain this but in this case heterolytic fission occurs.

Rate is a kinetic issue where iodide is better nucleophile than chloride in an SN2 displacement of protonated propan-1-ol.

Sorry, I meant a given nucleophile like hydroxide attacking for example 1-iodopropane compared to 1-chloropropane. Textbooks give explanations of C-Cl has higher BDE than C-I but as you pointed out BDE is a measure of homolysis but C-X would break by heterolytic fission.


text:
Understanding Scatchard Plots
Im having trouble understanding Scatchard plots.
Y Axis = Bound/Free Ligand
X Axis = Bound Ligand
The graph has a negative slope.
Why when there is almost no Bound (Y axis = 0)  do we get a high positive number for Bound ligand on the X Axis corresponding to Receptor concentration? if there is no bound how is there a positive bound number?
it looks like the less bound ligand on the Y axis, the more bound on the X axis!

The Scatchard eqn can be written as $\displaystyle \frac{Y}{[L]}=\frac{n}{K}-\frac{Y}{K}$ with $[L]$ the free ligand concentration and $Y$ the fraction bound, (i.e. $[L]$ bound /total protein) so the x-axis intercept (when $Y/[L]=0$) is $n$ the number of ligands bound, the slope is $-1/K$ and the y-axis intercept is $n/K$. Your plot confirms this equation, the interpretation of intercepts and slope follow from the equation.
(Your plot seems to plot bound concentration not fraction bound so needs correcting with total protein to make $Y$ and this is presumably why $n$ comes out to be a small value)

Scatchard plots are intuitively difficult to grasp because the X-axis is not the "independent variable" of the experimental measurement.
When we do a binding isotherm, we measure B (bound, the dependent variable) at increasing concentrations of free (F, the independent variable). But in the Scatchard plot, F doesn't appear as the x-axis variable, and instead appears in the denominator of the y-axis variable, which is counter-intutitive. So the answer to this question is best explained by thinking about the hyperbolic binding isotherm for the reaction in three segments.
At very low F (below Kd) and very low B, B increases linearly with F. Our calculated B/F is high, and stays roughly constant as F increases. These are the points in the extreme upper left of the Scatchard plot.
As F rises to and passes Kd, larger increases in F are needed to get the same increase in B; the slope of the hyperbola is starting to level off, F is increasing faster than B, so B/F decreases. These are the points in the main part of the Scatchard plot.
At F much greater than Kd, the isotherm approaches and reaches saturation. Large increases in F are needed to produce only small increases in B. F increases much faster than B, so B/F gets even smaller.
Finally, as F theoretically approaches infinity, B reaches Bmax and can increase no further, and B/F approaches zero. These are the points in the lower right of the Scatchard plot, where the x-intercept is Bmax.


text:
Whatman 41 filter uses for?
I was reading a research article related to winter fog, so the researcher collected aerosols on Whatman 41 filters every 12h, and we know that Whatman 41 filters are usually used for Qualitative analysis than how come he find out the concentration of Aerosols through Whatman 41 filter? I am confused in the purpose of Whatman 41 filter. Why we use it and for what purposes it can be used and how can we use it?
Note: I need a basic explanation.

Whatman number their filters according to the type of material e.g. glass fibre, paper, wet strengthened, hardened, ashless paper and also the porosity e.g. 6 micron,  2.5 - 5 micron. These are also related qualitatively to speed, e.g. fast, medium. So 54 is fast speed (because it's larger pore size) and hardened paper.
In this case, 41 is a fast (20 micron) ashless paper filter for coarse particles, but can also be used for pollution analysis as it can be purchased as sheets rather than circles for Buchner filter assemblies.
I'm no air pollution expert, but if the 41 paper was used in combustion analysis, one might get the particle measurements that way.

I add some speculation. Filtration means to collect a product. The latter is forced through at least by weight. Different is to evaluate pollution or similar issue. One can well create a scale or compare different cases without stopping all or most of the particulate. Porosity will play a role of course, but processes of adsorption, electrostatic interactions, or simply stopping some of the paths could work as well. Say, one let the filter a fixed amount of time and compare. But, I repeat, this is just reasoning. I am sure it is but not of its relation with protocols, common practice, etc.


text:
Conventional unit cell for a hexagonal crystal system

The figure shows a unit cell of a hexagonal crystal system. Drawn in bold, is the unit cell. The lightly shaded one is a unit cell as well and has a six fold symmetry along an axis, hence is more symmetrical than the one in bold.
Why then do texts define the unit cell of a hexagonal system by the one drawn in bold? For example this this

By definition, the unit cell must be a parallelepiped. The gray thing isn't.

Thank you Mr Ivan Neretin . Isn't a unit cell an object whose repetition will result in generation of the lattice? That's the definition I've been taught. Furthermore why do the texts add the faded line in the unit cell?

True, a unit cell is an object whose repetition by translations will produce the entire lattice. As for the faded lines, they are added just to mess with you. At least, thinking of it this way will make your life easier.

So the hexagon shown in faded cannot be a unit cell because its not a parallelopiped. For the life making hard part I don't know what to make outb of it :)

No.  A unit cell need not be a parallelepiped.  The real requirement is that each pair of faces opposed to each other relative to the center must be congruent and parallel.  Parallelepipeds meet that criterion, but so does a regular hexagonal prism.  It's even possible to draw perfectly good, albeit nonregular, hexagonal-base unit cells for orthorhombic or monoclinic crystals.

But... I found on the internet that it has to be a parallelopiped. But why then don't we take the unit cell of a hexagonal crystal system to be a hexagonal unit cell as is shown above in faded lines? I'd be really grateful if you could post an answer if you have time to spare. :)

@OscarLanzi It is possible, but maybe not advisable unless you have a specific goal in mind (like e.g. with Voronoi diagrams).

A crystal repeats contents of a unit cell in three dimensions. If you choose the unit cell to be a parallelepiped, the translation vectors are simply the edges of it. This is an easy way to describe a hexagonal crystal (and it is the conventional way given in the International Tables of Crystallography).
You could also choose a hexagonal prism (if you look at the figure in the question, you will realize its volume is 3 times bigger, so there is translational symmetry within the cell now). You can also repeat that in three dimensions to fill up the entire space. However, the translation vectors will be different.

The smaller unit cell is more common, probably because the math is easier and more closely related to that of other crystal systems.
The relationship between the two ways of dividing up space into repeating units is shown below. Note that two of the green unit cells complete fit into a hexagon, with the remainder of the space taken up by two halves of green unit cells.

For face-centered or body-centered lattices, we do (conventionally) choose the larger unit cell because it makes many things easier. In this case, choosing the hexagonal prism would make a lot of things harder.

Interestingly enough, the reason why the cell in bold is the unit-cell for the crystal is actually because the hexagonal cell has this higher symmetry you mention.
To understand why, we have to scratch at the true meaning of symmetry: not just in terms of crystal cells, but more broadly what symmetry is as a mathematical idea. When we say something is symmetric, we think of it as "looking exactly the same" in some way (maybe specifically a reflection or rotation, perhaps.) A reflection makes things look the same, but backwards in a way; a rotation makes things look the same, but just spun around in a way. In mathematics, symmetry is defined as any action you can perform to some object where—after the action—the object remains fundamentally unchanged. Translations are types of symmetries; we can move things around, and it doesn't fundamentally change what they are. Sounds simple enough.
This idea is extremely powerful when we start analyzing systems of crystals—as we add more atoms to what we are trying to study, it becomes exponentially more difficult to keep track of what everything is doing. Chemists and physicists, in order to cope with all the excess information, have to encode that information into something that is possible to understand: in quantum chemistry, it's the wavefunction; in gas chemistry, it's pressure-volume-temperature coordinates; and in solid-state chemistry, chemists leverage the symmetry of crystals to encode that information in the unit cell.
How do we do this? Instead of focusing on all $6.022 * 10^{20}$ atoms or whatever in our sample, we instead have two things we focus on: we have to find the arrangement of atoms in the smallest volume of atoms we can possibly can without losing information, and then we have to describe all of the symmetries of the unit cell (a.k.a. all of the reflections, translations and rotations) that allow the unit cell to fit with itself and—in the mathematician speak—"tile the plane."
Because the atoms are arranged in a 3D-lattice, you can use the mathematics of matrices and linear algebra to prove that the smallest amount of information you need for atomic positions in a crystal unit-cell is three vectors that describe a parallelepiped (hence how all the forms are parallelepipeds.) The "hexagonal" in hexagonal crystal packing is not a statement about the shape of the unit-cell necessarily, but is instead a description of the types of symmetries necessary to know how unit-cells can be sewn together to make the entire crystal. The unit-cell describes all the atomic positions you need to know, and the type of unit-cell it is implies some of (but usually not all) of the information you need to know about the symmetries.
You can tile a crystal with larger boundaries than a unit-cell (a hexagonal boundary, e.g.,) and perhaps you could use an expanded definition of unit-cell to encapsulate cells larger than the smallest necessary... but there really is no need to conflate the unit-cell with anything larger than it needs to be—it would get muddy and confusing, and I would opt to call it a different word (I'm sure there is one in crystallography that I am just not aware of at the moment.)


text:
If a good choice of a unit cell should be the one of most symmetry ,then why keep body centered tetragonal if face centered cubic exists?
For any lattice a good choice of a unit cell is the one that is the most symmetric.
Now if every lattice made by a body centered tetragonal unit cell can be made by a face centered cubic unit cell ,which is more symmetric ,then why keep the body centered tetragonal unit cell as a class of Bravais lattice ?

Because not every body centered tetragonal cell is equivalent to a face centered cubic cell.

Body-centered tetragonal is face-centered cubic only if $c/a=\sqrt2$.  If you try your transformation with a $c/a$ value greater/less than $\sqrt2$, your "cube" will have lateral edges that are longer/shorter (resp.) than the basal edge and so really remains tetragonal.
You can also read this the other way.  If you are presented with a body-centered tetragonal lattice and $c/a$ seems to be close to $\sqrt2$, maybe it's really fcc.


text:
Could someone advise me on sources for crystallography and surface sciences?
I'm a physicist learning crystallography and surface science on the side of my research and I came across the concepts: surface reconstruction in hexagonal materials, unit mesh of surfaces, etc.
I was given problems to solve but they are too advanced for me. These are based on the book Modern Techniques of Surface Science 3rd ed. by Dr. Woodruff.
Can someone advise me with some other sources of information to better understand Lattice Planes and Miller Indices, to the level shown in the Cambridge website.

Recommend you start with something simpler: bulk crystallography. ;)

Would that be the title of the book? Are there any online sources for bulk crystallography?

Bulk was meant as a synonym for "basic", "general", "simple". That's still no picnic, I promise. Take out your old physics 101 textbook and revise what's in there, then go on.

'Crystallography'  W. Borchardt-Ott, publ Spinger for lots on space groups etc. and with a more on x-ray diffraction as well as crystallography, 'Essentials of Crystallography' , D. McKie & C. McKie, publ Blackwell.

Thank you very much, so far I had only found "Fundamental of Material Science and Engineering by W. Callister Jr." .


text:
hydrogen spectrum
In the spectrum of atomic hydrogen, several lines are generally classified together as belonging to a series (for example, Balmer series, Lyman series, Paschen series), What is common to the lines within a series that makes grouping them together logical?
The answers say that each series has a different lower energy level, can anyone pls explain

The Lyman series is made of lines in the spectrum corresponding to the return of the electron from a higher level ($n = 2, 3, 4,$ etc.) to the first and deepest level ($n = 1$) .
The Balmer series is made of lines in the spectrum corresponding to the return of the electron from a higher level ($n = 3, 4, 5$, etc.) to the second level ($n = 2$)
The Paschen series is made of lines in the spectrum corresponding to the return of the electron from a higher level ($n = 4, 5,$ etc.) to the third level ($n = 3$)

Because the energies go with one over n squared, the energy of the transition is mostly determined by the lower level (e.g. the transition 2 to 1 has 75% of the energy compared to the transition infinity to 1). For that reason, the spectral lines are grouped by the lower level and not the higher level.


text:
Precipitation of a lead(II) sulfate

Lead (II) sulfate precipitate can be dissolved if:
a) We add an acid
b) We add EDTA
c) None of the above is true.

I think EDTA is a ligand, and if you want to dissolve the sulfate, then Pb sticks with EDTA to form a complex. And so more Pb will be formed to balance the reaction.

The stability constant of $\ce{Pb−EDTA}$ complex is $\ce{10^{18}}$, which is huge. The solubility product of $\ce{PbSO_4}$ is $10^{−8}$
, which cannot be compared to the effect of $\ce{EDTA}$. So the precipitate $\ce{PbSO4}$ must be soluble in $\ce{EDTA}$ solutions. These numerical data have been taken from J. G. Stark, Chemistry Data Book, John Murray, London, $1994$


text:
Reference for molecular orbital theory
I am studying electron collision with molecules, and came across with notation like this $\left(\mathrm{B}^{1} \Sigma_{u}^{+}, \mathrm{c}^{3} \Pi_{u}, \mathrm{a}^{3} \Sigma_{g}^{+}, \mathrm{C}^{1} \Pi_{u}\right)$ for the $H_2$  molecules and I have no idea what they are.
Is there any textbook that explains what this symbols are?

The question has been answered on SE (the first two links), but also please refer to the Resources post (also linked): the books by Atkins and Levine will probably have more information. Consider also Hollas's Modern Spectroscopy, which is referenced in my answer to the second linked question. In a comment, porphyrin also suggests Herzberg's Molecular Spectra and Molecular Structure series.


text:
Is there a room temperature ionic liquid permanganate
I need to figure out a permanganate based ionic liquid that is usable at room temperature. Unfortunately, information on which compounds could be usable and which ones aren't is scarce. That's why I'm asking it here.

Why do you want a room temperature liquid ionic  permanganate?

Do you recall that permanganate is a very strong oxidizing agent?

Permanganate tends to destroy organic cations at rt or just above. I once made that mistake with tetrabutyl permanganate.

@waylander boom?

Not enough to go boom but spontaneous combustion on drying

You asked for assistance getting information about “permanganate based ionic liquid” that can be used at room temperatures. The following two articles(1,2) may qualify as “permanganate based ionic liquid”  Unfortunately, I do not have access to the full text of these articles so I do not know which ionic liquids are being used.  However, that fact that these are organic oxidation process suggest that the reactions might be done in a temperatures range which may of be useful to you.
Nevertheless, it looks like permanganate can be used in the presence of ionic liquid for some oxidation reactions.

Abstract: The selective oxidation of primary and secondary benzylic
alcoholsnwith potassium permanganate in ionic liquids to give corresponding
aldehydes and ketones in 83–97% yields has been described.
Anil Kumar, Nidhi Jain & S. M. S. Chauhan (2004) Oxidation of Benzylic Alcohols to Carbonyl Compounds with Potassium Permanganate in Ionic Liquids, Synthetic Communications, 34:15, 2835-2842, DOI: 10.1081/SCC-200026242

Abstract: A ligand-free catalytic epoxidation system using
permanganate in a surfactant-like ionic liquid (IL) medium was
developed. The results indicate that the IL takes crucial effects in
the epoxide selectivity. The loading of permanganate is also found
critical in preventing over-oxidation of epoxides. The system with 0.3
mol% permanganate and 3.5-equivalent CH3CO3H is able to achieve
excellent yields and selectivity of epoxides. The study of epoxidation
with KMnO4 in IL medium reveals an unusual oxidation behavior of
permanganate not found in traditional solvents
Lu, Yu-Jing & Wong, Wing-Leung & Chow, Cheuk-Fai. (2015). A surfactant-like ionic liquid with permanganate dissolved as a highly selective epoxidation system. Catalysis Communications. 69. 25-28. 10.1016/j.catcom.2015.05.020.

No, I am afraid this is not correct. Here the ionic liquid is being used as solvent and potassium permanganate as a reagent.

Fair enough.  Thank you for point out the reagent nature of the permanganate in the examples I provided.  Does this suggest there is an a priori reason that KMnO4 can not be an ionic liquid constituent?  Is permanganate simply too oxidative to allow the existence of the other component of an ionic liquid?

Usually the required anion is incorporated in ILs by ion-exchange resins. I think KMnO4 will easily oxidize these resins too. The anion is incompatible with most organic substances.


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What is the emf of a galvanic cell?
I am getting confused between EMF and potential difference of a cell. But i know that when a charge moves against an electric field it's potential energy increases which can be used to do some work. So can i say EMF of the cell is the work done in order to get the required potential difference between the electrodes. And if it is so can you explain the difference between PD and EMF

See wikipedia - Electromotive_force. You can see EMF is not work.

Chemistry SE site expects  you to provide your elaborated results of prior thinking, analysis and searching,
aside of just asking the question. Elaborated questions attract elaborated answers.
The opposite may lead even to the question closure, in case of lack of the own effort.

Potential difference and EMF are equal if the galvanic cell does not produce any currant. When the cell works and produces even a small currant $I$, the potential of the cell decreases a bit, because of the internal resistance $R_i$ of the cell. If $E_o$ is the the EMP, the potential $E'$ measured between the plus and the minus poles of the working cell decreases a little bit according to Ohm's law :  $E'  = E_o - R_i I$

Currants are small dried grapes or berries. I'm not sure how galvanic cells produce them. ;-)

@Maurice if suppose I have taken a Daniel cell and if the circuit is not complete then no current would flow. In that case we can say that the redox reaction has not taken place . So in that case i dont think there would be any potential difference between the electrodes.

@Matt_black. OK. Thank you. Wonderful ! Next time I'll check my typing vowels, and print current instead of currant.


text:
Consider difference of pressures as weight of piston
Let us say there is a container with two ideal gases in it divided by a piston.Volume of ideal below the piston is 2L and above is 8L.Total volume of container is 10L.
n of both is 1 mole and T is 300 for both.
Find the Temp when the ratio between the volume of gases is 4L and 6L.
How I solved it :
$\frac{R*1*300}{2L}$ - $\frac{R*1*300}{8L}$ since the gas below has more pressure.
Now my sir told me that difference in pressures is equal to weight of piston.The equation that I wrote above.My sir told me this because He said it is important into know that difference in pressures is equal to weight of piston .How is that possible?
Please explain
Please do share if there is any other way to solve this if possible.

The pressure in the lower chamber is equal to the pressure in the upper chamber plus the pressure exerted by the piston.  This is because both the gas in the upper chamber, and the weight of the piston, are exerting a compressive force on the gas in the lower chamber.  And the pressure exerted by the piston is the weight of the piston divided by its area.  Your expression correctly calcuates the pressure exerted by the piston as the difference between the pressure in the upper chamber and the pressure in the lower chamber.

The lower chamber gas is applying pressure on the upper chamber and piston.

You mean to say that P exerted by upper + piston = P by lower chamber.

Approach your tasks by algebraic solution first and plug in the numbers with units in the end.  It has several benefits.
1/ Better grasp of underlying principles.
2/ Both of you  and the readers can more easily spot a solution error or conceptual mistake.
3/ It gives to the Q/A the higher and more permanent value to be reused/applied for similar questions.
4/ It follows the site policy to focus on explaining principles and procedures, rather than solving of particular tasks with the one time value for 1 person only.
5/ It keeps you from intermediate rounding errors.

Ok for sure from next time.

Ok @Poutnik I’m doing.My net is slow.Just give me some time.

I am editing now.But it is getting hanged in between and suddenly refresh,

No hurry. Take your time. analyse scenario and write the final equation T=f(arguments). Then plug in the particular numbers. Remember the 1st Newton motion law.  BTW, what tou wrote is not an equation.

We know from the first Newton motion law, that the net force acting on an object in rest must be zero. The forces acting on the piston are gravity and gas pressure:
$$\vec F_g + \vec F_\mathrm{p,down} + \vec F_\mathrm{p,up}=\vec 0 \tag{1}$$
If $V$ is the given bottom gas volume, $V_0$ is the total gas volume, $n$ is the molar amount of each of gases, the gas constant $R \approx \pu{8.314 JK^-1mol^-1}$, $T$ is absolute temperature. $m, g, A$ are piston mass, gravitational acceleration and piston area, then:
$$ m \cdot g + p_\mathrm{top} \cdot A =p_\mathrm{p,bottom}\cdot A\tag{2}$$
$$ p_\mathrm{bottom} - p_\mathrm{p,top} =  \frac{m \cdot g}{A}\tag{3}$$
As we know from the ideal gas state equation that
$$p=\frac{nRT}{V}\tag{4}$$
then the pressure difference between the bottom and upper gases is
$$\Delta p = nRT ( \frac 1{V} - \frac 1{V_0-V})=\frac {mg}{A}\tag{5}$$
As the right side is constant, we can eliminate it, writing
$$nRT_1 ( \frac 1{V_1} - \frac 1{V_0-V_1}) = nRT_2 ( \frac 1{V_2} - \frac 1{V_0-V_2})\tag{6}$$
$$T_2 =  T_1 \cdot \frac  {\frac 1{V_1} - \frac 1{V_0-V_1}} {\frac 1{V_2} - \frac 1{V_0-V_2}}\tag{7}$$
where $T_1, V_1$ are the initial temperature and the volume of the bottom gas,$T_2, V_2$ are the final temperature and the volume of the bottom gas.
Now we finally plug in the numbers. Doing it before would be just distraction.
$$T_2 =  \ce{300 K} \cdot \frac  {\frac 1{\pu{2 L}} - \frac 1{\pu{8 L}}} {\frac 1{\pu{4 L}} - \frac 1{\pu{6 L}}} = \ce{300 K} \cdot \frac  {3/8\  \pu{L^-1}} {1/12\  \pu{L^-1}}= \ce{300 K} \cdot \frac 92 = 1350 K\tag{8}$$


text:
Is 18:82 a valid ratio for aqua regia?
Here it writes that nitrohydrochloric acid (possibly Aqua Regia) is of the ratio of $18:82$, which is $1:4.56$ in a simpler form. On Wikipedia and every other source i found the ratio $1:3$. So what's the truth behind this?
Also is Aqua Regis a valid way to call the mixture? On Wiktionary it says that this is a synomym of Aqua Regia but why would it have such a different ratio then?

Having made aqua regia fairly regularly during my (short) time in the lab - I never measured it out precisely, and I don't think the exact ratio matters too much... You get a slightly different colour (more nitric acid = more brown), and the exact composition won't be the same, but it'll probably still dissolve what you throw in it. BTW, 18:82 in what units -- volume, mass, etc.? 1:3 refers to the stoichiometry.

"Aqua regia" is "royal water", or more literally "water of kings". "Aqua regis" is "water of the king". I would suspect the latter has occurred because people didn't know Latin, or translated the term from their own language back into Latin. (For example, in German it's "Königswasser" which means literally "king's water", though it could also be interpreted as "water of the king" or even "king water"; hence also similar terms in languages spoken around the Baltic Sea and in Scandinavia, where German cultural influence was significant for many centuries.)

@tripleee, Interesting account. You are right. I checked the etymology it was indeed aqua regis until 1612s and by 1650s aqua regia appeared.

thank you for the answers! idk what units is the 18:82 as i just found it on the net
@M. Farooq could you tell me which etymology dictionary you used for checking the word?

It is called the Unabridged Oxford English Dictionary (www.oed.com). I think it is 22 volumes in print if I remember correctly. Unfortunately, it is not free and you have to pay for subscription or a library has a license. The ordinary OED does not have dates with usage examples.

I think without units it is hard to come up with a sensible answer to the ratio question, unless someone can figure out / reverse engineer the significance of the "magic" numbers 18:82 and how it relates to the usual ratio 1:3.

Aqua regia does not have that rigid fixed composition as the OP is suggesting. There is no magic in his 18:82 numbers.

@orthocresol There is a footnote in the wikipedia article of aqua regia which says "The relative concentrations of the two acids in water differ; values could be 65% w/v for nitric acid and 35% w/v for hydrochloric acid – that is, the actual HNO3:HCl mass ratio is less than 1:2".


text:
Origin of the 15% difference limit between solute and solvent atom radii in the Hume-Rothery rules
According to the Hume-Rothery rules for the substitutional solid solutions the atomic sizes of the the solute and solvent must not differ by more than 15%.
Why should it be so?

Possibly, there are two extrema to account:
a) the atomic radii of metal A and metal B do not differ significantly, and thus there are chances both metals blend statistically with each other.  Which would be covered by the $< 15\%$-rule, a situation similar to mixing apples and oranges in one crate.
b) the atomic radii differ a lot.  If we assume atoms as hard spheres, even the closest arrangement of spheres of one sort does not occupy fully the space available (see e.g., fcc and hcp packing).  If the diameter of the other metal(s) is/are much less than the one of the host, it is possible for some of these pockets to be filled more, and for others to be filled less. Overall, the distribution of metal B and metal A in the blend of (A .and. B) no longer is even; the smaller atoms may segregate.  As if you would transport a crate of filled with melons and hazel nuts.  This non-regular demixing is possible because -- contrary e.g., to spinels -- you may assume absence of ions charged oppositely charged.
For the purpose of illustration, both the examples as well as the figure below intentionally exaggerate the contrast between case a) and b).


text:
Silicone based polymer that can be functionalised on one end?
I am looking for a silicone based polymer (think PDMS but I need to do some chemical reactions with it) where I can attach on one side (or both, I am not sure yet) another molecule that has a carboxylic acid at the end or an amine. Maybe some of you know of a PDMS derivative that's functionalised at the end?
Many thanks in advance.

That you can buy off the shelve. https://www.sigmaaldrich.com/materials-science/material-science-products.html?TablePage=113790985 And google has 400 000 hits for "end group functionalized PDMS".


text:
Why do amides require much harsher conditions for hydrolysis than esters?
Amides require much harsher conditions to hydrolyse than its ester homologue. An explanation given is that the orbitals holding the lone pair on the nitrogen overlaps with the C=O π-bond to give conjugation, thus introducing a "partial" π-bond between nitrogen and the carbonyl carbon. Consequently, this makes the bond harder to break.
However, for the ester homologue, doesn't this conjugation also happen? An explanation given is poorer overlapping of the orbitals. However, how is that true?
In addition, such explanation is given to show that C=O is shorter in ester than it is in amide. However, I still don't get why this is the case.

Amide = crummy leaving group, maybe?

If harsh conditions are needed because of kinetics rather than thermodynamics, the main question is: what is the rate-determining step (rds)? Anything that facilitates that step should make the reaction faster. For acid-catalysed amide hydrolysis, it appears that the rds is the attack of water. Compare a protonated ester and a protonated amide: which one has the highest electron-density at the carbon that water must attack? See also this.

Dimethyl acetamide (DMA) is less susceptible to hydrolysis than ethyl acetate (EA). This statement is true for aqueous acid or base. The infrared (IR) frequency for the carbonyl group of DMA is 1662 cm-1 while EA is 1742 cm-1. The lower the frequency, the lower the energy and the more, relatively speaking, single bond character in the carbonyl group. Thus, the carbonyl group of DMA has less double bond character than that of EA. Because nitrogen can sustain a positive charge better than oxygen, resonance is more pronounced in DMA than in EA. Because resonance is stronger in DMA, its carbonyl group is less susceptible to nucleophilic attack. The carbonyl bond length of EA would be expected to be shorter than that for DMA because the more double bond character, the shorter the bond. It is no different than the carbon-carbon bond length in the series ethane, ethylene and acetylene.

Please would you elaborate on why nitrogen can better sustain the charge than oxygen, if oxygen has higher electronegativity? In this case, nitrogen is better because of the inductive effect due to the 2 methyl groups, and also possible sigma conjugation? would you explain by comparing a primary amide with its homologue? Thanks @user55119

The methyl groups do not have to be present. Amides are less reactive than esters in general. The more electronegative, oxygen, is less willing to bear a positive charge than nitrogen. CH3OCH2Cl is a liquid that reacts with nucleophiles by prior ionization to [CH3O=CH2]+ Cl-.  OTOH, Me2NCH2Cl exists as a salt, [Me2N=CH2]+ Cl-. Now you can explain why acyl chlorides are more reactive than esters and amides.

Both Amide and Ester groups possess resonance stabilization, Amides however are "better" at it then its cousin esters. This is due to Nitrogen from the amide group having less electronegativity than Oxygen from ester and thus, it gives up its electron to the carbonyl Carbon more easily. The easier this happen, the more often overlapping of electrons between the members (C,O,N) occur, increases the stability of the amide group as a whole. Amides are thus more stable and are generally less reactive then esters.

Think about the stabilities of the resonance structures. Nitrogen tends to be stable in its cationic form compared to oxygen; a protonated amine has a pKa around 9.5, whereas an alcohol has a pKa of around -2.4. Nitrogen's lower electronegativity results in higher nucleophilicity, allowing it to donate its electrons into the amide much more readily than the ether oxygen of an ester. As a result, the amide tends to be much more stable and, as a result, more resistant to hydrolysis.


text:
Is there an error on Wikipedia's page on characteristic rotational temperature of HF?
I am referencing this article: https://en.wikipedia.org/wiki/Rotational_temperature
So, I see that $\theta _R = \frac{(h/2\pi)^2}{2Ik_b}$.
I took the value of $I$ from https://cccbdb.nist.gov/mieigen2x.asp.
However, when I do the calculations for $\theta _R$ by plugging in the numbers, I get
$$\theta _R = 46.8324 K$$
not $30.2K$.
Here is my Mathematica snippet

My question is, has Wikipedia gone wrong, or am I messing this up somewhere?

The units of the inertial constant I are reported in $\pu{amu \cdot Å^2}$. One unified atomic mass unit (or Dalton) is approximately $\pu{1.660e-27 kg}$. Using the correct formula in MATLAB:
>> amu=1e-3/6.022e23;
>> h=6.626e-34; kb=1.3806e-23; tr=@(I) ((h/(2*pi))^2)/(2*I*kb*amu*1e-20);
>> tr(0.86)
ans =

 2.820249699307124e+01

Using instead $\pu{I= 0.80 amuÅ^2}$ from the NIST website returns a value closer to that on the Wikipedia table.

I realized i made a simple unit conversion error. thank your for your answer!


text:
How does steam prevent oxidation in tea leaves?
I recently started branching out into different "flavors" of tea. In this endeavor I've been learning more and more about different varieties of tea and was wondering if anyone could offer an explanation as to how steam can prevent the process of oxidation in tea leaves.

Who says it does?

Don't know why people are voting to close based on lack of clarity.  The question seems perfectly clear to me, as evidenced by my answer.  My only critique would be that, as Karl suggested, the person should have provided a reference support the basis of the question, but I didn't actually need that to understand it.

Note that steaming is Japanese way. The Chinese way is dry heating on a pan.

@Poutnik.  While that's generally true, there are exceptions.  Some Japanese teas, like Hōjicha, are roasted.  And some Chinese greens, like Enshi Yu Lu ("Jade Dew") are steamed.  Also, some Chinese tea are cooked in ovens rather than roasted by dry heating on a pan.

According to Tony Gebely at Tea Epicure (https://teaepicure.com/tea-leaves-oxidation/), it's not steam per se, but rather heating that slows oxidation, by denaturing two enzymes (polyphenol oxidase and peroxidase) that are responsible for it.  Steaming is simply one way to do this:

Much of the oxidation process revolves around polyphenols, particularly the enzymes polyphenol oxidase and peroxidase. When the cells inside tea leaves are damaged and the components inside mix and are exposed to oxygen, a chemical reaction begins. [Specifically when polyphenols in the cell’s vacuoles and the peroxidase in the cell’s peroxisomes mix with polyphenol oxidase in the cell’s cytoplasm. From Michael Harney’s 2008, “The Harney and Sons Guide to Tea.”] This reaction converts the polyphenols known as catechins into flavanoids called theaflavins and thearubigins (which are also polyphenols).

Fixing works by denaturing polyphenol oxidase and peroxidase – the enzymes primarily responsible for oxidation using heat. Fixing is also commonly referred to as de-enzyming, denaturing or kill-green.

The term kill-green is derived from the Chinese term shaqing (杀青), which translates to killing the green. The tea leaves must be heated to approximately 150 degrees Fahrenheit to “halt” oxidation. Oxidation is further slowed by drying the leaves, but it never completely stops. At temperatures over 150 degrees Fahrenheit, oxidation continues to occur at an extremely slow pace.

The process of fixing requires precise control of the temperature and length of heating; each has to be adjusted depending on the size and thickness of the leaves and the amount being processed.

Note also that, for teas, oxidation is not a bad thing.  Rather it is a parameter that is precisely controlled in order to achieve the desired type of tea.  Green teas are produced by minimizing oxidation. Oolongs are produced by allowing oxidation to proceed part-way (and they are thus known as "semi-oxidized teas").  Black teas are produced by allowing oxidation to proceed completely, or nearly so:

Oxidation refers to a series of chemical reactions that result in the browning of tea leaves and the production of flavor and aroma compounds in finished teas. Depending on the type of tea being made, oxidation is either prevented altogether or deliberately initiated, controlled and then stopped. [emphasis mine]

When oxidation is prevented altogether, the catechins are left largely intact. The tea leaves keep their green color and vegetal characteristics.

When a semi-oxidized tea is being produced, some catechins convert to theaflavins and thearubigins, resulting in a slight browning usually along the edges of the leaves and yellower liquor. Lipids, amino acids, and carotenoids also begin to break down into flavor and aroma compounds.

When oxidation is allowed to run its course, the leaves exhibit an aroma and taste profile completely unrecognizable from a finished tea that was exempted from oxidation.


text:
What's special about the purine scaffold?
Purine is a remarkable substance, given Nature has chosen it as the scaffold for two nucleobases from DNA/RNA: adenine (A) and guanine (G). Its structure also appears in several other substances of biological interest, by substitution on the fused ring system, like these examples shown in Wikipedia:

The base ring system has 4 nitrogen atoms and 5 carbon atoms. Then I wonder, it's not hard to imagine isomers or at least similar compounds just by changing the positions of the nitrogen atoms around the rings, with a few constraints. The five-membered ring needs at least one N to maintain aromaticity in the pyrrole ring. It's better to have this N at the position where a glycosidic bond is expected, if we want to incorporate it in some kind of nucleic acid. Along the edge connecting the two fused cycles, it's better to have C, to avoid N atoms with four bonds. So we are left with the following template:

We can distribute the remaining 3 N and 3 C atoms among the 6 free positions in $6!/((6-3)! \cdot 3!)  = 20$ different ways.
So my question is, what's special in purine, that caused its scaffold to be picked among all the other possibilities?

I am not sure it is 20 as there is some kind of symmetry with respect of the 6 permutating atoms. Still, this comment "per se" does not invalidate the question. I can add that the natural selection of the scaffold may involve interactions, and not only that entity.

It is important to remember that the purine scaffold is one of the most fundamental organic structures in all lifeforms. From an evolutionary standpoint, this means that there must have been a simple, effective way for this scaffold to be prepared in the pre-biotic earth. After all, the first organisms to contain purine scaffolds did not have "access" to complex enzymes, proteins, etc.
Take a look at the molecular formula for adenine, a simple purine derivative. C5H5N5. We could rewrite this as (HCN)5. As the molecular formula alone might lead you to conclude, adenine can be made from only one chemical: hydrogen cyanide. Interestingly, hydrogen cyanide was a very abundant chemical in the earth's primordial atmosphere. Roy et al.1 replicated the atmospheric conditions of the primordial earth and found that substantial quantities of adenine could be produced! The mechanism for this reaction is quite beautiful, and I will include Peter Jacobi's rendition of it below. The mechanism for the formation of adenine also explains why other constitutional isomers are not produced. We must consider the alternating nature involved in these nucleophilic-electrophilic interactions (e.g. two nitrogen atoms will never "decide" to couple).

So, to conclude, the purine substructure is "special" because it not only has the unique ability to hydrogen bond and form a base pair for DNA replication but was also incredibly easy to synthesize in the primordial earth's atmosphere. Although life forms do not use HCN to synthesize the purine substructure anymore, its synthesis is still incredibly simple.
1 Roy, D.; Najafian, K.; Schleyer, P. V. R. Proc. Natl. Acad. Sci USA 2007, 104, 17272-17277.


text:
Does migraine medicine Topiramate work by supplanting Pyridoxal phosphate in enzymes?
I have seen it said that the precise mechanism of action of migraine medicine Topiramate is not known. But I certainly see a resemblance between that molecule and PLP (Pyridoxal phosphate, the activated form of vitamin B6).

A significant similarity is that both have an "offshoot chain" (Topiramate's sulfamate taking the place of PLP's phosphate) where PLP's phosphate is what binds it to enzymes in the first place, and both have a more electronegative element in the middle of what would otherwise be a six-carbon ring, apparently important in the typical PLP enzyme reaction.

I have been wondering if Topiramate supplants PLP in the enzyme glutamate decarboxylase, changing the reaction into a deamination. This would deplete glutamate, causing the pill's migraine relief, and the deamination could even cause Topiramate hyperammonemia.
If it does the same or similar to serine hydroxymethyltransferase, which is also PLP dependent, it could cause glycine deficiency, thus collagen deficiency - and this would even explain Topiramate hair loss.
When you look at it this way, it's frankly hard to imagine the mechanism of action is anything different. Have I stumbled upon it? Or has this been suggested already?

Background
Topiramate (TPM) is an anticonvulsive and antimigraine drug that helps prevent or reduce the frequency of epileptic seizures and that is used as a prophylaxis of migraine. (Shank and Maryanoff, 2008)
TPM not only inhibits kainate subtypes and α-amino-3-hydroxy-5-methylisoxazole-4-propionic acid (AMPA) subtypes of glutamate activated ionotrophic receptors, the modulatory effects on GABAA receptors and voltage-gated Ca2+ and Na+ channels (Gibbs et al., 2000), but also inhibits carbonic anhydrase enzymes (demonstrated in electrophysiologic and biochemical studies). see drugs.com
OP question
It is an intriguing possible correlation that “Topiramate supplants PLP in the enzyme glutamate decarboxylase [GAD]”, notably based on structural similarity of the alkyl-sulfamate to the alkyl-phosphate as well as the linked rings.
However, the literature does not seem to align with this premise:

Shank and Maryanoff et al.(2008) noted that “[Topiramate’s] sulfamate moiety is essential, but not sufficient, for its pharmacodynamic properties.”
Whereas PLP is a substitute pyridine (planar) derivative, TPM is a sulfamate, substituted derivative of β-D-fructopyranose (Maryanoff et al., 1987) where its substituted ring is in the boat conformation and its ring is puckered. See the crystal structure as entry SEQKAA02 at CCDC

TPM does not seem to inhibit glutamate decarboxylase. “Single-dose and repeated treatments with TPM were without effect on the activities of GABA-T and GAD (Fig. 2) in mouse brain at 4 h after administration.” (Sills et al., 2000)
GAD inhibition and Topiramate treatment have opposite effects. “Clinical evidence shows that lowered GAD activity is associated with several forms of epilepsy which are often treatment resistant.” (Zhang et al. 2017; Lloyd et al. 1986) Furthermore, Zhang et al. synthesized an ethyl ketopentenoate (EKP), a lipid-permeable GAD-inhibitor, and used it to induce refractory seizures in zebrafish larvae. (yes, that’s mean.. but informative here.)

Molecular basis of carbonic anhydrase inhibition
Topiramate is a carbonic anhydrase inhibitor (multiple classes), a Zn-dependent metalloenzyme. Below is how Topiramate (green) binds to human carbonic anhydrase II. The zinc ion is shown as a silver sphere. RCSB entry 3HKU.pdb DOI: 10.2210/pdb3HKU/pdb and Lopez et al. (2009):

Conclusion
Given the above, I wouldn’t think the apparent structural similarity of Topiramate to PLP is relevant to epilepsy and/or migraine treatment.
References
Lloyd, KG, Bossi, L, Morselli, PL, Munari, C, Rougier, M and Loiseau, H (1986) “Alterations of GABA-mediated synaptic transmission in human epilepsy” Adv. Neurol. 44:1033–1044. source
Lopez, M, Paul, B, Hofmann, A, Morizzi, J, Wu, QC, Charman, SA, Innocenti, A, Vullo, D, Supuran, CT and Poulsen, SA (2009) “S-Glycosyl Primary Sulfonamides−A New Structural Class for Selective Inhibition of Cancer-Associated Carbonic Anhydrases” J. Med. Chem. 52:6421-6432. DOI: 10.1021/jm900914e
Maryanoff BE, Nortey SO, Gardocki JF, Shank RP, Dodgson SP (1987) “Anticonvulsant O-alkyl sulfamates. 2,3:4,5-Bis-O-(1-methylethylidene)-β-D-fructopyranose sulfamate and related compounds” J. Med. Chem. 30:880–887. DOI: 10.1021/jm00388a023
Shank, RP and Maryanoff, BE (2008) “Molecular pharmacodynamics, clinical therapeutics, and pharmacokinetics of topiramate” CNS Neurosci. Ther. 14:120-142. DOI: 10.1111/j.1527-3458.2008.00041.x
Sills, GJ, Leach, JP, Kilpatrick, WS, Fraser, CM and Brodie, MJ (2000) “Concentration-Effect Studies with Topiramate on Selected Enzymes and Intermediates of the GABA Shunt” Epilepsia 41 (Suppl. 1):S30-S34. DOI: 10.1111/j.1528-1157.2000.tb02168.x


text:
How is silicic acid formed?
My book gives the following reaction:
$$\ce{SiCl4 + H2O -> HCl + Si(OH)4}$$
Since equilibrium favours the formation of weaker acids. How is this reaction able to proceed if a strong acid like HCl is being formed? Am I missing some point here? Or is my interpretation or equilibrium in this case is faulty?

Rather than strong versus weak acid, it is affinity Si-Cl versus Si-O.

Related: here, here, here, here.

"Since equilibrium favours the formation of weaker acids." Not a very fundamental rule. Also SiCl4 is a strong Lewis acid.

Thanks for the replies, so what i am to basically look for is the overall stability of the components and the favourability of one substance over another in a thermodynamical sense

A good exercise would be to find out enthalpy/free energy of this reaction (using standard enthalpy of formation of these compounds) and you'll most likely notice that it's negative

As this video clearly confirms, the hydrolysis of SiCl4 is highly exothermic from breaking of chlorine bonds. The solution boils and solid insoluble Si(OH)4 is produced.
Also, per Wikipedia on the hydrolysis, to quote:

The reaction can be noticed on exposure of the liquid to air, the vapour produces fumes as it reacts with moisture to give a cloud-like aerosol of hydrochloric acid.[5]

As such, the balanced reaction is likely best expressed as:
$\ce{ SiCl4 (l) + 4 H2O (l) ⟶ 4 HCl (g) + Si(OH)4 (s)}$
where the gaseous and solid precipitates formed can push the reaction to the right.
There is an interesting comment in Atomistry, to quote:

Consider, for example, the two reactions:

$\ce{Ca(OH)2 + 2HCl → CaCl2 + 2H2O}$

$\ce{Si(OH)4 + 4HCl ← SiCl4 +4H2O}$

The difference between the metal and the non-metal is signalised by the direction of the reaction. This difference, however, is seldom absolute, and with elements of intermediate character it disappears. In the case of stannic tin, for example, the reaction is reversible:

$\ce{Sn(OH)4 + 4HCl ⇔ SnCl4 + 4H2O;}$

and stannic chloride can be prepared either by dissolving stannic hydroxide in hydrochloric acid, or by passing chlorine over heated tin or a mixture of stannic oxide and carbon.

So, part of the answer also relates to the non-metal characteristics of Silicon as well.


text:
Synthesis of Pt(II) complex using trans effect

By making use of the trans-effect in platinum(II) complexes, show efficient routes for the synthesis of all the possible isomers of $\ce{[PtCl(NH3)2(py)]+}$ (where $\ce{py}$ = pyridine) from the starting material $\ce{K2PtCl4}$.

So far, I've had problems with trying to substitute ammonia and py in the right places. Since $\ce{Cl-}$ is higher on the trans effect list than the two ligands I want to substitute with, I found that it's just not possible to do that.

So, exactly what possible routes have you contemplated?


text:
Electron configuration of non-hybridized carbon
I'm currently learning about orbital hybridization in carbon.
I see that carbon has an electron configuration of 1s2 2s2 2p2.
However, there are actually three p-orbitals in the second shell, namely px, py and pz, which could each hold two electrons. However, there are only two electrons to be distributed on these three orbitals. Which orbitals do they occupy? Since their energy levels are identical, is it random?
I also learned that the three p orbitals can be combined with the 2s orbital (sp3 hybridization) to form four new hybrid orbitals. It then makes sense that the four electrons available each occupy one of the four new orbitals (so that each of them holds one electron), because this way, they can have the greatest spatial distance to each other (I assume they "want" this as they repel each other).
However, how are the second shell electrons distributed in a non-hybridized carbon atom? Does such an atom even naturally occur or does it always hybridize?

Update: I found this question and this answer which both state that non-hybridized carbon has the configuration 1s2 2s2 2px1 2py1.
Why is this the case, and not, for example, 1s2 2s2 2py1 2pz1?

The hybridization happens when and only when a Carbon atom ¡s linked up to another atom.

https://chemistry.stackexchange.com/questions/105045/connection-of-term-symbols-with-specific-microstates-for-atomic-carbon

The choice of coordinate system is arbitrary, you cannot distinguish the axes and you can relabel then as they fit your choice.

@Martin-マーチン Thank you for your comment. I think it does answer my question; you might consider posting it as such. I than assume that saying that the $p_x$ and $p_y$ orbitals are filled first is just convention?

Unfortunately this is maybe an approximation to an answer. This actually requires a deeper dive into quantum mechanics. There is a lot to unpack, and I don't have the time for it to write it down. But someone else might. In the meantime I'm happy my comment helped a bit.

The answer comes in several layers, as Martin has alluded to. I'll try and give a short summary: each point goes slightly deeper than the previous one.

For an isolated atom, the labelling of the orbitals is arbitrary. That is to say, the $p_x$, $p_y$, and $p_z$ orbitals are all interchangeable. More formally, this reflects the isotropy of the system, which means that all directions are equal. There is nothing that can physically differentiate the $x$-axis from the $y$-axis or the $z$-axis.

You can still arbitrarily assign $x$-, $y$-, and $z$-labels to three perpendicular axes. If you kept these coordinate axes consistent between different atoms, such that every $p_x$ orbital pointed in the same direction, you should not expect to find in all molecules that the $p_x$ and $p_y$ orbitals are always occupied. Instead, on average, you would have one-third of them with the configuration $(p_x)^1(p_y)^1$, one-third with $p_xp_z$, and one-third with $p_yp_z$. Again, this symmetry stems from the fact that there is no physical factor that favours any of the three coordinate axes over each other.

The previous point assumes that you have made a measurement of the electrons, such that you can definitively say that they are in a specific set of orbitals.* But prior to this measurement, there is no reason that they should already be in that state. It's better to posit that an isolated carbon atom is in a superposition of those three states, i.e. $p_xp_y$, $p_xp_z$, and $p_yp_z$. My old question: What constraints are imposed on a wavefunction by the symmetry of the system? is somewhat related, except that it deals with boron (which has the same issue).

Having gone through this, there is still the question of why we write $p_xp_y$ when it's not really accurate. The answer is that it's far easier than dealing with a superposition of states (in (3)), and since space is isotropic, $xy$ is the same as $xz$ and $yz$ as long as you relabel the axes, which means that (2) is more of a technicality rather than a real "gotcha". Also, it's perhaps not worth overthinking this: once you form a bond, all of this ceases to have any meaning. In a molecule, electrons don't remember "where" they come from, so it becomes pointless to quibble over whether it was in the $p_x$ or $p_y$ orbital.

* You can't measure individual electrons in an atom, you can only measure their collective properties. Also, I'm skeptical about whether you can measure a state of $p_xp_z$ given that $p_x$ and $p_z$ are eigenstates of non-commuting operators, but let's leave this aside for now (or assume, perhaps, that we're dealing with $p_{-1}$, $p_0$, and $p_{+1}$, which are all eigenstates of $\hat{L}_z$).

That's a good answer, thanks for writing it up. I'm afraid I can't add anything else useful to it. So here's a fun fact: We often use the z-axis as the bond axis, which may lead to some confusion (see https://chemistry.stackexchange.com/q/57125/4945). Weirdly we also like to use the z-axis for empty orbitals (and lone pairs), like in carbocations (or anions), which may lead to even more confusing situations. Drag and drop, and mix and match. And boy, these things may become weirder ... As Ivan would say: so it goes.


text:
Does this general hybridization rule really apply?
So this rule is known about hybridization that:
if 4 groups (atoms/pairs of electrons) are attached to the atom -> sp3
if 3 -> sp2
if 2 -> sp
but isn't that true only to specific elements only like carbon/oxygen/nitrogen..?
If you take CCl4, C is sp3, but chlorine has 4 groups around it but it's still not sp3 nor hybridized, it's just an sp3-p bond.
Even for nitrogen like in pyrrole, nirogen has 4 groups around it but it's sp2 because of resonance.
but can someone make it clear? to which elements does that rule apply? what elements undergo hybridization? what elements can they ask about in college regarding that?
Thank you

It basically works for carbon. Even O and N have a number of exceptions.

This rule is valid for atoms which are in the center of a molecule like Carbon in $\ce{CCl4}$ , and not for outer atoms like here Chlorine in $\ce{CCl4}$.

what should i do about outer atoms? i mean nitrogen can be outer atom or chlorine... for instance in
H-CN, it's S-SP-SP

In the case of HCN, well, H is the easy one; it has nothing to hybridise. C follows the standard rules. In this case, N behaves much like C. It needs to form two π bonds, which means that it needs to have two p orbitals; that's only possible if it is sp-hybridised. The rule works here, but you already know there are a number of exceptions, mainly because of resonance effects. For things like halogens, I'd suggest to throw this out of the window, as whatever it predicts can be refuted by more sophisticated theories (see e.g. https://chemistry.stackexchange.com/a/76954/16683).


text:
Measured g-factor of a free-radical changing as a function of something. But I cant work out what?
I'm looking at EPR of a paramagnetic organic molecule in solution that has a very stable free-radical signature with gfactor ~2.0035. All good. Except when watching one particular experiment I saw that the g-factor was drifting [the zero-crossing was slowly moving to the right over the course of 24hrs, i.e. the g-factor was dropping]. The Vpp was not changing. I am happy to believe that the molecule was self-aggregating. I would be even happier if there was a theory that related g-factor to particle size (e.g. via tumbling rate) because then I could estimate aggregation by looking at the gfactor. And that is where I get stumped. Are there any theoretical papers that discuss rotation effects on organic free radicals?
Or maybe my EPR machine is just drifting :-(


text:
Does cyclooctatetraene also have a diradical anti-aromatic electronic state?
I read that the molecular orbital diagram of cyclobutadiene predicts that it should have a diradical nature as it has two unpaired electrons in two degenerate non bonding orbitals (one in each).
Why does this diradical state exist only in cyclobutadiene? Shouldn't other anti aromatic compounds, like cyclooctatetraene, also exist in a similar diradical state?

It predicts that if cyclobutadiene was square it would be diradical, but it's not.

This isn't a duplicate of Cyclobutadiene - Jahn–Teller effect or not?, although the answer there may be helpful.

And C8H8 isn't even nearly flat. There's just nothing antiaromatic about it.

Anti-aromaticity is often not taught very clearly. Let me start, then, by emphasising that this anti-aromatic diradical state should not be taken as a real thing. It is a purely hypothetical state that may arise if the molecule adopted the shape of a planar regular polygon (i.e. square for $\ce{C4H4}$, octagon for $\ce{C8H8}$).
Because of various reasons, this state never truly arises in real-life $\ce{C4H4}$ or $\ce{C8H8}$.

$\ce{C8H8}$ never adopts an octagonal geometry. See, e.g., Why is cyclooctatetraene non-planar...?
 or this image for a diagram of its conformation.

$\ce{C4H4}$ is not really square either; it is more rectangular.

So cyclobutadiene is not really special in this regard. Both $\ce{C4H4}$ and $\ce{C8H8}$ have the same status: if they were to have a regular polygon structure, they would possess a diradical electronic state.* However, neither of them have such a structure, so neither of them have such an electronic state.

* For advanced readers: even square $\ce{C4H4}$ does not really possess a diradical state either, as is explained in Cyclobutadiene - Jahn–Teller effect or not?. This is because the MO theory used to predict the diradical state is not sufficiently advanced. (Technically, the diradical state arises from Hückel theory, and the proof of its non-existence for square $\ce{C4H4}$ arises from post-HF theories, such as configuration interaction.)


text:
Does the process of chemically strengthening glass change its volume?
The process of chemically strengthening glass replaces sodium ions in the glass's surface with potassium ions through a process that uses a potassium nitrate bath solution. These potassium ions are larger than the sodium ions and therefore wedge into the gaps and create pressure on the surface of the glass which can make it 6-8 times stronger than glass that is untreated. My question is whether this process would be expected to cause the glass to swell or change volume.
I have conducted experiments with chemically strengthening blown glass globes and have measured a consistent volumetric increase in the diameter of the globes post process. I am trying to determine how to predict and estimate this change. I can't determine if this is the result of the added mass from the potassium ions or if the surface pressures themselves are stretching the material.
I have talked to a few chemical engineers and looked through textbooks and online journals but much of the research done on this topic seems to be proprietary. I am posting my question here in hopes of finding someone who can point me in the right direction with my research.
This is my first question asked in this Stack Exchange Chemistry Site. Please let me know if there is anyway I can improve my question.

So you know that there is a volume change in the near-surface region that leads to compressive stress. Knowing the volume change in unstressed material, and an estimate of the Youngs modulus, you can come up with the expected near-surface volume expansion.

related https://chemistry.stackexchange.com/questions/77249/what-makes-gorilla-glass-more-durable-with-each-generation

If I remember correctly, the technique of strengthening glass by ion exchange was discovered accidentally, when a glass maker noticed that preforms had shrunk noticeably after tempering... but I have not been able to verify that on the Web.


text:
Why is a third body needed in the recombination of two hydrogen atoms?
In the article Discuss. Faraday Soc. 1962, 33, 205, the authors say that in order to form molecular
hydrogen from collision to atom we need a third body to remove the excess energy. That is we have the reaction $\ce{H + H + M -> H2 + M}$.
I suppose the reason is because if we have the excess this energy will break the bond. If it this the case why the energy can not be transferred to the electron and so would have $\ce{H + H -> H2^*}$, where $\ce{H2^*}$ denotes an excited state of $\ce{H2}$?

You can use the \ce{...} macro to typeset chemical equations; it's easier and also provides the correct output (chemical symbols should be upright, not in italics). Check out my edit for some examples.

An excited state of H2 would still have the excess energy.

even at low temperature ?

Temperature is irrelevant. Or I'd better put it this way: the rest of the gas might be as cold as you like, but this one molecule that has just formed from two atoms is very, very hot.

what I am not understanding how this heat is produced

All right, speaking of the heat of one molecule is an abuse of language. Would it be better if I said that it has very high energy?

could you give a reference please

@amiltonmoreira - what is the binding energy of H2?

related https://chemistry.stackexchange.com/questions/14073/kinetics-of-interstellar-chemistry-applying-steady-state-to-formation-of-h%e2%82%82

Does this answer your question? How does the energy released during a bond formation typically manifest itself on atomic level?

https://chemistry.stackexchange.com/questions/111936/how-is-energy-released-when-a-chemical-bond-is-formed https://chemistry.stackexchange.com/questions/55914/how-is-heat-physically-released-in-an-exothermic-process

see https://chemistry.stackexchange.com/questions/140186/can-single-molecules-of-c-and-o2-react-in-isolation-and-if-so-how-will-momentum/140377#140377

Yes there is validity to the finding that two atoms of hydrogen form molecular hydrogen at very slow rates.  Review Gould, R.J., and Salpeter, Edwin E., Astrophysical Journal, Vol 138, pg. 393.  It is publicly available at: http://articles.adsabs.harvard.edu/pdf/1963ApJ...138..393G
The exact answer to your question is found on page 138 of that work, under the heading “II. Mechanisms for the Formation of Molecular Hydrogen.”
In space, molecular hydrogen is destroyed (by cosmic radiation) at rates that exceed its formation without some type of surface catalyst.  This conundrum motivated significant research to find the actual mechanisms of molecular hydrogen formation in natural settings.  More recent work can be found by Gianfranco Vidali, et.al., Journal of Physics: Conference Series:6, 2005:
https://iopscience.iop.org/article/10.1088/1742-6596/6/1/003/pdf
 This more recent work cites the Gould and Salpeter work. It appears to be still accepted as the explanation for the slow rate of formation of molecular hydrogen from atomic hydrogen.  According to this explanation, after ground state hydrogen atoms collide, the new hydrogen molecule will be in a triplet state, and it cannot decay into a singlet state because this is a forbidden transition.  A cursory search on the PES's of triplet states of diatomic hydrogen show that indeed these are shallow curves which would lead to rapid dissociation.

Indeed, without a way to release energy the two atoms are perfectly happy to go back to being separate hydrogen atoms, since that is what they were. They have to get rid of energy to be in a bound state.

The important point is that this triplet state, while being perfectly bound, has a positive formation enthalpy. That means it has a finite lifetime. It just breaks apart after a millisecond or a day, and the two atoms take the energy (+impulse) with them again.


text:
Probability of forming dihydrogen with molecular weight 3
During a lecture, my professor told us that the probability $P$ to form a hydrogen molecule $\ce{H2}$ with mass number 3 could be calculated out of the abundances, $\gamma$, of the isotopes of this element: $\ce{^1H}$ (normal hydrogen), $\ce{^2H}$ (deuterium), etc. So,
$$
\begin{split}
P[\ce{H2}] &= P[(\ce{^1H} \cap \ce{^2H}) \cup (\ce{^2H} \cap \ce{^1H})] \\
%
           &= P[\ce{^1H} \cap \ce{^2H}] + P[\ce{^2H} \cap \ce{^1H}] -
%
              P[(\ce{^1H} \cap \ce{^2H}) \cap (\ce{^2H} \cap \ce{^1H})]
\end{split}
\tag{1}$$
But since
$$(\ce{^1H} \cap \ce{^2H}) \cap\ (\ce{^2H} \cap \ce{^1H}) = \ce{^1H} \cap \ce{^2H}\tag{2}$$
and
$$P[\ce{^1H} \cap \ce{^2H}] + P[\ce{^2H} \cap \ce{^1H}] = 2P[\ce{^1H} \cap \ce{^2H}]\tag{3}$$
we end up with
$$
\begin{split}
P[\ce{H2}] &= P[\ce{^1H} \cap \ce{^2H}] \\
       &= \gamma(\ce{^1H}) \cdot \gamma(\ce{^2H}) \\
       &= 0.99972\cdot 0.00028 \\
       &= 0.0002799
\end{split}
\tag{4}$$
This is my result while for my professor, it should be twice that value. It's like he's forgetting the term
$$
P[(\ce{^1H} \cap \ce{^2H})\ \cap\ (\ce{^2H} \cap \ce{^1H})]
$$
Am I right?

I think you may need to explain exactly to us chemists what your notation is trying to express. I wonder, though, if this is also overcomplicating it. From your other posts on SE, you obviously know plenty of maths. This is nothing more than a simple combinatorics problem: if you draw a red ball with probability 99.97% and blue ball with probability 0.03%, what are the chances that you will get a red and a blue ball if you draw 2 balls.

It would be $99.97\cdot 0.03$ %, which is my answer because you'd be looking for $P[red \cap blue] = P[red | blue]\cdot P[blue] = P[red]\cdot P[blue]$ since $P[red | blue] = P[red]$. Now, if $red = ^{1}H$ and $blue = ^{2}H$, then my answer holds but not the one from my professor

The order doesn't matter. Not for the hydrogen molecule, anyway.

Yes, and that's why I wrote $P[^{1}H \cap ^{2}H] + P[^{2}H \cap ^{1}H] = 2P[^{1}H \cap ^{2}H]$ but you still have the term in Eq. (3)

Please do use symbols that may be understood by simple chemists ! What is the meaning of these (sometimes squared) round bridges ? What is gamma ? How do you obtain 0.056% by multiplying 0.9997 by 0.028% ? A factor 2 is missing ! Is it due to these strange squared bridges ?

There's no 2 over the round bridge, it's latex edition. That 2 is an upper index for $H$ of mass number 2 (deuterium). And yes, I added a 2 factor by mistake, I was copying my professor result by mistake. Gamma symbol is explained in the post, please read it before asking about notation introduced in the first lines

@Maurice Vicky uses set symbols describing probability. $A \cup{} B$ describes that either event $A$, or event $B$ takes place (addition of individual probabilities). $A \cap{} B$ describes if events $A$ and $B$ coincide.  E.g., $A \cup{} B = \varnothing$ states both events are incompatible.  You may state $A \cup{} \bar{A} = \Omega$, the event of either $A$, or the opposite of $A$ is sure to happen, probability = 1; and you may state $A \cup{} \bar{A} = \varnothing$ because coincidence of  $A$ and the opposite of $A$ is zero. Maybe you are more used to $\vee$ and $\wedge$.

@Vicky I understand your problem editing two concepts at once.  The irritation from the chemists side is that the standard mathematical notation about the probabilities  partially overlaps with the chemical notation of superscripts denoting the two most abundant isotopes of hydrogen.  Plus that the first expression spans over multiple equal signs as a one-liner (which often is not helpful at all).

@Vicky: For future reference: please familiarize yourself with MathJax ChemSE uses in general (https://chemistry.stackexchange.com/editing-help#latex) and mhchem in particular (https://mhchem.github.io/MathJax-mhchem/) which users of ChemSE use to format chemical formulae (including isotopes!) with less struggle in questions, answers, and comments.  Note, however, because MathJax and mhchem are special (= not universally understood by all interfaces), you shall not use them in the title of the questions / answers.

I think equality (2) is incorrect - shouldn't it be zero?

Why not writing D instead of $\ce{^2H}$ ? In that case, the problem gives the following answer :  $\ce{P[HD] = 0.00056}$

Extending / reformulating Karsten Theis' answer: A probabilistic approach is to draw a tree about drawing two spheres where event $A$ (normal hydrogen, $\ce{^1H}$) dominates, but event $B$ (the next abundant isotope, $\ce{^2H}$) complements, $P(\ce{^1H} \cup{} \ce{^2H}) = 1 = \Omega{}$ and eventually to multiply the probabilities for each path (e.g., $P(\ce{^1H^2H})$ as shown on the left hand side).

Yet because in your case both paths $P(\ce{^1H^2H})$ and $P(\ce{^2H^1H})$ yield a dihydrogen molecule of mass 3, these individual probabilities need to be summed up.

You are wrong. There are only four cases (first atom is proton or deuteron, second atom is proton or deuteron), and you have to add up two cases to get all molecules with mass number of 3, i.e. the two light blue areas in the schematic diagram below (not to scale):


text:
What can be used as substitute to dichloromethane for breaking optical cement bonds?
An optics supply company recommends using a dichloromethane solution to break bonds between optical elements joined with its UV-cured optical cement.1 The page mentions that dichloromethane-based paint strippers also work. Use of dichloromethane in paint strippers has been phased out in the USA  as of last year. Is there an obvious chemical property that dichloromethane possesses which causes it to break UV cement bonds and if so are there any more readily available hardware store chemicals which could be used as substitute?

Does not optics deserve pure chemicals ?

I would think so too but I am sure the manufacturer itself would understand. It is also probably true that such strong solvents also do away with any coatings on the lens elements.

Acetone or Tri/tetrachloroethylene.  Chloform would be probably the same case as DCM.

If you follow the supplier instruction is fine. Otherwise, optical element it is a vague term. There are optical elements which can be washed in organic solvents and other which will be ruined. In your case is probably lenses (or anyway metal, or inorganic glasses) and you should do what they said. I do not think that chlorinated solvents are banned for every use. Try also acetone anyway, as Poutnik said.

Write them an email asking for a non-chlorinated alternative? They know about the problem.

Chlorinated solvents or acetone are usually ok. AR coatings are normally inorganic materials such as MgF2 so should be ok with solvent. The solvent may leave residue on the lens/mirror if not pure, clean this off with iso propyl alcohol, use a proper lens tissue not any old tissue as these will scratch at microscopic level. Drag tissue across surface, do not 'scrub'.

If acetone doesn't work so well, cyclohexane or petrol ether / benzine are the next guess.

Just because dichloromethane has been phased out in paint stripper (it was never used in paint thinner and I'm surprised it has been phased out) doesn't mean you can't get the pure chemical. It is fairly safe in small amounts and probably safer than alternatives like acetone.

@matt_black I think the way that comment is written could cause a false sense of security. DCM is a volatile, permeable carcinogen. It can be used safely, but only if its dangers are respected.

@electronpusher dCM is not a carcinogen in people. It was suspected to be a human carcinogen (because its congener chloroform clearly is) and widely tested. But isn't. The major hazard of DCM is from displacement of air in confined spaces leading to suffocation (plenty of accidents from that but you need a large amount of DCM). Toxicity is not a major hazard for users of small amounts who treat it with caution.

@matt_black It is also suspected to be a human carcinogen because it has shown to be carcinogenic in mice, rats and hamsters. Human data is insufficient for a full conclusion, but the current recommendation is to classify DCM as a suspected human carcinogen (category 2). I would be interested to see what evidence informs your claim that DCM is not carcinogenic in humans (are you so sure there's enough data for such a bold claim of safety?) https://pubmed.ncbi.nlm.nih.gov/33387565/

@electronpusher The link between animal studies and human carcinogenicity is suspect because some of the key metabolic pathways in rodents are not found in people. actual human epidemiology studies are weak and inconsistent with no clear conclusions. I can't claim DCM certainly isn't a problem (but I can't claim that for most chemicals either). If it were a notable problem the results from human exposure studies would be stronger and clearer than they are.

According to an email i got back from Norland, "It will work if the paint stripper has a chlorinated ketone in it"

Careful here most simple chlorinated ketones are good lachrymators aka tear gas.


text:
How does flocculation in paint work?
I recently learned that you can clean up latex paint wash-up water by flocculating it: adding aluminum sulfate and hydrated lime. I vaguely remember how flocculation works in clay slurries (change in pH changes the electrostatic charge on kaolin plates, causing them to attract and then precipitate). Does flocculation in paint work the same way? (Is it exactly the same? Is there kaolin in paint?) What exactly is happening when you add aluminum sulfate and hydrated lime to the paint?


text:
Help on Research Proposal
I am a freshman majoring in chemistry. I am aware that one of the tasks required in this field is extensive research, and I am encountering a big problem. I am lost as to how to think of a plausible paper— I am not sure how to think of proper "topics" that would suffice the requirement in this field. I am once advised to look for societal problems and how my field can potentially propose solutions to it, but as of now, I'm still blank to any topics and/or ideas. I hope any of the people here can guide me through this so that once that I'm actually assigned to write, I can think of a topic without any problems.

A few suggestions: antibiotic resistant pathogenic bacteria, carbon capture technology, fossil fuel replacements, novel antivirals, the impact of CRISPR techonlogy on genetic manipulation.

Perhaps this reference will give you some ideas to explore, see Zou, P.; Leddy, J., 'Magnetized Nickel Electrodes for Improved Charge and Discharge Rates in Nickel Metal Hydride and Nickel Cadmium Batteries', Electrochemical and Solid-State Letters 2006, 9, A43-A45. To quote: "magnetized electrodes yielded higher currents than the corresponding nonmagnetic electrodes by 70 and 230 %, respectively." at https://www.electrochem.org/dl/ma/206/pdfs/0541.pdf .

You're posing a question that many advanced graduate students have trouble answering...