text:
What defines a proton as "labile" in NMR?

What makes some protons labile and others not? For example the OH and NH2 protons get exchanged but CH ones don't.
When we use the D2O test, we say the labile protons get exchanged for D so their peaks disappear. But can't the H be re-exchanged? And why doesn't the H now in the solvent give its own peak?

You've never seen https://chemistry.meta.stackexchange.com/questions/524/i-question-the-use-of-question-in-question-titles I guess... And this looks like borderline case as far as homework  policy violation is concerned.

For your first question you may want to look up "proton exchange NMR", as partly addressed in for instance this post:
Proton attached to N or O in HNMR spectroscopy
$\ce{^1H/^2H}$ exchange is a solvent catalyzed process that results in rupture of the original O-H or N-H bond with formation of a new bond to previously H-bonded solvent $\ce{^2H/^1H}$. What keeps the exchange process from occuring is an activation energy barrier.  Available nonbonding electrons on O or N atoms in the solute or solvent which can interact with $\ce{^1H}$ or $\ce{^2H}$  affect the process, as does the pH (exchange can proceed through acid or based-catalyzed mechanisms). Catalysis of C-H bond rupture with formation of a new C-H bond has a larger activation energy barrier or may lead to chemically nonidentical products and reactants.
As regards the second question, if the (protonated) solute is dilute and the solvent is deuterated, there isn't a significant concentration of solvent $\ce{^1H}$ to drive reversal to the protonated state. For instance, if your molecule is present in water at 1 mM concentration and contains one exchangeable proton, and if all of the protons exchange and the solvent water was originally 100% deuterated, then the final concentration of protons in the solvent is $\pu{\rho = ( 0.001/110) mole/mole = 9 \mu mole/mole (or 9 ppm)}$ and the solute will exhibit a similar final isotopic concentration ratio (ignoring isotope effects).


text:
What happens when an ionic compound is boiled?
Do ionic compounds such as NaCl become individual ions such as one Na ion and one Cl ion or are they hold together by ionic bonds even when they are boiled. Thanks in advance.

See: https://chemistry.stackexchange.com/questions/2511/will-gaseous-ionic-compounds-be-free-moving-ions?noredirect=1&lq=1

Ah! The question isn't clear .. By the way , how can you boil NaCl !! Until and unless you melt and then boil it to vaporize it...

Some of the most celebrated experiments in Chemistry  have been done on NaI molecules in the vapour phase by Zewail and coworkers. They observed that the dissociation of NaI molecules  into Na + I atoms takes place in steps as the molecule vibrates. Zewail was awarded the Nobel Prize for these experiments.

@porphyrin, Zewail must be using very powerful lasers.

https://chemistry.stackexchange.com/questions/14174/what-is-sodium-chloride-like-in-gas-state

Zewail did not use very powerful ones in that sense but ones that produce very short pulses typically $10^{-13} seconds long. Such pulses are actually not that herd to produce in the lab and are produced in a laser oscillator by the process called mode-locking.

Interesting question with no generic answer. Boiling the salts, or ionic compounds, makes them a plasma or freely moving ions is nothing but a misconception. The temperature is too low to convert a substance into a plasma.

Let us start with ionic liquids which are room temperature cations and anions but made up of organic molecules. You heat them, to the state of boiling in the absence of oxygen. They decompose into various products. So nothing special, this is as expected for organic compounds.
Heating of non-decomposable salts at high temperatures, for example, NaCl, KCl: What will happen when we heat them to the point of boiling? Answer is nothing. They remain in vapor phase as NaCl "molecules". The vapors behave as gases. People have developed equations of state. Interestingly enough $\ce{Na2Cl2}$ (dimers) are found in gas phase.

See: Critical point and vapor pressure of ionic fluids including NaCl and KCl, Chemical Physics Letters, 105, 1984, 484-489

Ionic materials, KCl or NaCl, when introduced into flames or plasma convert into gas phase "atoms". For example $\ce{Na+}$ would be converted into a neutral Na atom. This is utilized in a completely different field in chemistry, called flame atomic absorption or flame atomic absorption spectroscopy.

In short, these complexities make science very interesting.

The user have asked about ' boiling ' , heating and boiling are completely two different terms . Isn't?

To be precise, it is heating to the point of boiling. Note vaporization occurs even before boiling, just like water. There is nothing special in heating, melting and vaporization of ionic salts.

Boiling is the rapid vaporization of a liquid, NaCl in aqueous phase can only be boiled.. or it can be melted and then can be boiled to the state of vaporization

I think you misunderstood the question. The poster is asking about "solid" NaCl which has been heated above its melting point (>800 oC) and heating is continued to the extent that liquid NaCl becomes a vapor. There is no discussion of water anywhere.

Hold on , points you are stating haven't been stated in the poster .. mere assumptions.. question is itself unclear

Depends on an individual's interpretation! No need to worry about the poster. He would clarify if needed.

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text:
Synthetic steps for the interconversion between aldehydes and ketones
I am learning about carbonyl chemistry at the moment. So far, we have been able to make aldehydes and ketones from alcohols (via oxidizing agents). I can't help but wonder if there is possibly any way to make aldehydes from ketones or vice-versa through a series of synthetic steps. I would appreciate it if the synthetic steps are in the level of an undergraduate organic chemistry course.
Question: Synthetic steps for the interconversion between aldehydes and ketones?

You could reduce an aldehyde to a primary alcohol, dehydrate the alcohol to an alkene, hydrate it again with the hydroxyl on the other side (secondary alcohol), and then oxidize to get the ketone.

@KarstenTheis yes, I was able to figure this out with the given answers, but I am still having trouble going in the reverse direction. Ketones to aldehydes.

Check out hydroboration vs. oxymercuration, e.g. by looking at this cheat sheet. Once you have the hydroxyl group on the carbon where you want it, you just have to make sure your oxidation does not go all the way to a carboxylic acid.

@JamesBond For converting Ketones to aldehydes, what you can do is to perform a Bayer-Villiger Oxidation  on the ketone to get an ester. The ester what you got can now be reduced with DIBAL-H (Diisobutylaluminium hydride) to get an aldehyde and the other part will leave as alcohol which you can again oxidise using PCC.

@SoumikDas Thanks! Makes perfect sense since I already know about the ester to aldehyde via DIBAL-H. The Bayer-Villiger Oxidation was the key to getting to the ester. Thanks a lot. If you want you can post your answer so I can accept it, I like this more.

You can easily convert aldehydes to ketones and vice-versa by using simple organic synthetic steps which are very much well known.
Converting aldehydes to ketones
For converting aldehydes to ketones there are numerous pathways. Here I am mentioning the simplest two ways, one of which is by using Grignard reagent and other by using thioacetal intermediates.
You can react aldehydes with Grignard reagents ($\ce{R^2 -MgBr}$) and perform acidic workup to generate secondary alcohols. Then you can oxidise the alcohol to get a ketone by commonly used oxidising agents like PCC (pyridinium chlorochromate).
The other way is to react the aldehyde with a propane-1,3-dithiol to generate a cyclic thioacetal . Now, interestingly, the proton attached to the carbon now can be removed by strong base like n-Butyllithium, and generate  Carbanion which can perform a $S_N2 $ reaction with an alkyl halide to generate cyclic thioketal. After that the thioketal can be hydrolysed using $\ce{HgCl2/CdCO3/H2O}$  to generate your desired ketone.

Converting ketones to aldehydes
The easiest step for this conversion is to perform a Bayer-Villiger Oxidation reaction on the ketone to get an ester. Then that ester can be reduced by Diisobutylaluminium Hydride (DIBAL-H) and an aqueous workup will generate desired aldehyde (along with an alcohol). (If you wish you can further oxidise that alcohol also in case you want aldehyde of that part).

There are two distinct ways to convert aldehydes into ketones and vice versa
An aldehyde is a carbonyl group on a primary carbon while a ketone is a carbonyl group on a secondary carbon. To interconvert the two, you can either keep the functional group on a given carbon and make/break carbon-carbon bonds. Or you can move the functional group along the existing carbon backbone without breaking carbon-carbon bonds.
Moving the carbonyl functional group
Enzymatic conversion from aldehyde to ketone and back is common in the biochemistry of carbohydrates. In glycolysis, for example, phosphorylated dihydroxyacetone is converted to glyceraldehyde. Sugars have a hydroxyl group alpha to the carbonyl group. In the enzymatic reaction, isomerization proceeds via an enediol intermediate (source):

A similar acid/base mechanism is used in the interconversion of glucose to fructose. This is a commercially relevant reaction for making high fructose corn syrup.
Turn a primary carbon into a secondary carbon and vice versa
See Soumik Das' answer.

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text:
Rate of hydrolysis of ester in presence of an acidic medium
Does the rate of hydrolysis of an ester depend on the concentration of both the ester and the acid, or only the ester?
I was told by my professor that hydrolysis of ester follows a pseudo-first order reaction: as the amount of water present for the reaction is very large, the change in concentration of acid/base will result in very negligible changes in concentration of $\ce{H+}$/$\ce{OH-}$ ion. So, the rate is only proportional to the concentration of ester taken.
Can someone explain if the above is in presence of basic medium or acidic medium?

If the reaction is acid catalyzed, the rate will depend on the hydrogen ion concentration, probably first order in that concentration. However, the overall reaction will be pseudo first order because neither hydrogen ions (catalyst) nor water (solvent) change in concentration over the course of the reaction.

We need to consider 4 quantities for the equilibrium reaction:
$$\ce{R1COOR2 + H2O <=> R1COOH + R2OH}$$

The equilibrium constant
The reaction quotient
The forward reaction rate
The reverse reaction rate

The forward reaction above is of the pseudo first order, as the professor says.
But its rate constant depends on acidic  catalytic factors and the net reaction rate is affected by the reverse reaction rate
The concentration of products- the organic acid and the alcohol - affects the net reaction rate by increased rate of the reverse reaction
and by shifting of the reaction quotient.
The content of a mineral acid  catalyzes the reaction, what means speeding up both reactions, therefore keeping the same equilibrium constant.(required as consequence of the energy conservation law).
The reaction is still the pseudo first order, but the rate constant depends on concentration of the catalysing acid.
The alternative approach is considering the mineral acid as self-regenerating reagent. But it must be then evaluated as reaction chain.
The content of a mineral base decreases reverse reaction rate and shifts the reaction quotient by reacting with the organic acid. It also directly reacts with the ester.
$$\ce{R1COOR2 + OH- -> R1COO- + R2OH}$$
This reaction is of the 2nd order.

so, the reaction will be 2nd order in both cases?

See the update.


text:
Does reduction/ oxidation take place separately at a half cell electrode?
In the redox reaction chapter, we learn that oxidation and reduction go hand in hand.  The oxidising  agent oxidises by accepting the electron released by the reducing agent and gets reduced in the process.
However, in the half cell electrode (say, $\ce{Zn}$/$\ce{Zn^{2+}}$) only oxidation is  taking place and no simultaneous reduction is happening in the scene. Also, in a $\ce{Cu}$/$\ce{Cu^{2+}}$ half cell electrode, only reduction is taking place. Until  the half cells are joined by a  saltbridge, oxidation and reduction go on violating the basic  principle of  redox reactions.

Half reactions cannot happen independently and (as you have already stated in the first line) a reduction is simultaneous with an oxidation. Half cells are just representations. Nothing violated. Reaction occurs only when 2 half reactions are present, and the sum of their potentials is positive.

Is that true?  In liquid ammonia solutions you can get electrons pulled out at the anode but they just pour into the liquid through the cathode.

I think the liquid ammonia situation is a plain reaction, not a combination of two half cells.

Redox reactions consist of oxidation and reduction processes occurring simultaneously. What does this actually mean?
Redox reaction involves the transfer of electrons between species (Easy way to remember: increase in oxidation number = oxidation, hence decrease in oxidation number = reduction). These processes occur together at the same time at different interfaces (boundaries).
Considering your cell (Galvanic Cell): The main reacting species (excluding the spectator ions) are Copper and Zinc. The overall reaction that is taking place is:
$$ \ce {Zn + Cu^{2+}  \rightarrow  Zn^{2+} + Cu}$$
Applying the previously mentioned trick, you can see that zinc is being oxidised from 0 oxidation number to 2+ (increase in oxidation number = oxidised), whereas copper ion is reduced from 2+ to 0 oxidation number (decrease in oxidation number = reduced). You can think of a half-cell equation as small components that make the overall reaction in redox reaction of Galvanic cell. They provide vital information on what is happening to the species in terms of electrons involved and the standard potential can be used to identify if the species is capable of operating as reducing agent or oxidising agent.
The half-cell equations is usually tabulated against standard hydrogen electrode as being the reference.(Note: half-cell equations are usually written as reductions).
In order to identify which species is oxidising agent, you can think of it this way: the more positive the potential $\Rightarrow$ stronger oxidising agent $\Rightarrow$ oxidising agents by definition are reduced (reduction occurs at the cathode).
The reactions are only taking place when the circuit is closed (cells are connected). This leads to the flow of the electrons around the circuit and redox reaction can take place.
To qualitatively prove that your reaction occurs, you have to consider the cell potential:
$$E_{cell} = E_{cathode} -E_{anode} $$
applying the concept of electrical work = Gibbs energy, thus
$$G=-nFE_{cell}$$
(Important: provided the concentration 1 M, if they are not you have to apply Nernst Equation to identify new cell potential) .

Could the reaction $\ce {Zn + Cu^{2+}  \rightarrow  Zn^{2+} + Cu}$ happen in a solution with the wire and the connecting electrodes. I mean if we just mixed a salt of $\ce{Cu^2+}$ with solid $\ce{Zn}$ will the reaction still happen?

Yes The purpose of the cell is either to keep the materials separate or to use the energy of the reaction to do work. such as running your cell phone or your Tesla

In a (standard) Zn/Zn++ half cell, no net oxidation (and no net reduction) takes place. A piece of zinc sits in a 1M solution of Zn++ ions (with their accompanying anions). On a micro scale, an equilibrium is occurring (could be occurring!) where Zn + Zn++  --> Zn++ + Zn. No net change occurs; this is a reference point. If it were not stable, some process would degrade the half cell and it would be react itself away. In reality, half cells are only useful to the extent that they don't react themselves away until they are connected to the other half of the overall cell.
Same for Cu/Cu++. However, when the half cells are connected (thru a salt bridge so ions can flow, and a wire so electrons can flow), two processes occur: electrons leave the zinc (the zinc oxidizes) and electrons flow to the copper ions (reducing them). No net "process" or reaction occurs in the half cells until they are connected.
The reality of net oxidation and reduction require both to occur simultaneously - WHEN THEY OCCUR - but we can speak of each process independently of the other. When electrons leave an atom (oxidation), they have to go somewhere (reduction); if electrons are not flowing thru a wire (no actual reaction) there is no net oxidation or reduction. The half cell potentials then define places from which electrons will flow, or locations to which they will flow.

Nothing is simultaneous; oxidation and reduction can be and are separated in time and distance There probably are electrons from the big bang that have yet to find a proton

If both oxidation and reduction are not essentially complete (~99%), we don't have equilibrium, we have a reaction in progress. Simultaneity here means that when we look at the system, it has stopped changing, so the total oxidation and the total reduction have both been accomplished to give an equilibrium by that time (however long it is). Many electrons cannot be lost (without finding a home) because the potentials would become incredibly high.

One of the requirements for equilibrium is propinquity. At equilibrium forward and reverse reactions are equal not 99%.  To be reasonably confident it must be perturbed from both directions. If separated in distance it becomes hard to tell.

I edited the answer to emphasize net reaction.
A metal dipped into a standard electrolyte reaches its standard potential by donating electrons to necessarily propinquitous (nearby) acceptors until the metal acquires its standard negative potential and then the net reaction stops. No net measurable oxidation or reduction occurs in an unconnected half cell.
If further net oxidation/reduction occurs, the metal is corroding; an equilibrium potential cannot then be measured. An example would be: Na in an aqueous solution of NaNO3.

Let's go back to the original text,  that I copy here in italics :
Also, in a Cu/Cu$^{2+}$ half cell electrode, only reduction is taking place.
No ! Sorry to say no ! The $\ce{Cu/Cu^{2+}}$ half cell can work in the oxidation mode, if it is coupled to a more positive half cell, like the $\ce{Ag/Ag+}$ half cell. In the whole cell, both half reactions are : $$\ce{Anode : Cu -> Cu^{2+}+ 2 e- ; redox potential +0.34 V}$$ $$\ce{Cathode : Ag+ + e- -> Ag ; redox potential + 0.80 V}$$ so that the whole cell delivers : $0.80$ V - $0.34$ V = $0.46$ V


text:
Are these calculations valid enough?
I am not able to understand how these calculations are being done in this book. Up until this section, the book is going smoothly. However, how are these calculations accountable? If anyone can explain this to me then it will be valid enough for me.
.

What's the book? What part exactly do you think is flawed and why?

BOOK NAME : INTRODUCTION TO THE THERMODYNAMICS OF MATERIALS
How are the moles calculated and what is the exact basis of the addition of 1 mole for calculations?

You should include this into your question and add proper bibliographical reference including not just the name of the book, but the authors/editors, edition, publisher, year and ISBN. Besides, it's not moles that are calculated, it's amount of substance.

I am learning sir, from the next time it will be even more specific for all of us.

I'm not criticizing your question, I'm just making it easier for others to quicker understand what it's about; and there is no need to call me sir, that's internet after all:)

Equations (iii) through (v) are based on the relation
$$
\Delta G = \Delta H - T \Delta S
$$
which holds as long as $T$ is constant (which it is in the problem). The values of $\Delta H$ and $\Delta S$ have been looked up and substituted, likely in tables provided in the book.
The next four (unnumbered) equations are based on the relationship
$$
K_p = \exp \left( \frac{-\Delta G}{R T} \right)
$$
where $K_p$ is the equilibrium constant and $\Delta G$ is a function of $T$ evaluated using the previous relationships.
The fractions discussed in the text relate to the stoichiometry. For example, consider the first ratio:

$K_p = 19.1$, so at equilibrium there are 19.1 H$_2$O molecules for every H$_2$ molecule
Assuming that the initial state before the reaction had no H$_2$O, all of the H atoms which end up in H$_2$O molecules must have originally been in the form of H$_2$. The balanced chemical equation shows that producing a H$_2$O molecule requires exactly one H$_2$ molecule, so for every H$_2$ molecule in the final state there were 20.1 in the initial state
Since 19.1/20.1 of the H$_2$ molecules react an 1/20.1 do not, and since one molecule of CoO is consumed per molecule of H$_2$ consumed, it follows that each mole of H$_2$ provided can react with 19.1/20.1 = 0.95 moles of CoO.

Thanks for your valuable time and attention to this question. @user1476176


text:
Calcium Nitrate - Why only 18.8% Calcium for fertiliser?
I brought a kilo of Calcium Nitrate for my hydroponic setup.
The chemical formula for Calium Nitrate is Ca(NO3)2. I did a quick calculation of its weight:
Ca = 1*40.078
N  = 2*14.006 = 28.012
O  = 6*15.999 = 95.999

Total = 164.085
Hence the % in weight of calcium is ~24.42%.
BUT I see on the fertilizer pack that Calcium is only 18.8% and Nitrogen is 15.5%
Eg: https://www.fertinagroindia.com/calcium-nitrate.html
Why is this?

Your math seems fine. The manufacturer, on the other hand, doesn't know how to market the stuff properly. The URL suggests "The content of Ca in our fertilizer is 24% that exceed our foreign analogues on 5%, their calcium nitrate contains only 19% of Ca.", whereas label on the package says "Ca–18.80%". If it's really  98% Ca(No3)2 as they claim, then it's about 24% Ca, as you suggested.

Also note that the element content in fertilizers is sometimes given for the corresponding oxide – this is a common practice for potassium (K2O).

I understand your point. But even other suppliers,  I think this is generally the case. ex: https://www.amazon.com/Calcium-Nitrate-15-5-0-0-Fertilizer-POUNDS/dp/B071YS779G

That's due to impurities. They list a bunch of extra stuff even on the label (chlorine, ammonia salts etc.) and it's easier to claim slightly lower content of the main product to avoid a lawsuit if the product doesn't meet the standards (for instance, gets moisturized due to improper storing conditions). If you really want to know the exact content, don't rely on what manufacturer tells you and do quantitative analysis for calcium yourself.

Also note the the calcium nitrate is hygroscopic and is usually provided as tetrahydrate. https://en.wikipedia.org/wiki/Calcium_nitrate

Oh Yes, I see now. @Poutnik. Wiki says: The fertilizer grade (15.5-0-0 + 19% Ca) is popular in the greenhouse and hydroponics trades; it contains ammonium nitrate and water, as the "double salt" {\displaystyle {\ce {5Ca(NO3)2.NH4NO3.10H2O}}} {\displaystyle {\ce {5Ca(NO3)2.NH4NO3.10H2O}}}.

@Nujufas Well, I guess now you can answer your own question:)

@Nujufas You are welcome. I have noticed the double salt as well, but was not sure if it was your case.


text:
2-step reaction mechanism for the decomposition of NO2Br
I need to propose a 2-step reaction mechanism for the following reaction:
$$\ce{2NO2Br -> 2NO2 +Br}$$
And prove the mechanism is consistent with: $$v=k[\ce{NO2Br}]^2$$
For the mechanism to be consistent with the rate equation, the first equation would have to be the limiting one and also be in this form:
$$\ce{2NO_2Br -> Products}$$
I've tried using:
$$\ce{2NO2Br -> N2O4 +Br2}$$
$$\ce{N2O4 -> 2NO2}$$
But  the problem with this mechanism is that $\ce{N2O4}$ is not unstable enough to be consumed immediately.
Any help with this problem?
Would also appreciate some advice on proposing reaction mechanisms. Thank you.

Note that there is ongoing equilibrium $\ce{N2O4 <=> 2 NO2}$. Also, there is no rule the 2nd reaction must be fast.

Can a mechanism be found so that the reaction rate depends solely on $\ce{[NO2Br]}$?

The reaction kinetics of multiple reagents is usually based on the slowest reaction of the reaction chain.But as there is no other reagent, it is quite possible. But I am not currently aware of the actual reaction mechanism.

It is more probable  like
$$\begin{align}
\ce{NO2Br &-> NO2 + Br} \\
\ce{NO2Br + Br &-> NO2 + Br2} \\
\ce{2 Br &->  Br2} \\
\end{align}$$
The last reaction is a minor one in case concentration of $\ce{Br}$ is low.
The reaction rate order can be concentration dependent and need not be the integer.
In fact, it is rather mathematical parameter, related to solution of differential equations for a complex reaction system.
If the 2nd reaction is fast enough, the overall reaction rate is given by the slow rate of generation  of $\ce{Br}$, which fast reacts to form $\ce{Br2}$
If the 2nd reaction is slow enough,
it's rate $$k_{\rm 2}\cdot [\ce{NO2Br}][\ce{Br}]$$ can be written as $$k_{\rm 2a}\cdot [\ce{NO2Br}]^2$$
The exact solution is to solve system of differential equations for the rates of the concentration changes.

$$\frac{\mathrm{d}[\ce{Br}]}{\mathrm{d}t}=k_1.[\ce{NO2Br}] - k_2.[\ce{NO2Br}][\ce{Br}] - k_3 [\ce{Br}]^2$$
For the dynamic equilibrium of the steady concentration of $\ce{Br}$:
$$\begin{align}
0&=-k_1.[\ce{NO2Br}] + k_2.[\ce{NO2Br}][\ce{Br}] + k_3 [\ce{Br}]^2 \\
[\ce{Br}]&=[ -k_2.[\ce{NO2Br}]+\sqrt((k_2.[\ce{NO2Br}])^2+4.k_3.k_1.[\ce{NO2Br}])]/(2.k_3) \\
\frac{\mathrm{d}[\ce{NO2Br}]}{\mathrm{d}t}&=-k_1.[\ce{NO2Br}] - k_2.[\ce{NO2Br}][\ce{Br}]\\
\end{align}$$

How is the order of reaction related to which reaction is the fast one?

@jcgereda See the update and the link.

Why does the constant change when [Br] is substituted to [NO2Br]? Also: How can I relate mathematically [Br] consumed by the second reaction with [NO2Br] consumed by the first one?

1/ Because concentrations of NO2Br and Be are not the same. 2/ see the update provided in a while..


text:
Can a double bond extend beyond the plane and if so, how would it be drawn?
Can double and triple bonds, etc., extend beyond the plane in a Lewis structure, and if so, how are they represented? I.e., how would 1 and 2 in this image be drawn if they were doubles or triples?

It is unlikely that you need this level of accuracy. Note that even the notation you have posted is only used when the stereochemistry aspect is of importance. Structure are otherwise to be seen as a projection* on the paper plane. In other words, suppose to represent CH2=CXY. You can always choose = to be in plane. Longer and complicated molecules would be projections. *deformed, as bond of the same type are draw with the same length, no the length of their projection. To resume: methane is just cross like, most of the times.

I've rarely seen one go into the plane, but on occasion, you see one coming out the plane where one of the two lines in the double is a wedge.

It is usually possible to define the "direction" of the multiple bond relative to the plane of the drawing implicitly by representing the correct orientation of the single bonds in the rest of the structure. Complex structures like strychnine, though, are better understood by building proper 3D models even if it is possible to draw the 2D structure on paper.

Yes, chemical bonds of higher order can be extended from the plane in some rare cases.
There are IUPAC recommendations existing at least for double bonds suggesting to depict them simply as doubled wedge bonds if it's really necessary.
From [1, p. 1930] (emphasis mine):

ST-1.5 Tetrahedral stereocenters including higher-order bonds
Tetrahedral configuration requires a tetrahedral arrangement of ligands around a central point. It does not require that all of those ligands be connected to a central atom using single bonds. It is perfectly reasonable to have one or more ligands connected to the central atom using higher-order bonds. This is most commonly seen with chiral sulfoximides, phosphates, and related compounds, each of which is
  commonly drawn with one or more double bonds. Stereocenters of this type should be indicated by the appropriate placement of bold and/or hashed wedged bonds connecting the central atom to the singly bonded ligands. Although wedged and hashed double bonds have appeared in the literature, they are rarely required for chemical clarity, and their use is infrequent enough to be confusing to the casual
  viewer (see ST-2.5).

Further from [1, pp. 1940-1942] (emphasis mine):

ST-2.5 Trigonal pyramidal Polyhedral symbol: TPY-3 Number of ligands: 3
[…]
In rare cases, a trigonal pyramidal atom with at least one double bond may be flanked by other stereocenters at the other ends of all of its single bonds. In such cases, one of the plain bonds should be changed to a wedged bond even though it joins two stereocenters. It is also acceptable to use a solid wedged or hashed wedged double bond, but that would not be preferred because wedged double bonds are so rarely seen in the literature.

References

Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. https://doi.org/10.1351/pac200678101897. (Free Access)


text:
Change in electronegativity order down a group from groups 13-16 to group 17

In general, going down a group Zeff initially increases but then becomes approximately constant, while electrons are in higher n orbitals, hence valence electrons on average further from nucleus, thereby size of atom increases, and electronegativity decreases. An exception is present in groups 13-16, going from row 2 to 3, where atoms of elements in row 3 have greater electronegativity than atoms of elements in row 2. This is because the row 3 element follows the d-block, where 10 extra electrons have been added in the next n down, so don't fully shield the valence electrons from the nuclear charge. That's a lot of electrons not doing much shielding, yet the nuclear charge has increased by +10 with 10 extra protons. Accordingly Zeff increases from row 2 to 3 more than expected, so electronegativity increases.
Why then, by the time you get to group 17, does this trend no longer hold i.e. electronegativity decreases from Cl to Br (see graph above)?

I "smell a rat" here. What's the source of your data and what electronegativity scale?

The source is my lecturer. I don't know which electronegativity scale he used to create the graph.

Then you should check valid data. Different scales use different properties for calculations. For example as far as Pauling (most common) scale goes S is slightly more electronegative then Se. Another point is for F it should be precisely 4.

This can be simply explained by the electronic configuration.
The electronic configuration of bromine is
$ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5 $
Thus we can see it has 5 electrons in the 4p subshell which provides shielding effect so it's electronegativity is less than that of chlorine . It has also decreased due to the increase in size from chlorine to bromine.

The Effective Nuclear Charge of an atom (Zeff) is the net charge an electron experiences in an atom with multiple electrons.5 It increases down a period (as an extra proton is being added for each incremental move down a period) and generally stays constant down a group (as the previous shell is now shielding some charge as well). I think you meant to say down a group rather than down a period. However, this change in Effective Nuclear Charge down a group is relatively insignificant.(1)
As one moves down a group, higher energy levels are being occupied by the valence electrons, the number of protons increases and there are more valence electrons as well. Consequently, not only is the atom more dense, but the atomic radii increases too, due to the increase in occupied shells and the increase in electrostatic repulsion by the increased number of valence electrons.(1)
Electronegativity is a measure of an element's ability to attract electrons in a covalent bond. According to periodic trends, it increase up a group and across a period (from left to right).(1) So in theory, the electronegativity values should decrease from Si to Ge (although Ge is not shown in your graph) or from Al to Ga, but this is not the case. Why is this not the case? Well, as you said, it has something to do with the d-block. To be more concise, it is due d-block contraction, which, in our case, refers to the "unexpectedly small increase (or even decrease) in atomic/ionic radius when moving down a group". (2)(3) For example Al and Ga, who have atomic radii of 184 pm and 187 pm respectively.(4) One would expect this increase in radius to be much larger as Gallium occupies an additional energy level. However, Ga's electron configuration is [Ar] 3d10 4s2 4p1. As one can see, it has the 3d orbitals occupied with electrons– these electrons are ineffective at shielding . Hence, the valence electrons of Ga experience a stronger attractive force from their nucleus. Consequently, Ga has a smaller atomic radius than expected and as this is the case, its electronegativity increases as well as a result of the nucleus having a stronger attraction for electrons than if it would have a larger atomic radius.
All in all, these inconsistencies in periodic trends for electronegativity can be attributed to the d-orbitals' electrons being ineffective at shielding, resulting in a stronger attractive force from a nucleus than expected. Thus, I think your question could be reformulated to why is there this disruption in electronegativity trends instead of why group 17 does not follow these trends– although in the end, these questions will result in the same answer.
Note: I recommend going through the references I have listed below to gain a better idea of what is being discussed.
References:

Pearson Higher Level Chemistry Textbook, 2nd Edition. By Catrin Brown and Mike Ford. Chapter 3: Periodicity
https://www.quora.com/What-is-a-detailed-explanation-of-d-block-contraction
https://en.wikipedia.org/wiki/D-block_contraction#/media/File:D-block_contraction--EN.png
https://en.wikipedia.org/wiki/Atomic_radii_of_the_elements_(data_page)
https://www.thoughtco.com/definition-of-effective-nuclear-charge-605056

Other helpful references:
https://www.quora.com/What-causes-Ga-and-Ge-to-have-higher-electronegativity-than-the-elements-over-them

Your definition of effective nuclear charge is incorrect. It is in fact the net positive charge experienced by any electron in an atom.

Good point. Changed.


text:
Can amorphous polymers have a melting temperature as well as their glass transition transition temperature?
Above the Tg, amorphous polymers are described as viscous liquids. What then would they be likened to above their melting point, should they have one?


text:
Does the main washing effect of soap come from foam?
I am curious if the main cleaning effect of soap comes from its foam (or, maybe, it is just a by-product).
Here are some  points, that might clarify what I mean:

Does foam sort of "carry" dirt and germs away from the surface?
If it is not the "main motor", which makes the soap function: is it possible to make a soap which will still be as effective without it producing any foam?
What percent of the cleaning work, if any, does foam undertakes (i.e. how effective would soap be against germs and dirt without its foaming ability)?

Foam is a side effect of using a tensio-active agents (though some tensio-active agents are engineered to produce as little foam as possible, e.g. in washing machines). In fact, you could consider that, if soap has nothing better to do, it will form foam. This answers some of your questions: foam does not (really) carry away dirt and its percentage of efficiency in the whole process is low.
Molecules of soap are made of 2 parts: a hydrophobic tail (which loves fat, grease, lipids, whatever you call them) and a hydrophilic head which loves water, and hate lipids.
When the hydrophobic tails are busy linking to a greasy stain or grease on your hands, they no longer participate into forming bubbles/foam.
The easiest way to verify that is to pour a little bit of oil in your hands and rub them. Wash your hands with soap a first time: there will be almost no foam. Now rinse your hands and wash them again: there should be much more foam!
Note that it does not mean that your hands are cleaner, regarding bacteria, but just that they are less greasy!
If you want to remove bacteria, rub your hands with soap for at least 30 seconds. So-called "anti-bacterial" soaps do not achieve better and promote bacterial resistance.

Anti-bacterial soaps are better at killing bacteria. It's just that normal soap is already amazing at killing bacteria, and humans are resistant enough that the difference is essentially non-existent for a normal healthy human. It's certainly very noticeable where people aren't healthy.

@Luaan Normal soap doesn't "kill" bacteria at all. Rather, the surfactants in soap allow the bacteria to be washed away. This is why hand-washing technique is also so important, and why water alone (no soap) is still surprisingly effective when no other options are available. If you want a biocidal effect, you need an anti-bacterial soap. But, as pointed out here, the efficacy levels are not significantly higher (although they do increase the longer you rub the soap around, unlike the surfactant effect, which peaks after a certain minimal amount of time), and they promote resistance.

The foaming action of soap was put there as a sort of "indicator" that something is happening. It's a marketing strategy because people are ignorant about how soap works. The bubbles are just air pockets and have no effect on cleanliness.

This is not true for "normal" soap, but is true for some "detergents".

Actually, it is. You can make some soap yourself and it'll only foam slightly, if you don't add foaming agent like producers do.

Made soap yesterday and have been making soap for personal use for several years. I use the standard oil/lard process. This soap foams quite well, in my experience. The bubbles seem finer, and the froth seems "creamier".


text:
A sample must not contain more than 2% contaminants to be pure. Are there instances where 2.12% is ok?
I do not have any background in chemistry, I am asking this on behalf of someone else, twice translated, so I really hope it makes sense.
"A maximum of 2%" seems quite a clear specification, but since  practice involves rounding of measurements and significant decimals, that might not be the case.
It may seem silly but which of the following options are valid work protocol?

at most 2% means less than 2.5% after rounding
at most 2.0% means less than 2.05% and option 1 still holds
at most 2% is the same as at most 2.0% and means at most 2.000...%

Off topic: An optimal answer would be in the context of the EuAB (Europäisches Arzneibuch).

One may also consider the confidential interval of the contaminant content should lay below 2%.

It depends mainly on the accuracy of your measuring instrument in practice and the accuracy required by your work.
For example:
If you want to make a solution with 2g solute in it, and to measure amount of solute you have an measuring instrument that have least count of 0.1g, then 2.05g is acceptable as the instrument will show 2.0g only but if the least count of intrument is 0.01g then 2.05g is not acceptable if you want 2g.
In real life the least count varies for different situations.
For example if you are manager of a water bottle manufacturing unit you won't need much accuracy to fortify the water with minerals, but if you are carrying out a titration in lab to find strength of some oxidising agent you need very high accuracy that is why burette are made such accurate.

Nice answer. It's only bad if you know you're lying. :-)

It depends....
The statement "A sample must not contain more than 2% contaminants to be pure" is ambiguous as you have pointed out. To understand exactly what that means requires further clarification.
First if expressed as 2% contaminants I'd expect that anything less than 2.5% would be acceptable. The point being that 2.0% contaminants is really a more stringent quality goal.
The statement "at most 2% contaminants" brings in a different sort of flavor to the problem, and gets closer to reality. As a supplier I don't to buy or manufacture 99.9% pure chemicals and then sell them to you as a lower grade 98% pure. As a buyer I don't want to pay a premium for 99.9% purity if I could buy 98% purity for less and that purity would be acceptable to my processing. This is the classic buyer/seller problem related to "Acceptance sampling."
The gist of Acceptance sampling is that generally the whole batch can't be analyzed. In other words samples are taken from a batch of some chemical and just the samples are analyzed, not the whole batch per se. (After all you want to have something left to sell or use.) Now there are obviously errors due to sampling and errors due to analysis. The gist is that I find that the sample has 1.92% contaminants. However let's say that the standard deviation is 0.33%. So it is probable that the batch does have less than 2% contaminants, but there is a some chance that the batch actually exceeds 2% contaminants. So the confidence interval of any particular analysis should also be specified for a more complete specification.
In fact if I were buying a lot of a chemical I'd probably have a contract that detailed how my acceptance testing would be done and under what conditions I could reject the whole lot at no cost to myself. So to write a less ambiguous specification requires some careful consideration.


text:
Determining amounts of product and excess reagent remaining post reaction

$$\ce{2FeS2 + 11/2 O2 -> Fe2O3 + 4 SO2}$$
Mass of $\ce{FeS2}$ is $\pu{600 g}$ and mass of $\ce{O2}$ is $\pu{800 g}$. Find the amounts of $\ce{Fe2O3}$ and $\ce{SO2}$ and the remaining amount of the excess reagent.

My Attempt
$$\frac{n_\ce{FeS2}}{2} = \frac{600/120}{2} = \frac{5}{2} \lt \frac{n_\ce{O2}}{11/2} = \frac{800/32}{11/2} = \frac{50}{11}\implies \ce{FeS2}~\text{is limiting reagent}$$
By Mole-Mole Analysis, we have
$$\frac{n_\ce{FeS2}}{2} = \frac{n_\ce{Fe2O3}}{1} = \frac{n_\ce{SO2}}{4}\implies \begin{bmatrix}n_\ce{Fe2O3} \\ n_\ce{SO2}\end{bmatrix} = \begin{bmatrix}5/2 ~\text{mol} \\ 10~\text{mol}\end{bmatrix}$$
$$\text{Excess reagent remaining} = \frac{800}{32} - \dfrac{5}{2}\cdot\dfrac{11}{2} = \pu{11.25 mol}$$
However, my answer does not match with my other classmates. What am I doing wrong?

At first glance your answers look reasonable to me, and good job formatting the question. I edited the MathJax a little, if you want to know more about advanced formatting on Chemistry.SE, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Were your classmates converting the answers to grams instead of moles? // IUPAC went to this crazy idea that you don't calculate the number of moles, but rather the "amount of a substance." I'm sure that a lot of "old" books that don't follow IUPAC's syntax used the word "amount" in a number of problems.

@MaxW As an adept of the "crazy idea" you mentioned, I edited the "amount" term in as a more appropriate one. Guess what, originally there were good old "numbers of moles" (poor animals).

@andselisk - If you're going to edit a problem like that I'd suggest that you note the change and the reason for it at the bottom of the problem. A lot of the questions seem to be posed by students from India who are using books with old condiations for STP (ie 22.4 liters/mole @STP) and "number of moles" instead of "amount."

@MaxW Probably I'd just drop a link to Meta: Buzzwords Season 6 - Questions Posts using Phrase Number of Moles. But yeah, that's a good point, I have very little clue about chemical education in India.

@andselisk, "But yeah, that's a good point, I have very little clue about chemical education in India" ...There is huge variation from place to place from very obsolete to quite modern. It is catching up I guess. Depends on the university. Some IITs are as good as Ivy League schools. Recently, I heard an Indian PhD student say "nascent hydrogen" does the reduction. This was a concept taught in 1940s. Similarly normality is still taught, as well as Sommerfeld model, parachor values (if you ever heard of it!) etc. Hard to generalize.

Thanks for taking part in the discussion and resolving my issue. @andselisk Cheers :)

Note the following relations from the equation you wrote:
1 mol $\ce{FeS2}$ = (11/4) mol $\ce{O2}$;
1 mol $\ce{FeS2}$ = (1/2) mol $\ce{Fe2O3}$;
1 mol $\ce{FeS2}$ = 2 mol $\ce{SO2}$.
You already determined that $\ce{FeS2}$ is the limiting reactant because we have 600/120 mol of iron sulfide and 800/32 mol of $\ce{O2}$;
This implies that 5 mol $\ce{FeS2}$ will require =5x (11/4) mol $\ce{O2}$.
Remaining mol oxygen = 25-13.75=11.25
Now use the relations established above: 5 mol iron sulfide would produce (5/2) or 2.5 mol iron oxide, and 5 mol iron sulfide would produce 5x2 = 10 mol sulfur dioxide.

But this is pretty much what OP did, isn't it?

True but just for confirmation only-parsed it for his classmates who are doing it wrong :-) The boy did a very good job in solving this relatively difficult problem for a 10th grader.

OP did very well indeed, how about showing him support with upvotes?;)

Slight off topic Q: How does this reputation matter in StackExchange? I noted some people across the board (not just chem) have very high scores. Is there any commercial aspect behind that? The scoring system became an issue in ResearchGate (I am sure you know it). They finally removed upvote or downvote option. After that some academics made an issue that some universities in developing countries use RG score for faculty promotions. In short how how does this reputation score help or is it just for self-satisfaction? In RG nobody can be anonymous. They need official academic address.

UPVOTING - @andselisk I consider that up voting the question indicates that it is an interesting question that I'd like to see answered too, rather than giving the OP a pat on the back. In other words for these homework problems there isn't a way to vote for the poster's answer.

There is no commercial aspect AFAIK, at least I got precisely $0 for all the stuff I did on Chemistry.SE, so it's more like an IRL RPG game to me. Votes help to quickly determine how much effort was put into Q&A and how valuable they are overall. As for RG, I tried to participate there too, but I was surprised how poor the quality of discussions and Q&As overall was there despite all the affiliations and honors and "official" profiles, so I left it for good and never looked back.

RG has gone down for sure. It has become more popular among graduate students. Nobody asks homework Qs like here (some questions are really good too). RG has boring simple research questions which Google can answer. I joined StackExchange because of Sci History & Math. I found answers which I was looking for ten years on group theory symbolism.

Thanks for confirming. Although I am yet to edit my bio. I'm in $11^{\text{th}}$ grade now. Cheers :)

@Paras, still pretty good if this is grade 11 stuff. This stuff is taught in the beginning of first year courses of North American universities.

Given the problem as stated, you did the problem correctly. However I would have done it a bit differently to make it easier to check. I like to do the problems in steps. I also dislike carrying a lot of fractions in intermediate calculations since I get confused easily. (This is basically M. Farooq's answer with some explanation.)
Given

$$\ce{2FeS2 + 11/2 O2 -> Fe2O3 + 4 SO2}$$
Mass of $\ce{FeS2}$ is $\pu{600 g}$ and mass of $\ce{O2}$ is $\pu{800 g}$. Find the amounts of $\ce{Fe2O3}$ and $\ce{SO2}$ and the remaining amount of the excess reagent.

Molar Quantities
For $\ce{Fe}$:
$n_\ce{FeS2} = \frac{600}{120.} = 5.0000 $
For $\ce{O2}$
$n_\ce{O2} = \frac{800}{32.0} = 25.000$
Here I calculate the decimal amounts rather than carrying fractions. Since it would seem that the problem has 3(?) significant figures, I carried two extra for the intermediate calculation.
Stoichiometric Quantities
For $\ce{Fe}$:
$\frac{n_\ce{FeS2}}{2} = \frac{5.0000}{2} = 2.5000 $
For $\ce{O2}$
$\frac{n_\ce{O2}}{11/2} = \frac{25.000}{11/2} = \frac{50.000}{11} =4.4545$
Thus $\ce{FeS2}$ is limiting reagent
Here I could have reached the conclusion in my head, but I do the math anyway so that I don't make a stupid mistake and also so that I can check the problem easily.
Rewrite the chemical equation
Since we now know that $\ce{FeS2}$ is the limiting reagent, rewrite the chemical equation in terms of $\ce{FeS2}$.
$\ce{FeS2 + 11/4 O2 -> 1/2Fe2O3 + 2 SO2}$
Now for each mole of $\ce{FeS2}$:

You use 11/4 mole $\ce{O2}$ so $\ce{O2}$ remaining is:

$25.0000 - (5.0000*11/4) = 25.0000 - 13.75 = 11.25$

You get 1/2 mole $\ce{Fe2O3}$ so:

$5.0000 / 2 = 2.50$

You get 2 mole $\ce{SO2}$ so:

$5.0000 * 2 = 10.0$
Now for a final check I note that at the start there were 25.0 moles of $\ce{O2}$, but I have only 11.25 left. That is good since some was consumed. If I had ended up with more oxygen I obviously would have a mistake...

LOL - When I started doing these problems in high school I had a slide rule which was a step up from using 3 place log tables. I didn't a calculator until I was in grad school.

I have not used slide rule but my boss showed slide rule with pride. However I have used log tables in late 1990s :-). The trouble of finding characteristic and mantissa from printed tables with fine print ...good old memories. How many students even know what a mantissa is.

@M.Farooq - Gave my Post Versalog (which was a professional tool, not a toy) slide rule to my sister who wanted it to play with. I have thought since that it would be nice to have it in a shadow box on the wall. The glass on the shadow box would read: In case of power failure, break glass.

Good one. As I told someone yesterday who was asking here how I survived without learning LaTex for writing papers. My answer was that very good science was also done when we had typewriters. Slide rule produced prodigies and geniuses, this was the time when Americans and Russians were also going to the moon. They had computers though, I believe.


text:
What happens when para-water ice is suddenly melted?
Background (hydrogen)
In the case of recently liquified hydrogen (which is quite cold of course) it must be re-equilibrated before loading on to a rocket as fuel to avoid a sudden exothermic equilibration of the ratio of the ortho- and para- forms. This is because the nuclear spin degree of freedom (singlet vs triplet) initially remains hot even when the other degrees of freedom of the molecule are cold.
The question What is the ortho/para issue with LH2 as a fuel? quotes Hydrogen Fundamentals on a hydrogen safety website:

Liquid hydrogen (LH2) has the advantage of extreme cleanliness and the more economic type of storage, however, on the expense of a significant energy consumption of about one third of its heat of combustion. Another drawback is the unavoidable loss by boil off which is typical to maintain the cold temperature in the tank. The evaporation rate is even enhanced when ortho hydrogen is stored. The heat liberated during the ortho-para conversion at 20 K is huge with 670 kJ/kg compared to a figure of 446 kJ/kg for the latent heat of vaporization at the same temperature. This represents a safety issue requiring a design of the hydrogen loop which is able to remove the heat of conversion in a safe manner.

Water
If instead I started with extremely cold water ice (we've now switched from $\ce{H2}$  to $\ce{H2O}$) say at liquid helium temperature, and add a enough heat to quickly raise it to 0°C and just melt it to liquid, it would be mostly para- water. This will be out of equilibrium, and the nuclear spins will still want to warm up as well.
What would happen next? Would the water quickly refreeze in a fraction of a second? Or would it take hours or days to quietly re-equilibrate?

Tikhonov and Volkov https://science.sciencemag.org/content/296/5577/2363 say it takes quite some time for the spin states to re-equillibrate, months in solid, minutes in liquid state.

@Karl there may be enough for an answer there. It seems the timescale is quite long, so long that there's probably not much chance of a re-freeze without extreme thermal insulation.

Good question!  I found this link interesting.  http://www1.lsbu.ac.uk/water/ortho_para_water.html  One thing I don't understand is how, if in liquid water there is extremely fast hydrogen atom exchange between water molecules, the equilibration between ortho and para can take so long.  Acid-base chemistry happens in water at (very fast) diffusion-limited rates because H exchange is so facile.  So why doesn't para / ortho equilibration take the same amount of time?

The Volkov and Tikhonov paper (sort of) addresses my question. They say: *[the rates in liquid water] are 10$^6$ greater than what could be expected from the rate of proton exchange between $\ce{H2O}$ molecules in liquid water (5). We conclude that fast proton exchange obviously does not lead to the fast OP conversion. The exchange without OP transitions, i.e., without change of energy (of resonance character), distinctly dominates.

Is this a good article to address your question?

@uhoh I'm not sure but, a bunch of sources online state that the OPR (ortho-para ratio) is usually 3:1 at room temp. and above, and 1:1 at the temp. of liquification of air (-196 degrees C), so at 0 degrees you should about 3:1 - so more ortho instead of para right?

Note that this answer is not complete. Note that wherever a "*" is placed in the answer, it means it requires figures or explanations to support it. If anyone found them or anything related, please share them! :)
Now unfortunately I do not have a complete and reliable answer to this, so I strongly welcome any comments objecting or adding on to my answer.

TLDR;
It should take a few hours to re-equilibrate*.
Longer answer

... and just melt it to liquid, it would be mostly para-water

To those who don't understand why this is so I'll add a short explanation. At lower temperatures, especially at around 0K-50K, para-water is more stable than ortho-water as shown by Statistical ortho-to-para ratio of water desorbed from ice at 10 kelvin:

As shown with Fig. 1, curve A, $T_{spin}$ may be used as a probe for low-temperature regions because para-$\ce{H_2O}$ ($J_{K_a,K_c=0_{00}}$) is more stable than ortho-H2O ($J_{K_a,K_c=1_{01}}$) in the gas phase owing to the $\ce{23.8 cm^{−1}}$ (34.2 K) rotational energy difference between them.

Now back to the main question. We know that it takes energy (likely a lot of it*) for the ortho-para transition to occur based on A Mechanism for the para-ortho Conversion of Hydrogen by Diamagnetic Substances:

According to the Wigner theory (I) non-dissociative conversion of the hydrogen
molecule between pm-u and ortho states can occur only in the presence of an external inhomogeneous magnetic field.

We also know that all that energy cannot be obtained and used in the ortho-para transition this quickly, and many sources suggest it would take hours to re-equilibrate*, two of such sources are here on the Natural Ortho-Para Conversion rate in liquid and gaseous hydrogen and Separation of Water into Its Ortho and Para Isomers.
Note that from Wikipedia - Spin Isomers of Hydrogen:

This is the T = 0 intercept seen in the molar energy of orthohydrogen. Since "normal" room-temperature hydrogen is a 3:1 ortho:para mixture, its molar residual rotational energy at low temperature is (3/4) × 2Rθrot ≈ 1091 J/mol,[citation needed] which is somewhat larger than the enthalpy of vaporization of normal hydrogen, 904 J/mol at the boiling point, Tb ≈ 20.369 K.[10] Notably, the boiling points of parahydrogen and normal (3:1) hydrogen are nearly equal; for parahydrogen ∆Hvap ≈ 898 J/mol at Tb ≈ 20.277 K, and it follows that nearly all the residual rotational energy of orthohydrogen is retained in the liquid state.

One source states from Iopscience - The Ortho-to-para Ratio of Water Molecules Desorbed from Ice Made from Para-water Monomers at 11 K:

There's some very interesting information on the ortho-Water and para-Water page of Martin Chaplin's water website. Since we're making progress and the bounty's grace period is expiring I'll award it now. I think this is a really challenging question and can use all the help it can get. Thanks!

Ya I don't know why but that site is http and not htttps. Luckily, per answer(s) to What happened to Martin Chaplin's famous Water Structure and Science website? we can find a backup copy in the Way Back Machine, yay! https://web.archive.org/web/20220215163214/https://water.lsbu.ac.uk/water/ortho_para_water.html


text:
Why does magnesium deposit during electrolysis of a molten mixture of MgI2 and AlBr3?
During the electrolysis of mixed salts, I read, and was taught, that the cation with a higher ionization energy would deposit at the cathode. I was also told that you cannot use the standard electrode potential to determine which cation would be reduced.
Is this ionization energy referring to the first ionization energy? If it is, why doesn't the second and third ionization energy factor into this problem seeing that magnesium would have a 2+ charge and aluminum would have a 3+ charge?

Yes, it is true that you cannot use the standard electrode potential to determine which cation would be reduced and vise versa during an electrolysis of mixed molten salts, in your case, $\ce{AlBr3}$ and $\ce{MgI2}$. That's because most tabulated values of standard electrode potentials are determined for aqueous solutions. Therefore, they can be used only as approximate guidance. Instead, electronegativity values (EN) can be used to estimate the stronger oxidizing and reducing agents. For example, EN values of $\ce{Br}$ and $\ce{I}$ are 2.8 and 2.5, respectively. Therefore, we can conclude that $\ce{I-}$ is the stronger reducing agent between two elements (because $\ce{I}$ is less EN than $\ce{Br}$). Similarly, between $\ce{Al}$ and $\ce{Mg}$, $\ce{Al^3+}$ is the stronger oxidizing agent because $\ce{Al}$ is more EN than $\ce{Mg}$ (1.5 vs 1,2, respectively), so $\ce{Al^3+}$ gains electrons easier.

Therefore, possible cathode half-reaction (reduction): $\ce{Al^3+(l) + 3e- -> Al(l)}$
And Possible anode half-reactions (oxidation): $\ce{2I-(l) -> I2(l) + 2e-}$

I suppose it does not address the title ( not repeated in the body ), why magnesium and not aluminium.

I suppose a contributing factor can be the tendency of aluminium to form halogen anions, like $\ce{AlF6^3- AlCl4^-}$. I see as an option to exist $\ce{MgX+ + AlX4-}$ or $\ce{Mg^2+ + 2 AlX4-}$

@ Poutnik: Thank you for your input. Yet, I didn't see the existence of your version in molten mixture. So for I found is simple molten $\ce{AlCl3}$ giving only $\ce{Al^3+}$ and $\ce{Cl-}$ ions, which converted to $\ce{Al(l)}$ and $\ce{Cl2(g)}$ only. So, I believe $\ce{Al^3+}$ the one reducing at cathode. Also, I'd like to see where is OP get " magnesium deposit during electrolysis of a molten mixture.." part, which is not clear in the text.

Another surprising point is the fact that $\ce{MgI2}$ does not melt and get decomposed at $700$°C.  On the other hand, $\ce{AlBr3}$ melts at $97$°C and boils at $263$°C. This electrolysis must be carried out between $97°C$ and $263°C$ and it should be admitted that $\ce{MgI2}$ is soluble in liquid $\ce{AlBr3}$


text:
Why are “diene” structures in the C-ring of anthocyanins singled out for their singlet-oxygen quenching ability?
Anthocyanins are a class of water-soluble vacuolar pigments. They are sensitive to pH, and can undergo a series of structural changes according to a pH-dependent equilibrium as illustrated in Figure 1. For this question, the flavylium cation (the form predominant at pH 1-4, which is also generally used to represent anthocyanins) and the quinonoidal anhydrobase (the form predominant at a neutral pH) are of particular interest.

Figure 1: The generalised structure of the anthocyanin flavylium cation, indicating the positions and numbering of the A, B and C ring of anthocyanins; the pH-dependent equilibrium between the different forms of anthocyanins.
I was initially under the impression that the antioxidant strength of the various forms of anthocyanins would mainly be dependent on the extent of conjugation (and thus electron delocalization) across the entire molecule, with more extensive delocalization stabilizing the anthocyanin radical formed upon interaction with singlet oxygen or some other reactive oxygen species. By this logic, the flavylium cation would have the highest antioxidant activity not only because it retains its hydroxyl groups for hydrogen donation to radicals, but also because there is conjugation across the entire flavonoid skeleton, even the oxonium function at position 4. So I was surprised when some papers singled out conjugated dienes (as opposed to the extent of conjugation across the entire molecule) to explain singlet-oxygen quenching ability.
In a paper$^1$ which argued that the quinonoidal anhydrobase is a better singlet-oxygen quencher:

In our experiments carried out at neutral pH, the anthocyanins exhibited a blue color, indicating that the quinonoidal anhydrobase likely dominated. This form is the most efficient for the quenching of nonradical singlet oxygen because its unsaturated diene conjugation (double-single-double bond arrangement) allows for the addition of singlet oxygen and the formation of endoperoxide at these electron-rich double bonds.

Figure 2: The diene structure (arrows) on the quinonoidal anhydrobase 2 is an excellent substrate for singlet oxygen.
In a follow-up paper$^2$ by another group arguing for the efficacy of the flavylium cation as a quencher of singlet-oxygen:

In the case of the flavylium cation form, AM1 semiempirical molecular orbital calculations indicated that 4’-OH substitution in the ring B causes resonance increment with C-ring allowing strong delocalization of the pi electrons and the positive charge producing also a 1,3-diene system in the ring. C [29] (Scheme 2). Therefore, it can be expected that increasing the number of electron-donating groups in the anthocyanin skeleton stabilizes the resonance structure of the diene system form in the flavylium cation favouring a [4+2] type interaction with the electrophilic species $\ce{^1O2}$.

Why these dienes? It appears that there are loads of other conjugated “dienes” in the anthocyanin molecules. Groups of double-single-double bonds on the C-ring which include the O at position 4 as one of the 4 atoms are automatically eliminated because a diene is defined as “a molecule which contains two alkene linkages”.  But why aren’t the “dienes” on the A-ring of the flavylium cation and the B-ring of both the flavylium cation and the quinonoidal anhydrobase considered as substrates for singlet oxygen? With reference to the 3,5 diene of the quinonoidal base and the 1,3 diene in the resonance structure of the flavylium cation, there are 2 possible key differences I can think of:

It is necessary for one of the carbon atoms of a singlet-oxygen quenching diene to be connected to the electron withdrawing oxygen at position 4 (which does this through inductance, being an [![electronegative atom)  to reduce the electron electron density of the conjugated pi system. (Is a system that is too “electron-rich” somehow a bad thing?) Wouldn’t this reduce the reactivity of the diene towards the electrophilic singlet oxygen, however? Is there something I am missing here?
The dienes that occupy these positions have unpaired electrons, thereby making them more reactive towards singlet oxygen, whereas dienes in other positions do not have unpaired electrons. I can’t give a source, but I remember hearing from somewhere that dienes can have unpaired electrons, depending on the molecular orbital diagram (I have no idea how this works though, if it is relevant an explanation would be appreciated).

In summary: is the reference to the presence/ creation of a “diene” system simply a convenient way of referring to the positions of the anthocyanin which are somehow more likely to react with singlet oxygen (with one of the two scenarios above being a possible explanation), or is this system somehow isolated from the rest of the conjugated pi-system? (The latter seems unlikely to me.) If the former is true, which explanation, if any, is correct?
References
1)  Jang, Y. P., Zhou, J., Nakanishi, K., & Sparrow, J. R. (2005). Anthocyanins protect against A2E photooxidation and membrane permeabilization in retinal pigment epithelial cells. Photochemistry and photobiology, 81(3), 529–536. doi:10.1562/2004-12-14-RA-402
2)  De Rosso, V. V., Moran Vieyra, F. E., Mercadante, A. Z., & Borsarelli, C. D. (2008). Singlet oxygen quenching by anthocyanin's flavylium cations. Free radical research, 42(10), 885-891.
(Accessible via ResearchGate)


text:
Shouldn't the "law of conservation of mass" prevent equations from being unbalanced in the first place?
How is it that a chemical reaction produces an equation, that is essentially an impossibility, according to the law of conservation of mass? If the law is true, shouldn't all chemical reactions balance the atoms in every single equation?
It's like you would add 2 balls together and come up with 1 + 1 = 3 and then say that we "need to balance the equation" so it will make sense. Hopefully my question is clear.

I'm sorry, but I have read your question five times, and probably will read it more often, but I don't have a clue what you are asking.

You've most likely misunderstood the law or its implications on chemical equations. I would suggest reading this introductory article. A chemical reaction does not produce an unbalanced equation, but we need to balance equations we create to match the stochiometry of the real reaction.

You are indeed balancing equation according to the conservation of mass and the chemistry that is going on. And furthermore there are not unbalanced equation by definition (though balancing equation is surely commonly said).

Could you provide an example reaction?

How is it that a chemical reaction produces an equation, that is essentially an impossibility, according to the law of conservation of mass?

You find unbalanced equations as exercises and as a step in writing a balanced equation when there is insufficient information that still needs to be discovered.
Balancing equations as an exercise
The observable chemistry underlying a chemical equation is always balanced. Presenting an unbalanced equation and asking students to balance it is an exercise. Similar exercises are given in many other disciplines:

Balance the following equations:
$$\tag{Maths} (a + b)^2 = a^2 + ab + b^2$$
$$\tag{Physics}E = m c$$
$$\tag{Genetics}\text{Fruitfly = Heads and Legs and Wings and Eyes}$$
$$\tag{Accounting}\text{Dollar = Nickel + Dime}$$

Balancing equations as a step in research
Sometimes, we don't know the formula of a product yet, but we know what reactants were used. For example, if somebody serves me a concoction for desert. I don't know what it is, but it seems be made from eggs, apples and marshmallows (yuk!). I would write:
$$\ce{Egg + Apple + Marshmallow -> Concoction}$$
Then, as I do more research, I find out what recipe was used:
$$\ce{3 Egg + Apple + 5 Marshmallow -> Concoction}$$
What are possible misconceptions?
In a balanced equation, absence of a coefficient means that the coefficient is 1. In an unbalanced equation exercise, absence of a coefficient means that we have to find what it is. So the absence of coefficients does not necessarily means it needs to be balanced, e.g.
$$\ce{NaCl(s) -> Na+(aq) + Cl-(aq)}$$
is balanced, while
$$\ce{MgCl2(s) -> Mg^2+(aq) + Cl-(aq)}$$
is not and does not represent anything that would actually happen. If students learning chemistry already had PhDs in math, we could pose the exercise simply as:

$$\ce{\nu_{MgCl_2} MgCl2(s) -> \nu_{Mg^{2+}} Mg^2+(aq) + \nu_{Cl^-} Cl-(aq)}$$
Determine the minimal integer values of nu (i.e. adding an additional constraint so the set of linear equations has a single solution in the simplest case). Omit the trivial solution of all values of nu vanishing. (Of course, the entire exercise is trivial if you have a PhDs in maths.)

In most cases, students learning chemistry don't have a PhD in math, so we avoid the greek symbols and keep the subscripts to a minimum.

Using the recipe analogy, I understand that when we're presented with an unbalanced equation exercise, we're actually being asked to find the minimum number of atoms needed in the reactants and the actual number of atoms in the product, such that this chemical reaction would be possible? That means that unbalanced equations are really just a generalized chemical reaction, not precise enough in terms of number of atoms. Am I getting this?

Yes the stoichiomtric coefficients are the unknowns you are figuring out.

First, provided the ball is an atom, you are not writing a chemical equation for a dozen of balls, rather a dozen times at least $10^{16}$ (to speak of equilibrium), usually a dozen times $10^{23}$ (macroscale that normal people operate on).
That's not the level you want to use addition for; smaller numbers are easier to operate with.
Second, it's not always single "balls".
Sometimes "balls" are "glued" together so that they cannot be separated prior to addition (the reactant is a polyatomic molecule).
Sometimes it's a tremendously huge, but ordered set of "glued balls" (bulk solids, crystalline matter) so that counting the absolute value of all sorts of balls is nearly impossible, but the ratio between, lets say, green and yellow balls can be easily determined and, lets say, is exactly $1:1$ (e.g. sodium chloride NaCl).
Third, the "ball" can carry additional label that is also to be accounted when it comes to chemical reactions (e.g. charges and isotopes), whereas you can neglect those while counting the balls in Modell's Sporting Goods.
All these principles is the basis of stoichiometry.

Different compounds always have to have the same number of each type of atom in them. For example, in respiration:
$$\ce{C6H12O6 + O2 -> CO2 + H2O}$$
The chemical formulas for these compounds must always remain the same else it is no longer that compound. With this in mind balancing increases the number of molecules of each compound you have to ensure that there is the same number of atoms on each side of the equation, so it becomes
$$\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6H2O}$$


text:
Reasons for low melting point of gallium
J.D. Lee writes in Concise Inorganic Chemistry:

Gallium has an unusual structure. Each atom has one closest neighbor at a distance of 2.43 Å.
This remarkable structure tends towards discrete diatomic molecules rather than a metallic structure. This accounts for the incredibly low melting point of gallium at 30 °C.

That's not all, this surprised me as well:

Note that the boiling point for Ga is in line with the others, whereas its melting point is not. The very low melting point is due to the unusual crystal structure, but the structure no longer exists in the liquid.

Why does gallium exist as diatomic units, that too in a solid crystal, instead of forming the usual metallic crystal structure? And why not in liquid state (where it seems to be more free to make diatomic molecules)?
I might guess that $\ce{Ga2}$ might form stable MOs, but why do the other members of group 13 abstain from such behavior?
It is quite weird to see an element in the middle of a group staging a drama of exceptions like this, it's usually the 2nd period elements which do that.

For your interest, I quote a paragraph from a book "Handbook of Nanophysics: Nanoparticles and Quantum Dots" by Klaus D. Sattler (Ed.). Section 12-8. I don't think people know the reason as to why Ga likes to exist as a dimer (as you know $\ce{S}$ is $\ce{S8}$ and so on).

The cohesive energy of the elemental solids is defined as the
  difference between the energy per atom of the solid and the energy of
  an individual atom in isolation. However gallium is rather unique
  among metals in that in the solid state it has a very unusual
  structure—an orthorhombic 8-atom unit cell composed of dimers which
  have a bond distance 15% shorter than the next nearest neighbor
  distance. As such, it has been called a “molecular metal,” which
  really contains two distinct kinds of bonding: covalent bonds within
  the dimers which result in a pseudogap at the Fermi level, and
  metallic bonding formed by overlap of the electronic wavefunctions
  along the buckled planes perpendicular to the molecular bond of the
  dimer. As a consequence, conduction in gallium is highly anisotropic,
  with a much greater conductivity along the buckled planes than
  perpendicular to this direction (Gong et al., 1991; Heine, 1968).
If we estimate the “metallic” cohesive energy of the solid, by
  calculating the cohesive energy relative to the dimer, we immediately
  see that gallium moves much closer to the other elements, as shown in
  Figure 12.6. In fact, in this case we are still overestimating the
  cohesive energy relative to the melting temperature, since the dimers
  in the bulk will be distorted from the free dimer geometry. This
  argument certainly explains the low melting point of gallium relative
  to other metals, and due to the anisotropy of the lattice most likely
  relates to the surface superheating of gallium that was observed very
  early on (Volmer and Schmidt, 1937).

@WilliamR.Ebenezer there's nothing wrong with leaving an answer unaccepted, if you think that there is still more to be said.

@orthocresol Alright, I guess that'd be better, as it would still show up as unanswered.


text:
Why is molybdenum that important in inorganic synthesis as well as biosynthesis of ammonia?
We know that Mo (Molybdenum ) is used as  a catalyst for Haber's process, during synthesis of Ammonia from Nitrogen. Again, in biological systems, Nitrogen fixation (ALSO forming Ammonia from Nitrogen) uses Mo as activator of Nitrogenase enzyme. My question is "why Mo is so special for Nitrogen?"

Helpful: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2999624/

Two processes running at very different temperatures and pressures. I'd wager this is coincidence.

Why Molybdenum is Used in the Manufacture of Ammonia?
Molybdenum tubing has particularly good resistance to corrosion by
mineral acids, provided oxidizing agents are not present. It is also
resistant to many liquid metals and to most molten glasses. Molybdenum
is relatively inert in hydrogen, ammonia, and nitrogen up to about
1100 C, but a superficial nitride case may be formed in ammonia or
nitrogen.(Ref. 1)

Also, molybdenum has a special capability to form stable complexes with nitrogen atoms and hence molybdenum goes very well in biological reactions as well(See the link @user79161 has posted in the comment section). Recently, a group of scientists has developed a method for synthesis of ammonia using a dinitrogen-bridged dimolybdenum complex as catalyst and under very mild conditions of room temperature and at atmospheric pressure. However, the reaction mechanism of this new method is still unclear.

Reference

https://samaterials.podbean.com/e/why-molybdenum-is-used-in-the-manufacture-of-ammonia/
https://www.u-tokyo.ac.jp/focus/en/articles/a_00259.html (Picture)


text:
Is position of equilibrium path independent?
I know that if a reaction is at equilibrium, then  changes in condition may change the position of equilibrium and the direction is such shift can be predicted by Le Chatlier's rule. But now we're studying the Haber process and my teacher applied this rule to two reactions happening at different pressures in two different containers (initial amounts added, temp were kept constant for both) . Can Le Chatlier's be used this way? I understand the idea that increasing the pressure will lead to a relatively higher mole fractions of the products in the Haber reaction ( I got this using Kp) but don't see how this is a consequence of Le Chatlier.
So basically, will the position of equilibrium be the same if I let my system come to equilibrium at a pressure A and then increase it to B or if I directly come to B?

If the position of equilibrium were path-dependent, you'd be able to build a perpetual motion machine out of it.

@Ivan, could you please elaborate?

What's the point? You won't build it anyway.

@Ivan, haha no I won't but I'd like to know why position of equilibria must be path independent.

If it was not independent, you could create energy by cyclical travelling between 2 states via 2 different paths.


text:
pH of CH3COOH + HCl solution (shouldn't be a buffer)

$\pu{0.05 L}$ of a solution containing $\pu{0.3 M}$ acetic acid $(\ce{CH3COOH}$, $K_\mathrm{a} = \pu{1.8e-5})$ are added to $\pu{75 mL}$ of a solution containing $\pu{0.02 M}$ hydrochloric acid ($\ce{HCl}$, a strong acid). Calculate the $\mathrm{pH}$.

Can I simply find the $[\ce{H+}]$ released by the reaction of $\ce{CH3COOH}$ with water and add them to the $[\ce{H+}]$ released by $\ce{HCl}$, add the two concentrations and then calculate $\mathrm{pH}$ that way? Or does some reaction happen that I'm not aware of?

Yes there is a reaction, one that you can't possibly be unaware of: the dissociation reaction. Acetic acid is weak, its dissociation is incomplete (that's why you'll need Ka), and HCl affects this process more than a little.

This is a simple case of COMMON ION EFFECT. Where you just have to take care of the Ka being constant (as isothermal process).

In other words, if you treat the components separately, and then combine them, the following net reaction happens: $$\ce{CH3COO- + H+ -> CH3COOH}$$

Or does some reaction happen that I'm not aware of?

There are three reactions, and you are probably aware of all of them:
$$\tag{1}\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$
$$\tag{2}\ce{HCl(aq) -> H+(aq) + Cl-(aq)}$$
$$\tag{3}\ce{Ch3COOH(aq) <=> H+(aq) + CH3COO-(aq)}$$
The concentration of $\ce{H+}$ in those three reactions is the same because it all happens in the same phase. So what happens in reactions (1) and (2) influences the equilibrium of reaction (3).

Can I simply find the [H+] released by the reaction of CH3COOH with water and add them to the [H+] released by HCl, add the two concentrations and then calculate pH that way?

No you can't because if you go back to reaction (3), you are no longer at equilibrium. To avoid trying to fix one equilibrium while messing up another, there are two strategies:
a) Put everything into a system of equations, and solve them in one swoop (preferred method if you are using an equation solver).
b) Start with the major species ignoring minor species and reactions that don't influence major species much. Then, move on to those other reactions and the minor species. This is preferred when you have to do the calculations on paper and don't need the exact solution.
The approximate answer
After mixing and ignoring all acid dissociation reactions, the concentrations are the following:
c(acetic acid) = 50 / 125 * 0.3 M = 0.12 M
c(hydrochloric acid) = 75 / 125 * 0.2 = 0.12 M
So here are the steps:

Let hydrochloric acid dissociate and get hydrogen ion concentration and pH
Check whether acetic acid dissociates appreciably
Check whether water dissociated appreciably

For the first step, we get [H+] = 0.12 M, and pH = 0.92. These are tentative because we did not let the other acids (water and acetic acid) dissociate.
For the second step, we are not at equilibrium yet (no acetate yet). However, the pH is very low compared to the pKa of acetic acid, so unless the pH changes a lot, it won't dissociate much. Let's try to calculate the acetate concentration assuming that the change in acetic acid and hydrogen ion concentration is negligible.
$$ [\text{acetate}] = K_a * [\text{acetic acid}] / \ce{[H+]} = \pu{1.8e−5} * 0.12 / 0.12$$
So we are not making a big mistake if we say the concentrations of acetic acid and hydrogen ions did not change much due to reaction 2. If we want, we can update the hydrogen ion concentration from 0.12 M to 0.120018 M.
For the third step, we do what we always do when the pH is substantially acidic. We just calculate the hydroxide concentration assuming the hydrogen ion concentration is not much affected by water dissociation. It comes out as $\pu{8.3e-14}$. If we want, we can update the hydrogen ion concentration from 0.120018 M to 0.120018000000013 M and a pH of 0.92075. (None of this makes sense because we only had 2 significant figures for the dissociation constant of acetic acid.)
Because we assumed a hydrogen ion concentration of 0.12 M for the second step and 0.120018 M for the third, even though this is not quite true, reactions (2) and (3) are not quite at equilibrium for the concentrations we calculated. The good thing is that in most cases, that does not matter.

I'll add that the point about significant figures is VERY important. If you try to solve all three equations simultaneously you end up with a very nasty cubic equation which is impossible to solve by hand. No problem for a computer, but 32 digits of precision makes no sense when the equilibrium constants aren't define to better than 1%. // -- The gist -- In chemistry it is all about finding approximate solutions, not mathematical exactitude. So $20.0002 \ne 20$ in a math class, but in a chemistry class they are equal.


text:
Particle-wave duality for atoms
Hmm Hello . I was wondering if we could apply the particle-wave duality of a subatomic particle for an entire atom.I know that atom is better described as space
but could we apply the de Broglie duality of state for a whole atom I mean something like (atom-wave duality).I was wondering because the strength of weak acids dissolved in water look very similar to the propability cloud for electrons(One electron can have more propability of being in a place than in another place).So for example the H+ cation would get out of a water molecule and since we add more and more energy increase the propability of the H+ cation to get out of a water molecule and leave OH- and go to stick to another water molecule forming H3O+.

Well, this question could be good if it asked about tunnelling of protons, but right now it isn't clear.

Can an atom behave as a subatomic particle?Can it have wave properties , can the strength of a weak acid to be described as the propability of an H+ cation to be cut off from an H2O molecule and stuck into another H2O forming OH- and H3O+?

It's fine to ask if atoms behave quantum mechanically, but I really don't understand the second half of your question and how acidity has anything to do with it.

I mean the propability of a hydrogen cation to get out of an H2O molecule and go to another forming OH- and H3O+

I was wondering if we could apply the particle-wave duality of a subatomic particle for an entire atom.

Yes.  However for such a relatively massive object it's behavior is on the border between where we can "cleanly" apply quantum theory and cleanly apply classical theories.  It is closer to the quantum world and to see that you need to check into the details of bonding between atoms.  You cannot properly describe bonding of atoms without quantum theory, which means the particle-wave duality applies in a sense.

could we apply the de Broglie duality of state for a whole atom

If you fire atoms at a diffraction grating small enough you will get the patterns you expect from a wave, so yes.  More precisely particle wave duality has been tested for relatively large molecules.

Can an atom behave as a subatomic particle ?

Any object can be (in theory) described and has behavior governed by quantum theory.  The larger the object at the more moderate the energies the fewer quantum effects you will see.
Atoms are about the smallest object we can directly and individually manipulate and "see", but to do this requires using quantum effects, e.g. a tunneling quantum microscope.  So while they can be considered as classical objects in some ways, they do require treatment as quantum objects in many ways.

can the strength of a weak acid to be described as the probability of an H+ cation to be cut off from an H2O molecule and stuck into another H2O forming OH- and H3O+?

The chemical properties of atoms are governed almost entirely by the behavior of electrons around and between nuclei.  Any theory describing any reaction at the level you suggest would have to involve quantum theory (model the electronic behavior).  In practice this is not a productive approach - the calculations for full quantum theory are extremely difficult - and more practical methods and approximate rules work just as well in practice.

So can the strength of a weak acid/base be described with quantum theory?

@DonaldDuck Yes, and in an accurate way though the problem would be very complicated, even with approximations.

You can apply matter wave  concept on HUMAN BODY also , but  the particle nature of human body is more dominant due to its large mass and large size. Actually every wave has a particle nature and every particle a wave nature too!... If you consider Heisenberg's Uncertainty Principle (H.U.P)  you would see that when we try to specify a subatomic particle's momentum, its position becomes uncertain, AGAIN  for macro world you would see that uncertainty in position is negligible even if we  try to specify its momentum. Hope it is clear enough!

So it is possible for example quantum tunneling of atoms

@SteohenG has provided a very clear answer, this one just provides some other relevant information.
The quantum theory predicts behaviour of various particles individually and combination of particles (many-body problem) as well.
Though we could, in principle, apply the principles of duality and probabilitistic nature of events, the problems involve very complicated methods to solve even, with simple systems and even after several approximations, we may not be able to arrive at a useful result.
Therefore, even if the principles are applicable to atoms, the several components make a solution which is practical or on which we can furthur build upon is difficult to achieve.

PS I did not know if appending this to StephenG's answer was better so I posted a separate answer.


text:
How to solve constants out of the internal energy equation?

Imagine we deal with a new kind of matter, whose state is described by:
$$PV = AT^3$$
Its internal energy is given by
$$U = BT^n \ln\left(\frac{V}{V_0}\right) + f(T)$$
where $A$, $B$ and $V_0$ is a constant and $f(T)$ is a polynomial function.
Find $B$ and $n$.

What I know
The given expressions remind me of adiabatic compression/expansion. If we assume quasistatic adiabatic compression/expansion we know that heat won't get out/in the system.
$$\Delta U = -W$$
And work is:
$$W = -P\Delta V$$
Some thoughts on how to solve the problem
We notice here that we are dealing with a non-ideal gas. Assuming that the above equations are correct and using first thermodynamics law one gets:
$$\mathrm{d}U = \left[nBT^{n-1}\ln\left(\frac{V}{V_0}\right) + f'(T)\right]\mathrm{d}T$$
$$\left[nBT^{n-1} \ln\left(\frac{V}{V_0}\right) + f'(T)\right]\mathrm{d}T = \frac{AT^3}{V}\mathrm{d}V$$
I do not see a solution to this differential equation. There has to be a easier way to get both $B$ and $n$, but how?

Your differential equation doesn't make sense to me: you have infinitesimals ($\mathrm dU$ and $\mathrm df$) and finite quantities ($nBT^{n-1}\ln(V/V_0)$) being added together. I gather you were trying to differentiate by $T$ throughout?

@orthocresol that is a typo let me fix it.

Yes I differentiated $U$ with respect to $T$. The idea is to set up a differential equation that relates the change in temperature and volume during the compression/expansion process. I assumed it will be adiabatic (based on the given equation: $PV = AT^3$)

By using the equation $ \big( \frac{\partial U}{\partial V}\big)_T = T\big(\frac{\partial P}{\partial T}\big)_V - P $, you will get $B$ as $2A$ and $n=3$.

You need to use the equation $$\left(\frac{\partial U}{\partial V}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]$$

Thanks for the answer Chet Millet. However, I still do not see how to derive it. I know the equation you propose comes from the thermodynamic identity: $dU = TdS - PdV + \mu dN$. I have been trying to derive it with respect to $V$ but I do not get rid of the entropy term.

The derivation is in every thermo book.  You are correct about the starting point.  The next step is to express dS in terms of dT and dV.  Then, for the dV term, you derive a Maxwell relationship from dA=-SdT-PdV.

I use to study from two books I really like: Physics for Scientists and Engineers (Serway and Jewett) and Thermal Physics (Schroeder). The former is more general and has a good part on Thermodynamics but it doesn't deal with thermodynamic identities. The later does (pages 156 and 157) but it is not enough for me. May you recommend a source to study thermodynamic identities deeper (where all the formulas are derived and so on)? Please let me know if I should ask this in a separate post. Thanks

Fundamentals of Engineering Thermodynamics, Moran et al.  Introduction to Chemical Engineering Thermodynamics, Smith and van Ness. Principles of Chemical Equilibrium, Denbigh.

I am sincerely SORRY that I couldn't provide a LaTeX markup for the answer , but I am busy preparing for one of the biggest exam of my life , I would surely update the answer as soon as the exam is over , any help with OCring image would be appreciated.

Edit 1 : theirs a typo in the equation of dU it's PdV instead of VdP


text:
Can tin(IV) sulfide be synthesized from tin(II) chloride via oxidation with sulfite in acidic media?
Will the following reaction occur?
$$\ce{2Na2SO3 + 6SnCl2 + 12HCl → 4NaCl + 5SnCl4 + SnS2 + 6H2O}$$
I'm going to synthesizing $\ce{SnS}$ and $\ce{SnS2}$ NP's, but our supply is runned out and deadline of assignment is coming to soon, so will this reaction  occur? If it is possible, what is are the conditions of this reaction?

You are using a reducing agent, sulfite, to effect an oxidation.

@user55119 "reducing agent" is a relative term. Sulfite might act like one with strong oxidizers, but SnCl2 is even better reducing agent in comparison, making sulfite act as oxidant, although the oxidizing effect is even more pronounced among selenites and tellurites.

Do you not have hydrogen sulfide?

Yes, the reaction
$$\ce{2 Na2SO3 + 6 SnCl2 + 12 HCl → 4 NaCl + 5 SnCl4 + SnS2 + 6 H2O}$$
is described in chapter 35.6.4 Reactions of chalcogen oxo-compounds of the Russian translation [1, p. 526] of Blumenthal's Anorganikum 10th Ed. (in German).
Unfortunately, I don't posses original German edition, so the following is my English translation of the procedure:

To the 0.5 M solution of $\ce{SnCl2}$ in 12 M $\ce{HCl}$ in three test tubes 0.2 M solutions containing $\ce{SO3^2-}$, $\ce{SeO3^2-}$ and $\ce{TeO3^2-}$ are added. In all three tubes precipitations of $\ce{SnS2}$, $\ce{Se}$, and $\ce{Te}$, respectively, are formed.

Original Russian text:

К 0.5 M раствору $\ce{SnCl2}$ в 12 M $\ce{HCl}$, находящемуся в трех пробирках,
  приливают 0.2 M растворы, содержащие $\ce{SO3^2-}$, $\ce{SeO3^2-}$ и $\ce{TeO3^2-}$. Во всех трех пробирках наблюдается выпадение осадков, которые представляют собой $\ce{SnS2}$, $\ce{Se}$ и $\ce{Te}$ соответственно.

References

G. Blumenthal, S. Engels, I. Fitz, W. Haberditzl, K.-H. Heckner, G. Henrion, R. Landsberg, W. Schmidt, G. Scholz, P. Starke, I. Wilke, K.-Th. Wilke Anorganikum: Lehr- und Praktikumsbuch der anorganischen Chemie mit einer Einführung in die physikalische Chemie; Ed. of Russian translation: L. Kolditz; Mir: Moscow, 1984; Vol. 1.

Now that's something I didn't expect.

@IvanNeretin "Anorganikum: Überraschend anders!" :D


text:
why is white phosphorous given to have ZERO standard heat of formation
why is white phosphorous given to have ZERO standard heat of formation??
while it is well known fact that white phosphorous is highly reactive and is obviously less stable than red and black phosphorous
also here=Which allotrope of phosphorus is kinetically the most stable?
it says that black phosphorous is thermodynamically more stable
also what is the relative stability of the allotropes [black red white purple]

Why do we measure height above sea level and not height below Mount Everest?

@orthocresol cause the sea surrounds the world around and Mount Everest is only a single entity which is subject to change

All elements have one defined standard thermodynamic  state.  If the pure element has multiple allotropes the choice is somewhat arbitrary. For instance diamond has a more consistently defined structure than graphite, but diamonds are expensive to burn in a bomb calorimeter. So graphite is consistent enough from sample to sample, it is stable, and it is cheap. Hence the chosen standard state of carbon is graphite.

Related: https://chemistry.stackexchange.com/questions/73951/why-is-the-standard-enthalpy-of-formation-of-black-phosphorus-not-zero/73953#73953


text:
Resonance effect of amide group
How will an amide group behave while being atrached to a benzene ring?Will it be donating due to lone pair of nitrogen or withdrawing due to carbonyl carbon?

As you can see, the positive charge comes on the benzene ring, therefore the group draws electron cloud away from the benzene ring.
Edit: I have assumed it is attached through the C.
Sorry about the untidness.

Thsnks user79161.Can you show me the effect of ester group in a similar way?


text:
Finding the sum of coefficients of the reaction

What is the sum of the coefficients $(a + b + c)$ in the following reaction?
$$\ce{x Zn + y HNO3} -> {a Zn(NO3)2 + b H2O + c NH4NO3}$$

Here if the equation is balanced we get 1 mol of $\ce{Zn}$ and 4 mol of $\ce{HNO3}$. As  $x + y = a + b + c$ we get sum of coefficients $a + b + c = 5$. According to me, this is the correct answer, but in the actual answer key the answer is 8.
I may be wrong, but can someone help me clarify the right answer?

So for the nonsensical reaction what is your balanced chemical formula? In other words the Zn, H, N, and O atoms have to balance on both sides of the reaction. // Obviously $x = a$, so $$\ce{aZn + yHNO3 -> aZn(NO3)2 + bH2O + cNH4NO3}$$

It's 4 ,10,4,1,3 . I was wrong earlier . The answer is 4+3+1 which is eight

Curious, did you pound your head on the wall guessing, or did you use the set of linear equations for the atoms?

Linear equations

It's about 2:00 in the night hene mixing things up

Good. I just wanted to be sure that you knew there was a systematic way to solve this - at least mathematically. As I said, the chemistry makes no sense at all.

I sincerely don't understand why the question is downvoted. The question itself looks reasonable, the reaction is chemically correct, and OP demonstrated their efforts and thoughts (partially in the comments, but still).

Linear equations aside, note that this is a redox reaction and you can use half-reactions method as well (BTW the reaction looks fine to me):
$$\ce{$x$ \overset{0}{Zn} + y H\overset{+5}{N}O3 -> $a$ \overset{+2}{Zn}(NO3)2 + b H2O + $c$ \overset{-3}{N}H4NO3}$$
$$
\begin{align}
\ce{\overset{+5}{N} + 8 e- &→ \overset{-3}{N}} \tag{red}\\
\ce{\overset{0}{Zn} &→ \overset{+2}{Zn} + 2 e-} &|\cdot 4 \tag{ox}\\
\hline
\ce{4 \overset{0}{Zn} + \overset{+5}{N} &→ 4 \overset{+2}{Zn} + \overset{-3}{N}} \tag{redox}
\end{align}
$$
So that the complete balanced equation is
$$\ce{4 Zn + 10 HNO3 -> 4 Zn(NO3)2 + 3 H2O + NH4NO3}$$
leaving $a = 4$, $b = 3$, $c = 1$ and the sum $a + b + c = 4 + 3 + 1 = 8$.

I don't understand why it was commented that this is a nonsensical reaction? Nitrate  ion can be easily reduced to ammonium ions.

@M.Farooq You should ask the one who commented. As I mentioned in the comments and in answer itself the reaction looks completely fine to me.

What is the sum of the coefficients $(a + b + c)$ in the following reaction?
$$\ce{x Zn + y HNO3 -> a Zn(NO3)2 + b H2O + c NH4NO3}$$

This is a redox reaction. During OP's comments, it was clear that he/her got the correct answer, but it wasn't clear that OP understand this is a redox reaction. To balance the equation, it may be better work on with two relevant half reactions. It is also easier to work with ionic equations and add counter ions at the last step. First consider reduction half reaction:
$$\ce{NO3- + d e- -> NH4+}$$
Now, balance $\ce{O}$ with $\ce{H2O}$, $\ce{H}$ with $\ce{H+}$ (since this is in acidic medium), and lastly, charge with $\ce{e-}$ (now you see $d=8$).
$$\ce{NO3- + 10 H+ + 8e- -> NH4+ + 3H2O} \:\text{---} (1)$$
Now, consider oxidation half reaction:
$$\ce{Zn -> Zn^2+ + 2 e- } \:\text{---} (2)$$
Now, solve equation (1) and (2) by cancelling the electrons:
$$\ce{4Zn + NO3- + 10 H+  -> 4 Zn^2+ + NH4+ + 3H2O}$$
Now, to balance by counter ion, add $\ce{9NO3-}$ to both side of last equation so it keep balanced:
$$\ce{4Zn + NO3- + 10 H+ + 9NO3- -> 4 Zn^2+ + NH4+ + 3H2O + 9NO3-}$$
Rewrite the equation with tight pair ions:
$$\ce{4Zn + 10 HNO3 -> 4 Zn(NO3)2 + NH4NO3 + 3H2O }$$
Now you see: $x=4; y=10; a=4; b=3;$ and $c=1$.
$$\therefore a+b+c=4+3+1=8$$