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How is it Possible to Measure Electrode Potential?
It is known that potential difference on metal/electrolyte or any kind of interphase can't be measured since voltmeter needed to measure it is made of metal which would in contact with metal and solution develop its own potential difference at interphase and as such affect the measurement. To solve this problem, potential is measured with respect to SHE. What I don't understand is that if you want to measure the electrode potential with respect to SHE, voltmeter needs to come in contact with two metal electrodes (one is electrode which potential you want to measure and the other is SHE) which will again create potential difference between the electrodes and voltmeter since two different metals in contact create potential difference between them. Can you explain?

When two metals are in contact, they are at the same potential

Seems well explained here: https://en.wikipedia.org/wiki/Galvani_potential.

@Maurice  Thermoelectric Seebeck effect is based on changes of this non-zero potential difference with temperature.

Generally, potential difference measurement and its errors and biases belong to domain of metrology, in spite of being essential to many domains of physics, chemistry and electronics.

It is the same as measurement of voltage difference at any electronic circuit. The fact the metallic measuring points are attached to electrochemical electrodes is not relevant here.
Yes, there is metal contact potential difference, but it very low and very probably mutually canceling.

Yes, maybe potential differences developed between different metals isn't too big so when comparing two redox systems these cancel each other. Another solution to this problem may be using voltmeter with probes made of same metals as electrodes, so that no potential difference develops.

Even then, it is not related to electrochemistry, but rather to electronics or metrology.

I wouldn't agree since measurements of electrode potentials are very important in electrochemistry

It is important, indeed. But mentioned issues have nothing to do with electrochemistry and are very minor. The measured voltage offset is much less than the measured voltage. The same metal would not help, as different metal contacts would just move elsewhere withing the measuring circuit.

Yes, I  don't agree with what you said in the beggining for two reasons: 1) Contact potential differences of metals are actually electrochemical in nature, 2) Measurement and determination of potential differences on electrode/electrolyte interface is one the most important problems in electrochemistry and as such it is important to have good understanding of this problem. I agree with you that this problem is not something crucial since as you said these potentials are small compared to what you measure.

If I attach a piece of gold to a piece of platinum, most of chemists would not say it is about electrochemistry. Rather physics of solid phase. Electrochemistry is meant in context  involving electrolytes.

Yes, that may be true, depends how we define electrochemistry. In my opinion, we shouldn't really bother ourselves with definitions, but it is important to develop understanding of phenomena. Regardless of how we think of potential difference between metals, it is a phenomenon which is described in many electrochemistry textbooks since it has its influence on measurements of electrode potentials. Electrochemistry isn't only about electrolytes, in my opinion, there is more about it which you can see if you read some interesting books on it like Bockris's Modern Electrochemistry.

It is measured with the help of a reference electrode known as the standard hydrogen electrode (abbreviated to SHE). The electrode potential of SHE is 0 Volts. The standard electrode potential of an electrode can be measured by pairing it with the SHE and measuring the cell potential of the resulting galvanic cell.


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How to determine the number of π-electrons in a conjugated system from UV-Vis data?
Is there a way to determine the number of π-electrons in a conjugated system from UV-Vis data? I am modelling the electrons as particles in an infinite box of length $L$ equal to the length of the conjugated system. I am unsure how to interpret the spectra to obtain the number of π-electrons, including the systems with more than one peak.

I don't know if you can determine the number of pi-electrons from UV-vis, because UV-vis is basically the HOMO-LUMO transition. In some cases, there are additional bands, but they might come from sigma or non-bonding electrons being excited. So, it's hard to say what the total number of pi electron is, even if the peak comes from pi->pi* excitation. Aside from that, particle in a box is very inaccurate quantitatively, so I don't know what you are trying to achieve with that.

Yes and no. I might come back to this but it depends on the level of prediction you have in mind and on the classes of compounds. In all cases a good amount of knowledge is required as starting point. The peak to look at would be that associated with the pi-pi* transition. You can right "when the particle in a box is somehow a decent model".

I think the point of the exercise I’m doing is to show how inaccurate this is. How would I go about finding the number of pi electrons? I just don’t see it with this theory

Not really. A uv-vis spectrum normally contains peaks due to vibrational transitions on top of an electronic transition, but your particle in a box calculation only produces electronic transitions and in any case is a very crude approximation. If you can see two or more  electronic transitions and work out the transition energy that would be better. So you can only have an inspired guess at best, not a proper prediction.

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Flexible urethane additives in epoxies
For those who understand plastics and epoxies in particular, curious to know if I can add a correctly proportional amount of flexane urethane to epoxy paint without worrying about compatibility issues. IOW is there any non-typical or particular ingredient in epoxy that might prevent successful mixing of the two?
For clarity, I am using epoxy paint for a waterproof top coating on my rv roof, but would like to ensure it is a bit rubberized and not too brittle. Devcon flexane addition seems to be the answer but I can't find any guidelines for the DIYer.
Please go easy if this question is too general in the area of chemistry, seems like this is the correct forum, and thanks for any info. If you have personal experience working in the chemistry of urethanes, you may have valuable knowledge on this.
Product links:
https://www.amazon.com/Olympic-Poxolon-Two-Coat-Epoxy-Swimming/dp/B07BLMHKBK
https://www.reynoldsam.com/product/flexer-epoxy-flexibilizer/

Links to what you wanna use? Anyway, it doesn't sound like a good idea.

Added the links, thanks for pointing out and for the input. I'd like to know if it's chemically fine or a gamble to do this. Seems it could reasonably go either way.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Why do electrons have to constantly be moving?
As a pretense, my understanding of chemistry extends only to a high school honors level.
As I know it, electrons occupy orbitals in which they could possibly be situated, and constantly move around those orbitals by their attraction to the nucleus and repulsion to each other.
Why is it that this specific arrangement should happen? Wouldn't it make more sense that, since they are attracted to the nucleus and repulsed by each other, they find some sort of arrangement where they are all static and the same distance away from the nucleus and an equal distance away from each other (IE, they form one big s-shaped orbital and occupy it such that they are all equally as far away from each other as possible while also being equally as close to the nucleus as possible)?

It is all very different from what you are imagining, and different from its opposite, too. There is no motion, no arrangement, not even distance. Trust me, there is no point even to think of it before you start learning real quantum mechanics.

Is chemistry about what makes electrons, well, electrons? Not really...

Ignore Ivan's comment, in which he's effectively saying "don't worry your pretty little head about it" until you're taking upper-div QM. He's essentially saying it's all so mysterious and over-your-head that, as a h.s. student, there's no point in even trying to think about QM. That's both condescending and patently false. There are many ways that h.s. students can start developing an intuition about QM, using analogies like standing waves on strings. So don't let comments like this discourage you.  It's a disservice for this site to be telling curious students to put their curiosity on hold.

For instance, here are youtube videos in which experts show that both higher dimensions and quantum computing can be explained at multiple levels, from child to teen to college student to grad student to expert: https://www.youtube.com/watch?v=3KC32Vymo0Q and  https://www.youtube.com/watch?v=OWJCfOvochA

According to Earnshaw's theorem, a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges.

I want to be sure that you get the answer and that is the comment above (Poutnik's one).

The uncertainty principle and the impossibility of stable static structures
I could say "it's all quantum stuff, wait until you are taught that" but there are good ways to grasp the key ideas without all the highly mathematical detail that comes with that level of understanding.
Two things matter. Even if we assume no quantum effects there is no arrangement of static electrons that can be stable. There is a mathematical derivation of this (as @Poutnik points out this is Earnshaw's Theorem) but the intuition is similar to the observation that a planetary system with no motion can't be stable (though gravity only attracts). Or the observation that a mechanical system with a rigid rotating object (say a hula hoop) won't stay stable unless there is both pulling a pushing from a central point (it looks stable until there is any fluctuation away from a perfect circular orbit when the fluctuation is amplified by the pull to the centre).
The second thing that matters is the Heisenberg uncertainty principle. It is possible to understand the implication without needing the grasp the math. In simple terms there is a limit to the precision of knowledge about both momentum and position: if we know exactly where something small is our knowledge about how fast it is moving must be very imprecise (strictly speaking this is more than a limit on our knowledge: it is a fundamental limit on reality). We can't have static electrons: if we knew where they were, they would be moving at some indefinite speed. The implication of the math is that only certain combinations of momentum and spatial coordinates are possible (we call them orbitals and they represent regions of probability of finding electrons). Like a vibrating string, only certain specific modes of vibration are possible. Unlike a string there are no modes where everything is static.

So what you're saying is that there's no use visualizing it, it's based off of theoretical reasoning and math?

@Paulemic there are some physical analogies as to why some systems can't be stable. But heisenberg stuff doesn't have many good analogies on the macro-scale (unless you understand fourier transforms in which case vibrating strings illustrate the trade-off between frequency specificity and time resolution well: a string has to vibrate for an infinite time to have a very precise frequency; a short time slice inevitably has a fuzzy frequency in a way very similar to the momentum/location tradeoff for quantum systems)...

@Paulemic ...but it is wrong to say it is all about theoretical reasoning and math. This is what we observe to be true for very small things. The theory is subordinate to the observations.

Wouldn't it make more sense that, since they are attracted to the nucleus and repulsed by each other, they find some sort of arrangement where they are all static and the same distance away from the nucleus and an equal distance away from each other (IE, they form one big s-shaped orbital and occupy it such that they are all equally as far away from each other as possible while also being equally as close to the nucleus as possible)?

What you are suggesting here is a static equilibrium of many particles. It would be similar (but not equivalent to, because of all sorts of issues) to a pebble tower or balancing a razor-thin object on its razor-thin side: inherently unstable. As soon as you accidentally as much as move a nucleus further away, the entire system would no longer be in equilibrium as the electric field changes; the first particle would start moving, changing the electric field again and so on until complete chaos.
This isn't a stretch of imagination either: with every movement my fingers are making, tons of atoms are moving closer to and further away from each other meaning any type of static equilibrium wouldn't last a femtosecond.

According to the Heisemberg principle, it is not possible to know with an arbitrary precision the values of two observables whose operators do not commute.
According to QM, every physical property, like energy, is associated to an operator. An operator is not black magic, it is a mathematical being that does things, like the addition operator $+$.
Two operators $\hat{A}$ and $\hat{B}$ do not commute is $\hat{A}\hat{B} \ne \hat{B}\hat{A}$. Like with matrices.
If you do some paper and pencil calculations, momentum, therefore velocity, and position of a particle are observables whose operators do not commute. Then, from a micorscopic point of view you cannot know at the sime time both velocity and position of an electron. In fact, an orbital is a probabilistic concept: it is the space region in which the electron has the 95% probability to be found. The "continous motion" of an electron is just a consequence of the Heisemberg principle.
I suggest you the magnificient divulgative book of Paul Halpern "The quantum labyrinth".


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Solvent + Photopolymer Waste Disposal
I have a 3L jug of SLA printer-related waste that's been sitting under my desk for a while now and I'd really like to finally figure out how to get rid of it. It contains:

Ethanol (primarily) and isopropanol, saturated (in the literal sense) with uncured resin (photo-polymers, photo-initiators, binders). Note there is likely very little water left (started with 90% ethanol and 99% iso, sieve-dried multiple times).
About 10% by volume undissolved waste, a slimy gray slurry of: uncured resin, a significant concentration of glass fibers, and some uninteresting filter waste (charcoal, sodium bentonite, probably some sieve dust, etc.)

Also it smells awful when unsealed, numerous recyclings through filters have left it disproportionately saturated with god knows what. It smells like ... pine and cancer.
How do I get rid of this? Is there some way I can process it myself to make it environmentally friendly? The uncured photopolymers are the worst offender, I believe. Plus the whole thing is very flammable.
Location: NYC (Kings County), NY USA
Also: I don't mind going through the effort of processing it myself if it keeps it out of the ocean or some landfill somewhere. Plus, I'll be continuing to generate this waste in the future (low volumes; probably about 1L per month at the highest rate) so I don't mind investing in a processing setup.

Normally you confer wastes to an appropriate service/company

In which country are you living ? In my country, such wastes are sent to waste collection centres which are organized in every little town.

@Alchimista Do they generally accept such small amounts? When I think of disposal services for some reason I always think industrial scale.

@Maurice I've added location info. NYC, NY USA.

@Alchimista Also; do disposal companies usually dispose of things in some responsible way, or do they end up being a USCSB case study some day (lol)? I don't mind going through the effort of processing it myself if it keeps it out of the ocean or some landfill somewhere. Plus, I'll be continuing to generate this waste in the future (low volumes; probably about 1L per month at the highest rate) so I don't mind investing in a processing setup.

@Jason C. In my country, Switzerland, waste materials are burnt in big waste collection  centers, and the heat produced is used for heating the houses and buildings in the vicinity. I am pretty sure it should be more or less the same in New York, NY. They may have thrown away household wastes down to the ocean. But it is simply not possible to still do it today.

@JasonC even university department find more convenient to pay for regulated disposal. The volume / mass are considerably more than your projected one. Of course, we have to trust our institutions...For the scale they do accept I have honestly no idea. There might even be a city run service to which you physically go when amount are so small. In all cases you have to look for rules in your county so you will find out how to proceed.


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Why are electrons pictured to be revolving around the nucleus in Bohr's model
Why exactly was it guessed by Bohr that electrons do revolve around nucleus in orbits than stay stationary around it in some discreet locations corresponding to a specific energy, or go around some other point or path. I believe stationary electrons in the electric field of the nucleus should also posses energy.

... and would accelerate towards each other until collision?

I took the liberty of changing the tag to "history of chemistry", because that's the only context in which Bohr's model is still relevant. With that in mind, let's move on.
Sure, a bunch of stationary electrons would have energy and all. What forbids it is the fact that a static configuration of charges can't be stable (that is, it is guaranteed to fall apart if disturbed). That's a mathematical result known as Earnshaw's theorem. So it was obvious to Bohr that the electrons must be moving in some way or another.
Now what about the orbits? Well, a single particle in a bound system with Coulomb-like potential can only go in circular (or, more generally, elliptical) orbits. It was known for centuries thanks to Newton and celestial mechanics, which just happens to use a potential of the same form. None of this was left for Bohr to guess. He used a well-known kind of motion in a different setting. (Elliptical orbits appear in the so-called Bohr–Sommerfeld model, but that's not much of a thing.)
Well, now what if we have more than one electron? Nothing. Bohr's model does not describe such systems.
So it goes.

RE: changing the tag to "history of chemistry", because that's the only context in which Bohr's model is still relevant. -- That seems a bit harsh. I'd assume that the Bohr model is still used to introduce quantization in a general chemistry course along with atomic line spectra. Then that model would be further built on.

I'd put it differently. The Bohr model is used to mentally break the students and thus prepare them for the worst. Then it is discarded and never further built on. The real quantum theory is then built from scratch.

On the contrary, I think the Bohr model can still give us useful basic intuition. As an example, Batty uses it as an effective theory to describe exotic atoms at CERN "We now discuss, using the classical Bohr model, the general characteristics of antiprotonic-hydrogen atoms". (C.J. Batty, Rep. Prog. Phys. 52 (1989), p1169)


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Why should the conical flask (apparatus on the right) used in Fractional Distillation be left open (without a bung)?
With context to the image shown, is there a reason as to why the conical flask shouldn't be closed with a bung and is there a way to describe this error?

Who said  the receiving flask should be closed? What "error" are you talking about? Sure, for the distillation of, say, nitric acid, the receiving flask is connected with an adapter and a gas outlet, but there is no need to do so for the non-volatile fractions.

Also, you forgot about attributing the illustration.

There needs to be a release in the system somewhere otherwise you are heating a closed system and that is not a good idea. Another problem is using a Bunsen burner for this.

If you want to find "errors" one is the lack of a fire spreader or a bath and perhaps the lack of some piece of glass or pomix within the liquid to be distilled.

If a bung does close the flask, the pressure inside the apparatus will grow. The bung will soon or later explode.

@Maurice - well, you hope the bung explodes and not the glassware...

Many of the details in the that diagram are obsolete but the lack of a bung is not one of them
This is probably a very old illustration. Many aspects have been uncommon for decades. Bunsen burners are rarely used in labs now (too unsafe where volatile solvents are present). Most modern apparatus uses ground glass joints not rubber bungs for connections between components. And the specific types of flasks are irrelevant (it doesn't need to be a conical collector flask, for example).
But the lack of a bung on the collector is important whatever type of flask is used. The reason is simple: you don't want pressure to build up. The diagram shows a fractional distillation at atmospheric pressure and that means you have to leave the system open to the atmosphere. A bung would prevent that meaning that, as the system was heated, pressure would build, the distillation might not even happen and, worse, something would explode.


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What signal is measured at the detector in atomic absorption spectroscopy?
I understand that a hollow-cathode lamp (HCL) with a specific target is used to excite the sample. The atoms in the sample absorbs the photons and re-emits at the same wavelength and the signals are measured by the detector.
What about the signal from the source? Let's say only 20% of the energy from the source are being absorbed, wouldn't the other 80% also travel through the flame to the detector?

You raised an excellent point, which is not as simple a UV-Vis spectrophotometry because most molecules do no fluoresce or emit light, whereas atoms do. This issue makes an atomic absorption experiment slightly more challenging (at least in principle). The critical question is "what" signal reaches the detector. How does one distinguish the contribution of flame emission from the hollow cathode lamp (HCL) beam? Last but not least, is there is a way to distinguish atomic emission due to HCL?
Have a look at a typical AAS instrument. The critical component is the chopper, which is nothing but a rotating mirror with holes. It merely blocks the light from reaching the flame while rotating at an exact frequency.

You can see only two signals are needed in an atomic absorption spectrophotometry experiment: a) Light beam intensity when there are no atoms in the flame and b) light beam intensity when analyte atoms are present in the flame.

You aspirate a blank (no analyte solution) into the flame- this sets the 100% transmission or absorbance = 0 at the chosen wavelength. Ignore the reference beam; it is for ensuring that the lamp is stable. We are only interested in the beam passing through the flame.

Next, you would aspirate your analyte solution into the flame. Let us take Na as the most straightforward example. There is light absorption by Na atoms in the flame, leading to a reduction in the HCL beam's light intensity after passing through the flame. The ratio of light intensity in steps 2 and 1 gives the transmittance, which is easily converted into absorbance.

However, there is an almost continuous emission spectrum from the flame too.
Here is the smart idea. I do not know who came up with it. The detector continuously sees the background emission from the flame. However, the light beam from the hollow cathode lamp reaches the detector alternately at a particular frequency due to the rotating chopper. The electronics are tuned to register only the light, which is oscillating due to the chopper

The critical but no so critical problem: When atoms absorb a specific wavelength, they emit precisely the same wavelength as you noticed in your post in the flame (called atomic fluorescence). the chopper cannot stop that light. Hence it is part of the transmitted light beam.

The good news is that atomic fluorescence is not directional, unlike the light beam from the HCL. Secondly, the intensity of atomic fluorescence light is so small compared to the HCL beam intensity that it is practically negligible. There is no way to prevent atomic fluorescence from reaching the detector in an atomic absorption experiment, to the best of my knowledge.
Can we see stars during daylight? Stars are always there and "part" of the sunlight. However, stars contribute very little to the daylight. In the same way, atomic fluorescence contribution in an atomic absorption experiment is almost null because the hollow cathode lamp beam is quite strong as compared to the atomic fluorescence in random directions.

Another point that should be made is that when the sample atoms emit photons there is no preferred direction. So the emitted photons are emitted spherically and only a small fraction will be oriented towards the detector.

There is measured the relative portion of the passed, not absorbed, eventually modulated light. It is the same principle as measurement of absorbance of coloured solutions by classical UV/VIS spectrometer.
$$I = I_0 \cdot 10^{-A} = I_0 \cdot 10^{-\epsilon \cdot c \cdot l}$$
The difference is the atomic absorption bands are very narrow and intense, so measurement is very selective and sensitive.
Re-emission or fluorescence occurs, but as it is omdirectional, it is negligible in directions of the primary ray. Additionally, fluorescence happens at not measured wavelength.


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Reaction between phosphorus(III) bromide and alcohol
Alcohol acts as a nucleophile in a reaction with $\ce{PBr3}.$ It attacks the $\ce{PBr3}$ leaving $\ce{Br-}$ behind. The alcohol forms a bond with the $\ce{PBr2+}.$ How the oxygen having three bonds makes the hydrogen more acidic?

Think of $\ce{H3O+}$.

You may also realize that the $\ce{P}$ atom "prefers" to make a $\ce{P-O-H}$  bond than a $\ce{P-Br}$ bond, because $\ce{P-O}$ bond dissociation energy is bigger ($\pu{351 kJ/mol}$) than $\ce{P-Br}$ ($\pu{272 kJ/mol}$)

@Maurice Can you please explain about it further? I could see the logic in your answer but a deep explanation would be better to understand more clearly.

See: https://chemistry.stackexchange.com/questions/26996/why-does-phosphorus-tribromide-act-as-a-lewis-acid-electron-acceptor ... https://chemistry.stackexchange.com/questions/51891/does-halogenation-by-phosphorus-tribromide-lead-to-an-inverted-product/51894


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Effect of coffee brew water concentration on further extraction in immersion brewing
In immersion brewing coffee is extracted with an increasingly saturated solution, as opposed to percolation brewing, where the extraction is performed with a continous supply of fresh water.
Because water can only extract a certain amount of solubles from the coffee and the fastest extracting compounds are dissolved first, I wanted to ask if the increasing concentration of fast extracting compounds (caffeine, volatile oils) hamper the ability of the remaining brew water to dissolve slower extracting components of the coffee bean (e.g organic acids).

I doubt that this would be the case, but it is not impossible. Consider the following from the Wikipedia (unfortunately unreferenced): " In Turkish coffee, the extremely-fine grounds are not removed and left suspended in the final beverage – resulting in an extraction of 100%, which would be considered over-extracted by Western standards. "

OTOH see the part in the Wikipedia article (https://en.wikipedia.org/wiki/Coffee_extraction) on extraction from espresso


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How do you simplify for Q+w=m(h1-h2)?
So, I'm given the following equation:
$$\frac{∂Q}{∂t}+\frac{dm_i}{dt}(E_i+P_iV_i)=\frac{dE}{dt}+\frac{dm_{fi}}{dt}(E_f+P_fV_f)-\frac{∂W_s}{∂t}$$
Using that equation, I am supposed to then simplify it into
$$\overset{.}Q+\overset{.}W=\overset{.}m(h_1-h_2)$$
However, I don't know how to start, and so far I haven't found anything similar that I can use as a basis.

It wouldn't hurt if you define every variable in the posted equation, too. I'm pretty sure you are using inconsistent labelling, e.g. with "f", "2" and "fi" for "final", and the derivative shouldn't be a variable. A general note: in chemistry and physics, boldface usually refers to a vector.

Please rewrite ! Write $E$ instead of $e$, $ H$ instead of $ h$. Write $dt$ instead of $d_t$. Write the indexes as proposed by Andselisk. Explain what is $m$ ! Write $dE$ instead of $E_{t+dt} $- $E_t$ Explain why the mass and the work are supposed to change in time.

The first equation is the general form of the open system (control volume) version of the 1st law of thermodynamics.  The 2nd equation is what this reduces to under a certain set of circumstances.  In determining how this comes about, I would consider:

Steady state versus transient operation

Expression for e in terms of specific internal energy u, kinetic energy per unit mass, and gravitational potential energy per unit mass.

Importance of internal energy change relative to kinetic energy change and gravitational potential energy change.

Expression for h in terms of u and Pv.

The need for a subscript of S on the W of the 2nd equation, and what this means physically.


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What are the additional chemicals required to make polyurethane foam more viscoelastic?
I am currently experimenting with polyurethane forms for a science and research project I am working on. I need to make the polyurethane foam more viscoelastic, almost like memory foam. I am unable to proceed and would be great if I could get some guidance.


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Relative acidity of acetic acid derivatives with strongly withdrawing substituents

What would be the order of $\mathrm{p}K_\mathrm{a}$ values of the following:

$\ce{O2N-CH2-CO2H},$
$\ce{Me3\overset{+}{N}-CH2-CO2H},$
$\ce{NC-CH2-CO2H},$
$\ce{HO-CH2-CO2H}?$

Since only the groups attached to $\ce{-CH2-CO2H}$ are changing, we must look at the point of difference in each group and I was able to figure out that since there is no conjugation, we cannot compare these groups on the basis of the mesomeric effect.
So, we must compare them on the basis of negative inductive effect as the more stable the conjugate base will be, the more acidic our compound is and hence the lesser is its $\mathrm{p}K_\mathrm{a}$ value. But I am not able to come to a final conclusion which matches with what's given in my book.
Is there some other factor playing role in this as well? Because if we just look at it on the basis of inductive effect, the order for increasing values of $pK_a$ should be
$$\ce{Me3\overset{+}{N}-CH2-CO2H} < \ce{O2N-CH2-CO2H} < \ce{NC-CH2-CO2H} < \ce{HO-CH2-CO2H},$$
which seems to be wrong.

The order for the increasing $pK_a$ values given in my book is:
$$\ce{O2N-CH2-CO2H} < \ce{Me3\overset{+}{N}-CH2-CO2H} < \ce{NC-CH2-CO2H} < \ce{HO-CH2-CO2H},$$
Note that the order given in my book may be subject to errors.

EDIT:
After pondering over this for a lot of time and with no convincing answers or explanations, I finally came to a potential explanation for this. Could Hydrogen bonding be the reason for the compound with $-NO_2$ to be more acidic than the one with $-Me_3\overset{+}{N}$ group ?

EDIT-2:
Thanks to Buck, it's now clear that my book's answer is right. Could someone please give a possible explanation to this from organic point of view?

Hmm, maybe the Me3N+ group draws in a counterion? I am not sure why the NO2 would be more electron withdrawing than a positively charged group. I am confused too

I assume induction effect of $\ce{(CH3)3}$  would diminish the effective nitrogen charge.

@Poutnik Hmm... But according to https://en.wikipedia.org/wiki/Inductive_effect#Relative_inductive_effects  $Me_3\overset{+}{N}$ is one of the highest negatively inductive group

Rather strange... It looks like R is inconsistent in its sign of the induction effect. 1 CH3 decreased acidity of formic acid by 10.

@Poutnik please check the edit that I have made in my post and let me know your thoughts about that...

@ShoubhikRMaiti please check the edit that I have made in my post and let me know your thoughts about that...

Have you looked up actual experimental data? Also, consider that counterions must be present in a solution of TMA salt.

@BuckThorn Nope... I wasn't able to find the data for the compound with $Me_3\overset{+}{N}$ and thats whats causing all the trouble. I'm just really confused between the $NO_2$ and the $Me_3\overset{+}{N}$ and I have no idea what the second part of your comment means :')

Imagine you want to prepare a solution of the Me3N+ compound. What are the reagents?

Some cheminformatic descriptors for the puzzling compounds can be found on the PubChem website:

SMILES    N+(=O)O ;
Nitroacetic acid; 625-75-2; alfa-Nitro
acetic acid; 2-nitroacetic acid; ALFA-NITROACETICACID;
CID: 43581;
MF: C2H3NO4;
MW: 105.05g/mol;
InChIKey: RGHXWDVNBYKJQH-UHFFFAOYSA-N   ;
IUPAC Name: 2-nitroacetic acid   ;

and

SMILES N+(C)(C)C;
TRIMETHYL GLYCINE;
N,N,N-trimethylglycinium; Cystadane;
CHEBI:41139;
carboxy-N,N,N-trimethylmethanaminium;
CID: 248   ;
MF: C5H12NO2+   ;
MW: 118.15g/mol;
InChIKey: KWIUHFFTVRNATP-UHFFFAOYSA-O  ;
IUPAC Name: carboxymethyl(trimethyl)azanium

With glycine betaine it is usually meant the chloride salt, which would result in chloride counterions. For dilute solutions presumably those chloride ions don't matter, but they would at higher concentration.
Nitroacetic acid even has its own Wikipedia page. The acidity reported there is pKa 1.68, which is similar to that found in other sites such as in ChEMBL, pKa 1.75.
Glycine betaine or trimethylglycine also has Wikipedia and ChEMBL entries. From Wikipedia, acidity (pKa) is 1.84. From ChEMBL, 2.26.
So from this data alone it would certainly seem like the book is correct. Still, those pKa values are very very similar.

hmm so how can we explain this from the organic point of view? What factor is responsible to make nitroacetic acid even more acidic than trimethyl glycine even though $Me_3\overset{+}{N}$ has one of the highest negative inductive effects which plays a role in stabilising the conjugate base of the acid and hence making it more acidic?

@PrajwalTiwari I can't explain it and hesitate to include a guess in my answer. That does make this an incomplete answer, but it does support the answer in your book with quantitative data.

Alright! At least now I know that my book is right so I can try and figure out how to justify what my book says. Thank you so much! :)


text:
Pressure vs electric conductivity in solids
If we apply pressure to a solid, does it change its electric conductivity? I am asking this because in 2020 they announced a superconductor at room temperature but under a pressure of 3/4 of the pressure in the core of the Earth. Decreasing pressure decreased the critical temperature. So I was wondering if pressure affects the electric conductivity of other materials as well? And if it does is there an intuitive explanation, why?

Superconductivity is almost unrelated to ordinary conductivity. As for the pressure that great, sure, it changes the structure of compounds and pretty much all their properties.

Regardless of the change in the structure of compounds does it influence the motion of electrons(or holes) or not?

https://chemistry.stackexchange.com/questions/83355/what-is-chemical-pressure

Well, a change in structure (crystal type) will definitely change the band structure. Further, even in the same crystal structure hydrostatic pressure does change the inter-atomic spacings which also changes the band structure. For example, with increasing pressure the band gap of silicon decreases which changes the intrinsic carrier concentrations at a given temperature.

This is a very rich area. If you Google "effect of pressure on electrical conductivity" you'll find many articles on the subject. With some materials the conductivity goes up with pressure, with others it goes down, and with still others it goes in one direction and then reverses. And the conductivity is much more pressure-sensitive in some materials than others. So you can infer from this that the mechanism by which pressure affects conductivity is very material-specific. Alas, I don't know the field well enough to offer a particularly nice specific example to illustrate the phenomenon.

Does a change in pressure changes the electric conductivity depending on the material (conductor,semiconductor-insulator) , just like temperature does(for conductors increased temperature->increased resistivity and for semiconductors and  insulators increased temperature->decreased resistivity) or is it more complex?

@JonCuster in the case of silicon decrease of the interatomic spacing means increase in energy of the valence band which decreases band gap?

@MissMulan - the bands change, yes. See https://journals.aps.org/prb/abstract/10.1103/PhysRevB.40.12516 as a place to jump into the literature.

Increase in pressure decreases the interatomic distance which increases the energy of each covalent bond->valence band.By why is the energy of the conduction band increased?


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IUPAC: Would I prioritize low numbering to highest-priority group, OR try to assign lowest numbers overall?

I made up a pretty silly example of a molecule here, but I hope it illustrates my point. It is a dial, but which is the scientifically-preferred order of numbering: left to right, or vice versa?
As you can see in this example, going from left to right gives the -OH group — highest priority after the aldehydes — the lowest possible number (19 versus 21, if we started from the right). However, numbering from right to left gives us the lowest numbers cross the board, but we’re not prioritizing the lowest possible number to our -OH, and instead give it to our lowly methyl group.
I have a feeling it’s right to left, because 4- and 21- is so much more reasonable than 19- and 36-. Would you mind sharing any thoughts, please?

There is no need for such a long chain. You could have used 4-hydroxy-2-methylhexanedial to make your point.

Very loosely speaking, if your OH group is being named as a prefix, i.e. "hydroxy-something", then it's no longer "higher priority" than the methyl group, which is itself also a prefix. The "priority" only matters when you're deciding on the suffix, and the aldehydes already occupy that spot.
Yet, that rather informal and unsubstantiated claim probably won't satisfy some readers, and it shouldn't satisfy you, as I could be any useless person on the Internet feeding you false information (which is sadly common). Let's just say that if you had to speak to your teacher about it, you would probably be embarrassed to tell them that "somebody on the Internet wrote a comment that said this".
So, I will quote a few rules in the IUPAC Blue Book (Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013) which need to be looked at to assign the name. The first step is to decide on the  "principal characteristic group". The seniority order is listed in Table 4.1, P-41:

1 Radicals
2 Radical anions
[...]
15 Aldehydes and chalcogen analogues
16 Ketones [...]
17 Hydroxy compounds and chalcogen analogues (includes alcohols and phenols) [...]

The principal characteristic groups here are obviously the aldehydes. Notably, P-33.1 states that "only one [suffix] can be placed at the end of a name to indicate the principal characteristic group", so the alcohol is now relegated to being a prefix, i.e. "hydroxy-". The next step is to decide on the "parent structure":

P-44.1.1
The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

Again, it is quite clear that the parent structure here is the chain containing both aldehydes. Lastly, we come to the numbering:

P-14.4
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
[...]
(c) principal characteristic groups and free valences (suffixes);
[...]
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
[...]

(c) doesn't help us resolve the ambiguity at hand, because the aldehydes are at both ends of the principal chain whichever way round we count. So (f) is what matters, and note that it says "all considered together": it no longer distinguishes between any kind of "priority" between different prefixes. The locant set (4, 21) is lower than the locant set (19, 36), and is consequently the preferred numbering.
If you search Chem.SE for "P-14.4" there will no doubt be similar questions and answers, which may help to clarify the situation with similar examples.
Note that the "lower" locant set is obtained by point-by-point comparison i.e. comparing the first locant of both at the same time, then the second locant of both, etc. until any ties are broken (P-14.3.5). In this case, (4, 21) is lower than (19, 36) because 4 is lower than 19; there is no need to compare the second locants. A common and baseless misconception is that the smallest "sum of locants" wins. This is not true and is not written anywhere in the IUPAC Blue Book.


text:
Can a previously neutral mix of chemicals become acidic or alkaline somehow?
This question relates to
Is there any law for pH conservation in chemical reactions?
I had basically the same question, but I did not get the answer I was looking for from the above link.
For me this comes from a cooking background. If I add some acid to ph-neutral food, is this acidity preserved? Of course, when I add water, the ph goes up, but the overall acidiy (for the lack of a better term) stays the same. To get ph back to neutral I will have to add some alkaline substance, right? Or can I do anything else to change the ph back to neutral.
Equivalently: when I eat something acidic, does my body somehow neutralize it, without me needing to eat something alkaline?
Furthermore, when I breath I loose carbondioxide, which should make my body a little more alkaline each breath, such that I have a natural desire for some acidic foods, right?
I hope your getting what I'm after.
An example would be great, something along the lines:
When you add vinegar to flour-dough and cook it, something happens and after that, the total acidity is less than it was before some specific reaction happened.
Thanks

There is no pH conservation law.

Also, no law of "conservation of acidity" in any form.

Yes, I know, but a few examples where my supposed conservation law is broken would help me, as the only way I know to neutralise some acid is to add some alkaline substance

"Furthermore, when I breath I loose carbondioxide, which should make my body a little more alkaline each breath, such that I have a natural desire for some acidic foods, right?" Careful, you're almost veering into pseudoscience territory there. Carbon dioxide gas isn't acidic: only dissolved carbon dioxide is (as it dissociates to $\ce{H+ + HCO3-}$), and that is regulated very strictly by buffer systems in the blood system. You know what, maybe look up a buffer? That's a good example which breaks your hypothesis: you can add strong acid or strong base, and the pH barely changes.

There is typical natural example of fermentation of ethanol in wine to acetic acid. Or, the opposite, in distilled spirits traces of created acetic acid react with ethanol, forming ethylacetate. Another case is ammonia or hydrogen chloride synthesis from elements.

I think the source of confusion is connecting the general notion of neutrality with pH=7.0. pH is just a scale for representing the concentration of hydrogen ions in water =-log [$\ce{H+}$]. Unfortunately, companies have misused the term to a large extent by marketing.
What do you think about ethanol? Is it neutral? I cannot even measure its pH. What is the pH of cooking oil or dough? Yet cooking is an acid from a chemist's perspective.
In a strict sense you cannot even define the pH of "food". Our food is horribly complex mixture of simple to very complex molecules. There is no pH-neutral food. Food juices are acidic, baking soda is slightly basic, when they all go to the stomach, they all meet hydrochloric acid, which helps in breaking down the food.


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Why would sodium ions want to go down just because of one side is more positive than the other?
From Khan Academy starting from 1:25, the narrator, while talking about the Electrochemical gradients, said that sodium ions would naturally want to diffuse
down when having a high concentration up here and a low concentration down there.
My question is Why would sodium ions want to go down just because of one side is more positive than the other, I mean, isn't the same charges repelling each other? I don't think the degree of positivity will enable them to be attracted to each other.
Could anyone help me explain this?

Please make sure the external links are supplementary and not essential parts of your question. Links rot over time, and your question becomes incomprehensible. Also, very few people have inclination of clicking random URLs, and even smaller number of them would watch a video, which is an objective waste of time and bandwidth.

This has more to do with the diffusion gradient. Sodium ions move from the region of their higher concentration to a region of their lower concentration.
In the example given by you, I believe, there are additional negatively charged molecules present in the inner side of the membrane, resulting in an influx of sodium ions to balance the same. However, like charges do repel, so this influx is not indefinite, but only up to a certain extent- until the electrical potential difference across the membrane exactly balances the concentration gradient. This point is known as the equilibrium potential.
(This question and answer might be more appropriate in the Biology SE, I think).


text:
If electrons absorb photons to reach higher energy levels in Bohr's atom, why do they have lesser kinetic energy?
In Bohr's atomic model, an electron can jump to a higher energy level by absorbing a photon with energy equal to the difference in energy between 2 energy levels. This should mean that the electron now has more kinetic energy than it had in the previous orbit. But the velocity of an electron in an orbit is inversely proportional to the orbit number (the formula for velocity given as $\frac{2\pi ke^2}{h} \times \frac{Z}{n}$), which means that the electron has lower kinetic energy.
So my question here is, if the electron absorbs a photon to go into a higher energy level, why does it have lesser kinetic energy than it had before?

Beside that you already know that Bohr model is just a passage, the same can be asked about a satellite. This is how mechanics works. The energy input goes into kinetic and potential energy. You said it before asking. When you rise a weight, its potential energy rises, independently of its motion.

@Alchimista oh so it gets converted to potential energy, got it, thank you :)

By the way, Bohr's atomic model is not very accurate.

The Bohr model, in its initial form, is still based on the circular motion around a central object acting by a quadratic force $F = \frac{k}{ r^2}$.
In such a system, objects with higher total energy have lower kinetic energy, as the increase of their potential energy is 2 times greater than the loss of the kinetic energy.
$$E_\mathrm{tot} = E_\mathrm{pot} + E_\mathrm{kin}$$
$$E_\mathrm{pot} = - 2 \cdot E_\mathrm{kin}$$
so
$$E_\mathrm{tot} = - E_\mathrm{kin}$$
this is also the reason, why $$v_\mathrm{escape} = \sqrt{2} \cdot v_\mathrm{circular}$$
If Earth had been orbiting Sun at the distance Venus has, it would have  moved faster than now, with the total ( kinetic + potential ) energy lower. Similarly, satellites need to gain energy to reach a higher orbit with a lower speed.


text:
Hydrolysis of one mole of peroxydisulphuric acid with one mol. of water produces?
I am having trouble in this question:

Hydrolysis of one mole of peroxy disulphuric acid with one mole of water produces :

a)two moles of sulphuric acid only
b)two moles of peroxymonosulphuric acid
c)one mole of sulphuric acid and one mole of peroxy monosulphuric acid
d)one mole of sulphuric acid, one mole of peroxymonosulphuric acid and one mole of hydrogen peroxide

when hydrolyzing H₂S₂O₈ cleaving the S-OH linkage and S=O is of no use since it returns the same compound and peroxide linkage is non-polar hence can't be cleaved. So only S-O linkage should be cleaved that would result in the formation of 1 mole of H₂O₂ and 2 moles of H₂SO₄. But that doesn't match to any of the options. Can someone please explain me how to solve it?

The problem specifies a 1:1 ratio of peroxy disulfuric acid to water. So what do you get if you cleave one S-O linkage?

Without speaking of mechanism, here is an explanation valid for the average high school students, not specially inclined to chemistry. Suppose the bond $\ce{O-O}$ is broken, producing two identical radicals $\ce{HSO4·}$. Suppose one mole of $\ce{H2O}$ is also broken into $\ce{H}$ and $\ce{OH}$. Then, $\ce{H}$ gets attached to the first radical, and $\ce{OH}$ to the second. It produces one mole sulfuric acid and one mole peroxymonosulfuric acid.


text:
Confusion regarding $\delta$ charge compared with electronegativity
I heard on a lecture of an online site that, let us say, we have two orbitals (1 and 2). Now, it will be like a spherical cloud. Now, what they say is that

Now, we can notice that for orbital number 1. There is a dark region and a light region (right side). Now, what they say is that the darker region means the electron finding probability is more there. Similarly, for the number 2, electron finging probability is more on left side.
Therefore, we say that 1 develops $\delta +$ charge and 2 develops $\delta -$ charge. Therefore, the two will attract each other.
Now, my teacher on the other hand told it a bit differently:

The more electronegativity means more $\delta -$ charge. But, I have a confusion in it. Between $\ce{H}$ and $\ce{F}$. I have drawn a line. Fluorine has 7 electrons. My teacher says that $\ce{F}$ and $\ce{H}$ share that one electron where $\ce{H}$ has one and fluorine 7th electron. Now, since $\ce{F}$ is more electronegative. The electron (by hydrogen) is more pushed towards fluorine. My confusion is that if $\ce{F}$ is more electronegative than $\ce{H}$. Then, the electron should be present more towards the $\ce{H}$ side.

Please do not use (+)ve or (-)ve notations in your text body.

Although, OP indicated the confusion regarding $\delta^\pm $ charge compared with electronegativity, the text body indicated whole lot of confutations over interatomic/intermolecular forces. Thus, I start with the electronegativity.
According to IUPAC Gold Book, electronegativity is the concept introduced by Linus Pauling as the power of an atom to attract electrons to itself. There are several definitions of this quantity. However, Wikipedia gives a simple to understand version better suited for OP:

Electronegativity is the measurement of the tendency of an atom to attract a shared pair of electrons (or electron density).

The relative electronegativity of $\ce{H}$ atom has given a value of $\pu{2.1 eV^{-\frac{1}{2}}}$ and the electronegativity for each of other atoms in the periodic table have been calculated relative to this value for $\ce{H}$ atom (IUPAC Gold Book). These calculations are beyond OP’s field of study so should be disregarded. The important point for OP to consider is the higher the associated electronegativity, the more an atom or a substituent group attracts electrons. For instance, the electronegativity of $\ce{F}$ is $\pu{4.0 eV^{-\frac{1}{2}}}$ and hence, shared electron fair in $\ce{H-F}$ bond stays more closed to $\ce{F}$ than to $\ce{H}$. As a result, $\ce{F}$ carries permeant $\delta^- $ charge while $\ce{H}$ carries   permeant $\delta^+ $ charge in the $\ce{HF}$ molecule, creating a permanent dipole. Hence, $\ce{HF}$ is a polar molecule. Therefore, among $\ce{HF}$ molecules, there are always dipole-dipole interactions (the image 2 of OP's text). Since the positive end of these dipole is $\ce{H}$ atom, the relevant dipole-dipole interaction is called $\ce{H}$-bonding, the strongest intermolecular force among dipole-dipole interactions. These dipole-dipole interactions occur only in molecules.
Now, I think I have addressed the OP's main confusion:

My confusion is that if $\ce{F}$ is more electronegative than $\ce{H}$, then, the electron should be present more towards the $\ce{H}$ side.

According to the above explanation, PO's understanding of electronegativity is not correct. In reality, if an atom is relatively more electronegative than that of the other atom it attaches to, that means it attracts the shared pair of electrons, increasing its electron density compared to its protons (gaining $\delta^- $ charge). Meanwhile, the other atom feels decreased electron density compared to its protons (gaining $\delta^+ $ charge). This is completely opposite of what OP's indicated understanding.
On the other hand, the London dispersion force is also one of the intermolecular forces, but is the weakest among them and occurs between atoms as well. For example, why do you think gaseous $\ce{He}$ become a liquid when the temperature is lowered sufficiently? You can find the answer in chem.purdue.edu:

The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms/molecules occupy positions that make the atoms/molecules form temporary dipoles. Thus, this force is sometimes called an induced dipole-induced dipole attraction. London forces are the attractive forces that cause nonpolar substances to condense to liquids and to freeze into solids when the temperature is lowered sufficiently.

An atom or a molecule can develop a temporary (instantaneous) dipole when its electrons are distributed unsymmetrically about the nucleus because of the constant motion of the electrons (e.g., the image 1 of OP's text). The explanation can be depicted in an image:

For instance, assume two helium atoms approaching each other. When they get closed to each other, the electrostatic attraction force between the nucleus of first $\ce{He}$ atom with two positive charges and the two electrons in the second atom causes each atom to polarize  (creating an induced dipole in each atom). Note that at the same time, there is  an electrostatic repulsion acing as well between electrons of each other as depicted in the image above.
Dispersion forces are present between all atoms or molecules. In molecules, these forces are present regardless of whether they are polar or nonpolar. London dispersion forces depends on few factors:

Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones. In a larger atom or molecule, the valence electrons are farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles (e.g., $\ce{Cl2}$ is a gas at room temperature and $\pu{1 atm}$ pressure, while $\ce{Br2}$ is a liquid and $\ce{I2}$ is a solid under the same conditions).
The strength of London dispersion forces varies according to molecular shapes. For example, $n$-$\ce{C5H12}$ and $neo$-$\ce{C5H12}$ are both pentanes, but former is more cylindrical shape while the latter is more spherical shape. Although, both have the same molar mass, $n$-pentane is a liquid (average boiling point of pentanes: $\pu{36.1 ^\circ C}$) at room temperature and $\pu{1 atm}$ pressure, while $neo$-pentane is a gas (boiling point: $\pu{9.5 ^\circ C}$) under the identical conditions.

Also, London dispersion forces are depend of the molecule's polarizability (The ease with which the electron distribution around an atom or molecule can be distorted is called the polarizability). London dispersion forces tend to be stronger between molecules that are easily polarized, but weaker between molecules that are not easily polarized.

Ohk. Thank you very very much for the efforts in solving my confusion . I would just like to give a short summary. 1 London dispersion forces can be between atoms to atoms or molecules  with molecules. 2 If F atom is more electronegative, that means it has a stronger electric field to attract the H atom. Therefore , the Electron of H atom shifts towards F. It shifts at such a point where the forces by nucleus to F electrons and Forces by F’s electric field to the electron of H atom cancels out. So ,electronegativity does not mean that the electron is present more on the right side or left side

Now , since the F atom shifted towards electron . It carries permeant $\delta$ + ve charge and F carries negative charge. Why ? Electron is still present more on the right side for H atom but it still is far away so it carries $\delta$ +ve charge.

Also , as you pointed that it is a temporary dipole attraction. Does it mean it has a probability of it to happen or not ? Moreover , you took two He atoms which have same electronegativity difference. So , they won’t form LDF right. It would be a non polar covalent bond?

How did induced dipole in each formed. Just when a Cl gets closer to H atom . It forms it automatically?

@Rider: You still did not resolve your confusion. Your understanding is some what different from that before. Yet, get tangled with two different concepts. Please read my post carefully from beginning to finish. You may get some clues. Electronegativity applies in specific bond within the molecule. My example is $\ce{HF}$ molecule, and the bond in concern is $\ce{H-F}$, which is a covalent bond in gas phase. The $\ce{H-F}$  is made by $\ce{H^.}$ and $\ce{F^.}$ sharing the each valence electron.

$\ce{H^.}$ or $\ce{F^.}$ does not have a charge. And, $\ce{H-F}$ also does not have a charge if the sharing pair is right in the middle of $\ce{H}$ and $\ce{F}$ nuclei. However, that is not the case in reality. That sharing pair is more close to $\ce{F}$ nucleus than to $\ce{H}$ nucleus due to high electronegativity of $\ce{F}$. Because of that, $\ce{F}$ feels more electron than it needed to stay in neutral state (it has part of sharing electron of $\ce{H}$).  That means it has $\delta^{-}$ charge.

Similarly, since the sharing pair is more close to $\ce{F}$ nucleus than to $\ce{H}$ nucleus due to high electronegativity of $\ce{F}$, $\ce{H}$ feels less electron than it needed to stay in neutral state (it has lost part of its sharing electron to $\ce{F}$).  That means it has $\delta^{+}$ charge. As long as $\ce{H-F}$ bond didn't break, these two $\delta^{\pm}$ charges stay permanently with the molecule, making a permanent dipole.

Now , since F is more electronegative . The electron(by hydrogen ) is more pushed towards Fluorine. My confusion is that if F is more electronegative than H. Then , the electron should be present more towards the H side.

Electronegativity is a measure of an atom or an ion to attract electrons in a covalent bond and by definition both electrons of the covalent bond will be closer to F.

Now , we can notice that for orbital number 1. There is a dark region and a light region(right side). Now , what they say is that the darker region means the electron fining probability is more there. Similarly , for the number 2. Electron fining probability is more on left side.

This is not a London Dispersion Force.We use London Dispersion Forces to understand intermolecular forces of molecules which are non-polar.

You are right that what OP needs is a definition of Electronegativity. For the second part, the sketch could be well one to speak about London forces. It might represent an instant, we weren't present where and when it was sketched at what was the accompanying speech/text. This is just to no confuse even further the OP.

London forces are between molecules not atoms.The OP clearly states he believes it is between atoms so I added it to get rid of the confusion.

well there is no reference if not electron clouds influencing each other. We can't really reteach from scratch. It seems that OP is indeed confused and mixing too much things. For instance the first sketch is unrelated to the question about El Neg. Note that I have nothing against your answer.

I checked this on Khan academy @MissMulan

https://www.khanacademy.org/video/london-dispersion-forces@MissMulan

@MissMulan The second yellow line that you have marked. The attraction between orbital 1 and 2 is London dispersion force according to the video I have sent.

@MissMulan Also, their attraction is due to electric fields caused by them which is electrostatic force of attraction . Am I right ?

@Alchimista You can check sir.

Op: keep things easy. Electronegativity is the tendency of an atom to attract towards itself the electrons of a covalent bond. Look at chapter about Electronegativity. If you are limiting the scenario to biatomic molecule X-Y, the difference between El Neg of X and Y dictates the electrical dipole moment. Molecule having a relevant one, means the molecules can interact each others via electrostatic interactions. Besides, the electron clouds can mutually influence each other leading to dynamical induced dipoles. Also leading to electrical interaction. Those are relevant in molecule without..

@user883022..... permanent dipole.


text:
Why are hydrides of transition metals called hydrogen-deficient?
My textbook states:

Metallic hydrides are formed by many d- and f-block elements. Unlike saline hydrides, they are almost always non-stoichiometric, being deficient in hydrogen, for example, LaH$_{2.87}$, YbH$_{2.55}$, etc.

After some research online, I found that non-stoichiometric compounds are so called because they don't follow the law of definite proportions, that is, "their elemental composition cannot be represented by a ratio of well-defined natural numbers."
From what I understand, some transition metals form hydrogen bonds with H atoms that fill interstitial spaces between metal atoms in crystal lattices. However, what does "hydrogen-deficient" mean in this context? Apologies if this is an elementary question, but I could not find an answer online.

It means not all the available spaces are filled with H.

As you note, many hydride phases have the hydrogen occupying interstitial sites within a metal phase, not unlike many carbide/nitride/oxide phases as well. In theory this could range from no hydrogen at all up to full occupancy (and perhaps even double occupancy).
Since you mentioned lanthanum hydride, I pulled up a Calphad analysis for the H-La-Ni system, focusing on the H-La part. Hydrogen has almost no solubility in dhcp-La, yet shows a wide solubility range in both fcc-La and bcc-La phases.
Now for the hydride phase, where I'll quote from section 3.4 in the paper above.

The binary LaH$_{x}$ hydride has an fcc structure... Usually, La atoms occupy the 4a sites, while hydrogen atoms occupy the tetrahedral 8c and octahedral 4b sites. Experimental determinations using using several different techniques ... show that hydrogen atoms occupy tetrahedral sites first, for concentration up to $H/M = 1.95$, and only above this concentration value octahedral sites begin to be occupied.

Their calculated diagram is:

So, hydrogen happily fills in to interstitial spaces, but prefers some more than others. Eventually (perhaps strain, perhaps getting too close to each other) other sites become more desirable for the additional hydrogen. In any event, often a quite broad range of compositions are stable, and perhaps (depending on the metal) you never make it to the ideal stoichiometric composition.


text:
Orientation of external cell when voltage is supplied by it to voltaic cell so as to oppose its cell potential
Given below is a page from our Chemistry Textbook:

As it can be seen, the direction of $E_{ext}$ in figure $3.2(c)$ is opposite to the direction in figures $3.2(a)$ and $3.2(b)$.
I understand that $E_{ext}$ is supposed to oppose potential of the voltaic cell in figures $3.2(a)$ and $3.2(b)$. But in figure $3.2(c)$ since $E_{ext}$ is flipped, isn't $E_{ext}$ supporting the electrical potential of the voltaic cell?
Is there is reason for the flipping of the direction, or is it a mistake on part of the writer?

You are correct: figure 3.2(c) should not have the battery reversed. Whomever drew that figure messed up.

(+1) I am giving you an upvote to get you 10 rep points: anyone who throws down a 50 bounty, when they are barely over a 100, deserves respect.

I am going to be blunt. This is NCERT, what else do you expect? That's the only reason I never read this book even after how many times I was forced to read it by the school. There arrangement works but obviously it's not the arrangement the text is referring to. Surely the diagram making team knew nothing about chemistry and so when they had to reverse the direction of current they foolishly also reversed the direction of the batter. They must have thought that they are smart.

@NisargBhavsar What's more frustrating is that reference book I decided to use blindly follows NCERT. This same mistake is in Modern's ABC Chemistry.

Thanks for the green checkmark and the bounty! Happy to be of assistance!

The figures in the OP's post are nicely drawn, as expected in a modern textbook, but figure 3.2(c) has the external battery reversed, which is incorrect. Here is how it works, without the needless complication of the potentiometer.
First, start with a standard Daniell cell with standard assumptions, i.e., negligible internal resistance, unimolar concentrations, and so on. Then the open circuit voltage is 1.100 V and, under light load, essentially the same. This is shown in Fig. 1 below:

As shown, electron flow is from the zinc anode to the copper cathode, via the external load resistor. Note that DMM means digital multimeter, used in voltmeter mode, and DVM means digital voltmeter. The current flow is $11 \mu A $.
Now cut the wire to the cathode and insert an external DC voltage supply that is turned on, but set to supply zero volts between its terminals. This is shown Fig. 2 below:

The external DC supply is schematically depicted as a battery supplying 0.000 V between its terminals. Its internal impedance is assumed negligible. Effectively, this 0 V battery is the same as a piece of wire: the situation is the same as in Fig. 1. The current flow is still $11 \mu A $.
Now start making it interesting. First, set the external DC supply to produce +0.500 V, as shown in Fig. 3 below:

Note how the external DC supply is connected: its positive terminal connects to the copper cathode. The Daniell cell potential is opposed by the external supply voltage and the DMM shows that the difference, which is +0.600 V, is across the load resistor. Therefore, the current flow is only $6 \mu A $.
Next, set the external DC supply to produce +1.100 V, as shown in Fig. 4 below:

No current flows because the external DC supply voltage nulls (exactly opposes) the Daniell cell voltage. Both ends of the resistor are at -1.100 V with respect to the copper electrode, so no current flows and there is no oxidation or reduction taking place in the cell reservoirs. There is no anode or cathode, just electrodes. This is the dividing line between voltaic cell operation and electrolytic cell operation.
Finally, set the external DC supply to produce +3.000 V, as shown in Fig. 5 below:

Now the electron flow is from the negative terminal of the external DC supply, through the load resistor and into the zinc electrode, where reduction will take place in that cell reservoir. This is the electrolysis mode of operation. Note the voltage across the resistor is -1.900 V, i.e., -3.000 V minus -1.100 V. So the zinc electrode is now the cathode and the copper electrode is the anode.

Thanks. You have a minus before 1.100V in the second last line which is unnecessary. I can't suggest an edit because your answer is so good that I can't find the other 5 characters required to do so.

The DMM reads -1.900 V because the voltage across the resistor is -3.000 V - (-1.100 V) = -1.900 V. The voltages are with respect to the copper anode. In the Daniell cell itself, as in Fig. 1, the DMM probes would be black (reference) on the zinc and red (input) on the copper, so the cell voltage would be +1.100 V.

It's a shame the book is wrong. It's a staple book used by almost every student in my country. The mistake has been around for a long time and still hasn't been rectified in the latest edition.

Sorry. My bad. I missed the minus before 3.000V

Yeah, textbook mistakes happen and are often hard to catch and fix. I think this one was caused by someone getting a little tricky with the potentiometer. Had they simply Thevenized the battery and pot, it would have reduced to the Thevenin equivalent voltage, i.e., the voltage divider voltage, in series with the Thevenin equivalent resistance, which is just the parallel resistance of the two pot arms. By design, to have a stiff voltage source, the Thevenin equivalent resistance would be negligible compared with the load resistor I used.

Suppose two identical Daniell cells are connected with the zinc electrodes electrically connected through an amperemeter, and the two copper plates also electrically connected. Of course no currant will be measured by the ammeter. But the tension between the plate is independent on the connections. Now if one of the cell has a higher nominal voltage, maybe because of aa lower zinc ion concentration, this "stronger cell" imposes its capacity of producing electrons to the other one. Electrons will be flying through the ammeter. So the weakest cell has to reverse the sense of its reactions on its plates. The cell will work as in an electrolysis. Zinc ions $\ce{Zn^{2+}}$ will be reduced into metallic zinc in this cell. Slowly the two zinc ion concentrations will tend to equalize themselves in both cells.

I must admit that figure $3.1$ has something puzzling. There is an arrow coming out of the copper plate. This arrow has no meaning. Nothing is coming out of this plate. On the contrary, some ions $\ce{Cu^{2+}}$ are attracted by the copper plate and gets discharged as metallic copper. The only thing that is leaving the copper plate are the sulfate ions $\ce{SO4^{2-}}$ which are pushed through the slat bridge. But this ions are not leaving the plate ! The same stupid arrow is reported on the zinc anode, as if something was deposited on the zinc plate. In the contrary, Zinc metal is dissolved.

Right now, you have one up vote, which I gave you previously. There is nothing wrong with your answer, but I will double check. ;-) I think the drawings are a little too cute and the use of the potentiometer just reduces clarity. It would have been better, in my opinion, to just show a simple opposing DC voltage source.

@Maurice The yellow arrows show the direction of electron flow. Electrons released by zinc's oxidation flow through the wire, through the copper electrode into the solution, which are used by $\ce{Cu^{2+}}$ ions to undergo reduction.

@Alpha Delta. Electrons do not go into solution on the copper plate !!! They do not go out of the solution in the zinc compartment !!!

@Maurice That's what the book is implying. I cannot vouch for its accuracy but that's what its trying to say.

@Alpha Delta. We are all doing mistakes. Even a book may be wrong !


text:
Haworth structures of sucrose and lactose
The haworth structures of lactose and sucrose  is

My question is regarding the glyosidic linkage, in lactose it is represented as a zig zag and in sucrose it is a "V". What does that mean. Also, in sucrose why is the the glyosidic linkage displaced down w.r.t the compound

The link in lactose structure is beta (up position)-1(1-position around the ring)-4(glucose's 4-position around its ring, in which the OH is down for glucose). The link in sucrose is alpha (down)-1(glucose's 1-position)-2(fructose's 2-position which is at the down position because fructose is "facing" the glucose unit and it is upside down).

Haworth structures are cumbersome to draw in this regard. The bridging oxygens are attached directly to the rings and not to CH2 groups as the "zig-zag" and "V" might imply.


text:
Ethanol and water resistant adhesives
I need an adhesive that can bind to ABS plastic but is highly resistant to ethanol, isopropanol, and water. It's for use in an environment where it will spend a significant amount of time soaking in >95% alcohols, with agitation.
Temperature and pressure requirements aren't a big deal, this is for use at near room temperature and pressure. Strength requirements aren't anything crazy either, it just needs to survive light vibrations and light fluid flow. UV exposure is also a non-issue. However, I do need the adhesive to be watertight (where "water" = the solvents too).
My default is a cyanoacrylate-based adhesive but according to an incidental note on this page (emphasis mine):

The bottom line is cyanoacrylates have poor resistance to polar solvents and better resistance to non-polar solvents.

So it seems like CA glues are out. What classes of adhesives could I use here? My gut says maybe a silicone-based adhesive, although I don't like working with those when possible (messy cleanup). Or maybe some type of epoxy adhesive?

I hope this question is OK here. I'm just looking for adhesive classes and types, not specific products.


text:
Can water oxidize zinc?

According to this textbook, H₂O has lower reduction potential (-0.83 V) than Zn (-0.76 V). This mean zinc is unable to reduce water to hydroxide. If so, why this video demonstrate that by heating zinc in water, a reaction occurs producing hydrogen gas from the bubbles while also turning wet red litmus paper blue?

The given value of −0.83 V corresponds to pH = 14. You should also be able to find the value for pH = 0 in your list, and then estimate what happens with water at a roughly neutral pH

Ok so i found out the answer, zinc cannot directly reduce water to form zinc hydroxide because it has lower reduction potential (-0.76V) than water (-0.83V). Zinc does not react with cold water, however it reacts with hot steam or boiling hot water. This is because high temperature increases water autoionization, thereby increasing the presence of H⁺ and OH⁻. So, zinc is actually reducing H⁺ ion to H₂ because H⁺ ion has higher reduction potential. The Zn²⁺ then reacts with OH⁻ to form Zn(OH)₂

I've seen similar demonstrations with magnesium, and in both cases the explanation is the same.  The metal reacts to produce the hydroxide plus hydrogen; with zinc this would be
$\ce{Zn + 2 H2O -> Zn(OH)2 + H2}$
But, the zinc hydroxide is essentially insoluble in cold water and seals up the metal before the reaction gets very far.
Boiling the water speeds things up in two ways.  First, the metal hydroxide, with either zinc or magnesium, is inherently more soluble.  Second, the turbulent action created by the boiling (the bubbles you see are actually steam, not hydrogen) acts as a stirring mechanism, speeding up the mass transfer so the slight concentration of dissolved hydroxide gets to the litmus paper faster.
Magnesium differs from zinc because the former hydroxide is somewhat more soluble.  Therefore, the basic nature of the product with magnesium can be demonstrated with phenolphthalein, which requires a slightly higher pH to react than litmus.  Some demonstrations with magnesium show a rose coloring of the magnesium metal in cold water with the phenolphthalein showing that the magnesium hydroxide remains essentially on the surface.


text:
Can potassium carbonate be used to calibrate hygrometers?
Context:
In the process of trying to calibrate a number of hygrometers, with Sensirion MEMS electronic sensors, I came across the method of using wet salt: an airtight container will stabilize at a predictable relative humidity over time, given a particular salt and temperature.
For a list of salts and their RH values at particular temperatures see this page on the website of Vernier, a maker of STEM education equipment. The various salts suggested for this method repeatably produce a range of low to high relative humidities, also allowing for two-point calibration.
I require my hygrometers be calibrated close to 40% RH. With a predictable 43% RH that makes potassium carbonate the ideal candidate. However, when I wettened potassium carbonate with water it underwent an moderate exothermic reaction. Googling what was going on, I read "potassium carbonate dissociates completely in water into potassium (K+) and carbonate ions (CO32-)."
Which makes me wonder:
Can potassium carbonate be used at all for the wet salt calibration method, or does its swift dissociation render it useless for this application?

While there are guides available suitable for checking this at home (example), the question does not mention if the device is electronic, or mechanic. The later will take more time to settle, than the former. In addition if mechanic be aware the ones with natural hair (non-linear scale, especially in the lower values) differ from those based on a synthetic fiber (almost linear scale). Don't forget the natural hair hygrometers need recurrent exposure to high humidity (e.g., bath with shower, not a sauna) to work reliably.

Updated the original post: the sensors are electronic.

@Buttonwood Czechoslovak civil and military stations – I served as military airfield meteorologist, being enlisted in 1989-90 – used in late 80s early 90s as the humidity sensor a mechanically primed animal-origin membrane. It had reportedly much more linear response, compared to hairs. Hair sensitivity  is highly nonlinear and dramatically decreases toward 100%.

Note that all salts, used in humidity calibration, dissociate to their respective hydrated cations and anions when dissolved.
From the solution crystallizes the sesquihydrate $\ce{K2CO3·\frac 32 H2O}$ ("potash hydrate"). Heating this solid above 200 °C gives the anhydrous salt (Wikipedia).
It seems you have anhydrous carbonate, that hydrates itself exothermically. You can still use it, but you need to prepare rather the saturated solution with excess of carbonate, that would integrate the crystal water into itself.
Or, let the prepared saturated solution partially evaporate and crystalize.

What is important is maintained equilibrium.
$\text{Air 43% humidity }\ce{ <=> K2CO3} \text{ saturated solution } \ce{<=> K2CO3 . 3/2 H2O}$
Anhydrous $\ce{K2CO3}$, wetted by water, rather just absorbs water into its crystall lattice. If you add too little water, no solution would left, similarly what would happen to the plaster of Paris.
You need the carbonate in hydrate form, together with its saturated solution. Take some water, add there carbonate to dissolve as much of as it can, and then some. Let is settle.
If you had the hydrate, you could just moisten it. But you do not have it.
There is the third option - moistening  it enough so some liquid phase always remains. For $\pu{100 g }\ce{K2CO3}$, you need about $\ce{20 mL}$ of water just for the crystal water, so it becomes the part of the solid.
The carbonate solubility is $\pu{110.3 g/100 mL}$, so I would carefully mix some $$\pu{100 g}\  \ce{K2CO3} + \pu{50 mL}\ \ce{H2O}$$
or recalculated to  more suitable amounts.

The preparation should become viable naturally, if left exposed to air until the water partially evaporates?

I do not have the scientific framework to understand your answer. Do you mean the salts should be moistened, not dissolved?

See the answer update.


text:
Determine the normality of sulfuric acid solution, given that it gives a yellow color with methyl orange and a red color with methyl red
Determine the normality of sulfuric acid solution, given that it gives a yellow color with methyl orange and a red color with methyl red.
can someone help me with this? what should do first?
i know that normality = (grams of solute)(factor) / (MW of solute)(liters of solution)

Please use proper names and symbols for physical quantities and make sure it is clear what belongs to numerator and what to denominator. For example, "g of solute" is illiterate and should be replaced, I presume, with mass $m.$ What is "factor"? Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown.

It is an incomplete question. If your teacher wrote this question, ask him/her what is in the flask and what is in the burette? Only then we can decide which indicator is to be used. Also you need a balanced equation, you also need to know the volume of the acid and the base.

@andselisk, factor is the same 18th century conversion factor between molarity and normality. I don't know when this syllabus would be updated.

Right, so OP probably should be aware that IUPAC, SI and ISO consider this "factor" (which should really be called an equivalent) deprecated for decades. If you need a brief, easy-to-access and authoritative document to use in discussion with the normality apologists, feel free to use https://doi.org/10.1515/pac-2017-0410.

As the first step, determine [H+]

The pH of the solution should be $4.4$. So now do calculate [$\ce{H+}$] and here you are.

thanks! can you help me check if I did it right? here's the link to my solution:
https://preview.redd.it/wr9rz7pv6em61.jpg?width=355&format=pjpg&auto=webp&s=75ecd0f5694d78a00f85e9644714a88a9a2f1f80


text:
How does the treatment of resin R-SO3H with sodium chloride lead to formation of RNa?
I am studying the methods of removal of permanent hardness of water, and came across the "synthetic resins method". From my textbook,

Cation exchange resins contain large organic molecules with $-\text{SO}_3\text{H}$ groups and are water insoluble. Ion exchange resin $\text{RSO}_3\text{H}$ is changed to $\text{RNa}$ by treating with $\text{NaCl}$. The resin then exchanges $\text{Na}^+$ ions with $\text{Ca}^{2+}$ and $\text{Mg}^{2+}$ ions present in hard water to make it soft.

I don't understand how the treatment of $\text{RSO}_3\text{H}$ with $\text{NaCl}$ to form $\text{RNa}$ works. For one thing, in the former compound, since $-\text{SO}_3\text{H}$ is an anion, $\text{R}$ is a cation, but in $\text{RNa}$, $\text{Na}^+$ being a cation, $\text{R}$ is an anion. What is the redox reaction that is responsible for the conversion here?

It is a typing mistake. Cation exchange process can be written as
$$\ce{R-SO3^-H^+ (s) + NaCl (aq) -> R-SO3^-Na+(s) + H^+(aq) +Cl^-(aq)}$$
You can immediately see why the resin must be in sodium form for water softening. If the acid form of the resin is used for softening water, what would you get? Extremely strong acids!!
For a divalent ion, the sulfonic acid groups should be in close proximity to "catch" ions such as $\ce{Ca^{2+}}$. Each calcium ion would quantitatively release two hydrogen ions from the resin.

Thanks for your answer. Is there a reason for writing sodium chloride as NaCl (aq) but writing the ions evolved, H+ and Cl-, separately and not as HCl (aq)?

No, just a matter of convenience, you can write HCl(aq) as well.

There is no redox reaction here. This is acid-base chemistry. RSO$_3$H dissociates into RSO$_3^-$ and H$^+$ and then the H$^+$ gets replaced by Na$^+$. RNa is an inaccurate way of writing this. Washing a cation exchange column with NaCl solution does not remove the sulfonic acid groups.


text:
Fluorescence quenching by palladium(II) vs. platinum(II) complexes
Why do we observe much stronger (in most cases complete) fluorescence quenching by $\ce{Pd^2+}$ than $\ce{Pt^2+}?$ I work with complexes of both metals and there is no real 100% explanation I can find why this is the case. I am looking for a full quantum mechanical explanation (if possible). If you know a piece of literature, please point me towards that.

Quenching of what molecules fluorescence ? Do you mean fluorescence from complexes with these metals  or diffusional quenching?

@porphyrin yes I mean quenching of complexes formed with these metals

It would help if you indicated what these compounds are, but I suspect that the difference is due to relative positions of energy levels such as eg & t2g relative to those in the  (presumably) aromatic ligands.

@porphyrin yes you guessed right; if it’s due to eg and t2g positions: how exactly this plays a role would interest me.

The theory of this quenching effect was developed about $70$ years ago by Förster, who showed that the quenching is not due to diffusion or collision. It is due to a resonance phenomena. The fluorescence yield is quenched by a solute whose mirror image of the adsorption spectrum overlaps the emission spectrum of fluorescence. It is is proportional to the overlap integral of these two spectra.
The quenching effect due to an agent X on the fluorescence is given by Stern-Volmer equation : ${I_0/I = 1 + \gamma [X]}$. Now T. Förster has shown that the quenching constant $\gamma$ is equal to $\gamma = \frac{2\pi ^2}{3} \frac{N}{1000} R_0^3$, where $N$ is Avogadro number, and $R_0$ is the critical Förster distance, where $R_0^6$ is proportional to the overlap integral $J(\nu )$ of the two spectra, and to the sum of all energy transfer rate constants (solvent to solutes and solute to solute).
This theory will not be developed here.
But, as you are looking for a piece of literature, I recommend the following references, where this theory is explained.
See : T. Förster, Discuss. Faraday Soc. $27, 1, (1959)$ and : T. Förster, Ann. Physik $2, 55, (1947)$.

Is this also true for the complexes formed with these metals?

@Maurice, Resonance/Forster/dipole-dipole  quenching can also be diffusional controlled, and it is not the only type, there is also Dexter, electron transfer and heavy atom quenching.

@porphyrin. The OP wanted references. I have given him what he wants. I should have said that Förster theory is well adapted to compounds having an absorption and/or emission spectrum in the region between$ 250$ nm to $300$ nm. Diffusion quenching also happens in the other situations, but it is weaker. Please give him references for other types of quenching.

@porphyrin yes, I so understand quenching but why the difference between Pd and Pt. There is obviously the relativistic effect which is stronger for Pt than Pd but I doubt that is the real reason.

@Maurice, All forms of quenching can have an equally large or small rate constant, diffusion or no diffusion. The OP is asking why one set of compounds is less luminescent that another set, so quenching is the not best term to use here.

@porphyrin Why do you think the word "quenching" is not the bet term to use here ? This word is in the title of the question, and also in the first sentence of the text.

because quenching is  a before and after measurement, i.e molecule yield then molecule yield plus quencher. The OP has clarified that these are complexes so we compare different molecules not the same molecule with and without a quencher.


text:
E1 or E2 Elimination? Which Elimination occurs first?
In this reaction scheme, two dehydrobrominations occur. A bulky base is used but both hydrogens are sterically hindered. How is it possible to determine, which one is removed first? Or is this an $E_1$ elimination and the tertiary bromine leaves first?


text:
Why is hydrated barium peroxide used in the preparation of hydrogen peroxide?
I read about the laboratory method of producing hydrogen peroxide. The chemical reaction in question is: $$\text{BaO}_2.8\text{H}_2\text{O} + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4+\text{H}_2\text{O}_2 + 8\text{H}_2\text{O}$$
I wanted to find out why the hydrated form of barium peroxide is used. Apparently, what made barium peroxide so attractive to use in this reaction was the fact that the product, barium sulphate, being insoluble, is easy to filter out after the reaction. However, the side effect is that it precipitates on the barium peroxide sample, preventing further reaction. Using the hydrated form of barium peroxide solves this problem as the salt molecule is "surrounded" by water molecules.
I don't understand exactly how this works. What does a hydrated $\text{BaO}_2$ molecule look like in space, and how does its geometry help avoid the problem of a precipitate forming on its surface?

What book are these statements from? Sounds quite detailed.

I'm paraphrasing what I read and watched online. My high school textbook contains only the chemical reaction at the top of the question.

Both peroxides are nearly insoluble in water. Probably the eight water molecules from $\ce{BaO2·8H2O}$ help insoluble $\ce{BaSO4}$  produced by he reaction quit the surface of the grain, which allows new sulfuric acid to touch and react with the inner part of the grain of peroxide. If $\ce{BaO2}$ is not surrounded by water molecules, $\ce{BaO2}$  is simply replaced by $\ce{BaSO4}$ at the surface of the grain, which produce a protective layer of insoluble $\ce{BaSO4}$ and protects the inner part of the grain from further reaction with the sulfuric acid solution.

Thanks for your comment, that explanation sounds convincing. Although I think I understand what "grain" means in this context, I've never heard the term used before so I can't be sure. Could you explain what you mean by it please?

@Ray Bradbury I might have said a piece, or a bit, or a crystal, instead of a grain.

Why hydrated? If you check Wikipedia, you'll see that anhydrous barium peroxide was once used to produce hydrogen peroxide. $\ce{BaO2 + H2SO4 → H2O2 + BaSO4}$.