text:
Why metal hydrides have low lattice enthalpy
Lattice enthalpy is inversely proportional to the distance between the ions. And the trend of lattice enthalpies is: $\ce{MF>MCl>MBr>MI}$. But why do metal hydrides have lesser lattice enthalpy than flourides?
Data

I guess they are less ionic.

Hydride is HUGE - H- ionic radius 208pm, F- 119pm, Cl- 181pm (from https://en.wikipedia.org/wiki/Ionic_radius) . Hence the lattice energy is lower.

OK, slightly more complicated as H- is rather squishy (technical term), see https://chemistry.stackexchange.com/questions/125814/radius-of-hydride-h%E2%81%BB-ion but even if you accept the lower value of 139.9pm it is still somewhat larger than F-

As suggested in the comments, hydride ion size is founded on shifting sands, and even the size of the cation is not really fixed. Everything in ordinary chemistry is at least somewhat squishy.
We should be looking at distances between ion centers and crystal structures. Where the crystal structure is the same we can assess the distance from Lattice constants, which are themselves a little squishy but at least accessible by direct (X-ray diffraction) experiments.
Let's look at sodium compounds. Lattice constants in Angstroms (Å) are from the Wikipedia articles on the respective compounds, lattice energies in kJ/mol are quoted from https://www.wiredchemist.com/chemistry/data/lattice-energies.
$\ce{NaH}: \pu{4.98 Å}, \pu{811 kJ/mol}$
$\ce{NaF}: \pu{4.62 Å}, \pu{904 kJ/mol}$
$\ce{NaCl}: \pu{5.64 Å}, \pu{769 kJ/mol}$
We would expect that for the same ion charges and crystal structure (the latter true of all the compounds cited above), we would expect smaller lattice constant to correspond to more lattice energy, and this we see. Essentially, the hydride ion holds its valence electrons so loosely that they wander over slightly more volume than the greater number of, but more tightly held, electrons in fluoride ions.
The similar lattice constants for alkali and alkaline earth fluorides versus hydrides facilitates the formation of solid solutions, which have been considered for hydrogen storage applications.


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Weird needle like crystals formed in galvanic cell after drying
So I set up a series of galvanic cells for fun; had a brass nut in a glass container and a chunk of zinc in another. I immersed the brass nut and zinc in a solution of saltwater, and connected them with a saltwater-soaked paper towel.
When the side with the brass dried out, a bunch of salt crystals, as well as blue copper carbonate formed. However, a bunch of long, white, needle-like crystals also formed. What could these be?

A picture is worth a 1000 words. Share a photo if possible.

What is the meaning of the words "when the side with the brass dried out ? Does it mean, that you had waited long enough to let both solutions evaporated ? During this time, did the cell function, and deliver enough currant to be discharged ? Did you see the salt crystals, the copper carbonate and the long needle-like crystals on the same electrode ?

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How exactly is heat supplied in an isothermal process
How exactly is heat supplied in an reversible isothermal process. If the temperature of system and surroundings always remains the same, then how come surrounding is supplying heat?
I studied that T of system and surrounding is the same while studying entropy.

Reversibility has two distinct meanings.

Bidirectionality. You can melt ice to water and freeze it back to ice.

Being in equilibrium along the whole process path. This can be just approximated in real processes by being near equilibrium.

The second meaning requires that reversible process is returned to its initial state by returning the neighbourhood to its initial state.
If the above cannot be done, the process is not reversible(2), even if it is reversible(1).

The heat is in your case provided as usually, but infinitely slowly for infinitely long time, with infinitely small temperature difference between the system and its surrounding.
A reversible isothermal  process is an idealised process lasting infinite (long enough) time, so heat can be provided at zero (small enough to neglect) temperature difference.
$$Q \propto \Delta T \cdot \Delta t$$
$$\lim_{\Delta t \to \infty}{\Delta T} = 0$$
E.g., you need 1 hour to transfer heat Q at temperature difference 1 K.
You need 2 hours for 0.5 K difference.
4 hours for 0.25 K
etc.
For time going to infinity and temperature difference going to zero, systems converge to being reversible.
With infinite time, zero temperature difference is needed to provide finite heat.
In reality, scientists choose "good enough" approach with small enough differences and long enough time.
Truly reversible processes do not exist in reality, as they are assumed to be in equilibrium all the time, having infinite time to achieve anything.

Sir I am asking that how Heat is provided are we providing the heat then how surrounding and system are at constant temperature or  does the system automatically absorbs heat from the surrounding but that doesn't seems the case could you please clarify.

Better is to write, what particularly you do not understand in my answer, as I did answer your question.

Ice fusion at $0°$C in "hotter" bath (just hotter than $0°$C) is a good example of a nearly reversible isothermal process. Heat can be introduced in such a system by having a mixture of ice + water in a metallic container, and dipping it into a bigger container full of water at $+1$°C.

If we consider 0 versus +1 Deg as approximate thermal equilibrium, than it is approximately reversible process.

@Poutnik. Yes. In my last message, 2 hours ago, I have explicitly written "nearly reversible".

@Maurice I have not said you have not. It was an addition, not objection.

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Predict who is the acid and who is the base in an acid-base reaction using pKa
In my book (Bruice) it is specified how it is possible to compare the pKa of two substances to understand who acts as an acid and who as a base. Here's the example in the book:
$\ce{NH3 + H2O}$.
$\mathrm{p}K_\mathrm{a}$(NH3) = 36
$\mathrm{p}K_\mathrm{a}$(H2O) = 15.7
Now it is explained that, being the $\mathrm{p}K_\mathrm{a}$ of water lower than that of ammonia, water will be the substance that will behave as an acid, so $\ce{NH4+}$ and $\ce{OH-}$ will be the products.
But this reasoning, in my opinion, does not make much sense, as the $\mathrm{p}K_\mathrm{a}$ of two different reactions are compared: the acid hydrolysis reaction of ammonia, which generates $\ce{NH2-}$ and $\ce{H3O+}$ and whose $\mathrm{p}K_\mathrm{a}$ is actually 36, and the autoprotolysis reaction of water, which generates $\ce{H3O+}$ and $\ce{OH-}$ and whose $\mathrm{p}K_\mathrm{a}$ is actually 15.7. So, I wanted to understand if it actually makes sense to compare the $\mathrm{p}K_\mathrm{a}$ of two different equilibria to predict who the acid is.
Furthermore, also $\ce{NH3}$, and not necessarily $\ce{H2O}$, can act as an acid. Indeed, ammonia can give both acid hydrolysis and basic hydrolysis to which it is possible to associate a $\mathrm{p}K_\mathrm{a}$ and a $\mathrm{p}K_\mathrm{b}$.
Reference (1) Paula Yurkanis Bruice, Organic Chemistry, 8th ed.; Pearson Education, Inc., 2011, pp. 58.

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

If you already know all that, then why ask?

Because i was in doubt with what my book said.

But this reasoning, in my opinion, does not make much sense, as the $\mathrm{p}K_\mathrm{a}$ of two different reactions are compared: the acid hydrolysis reaction of ammonia, which generates $\ce{NH2-}$ and $\ce{H3O+}$ and whose $\mathrm{p}K_\mathrm{a}$ is actually 36, and the autoprotolysis reaction of water, which generates $\ce{H3O+}$ and $\ce{OH-}$ and whose $\mathrm{p}K_\mathrm{a}$ is actually 15.7. So, I wanted to understand if it actually makes sense to compare the $\mathrm{p}K_\mathrm{a}$ of two different equilibria to predict who the acid is.

You are absolutely right, the argument does not work like that. It is possible to make a different, correct argument based on $\mathrm{p}K_\mathrm{a}$ values, though.
First, you write down the two possible reactions:
$$\ce{H2O(l) + NH3(aq)  <=>  OH-(aq) + NH4+(aq)}\tag{1}$$
$$\ce{NH3(aq) + H2O(l) <=> NH2-(aq) + H3O+(aq)}\tag{2}$$
I chose the order so that the reactant acting as acid is written first (on the far left) and the conjugate acid produced from the reactant acting as base is written last (on the far right).
Then, you compare the strength of the acids on the far left and the far right.
(1) $\ce{compare H2O with NH4+}$
(2) $\ce{compare NH3 with H3O+}$
For reference, here are the $\mathrm{p}K_\mathrm{a}$ values:
$\mathrm{p}K_\mathrm{a}(\ce{H2O})  = 15.7$
$\mathrm{p}K_\mathrm{a}(\ce{NH4+})  = 9.8$
$\mathrm{p}K_\mathrm{a}(\ce{NH3})  = 36$
$\mathrm{p}K_\mathrm{a}(\ce{H3O+}) = -1.7$
For (1), $\ce{NH4+}$ is a stronger acid than $\ce{H2O}$ (difference of about 6 in $\mathrm{p}K_\mathrm{a}$), so the equilibrium constant will be smaller than 1. For (2), $\ce{H3O+}$ is a much stronger acid than $\ce{NH3}$ (difference of about 38 in $\mathrm{p}K_\mathrm{a}$), so the equilibrium constant will be much smaller than 1. Neither reaction will be on the side of the products if both reactants are initially present at reasonably high concentrations.
Experience confirms this. The species $\ce{NH2-}$ does not exist in aqueous solution, it would get protonated immediately. Ammonia in water acts as a base, but most of the ammonia stays deprotonated (i.e. ammonia is a weak base). Of course, you can add acid to lower the hydroxide concentration and increase the ratio of ammonium to ammonia to the point where ammonium is the major species.

[OP in comments] when we compare the strength of the acids on the far left and the far right, we are still comparing the pKas of two different reactions

No, we are looking at the net reaction of one species donating and one accepting a proton, without showing the proton explicitly. For reaction (1), for example:
$$\ce{H2O(l) <=> H+ + OH-(aq)}$$
plus
$$\ce{NH3(aq) + H+ <=> NH4+(aq)}$$
gives (after cancelling the protons):
$$\ce{H2O(l) + NH3(aq)  <=> OH-(aq) + NH4+(aq)}$$
If you know the equilibrium constants of the "half reactions" (or the $\mathrm{p}K_\mathrm{a}$ values), you can get the equilibrium constant of the acid/base reaction.

Thanks for your answer! Therefore, since equilibrium (2) is shifted to the left more than equilibrium (1), the pKb is lower than the pKa, so that ammonia prefers to act as a base rather than an acid, but this does not exclude that acid hydrolysis can still take place.
Does this apply to all acid-base reactions? Even for example by reacting ammonia and a carboxylic acid?
Furthermore, when we compare the strength of the acids on the far left and the far right, we are still comparing the pKas of two different reactions. So Is It possible?

@Luckenberg You can't just compare the acid and base strength of a molecule, you have to know what it reacts with. Substances behave differently depending on whether they are in water, in glacial acetic acid or in liquid ammonia. However, with our usual bias towards aqueous solutions, you can say ammonia acts as a base rather than an acid. I would have to see the quantitative argument before I can say whether it is sound or not.

@Luckenberg "when we compare the strength of the acids on the far left and the far right, we are still comparing the pKas of two different reactions." No, I think we are just separating the acid/base reaction into half reactions, where one species gains a proton and the other loses it. I added a bit to the answer to illustrate.

brilliant! just one last quick example to evaluate if I understood correctly.
Consider the reaction between ammonia and acetic acid: if ammonia acts as a base, then the reaction is shifted to the right, since NH4+ is a weaker acid than CH3COOH [pKa(NH4+)= 9.8; pKa(CH3COOH)=4]. Conversely, if acetic acid were to act as a base, the equilibrium would be enormously shifted to the left, this because NH3 is an extremely weaker acid than CH3COOH2+. So this explains why ammonia actually acts as a base, since the other reaction is practically negligible. And this is true for every acid-base reaction.

@Luckenberg The discussion of ammonia and acetic acid is correct. They react almost to completion. CH3COOH2+ is such a good acid that it does not exist in water. In vacuum, doing mass spec on acetic acid, it might fly, though.


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How many carbon, hydrogen, oxygen, nitrogen, and phosphorus atoms are there in the observable universe?
If I could somehow reliably count all the carbon, nitrogen, hydrogen, oxygen, and phosphorus atoms in the observable universe, what number would I come up with?

A big one, that's for sure. Phosphate is not an atom. Other than that, see https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements#Universe

Nobody knows. Every number estimate will be a conjecture about the unknown worlds. Where does the universe start and where does it end? And who will define the entire universe?

@achem Astronomers. Whose job is partly observing how much stuff there is and what it is made of.

Yes but astronomers/astrochemists are as human as they can be. They are intelligent but nobody is yet good enough to know everything about "the entire universe." We do not know "everything" about the Earth and Sun, let alone the universe.

@AChem It isn't about knowing everything, it is about using actual observations to make a reasonable estimate. Yhey don't, for example, need to know everything about the sun to estimate its mass and composition very reliably.

If you could somehow reliably predict when the first nuclear fusion plant goes to production, what year will it be? // You contradict yourself in the single sentence. As the answer assumes it can done.

I changed this pesky "entire" as it indeed pretty much disqualified the question. Answer tells about observable universe and that's what question should ask (also no "phosphate atoms" in body...) If a user wants to answer a question, that is having serious issues, it should be fixed quickly.

Two key numbers can be estimated from known observable things in astronomy (and cosmology where current big bang models explain the processes that created the "light" elements and their abundance and observations broadly agree with theory). Spectroscopy allow astronomers to estimate the abundance of many elements in stars and galaxies.
There are also some estimates of how many nuclei there are (based on estimates of the mass of the observable universe.
We can combine the estimates about the number of nuclei with their relative abundance to give some approximate counts for elements.
The number of nuclei is estimated to be between $\pu{10^78}$ and $\pu{10^82}$ (which are both crazy big numbers though the upper estimate is 10,000 times the lower estimate so don't expect precision).
But we also understand the relative abundance, though most estimates count mass proportion (counting nucleons not atoms). Using the numbers in Wikipedia and adjusting the estimates for the mass of each elemental nucleus gives the following proportions:
H 92%
He 7.5%
O 0.08%
C 0.05%
N 0.01%
If we take the mid estimate of total nuclei as $\pu{10^80}$ then there are approximately $\pu{8E76}$ oxygen atoms, $\pu{5E76}$ carbon atoms and $\pu{1E76}$ nitrogen atoms in the observable universe (if I've done my maths right).

I think the number just changed, so you have to update it. Like the joke of dinosaurs going extinct 65 million and 5 years ago.

I gave the error bars of a factor of 10^4. Did it change by more than that? ;-)

Mid? Pretty sure the halfway point between those two estimates is still something like  $5 \times 10^{81}$ :)

@chepner I good point. I'm thinking logarithmically, as an astronomer probably would.

And yet, apparently the average density is only one proton every couple cubic meters.

Also, I feel the urge to mention one googol -- more than everything.

All these estimated numbers are a figment of human imagination. It is very audacious to even estimate the mass of observable universe. Should we start believing random websites and Wikipedia?

Even ChatGPT answered that "The mass of the observable universe is not an exact number and is hard to estimate. Different methods and data sources can produce different results. According to recent observations and measurements, it is estimated that the observable universe contains about 10^52 kg of matter. However, it is important to keep in mind that this is an estimate and there are many unknown factors that could affect the accuracy of this number. It's also important to note that this number only account for the observable universe, the actual universe could be much more massive."

@AChem 10^52 kg of hydrogen would be 6 * 10^78 hydrogen atoms. This is about the same as matt mentioned

@AChem There is a whole science called astronomy. It has a great deal of actual observational evidence and theory behind it. It is not merely "human imagination". And it has some basis for making estimates about the size of the actual universe. Those estimates have big uncertainties, but they are not imaginary or entirely speculative but are based on actual observations. and the actual numbers are not just random website guesses, but based on actual science.

@AChem Also, chat GPT does not make good judgements: it merely aggregates internet text without intelligent assessment. So the last sentence is not a sound basis for judging what the science of astronomy actually thinks.

You might like Feynman's essay (http://www.feynman.com/science/what-is-science)/, as Feynman wrote "Science is the belief in the ignorance of experts." The only point I was trying to convey was that there is no way to prove or disprove the accuracy numbers of such a massive system. Very little is known about it and little knowledge is dangerous to make big claims. Perhaps an advanced alien from a different galaxy can confirm those numbers. Ignore ChatGPT's satire.


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Why is the sp3-sp3 reductive elimination between the R1 and norbornene in the Catellani Cycle ''not favorable''?
I have recently started reading a book about C-H activation and I cannot figure out why in the Catellani reaction cycle the R1-norbornene reductive elimination side reaction which can be a major issue isn't favored.The author only briefly mentions sp3-sp3 coupling saying that it isn't favored but it doesn't provide an explanation (Because he probably assumes the reader already knows).
(Step 6.)

The book: C-H Activation (Topics in Current Chemistry, 292)

All most all organic reaction mechanisms are given according to the products received after doing variety of kinetic experiments. If you are familiar with reaction progressing energy diagrams, you know all intermediate reactions are reversible to give the thermodynamically most stable product(s). Thus, in the case of choice for $\ce{C_\mathrm{sp^3}-C_\mathrm{sp^2}}$ bond formation versus $\ce{C_\mathrm{sp^3}-C_\mathrm{sp^3}}$, the final product(s) rececieved go through more stable $\ce{C_\mathrm{sp^3}-C_\mathrm{sp^2}}$ bond formation (because reverse reaction require more energy to break).

This is the same situation when palladation of norbornene $(\bf{3})$ ocuurs to give $\bf{4}$ instead of the presence of the equally reactive $\ce{R^2-Y}$ in the reaction mixture, which is a terminal alkene.
Also keep in mind that, there is a possibility to receive tricyclic benzo-norbornane product by $\ce{C_\mathrm{sp^2}-C_\mathrm{sp^3}}$ reductiove elimination from $\bf{5}$, based on steric hindrence, solvent used, base used, etc. (Ref.1):

References:

Xiaojin Wua and Jianrong (Steve) Zhou, "An efficient method for the Heck–Catellani reaction of aryl halides," Chem. Commun. 2013,49(94), 11035-11037 (DOI: DOI  https://doi.org/10.1039/C3CC46381H).


text:
Cell notation of all aqueous solutions
If some redox reaction happened with no solid anode or cathode, like all aqueous solutions, something like the unbalanced reaction
$$
\ce {MnO4^- (aq) + C2O4^-^2 (aq)  -> Mn^+^2 + CO2 (aq)}
$$ What would an electrochemical cell notation represent in this situation? Is it even logical to try to get a cell notation? I'm really confused about that topic. I heard my professor saying something about a platinum electrode and I don't think this is rational by any means. Do things like this exists?

No, there is no use of a cell notation in this equation. However, the same reaction can be monitored via ORP (oxidation reduction potential) electrode which is indeed made of Pt. Search ORP electrodes.

Here is how to explain the use of the platinum electrodes.
If two solutions are prepared separately, one with acidic potassium permanganate, and one with sodium oxalate, they can be joined by a salt bridge. Now if one platinum plate is dipped into either solution, a potential difference can be measured between these two plates, or between these two solutions. Electrons are produced in the oxalate solution according to the half-equation :$$\ce{C2O4^{2-}-> 2 CO2 + 2 e^-}$$These electrons are collected by the platinum foil, and will go through the electric wire up to the other platinum foil dipped in the permanganate solution. Here they produce the following half-equation $$\ce{MnO4^- + 8 H^+ + 5 e^- -> Mn^{2+} + 4 H2O}$$ I should be mentioned that in the oxalate solution, the $\ce{Na^+}$ ions must quit the solution to maintain electric neutrality of the solution, and they do it by crossing the salt bridge joining the two solutions. When arriving in the permanganate solution they compensate the missing charges in solution. It takes some time to explain this movement, due to missing charges in the permanganate solution.
First, it should be mentioned that, in order to equilibrate the charges in the whole system, $10$ electrons must be produced by oxalate, and $10$ electrons must be consumed in permanganate. The corresponding half-equations become :
$$\ce{5 C2O4^{2-} -> 10 CO2 + 10 e-}$$ $$\ce{2 MnO4^- + 16 H^+ + 10 e- -> 2 Mn^{2+} + 8 H2O}$$
If the acid producing $\ce{H+}$ is $\ce{H2SO4}$, $16$ $\ce{H^+}$ ions are consumed in the half-equation. As a consequence, 8 $\ce{SO4^{2-}}$ ions are left in solution, after the arrival of the electrons through the platinum foil. These $16$ negative charges in excess are compensated by

$2$ $\ce{K^+}$ remaining after the consumption of the permanganate ions coming from $2$ $\ce{KMnO4}$,

$2$ $\ce{Mn^{2+}}$ ions produced by the second half-reaction

$10$ $\ce{Na^+}$ ions having crossed the salt bridge,

As a consequence, at the end of the whole process, the composition of the solutions are :

No ions in the former oxalate solution (just some $\ce{CO2}$), and
$\ce{2 Mn^{2+} + 2 K^+ + 10 Na^+ + 8 SO4^{2-}}$ in the former permanganate solution.

This is also the composition of the final solution one would obtain if the reaction had been done in only one phase, by mixing both solutions, with or without any platinum plates, which are now useless.

So in other words, if the reaction is carried out in two separate containers or in a single one the product will not change. However, we can make use of the electrons flow of the reaction only if we separate the two reactions with a Pt electrode in each one. And in that case the cell notation can be something like
Pt (s) | C2O4^2- (aq) | CO2 (g) || MnO4^1- (aq), Mn^2+ (aq)| Pt (s)

I am not familiar with the use of these vertical bars. But your description seems OK, but it seems to me that you have to add a vertical bar between $\ce{MnO4^-}$ and $\ce{Mn^{2+}}$

Note that an electrode(1) in the wider sense (electrochemical half cell) consist of

an electrode(2) in narrower sense(a conductor)
a redox system

The electrode(2) may be

the part of a redox system like $\ce{Zn(s)|Zn^2+(aq)}$
or may stay aside like $\ce{Pt(s)|H+(aq)/H2(g)}$

Without at least a half cell, a cell notation does not make sense.
If more redox systems exist in parallel in the same place, they have all at assumed equilibrium the same redox potential.


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Why is the conductivity of NaCl, NaBr and NaI decreasing with temperature?
I collected the conductivity of NaCl, NaBr, and NaI from 30 celsius to 80 celsius. I used the solutions mixed in water.
In all instances, the conductivity seemed to reduce as the temperature increased. I am unsure if it was an experimental error, or if my results actually make sense. Please state your reasons for thinking its an error or not.
Experiment Details: 50 ml, 0.1 molar of all solutions. I used vernier sensors for detecting the temperature and the conductivity.
For this experiment, I had a 50ml beaker on a hotplate and had the conductivity and temperature vernier sensors dipped in the liquid.
(I made sure the sensors are only touching the solution and not the bottom of the beaker). Then, I just waited for the hotplate to heat up the liquid to 80 Celsius.
From research, the opposite seems to be accurate, but I have conducted many trials giving the same trend of decreasing conductivity.
Source: https://link.springer.com/article/10.1007/BF02877571
(This is my first question, please let me know any other details that would be required to answer this question )

I assume you are talking about solutions if the salts, not the salts themselves, right? And you have to provide a lot more details about your experiment as well as the results of your literature search to make this question answerable.

The trend here (figure 1c, 0 to 30 centigrade) is a different one.  Yet your question doesn't state the salts' concentration, the equipment used, the design of the experiment (well yes, some replication is worth to estimate the significance of the results vs propagation of uncertainty, and accuracy vs precision of the readings.


text:
What is the highest volume change for common miscible liquid pairs, and common solvent/solid pairs?
When a solute is added to a solvent, the total volume of the solution is lower than that of the solute+solvent. A good example of this is the addition of salt to water, causing an overall volume decrease. What are the most dramatic example of this for commonly available solvents and solutes, both liquid+solid, and liquid+liquid?

Interesting question. My guess is the best examples will be pairs where mixing "turns on" some very strong intermolecular interaction which is not present in either compound separately. For example, choose a strong donor but weak acceptor of hydrogen bonds, and mix it with a strong acceptor but weak donor of hydrogen bonds. Maybe something like hexafluoroisopropanol and hexamethylphosphoramide.

What are the dramatic examples, to find the most dramatic one?

https://chemistry.stackexchange.com/questions/49913/what-solutes-will-give-the-largest-reduction-in-volume-when-dissolved-in-water https://chemistry.stackexchange.com/questions/23534/how-does-volume-contraction-in-solvent-mixing-work

https://chemistry.stackexchange.com/questions/167434/why-is-there-a-decrease-in-the-total-volume-when-naoh-dissolves-in-water https://chemistry.stackexchange.com/questions/83765/is-it-possible-to-have-volumesolution-volumesolvent-before-adding-solute

lower alcohols and water give ~10% decrease, if I remember correctly. In a plastic container, filled 50/50 with alcohol carefully floated atop water (might need a plastic-wrap separator), the container should get squeezed in the middle when shaken.


text:
Do electroluminescent quantum dot displays suffer from "burn-in" effect like OLED displays?
To my knowledge, compared to oled, elqd displays will be very cheap to mass produce using ink jet printing process. Quantum dots can have very narrow FWHM which allows displays to have wider color gamut of >90% rec2020. The response times of quantum dots are also faster than oled. Most importantly, it is also self-emissive like OLED.
However, one of the biggest concerns I have is if it suffers the same "burn-in" effect like OLED. To my knowledge, the challenge of commercializing ELQD displays is that the lifetime of the blue emitter currently only lasts a couple of hours at 1000 nits. However, I'm not sure if the term "burn-in" is separate from "lifetime." In an OLED display, when a static logo is shown for a really long time without any "burn-in" compensation technologies, the logo will burn in and the respective pixels will permanently shift colors which is very noticeable. I wonder if the same occurs with ELQD displays. In addition, I have also read somewhere that quantum dots are also not truly inorganic materials because the "ligands" are organic.

I’m voting to close this question because it's off-topic.

@Mithoron Why do you say so? https://chemistry.stackexchange.com/help/dont-ask

Well, everything is based on or uses chemical compounds and materials, but it does not mean chemists are experts in everything involving matter. This is highly applied science.

@Poutnik Most people conducting research on Quantum Dots and in companies trying to use these in display applications have PhDs in Chemistry. If I cannot ask a question here, I don't think there is a more appropriate place to ask this question. I do not think a question should be classified as "off-topic" just because some users here are not familiar with the topic.

I do not say you cannot ask here, I just think meeting an expert in the area would be accidental.

This question enters the interface between chemistry (like how the material quantum dots are made of [partially/fully] converts into an other) and physics (HOMO/LUMO, DoS). Because the later offers a tag dedicated to quantum-dots and an answered ELQD question , I suggest to keep this interesting question open and to relay the question to sibling physics.se.

In order to make the question more answerable here, it might be worth adding something about the chemistry of quantum dot materials and how they differ from traditional OLEDs. The audience who know something about the chemistry will be larger than the audience with a good familiarity with the technology.

@matt_black I agreee, https://en.wikipedia.org/wiki/Quantum_dot_display is a bit sparse, chemically speaking.


text:
What is the entropy change in isochoric process
I have studied that entropy increases with increase in temperature and it decreases with increase in pressure but in case of isochoric process both are happening at the same time but still the overall entropy of the system is somehow increased.
I want to know why the increase in entropy with increase in pressure dominates the decrease in entropy with increase in pressure since both are happening at the same time, resulting in an overall increase in entropy of system.

At an isobaric process, entropy increases with temperature, as you provide heat to the system.
At an isothermic process, entropy decreases with pressure, as you provide work to the system, which is converted to heat, leaving the system.
At an isochoric process, entropy increases with temperature (which increases the pressure), as you provide heat to the system.
With implied constant heat capacity $C_V$:
$\Delta S = \int_{T_1}^{T_2}{\mathrm{d}S}=\int_{T_1}^{T_2}{\frac{\delta Q}{T}}=\int_{T_1}^{T_2}{\frac{C_V\cdot \mathrm{d}T}{T}}=C_V \cdot \ln{\frac{T_2}{T_1}}$

For an arbitrary differential change of a single phase substance, we have $$dS=C_p\frac{dT}{T}-\left(\frac{\partial  V}{\partial T}\right)_PdP$$For an isochoric process, $$dV=\left(\frac{\partial  V}{\partial T}\right)_PdT+\left(\frac{\partial  V}{\partial P}\right)_TdP=0$$or$$dP=-\frac{\left(\frac{\partial  V}{\partial T}\right)_P}{\left(\frac{\partial  V}{\partial P}\right)_T}dT$$So, for an isochoric process, $$dS=C_p\frac{dT}{T}+\frac{[\left(\frac{\partial  V}{\partial T}\right)_P]^2}{\left(\frac{\partial  V}{\partial P}\right)_T}dT=\left[C_p-T\left(\frac{\partial  V}{\partial T}\right)_P\left(\frac{\partial  P}{\partial T}\right)_V\right]\frac{dT}{T}$$For all materials, the second term in brackets is always less than $C_p$, and is equal to $C_p-C_v$.


text:
If the effective nuclear charge of Beryllium(Be) is less than Sodium's(Na), how is the ionization energy of Be higher than Na?
According to Slater's rules the effective nuclear charge of Beryllium and Sodium is 1.95 and 2.20 respectively.
That means that the outermost electron of Na feels a stronger attraction from the nucleus than the outermost electron of Be.
But we  also know that the atomic radius of Be is smaller than Na and hence Be atom is smaller than Na. That means that the electrons of Be are closer to its nucleus so they must feel a greater attraction than the electrons of Na
The way i think about the above, somewhat contradictory statements, is that the pulling of the outermost electron in Na has to be bigger because there is also a greater shielding effect. Indeed the electron in the 3s orbital of Na feels a greater attraction than the 2s electron of Be but because of the inner electrons pushing it out it can't come closer to the nucleus and so Na ends up being bigger than Be.
Is my way of thinking this right?
My question though has to do with the 1st Ionization energy of Be being greater than Na. Why is that? It's the effective nuclear charge that tells us how much attraction the electrons feel from the nucleus.

$E_{pot} \propto Z_\mathrm{eff}/r$. For Na/
Be is the atomic radius ratio greater than Zeff ratio.

The attraction of alectron to the nucleus also depends on hoq far the electron is from the nucleus. For an outer eletron this is rendered as the atomic radius, which is bigger for sodium at the beginning of the third period versus beryllium in thw second slot of the second period. So sodium has its valence electron fartheraway from the nucleus than beryllium, weakening the nuclear attraction to it.


text:
How do you synthesize 1-fluoro-4-(tribromomethyl)benzene from toluene
I got this question from an o-chem II exam which kinda puzzles me a bit
My proposed path is:

which doesn't seem to be correct. Is there any other way within the knowledge of o-chem I & II?

Why do you think this is not correct? It is a bit old-tech but it will work. It is possible to directly fluorinate toluene, but you get a mixture of o and p  https://www.ncbi.nlm.nih.gov/pmc/articles/PMC8329385/

I am wondering that bromine reacts specifically with methyl, but my organic chemistry is very rusty.

@Poutnik under radical conditions it will

@Waylander I was being skeptical about the last step (bromonation), but seems like it was correct after all, thanks!

To complement a search in a reaction/literature database like Reaxys/Scifinder, there is an entry p-Fluorobenzotribromide with entries in sections literature and patents. Beside reference to the patent, double click on title of the patent reviews the patent (example) including section 7 / linked molecules (example). Although a search by patent can take long/be in part inaccessible by language/script.

This route will work.
p-Nitrotoluene is the major product of toluene nitration and may be isolated by distillation and recrystallisation, procedure here
This recent paper describes low temperature and light catalysed Balz-Schiemann conditions which are rather more user-friendly than some of the older procedures.
This patent describes a method of brominating the p-F-toluene to the benzotrifluoride


text:
How are the formulae of isochoric processes used in adiabatic processes?
We have the formula for an isochoric process:
$c_V = \frac{\mathrm{d}E}{\mathrm{d}T}$ at constant volume.
$c_V=$ molar heat capacity at constant volume.
But it's given for adiabatic changes,
$$\delta E=c_V\cdot \Delta T=W$$
How is this an equation for an adiabatic process when it's derived from equations of an isochoric process?
Is an adiabatic process also isochoric?
Where am I going wrong?

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

The heat capacity and internal energy of an ideal gas is independent of volume.  For cases where the gas is non-ideal, your equation for adiabatic changes is incorrect.


text:
Why are proteins made up of alpha amino acids and not beta amino acids?
Why are proteins made up of alpha amino acids and not beta amino acids? Or gamma amino acids? My idea on this would be that a world with beta or gamma amino acids would be too complicated? Is that true or are there other reasons?

Question could be made a bit less speculative, if you just asked what's the difference between such polyamides (arguably a "protein" made from beta amino acids wouldn't be a protein).

The real reason is : Nobody knows why life is made like it is.

Adding to Poutnik's useful answer, polyptides/proteins made of alpha-L-aminoacids are capable of participating in a number of secondary structures, such as alpha-helix, beta-pleated-sheet, &c. In turn these participate (along with the variety provided by aminoacid sidechains) in the construction of higher-order structural features with biochemical activity.
So one can say at least that things would be different if biochemical evolution had for some reason favoured beta or gamma aminoacids over alphas.
Also, as Poutnik suggested, the betas and gammas might more easily form unwanted cyclic lactams competing with any reaction (if still possible) to form (?pseudo)polypeptide chains. Then, even if such chains could still form in reasonable yield with beta and gamma aminoacids, the added methylene or dimethylene character of each aminoacide residue, (due to CaCb or CaCbCc carbon-chains taking the place of Ca), might restrict the overall range of physical-chemical properties in the pseudopolypeptide, irrespective of its side-chain -- perhaps, for example, it might make all aminoacid contributions more lipophilic than a corresponding alpha-amino-acid residue would do.
Really these considerations only add some approach towards exemplifying the difficulties of answer pointed out by poutnik. It may be doubted whether the subject holds enough urgency to persuade anybody to undertake the probably difficult course of synthetic and analytic chemical experimentation that would be needed to test the ideas.

Questions why? in context of various levels of life design are tricky and rather philosophical.
Beta/gamma  amino acids would not be optically active, neither (alternative) proteins formed from them. As nucleic acids are optically active due present ribose/deoxyribose, there would be problem with their synthesis.
Aside of that, considering the length of eventual lactame (cyclical amid) cycle, 3-member cycle of alpha amino acids (as possible unwanted side reaction) is much harder to form than 4/5-member cycle of beta/gamma amino acids.


text:
Stable thioesters in biological millieu?
It seems that most thioesters are highly reactive in cells and blood due to the high concentration of biological thiols. Are there any derivatives of thioesters that are stabilized at physiological conditions by certain functional groups? Fatty acid metabolism is mediated by transfer of CoA to the fatty acid, so I imagine this bond must be somewhat stable. Is this stabilized by the nitrogen that sits a couple carbons away from the thioester?

I think the most important thing is simply pH.

Stability between pH 4 and 7 (physiologically relevant conditions) is pretty much unchanged for all thioesters. I was thinking transthioesterification would have a greater effect on bond stability than pH.

can you give a source for your claim that "most thioesters are highly reactive in cells and blood"?  Biological thioesters that I'm familiar with are generally quite stable and only undergo transthioesterification at a high rate in the presence of enzymes that catalyze those reactions.

And my point is that at this pH their resistance to hydrolysis is good enough.

Sure, I understood your point, i was just explaining that my concern is not hydrolysis, its transthioesterification. There are countless reports showing extremely short half lives for thioesters in the presence of other thiols like N-Acetylcysteine and glutathione, which are present in high concentrations in cells and blood. The paper below shows S-methylthioacetate has a half life of 38 h in the presence of 1 mM of 2-
sulfonatoethanethiolate. Physiological glutathione conc is 5-10mM resulting in thioester half life of 1hr https://link.springer.com/article/10.1007/s11084-011-9243-4


text:
Is there a difference between $A_{1g}$ modes and $A_{g}$ modes?
I know that both are totally symmetric and that the '$1$', according to the Mulliken table, refers to symmetry around the $C_2$ axis but I was reading some papers and found that modes were sometimes referred to as $A_{1g}$ and sometimes referred to $A_{g}$, as shown below (difference is between undistorted and distorted unit cell).

Is $A_{g}$ equivalent to $A_{1g}$ here because the Raman tensors are identical for both (both are a 3 by 3 maxtrix with a leading diagonal of $a, a, b$) and yet the names are slightly different?

To answer your question of A$_{1g}$ and A$_g$ are the same? No, they are not,
because they correspond to irreducible representations of different symmetry point groups that differ in the allowed symmetry operations. However, they correspond to the same vibrational mode, since the quadratic functions associated with both these modes are same. If you see the Character Table for D$_{6h}$ and C$_{6h}$, you would see that A$_{1g}$ corresponds to the fully symmetric mode of vibration, and there are other modes of A, B and E types. However, once you lower the symmetry to C$_{6h}$ symmetry, you loose some elements of symmetry since C$_{6h}$ is a subgroup of D$_{6h}$. This would be more clear if you see the character tables available online D$_{6h}$ and C$_{6h}$


text:
Metaphosphoric acid - can it exist as monomer?
I read in my textbook that N2O5 can be prepared by reaction of HNO3 and P4O10 . A product HPO3 is also formed.
Then while discussing oxoacids of phosphorus , it said that metaphosphoric acid exists only in polymeric form.
So is there a contradiction ?

$\ce{HPO3}$ does not exist as a single entity. It is actually a cyclic molecule of formula $\ce{(HPO3)_n}$. The simplest such compound is $\ce{(HPO3)3}$ which is called trimetaphosphoric acid.

See previous discussions: https://chemistry.stackexchange.com/questions/163305/reaction-of-phosphorus-pentoxide-with-perchloric-acid, https://chemistry.stackexchange.com/questions/163305/reaction-of-phosphorus-pentoxide-with-perchloric-acid,https://chemistry.stackexchange.com/questions/71030/what-is-the-difference-between-metaphosphoric-acid-and-phosphonic-acid,https://chemistry.stackexchange.com/questions/137045/why-is-phosphorous-pentoxide-a-dehydrating-agent?noredirect=1&lq=1

https://chemistry.stackexchange.com/questions/127629/how-to-predict-the-products-of-an-inorganic-dehydration-reaction, https://chemistry.stackexchange.com/questions/71030/what-is-the-difference-between-metaphosphoric-acid-and-phosphonic-acid, https://chemistry.stackexchange.com/questions/153619/reactions-of-p2o5-and-p4o10-with-sulfuric-acid, https://chemistry.stackexchange.com/questions/54015/metaphosphoric-acid-and-water-in-cold

$\ce{HPO3}$ is an empirical formula for a material that is indeed polymerized. Phosphoric acid is actually any of several acids containing phosphorus(V); the formula $\ce{HPO3}$ represents a limiting case for long chains, or cycles containing a finite number (at least three) of $\ce{HPO3}$ units.
What is often called phosphoric acid, the monomer $\ce{H3PO4}$, is technically orthophosphoric acid.


text:
Equilibrium Constant: G vs. Go
I'm reading Clayden's Organic Chemistry textbook and in Chapter 12 on equilibrium he writes:
I am confused by the usage of $\Delta G$ as opposed to $\Delta G^\circ$. Isn't the equilibrium constant defined in terms of the standard-state Gibbs energy? Is there an unstated approximation in his definition of the equilibrium constant in terms of the bare $\Delta G$?

Yes, you are right. That equation should read $\Delta G^\circ = -RT \ln K_\text{eq}$. $\Delta G$ refers generally to the Gibbs free energy change for any process.

Suppose the reaction is $A\rightleftharpoons B$ then the equation is $\Delta G'=\Delta G^0 +RT \ln(P_B/P_A)$ and at equilibrium the pressures are the equilibrium values and $\Delta G'=0$ so that $\Delta G^0 =-RT\ln(K_p)$ where $K_p$ is the equilibrium constant.


text:
Do Boron and Silicon form metallic-type bonds in alloys?
For context I am a physics student currently doing a project that involves metallic glasses. I am trying to figure out if you can get Ionic bonds inside an alloy with metalloids. More specifically, I have 2 examples.

Iron,Nickel,Boron
Palladium, Copper, Silicon

I have recently found out about Arkel–Ketelaar triangles and assume they answer my question but I'm not entirely sure how to use them in the context of an alloy of three components. I am also not sure if I need to use a particular measure of electronegativity or if any will do.

"When in Rome, do as the Romans do." In a predominantly metallic environment, B and Si would act somewhat like metals.

@IvanNeretin This was my suspicion, but can you maybe link me to a source for this? Its ok if it's a book or something I can get from my library.

Silicon makes many binary metallic compounds (silicides) with most metals in the periodic table (Al, Ag, and Au being outliers). There are three Si-B alloys, SiB${3}$, SiB${6}$, and SiB$_{n}$ (it has some width to the phase field, and $n$ is about 23.39 at the peritectic point).

The two examples are been used to make metallic glasses. Look here.

@MathewMahindaratne I know they've been used to make Metallic Glasses, what I'm curious about is if the bonds between those atoms and the rest are of a metallic sort.


text:
Find percentage of carbon monoxide reacted using ideal gas equation
Question
A palladium or platinum catalyst was used in an automobile to convert carbon monoxide gas to carbon dioxide according to the following reaction:
$$\ce{2CO(g) + O2(g) -> 2CO2(g)}$$
A chemist researching the effectiveness of a new catalyst combines a 2.0 : 1.0 mole ratio mixture of carbon monoxide and oxygen gas (respectively) over the catalyst in a 2.45 L flask at a total pressure of 745 torr and a temperature of 552 K. When the reaction is complete, the pressure in the flask has dropped to 552 torr.
What percentage of the carbon monoxide was converted to carbon dioxide?
(a) 67.7 %
(b) 85.7 %
(c) 77.5 %
(d) 57.8 %
(e) 46.5 %

My attempt
I calculated the moles of gas in the original system
$$n = \frac{PV}{RT} = \frac{745 \times 2.45}{8.31 \times 552} = 0.398 \, \mathrm{mol}$$
Then I worked out the moles of $\ce{CO}$ and $\ce{O2}$ in the original system using the mole fractions.
$$\ce{2CO   +  O2  ->   2CO2}$$
moles of $\ce{CO}$ = $(2/3) \times 0.398 = 0.265\, \mathrm{mol}$
moles of $\ce{O2} = (1/3) \times 0.398 = 0.1326\,\mathrm{mol}$
moles of $\ce{CO2} = 0\, \mathrm{mol}$
Then I calculated the number of moles in the new system.
$$n-\frac{PV}{RT} = \frac{552 \times 2.45}{8.31 \times 552} = 0.295\, \mathrm{mol}$$
Then I worked out the moles of $\ce{CO, O2}$ and $\ce{CO2}$ in the new system using the mole fractions.
$$\ce{2CO(g) + O2(g) -> 2CO2(g)}$$
moles of $\ce{CO} = (2/5) \times 0.295 = 0.118\, \mathrm{mol}$
moles of $\ce{O2} = (1/5) \times 0.295 = 0.059\, \mathrm{mol}$
moles of $\ce{CO2} = (2/5) \times 0.295 = 0.118\, \mathrm{mol}$
Therefore percentage of CO reacted
$$\%_{\ce{CO}} = \frac{0.265 - 0.118}{0.265} \times 100 = 55.5\, \%$$
The correct answer according to the mark scheme is C) 77.5%. Is there any other method to get this answer?

You need not to compute molar amounts at all. You can compute the final/initial pressure ratio that is equal to final/initial molar amount ratio. From that, you can directly calculate the percentage of conversion from inverted function pressure ratio = f(conversion percentage).

Let:
A represent $\ce{CO}$
B represent $\ce{O2}$
C represent $\ce{CO2}$
So the reaction becomes:
$$\ce{2A + B->2C}$$
We begin by calculating the initial molar fraction of A.
Since only A and B are initially present:
$$y_{Ao}+y_{Bo}=1$$
Since the initial molar ratio between both species is 2:1, then:
$$y_{Ao}=2\;y_{Bo}$$
Using both equations to solve for $y_{Ao}$, we get:
$$y_{Ao}=\frac{2}{3}$$
Next, we observe that between the initial and final state, temperature and volume are constant, but there is a change in moles between reactants and products, so:
$$T=T_o$$
$$V=V_o$$
$$\Delta n=-1$$
If we divide the ideal gas law in the final state by the initial state, we get:
$$\frac{PV}{P_oV_o}=\frac{nRT}{n_oRT_o}$$
Cancelling all equal terms:
$$\frac{P}{P_o}=\frac{n}{n_o}$$
The molar ratio on the right hand can be expressed in terms of conversion $X$ and expansion coefficient $\epsilon$ after performing a molar balance of A, B, and C:
$$
\require{cancel}
\begin{align}
n_A=n_{Ao}-ax \\
n_B=n_{Bo}-bx \\
n_C=n_{Co}+cx \\
\hline
n=n_o+\Delta nx
\end{align}
$$
In this case, $\Delta n=c-a-b$, and small $x$ represents the non-normalized conversion with respect to the limiting reagent. In order to normalize it, we arbitrarily define A as the limiting reagent, and its conversion as:
$$X=\frac{ax}{n_{Ao}}$$
Solving for $x$:
$$x=\frac{n_{Ao}X}{a}$$
Substituting $x$ into the overall molar balance:
$$n=n_o+n_{Ao}\frac{\Delta n}{a}X$$
Dividing both sides by $n_o$:
$$\frac{n}{n_o}=1+\frac{n_{Ao}}{n_o}\frac{\Delta n}{a}X$$
Or equivalently:
$$\frac{n}{n_o}=1+y_{Ao}\frac{\Delta n}{a}X$$
Then, we define:
$$\epsilon=y_{Ao}\frac{\Delta n}{a}$$
And substitute it above to get:
$$\frac{n}{n_o}=1+\epsilon X$$
Which can also be substituted above:
$$\frac{P}{P_o}=1+\epsilon X$$
Solving for $X$:
$$X=\frac{\frac{P}{P_o}-1}{\epsilon}$$
Calculating $\epsilon$:
$$\epsilon=y_{Ao}\frac{\Delta n}{a}=\frac{2}{3}\;\frac{-1}{2}=-\frac{1}{3}$$
Plugging in all numerical values:
$$X=\frac{\frac{552}{745}-1}{-\frac{1}{3}}$$
So we get:
$$X=0.777$$
Or equivalently 77.7%

Hmm,  1-2 equations should be enough.

@Poutnik Elaborate answers is my style :) or at least in cases where it's likely the level implicated is high school/college.

After some point it may become intimidating, like more complex than it is. But yes, I know your style. :-)

@Sam202 This is great. Thank you so much!

@Sam202 However this step "The molar ratio on the right hand can be expressed in terms of conversion X and expansion coefficient e" is not something taught in high school chemistry or biology. Can you please expand on that or direct me to a resource that explains where the expansion coefficient comes from for a bio student?

@Poutnik Can you please show your simplified working well?

@Natalie See the answer.

@Natalie It is derived from a molar balance of all species that "enter" (reactants) and "leave" (products) the reaction. For example, for reactant A, the final amount remaining after reaction completion is: $n_A=n_{Ao}(1-X)$, where the conversion $X$ is defined as the ratio between amount reacted and initial amount: $X=\frac{n_{Ao}-n_A}{n_{Ao}}$. However, since A is arbitrarily defined as the limiting reagent, no molar coefficients appear in its equation. Conversely, when the total amounts $n_o$ and $n$ are involved, their equations must include all molar coefficients for $\Delta n$.

@Sam202 Alright… I’m getting everything except the following part. In your previous working we have n/no = 1+eX and in your comment we have na=no(1-X). I’m sorry if it seems obvious, but why the change?

@Natalie I updated my answer to include molar ratio derivation in terms of $\epsilon$ and $X$.

Oh I see it now. Thank you so much! @Sam202

@Poutnik Thank you! I can see now how to go with your working as well.

Using the ideal gas state equation as the starting point:
$pV=nRT \implies \frac{p_1V}{p_2V}=\frac{n_1RT}{n_2RT} \\
\implies \frac{p_1}{p_2}=\frac{n_1}{n_2}$
leading to the consequence that the gas pressure is proportional to the total molar amount of present gases.
As there is molar conversion in ratio 2/3, the final/initial pressure ratio 2/3 would mean 100 % conversion.
The rest below is just interpolation between the ratio 1 (0 %) and 2/3 (100 %)
If

$\alpha$ is the percentual conversion
$p_\mathrm{tot,ini}$ is the initial pressure of gases
$p_\mathrm{tot,fin}$ is the final pressure of gases

then
$$\frac{p_\mathrm{tot,fin} }{ p_\mathrm{tot,ini}} =  1 - \frac 13 \cdot\frac{\alpha}{100}$$
$$\alpha = (1 -  \frac{p_\mathrm{tot,fin} }{ p_\mathrm{tot,ini}})\cdot 300 \ \%=  (1 -  \frac{\pu{552 torr} }{ \pu{745 torr}})\cdot 300 \ \% \approx 77.7\ \% $$


text:
UV absorbance titration to determine binding stoichiometry
I previously posted a question but I think I did not explain correctly. I want to construct a Job plot using UV-Vis titration to find binding stoichiometry. I have 20uM of ligand solution and 20uM of DNA solution. Volume of each solution is 800uL. The total concentration of ligand and dsDNA remained unchanged(20uM) during entire titration. Ligand intercalates into dsDNA.
I want to know the protocol to do UV-Vis titration. Like how I should proceed. Do I need to prepare solution of different concentration everytime (for example ligand solution --- 20uM, take UV absorbance,
Now make a complex solution of say 18uM ligand and 2uM dsDNA and so on)
I have no idea what to do.
Thank you.

It is unclear what you want and therefore difficult to provide it. Unless you have troubles with equation y = x / ( 800 - x ). Regarding protocols, it is mixing liquids in desired ratio, waiting for required time if needed and then measuring absorbance.   Then you can interpolate points visually or numerically by 2 lines and taking the intersection as the equivalence point.

I want to do continuous variation method to construct Job plot.

As it is about the same thing, with additional info, it should be put in initial question, with a comment if needed.  // So you have 800 uL total for each solution, right? What total volume is needed for measurement?   //  Asking for answers and not providing enough relevant details, purpose, context or background of questions are contradictory decisions.

I closed previous question as a duplicate. As @Poutnik mentioned, please [edit] your question instead of re-posting it again with more details. This way comments, your editing history and narration are all preserved within one post.

I can advice only in general methodic, I cannot say what you can afford wrt the particular solutions, regarding their stability and interaction.
If the volume of available solutions is not the limiting factor, the easiest way is to prepare extra solution mixture for each measurement.

Take $x\ \pu{ \mu L}$ of the ligand solution
Take $a - x\ \pu{\mu L}$ of DNA solution
It gives $\frac{x}{a-x}$ ligand/DNA ratio.

Then you can measure the series of plot point absorbance vs ligand/DNA ratio.
If the volume of available solutions is the limiting factor, you may try the series of incremental additions of the ligand and following measurements.
You would take

$V_1\ \pu{ \mu L}$ of solution of the prior ratio $r_1$
add $V_2\ \pu{ \mu L}$ of the ligand solution to get the new ratio:

$$r_2 = \frac{r_1(V_1+V_2) + V_2}{V_1}$$
But there are 2 cons:

Higher cummulative experimental errors with excessive solution manupulations.
Possible effects of stability and interaction of incrementally mixed solutions.

When you have measurement done, you can visially or numerically approximate the poins by 2 lines and to take the intersection as the point of equivalence.

Thank you for the explanation. So is it same if I want to calculate for DNA/ligand ratio?

Yes, it is, you would have just mutually switch the relevant parameters.


text:
Why can't the product of an acid reacting with water react with water itself and produce OH-?
Why can't the product of an acid reacting with water react with water and produce $\ce{OH-}$?
Take the reaction below as an example:
$$\ce{H2CO3(aq) + H2O(l) <=> HCO3-(aq) + H3O+(aq)}$$
$$\ce{HCO3-(aq) + H2O(l) <=> CO3^2-(aq) + H3O+(aq)}$$
Why can't the $\ce{CO3^2-}$ react with water (not $\ce{H3O+}$) and produce $\ce{OH-}$ ions? If it can then why is the solution acidic and not basic?
PS. I am a high school student and would really appreciate a straightforward answer.

It can and does. They all sit together in a nice equilibrium.

The following hydrolysis reaction is a well known equilibrium : $\ce{CO3^{2-} + H2O <=> HCO3^- + OH-}$ But $\ce{CO3^{2-}}$ is not an acid. It is a base.

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

@Maurice I see. But if this reaction also happens in a container with water and H2CO3, why is the solution and acid and not a base?

Imagine CO2 solution with anions OH-, HCO3- and CO3^2-. As the only present cation is H3O+ and there is requirement of electroneutrality, there is more H3O+ than OH- and the solution must be then acidic.

It definitely does. Let's simplify it to a monoprotic acid (like acetic acid, $\ce{CH3COOH}$), which I'll just represent as $\ce{HA}$ for simplicity. In a solution of the acid, you will have all of the following species floating around: $\ce{HA}$, $\ce{A-}$, $\ce{H2O}$, $\ce{H3O+}$, $\ce{OH-}$. Every conceivable reaction between these species will happen, such as:
$$\begin{align}
\ce{HA + H2O &<=> A- + H3O+} \tag{1} \\
\ce{A- + H2O &<=> HA + OH-} \tag{2} \\
\ce{2 H2O &<=> H3O+ + OH-} \tag{3}
\end{align}$$
To obtain a completely accurate value for the pH, and the equilibrium position, it is necessary to take all of these equations into account.*
But often we can get away with less work, and only bother with the first. Why? One qualitative way of looking at it is: when we add a weak acid $\ce{HA}$ to water, we know that very little of it will dissociate anyway; this is kind of by definition. So, there isn't much $\ce{A-}$ in the system anyway; and the second reaction can pretty much be ignored.
That's not to say that there is no $\ce{OH-}$ in the system. There is still $\ce{OH-}$ floating around. For example, in a pH 5 solution (i.e. $[\ce{H+}] = \pu{10^{-5} M}$), we have that $[\ce{OH-}] = \pu{10^{-9} M}$.* It's still there, it's just that there's very little of it.
A more formal explanation would involve solving it properly and verifying that the true solution is very close to the approximate solution obtained by ignoring eqns. (2) and (3).
On the other hand, if $\ce{HA}$ is a strong acid (like $\ce{HCl}$), then the reverse reaction $\ce{Cl- + H2O -> HCl + OH-}$ is completely negligible because $\ce{Cl-}$ is a pitifully weak base. (Strong acids have very weak conjugate bases.)

* In fact, these equations are not independent; note that (1) + (2) = (3), so there are only two 'pieces of information' to be obtained from these reactions, one corresponding to $K_\mathrm{a}$ and one corresponding to $K_\mathrm{w}$. But that's a story for another day.

Wow, this is the best explanation I have ever received! Although I am still wondering if the acid is so weak that we can ignore the second reaction (due to the small amount of A-) how can the H3O+ make a difference? (considering the fact that the amount of A- and H3O+ is the same?)

In pure water, which is pH 7, there's only $\pu{10^{-7} M}$ of $\ce{H3O+}$. You really don't need a lot of extra $\ce{H3O+}$ to turn that into pH 4 or 5, so just the tiny bit that dissociates from the weak acid is enough.

if this is the case, then how can we ignore the reaction that A- will have with water and the small amount of OH- that gets produced?

@AuthenticMelody [H3O+]=[OH-]+[HCO3-] + 2[CO3^2-] -> [H3O+] >[OH-] -> pH <7.

@AuthenticMelody Because basicity (as judged by pH > 7) isn't caused by A-; it's caused by OH-. Reaction 1 starts from quite a lot of HA and generates a tiny bit of H3O+ and A-. Reaction 2 starts from a tiny bit of A-, and generates a vanishingly small amount of OH-. Of course, implicit in this is the assumption that HA is a weak acid and A- is a weak base. This is true, though: consider acetic acid, pKa = 4.76; that's a weak acid. The conjugate base, acetate, has a pKb of 14 - 4.76 = 9.24; that's also a weak base. So, reaction 2 pretty much just doesn't happen.

I really like this question because it digs at wanting to understand what's going on beneath the equations you've been given to know why that equation works.
Orthocresl answered it really well, but I might be able to get a little more to what pH and pOH are and give a little more understanding to what's meant by acidic and basic, and what equilibrium is. I feel like that might be as much your question as whether or not H2O can react with CO3 2- (apologies, first time answering a question here, and not sure how to format this correctly).
tl;dr version up front:
Pure water has an equal amount of H+ and OH-, both of them exist in water. We consider this neutral, neither acidic or basic. Anything that moves the concentration to favor more H+ is considered an (Arrhenius) acid, anything that moves the concentration to favor more OH- is considered a base. Even so, there will still be both H+ and OH- in the solution. Individual water molecules will continue to accept protons (H3O+, usually shown as H+) and donate protons (OH-), but there will be more water molecules accepting a proton to form H3O+ than donating them to form OH- at lower pH.
non tl;dr version:
Very, very few chemical reactions are truly one way/irreversible unless the products leave the system entirely. Almost all will be occurring in both directions at the same time. But, one will be occurring more/faster. Let's say reaction A occurs twice as fast as the reverse reaction B (A: a + b -> c; B: c -> a + b). Since A occurs twice as fast a B, then there will be more products, c, than reactants, a and b. However, c will still be producing a and b. Eventually you'll get enough c that the reaction will reach a point that it will look like the reaction stopped. If reaction A occurs twice as fast as reaction B, this will be when there's twice as much products as reactants.
Basically, if a + b produce c twice per minute, and c produces a + b once per minute, you'll eventually get twice as much c as you have a and b, and the reaction will look like it stopped from a macro point of view, but will continue to occur in your solution.
Take what I said there from a conceptual view. It's useless for practical purposes. If you want to know the math, this video explains it really well (https://www.youtube.com/watch?v=2PM1yc_z4Bk&list=PL2ub1_oKCn7qmUZ80MJDPaRTdgJm8ZenX).
I don't see a good way to not discuss different definitions of acids even though I feel it's a little beyond this question. Arrhenius acid is anything that lowers the pH of pure water (base is the opposite). But that's not a good definition. A much better one is the Bronsted definition, anything that donates a proton acts as an acid, and anything that accepts a proton acts as a base (there's a 3rd definition, but it's difficult to understand at first). In this case, your carbonic acid is acting as an acid when it donates a proton to water to form H3O+, and water acts as a base. However, the reverse happens as well. H3O+ will donate a proton to HCO3-. In which case H3O+ is acting as an acid. H2O can also act as a Bronsted acid and donate a proton to form OH-. So yes, CO3 2- can and does act as a base and accept a proton from H2O, and you get OH- floating around.
In any aqueous solution, you'll end up with both H3O+ and OH-. This is usually measured with pH. pH is actually the result of this equation pH = -log(10)([H+]). That equation means that pH is the negative base 10 logarithm of the concentration of protons (hydronium ions) in the solution. However, because we know there's an equilibrium between [H+] and [OH-], we can actually measure this in terms of the concentration of OH- ions instead, pOH. pOH is remarkably, the negative logarithm of the concentration of hydroxide ions, pOH = -log(10)([OH-]).
So let's say you have a pH of 5, a slightly acidic solution. You'll have a pOH of 9. The pOH measures in the opposite direction, pOH below 7 is basic, above 7 is acidic. These two measure will equal 14. A pH of 1 will correspond to a pOH of 13. So even at a very high acidity of pH 1, you will still have H2O acting as an acid and donating protons to form OH-. It will just be happening very slowly, and so there won't be many of them.

Thank you for the thorough answer. It really helped me gain a better understanding of the topic.


text:
What is the difference between differential yield and selectivity?
I have a hard time understanding the difference between the differential yield and overall selectivity of a reaction.
Levenspiel, Chemical Reaction Engineering writes:

For convenience in evaluating
product distribution we introduce two terms, p and a. First, consider the decomposition of reactant A, and let cp be the fraction of A disappearing at any instant
which is transformed into desired product R. We call this the instantaneous
fractional yield of R. For any particular set of reactions and rate equations cp is a function of CA, and since CA in general varies through the reactor, p will also change with
position in the reactor:
$ \varphi = \frac{-dC_A}{dC_S}$

and for selectivity:

The Selectivity. Another term, the selectivity, is often used in place of fractional
yield. It is usually defined as follows:
$$\text{selectivity} = \frac{\text{moles of desired product formed}}{\text{moles of undesired material formed}}$$

So how do they differ? How can I imagine both?
Aside from that, they also provide some nice plots:

One question I have:
How can I imagine the selectivity for a MFR (gray area) crossing the selectivity curve? The formula for the integral yield for a PFR/MFR make sense, but how do I imagine that in a practical sense?
Also: What does the curve describe?

First, these two concepts as you may already know are useful for describing various chemical reactions taking place or multiple reactions. Why Levenspiel wants you to know these concepts? Because you are interested, typically, in one product, not all of them.
As an example, consider the following reaction scheme, where you are interested in B; and we assume power rate laws for both of them:
$$ A \rightarrow B \hspace{1 cm} r_1 = r_{1,B} = - r_{1,A} = k_1C_A^2 $$
$$ 2A \rightarrow C \hspace{1 cm} r_2 = r_{2,C} = -r_{2, A} = 2k_2C_A^3 $$
The fractional yield of $A$ and selectivity are
$$ \phi_{B/A} = \dfrac{r_{1,B}}{-(r_{1,A} + r_{2,A})} = \dfrac{k_1C_A^2}{k_1C_A^2 + 2k_2C_A^3} $$
$$ S_{B} = \dfrac{n_B}{n_C} $$
So now we can answer your first question:

The fractional yield gives you, as a function of the concentration of species $A$, the rate of formation of the species you want with respect to the rate of consumption (with a negative sign) of the reactant $A$. As an engineer, what do you want to do? Maximize this quantity once we give you a type of flow within a reactor. This does not mean maximizing $ B $, but will give you the best exchange between the reactant $ A $ and your valuable product $ B $.
The selectivity gives you, as a function of time or position, the amount in moles formed of the species you want with respect to the amount of moles of the species you do not want. As an engineer, what will they ask you? Minimize this quantity once we give you a type of flow within a reactor. As before, this does not mean that you will generate a lot of $ B $, rather, it will give you the best exchange in terms of generating $ B $ and not generating $ C $.

Levenspiel states that this definition is not useful, because it may be hard to quantify in moles all of the species that you don't want (imagine 1 valuable product and 19 not valuable products). However, a more useful mathematical definition would be the one I give you below, analogous to the yield
$$ S_{B/C} = \dfrac{r_{1,B}}{r_{2,C}} = \dfrac{k_1C_A^2}{2k_2C_A^3} $$
Lets go now to the second part. The curves you are seeing there are indeed the fractional yield vs the concentration of $A$, i.e., $\phi_{R/A}(C_A)$ vs $C_A$. They only depend on the kinetic rate laws governing the reaction scheme. However, in this case, thay are so simple that they only depend on the concentration of only one reactant. Depending on the mathematical form, they may have a maximum, or not, or have rare behaviours (imagine how the rate law may dramatically change with a catalyst, etc.). In addition, they also give you the concentration of $R$, i.e., the species you want. For a MFR, it is a rectangle of base $ C_{A0} - C_A $ and height $\phi_{R/A}(C_A)$. For a PFR, it is the area under the curve bounded between $ C_{A0} $ and $ C_A $, as you said.
The MFR can't operate passing the curve, because its type of flow is such that its operating point is the exit; such that the whole reactor "exists" as a unique point in that curve.


text:
Is the proportion of molecules with sufficient energy to react *really* equal to $e^{-E_A/RT}$?
In A-level chemistry class today I was just told the following:

For a reaction with activation energy $E_A$, the proportion of reactant molecules with kinetic energy $\ge E_A$ is: $$e^{-E_A/RT}$$Where $R$ is the gas constant and $T$ is the temperature of the reactant(s?).

I find that very suspicious. We are told that the area under a Boltzmann distribution curve 'represents' the number of particles with energy greater than a given threshold, so the above statement should suggest: $$\int_{E_A}^\infty\text{Boltzmann}(t)\,\mathrm{d}t=Ke^{-E_A/RT}$$Up to some constant $K$ (we aren't taught quite what the distribution is nor any of the grisly thermodynamical details, so I threw in a constant $K$ just because I'm not sure whether it's computing proportions/probabilities or the total amount of substance).
This would imply that the Boltzmann distribution curve is exponential. However, if one sketches a Boltzmann distribution as we are taught to do in A-level it is clearly not exponential. It is vaguely skew-normal from what I can see. It does look exponential after a certain point (once you're over the central hump) but it isn't everywhere. I then suspect that what we are told in class is sort-of accurate and is applicable to most of the tame situations that could arise in A-level theory, but it is not always correct.
My question:
In terms as elementary as possible, could someone explain:

To what extent is the statement: "the proportion of molecules with sufficient energy is $e^{-E_A/RT}$" actually true?

Before someone says it, yes, I could look up precise definitions of the Boltzmann distribution and try to headache my way through this. But Googling mathematics definitions and trying to understand them is hard enough: in my experience, trying to do the same for any of the natural sciences is essentially impossible (everyone tells you a different thing up to a different level of expertise...) so I doubt I'd get anything out of that.
Many thanks for any response.

That's a good approximation. $E_A$ is always well above the central hump, otherwise most molecules would already have it and the reaction would just occur instantly.

You are confusing two things. When you say the Boltzmann distribution is not exponential, you are correct.  But the "number of molecules with energy greater than x" is not the distribution, it is the integration of the distribution from x to infinity. And that is exponential in most circumstances (essentially where the X is higher than the "hump" of the full distribution.)

It seems to me that you are confusing the Maxwell-Boltzmann distribution with the Boltzmann distribution sometimes also referred to as Boltzmann ratio.

@matt_black When you say "in most circumstances", do you mean that there are other factors at play but the (Maxwell?)-Boltzmann distribution is approximately (locally) exponential most of the time?

@Paul Perhaps. We don't learn the concept very precisely at A-level, we're just given a name for a distribution of number of molecules against kinetic energy and a rough outline of its shape and basic properties.

@FShrike For most real reactions the proportion of molecules with higher than X activation energy is small. In other words the X is far away from the "hump" in the distribution. The resulting integration of the distribution will look very close to an exponential in X (same point as Ivan Neretin's comment).

Ok. Thanks to all responders for the clarifications

In an ideal gas the Boltzmann distribution tells you the proportion of molecules in  given velocity range. This is purely exponential. The Maxwell-Boltzmann distribution does the same thing for the molecules' speeds. This is not a pure exponential, see https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution. The difference is important.

@IanBush That's subtle. Does the difference arise from there being a significant likelihood of many particles having the same speed but at different velocities, thereby changing the probability distribution? I suppose what we draw in school (the kinetic energy, so a constant multiplied by the square of the speed) is a squared and scaled version of Maxwell-Boltzmann then

Yes. Hit the nail on the head. A given speed corresponds to many different velocities due to the different directions a molecule can move in.

For an extended discussion, see this possible duplicate: https://chemistry.stackexchange.com/questions/108854/boltzmann-factor-vs-graph-of-maxwell-boltzmann-distribution

When $E_A/k_BT \gt 1$ then integrating the Boltzmann distribution from $E_A \to \infty$ does give a result $K\exp(-E_A/K_BT)$ where $K$ depends on $E_A/T$ so varies weakly compared to the exponential.  (The Maxwell-Boltzmann distribution is $p(v)\sim \exp(-mv^2/(2k_BT)$ in velocity so in terms of energy $E=mv^2/2$ we get $p(E)\sim \exp(-E/k_BT)$.)


text:
Redefine a Mixing Modulus based on new reaction scheme
I am studying the effect of diffusion on liquid chemical reactions.  The model is very simple in that it assumes that striations or slabs of fixed thickness of one reactant are next to the other.  Typical widths (L) of the striations or slabs are on the order of microns. However, the larger the value of L the greater the effect of diffusion on the overall reaction.  For the reaction schemes presented below, this means that more S is produced assuming that A is originally present in a slight stoichiometric excess to B.
Originally, the chemical reaction being studied was as follows:

The rate constants at 25 C are $k_1$ = 12,000 and $k_2$ = 2.07 both in $m^3/(mol*s)$.  If diffusion does not affect this reaction scheme, then the amount of S produced is very, very small (<0.1%). The first reaction is considered to be virtually instantaneous.  Thus when comparing the "characteristic" reaction time to the "characteristic" diffusion time of the reactants, the following relationship was used:
$$M = \frac{k_2c_{bo}L^2}{D}$$
where $k_2$ is the reaction rate constant of the second, relatively slow reaction, $c_{bo}$ is the initial concentration of B, L is the striation or slab thickness and D is the diffusion coefficient.  This has worked fine for analyzing the data.
Now, I want to update to the following reaction scheme:

where the rate constants are as follows: The rate constants at 25 C are $k_{1o}$ = 920, $k_{1p}$ = 12240, $k_{2o}$ = 1.84 and $k_{2p}$ = 22 all in $m^3/(mol*s)$.
My question is: can M be rewritten in terms of this new reaction scheme?  Let's assume that the reaction of A with B by both mechanisms (ortho and para) is still fast enough and that the reaction of R with B is still the rate limiting step for the reaction (or is this now a bad assumption?).
It looks like the 2nd pathway of the reaction of A with B is an order of magnitude faster than the first (12,240 versus 920).  The 2nd pathway then reacts with B with a rate constant $k_{2o}$ = 1.84.  So is it reasonable to just replace $k_2$ in the expression for M above with $k_{2o}$ for the 4 step reaction scheme?  Or do I need to use some kind of average of $k_{2o}$ and $k_{2p}$, which doesn't seem quite right.  Or something else.......
Here is some extra information.  Kinetic studies on the two reaction schemes under conditions where diffusion is negligible, but A and B are present in the same amounts initially, yield the following results (quantities are in mole/$m^3$):
Two reaction scheme:  R = 0.0333; S = 0.0004452
Four reaction scheme: p-R = 0.03095; o-R = 0.003144; S = 0.000731.
So for the four reaction scheme, the amount of S produced is higher.
Again, I am trying to redefine M based on the new 4 step reaction scheme.


text:
Why doesn't the Nernst Equation match my data at low values of Q?
I have formatted my data below in a table so my issue is more clear and asked this question again, I apologize for any confusion. I don't understand why the low values of Q are inaccurate while the high values are accurate. I saw online that the Nernst equation doesn't work for high concentrations, but why would high concentrations of ZnSO4 work while high concentrations of CuSO4 wouldn't? I also wanted to use high concentrations since when I did the Q = 1 trial with 2 0.01 M solutions instead of 0.1 M, I got around 1.08 V instead of 1.1 V which I assume is due to resistance. Other factors like the salt bridge's solution and concentration, volume of solutions used, and temperature were kept constant. I also spent a while waiting every time for the solution's voltage to slowly increase until it stabilized, although for the 3 trials that were accurate, they didn't take more than 10 seconds to reach the expected while the others would slowly increase for upwards of 2 minutes. I made sure to sand the metal before use. I couldn't find any other information online so I would greatly appreciate any help!

Increment #
Concentrations, M
Ratio of Concentrations, Q
Theoretical Voltage, V
Experimental Voltage, V

Increment 1
ZnSO4=0.01 M, CuSO4=1.00 M
Q = 0.01
1.158
1.115

Increment 2
ZnSO4=0.10 M, CuSO4=1.00 M
Q = 0.10
1.129
1.110

Increment 3
ZnSO4=0.10 M, CuSO4=0.10 M
Q = 1.00
1.100
1.098

Increment 4
ZnSO4=1.00 M, CuSO4=0.10 M
Q = 10.0
1.071
1.073

Increment 5
ZnSO4=1.00, CuSO4=0.01
Q = 100
1.042
1.041

Here is a graph to visualize the problem too:

Have you deleted your prior question?

Yes, but I still have it visible. I saw your answer saying that the concentrations were too high and you asked if I observed a drift. Can I ask what you mean by drift?

I can restore the question if you would like. I am wondering why the Nernst equation only fails for high concentrations of CuSO4 and not ZnSO4. Also, why does the equation predict the Q=1 trial when the concentrations of both solutions are  0.1 M but not 0.01 M even though it should be the opposite?

Generally speaking you should attempt to edit a question (there is a button at the bottom of the post for this) but do not reopen your older post now.

Ok, I have it closed now. Do I just wait for others to answer my question?


text:
More minerals in seawater at greater depths?
There are lists such as the one below detailing how much of different minerals seawater contains:
https://web.stanford.edu/group/Urchin/mineral.html
(For instance )
Would you expect this to be the same at all depths or should sea water at the greatest depths, several kilometers down, for some "mixing reason" contain more heavy elements than shallow water? Maybe there is a difference but miniscule?

To make the question more chemistry focused:
Take a column of pure water, several kilometers deep in a controlled environment. Add some heavy compound in the form of a fine grained dust of, for example, gold.
Given enough time and assuming the water does not move around significally the amount of solved gold is expected to be the same at all depths, at least as long as the temperature is the same everywhere in the column?

Deeper water is often colder, so solubility likely decreases.

Are you specifically interested in gravity separation, by any chance?

It is a topic rather for Earth Science SE site https://earthscience.stackexchange.com

There are huge differences in water composition (and temperature) where there are undersea feature such as hydrothermal vents. The heat and particulates can be "detected as far as tens to hundreds of kilometers away from the vent fields."
Also consider:

The solubility of substances in relationship to temperature (e.g., many solids are more soluble at elevated temperatures, while gases are less soluble)
The influence of precipitation, glacial melt and runoff from land and estuaries.

N.B. The biggest influence on "heavy elements", though, would likely be sources on land (e.g., coal-burning lofting mercury into clouds) and from hydrothermal processes.


text:
What do these arrows mean in inorganic chemistry?

I tried looking for these arrows in the 3rd molecule starting from the left(Pd) online and in my notes but I couldn't find anything. Is Pd actually bonded to something?

Aha! So the request for context was even more justified than usual, for it shed a whole new light on the entire picture. Now let me hazard a guess: this is a Pd atom surrounded by two allyl ligands, drawn by a chemist whose frivolous use of notation got them banished from the profession and severely beaten by fellow chemists on more than one occasion.

That is presumably an attempt to represent an allyl anion, which exists as a resonant structure. The double bonds are actually a delocalized double bond. See for instance https://en.wikipedia.org/wiki/Allylpalladium_chloride_dimer

Is this not typical use of notation in inorganic chemistry?

This is a typical representation of a delocalized system when dealing with organic systems. In fact, it is in complete analogy to the structure immediately two the left. The curves associated with an allyl system are just the circles for the cyclopentadienyl system.

I object; it is not typical even then. You don't draw delocalized systems with double bonds all around.

These are not arrows, but fragments of the molecule.
We suspect they are supposed to mean the allyl ligands.
This hardly qualifies as inorganic chemistry. Organometallic is probably the right word to use.
The notation is by no means standard; I would go as far as to call it wrong, and not just a slightly-raised-eyebrow kind of wrong, but a punch-in-the-face kind. Those who want to be understood would draw the same molecule in any of the following shapes:

(There are more possible variants.)
So it goes.

In deed, they are sandwiched metal complexes of allyl compounds. For example, looks here.


text:
What decides the number of water of crystallization molecules in a compound?
CuSO4.5H2O, FeSO4.7H2O, ZnSO4.7H2O are some well known compounds. Known as blue, green and white vitriol respectively. As we can see the first compound has 5 molecules of water while the latter two have 7 each. What seems to decide the number?
It is stranger if we look into CaCl2.6H2O and MgCl2.7H2O. These compounds are chlorides  of Ca and Mg, present in the same group. However they don't have the same number of water of crystallization molecules. But why? What dictates the number of water of crystallization molecules in a compound?
Ps: 1. I know that the hydrated form of FeSO4 is FeSO4.nH2O where n= 1,2,3,.. However, n=7 seems to be the most common. This is true for other examples in this question.

I thought about it for quite a while. I couldn't find a clear answer. However I have a guess.

FeSO4.7H2O has it's central atom Fe2+ actually is bonded with 6 water molecules via coordinate bonding.
Fe2+ = ... 3px2 3py2 3pz2 3dxy2 3dyz1 3dxz1 3dx2-z21 3dz21 4s0
Fe2+ * = ... 3px1 3py1 3pz1 3dxy1 3dyz1 3dxz1 3dx2-z21 3dz21 4s1 4px1 4py1 4pz1
and then 6 water molecules create coordination bonding with Fe2+* by donating electrons to these twelve orbitals and filling thsese orbitals. However, this is simply an educated guess and can't explain the examples.
I would much appreciate any help. I am open for a discussion. Thanks for reading!

Does this answer your question? What determines how much water of crystallisation a salt has?

@NilayGhosh sadly, it doesn't.

I kinda agree. That answer should have more details but the core question is already asked in that post. Maybe if we get a good answer here, we will close the other one.

Also see: https://chemistry.stackexchange.com/questions/60378/why-does-oxalic-acid-have-exactly-two-water-molecules-of-crystallization

Am I missing something or is your post missing something? You seem to have posted the same image twice.

@BuckThorn It was a technical problem, sorry for the confusion. Thanks for notifying. I was and still am on my phone, so I am facing a little problem writing. Cheers!


text:
Is ethylene more acidic than phosphine?
Yeah, it's straight forward. Is CH2=CH2 more acidic than PH3 or not? Why so or why not?
Any help or hint would be amazing. Thanks!
PS: I think ethylene should be more acidic. It has a carbon-carbon double bond, so the density of electron cloud would be higher in the areas of the double bond while the carbon-hydrogen single bond have way less electron density. This, plus the fact that phosphorus has lesser electron negativity than carbon should mean ethylene is more acidic.

It seems to be extraordinarily difficult to find a figure for the pKa of a phosphine proton (P-H). However the preparation of alkylsilylphosphines has been reported by nBuLi deprotonation of phosphine here and sodium dimethyl phosphide has been prepared by deprotonation of dimethyl phosphine using sodamide here. This leads me to feel that the pKa is somewhere beyond 30, but less than that of ammonia which is estimated to be 38 source . This is less than ethylene which is estimated to be ~45-50  source  here

Thank you, sir for this piece of information. However, I'm very curious about the theoretical aspect of this. Why is the pKa of ethylene so high? Due to the low density of electron present in the C-H bond in electron, shouldn't it be the opposite?

Maybe someone else can answer this, I cannot. Otherwise it would make a good follow-up question.


text:
Why do stable covalent compounds react with water
I am curious why a discrete covalent molecular substance, say CO2, would react with water to form an acid, what happens in such a reaction? I'm assuming it has something to do with disassociated ions in the water but why would CO2 want to bond?

Stability depends on the environment. Many many substances are just fine around themselves, mas will promply react when exposed to other substances.

Stability and reactivity are two different concepts. Elementary fluorine is infinitely stable as it does not undergo decomposition. Yet it is extremely reactive. // As stability is usually meant the tendency not to undergo decomposition in pure form or in defined environment. // As reactivity is usually meant the tendency  to react in given environment with explicit reagents or reagent classes.

What you meant by "stable" is relative. It is not some absolute. First we need to define "stability".
From what you wrote, I assume by "stable" you mean compounds that don't participate in chemical reaction in a normal environment. We need to define "normal environment" as well. We may define it as STP. But that is not absolute and so, "stability" will be relative.
Now, how do we know that CO2 is stable? One may say, well, it doesn't react with many elements/ compounds. For example it doesn't usually react with  dilute hydrochloric acid in normal environment. However, if we give enough hea, the temperature will rise enough that it will eventually react. Why will it react then?
Before giving heat, the sum of the bond energies of the products were more than the sum of the bond the bond energies of the reactants. So, Δh was positive. After giving heat and the heat energy being greater than Δh, the reaction starts.
When CO2 reacts with water, Δh is not positive, it's negative, and thus it will react. Here, carbon di oxide is stable but it will be more stable after becoming H2CO3 as it will have less bond energy and in our universe having the lowest energy usually leads to the most stable state. However, it won't become more stable after reacting with most other compounds. That's why it's relatively stable.
Again, stability is relative. Unstable compound will become more stable with the temperature drop, stable elements become unstable with the temperature increasing (relatively speaking).
Hope this helps!

Note that carbonic acid forms less than 1 % of dissolved carbon dioxide which stays mostly in form of hydrated oxide.

Can't help but wonder how you imagine its reaction with HCl... Other than that, you should check your post for typos and don't use bold too much.


text:
In mercury the o shell has 18 electrons but we know that second last shell cannot have more than 8 electrons?
The last shell has 2 electrons and the second last has eighteen electrons. Byt we know that second last shell cannot have more than 8 electrons. Then how is this possible. Please explain in simple manner.

How do you know that the second half-shell cannot have more than $8$ electrons ? Is this rule valid ? Never seen before..

It depends if by a shell is meant orbitals grouped by similar energy or the same main quantum number.  It is 18 for the former but 8 for the latter. E.g. for Rb with 5s1 electron, there are 8 electrons with main quantum number 4 but 18 if 3d electrons are involved energetically. For mercury, energetically, there come also 14  4f electrons.

The second last shell can have more than 8 electrons if it is M shell or above. According to the 2n2 rule (n= principal quantum number), O shell can theoretically can have upto 100 electrons.
Mercurys 5s, 5p, 5d subshells are full having 2,6 and 10 electrons respectively and there seems to be no problem.
In reality,  there is no rule preventing the second last shell having more than 8 electron. Look at Zinc! It also has 18 electron in its M shell.

It is actually the last shell that typically has no more than eight electrons in a typical neutral atomic ground state. The second to last shell can have up to eighteen in the ground state.
This is dictated by the manner in which shells and subshells are filled in typical ground state configurations of neutral atoms. Take the $n=3$ shell. It can hold a total of $2×3^2=18$ electrons, but only eight are there in argon and then when we move to potassium and calcium, the $4s$ orbital fills before we get any electrons into the $3d$ subshell. Then the last shell becomes $n=4$ instead of $n=3$. Only after it becomes the second to last shell does $n=3$ gain ten more electrons (filling the $3d$ subshell) as we go from scandium to zinc, and then it gets to eighteen.


text:
Why is phosphine more acidic than ethylene?
Shouldn't ethylene should be more acidic? It has a C=C bond, so the density of electron cloud would be higher in the areas of the double bond while the C-H single bond have way less electron density. This, plus the fact that carbon has higher electron negativity than phosphorus should mean ethylene is more acidic as C-H bond will be more weakened than P-H bond and thus, C-H bond will break relatively easily, releasing H+ ions more frequently.
However that seems to be not the case (as discussed in this question Is ethylene more acidic than phosphine?).
Why so? Or am I wrong? Any help or discussion is welcomed. Thanks!

Phosphorus, a third-period element, forms a weaker bond to hydrogen than carbon; see for instance https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_121/Chapter_8%3A_Chemical_Bonding_and_Molecular_Structures/8.6%3A_Bond_Enthalpies. Thus even though phosphorus has a relatively low electronegativity, the dissociation of a hydrogen ion costs less energy with phosphorus than with most hydrocarbons.


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Methanol/water mix freezing point vs specific gravity?
I need to determine the freezing point of the methanol-water mix in the ground loop of my heat pumps.  I have a hydrometer and have measured the specific gravity as 0.905.  How do I convert this to a freezing point?
The specific gravity was taken at about 16°C (60°F).
I found this chart for mass and volume, but not specific gravity.

You need a table of specific gravity vs concentration.

In case of methanol, the English edition of Wikipedia compiled a data page including a section Properties of aqueous methanol solutions: concentration of methanol by weight or by volume% in water, specific gravity data recorded at different temperatures, and freezing temperatures of these solutions. Is the range of protection (lowest entry for a solution at $\pu{−35.6 C^\circ{}}$) enough for your area?

@Buttonwood: I'm happy to see the concentration of the methanol/water solution with density of $\pu{0.905 kg/L}$ is about $57% \ (w/w) $ in both your chart and mine. Thus, I'm confident of my freezing point calculations.

@MathewMahindaratne Affirmative, Canadian Novosolution publishes a plot (pdf chart Freezing Points of Methanol Solutions) equally in vicinity of your results.  Since their business depends on using the correct data, arguably data backed by experience.

@Buttonwood  Yes, thank you.  Duh.  Not sure why that didn't come up in my searches; I should have just looked that up.

It is kind of interesting question so I have done few searches. According to this calculations, concentration of a methanol/water solution of which has the specific gravity of $0.905$ at about $\pu{16 ^\circ C}$ is roughly $57\% \ (w/w)$ (see density under the $\pu{15.56 ^\circ C}$ temperature column). Then, you can find the freezing point of that solution under this chart. Accordingly, it is approximately $\pu{-63 ^\circ C}$.
When I used the above calculator instead of the chart, I got concentration of $56.6\% \ (w/w)$ which gives the freezing point of $\pu{-62 ^\circ C}$ (not big different for your perspective).


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Is there an alternative to potassium iodide as an electrolyte in dye-sensitized solar cells?
I'm helping my son build a dye-sensitized solar cell for his school project according to the directions on homemade-circuits.com. We're currently in the process of gathering the necessary materials. So far, we salvaged the conductive gas from an LCD of a broken digital picture frame, purchased a small amount of titanium dioxide, and abstained from devouring the last of our frozen raspberries. Unfortunately, we're experiencing difficulty sourcing the final ingredient, the electrolyte.
The instructions call for an iodine solution of 65% ethanol, but don't elaborate on it other than to state: "results may be below average compared to commercially available redox electrolyte." Other resources we encountered suggest using potassium iodide combined with other chemicals. It's all daunting considering I have a high school understanding of chemistry.
Is there a relatively easy-to-find and easy-to-prepare alternative we can use instead?

Have you checked for tincture of iodine at your local pharmacy/chemist shop? Also, potassium iodide is available as pills to take in event massively evil people use nuclear weapons: they protect the thyroid gland by flooding it with iodide ions, thereby keeping it from taking in radioactive iodine (from said nukes).

I considered using iodine tincture, but wasn;t sure the formulation would work. Thanks.

From the article, that is exactly what they use: tincture of iodine is a solution of ethanol and water, with iodine and potassium iodide solutes. The tri-iodide ions are from the iodine molecules and iodide ions.

I am interested in "conductive gas from an LCD".....


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Does a higher concentration of Methanol to Water increase or decrease heat absorption?
I have a ground source heat pump with a relatively high concentration of methanol in the ground loop.  See this post.
When the system is running, the fluid in the ground loop comes in at about -3.3°C and exits at about -5.6°C.  As answered in the above post, the freezing point of the fluid is about -63°C, well below the operating point.
Will heat transfer be increased or decreased if the methanol concentration is lowered, for example, down to 20%?  Is the difference significant?
I see that the isobaric mass heat capacity at 25°C of methanol is 2.14 and water is 4.18, but I don't know if or how that relates to heat transfer.

Heat conductivity is probably not an issue if the heat exchanger is properly designed. Higher heat capacity means the substance will take up more heat to reach the T set point. Probably volumetric units would be more useful than specific (mass based), but you can probably assume densities are comparable.

You have data tables here for mixtures but need to convert it to your volume ratio and to a specific heat capacity: http://www.ddbst.com/en/EED/CPE/CPE0%20Methanol%3BWater.php

@Buck Thorn  Thanks, way over my head here.  Assuming operating temp of ~-3°C, those tables & graph are all for too high temps.  But the slopes of the curves are about the same, so if I seat-of-the-pants extrapolate to a lower curve for 270°K, the increase in heat capacity for a methanol/water ratio of .2 instead of .5 is about 2 to 3 J/mol*K.  That seems like a lot compared to what it is at 0.5 now; maybe double.  But higher heat capacity is useless if it doesn't transfer due to conductivity limits as a result of time below ground; which I don't have a handle on.

An educated guess is that if either the mixture or water equilibrate in the loop to the external T then the mixture will reduce the heat transport rate since it will absorb or release less heat on each pass. The difference could be as high as ~30% lower for a 50:50 mixture if the heat capacity is additive (which it might not be). If the liquid does not equilibrate with the external T, the water will absorb even more heat (>~30% difference) - the T difference with the surroundings will be greater with water and Newton's law of cooling says more heat is transmitted for a greater T difference.

Thanks.  The ground temp is ~7°C, the fluid ~-3°C, so a ΔT of ~10°C.  So not a lot, but not equillibrium and enough to work.