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Combining two high-VP chemicals to get a low-VP mixture
I'd like to do a class experiment as a metaphor for human relationships.  Two people each have a different chemical which evaporates quickly (at STP).  There is one glass between them.  If only one person pours his/her chemical into the glass, it quickly evaporates.  However, if both pour equal portions of their respective chemicals into the glass, it is a stable mixture which can fill the glass and not evaporate.
Are there two such chemicals that will do this?

I cannot immediately come up with an example for evaporating chemicals, but how about epoxy resin? It is usually sold as a two liquid components which quickly solidify upon mixing. As solid as friendship in a fairy tale.

Thanks.  Epoxy resin won't work for this analogy; because the idea is that, if one person doesn't add anything, the glass becomes empty fairly quickly.  I.E., the only way to fill the glass and keep it full is if both people are adding what they can.

Well, there's plenty of azeotropes which are somewhat less volatile then their constituents.

Silane and oxygen, although the transition to a low vapor-pressure state is, well, exciting...

Both components need to be liquid at STP; therefore, the Silane/Oxygen combo doesn't work.

Thanks Mithoron.  Can you suggest a good azeotrope with high vapor pressure constituents? -- something easily available?

With regard to common azeotropes, I've been through the entire list; and virtually every azeotrope has a lower bp than its constituents.  That's the opposite of what we need.


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X-ray bombardment of CaCl2 and its resultant chemical changes
If one were to take $\pu{1 mol}$ of pure liquid Calcium Chloride (at $\pu{600 ^\circ C}$) and bombard it with $\pu{7.65 \times 10^{14} Ci}$ ($\pu{2.83 \times 10^25 particles}$) of x-rays ($\pu{75 keV}$), how the calcium chloride become chemically altered in anyway? Would the radiation cause the calcium and/or chloride ions to further ionize into a form that would not bond with the other ions?
I ask this because my purpose requires that the liquid $\ce{CaCl2}$ (not aqueous) would be able to keep the calcium and chlorine from precipitating out of the liquid.

What's Cu?$\mathstrut$

@ivan neretin Cu is the symbol for Curies in this case. 1 Curie is equal to around 3.7 * 10^10 x-rays per second.

I thought Curie is Ci, but what do I know. Whatever. Well, then in effect your sample is sitting near the center of a decent nuclear explosion, so its temperature, phase state, shape, chemical composition, and pretty much everything else about it are rather irrelevant. A rapidly expanding cloud of hot plasma does not remember what it was before.

@IvanNeretin So effectively it will just become plasma?

Yes, many times over.

@JonCuster Greetings, I thought that the energy o the X-rays would have been 340 billion joules, as there would be 7.62*10^14 Curies of x-rays, each with an energy of 75 keV. That would total about 2.12 * 10^30 eV, which is 340 billion joules, rather than 6 joules. Was there an error in how I quantified this?

Ahhh, totally screwed up - I'm used to doses in flux... Sigh...

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What is the formula for vapor pressure of liquid related to surface tension of liquid?
What is the formula for vapor pressure of liquid related to surface tension of liquid? I know that if the surface tension increases, vapor pressure will decrease. Toluene evaporates faster than ethanol even though it is having more surface tension. How can I relate vapor pressure and surface tension between every solvent through a formula?

Who said there is a formula?

Please make sure to write your arguments coherently so we can follow them. Are you using the example of toluene vs ethanol as a counterargument to what your intuition tells you? Also, cite sources if possible.

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What is the relationship between enthalpy of atomisation and enthalpy of sublimation?

The $E^θ(\ce{M^{2+}}/\ce{M})$ value for copper is positive ($\pu{+0.34 V}$). What is possibly the reason for this? (Hint: consider its high $Δ_aH_V$ and low $Δ_{hyd}H_V$)

We know that we have to first break the metallic lattice to turn the metal into gas and hence there should be some sort of relation between the two.
My confusion is that when we split the lattice, aren't we technically subliming the substance as well?

tl; dr: In the case of a metal, sublimation and atomization can be regarded as synonymous when atomization refers to the process of converting the solid into a dilute atomic gas. In particular, the end points (states) of the process should be identical.
There are usually conventions on what you call "enthalpy of X". However the official source of chemical nomenclature - IUPAC - has not bothered to post a definition for enthalpy of atomization (or atomization enthalpy) in its Gold Book. There are entries for atomize:

To subdivide a liquid into very small particles; methods include: impact with a jet of gas, use of a spinning disk generator, vibrating orifice generator, etc.

and for atomization (in analytical flame spectroscopy):

The conversion of volatilized analyte into free atoms

These definitions do not refer to the same process, which leads to some ambiguity. Are you speaking of a transition from a condensed into the gas phase that preserves molecular structure, or of a chemical process that converts a sample in some phase into an atomic gas?
The wikipedia meanwhile has an entry for enthalpy of atomization which provides the following definition:

The enthalpy of atomization (also atomisation in British English) is the enthalpy change that accompanies the total separation of all atoms in a chemical substance (either a chemical element or a chemical compound)

There is an entry in the IUPAC gold book for bond energy (theoretical):

The energy required to break a given type of bond between atoms in certain valence
  states. An averaged bond energy is commonly derived by dissecting the heat of atomization of a molecule into contributions of individual bonds. For molecules with localized bonds, the heats of atomization (formation) are usually well approximated by the sum of pertinent averaged bond energies.

This reference to "heat of atomization" (which is not otherwise defined in the Gold Book) is consistent with that in the wikipedia.
If you apply this last definition then yes, the sublimation enthalpy and the atomization enthalpy of a solid metal should be identical, since both refer to the process:
$$\ce{M(s)->M(g)}$$
(assuming identical P,T end points). Note that sublimation enthalpy means standard sublimation enthalpy when referring to the process at standard pressure (1 bar) and a fixed temperature. Similarly for atomization enthalpy.
As regards the question:

The $E^θ(\ce{M^{2+}}/\ce{M})$ value for copper is positive ($\pu{+0.34 V}$). What is possibly the reason for this? (Hint: consider its high $Δ_aH_V$ and low $Δ_{hyd}H_V$)

Different uses of the word "atomization" render the meaning of "enthalpy of atomization" ambiguous. This is why I open my answer with an exposition on the definition of "atomization". Your problem refers to an ideal solution, but I assume here atomization refers to sublimation since reference to hydration would not seem to make sense otherwise. The question asks to break down the reaction into steps including atomization (sublimation), ionization and solvation. Ignoring the ionization step, a larger atomization enthalpy and smaller hydration enthalpy is consistent with the metal being more stable than the solubilized ion, ie the reduced form is favored (relative to standard hydrogen electrode).
A libre text and the accompanying diagram explain some of this

I got the big picture of this but I still have a question of atomisation energy of copper, it'd be great if you could check it out https://chemistry.stackexchange.com/questions/123802/why-does-copper-have-a-high-enthalpy-of-atomisation-when-it-has-a-fully-filled-d

Sublimation is not exactly  the same as atomisation, as many metals form diatomic molecules, especially this with lower boiling point. Rather an exception is mercury, which does not, what has close relation to its low melting point

@Poutnik That seems an important caveat. Feel free to add an answer, else I'll edit mine.

You can add it to your answer,it was just minor comment. See e.g. dirubidium


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How is taste of chemical compounds determined?
How do we know the taste of compounds, and their smell too, when some of them are poisonous. My textbook had that Lead is sweet in taste, but it's poisonous. So, how exactly do we determine their taste and odour of such compunds?

We don't. It was done by the chemists of the old times. Coincidentally, they also determined what is poisonous and what isn't.

How do people know HCN smells like almonds?; How do we know a nerve agent is tasteless?; Do all salts taste salty?

@IvanNeretin Lucky for us, these chemists of antiquity survived just long enough to report the sweetness of substances that were also poisonous.

Related for elements: https://chemistry.stackexchange.com/questions/21911/what-defines-an-elements-taste?r=SearchResults

CuSO4 "Nauseous and metallic".  Delicious. Manual of Mineralogy, Dana, J.D., 1878.  The thing with lead is in many forms it's not acutely poisonous, but Lead acetate was both known to be poisonous, and used as a sweetener.

When a compound is synthesised or isolated, it needs to be characterised. This characterisation serves to determine whether something has been synthesised/isolated before and as a template for later generations to compare their synthesised/isolated compounds to the one you have.
Nowadays, one reports NMR and mass spectrometry data; for chiral compounds specific rotation is measured. Sometimes, elemental analysis is included. Depending on the journal’s requirement you might also need to measure IR and/or melting points. Furthermore, you typically describe your compound (is it a yellow oil or white needle-like crystals?) although the description is considered secondary and the NMR/mass data is what you look at for matches.
In the old days, these analytical methods were not available. Instead, analyses would rely heavily on what a compound looks like and how it behaves. Thus, it was attempted to generate various crystalline forms and reactions with various standard reagents were investigated. Additionally, melting points were measured and chemists used all available senses to characterise. Thus, when the paper says that boron phosphate is a white, odourless and tasteless compound, the chemist in question not only sniffed the flask but also put a bit on his tongue to assess taste.
The problem with this approach is that you have no idea whether something is toxic/harmful or not. As there are many mechanisms by which a compound can interfere with the tightly controlled reactions in our body and cells or the macromolecular structures therein, many (if not most) compounds will be harmful in at least one way. While the common approach was to taste as little as possible, the risk remains as sometimes the $\mathrm{LD_{50}}$ is in the $\pu{\mu g/kg}$ range so even the smallest sample could kill you (although compounds that toxic are thankfully rare). Furthermore, a lot of compounds are able to induce DNA mutations which likely led to increased cancer rates and an overall lower life expectancy for chemists.
This is, quite obviously, unacceptable from a workplace safety point of view. Thus, tasting chemicals was prohibited practically everywhere. Furthermore, to prevent accidental contamination or confusion of glasses, not even the consumption of food and drink is permitted in most laboratories. Therefore, for anything that has been synthesised or isolated for the first time after these measures were in effect, we typically do not know taste any more. In fact, unless we are certain that we have evidence a compound is harmless, it is to be treated as highly toxic in the lab with all according security measures.
A compound’s odour is a slightly different story. It is said that the chemist working ideally will never get a wiff of their compounds because they will make sure that none access the common lab air. In practice, that is practically unattainable especially since lab scales are often in the middle of the room and most solids are in glass bottles that need to be opened fully to take out a sample. Thus, it is unavoidable that chemists pick up the smell of a number of (typically particularly volatile or particularly odourous) compounds. Systematic odour examinations are no longer performed but where the odour is characteristic, it is sometimes included in a compound’s characterisation (e.g. ‘sulphurous odour’). Some chemists take this a step further: a former colleague once sniffed at the flask of another former colleague and exclaimed ‘it smells like your product’, explaining that tertiary alcohols apparantly have a distinctive odour.

"new compounds have a high probability of being poisonous" Could you please elaborate on this, or add a source? For a layman, this claim is far from obvious. Is this because all compounds (statistically) are more likely to be poisonous, or something due to the nature of yet-undiscovered compounds?

@doldt Good point you’re raising. In part, a number of newly discovered compound are highly reactive (which is why they weren’t discovered before) and would obviously harm everything including your tongue. But I’ll have to step back and rethink the intention of what I wanted to say and how to phrase it. Expect an edit in the next couple of days.

@doldt I think the more defensible claim is that the probability of toxicity is higher than most people would find acceptable for something you're going to ingest (with notable exceptions such as the health & safety officer's nightmare that is cody'slab on youtube)


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Why does copper have a high enthalpy of atomisation when it has a half filled d sub shell?
If I understood correct, the high energy of atomisation for d-block element comes from the ability to make many metallic bonds due to half electron filled d-orbitals, but it seems here that copper has no free d-orbitals yet has high enthalpy of atomisation? Is there a mistake in my understanding or is there a reason for this anomaly?

It's s orbital is half filled...

yes but other elements in coppers row has half filled s orbital+many half filled subshells in d-orbitals yet it has lower enthalpy then coppers. I am of the understanding that more half filled orbitals=moer metallic bonding

There is no filled $3d$ subshell in metallic copper.  Metallic band structure makes combinations of initially filled and empty valence orbitals overlap each other in energy.

Copper, with a nearly ideal Fermi surface, contributes one electron per atom to the conduction band in the crystal state. I'm unclear what the question is about.

Oh lol I asked this qsn after reading my high school text, in it says that transistion elements is able to create multiple bonds due to presence of half  filled d -orbitals

When we look up the Wikipedia articles for various elements, we find that the heat of fusion and heat of vaporization of copper are modestly below those quantities from iron, cobalt and nickel, but hugely exceeding zinc.
In the first-row transition elements when they're condensed as metals, we don't really have separate $3d$ and $4s$ subshells because the metallic bonding spreads these subshells into bands and the bands overlap in energy.  This article in Chem Libretexts shows this overlap between $s$ and $p$ orbitals in sodium; a similar concept involving $s$, $d$ and even$p$ orbitals in transition metals.  So we can't separately consider $3d$ and $4s$ subshells in a first-row transition element such as copper.  With contributing orbitals from different subshells mixed together we have a significant number of filled and empty states mixed together in the valence band, leading to relatively strong metallic bonding as long as we keep that energy overlap.  Among elements in this period, only when we get to zinc do we see the $3d-4s$ band overlap drop off and the nonbonding character of the filled $3d$ subshells emerge as a major factor.  Therefore, metallic bonding and with it, enthalpy of atomization drop off sharply at zinc, not so much at copper.


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Enol percentage comparison between 1,3 diketones and 1,2 cyclopentadione
To what extent does 1,2-cyclopentadione exist in enol form? Will it be more than for 1,3 diketones? 1,3-Diketone with active methylene groups can form hydrogen bonds with a 6-membered ring structure. 1,2-cyclopentadione could also form hydrogen bonds, but it forms only a 5-membered ring.

Hi! Welcome to Chemistry.SE. I edited your question to hopefully improve the clarity of what you are asking.

Also, this answer explains some of the factors that contribute to increased enol percentage, although it does not answer your question:https://chemistry.stackexchange.com/questions/18904/which-is-the-more-stable-enol-form/18917#18917


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Last step of proposed cubane synthesis
I was watching an online lecture on decarboxylation by soda lime and the first few examples were straight forward but after a while a felt that the questions couldn't simply have one effect and maybe other factors should also come into consideration as complexity of structures increases.

In the above reagent soda lime was added and the teacher said that cubane will be produced however though it might be straight forward based on the mechanism as a carbanion is produced in the intermediate is it correct to directly say that the mechanism should be followed as expected by one pathway without considering any other possibilities such as attempt to relieve angle strain?
Over the course of studying organic chemistry I have realized things often aren't as simple as they appear to be as many factors come into play such as ring closing or bond breaking to relive angle strain.
According to the video the product is Cubane
I know that cubane itself is kinetically stable (primarily due to lack of availability of pathways for decomposition) but the intermediate formed can it not provide any pathways for decomposition?

What was the yield on the reaction according to the video?

Sodalime forms the dianion of the carboxylic acid group and SN2 ring closure occurs. Decarboxylation of the anion of cubane carboxylic acid is facile given the kinetic acidity of cubane.  https://doi.org/10.1021/ja00001a051

@user55119 i didn't get your point about dianion. Isn't a simple anion formed? Also I would appreciate if you elaborated a but more on the kinetic acidity point as it is a relatively new term for me. Maybe formatting as an answer would be a good choice

Sodalime conditions are quite vigorous. The acidic hydrogen is removed from -CO2H but the alpha hydrogen next to the carboxylate is also deprotonated, thus a dianion. The carbanion displaces chloride. The smaller the C-C-C bond angle the more acidic the C-H bond of the central carbon. The authors claim that cabane is ~63K times more acidic than cyclohexane. Also, cyclopropane is more acidic than methane.


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What is the absolute configuration of carbon 5 in 5-bromo-2-methylhex-2-ene
I am trying to work out the absolute configuration around carbon atom 5 in 5-bromo-2-methylhex-2-ene. Could someone please let me know if my working is correct? Does the wedged bond from C4 to C5 affect the structure of the molecule and its configuration?

Your approach is correct.

you already answered your own question. small tip when determining absolute configuration: make sure the substituent with the lowest priority is facing AWAY from you (so it points into the piece of paper).

You are correct.


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How is mole fraction of a solution different in vapour state vs in solution vs vapour-liquid equilibrium?

'x' grams of benzene and 'y' grams of toluene, it is said that the vapor pressure of benzene is 75 mmHg and that of toluene is 22 mmHg. Find mole fraction in the vapor state,

What I do not understand is how different states can have different values for mole fraction? Do the moles change with a change of state?
More precisely how does the fraction of mole in each state look at transition states? As in say I started with 100 moles in the liquid phase, when it reaches the point where it is about to turn to gas (b.p) how much fraction of moles are gas and how much are liquid?

HINT - Assume that the liquid solution is an ideal liquid mixture.

how does the answer the title question?

As benzene is more volatile than toluene, it should be clear to you the vapour composition must be enriched by benzene, compared with liquid composition. If ideal behaviour is implied , vapour composition is interpolation of vapour composition over pure liquids, if liquid molar fractions are considered.

When you have a liquid mixture, there is a vapor phase that is made up of those same molecules in equilibrium with the liquid phase. There is no rule that says the moles will split 50/50 in the vapor and liquid phase or any other fraction. The amount that stays in any one phase is influenced by many variables (e.g. molecule to molecule interactions, temperature, pressure, volatility) so we use many different models and assumptions to calculate the amount in either phase or we conduct experiments to measure the amounts.

I can not grasp how the number of particles change in each state,where does all the other moles go to?

@DDD4C4U See Poutnik's comment for some further clarification. Can you leave a comment describing the mental picture you have about the situation? At a high level, you should consider how molecules are able to form the vapor phase - some molecules from the liquid have enough energy to escape the molecular interactions within the liquid. The number of those molecules that have that energy dictates the molecule split. The total number of molecules in the vapor and liquid phase, in a closed system, will be constant.


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Why must there be degeneracy in the orbitals to have a Jahn-Teller effect?
Why is it the case that for Jahn-Teller effects to occur in transition metals there must be degeneracy in either the $\mathrm{t_{2g}}$ or $\mathrm{e_g}$ orbitals?

Jahn-Teller effect is way broader than transition metals, or t2g, or eg.

What do you mean? But for transition metals there must be degeneracy

I mean just what I said, Sure, there must be degeneracy in any case. But even when occurring in transition metals, Jahn-Teller effect is not restricted to t2g/eg thing.

A nice non-inorganic case of JT distortion, let me mention that, is that of cyclobutadiene: in its equilibrium structure (singlet state), it "escapes" antiaromaticity (and degenerancy) with asimmetry. The reason of JT distortion is pretty much the same in inorganic chemistry: an asymmetric (so, non-degenerate) state is energetically favoured compared to the degenerate and symmetric alternative

In the abstract to the original Jahn-Teller paper it is stated:[1]

We shall show that stability and degeneracy are not possible simultaneously unless the molecule is a linear one.

Thus the Jahn-Teller effect is specifically about the removal of a degeneracy, and that that will always lower the total energy of the system. Thus you can not have a Jahn-Teller effect without a degeneracy in the reference, fully symmetric state.

Reference:
[1]: H. A. Jahn, E. Teller, Proc. Royal Soc. A 1937, 161, 220–235. DOI: 10.1098/rspa.1937.0142.

That is correct. But the JT theorem concerns the degeneracy of a state, not the degeneracy of two (or more) orbitals, which is what OP is asking about. There is almost certainly a direct link, which would be insightful to flesh out (I am unwilling to think about it properly now, although I think it should be easy to show that an uneven population of degenerate orbitals leads to a degenerate state).

Good point - I mentally missed the word orbital


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Lewis Structures and electron distribution with oxygen atom
I am a novice in chemistry.
I don’t understand the way Lewis structure and how electrons are shared between elements.
For instance
As we know, Oxygen has six valence electrons that means it has 3 lone pairs.
So when oxygen wants to form a covalent bond, one of its valence electron should participate in the covalent bond.
In the picture above, however, even though that oxygen use one of its valence electron, we can clearly see that it still has 3 lone pairs. Well, they should be five electron Instead of 6.
In my interpretation, it should be like this ⬇:

Could you explain this?

Oxygen is a special case, there are so many Youtube videos on Lewis dot structures. Just use the keywords on Youtube.

In the Lewis formalism, an electron has been added to the valence shells of two oxygens in $\ce{NO3-}$, meaning they have seven electrons at their disposal. Likewise, the central nitrogen has one electron removed leaving it with four. To make this clear, a proper Lewis depiction should include formal charges; the two oxygens on the end of an $\ce{N-O}$ single bond should be $\ce{O-}$ while the nitrogen should be $\ce{N+}$. The sum of formal charges must equal the compound’s overall charge.


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Can acid-base reactions occur in a non-aqueous medium?
To better illustrate my question, allow me to refer to the case of ammonium chloride, formed through a classic acid-base reaction.
The reaction between ammonia and hydrogen chloride to give ammonium chloride is a neutralization reaction, since ammonia acts as a base and hydrogen chloride acts as an acid.
The reaction is usually written as:
$$\ce{NH3 + HCl -> NH4Cl }\tag{1}$$
Usually, in the synthesis of ammonium chloride, the reagents in aqueous solution (ammonium hydroxide and hydrochloric acid) are used. In aqueous solution, they behave differently than when they are in anhydrous form.
On the one hand, when ammonia dissolves in water, the ammonium cation, its conjugate acid, is formed:
$$\ce{NH3 + H2O <=> NH4+ +OH-}\tag{2}$$
On the other hand, a similar, but opposite, reaction occurs when hydrogen chloride dissolves in water, forming the chloride anion, its conjugate base:
$$\ce{HCl + H2O <=> H3O+ + Cl-}\tag{3}$$
Then, when mixing both solutions, the following reactions occur:
$$\begin{align}\ce{NH4+ +Cl- &-> NH4Cl}\tag{4.1}\\[0.5em]
\ce{H3O+ +OH- &-> 2H2O}\tag{4.2}\end{align}$$
However, apparently, it is not necessary to mix the solutions to obtain ammonium chloride, just by mixing the vapors of the solutions, it is enough to give the expected product.
But when the ammonium hydroxide evaporates, the ammonia ceases to be in the form of ammonium and returns to its anhydrous form. The same goes for hydrochloric acid. Then, the vapors would be ammonia and hydrogen chloride, but wouldn't this affect the reaction since they are not in the form of conjugate acid or conjugate base now?
I still don't understand acid-base chemistry very well and I would like to know if such reactions, such as that of ammonia with hydrogen chloride, can occur without the reagents being in the form of ions in aqueous solution.

Note that chemical equations are used to be written via MathJax mhchem: like $$\ce{NH3 + H2O <=> NH4+ + OH-}\tag{2}$$  (  $$\ce{NH3 + H2O <=> NH4+ + OH-}\tag{2}$$ )

Ion reactions do not require water and acid base reactions are not limited to ions.  Btw, you have forgotten $\ce{H3O+ + OH- -> 2 H2O}$ and $\ce{NH3 + H3O+ <=>> NH4+ + H2O}$. See https://en.wikipedia.org/wiki/Nonaqueous_titration   and https://en.wikipedia.org/wiki/Lewis_acids_and_bases

If you dissolve $\ce{NH4Cl}$ in water, your solution contains $\ce{NH4+}$ and $\ce{Cl-}$ ions. If you dissolve $\ce{NaOH}$ in water, your solution contains $\ce{Na+}$ and $\ce{OH-}$ ions.
Now if you mix those two solutions, nearly all $\ce{NH4+}$ and $\ce{OH-}$ will react and produce $\ce{NH3}$ and $\ce{H2O}.$ This shows that $\ce{NH4OH}$ cannot exist. It is decomposed into a solution of $\ce{NH3}$ in water, with only a few percent of $\ce{NH4+}$ and $\ce{OH-}$ ions.
You will certainly object and say that $\ce{NH4OH}$ is a commercial substance, that you can buy anywhere in the market. You are right. But you may not know that the reason of this mistake is commercial and goes back to the 19th century. In the 19th century, the ions were unknown. All substances had a chemical formula, whatever pure or in solution. $\ce{NaOH}$ in solution was made of $\ce{NaOH}$ and that's all. Charged ions like $\ce{H3O+}$ and $\ce{OH-}$ did not exist.
At this time, chemists were often studying the neutralisation reaction
$$\ce{Acid + Base -> Salt + Water}$$
They succeeded in carrying out this reaction with mineral or organic acids, and with many hydroxides (hydroxide was synonymous to base). When a new organic acid was discovered, they tried to make salts with all available hydroxides. Each new salt deserved a publication. And this was fine!
Unfortunately, ammonia was a problem: $\ce{NH3}$ reacts with any acid, but it does not produce any water. $\ce{NH3 + HCl}$ makes $\ce{NH4Cl}$ and that's all: no water! This was difficult to admit. So the chemists decided that, when dissolved in water, $\ce{NH3}$ is transformed into $\ce{NH4OH}.$ This makes sense, because $\ce{NH3}$ was very soluble in water, as if a chemical reaction happens. It was fine. When neutralizing a solution of ammonia by reaction with any acid, the equation of the reaction was:
$$\ce{NH4OH + HCl -> NH4Cl + H2O}$$
$\ce{NH3}$ was only existing in a gaseous state. It was considered as a sort of "dehydrated ammonium hydroxyde". A little bit like other hydroxides, which can easily lose water. At high temperature, $\ce{Cu(OH)2}$ becomes $\ce{CuO}$ and $\ce{Al(OH)3}$ becomes $\ce{Al2O3}.$
So the merchants started to sell ammonia solution with the label $\ce{NH4OH}.$ It was soon discovered that this solution is helpful unblocking blocked up pipes. Everything was OK!
Unfortunately at the end of the century, Arrhenius discovered the ions, and showed that $\ce{NH4+}$ and $\ce{OH-}$ ions cannot exist simultaneously in solution. Disaster! The so called "ammonium hydroxyde" was mainly containing $\ce{NH3}$ and $\ce{H2O},$ with only a few percent of remaining ions $\ce{NH4+}$ and $\ce{OH-}.$ So the merchants changed their labels and started to sell "concentrated ammonia solution". It was OK for the chemists.
But this change was not accepted by the caretakers, managers and contractors. They protested to the manufacturers against this new "ammonia solution" which was useless to unblock blocked pipes. They said they have tried the new and the old solutions. They demanded to get the old ammonium hydroxide, which was "much better" than this stupid ammonia solution.
Well! The merchants are not stupid. They want to sell and get money. So they renounced to modernize their labels, and continued to sell $\ce{NH4OH}$ solutions. This was accepted by chemists and non-chemists.
That is why $\ce{NH4OH}$ is still on sale today. I am an old man, 82. This story was reported to me by my father, who was a chemist who remembered this old time.

Related (I've included a link to your answer): Why NH4OH? Why not NH3(aq)?


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Aluminium Oxide as Dehydrating agent
According to this paper on dehydration by $\ce{Al2O3}$,

Dehydroxylation causes a surface defects as Lewis acidic centers (unsaturated $\ce{Al^3+}$ ions) and basic Lewis centers ($\ce{O^2-}$ ions) creation. Water molecules deactivate Lewis centers and create strongly acidic Brönsted’s centers. Alcohols dehydration reaction performs on Brönsted’s active centers where hydroxyl groups disconnect proton from alcohol molecule. During reaction alcohol molecule adsorbed on aluminium oxide surface reacts with proton.

But I am not able to understand what they are trying to convey here.

I have revised my question and added other points also. It is now clearer than before.

Regarding the flag you've raised on your own question as to why it has been closed as HW, I think the reason is obvious if you read the information in the reason for closing. The question itself is all right, but you are supposed to add your thoughts and/or demonstrate the research done prior to asking.

Simply stating "I don't understand X" isn't constructive. Something like "I've read this paper [ref] and it claims X, whereas my textbook [ref] says Y, but I think it's Z because of this, that and the other" would be much better and help others help you since we would have better understanding what exactly you are having difficulties with.

Alumina absorbs water to give aluminium oxide hydroxide, a solid which removes the water from the reaction environment.
According to the paper by Pethrick et al. [1]:

Aluminium oxide exists in several forms which on addition of water give a complex range of oxide-hydroxide; boehmite, bayerite and gibbsite.

References

Pethrick, R. A.; Hayward, D.; Jeffrey, K.; Affrossman, S.; Wilford, P. Investigation of the Hydration and Dehydration of Aluminium Oxide-Hydroxide Using High Frequency Dielectric Measurements between 300 KHz–3 GHz. J Mater Sci 1996, 31 (10), 2623–2629. https://doi.org/10/chcqs8.

Is there any kind of mechanism involved in the reaction?

There is a reaction between the alumina surface and free water with water adding to the aluminium atom, but afaik that is as much as can be said.


text:
Can stimulated Raman processes be strong enough to drive out of equilibrium?
In the (spontaneous) Raman process, incident light $\hbar \omega_1$ scatters and transfers some energy $\hbar \omega$ to a vibrational excitation of molecule or solid.
Typically this is a very rare process, and only happens to one of every $10^9$ photons or so. So even if a sample is irradiated with a ultrafast pump laser of high intensity, Raman processes tend to not be significant enough to move the system out of equilibrium. This is in comparison to something like direct absorption processes which easily bring the system out of equilibrium and often dominate the response.
But the discussion above is for spontaneous Raman processes and makes me wonder, what about stimulated Raman processes? In the case of stimulated Raman, two photons come in with $\omega_1-\omega_2=\omega$, causing vibrational excitations to be much more efficiently created.
So my questions are

In practical cases, what is the efficiency of Stimulated Raman processes? In other words, for a given number of pump photon $n_1$ and Stokes photon $n_2$, what are some ballpark numbers for the number of vibrational excitations created?
Can stimulated Raman processes be so strong as to move a system out of equilibrium in a pump-probe setup?

Vibrational relaxation is, except in a rather dilute gas, extremely fast. I`ll say no chance.

@Karl, what about in a pump-probe setup like in Q2? Usually they look at femtoseconds there, which is also pretty fast.

even the faintest absorption drives a system out of equillibrium. Do you mean you want to get a significant saturation effect?

@Karl yes that's more along what I was thing, some significant saturation on the ultrafast time scale


text:
Why could it be assumed that the standard enthalpy change is independent of temperature down to 25 °C?
For the reversible reaction of pyruvic acid to 2,2-dihydroxypropanoic acid, we calculated enthalpy using NMR and extrapolated from area (integration) and peaks etc.
We then plotted $\ln K$ vs $1/T.$ I am confused by this questions because I thought enthalpy was temperature dependent.
I guess my question was unclear. The exact question we are asked is:

With the assumption that the standard enthalpy change is independent of $T$ down to 25 °C, calculate an extrapolated value for $K$ (25 °C) and compare this to your experimental $K$ value determined at 25 °C. Is this a fair assumption? Why?

I do not understand if this assumption is fair and why?

What "questions" are you confused about, exactly? It looks like you are doing an Arrhenius plot; why do you think there is an evidence for enthalpy being independent of $T?$

So the specific question we are asked is "With the assumption that the standard enthalpy change is independent of T down to 25°C, calculate an extrapolated value for K(25°C) and compare this to your experimental K value determined at 25°C. Is this a fair assumption? Why?" So we do do an arrhenius plot with points at temperatures 45, 50, 55, 60 and 65, and they we are supposed to extrapolate down to 25 to get a K value

I removed the kinetics tag since the question does not involve kinetics.

It is assumed that, over the temperature range of interest, the enthalpy change is approximately constant.

@ Chet miller, but why exactly??? Why is that assumption valid?

I edited 2,2-dyhrydypropanic to 2,2-dihydroxypropanoic which is I guess what you meant, feel free to revert the post if that change is incorrect. Also, it would be nice if you avoided the ambiguous "etc" where it might not be self-evident what you did.

If the extrapolation works well, it suggests that the approximation of temperature independence is reasonable. If the extrapolation is well off the mark, the assumption is not good. Related:https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Physical_Chemistry_(Fleming)/3%3A_First_Law_of_Thermodynamics/3.4%3A_Temperature_Dependence_of_Enthalpy

Generally speaking you can write that
$$\Delta H^\circ(T) = \Delta H^\circ(T_\mathrm{ref}) + \int_{T_\mathrm{ref}}^T \Delta C_p \,\mathrm dT \tag{1}$$
where $T_\mathrm{ref}$ might be $\pu{25 ^\circ C}$ and the pressure is 1 bar (in your case, approximately 1 bar).
If your temperature window is not too large you may get away with assuming $\Delta H^\circ$ is constant. This amounts to assuming that the second term in the equation, dependent on $\Delta C_p$, is small compared to the first. If it isn't then rather than using the integrated van't Hoff equation
$$\ln\left(\frac{K_2}{K_1}\right)=\frac{-\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \label{eqn:2}\tag{2}$$
maybe you should attempt a polynomial fit to $\ln K_\mathrm{eq}$ and taking the derivative of the polynomial wrt $T^{-1}$ to evaluate $\Delta H^\circ$ as a function of $T$.
To test whether \eqref{eqn:2} is appropriate you might evaluate the quality of a fit to that equation or check the linearity of a plot of $\ln K$ versus $T^{-1}$.

The fit is 100% linear, the question is specifically about the validity of the assumption. I have the fit, and I have a value.

Well, probably linearity is enough to justify extrapolation to the T of interest. Still, there is the question of how far you had to extrapolate. There are details you don't provide such as T points.

For the high-temperature runs on sample 1, calculate K at each temperature from the integrated area values and plot In K versus the reciprocal of the absolute temperature. Determine the standard enthalpy change. With the assumption that the standard enthalpy change is independent of T down to 25°C, calculate an extrapolated value for K(25°C) and compare this to your experimental K value determined at 25°C. Is this a fair assumption? Why?

I understand, but what does "high T" mean relative to 25C? Also, I suppose you might want to provide the Keq values at 25C (from the extrapolation and the "experimental" one you got either from a separate experiment or?) and from their difference (and uncertainty) you can answer the question "is this a fair assumption". The assumption is fair if the data supports it, in the absence of more info regarding heat capacities. You mention that the fits to vant Hoff are linear, which supports small $\Delta C_p$ so I'd say that alone justifies the assumption of constant $\Delta H$

So at 25 experimentally you get aroun 0.7, when you extrapolate you get 147.6, so several orders of magnitude difference. You can fit a 3rd order polynomial to the van;t hoff plot but I am not sure what this means

I assume you are making sure to compare the same thing (say either ln(K) or K). I can't explain the discrepancy without looking at the data, and knowing that you could fit a 3rd order polynomial doesn't tell me much either. What did suggest a lot is your mention of a nice linear fit of lnK versus 1/T. That suggests that (at least for the high T data) $\Delta H$ is constant. However it seems the approximation may break down if the T window your experiments was far from 25C


text:
Intrahydrogen bonding
Is it okay to conclude that in 2-nitrophenol where intra hydogen bonding is experienced, two similar molecules are only held by the weak van der waal's forces and is the reason for it to have a relatively lower boiling point than other nitro phenols

Lower boiling point than what?  3 and 4-nitrophenol maybe?

than other nitro phenols like 3-nitrophenol or 4-nitrophenol

I agree with your analysis

I don't agree, there's plenty of pi-stacking/dipole-dipole and other interactions, vdW are like least important.

From the abstract to ref 1:

We investigate both experimentally and theoretically the structure and photodynamics of nitrophenol molecules and clusters, addressing the question how the molecular photodynamics can be controlled by specific inter- and intramolecular interactions. Using quantum chemical calculations, we demonstrate the structural and energetic differences between clusters of 2-nitrophenol and 4-nitrophenol, using phenol as a reference system. The calculated structures are supported by mass spectrometry. The mass spectra of 2-nitrophenol clusters provide an evidence for a stacked structure compared to a strong O–H···O hydrogen bonding for 4-nitrophenol aggregates.

The key words in the context of 2-nitrophenol are "stacked structure". This suggests that Mithoron's comment might be correct. However "stacking" is not evidence of a particular attractive interaction mechanism, it doesn't tell us of the relative importance of classical pi-stacking versus dispersion interactions between 2-nitrophenol molecules.
I'll throw up my hands and claim ignorance, deferring explanation to the Wikipedia:

In chemistry, pi stacking (also called π–π stacking) refers to attractive, noncovalent interactions between aromatic rings, since they contain pi bonds. These interactions are important in nucleobase stacking within DNA and RNA molecules, protein folding, template-directed synthesis, materials science, and molecular recognition, although new research suggests that pi stacking may not be operative in some of these applications. Despite intense experimental and theoretical interest, there is no unified description of the factors that contribute to pi stacking interactions.

References

Photochemistry of Nitrophenol Molecules and Clusters: Intra- vs Intermolecular Hydrogen Bond Dynamics. Kateryna Grygoryeva, Jakub Kubečka, Andriy Pysanenko, Jozef Lengyel, Petr Slavíček, and Michal Fárník. The Journal of Physical Chemistry A 2016 120 (24), 4139-4146. DOI: 10.1021/acs.jpca.6b04459

thanks for this fruitful information. so is there any chance of hydrogen bonding in 2-nitrophenol ?

@HonBisasoIsmail I don't see why not, it's evidently just not as stable as in 4-nitrophenol. Unfortunately I don't have hard numbers to share but I suggest searching the ref in my answer, the references it cites, and other articles that might cite the ref.


text:
How to get the enthalpy of a reaction from the enthalpy of formations with Hess' law?
Let's consider a generic reaction:
$$\sum_{i=0}^{i=I}n_{i}A_{i} \rightarrow \sum_{j=0}^{j=J}m_{j}B_{j} $$
where $n$ and $m$ are the
stoichiometric coefficients, while $A$ and $B$ are reagents and products. In order to compute the enthalpy of this reaction we use the enthalpy of formation of each product and each reagent. We know that for the reagents:
$$\sum_{s=0}^{s=S}w_{s}^{i}C_{s}^{i} \rightarrow A_{i} \qquad \Delta H^{A_i}_{f} $$
By multiplying for $n_{i}$:
$$n_{i} \sum_{s=0}^{s=S}w_{s}^{i}C_{s}^{i} \rightarrow n_{i} A_{i} \qquad \Delta H^{A_i}_{f}n_{i} $$
Summing over $i$ you get:
$$\sum_{i=0}^{i=I}n_{i} \sum_{s=0}^{s=S}w_{s}^{i}C_{s}^{i} \rightarrow \sum_{i=0}^{i=I}n_{i} A_{i} \qquad \sum_{i=0}^{i=I}\Delta H^{A_i}_{f}n_{i} $$
Note that all these passages are guaranteed by Hess law.
Let's pose for simplicity:
$$\sum_{i=0}^{i=I}n_{i} \sum_{s=0}^{s=S}w_{s}^{i}C_{s}^{i}=K_{1}$$
We can do exactly the same process for products obtaining:
$$\sum_{j=0}^{j=J}m_{i} \sum_{g=0}^{g=G}z_{g}^{j}D_{g}^{j} \rightarrow \sum_{j=0}^{j=J}m_{j} B_{j} \qquad \sum_{j=0}^{j=J}\Delta H^{B_J}_{f}m_{j} $$
Let's pose for simplicity:
$$\sum_{j=0}^{j=J}m_{i} \sum_{g=0}^{g=G}z_{g}^{j}D_{g}^{j}=K_{2}$$
With an easy passage we finally get:
$$\sum_{i=0}^{i=I}n_{i}A_{i}+K_{2} \rightarrow \sum_{j=0}^{j=J}m_{j}B_{j} +K_{1}
 \qquad \Delta H=\sum_{j=0}^{j=J}\Delta H^{B_J}_{f}m_{j}- \sum_{i=0}^{i=I}\Delta H^{A_i}_{f}n_{i}$$
Now, in my book it is written that the last enthalpy we got, is the reaction enthalpy of my initial reaction, but how you can see, it is the reaction enthalpy of the last equation, that it is not the same (there are $K_{1}$ and $K_2$ which do not cancel each other). Someone could explain this?

I am pretty sure that $K_1$ and $K_2$ have to be the same, otherwise your reaction will not balance. Unfortunately I don't have the time to prove this.


text:
Dissolving Metal Reduction of 4-Octyne to (E)-4-Octene and What Else?
I was reading Solomon's Organic Chemistry and came across the Dissolving Metal Reduction of 4-Octyne to (E)-4-Octene. (The same as can be found this question)
But the book states that the (E)-4-Octene is only produced in a 52% yield. This doesn't seem very impressive to me. What is the other 48%? If it's (Z)-4-Octene, can we really say that the reduction favors anti addition? If it's octane, I'd understand, but still would think you're losing a lot of product.
So does anyone know what the other products in the reaction are?

Without detail of the experimental conditions it is difficult to tell where the remaining mass balance is. Typical yields I've seen are much higher (e.g. here https://books.google.co.uk/books?id=P0uGAwAAQBAJ&pg=PA889&lpg=PA889&dq=Lithium+metal+reduction+of+Oct-4-yne&source=bl&ots=yqkA_z_tRW&sig=ACfU3U0Sw6N8kL-Ye-Vfh3ATASrQgrXKyA&hl=en&sa=X&ved=2ahUKEwj5nbfoje_lAhWqWhUIHcJFD9IQ6AEwDnoECAgQAQ#v=onepage&q=Lithium%20metal%20reduction%20of%20Oct-4-yne&f=false) It could just be an inefficient work-up and isolation procedure on the part of the experimenters.

@Waylander Those yields are better. I noticed the resulting alkenes are drawn with wavy bonds, does that imply that cis and trans products are produced? Because this link suggests that the mechanism also produces some cis product.

They are indeed drawn wavy, but the text refers to "reductions in good yields, and in a remarkable state of isomeric purity" . The link you highlighted seems confused as they offer the cis product as the major product in one diagram, yet (correctly) show the mechanism as giving trans. I'm inclined to blame an inefficient isolation procedure for the low yield, possibly losses on distillation.


text:
Is it possible to calculate the 'overall' redox potential of a system?
For a solution where many redox reactions occur (e.g. cytosol, soil solution), can we consider anything as 'overall' redox potential of a system? Is there a formula to quantify it in relation to the amount of each of the constituents?

I think this is a good question, especially the environmental and biological chemistry aspects of it. One issue (in real life and in simulation): you have to draw the line somewhere for your system boundaries, otherwise the answer(s) is/are ill-defined.

Related: https://chemistry.stackexchange.com/questions/124988/is-redox-potential-a-valid-term/125026#125026

Is it possible to calculate the 'overall' redox potential of a system?

No, it is impossible. You can have a situation where a cellular compartment contains a high ratio of $\ce{NAD+:NADH}$ (i.e. favoring oxidation) and a low ratio of $\ce{NADP+:NADPH}$ (i.e favoring reduction). This is advantageous for the cell because it can run oxidation of some metabolites in parallel with reduction of other metabolites, just by employing enzymes with distinct co-factor requirements.
This is different from pH. A solution is either acidic or basic, not both. The difference arises because acid/base reactions are usually fast, and water acts as acid or base, "communicating" acid/base status across the solution.

related: https://chemistry.stackexchange.com/questions/124988/is-redox-potential-a-valid-term/125026#125026

The question came to my mind while I was conducting a heme protein dithionite titration. I obtained a solution of photosynthetic bacteria chromatophores with redox mediators and constructed a cell using a platinum electrode.
If nothing as 'redox system potential' exists, then what exactly is responsible for the resultant potential of such half-cell? In a solution of so many compounds as a solution of biological origin, it is probably impossible to pinpoint a single reaction that would solely determine it.

Before answering i would like to mention what I was able to interpret from your question:
You want to know that is it possible to find the redox potential of a mixture basically (anything which has multiple non reacting constituents such as soil solution).
I think it is plausible to do so by using the same method which is employed to find out the redox potential of any cell i.e. to write down the cell reactions individually for each reacting species and then writing the overall cell reaction and then calculating the Cell potential keeping into consideration the individual changes in electron in each reaction.
At any instant only one cation and anion would be discharged so when one substance is used up then the lesser reacting substance will get a chance such as the case in aqueous solution of any salt where Hydrogen gas is liberated first and then the cation (if the are difficult to liberate as gases). Hence at different instances different cell reactions will be there and different values of redox potential will be obtained.
NOTE
I have tried to answer the question as per my understanding of Electrochemistry but as i was fairly recently introduced to the subject in case of any mistakes in my arguments I would deeply appreciate any form of advise or correction.


text:
Hess's Law and ΔH
Target equation: $\ce{Mg(s) + 1/2 O2 -> MgO(s)}$
$$
\begin{align}
\ce{MgO(s) + 2 HCl(aq) &-> MgCl2(aq) + H2O(l)} &\quad ΔH &= \pu{-1300 kJ} \tag{1}\\
\ce{Mg(s) + 2 HCl(aq) &-> MgCl2(aq) + H2(g)} &\quad ΔH &= \pu{-602 kJ} \tag{2}\\
\ce{H2(g) + 1/2 O2 &-> H2O(l)} &\quad ΔH &= \pu{-286 kJ} \tag{3}
\end{align}
$$
The target equation should have a negative $ΔH$ meaning that it's exothermic, but when I add it up, $ΔH$ ends up being positive. Is there anything I'm missing?

If you are calculating $ΔΗ$ as $(2) + (3) - (1)$, then it looks correct, however, you are right, it should be negative. Is there a typo for $ΔΗ_1?$ It looks way to high: $\pu{-130 kJ mol-1}$ would make more sense than $\pu{-1300 kJ mol-1}$ IMO (I assume molar enthalpy based on the value for (3)).

I got 1300 kJ mol−1 by taking 3.34 kJ (the q value for energy absorbed by HCl) and the number of mols of MgO used, 0.00257 mol, and dividing 3.34 by 0.00257. One thing I forgot to mention is that this is a lab. Is there anything in the q = mcΔT calculation for HCl that I messed up on?

It's hard to tell, but most likely yes, enthalpy of the first reaction has been determined incorrectly. You probably should [edit] in more details as to how it was obtained, as well as the value you expect and why (I already commented on that).

I agreed with @andselisk. I already give an answer, but I'm looking at where you have made mistake, as a part of following andselisk's suggestions. I assumed you made mistake on amount of $\ce{Mg}$ you have used. It should be $\pu{\approx 0.025 mol}$, if you have used $\pu{\approx 1 g}$ of $\ce{Mg}$. So please edit your post to include all these details.

Your target equation is: $\ce{Mg(s) + 1/2O2 -> MgO(s)}$. You have given following set of equations to get your target's enthalpy of formation:
$$\ce{MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l)} \quad ΔH = \pu{-1300 kJ} \tag {1}$$ $$\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)} \quad ΔH = \pu{-602 kJ} \tag {2}$$
$$\ce{H2(g) + 1/2O2 -> H2O(l)} \quad ΔH = \pu{-286 kJ} \tag {3}$$
Yet, if you added equations $(1)$ through $(3)$, you won't get your target equation because you have set equation $(1)$ in opposite direction so that $\ce{MgO}$ would ends up in reactant side (LHS). What you should have done is reverse that reaction. The reverse of Equation $(1)$ has identical $\Delta H$ but with opposite sign:
$$\ce{MgCl2(aq) + H2O(l) -> MgO(s) + 2HCl(aq)} \quad ΔH = \pu{+1300 kJ} \tag {4}$$
Now you can add equations $(2)$ through $(4)$, and you should get a negative $ΔH$ for your target equation (meaning the reaction is exothermic, and given $\Delta H^\circ_f$ for $\ce{MgO}$ is $\pu{−601.6 kJ/mol}$ according to Wikipedia). However, your given $\Delta H$ values are seemingly incorrect and I leave you to find the correct values (Note: The given $\Delta H^\circ_f$ for water is acceptable).
Late edition: According to OP's late comment to andselisk's suggestion on first comment, it seems OP has made a mistake on amount of $\ce{MgO}$ calculation, of which OP has used in the calarimetric experiment. It should have been $\pu{\approx 0.025 mol}$, if OP has used $\pu{\approx 1 g}$ of $\ce{MgO}$. Thus, $\Delta H$ for the reaction $\ce{MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l)}$ should have been $\approx \pu{3.34 kJ}/\pu{0.025 mol}= \approx \pu{134 kJ/mol}$. Thus, $\Delta H^\circ_f$ of $\ce{MgO}$ from the experiment is: $\pu{(+134-602-286) kJ/mol}=\pu{-754 kJ/mol}$ (literature value is $\pu{−601.6 kJ/mol}$).
Until the time of this edition, OP has not edited the question. Thus, I assume OP has also calculated the data for the reaction $\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}$, actual value of which has been around $\pu{−460 kJ/mol}$. Hence, approximate experimental value for $\Delta H^\circ_f{_\ce{MgO}}$ is: $\pu{(+134-460-286) kJ/mol}=\pu{-612 kJ/mol}$, which is better acceptable value for literature value, $\pu{−601.6 kJ/mol}$.

As I already mentioned in the comment, the first value for neutralization looks way off. Since you already posted an attempt to answer, probably it would make sense to add that $ΔH_1$ should be approx. double the value of standard enthalpy of neutralization, $\pu{-57.62 kJ mol-1}.$ That's why I assumed $\pu{-130 kJ mol-1}$. Still it would be nicer if you dig up actual literature values; in the absence of those it's just speculations.

@ andselisk: Couldn't fond real literature values for them, but approximate values were calculated using standard enthalpies of formation given in https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation.

Why do you add the three equations ?
By first adding (2) and (3) and then SUBSTRACTING (1), you find Mg + (1/2) O2  -->  MgO.
On the other end, I think some of these numerical values are wrong, because one should obtain -602 kJ/mol, which is the heat of formation of MgO. And the combination of the given values do not yield to this result, by far.

I got 1300 kJ/mol by calculating q of HCl divided by number of moles of MgO used, which ended being 3.34 kJ divided by 0.00257 mol. This was a lab by the way.


text:
What is the pKaH of pyrrole?
I am supposed to say what is more basic and I ran into a problem. I would like to ask if anyone has any idea what is the correct $\mathrm{p}K_\mathrm{a}\ce{H}$ value of pyrrole. I know it is easy to just google it or open a book to find the answer. I have done that, but got pretty interesting (confusing?) results:
Wikipedia: "Pyrrole is weakly basic, with a conjugate acid $\mathrm{p}K_\mathrm{a}$ of (−3.8)"
Human metabolome database (HMDB): "Pyrrole is a very weak base with a $\mathrm{p}K_\mathrm{a}\ce{H}$ of about (4)"
PubChem: "Pyrrole is a very weak base with a $\mathrm{p}K_\mathrm{a}\ce{H}$ of about (4)"
McMurry 9th edition: "pyrrole - $\mathrm{p}K_\mathrm{a}$ of ammonium ion: (0,4)"
University of Calgary: "pyrrole conjugated acid $\mathrm{p}K_\mathrm{a}$:  (-3,8)"

related https://chemistry.stackexchange.com/questions/48499/basicity-of-heterocyclic-compounds

Higher data are probably in DMSO.

BTW while Pubchem repeats what metabolome says it gives -3.8 value in separate section. I wager pKa you seek is about -3.8 to -4; which are like exact same thing for stuff with negative pKa. Quite possibly in DMSO value may be about +4.

Op's question is what is the correct $\mathrm{p}K_\mathrm{a}\ce{H}$ value of pyrrole. Thus, I'm not going to elaborate OP's findings, but would try to give a reasonable answer to the question. The most reasonable answer I found for $\mathrm{p}K_\mathrm{a}\ce{H}$ value for the pyrrole is $0.4$ (for the novices, this is the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of pyrrole).
It is well known that the best way to quantify the basicity of an amine is to examine the $\mathrm{p}K_\mathrm{a}$ of its conjugate acid (Master Organic Chemistry). The concept is the higher the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid, the stronger the base.
Let's compare two cyclic amine bases under this concept: Pyridine and Pyrrole. The following diagram illustrate the concept:

Accordingly, pyrrolium ion is about $10^5$ times stronger acid than pyridinium ion. Thus, pyrrole is about $10^5$ times weaker base than pyridine. Why? Read the note on the bottom of the box in the diagram. This is also well explained in chemistry.msu.edu with the diagram attached:

In each case the heteroatom has at least one pair of non-bonding electrons that may combine with the four π-electrons of the double bonds to produce an annulene having an aromatic sextet of electrons. This is illustrated by the resonance description at the top of the following diagram. The heteroatom Y becomes $\mathrm{sp^2}$-hybridized and acquires a positive charge as its electron pair is delocalized around the ring. An easily observed consequence of this delocalization is a change in dipole moment compared with the analogous saturated heterocycles, which all have strong dipoles with the heteroatom at the negative end. As expected, the aromatic heterocycles have much smaller dipole moments, or in the case of pyrrole a large dipole in the opposite direction. An important characteristic of aromaticity is enhanced thermodynamic stability, and this is usually demonstrated by relative heats of hydrogenation or heats of combustion measurements. By this standard, the three aromatic heterocycles under examination are stabilized, but to a lesser degree than benzene. Additional evidence for the aromatic character of pyrrole is found in its exceptionally weak basicity ($\mathrm{p}K_\mathrm{a}$ ca. $0$) and strong acidity ($\mathrm{p}K_\mathrm{a} = 15$) for a $2^\circ$-amine. The corresponding values for the saturated amine pyrrolidine are: basicity $11.2$ and acidity $32$.

Furthermore, for everybody's convenience, I attached a orbital representation of pyrrole and furan to show their resistance to protonation.
Comment on different $\mathrm{p}K_\mathrm{a}$ for pyrrole in literature: The $\mathrm{p}K_\mathrm{a}$ value of $3.8$ (and $\approx 4$) is probably for the protonation of pyrrole at $\ce{C2}$ position. For protonation at $\ce{N}$, $\mathrm{p}K_\mathrm{a}$ is most probably $0.4$, as described above. We can trust the value given in McMurray's Organic Chemistry textbook, since it is widely used in sophomore organic chemistry university course in USA.

That's a lot of writing for an answer that's not proving anything at all. 1) Did you see that there was "-" 3.8? 2) Any value in literature wouldn't be for protonation on nitrogen, because it's unfavourable 3) McMurry is a nice textbook for elementary org. chem. but hardly a proof for anything. Even a paper may have a wrong value. 4) For such well reasearched question I'd expect at least something like that: https://chemistry.stackexchange.com/a/85198/9961 to actually answer anything.

@ Mithoron: I have given the answer in my first paragraph. Also, I said I don't want to elaborate wrongful information given in internet as usual. This answer is for a elementary organic chemistry student so do not need any literature proof; commonly used textbook is good enough.  Finally, if you have a good answer, why don't you share with community?

I already said what I think in comments. What makes you think it's for "elementary organic chemistry student" and one like that should just trust his textbook? If this site has any penchant, it's debunking textbook mistakes, often with experimental data. Also you seem to not take my 2) which is about factual error...


text:
Does broader peak mean stronger absorber in UV/VIS?
I want to ask a question about peaks in a UV/VIS spectrum.
My lecturer in our introductory course mentioned the following:

Two molecules may have the same molar extinction coefficient, but if I have two molecules, A and B, and B has a peak that is broader than molecule A, it will be a stronger absorber.

He failed to qualify this further or provide any reasoning.
My thought was that if it is a broader peak, then molecule B will absorb a wider range of wavelengths in a UV/Vis spectrum, but I couldn't see what he meant by a "stronger" absorber.
Does he mean that molecule B absorbs more wavelengths, so it is "stronger" or is there another qualification for using the word "stronger" to describe B?

Tell him you are not paying tuition fees to be taught dubious facts which you then have to learn by heart.

The classical concept of  oscillator strength $f$ is useful here, but it should be used only in a qualitative way. This is defined as $f=a\int\epsilon_\nu d\nu$ where $a$ is set of units with value $4.3 10^{-9}$ if the extinction coefficient is in units  dm$^3$/mol/cm and $\nu$ is in wavenumbers. The maximum value of $f$ is unity and is close to this for an allowed transition in a dye molecule, such as rhodamine, but can be as small as $10^{-9}$ for a forbidden transition.
In your case if two molecules have similar $\epsilon$ at some wavelength (or frequency) but one has a narrow transition compared to the other, then the one with the narrow transition will have a smaller integral and hence the smaller oscillator strength. In this sense only is the transition 'weaker', i.e. it is not the probability of  absorbing at a given wavelength that is being considered but that over all wavelengths, so it really depends on how you want to define a 'strong' vs a 'weak' transition.  In QM the transition moment integral is used instead of oscillator strength and gives a proper description of absorption.

If a peak in UV/Vis is broader than another one, with the same absorption coefficient, the absorption is the same at the wavelength corresponding to the maximum of the peak. But at any other wavelengths, a bit different from the maximum of the peak, the absorption is larger for the broader peak. So the proportion of the total amount of light, integrated over the whole spectrum, is larger. Suppose the first spectrum is so broad as to be flat from 400 to 800 nm, and that the absorbance is equal to 1 everywhere. All wavelengths will be absorbed, and the total energy getting out of the cell is 10% of the entering energy. If the absorbance is equal to 1 at only one wavelength, the absorbance being zero at all other wavelengths, the total energy getting out of the cell will be nearly equal (about 99%) to the original energy getting in the cell.


text:
The physical intuition behind some simplifications in electrochemistry

Find the standard reduction potential of
$$\ce{AgIO3(s) + e- <=>  Ag(s) + IO3-}.$$
$K_\mathrm{sp} = \pu{3.1E-8},$ and the standard reduction potential of $\ce{Ag +e- <=> Ag(s)}$ is $E = \pu{0.799 V}.$

So, $E^\circ$ will be standard reduction potential and $E$ will be the nonstandard.
$$
\begin{align}
E(\ce{AgIO3/Ag}) &= E^\circ(\ce{AgIO3/Ag}) - 0.0257\ln\frac{1}{[\ce{Ag}]}\\
  &= E^\circ(\ce{AgIO3/Ag}) - 0.0257\ln\frac{[\ce{IO3}]}{\pu{3.1E-8}}
\end{align}
$$
$$\pu{0.799 V} = E^\circ(\ce{AgIO3/Ag}) - 0.0257\ln\frac{1}{\pu{3.1E-8}}$$
Now, my concern is that I understand the math and like most of the principles behind all this, but how come when we set $[\ce{IO3}] = 1,$ we can also say $E (\ce{AgIO3/Ag})$ is equal to $E(\ce{Ag})?$
tl;dr: how come setting the concentration of the anion to $\pu{1 M}$ allows us to assume that variable change for the electrode?

I took a liberty to format your question for better readability and proper notations; if you are interested further in technical details, feel free to visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
How does electrophilic bromination with 1,2-dibromoethane work?
Hi I am trying to figure out the mechanism of a reaction containing a lithium carbanion and 1,2-dibromoethane in THF. The lithium carbanion is generated by deprotonating using LDA. Then 1,2-dibromoethane is added to get the desired product (see image)

I think that the carbanion is brominated, otherwise this reaction does not make sense, but I really don't understand the mechanism. 1,2-dibromoethane can undergo a reductive elimination of the bromide-ions to form bromine and ethene. But in this specific reaction, I have no reductive agent... So how does the carbanion get brominated? Is it simply a fragmentation reaction, where the carbanion attacks the Br of 1,2-dibromoethane?
These are my thoughts on the intermediates.

Thank you!

Important thing to consider here is that organolithium there doesn't have carboanion - for an accurate mechanism Li has to be coordinated all the time.

related https://chemistry.stackexchange.com/questions/31094/alkylation-of-conjugated-nitriles-regioselectivity

The gamma picoline is the reducing agent and ethylene dibromide is the oxidizing agent. Gamma picoline is oxidized; ethylene dibromide is reduced.

@Mithoron I have corrected the mechanism. Does this seem correct now? The negative charge should be on the nitrogen after the picoline is deprotonated as it can accept the electrons.

Ah, you wrote just what I thought, but unfortunately a moment later I though that it can't work like that :( It's not a substitution, but transmetalation. Bromide would also be coordinated. Some digging for papers would be needed to answer this...


text:
Why acidity of p-methoxybenzoic acid is more acidic than p-hydroxybenzoic acid?
I know +M effect of $\ce{OCH3}$ is more than $\ce{OH}$, but in my book (ALLEN General organic chemistry), it is given that $\mathrm{p}K_\mathrm{a}$ of para-substituted benzoic acid containing $\ce{OCH3}$ is less than that of para-substituted benzoic acid containing $\ce{OH}$:

$\mathrm{p}K_\mathrm{a}$/Benzoic acid → $4.20$
$$
\begin{array}{lcccc}
\hline
\text{Substituent} & \mathrm{p}K_\mathrm{a} & \textit{o-} & \textit{m-} & \textit{p-} \\
\hline
\cdots \\
\ce{OMe} & - & 4.09 & 4.09 & 4.47 \\
\ce{OH} & - & 2.98 & 4.08 & 4.58 \\
\cdots \\
\hline
\end{array}
$$

I am in doubt whether these data wrong or correct. If correct, what can be an explanation?

Both -$\ce{OCH3}$ and -$\ce{OH}$ groups have exhibited two effects on the aromatic ring: (1) Electron donating resonance or mesomeric effect (+M) and (2) Electron withdrawing inductive effect (-I). For both acids in hand, the electron withdrawing inductive effect (-I) is almost same since both -$\ce{OCH3}$ and -$\ce{OH}$ groups are 4 carbons away from the acid center. However, when -$\ce{OH}$ group is attached to either the para- or ortho-position, it has a more tendency to delocalize its lone pair electrons towards the aromatic ring than that of -$\ce{OCH3}$ group (Order of activating:  -$\ce{O-} \gt$  -$\ce{OH} \gt$  -$\ce{OCH3}$; for additional learning, read articles on Hammett plots).
As a result, the electron density on the carbon para to -$\ce{OH}$ substitution group increases more than that on the carbon para to -$\ce{OCH3}$ substitution group. Thus, the electron density on carbonyl carbon of carboxylic group attached to those para-carbons also increases accordingly, and consequently, the polarity of $\ce{O-H}$ of carboxylic group decreases as a result. That polarity decrease causes the reduced acidity ($\mathrm{p}K_\mathrm{a}$s are $4.58$ and $4.47$, when para-substitutions are $\ce{OH}$ and $\ce{OCH3}$, respectively) compared to benzoic acid ($\mathrm{p}K_\mathrm{a}$ is $4.20$ when para-substitution is $\ce{H}$). The $\mathrm{p}K_\mathrm{a}$s of p-hydroxybenzoic acid is higher than that of p-methoxybenzoic acid here, because as explained above, p-$\ce{OH}$ group caused reduced polarity on $\ce{O-H}$ of carboxylic group than that caused by p-$\ce{OCH3}$ group.

I am directly going to address the source of your confusion: $\text{+M}$ of -$\ce{OCH3}$ is more than -$\ce{OH}$. You see, you are basically saying that the $\text{+I}$ effect of $\ce{CH3}$ in -$\ce{OCH3}$ is going to push more electron density towards the ring, hence, -$\ce{OCH3}$ is more activating than -$\ce{OH}$.
But look at the bigger picture, and let's evoke Bent's rule here.
You see, due to steric reasons,the $\ce{C_{(\text{benzene})}-O-C_{(\text{alkyl group})}}$ bond angle of $\ce{OCH3}$ is going to be more than the $\ce{C_{(\text{benzene})}-O-H}$ bond angle of the -$\ce{OH}$. Hence, the percent $\mathrm{s}$-character is going to go up for the central oxygen atom in $\ce{OCH3}$ relatively (to rationalize this, think of the bond angles associated with $\mathrm{sp^3, sp^2,}$ and $\mathrm{sp}$ hybridization and their respective $\mathrm{s}$-characters)
So now, due to increased $\mathrm{s}$-character on both sides of the $\ce{OCH3}$ oxygen, you get an overall more electronegative substituent for the ring as compared to $\ce{OH}$, as the electrons displaced towards the oxygen from the methyl in $\ce{OCH3}$ would still experience a lowering in the energy of the $\sigma_\ce{(O-CH3)}$ bond. Hence, as compared to $\ce{OH}$, the ring will be activated less by $\ce{OCH3}$. I feel that you can now complete the reasoning, by checking the stability of the conjugate bases.
Note: A better qualitative reasoning could have been provided via frontier MO theory, but I chose not to evoke it here. If the OP requests, I can supply the same as well.

Give me your point of the above explanation throught frontier molecular orbital theory also.


text:
What does higher electron gain enthalpy mean here?
While talking about ionic bonds my book states :

Higher the value of electron gain enthalpy of the atom, greater the ease of formation of the anion from it, i.e., other atom should have high value of electron gain enthalpy.

Now I think it should say more negative than just higher (as negative enthalpy means release  of energy and hence more stability). So what should it be? Why the chemistry books (I have to read)  seem to always avoid the term magnitude(Is there some hidden assumption which are generally in the field of inorganic chemistry  ?

I take for granted that what you call "electron gain enthalpy" is generally described as "electron affinity". This is the energy gained (or lost) when an electron is added to an atom, ion or molecule. In general the electron affinity is a measure for the tendency of elements to form anions. The trouble is that you may find tables giving these numerical values as positive or negative values. For example, for fluorine $\ce{F}$, the electron affinity is $\pu{338 kJ/mol}$. But you may find books where the same affinity is given as $\pu{-338 kJ/mol}$. It depends on the sense you consider the reaction $\ce{A + e- -> A-}$.


text:
NH4+ reaction with NaOH
Why does $NH4^+$ react with NaOH to give $NH_3+H_2O$ and not $NH_4OH$

Because there is no $\ce{NH4OH}$, except as the separate solvated ions which are minor species in an aqueous ammonia solution.  See this answer for the whole story.

$\ce{NH4+}$ is the weak acid and reacts with the strong base $\ce{OH-}$:
$$\ce{NH4+ + OH- <=>> NH3 + H2O}$$
by the same way as $\ce{HCN}$, an acid with similar acidity as $\ce{NH4+}$:
$$\ce{HCN + OH- <=>> CN- + H2O}$$


text:
Can compounds dissolved in different miscible solvents react?
For example, can crystal iodine dissolved in ethanol react with sodium thiosulfate dissolved in water when mixed together?

Yes, used widely in chemistry

As ethanol and water are miscible in any ratio, it is not the best example. Additionally, for nonmiscible solvents, compounds have distribution constants, so they are present in both solvents, even if in  some cases, they prefer one solvent almost exclusively.

@Poutnik, the question concerns miscible solvents, so the ethanol/water example fits.

I see, but I do not see a sense of the question, regarding miscible solvents. They mix themselves and compounds react.

@sekharm, Why wouldn't they react, assuming at least some of both solutes stay dissolved after mixing? Try it.

@sekharm You are mixing up the past with the present in chemistry, and consistently so. Iodine dissolved in ethanol is not crystalline; maybe it was crystalline at some point in the past, but that's irrelevant now. As to the point of your question: can the two compounds in different jars react? Why, sure, as soon as they are put together, at which point they won't be in different jars anymore. Same thing with different solvents.

There are two ways to add two miscible solvents to each other. One is the bartender skill, whereby you attempt to create two different layers of solvents and then try to keep the flask motionless to prevent mixing. The other way is to mix them all along, whether by shaking, careless pouring, or a stirring bar. I will concentrate on the second as that is, in my opinion, the far more common use in chemistry.
Mixing to miscible solvents, even with compounds dissolved in them, results in a mixed solvent that for all intents and purposes of this question can be considered a new solvent. The question is no longer whether the individual solutes dissolve in their respective solvents but whether they dissolve in the mixed solvent. In many reactions, pouring the crude reaction mixture into a different solvent (miscible with the reaction solvent) serves as an easy way for the product to precipitate ready for isolation. For example, when xanthones are synthesised by dehydration in concentrated sulphuric acid, the crude mixture is then poured into water. Xanthones are not soluble in water but also not in water/sulphuric acid mixtures if the acidity is not sufficiently high so the xanthones precipitate and can be filtered off.
Obviously, there are three cases here: both compounds remain soluble, neither compound remains soluble and one of the compounds remains soluble. The first two cases are pretty trivial: if both compounds are soluble they can react as they would in a single solvent. Likewise, if neither compound is, they will both precipitate and not react (disregarding very slow, very minor solid state reactions if there are any relevant interfaces between the precipitates—but I digress).
The interesting case is the third case: what happens when one compound remains soluble but the other does not. Interestingly, with sufficient stirring there is typically enough contact surface exposed to molecules in solution that these two compounds will still react. Furthermore, precipitation doesn’t always mean entirely insoluble; the solubility might be low but sufficient to sustain a reaction. Thus, in most cases if one reactant is not sufficiently soluble, a reaction will still occur.

I will spare a few words on the bartender skill as there is an example reaction undergrad students may well perform as part of their lab courses: a reaction to test for the presence of nitrate. At first, nitrate is reduced by iron(II) salts in diluted sulphuric acid giving nitrogen oxide:
$$\ce{3Fe^2+ + NO3- + 4H+ -> 3Fe^3+ + NO + 2 H2O}\tag{1}$$
Then, the solution is layered with concentrated sulphuric acid whose greater density means it will drop to the bottom of the test tube. At the interphase, water is drawn into the sulphuric acid leading to the formation of pentaaquanitrosyliron(II) ions $\ce{[Fe(H2O)5(NO)]^2+}$ from unreacted iron(II). This complex has a dark-red, violet or brown colour and forms in a ring-like manner at the bottom of the water phase as shown in the image from Wikipedia. In German, this is known as the ring test.
$$\ce{[Fe(H2O)6]^2+ + NO <=>> [Fe(H2O)5(NO)]^2+ + H2O}\tag{2}$$

This shows how the compounds may still be able to react.


text:
Perturbation Theory in Orbital Interactions in Chemistry
In the entire book, perturbation theory is used as a qualitative tool to rationalise some chemical phenomena. The authors write that
$$\psi_i = \sum_\mu{T_{ji}\psi_j^\circ}\tag{1}$$
The proof of the derivation for $T_{ji}$ involves the following statement:
$$T_{ji} = \left(\mathbf{C}_j^\circ\right)^\mathrm{T}\mathbf{S}^\circ\mathbf{C}_i \tag{2}$$
where
$$\left(\mathbf{C}_j^\circ\right)^\mathrm{T} = \left(C_{1j}^\circ \quad C_{2j}^\circ \quad \cdots \quad C_{mj}^\circ\right)\tag{3}$$
(indexing the coefficients for the jth unperturbed MO), and
$$
\mathbf{C}_i =
\begin{pmatrix}
C_{1i} \\
C_{2i} \\
\vdots \\
C_{mi}
\end{pmatrix}
\tag{4}
$$
(indexing the coefficients for the $i$th perturbed MO); $\mathbf{S},$ the overlap matrix, is defined as usual.
I'm not quite sure how the aforementioned statement is obtained. Could someone please walk me through how
$$T_{ji} = \sum\sum C_{nj}^\circ S_{nm}C_{mi}?$$
Source: Orbital Interactions in Chemistry, 2nd edition, page 794-802

Please utilize MathJax for typesetting math and chemical formulas whenever possible: this makes them searchable and reduces traffic (especially if images are large). Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also, see MathJax basic tutorial and quick reference.

This equation (as written) is wrong; it's not actually in the book. The (correct) formula is inversely proportionate to $E_j^{(0)} - E_i^{(0)}$. It is a standard result in quantum mechanics and I suggest looking in a QM textbook of choice. Or the Internet. There are lots of good websites with the full derivation - search "perturbation theory".

Also, please add a complete reference for the textbook you are quoting from, and format quoted portions of the text as such with >.

@orthocresol it's in Appendix 1, and I've reflected that in an edit.

Ah, ok, I take that back then, sorry. It is confusing because in the actual text they have $t_{ji}$ which has a different meaning...

@orthocresol it's actually the same meaning as in the text; the Appendix simply tries to derive the expressions for intermolecular, electronegativity and geometrical perturbation from first principles.

Rather than focusing on the individual matrix entries, it may be more intuitive to look at the matrix $\mathbf{T}$ formed by putting together all entries $T_{ij}$. In this case, $\mathbf{T}=\mathbf{C}^T\mathbf{S}^0\mathbf{C}$, where $\mathbf{C}$ is the coefficient matrix with column $k$ denoting how atomic orbitals mix together into molecular orbital $k$.
Now the point is that molecular orbitals, or the columns of $\mathbf{C}$, are orthonormal and are eigenfunctions of $H_{eff}$ (see page 34). This orthonormality makes sense if you think of the Hamiltonian matrix, which is certainly self-adjoint, and solving the eigenproblem is equivalent to diagonalizing the matrix with an orthonormal basis of eigenvectors.
Hence, $\mathbf{C}$ is unitary, and $\mathbf{T}=\mathbf{C}^T\mathbf{S}^0\mathbf{C}$ can be simply seen as a change-of-basis transformation done on the overlap matrix from atomic basis to the molecular orbital basis.
I hope this rather mathematical perspective may be of help to you. Thanks!

It look like the expression is just the equivalent way of expression the matrix product. I've attached two pictures that might help.

I think you've misunderstood my question; I don't understand the rationalise behind the matrix expression itself, and had simply restated it in terms of the double summation. Thanks for the help, though!


text:
Redox couple notation

How is the redox couple as it is written for the oxidation of $\ce{MnO4-}$, as $\ce{MnO4-,H+/Mn2+,H2O}$ correct?
I agree that $\ce{MnO4}$ is being oxidized as it loses electrons (going from -1 charge to +2) but I don't see why $\ce{H+}$ is also being oxidized. Could someone explain?

H+ isn't and can't be oxidized. It is already in the highest possible oxidation state.

Can you help me understand why it should be on the left of the /, meaning how do I see that H+ is being oxidized to H2O? Because oxidation is loss of electrons but H+ goes from positive to neutral, in my head I see that perhaps too simplified as being reduced @IvanNeretin

The charge on ions is irrelevant. It is oxidation state of elements that matters.

My take is: $\ce{MnO4-}$ can be reduced in two media, namely acid or base/neutral with different electrode potentials. The equation given here is for acidic medium half-reaction. So, $\ce{H+/H2O}$ represent the medium.

@ Mathew, It is still a poor notation, if several PhDs have to make a guess or an deliver an opinion on the meaning of certain notation which is not worth the complication, that notation is indeed useless. The author is unnecessarily making the redox notation complicated.

The H atom is neither oxidized nor reduced in the redox semi-equation. It remains at +1. Only Manganese reacts, as it is reduced from +7 to +2. The redox couple should be written :  MnO4-/Mn2+, without mentioning Hydrogen
And MnO4 does not lose electrons, as you say. On the contrary, MnO4 fixes or consumes 5 electrons in its reduction to Mn2+.


text:
Are these numbers Madelung constants?
I have no idea of chemistry, just a little bit what I know from school, but I found through some search some connection between entropy (Shannon entropy) of the divisors of some natural numbers and the Madelung constants.
My question is this:
Are these numbers Madelung constants?
1 -0.000000000000000
2 1.38629436111989 ( http://oeis.org/A016627 )
3 1.38629436111989
4 3.31573052232910
5 1.38629436111989
6 5.54531864369184  ( http://oeis.org/A257872 )
7 1.38629436111989
8 5.57849454296371
9 3.29975689063759
10 5.54521210457048
11 1.38629436111989
12 10.7943529563000
13 1.38629436111989
14 5.54518969515687
15 5.54519368543309
16 8.08924328274917
17 1.38629436111989
18 10.7599543952466
19 1.38629436111989
20 10.7921021686465

Thanks for your help!
Related question:
https://mathoverflow.net/questions/346239/entropy-magnitude-diversity-of-finite-metric-spaces-in-number-theory
Edit:
I think I found an explanation for the numbers ( entropy ). It is approximately:
$\tau(n) \log(\tau(n))$, where $\tau(n)$ is the number of divisors of $n$.
So the $1.386$ and $5.545$ are just coincidences with Madelung constants I think.

1.386 is 2log2 but what about 5.545 the closet approximation is 7log7 which is still 5.91. Am i missing something?

$5.545=8 \log(2)$ see the oeis link which relates this number to one Madelung contant

Ok i was mistaking that both the numbers inside and outside the log must be the same

I don't know if this would be very helpful or not but as i had some free time on hand I got a list of Madelung Constant Values from the Internet
In the list most of values were reported to 3-4 decimal places so I have Compared only the first decimal places as once the second decimal place was used none of the values matched. Maybe this would be a helpful start.
$$M_a$$ of ZnS (Wurtzite Hexagonal) is $10.7153$ (nearby the 12th and 18th entry in your table)
As Oscar Lanzi has commented $1.386$ is approximately $$2\log(2)$$ which is the Madelung constant for a one-dimensional line of alternating charges
As the question has been edited to add further details maybe the second one is a coincidence.
Only Nearby Matches So Far

The $1.386$ figure is $2\log 2$, which is the Madelung constant for a one-dimensional line of alternating charges.

@OscarLanzi should i add this into my answer or are you going to write one on your own?

I say add it in.

@OscarLanzi Done


text:
Why is the cathode of a lithium-ion battery metallic, which is the opposite of most other batteries?
In most types of batteries, primary and secondary, the metallic electrode is the anode, and a non-metal acts as cathode.  Or a pure metal is the anode, and the cathode is a metal oxide.
Why is the opposite true in a lithium-ion battery?

This is not entirely true.  Wikipedia lists several possible cathode materials and they are oxides.  They may be "metallic" is the sense of being metallic conductors (some oxides are so), but they are ceramic compounds.
In secondary batteries the anode is typically Li-intercalated carbon (which is also a good conductor).  Lithium intercalates readily into carbon, allowing a safer and less side-reaction prone choice than pure lithium, whereas most other metals do not.

Perhaps I should have said that the anode in most batteries is 'more metallic' than the cathode, with the opposite being true in a lithium-ion battery.....

Then how do you define "more metallic"?

I don't agree that the cathode is more metallic. A Li-ion battery with a lithium metal anode is a common chemistry for primary batteries.


text:
How do I separate a graphene/graphite oxide and NaOH solution without centrifugation? (filtration didn't work)
The graphene oxide solution was made from electrolytic exfoliation of graphite electrodes in a 10M NaOH solution. Some graphene powder deposited on the bottom of the recipient but a brownish solution remained, even after the powder settled. Graphene oxide is believed to be the source of the brown color.
The solution still has a lot of the hydroxide present so evaporating the solvent also leaves that compound behind along with the brown crystals, so it's hard to identify if it is actually graphene oxide or just impurities.
Right now I don't have a centrifuge, and filtration didn't work, so how do I separate them?

The literature is full with ways to wash and dry graphene.

@Karl Yes, I saw some methods using dialysis membranes but I can't afford these at the moment, but thanks for the answer anyway

Come on, you don`t need new medical grade membranes. Be creative!

@Karl Well I think I'll probably have to come up with a method for osmosis, since the solution itself is so concentrated. But anyway, thanks for answering

@Caju do the solution clear after a while with the material settled down at the bottom of the flask? If yes, decantation could work for you.

@FelipeS.S.Schneider Actually it doesn't, for a long while all it did was to decant the graphite particles, but not the supposed brown ones


text:
dissolving powder
I have small amount of rosemary herb powder mixed in a cup of water and it's only party dissolving. I was wondering if there is another powder that can be added to this water solution and dissolve the rosemary completely?
I would like to avoid using rosemary essential oil as it has a certain scent that is different than the scent of the full leaf powder. I wish to only use the necessary ingredients to make the rosemary powder fully dissolve. The goal is to make the rosemary powder not visible when holding the glass of water up to the light.

Im tempted to flag this as "homework". ;) Why do you think it should dissolve?

Do you know the difference between a suspension and a solution? Does flour dissolve in water? Sugar?

The goal is unattainable.

The dry and powdered parts of former bodies of living things are not soluble without destruction of their chemical and structural nature.
The only achievable way is to prepare such a fine powder that the difference between its suspension and clear liquid does not bother the end user. Surfactants and thickeners may stabilize the suspension, but it would be still suspension.
Alternative achievable way could be the powder extraction by high percentage food grade ethyl alcohol, keeping the filtered extract and adding it to water.

You could add a surfactant and/or thickener to make and keep a suspension more stable.

@James Gaidis  Sure, you can. But it would be still suspension.


text:
Volatile alkali metal compounds
Most elements are or have compounds that are gases or have a significant vapor pressure at room temperature. Fluoride is a powerful tool for producing highly oxidized covalent and volatile compounds. Tungsten hexafluoride is gaseous at room temperature despite being made with the element with the highest melting and boiling point. Refractory elements can also have volatile oxides such as $\ce{CO2}.$
Organometallics such as iron pentacarbonyl and dimethyl zinc are liquids at room temperature but have a decent vapor pressure, boiling at 103 °C and 46 °C, respectively. Other examples of "volatilized elements" include diborane and heaver analogous of ammonia such as phosphine.
All of the compounds I found of alkali metals, however, have very low vapor pressures at room temperature. For example, organolithium compounds tend to form lithium clusters or ionic crystals and decompose into lithium hydride when heated (which isn't volatile), rather then the covalent examples previously. Alkaline earth organometalics seem to suffer a similar problem with dimethylmagnesium forming a solid lattice.
Are there any known examples of an alkali metal compounds that has a reasonable vapor pressure, say 100 Pa at 300 K?

Is tert-butyllithium ok for you then?

@Mirhoron: Is the boiling point for the pure compound, and does it maintain stability when boiled and recondensed in an inert atmosphere?

As Mithoron says in the comments, tert-butyllithium is one choice.  Like pure organolithium compounds generally, this is pyrophoric and reactive to many common atmospheric gases or vapors. An article by Schwindeman et al. [1] recommends using a nitrogen or argon atmosphere.
In the case of alkaline earth metals, magnesocene provides a relatively volatile magnesium compound, subliming at $\pu{100 °C}$.  This is also air- and moisture-sensitive and again, should be handled under nitrogen or argon.
References

Schwindeman, J. A.; Woltermann, C. J.; Letchford, R. J. Safe Handling of Organolithium Compounds in the Laboratory. Chemical Health and Safety 2002, 9 (3), 6–11. https://doi.org/10/ctj2sj. (PDF mirror: chemistry.ucla.edu)


text:
Is it possible to prepare a 0.1 N solution of HCl and methanol?
In my lab guide I read: "Use a $\pu{0.1 N}$ solution of methanol-chloridic acid $(9:1)$ for the anthocyanin extraction", so I tried to calculate the volume of $\ce{CH3OH}$ and $\ce{HCl}$ to prepare it, but I don't know how to do it. I mean, if I push one mole of $\ce{CH3OH}$ and one mole of $\ce{HCl}$ in $\pu{1 liter}$ of water, do I make a $\pu{2 molar}$ solution?

What about 9 volumes of CH3OH  + 1 volume of 1 M HCl, giving 0.1 M HCl in ( approximately ) 90% (v/v) MeOH ? BTW, I have not ever seen the term chloridic acid.

You mean, a 1 M solution of CH3OH?

Why,  I have said 1 M HCl.

So, can I use 2 or 3 M solution of CH3OH? The solution ever going to be 0.1 .  Also, I think this is a very good answer.

I doubt they consider CH3OH molarity. In my understanding, the target is to have 0.1 M HCl in 90% CH3OH.(approx.)

Furthermore, if you dissolve 1 mole CH3OH and 1 mole HCl in one liter water, you don't obtained a 2 molar solution. You obtained a solution which is about 1 molar in HCl and 1 molar in CH3OH. I say "about", because the final volume may be bigger than 1 litre. To be sure to get a solution which is 1 molar in HCl and 1 molar in CH3OH, you should dissolve 1 mole HCl and 1 mole CH3OH in a 1 liter flask, half filled with water, and afterwards, add enough water to get exactly one liter.

Of course, mixing acids and solvents may be an ES&H issue depending on jurisdiction (it is explicitly not allowed where I work).

Of course there are such solutions.  Take, as an example, Sigma-Alrich's offer of methanolic solution of HCl (3 mol/L, e.g. here), or the dry ones in diethyl ether (e.g., 2 mol/L here), in cyclopentyl methyl ether (e.g., here), or in 1,4-dioxane (e.g. 4 mol/L, here).


text:
Why does eutectic point have three phases, but consolute point have only two phases?
At eutectic point, there are three phases: solid A, solid B and homogenous liquid phase containing both A and B.
But at consolute point, we are told it has 2 phases: liquid A and liquid B.
Why isn't there a homogenous phase containing both liquid A and B at consolute point just like an eutectic point?

It would make more sense to compare the eutectic point with the azeotropic point for a heteroazeotrope (not for a typically seen homoazeotrope). You can see the similarity between them in the following two phase diagrams, both showing three phases at the eutectic and azeotropic points.
The eutectic point is the point for a solid/liquid equilibrium while the azeotropic point is the point for a liquid/vapor equilibrium. At the eutectic point, in addition to the liquid phase, the solid exists in an eutectic lamellar structure of A and B, or in the case shown below, an eutectic structure of α and β solid solutions rich in Ag and Cu, respectively. Likewise, at the azeotropic point, in addition to the vapor phase, the liquid exists in immiscible layers of liquids rich in X and Y, respectively.
Therefore, in both cases there are three phases, leading to conditional degree of freedom (or variance) of zero, where the word "conditional" refers to a specific pressure and "zero" means a fixed point on the phase diagrams.

Panel (a) is taken from libretexts.
Panel (b) is taken from Wikipedia.

I just realised, the 2 phases present in a liquid-liquid solution system with temperature below its solution temperature is a) conjugate solution enriched in A
b) conjugate solution enriched in B
No pure liquid A or pure liquid B phase exists.


text:
What does "parallel in-register" mean for a protein structure in a fibril?
I read a paper by Ross and Poirier [1, p. S12] which has the following lines (see section Commonalities of amyloid structure) that describe the structure of protein in a fibril:

The most extensively characterized amyloid fibril is that formed from the β-amyloid (Aβ) peptide implicated in AD. Using SOLID-STATE NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY, the in-register, parallel β-sheet organization of fibrils formed by Aβ10–35, a fragment of the full-length 42-residue Aβ peptide, was first described.

I couldn't understand what "in-register, parallel β-sheet" means.
References

Ross, C. A.; Poirier, M. A. Protein Aggregation and Neurodegenerative Disease. Nat Med 2004, 10 (7), S10–S17. https://doi.org/10/dppb95.

Searching 'protein structure' gets 250 hits on chemistrySE, and 500 hits on biology SE.  Thus you might have luck on either site.  I'd give it a bit, and if you don't get a satisfactory answer here, ask it on biologySE (or perhaps ask that the question be moved there).

Have you looked for ref 43 in that paper, as this refers to the particular statement. Otherwise you could try emailing the authors.

There is a clear explanation and figure showing this in the following reference:
Anton Gorkovskiy, Kent R. Thurber, Robert Tycko, Reed B. Wickner, PNAS 2014, 111 (43) E4615-E4622.
A parallel beta sheet is one where the direction from N- to C-termini on adjacent strands run in a parallel direction (rather than antiparallel). In-register means that each residue in one molecule is aligned with the same residue in the neighboring molecule, producing lines of identical residues along the long axis of the filament.

Panel b of figure 1 shows the common position of a given amino acid in each aggregated peptide along the la-axis of the fiber; axes defined in panel a of the same figure.


text:
What is the IUPAC name of the given compound?

IUPAC name of the given compound is

(1) 3-Hydroxycyclohex-4-en-1-one
(2) 5-Hydroxycyclohex-3-enone
(3) 5-Ketocyclohex-2-enol
(4) 3-Ketocyclohex-5-enol

By what I understand by assigning the priority the parent is the ketone.
If I number carbonyl carbon as 1, then I should move towards the alcohol group because it has higher priority than the alkene and both are at number 3, so the alcohol should be assigned the number 3 and alkene as 4?
So, according to me, the answer should be (1), but in the key it is given as (2).

As for your other question, the simplified criteria for the numbering are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

For the compound that is given in the question, you have correctly identified the principal characteristic group as ketone. It is expressed as suffix “one”. Thus, a low locant is assigned first to the ketone according to Rule (c), which yields “…1-one”.
Next, a low locant is given to the double bond according to Rule (e), which gives “3-en-1-one”.
Finally, a low locant is assigned to the remaining $\ce{-OH}$ substituent according to Rule (f).
Thus, the complete name is 5-hydroxycyclohex-3-en-1-one.
Rule P-14.4 applies to chains as well as rings.