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Ester of glycerol and glyceric acid
What is the ester of glycerol ($\ce{C3H8O3}$) and glyceric acid ($\ce{C3H6O4}$)? I couldn't find any information about it online, so is this reaction even possible?
I'm not sure what would happen, from my understanding, the alkyl of the alcohol replaces the hydrogen in ($\ce{-COOH}$) the  carboxylic acid. But I can't work out this one.
I think it could be $\ce{C6H12O6}$ (mono ester) or $\ce{C12H23O15}$ (tri ester) , but I'm not sure. Is this correct? Is there a characteristic difference between the 3(mono, di or tri esters)? What dictates which one will be formed?
Also, do the compounds have any significance?

Since glycerol has 3 hydroxyl groups, you could have mono, di or tri-esters. How could it be C5? For a monoester it would be C3 + C3 = C6!

Oops! Missed that, could it be C12H23O15 (tri ester)?

There isn't enough information to make a decision. The triester will be C12 but if you use racemic glyceric acid there would be several diastereomeric triesters. PS: Can't have an odd number of hydrogens and the oxygens are wrong. Do you know what an ester is and the structures of the 2 reactants?

And don't forget, glyceric acid and react each other when its $\ce{COOH}$ is activated, depend on the method you'd use.

That issue is readily averted by using the acetonide of glyceric acid for the esterification with glycerol. The acetonide can be hydrolyzed without affecting the ester. @Mathew Mahindaratne

The formation of the triester of glyceric acid and glycerol cannot be done directly because of the poly-esterification of glyceric acid itself. This problem can be circumvented by employing the acetonide of glyceric acid. Subsequent to the esterification, selective hydrolysis of the acetonide in the presence of the esters can be effected.
Esterification with a single enantiomer of the acid would produce a single triester, either 1A or 1B. Esterification with the racemic acid could form, in principle, 16 (24) stereoisomers. However, this number is reduced to eight. Column A lists 6 permutations of the stereochemistry with two missing owing to the non-stereogenic nature of C2 in 1A and 4A. Column B lists the enantiomers of column A. Rows 2 and 5 are duplicate as are rows 3 and 6. The eight stereoisomers are within the black box (1A through 4B). While the enantiomeric pairs in each row (1, 2, 3, 4) are formed in equal amounts, the four racemates are not necessarily formed in equal amounts.
The assignment of the r/s descriptors at pseudoasymmetric center C2 in stereoisomers 2A, 2B, 3A and 3B are governed by the 2013 CIP Rules 1a and 5.


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t-Boc protection of amide (urea)
I'm trying to prepare protected urea and I'd like to protect it with some alkali stable PG. Is it possible to protect urea with t-Boc? Or is there another way to get boc-protected urea?

Its commercially available. Google "boc urea".

Thank you for the answer, but I'd like to prepare my own. We are not allowed to order from China (University rules) and I have all starting materials that I possibly need.

Reflux with BOC2O and DMAP in MeCN

Check this:Tetrahedron Lett. 2014, 55(4), 842–844 (but method is for mono-protected).

Thank you both for answers Waylander and Mathew, I will try to run this reaction tomorrow.

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Emergence of Lognormal distribution for the concentration of chemical compounds
I'm currently reviewing the literature about lognormal distributions describing/approximating the variability of a given chemical compound across different cells/ samples etc etc. The main argument pro lognormality is the multiplicative central limit theorem, that many authors mention without providing a clear example of a reaction pathway where a targeted concentration can be approximated by the multiplication of many factors. In https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5036630/ and http://web.mit.edu/jakebeal/www/Publications/ENB17-Lognormal-Biology.pdf they make the case of autocatalytic/ catalytic processes. What would be the simplest example of a chemical reaction system where the concentration of a target chemical (not the rate) would appear Lognormal?

I looked at the first paper quickly. It seems that they are measuring concentrations of chemicals within living cells. Living cells don't have a static equilibrium. Living cells are dynamic.

"concentration of target chemical appears lognormal". What exactly do you mean? Concentration when? Where?

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How to define the reduced mass for estimating Kinetic Isotope Effects
I am trying to think about radical reactions using zero-point energies (from the harmonic oscillator model). I have read that substitution of a bond with isotopes can alter the rate of a reaction.
For example, consider the reaction:
$\ce{CH4 + Br2 -> CH3Br + HBr }$
Suppose the 3 non reacted hydrogens are instead deuterated. The new reaction will be:
$\ce{CD3H + Br2 -> CD3Br + HBr}$
Now, based on the harmonic oscillator model, the zero point energy of the C-H bond in $\ce{CH4}$ can be calculated from $E_0=\frac{1}{2}h\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}$ where $k$=the force constant for the bond and $\mu$ is the reduced mass.
I read that "isotopic substitution does not change the potential energy surface of molecules" so the value of $k$ is unaffected by isotopic substitution. Thus in a kinetic isotope effect analysis based on zero-point energies, we need only consider the change in reduced mass.
Now my question is about the specifics of handling the listed reaction above. For calculating the reduced mass of the C-H bond, would you use the mass of the whole molecule (1) or just the mass of the carbon atom that the hydrogen is bonded to (2)?
In case the wording is confusing, in the example reaction the difference between the reduced masses would be:
$\mu=\frac{m_1 m_2}{m_1+m_2}$
(1) $\mu=\frac{15\cdot2}{15+2}=1.76$
This version of the calculation takes $m_1$ as the mass of the rest of the molecule ($\ce{CD3}$) beyond the C-H bond.
(2) $\mu=\frac{12\cdot2}{12+2}=1.714$
This version takes $m_1$ as the mass of only the carbon atom in the bond.
Which would be the correct way to carry out the calculation?
I kept the rest of the molecule simple in the calculation, but if the rest of the molecule had a much larger molar mass (say 200), then the reduced mass would become closer to 2, so I feel like the difference between the two methods could significantly affect the results of calculations on more complex molecules.

I'm not an expert but I dimly recall that $m_1$ and $m_2$ refer to the two "point" masses at either end of the "vibrating" bond.  Since the (complex) vibrational mode of the bond in question includes all the atoms, not just C, my guess is the first way you did it is right.


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Deuteration of following enone - mechanism?

What happens when the above compound is subjected to DCl in MeOD? As a result of tautomerism, I'd expect at least 7H atoms, two alpha to carbonyl carbon (right side), and five at gamma position (two on one position and three on the other); to be replaced with deuterium.
However, I'm not clear about the mechanism that'd be followed to achieve the same. And at the same time I'm not sure if there will be an 8'th replacement at alpha position to the carbonyl carbon (left side).
I know that the hydrogens on the right alpha position is replaced due to enolization and I think I got the mechanism for this, but the rest is a blank for me at the moment.
If someone could please help  provide a detailed mechanism, it'd be really helpful. Thanks a lot.

Take a look here. It is under basic conditions but it should serve as guide as long as you know how to form enols in acid. https://chemistry.stackexchange.com/questions/88672/deuteration-of-enone-mechanism/88827#88827

I do know how they form. It is mostly the deuteration I have problems with since the two books I have about organic chemistry doesn't really cover this subject. But thanks for the help I think I understand how to tackle this now. My only last question would be if there is gonna be 7 or 8 replacements? (as in the original question)

Here are acid and base catalyzed deuterations of a ketone: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens/Deuterium_Exchange.Your enone is simply an extended ketone. Eight maximum!


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Can Weinreb amides be used to form 1,3 dicarbonyl compounds?
Weinreb amides are useful in reducing acid chlorides (and esters) to form ketones due to the fact that the more reactive keto group is unmasked only after treatment with acid and water, which destroys the reducing agent. One example I have been given of use of Weinreb amides is the following:

We see here that the 1,3 dicarbonyl compound is formed. This implies that under these conditions, the ester lithium enolate does not attack the ester moiety in the Weinreb intermediate, presumably because it is not reactive enough. Is this a general case? If the ester was, say, replaced by a ketone, would we still form the 1,3 diketone species or would we have attack of this carbonyl centre to form an alcohol?

I think the steric effect of the tBu group is important in blocking possible enolate attack on the ester. Does this reaction work for methyl or ethyl ester enolates?

I don't really understand why this is better than just running a Claisen condensation. The $\beta$-keto ester product will tautomerize after the reaction, which it likes does here as well. Then it's no really prone to reacting again.


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Why do properties of colloidal solution depend only on dispersion medium's physical state?
According to a past year JEE question, properties of colloidal solution depends only on dispersion medium's physical state and not on nature of dispersed phase or dispersion medium, or temperature of system. I cannot find any further information on this subject. I would like some clearer guidance on this. Would someone please help?

Which properties? I guess that's only about things like melting/boiling point.

The assumption that properties of colloidal dispersion only depends on the medium is completely incorrect. The vague part of the question is "properties" in the question, because the examiner is not clear about what type of properties he/she is talking about. It is not uncommon to find open ended questions in public exams. The properties of colloidal solutions heavily depend on the temperature (since their viscosity is temperature dependent as well as flow rate dependent). Similarly, the size of colloidal particle controls the optical properties of the dispersion. Similarly the surface charge on the particles determines the sign/magnitude of the zeta potential. If you want to read further, read DLVO theory (named after four persons) on the stability of the colloids from Wikipedia. One should not worry about an incorrect question. If it were a true-false question, one can simply mark it false.


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A simple question about the direction of a chemical reaction
I do not know much about Chemistry and whatever I knew has become rusty. I have a very simple query. I take a beaker containing pure $\ce{H2SO4}$ solution and immerse a rod of pure $\rm Zn$ into it. Now, the $\rm Zn$ will react with $\ce{H2SO4}$ to form $\rm ZnSO_4$ and hydrogen gas.
$$\ce{H2SO4(aq) + Zn(s) -> ZnSO4(aq) + H2(g)}$$
Why does not the reaction spontaneously reverse once again forming pure $\rm Zn$ and $\ce{H2SO4}$ molecules?
$$\ce{H2SO4(aq) + Zn(s) <- ZnSO4(aq) + H2(g) ?}$$
If your answer contains any technical jargon of Chemistry, please explain that in simple terms.

Not quite right, the reaction forms hydrogen gas which bubbles off making the reaction irreversible.

Imagine a crowded bus during rush hour. Some people board the bus, others get off the bus. Why don't they spontaneously reverse? Well, sometimes they do, but most would eventually get where they want to be. Same thing here.

Dear @Waylander, I am afraid that logic "the reaction forms hydrogen gas which bubbles off making the reaction irreversible" is not correct. Assume we conduct this reaction at 10,000 psi; at this pressure we will not see any bubbles of H2, can we prevent zinc from dissolving in sulfuric acid or reverse this reaction? Most likely not. The formation of hydrogen is highly favored thermodynamically and its equilibrium constant (if it exists) must be extremely large.

@M.Farooq The OP asked for a simple explanation without jargon. Feel free to provide a full answer to the question

M.Farooq i'll have to agree. Zn2+ will not spontaneously return to metallic state. Hydrogen, even if contained in solution, will not change anything. @Waylander your statement is certainly not a simplification!

If the question were "why does the reaction go to completion", both Waylander and @M._Farooq offer valid points: a) It is an open system where  H2 is continuously removed from solution as the reaction goes forward (as long this goes on you can't reach equilibrium). b)The equilibrium constant is super-large so that even in a closed system most of the zinc will react. The OP's question is about reaction going in one direction and then - without any outside influence like doing work of the system - changing course. I don't think the specifics of the reaction are relevant for that question.

Before I talk about the chemistry, I will use an analogy from physics. If I drop a piece of chalk, it falls down. It does not at some point later "fall up" again. However, the atoms inside the piece of chalk do go up and down a bit because of thermal vibration. The chalk does not start jumping up because these vibrations are random rather than synchronized.

Why does not the reaction spontaneously reverse once again forming pure Zn and H2SO4 molecules?

If you are asking why at no time the mass of elemental Zn (metal) increases again, it is because there is a strong driving force toward turning the reactants into products for this specific reaction. If you are asking about individual particles, it turns out every now and then, the reverse reaction does occur (this is true for all reactions and is called micro-reversibility).
For some reactions, this back and forth results in the reaction not going to completion. Instead, these reactions may reach an equilibrium where reactant and products are present, and the forward and reverse reactions go on at the same speed so nothing seems to change (this state is called a dynamic equilibrium).
In the reaction you are asking about the driving force forward is too large, and one of the products continuously "escapes" from the reaction mix, so the reaction goes on until one of the reactants is used up (goes to completion rather than reach equilibrium).
In any reaction, reversible or not, once the reaction reaches equilibrium or goes to completion, it does not spontaneously go in the opposite direction unless something changes (add materials, change temperature, do work).


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How do I draw a curly arrow diagram for the oxidation of primary alcohol to an aldehyde and subsequently to a carboxylic acid?
I've determined the chemical equations for both the mechanisms, but I'm unable to draw curly arrow diagrams for the two reactions using potassium dichromate as the oxidising reagent. I'm not sure how to go about doing this, and I can't seem to find anything online related to do this.
From what I've studied,
oxidation of primary alcohol to aldehyde:
$$\ce{3CH3CH2OH + Cr2O7^2- + 8 H+ -> 3CH3CHO + 2 Cr^3+ + 7 H2O}$$
and, oxidation of primary alcohol to carboxylic acid:
$$\ce{3CH3CH2OH + 2Cr2O7^2- + 16 H+ -> 3CH3COOH + 4 Cr^3+ + 11 H2O}$$
I've found information pertaining to drawing curly arrow diagrams for the oxidation of secondary alcohols to ketones, but I'm unable to apply the same to my query.

What was the oxidation agent in the examples? Could you draw curly arrows for the reaction $\ce{CH3CH2OH -> CH3CHO + H- + H+}$? Or $\ce{CH3CH2OH + O2 -> CH3COOH + H2O}$? That way, you would not have to deal with the chromium chemistry and still see what is going on with the functional groups.

Do the same thing with the primary alcohol as you would for the secondary alcohol. As for the aldehyde, hydrate it first and then proceed.

https://chemistry.stackexchange.com/questions/49542/mechanism-of-the-oxidation-of-alcohols-with-kmno4


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Polyethylene terephtalate Biodegradability
I'm looking into plastics and how they break down. I have learned 3 ways oxo-biodegradation, photodegradation and thermaldegradation. Does anyone know how PET (polyethylene terephtalate) degrades?

It's oxo-biodegradation* with an asterisk. PET polymers are polyester meaning that the monomer units are link by ester functional groups. This ester group can be hydrolyzed in the presence of water to for the alcohol and acid that formed the ester. For PET the alcohol is ethylene glycol which is readily digestible by bacteria and the acid is terephthalic acid which is not as redily digestable but still digestable.
It is for this same reason that poly(glycolic acid) which is used in dissolvable stitches and poly(lactic acid) which is used in food containers (and 3D printing) are also biodegradable.


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Drug pH and pKa known, is it an acid or a base?
I want to answer the question if a drug which is analysed for pH and pKa is ionised or not ionised in a cow stomach with a pH of 6.
The drug has a pH of 8,52 and a pKa of 8,78.
Since the pH is 8,52 I considered it as a base and in that case the drug would be almost completely ionized at a pH of 6, which means it is not absorbed through the stomach wall.
However, the drug has also some acidic side groups (carboxylic and phenols), which could mean that it works as an acid. In that case the equation turns around and it would mean that it is unionized at a pH of 6, which means it is absorbed through the stomach wall.
Is there a way that I can tell if the drug, based on just the pH and pKa, is considered an acid or a base?

If it has multiple functional groups that can act as acid or base, it should have multiple pKa values. I don't understand what you mean when you say "the drug has a pH of...". A solution has a pH, a molecule has a pKa. The pH of a solution containing the drug (only, nothing else) would depend on the concentration of the drug. Does the drug in the solid form contain sodium or chloride counterions? Could you give an example structure of the kind of drug you are asking about?

Fist things first: a substance with multiple basic and acidic dissociation sites has multiple $pK_a$'s, not only one.
Normally, for basic sites the $pK_a$'s of the corresponding conjugated acids are given.
The highest $pK_a$ is the one of the 'most basic' site, because its conjugated acid is the weakest.
For acidic sites, the $pK_a$'s of the acids themselves are given. The lowest $pK_a$ is the one of the 'most acidic' site, i.e. the strongest acid.
Under some assumptions, you can study the behaviour of a molecule looking only at these two $pK_a$'s.
3-hydroxypyridine has 1 basic $pK_a$ (for the pyridinium $NH^+$, 4.79) and 1 acidic $pK_a$ (for the $OH$, not far from 8.75).
So it will have mostly charge +1 at $pH < 4.79$, mostly charge 0 (and no charges on any atoms) at $4.79 < pH < 8.75$, mostly charge -1 at $pH > 8.75$.
Note however that the most basic $pK_a$ is not always lower than the most acidic $pK_a$.
E.g. glycine (2-amino-acetic acid) has 1 acidic $pK_a$ (for the $COOH$, 2.35) and 1 basic $pK_a$ (for the $NH_3^+$, 9.78).
So it will have mostly charge +1 at $pH < 2.35$, mostly charge 0 (but zwitterionic, i.e. with 1 positive charge and 1 negative charge) at $2.35 < pH < 9.78$, mostly charge -1 at $pH > 9.78$.
So to answer your initial question, i.e. how to calculate the fraction with charge 0 of your drug at a specific pH, you need to know at least the most basic $pK_{a,MB}$ and most acidic $pK_{a,MA}$.
Obviously if there is no acidic or basic site, the molecule is always neutral, and if there is only one acidic or only one basic site you can ignore the other one.
In practice, for the calculation, if there are no acidic $pK_a$'s, you can set the most acidic $pK_a$ is 100; and if there are no basic $pK_a$'s, you can set the most basic $pK_a$ is -100.
Then at a given pH the fraction with charge 0 is:
$$f_{z=0} = \frac 1 {1+10^{-pK_{a,MA}+pH}+10^{+pK_{a,MB}-pH}}, if \ pK_{a,MA} > pK_{a,MB}$$
or:
$$f_{z=0} = \frac 1 {1+10^{+pK_{a,MA}-pH}+10^{-pK_{a,MB}+pH}}, if \ pK_{a,MA} > pK_{a,MB}$$
For instance for 3-hydroxypyridine, at pH 6:
$$f_{z=0} = \frac 1 {1+10^{-8.75+6}+10^{+4.79-6}} \approx 94 \%$$
At pH 3 (like in a human stomach) things would be quite different:
$$f_{z=0} = \frac 1 {1+10^{-8.75+3}+10^{+4.79-3}} \approx 1.6 \%$$
For glycine, at pH 6:
$$f_{z=0} = \frac 1 {1+10^{+2.35-6}+10^{-9.78+6}} \approx 100 \%$$
At pH 3:
$$f_{z=0} = \frac 1 {1+10^{+2.35-3}+10^{-9.78+3}} \approx 82 \%$$
As for your other question, whether a substance is an acid or a base, well as you can see from the explanation here, it depends on the presence of acidic and basic dissociation sites, and the pH we're considering as 'reference'.
Normally the reference physiological pH is $7.4$ (blood).
A substance that has mostly positive charge at the reference pH is usually considered a 'base', and if negative, an 'acid'. But you see how fluid the situation can be...


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Antiaromaticity can be explained by Hückel method?
Pentalene is one of the most famous antiaromatic molecules. But I obtained its energy of $\pi$ system as $8\alpha+10.46\beta$ by Hückel method (with a bit complicated but straightforward calculation). This value is lower than that of localized form $8(\alpha+\beta)$. So Hückel method looks to be not able to explain antiaromaticity. Are there any misunderstandings? It seems that many of antiaromatic molecules are predicted lower than their all-ethylene counterparts.
(I don't think radicality is a key for antiaromaticity, and I'd like to know a general, quantum mechanical explanation about antiaromatic molecules if it exists.)

Aromatic or antiaromatic, it is still conjugated, so no wonder it is better off than the all-ethylene analog. The general explanation is precisely where you don't think it is.

With Huckel MOT, aromaticity / antiaromaticity should be determined with reference to a linear conjugated analogue, so e.g. cyclobutadiene is compared with butadiene, not two ethylenes. I am not sure how this carries over to pentalene. Arguably, Huckel's rules were designed for monocyclic systems, not polycyclic ones.

I solved the secular equation of pentalene, whose matrix has two resonance integral in addition to that of cyclooctatetraene. The solutions have no degeneracy, so I think penetalene is not expected to be radical.

@sakai right.  Pentalene is not a radical.  But it will have a low-lying acceptor orbital and thus is highly reactive an an electrophile.  Which, of course, is what actually happens.  Fill the low-lying acceptor orbital and the resulting dimension is quite stable.


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Melting point in hydrocarbons
Checking out the molecular structure and the melting points of various hydrocarbons I concluded:
1)branching increases the melting point of a hydrocarbon molecule
2)symmetry increases the melting point of a hydrocarbon molecule

symmetry increases it way more than branching does.

ordering in terms of melting points:
2-methyl butane < pentane < 2,2-dimethyl propane
2-methyl butane is branched but has a bad symmetry, pentane has a good symmetry although not being branched, 2,2-dimethyl propane has a good symmetry and is branched.

2-methyl pentane < 2,3 dimethyl butane < 2,2 dimethyl-butane < hexane
so 2-methyl pentane has a bad symmetry
2,3 dimethyl butane has a bad symmetry as well but more branched compared to 2-methyl pentane
2,2 dimethyl butane has a bad symmetry but it is more branched than the previous two.
Then hexane has the highest melting point. it is not branched at all and not symmetrical. so why does it have the highest melting point?

I kept the text a bit long before asking the question because I wasn't sure if I understood the concept correctly. Can you please answer my question about hexane and correct me if I have some mistake in my interpretation of the whole thing?

Shouldn't that last example be hexane instead of pentane, since it is compared with 3 other $\ce{C6H14}$ compounds?

yes exactly, sorry. I just edited it.

For pentanes, the order is correct. But, I want to point out that explanation is not that simple. Basically, symmetry is not the only one playing a role here. For example, how about two hexane isomers (3-methylpentane and 2,3-dimethylbutane), you forgot to mention? In your standards, 3-Methylpentane should have better symmetry than 2-methylpentane, but look at their melting points. Same can tell about 2,3-dimethylbutane and 2,2-dimethylbutane, but 2,3-dimethylbutane has even lower melting point (Wikipedia).

yes you're right, my points seem to fail in those examples. Is there a simple way to understand which one has a higher melting point?

I think @MathewMahindaratne's point is that you can create simplistic rules for small hydrocarbons that don't work for larger ones. So such rules are really an oversimplification.

So there are no rules that we can use for larger hydrocarbons?

You can "think under pressure" but discussing boiling points would be more pertinent.

Another factor is geometry, especially for larger molecules. Longer linear molecules have better van der Waals intermolecular bonding, while branched ones rather "decrease their surface" for such bonding.


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The thermal decomposition of NH4Cl
When you heat $\ce{NH4Cl}$ it decomposes into $\ce{NH3 + HCl}$ gases. My question is why does it decompose into ammonia and hydrogen chloride gases on the molecular level. Is it because in the $\ce{N-(HCl)}$ bond, The $\ce{N-H}$ bonds breaks easily since it requires only $\pu{390 kJ/mol}$ and $\ce{H-Cl}$ requires far more heat energy? Thus the $\ce{N-H}$ bonds breaks first?

HCl isn't acidic in gas phase.

I don’t see how that answers my question? I specifically said hydrogen chloride instead of hydrochloride acid...

What is the alternate set of products you think it might have decomposed into?

That is simply not my question, I know it will decompose into ammonia and hydrogen chloride. But the question is on heating, how is the strength of bonds playing out? Is it that H-Cl bond is stronger and thus it will be more resistant? And in the N-H-Cl, the N-H requires less energy to break off? Consequently leaving HCl?

When you heat solid $\ce{NH4Cl}$ in a saturated aqueous solution, it will "decompose" (dissolve) to form $\ce{NH4+}$ and $\ce{Cl-}$. Is it because the H-Cl bond is weaker than the N-H bond?

If the H-Cl bond is weaker than the N-H bond, it will decompose faster since it requires less heat, right? And that’s not what happens

Are you asking why $\ce{NH4Cl}$ decomposes into $\ce{NH3}$ and $\ce{HCl}$ when heated as opposed to subliming as a $\ce{NH4Cl}$ molecule?

Look, the questions of "why is this" type are inherently flawed and can't be answered comprehensively. There are just too many alternatives. You'd better ask "why is this and not that". These often have a relatively short and meaningful answer.

We can also rephrase @Mithoron's original comment by saying that bond strength looks at a homolytic cleavage but acidity does not.

I have edited my question, if the mods could tell me which aspect of my question is unclear I might better be able to edit it

In the solid state of ammonium chloride, the nitrogen and hydrogen atoms are connected by covalent bonds, forming the ammonium ion. The bond length is about 98 pm. Each ammonium ion is surrounded by eight chloride ions (and in turn, each chloride ion is surrounded by eight ammonium ions). The distance between a chloride and the closest hydrogen atoms is about 237 pm. In contrast, the bond length in HCl is 127 pm.

In the reaction $$\ce{NH4Cl(s) -> NH3(g) + HCl(g)},$$
you lose the ionic interactions, and ammonium gets deprotonated while chloride gets protonated (lose an N-H bond and gain a H-Cl bond). The change in entropy for the reaction is positive, so it is favored at high temperature.
We might think that high temperature makes the H-Cl bond stronger and the N-H bond weaker. The bond dissociation energy for N-H is 390 kJ/mol while that of H-Cl is 432 kJ/mol. However, bond dissociation energies are not directly relevant because we are not transferring a hydrogen atom (making $\ce{NH3+ + HCl-}$). Instead, we are transferring a proton, so the pKa of ammonium and hydrogen chloride are more relevant but not directly either because this does not happen in aqueous solution.
The key is that the products are in the gas phase. If the products of the decomposition were ions and had opposite charges, they would not remain in the gas phase. In aqueous solution, you get $$\ce{NH4Cl(s) -> NH4+(aq) + Cl-(aq)}$$ because water is good at solvating ions. In the gas phase, the most stable combination of the atoms at hand is $\ce{HCl + NH3}$.

From your answer, what I inferred was that the increase in entropy is the best justification for the products obtained. Is that what you were trying to convey?

That's the rationale for gas phase products, even if the reaction is endothermic (losing all the ionic interactions). Which products are formed depends on the most stable species possible with that set of atoms. Unbound atoms have even higher entropy, and a plasma would be even better in terms of entropy, but its formation is highly endothermic.

Here we have a protolytic reaction like
H2O + HCl <=> H3O+ + Cl-
NH3 + HCl <=> NH4+ + Cl-
There no other products possible.

Sure other things are possible equations. For instance $$\ce{H2O + HCl(aq) ->[\Delta] H2O ^ + HCl ^}$$

This is only change from liquid to gaseous phase, no chemical reaction.


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Diazo Coupling reaction with para-substituted phenol?
Usually Diazo coupling occurs at para-position unless the para-position is occupied, in which case coupling occurs at ortho-position.
While solving questions I found in both the cases as shown, the coupling occurred at ortho-position of the $\ce{-OH}$ group:

So in case of phenol, is it possible that in all cases the coupling will occur at ortho-position with respect to the $\ce{-OH}$ group in phenol, irrespective of whether the other group has stronger +M Effect than $\ce{-OH}$?
I am not sure but I think this is happening because of Hydrogen bonding as shown below. But in that case it should also happen with the $\ce{-NH2}$ group. Is this the reason or there is some other?

If there’s hydroxide ion there, then the phenol will be deprotonated.

@orthocresol True ,  the Books Says Appreciable amount of Phenol will be converted in phenoxide ion . But Still the question and options all have -OH .

Don't get so caught up about what the question writes... You put in a phenol into the reaction, it gets converted into phenoxide, that's all there is to it. Just because somebody drew a phenol on a piece of paper doesn't mean that it can't be deprotonated.

I'm not sure what you're trying suggest with that hydrogen bond. This is probably an exergonic reaction, in which case, the Hammond Postulate would suggest an early transition state where the form of the product has little impact on the rate of the reaction...

@orthocresol: comparing the influence of steric hindrance, strength of activating resonance, and hydrogen-bonding stability, I concluded that(say the position ortho to the -OH is 1 and the other 2) for -OCH3, position 1 is favoured twice and hydrogen-bonding isn't a factor(so no ambiguity on the answer there), and for -NH2, all factors except steric hindrance favour position 2. Could you explain how the steric hindrance factor alone makes 1 more favorable? And while I appreciate the rigor in Zhe's comment, neither me nor the OP(at the time of posting) require that level of Chem applied here.


text:
Chugaev and Selenoxide Elimination giving different products
In class we are going over intramolecular elimination. In this question I will provide two similar reactants that provide a different mixture of products.
The chugaev elimination gives a 50/50 mixture of methylcyclohexene isomers while selenoxide elimination yields only one major product, but I do not understand why this is happening.
From what I can see, there is only one syn B-hydrogen available in both reactions. My professor went on to explain the significance of comparing the two reactions, saying that selenoxide elimination does not use heat while chugaev elimination does, causing a ring flip in one but not the other, which is understandable to me.

• The selenoxide reaction has a low activation energy proceeding via a syn-elimination giving only the less substituted alkene. The Chugaev requires temperatures of 200oC by a syn-elimination in acyclic systems. Although some syn-elimination may occur, the appearance of the trisubstituted alkene argues for an E1 process with this molecule. Formation of the trisubstituted alkene by a concerted syn-elimination would give rise to an E-methylcyclohexene (trans-cyclohexene) through a transition state high in energy. The foregoing implies kinetically controlled reactions and not equilibration.


text:
Why do surfactants lower the surface tension
Everybody knows that detergents, or generally amphiphilic substances, lower the surface tension. Of water, usually.
I wonder if that's true for any polar solvent (most likely), i.e. will the same detergents work in, say, ethanol or liquid ammonia?
What happens in a nonpolar solvent? The detergents aggregate on the surface there too, but now with the polar end sticking out. Does the surface tension go up, or down?
And what exactly is the reason for the effect on water anyway, quantitatively? Imagine a surface of water which is covered by exactly a single layer of surfactant molecules, e.g. in a Langmuir trough. Now go to half or double that amount. Is there a discontinuity in the change of surface tension? If yes or no, why?
(sorry for the jumble of questions, I'm not quite sure from which angle to tackle this. The effects seems quite obvious, but I'm not sure it is. My real question is how to explain the effect on a molecular level.)

related though typically for solids: isoelectric point

You might be interested in these: 1) https://physics.stackexchange.com/questions/150836/why-is-surface-tension-parallel-to-the-interface 2) https://research.utwente.nl/en/publications/why-is-surface-tension-a-force-parallel-to-the-interface and 3) https://michaelberryphysics.files.wordpress.com/2013/07/berry018.pdf

We usually say that at an interface, molecules will orient themselves at a "preferred" position, but we usually ignore that such a "preferred position" can actually be their "least unfavorable" position.
Energetically speaking, the interaction of a lipid chain with a gas is less unfavorable than interaction of a polar group with gas, because interactions from a lipid chain is mainly based on van der Waals interactions which are quite weak compared to polar (i.e. electromagnetic) interactions with a polar group. Also, they cannot make stabilizing bonds with the medium, i.e. water usually.
It must also be noted that the surface tension is linked to the interactions of molecules at the interface.
So "why do surfactants lower the surface tension"?
If you add some "surfactant", aka "tensioactive" agents, in water, its "least unfavorable" position would be at the interface between water and the other medium (usually gas). It will bring down the cohesive energy at the surface, compared to water alone, because it relies on van der Walls interaction. And that is why it lowers surface tension.

OK, so what's you prediction for soap on an oil drop? Surface tension up or down? I say up, because the polar ends have stronger interaction, so their cohesive energy is higher, and it's harder  to stretch them over a larger surface.

Here is the Answer from my friend's Ph.D. Thesis..hope it will help.
When two immiscible liquid phases are in contact with each other the molecules at the interface
experience an imbalance of various adhesive and cohesive forces. This leads to an accumulation of free
energy at the interface. This excess energy is called surface free energy and can be quantified by a
measurement of the energy required to increase the surface area of the interface by a unit amount. This
surface energy per unit area can be alternatively described as “interfacial tension” (IFT), which is a
force/length measurement. This force tends to minimize the area of the surface, thus explaining why for
example liquid drops and air bubbles are round. The common units for interfacial tension (IFT) are
dynes/cm or mN/m. [1]
A surfactant is a substance that, when present at low concentration in a system, has the property of
adsorbing onto the interfaces of the system and of altering to a marked degree the interfacial free
energy. A typical liquid-liquid immiscible system comprises of a polar (ex. aqueous) and non-polar (ex.
organic oil) component. Surfactants have a characteristic molecular structure consisting of a structural
group that has strong attraction for the polar solvent, known as a hydrophilic group, together with a
group that has a strong attraction for the non-polar solvent, called the lyophilic group. This is known as an
amphipathic structure. When dissolved in a two-phase liquid-liquid system such as oil-water, the
surfactant reorients itself by adsorbing to the oil-water interface, such that its hydrophilic group
interacts with the water phase while its lyophilic group interacts with the oil phase. This results in a net
decrease in interfacial free energy.


text:
What is the mechanism of platinum as catalyst poisons on alumina?
I'm wondering if anyone knows by what mechanism is platinum (or other noble metals) catalyst on alumina substrate poisoned?
Presently, I'm using platinum on alumina to decompose a hydrogen peroxide / $
\ce{KNO3}$ solution and have found the catalyst material poisoned.  I'm wondering if the poisoning is due to the $\ce{KNO3}$ or if my peroxide is contaminated.
I can imagine $\ce{KNO3}$ crystallizing and thereby occluding the active sites on the alumina but I'm not a chemist, so I've come here looking for better insight.


text:
Can a stoichiometric mixture of oxygen and methane exist as a liquid at standard pressure and some (low) temperature?
This answer to the question Pre-mixing cryogenic fuels and using only one fuel tank written by a non-chemist (me) begins with:

At STP:

LOX's boiling point is 90.19 K
Methane's freezing point is  90.7 K

This does not a priori prove that a solution of the two can not exist. However it does mean that they can not be handled as liquids at the same temperature, making mixing the two more difficult.
We know that liquid air exists which shows that LOX and LN2 can mix together. But methane is an organic molecules and we know that heavier $\text{C}_n \text{H}_{2n+2}$ hydrocarbons include oils and waxes don't like to dissolve in non-organic solvents.

A stoichiometric mixture of oxygen and methane would be 2:1 molar:
$$\ce{ 2O2 + CH4 -> CO2 + 2H2O }$$
Though the two can not be conveniently maintained as liquids at the same temperature, can a stoichiometric mixture of the two exist as a liquid at some (low) temperature and standard pressure?

I am not sure if both the solutions and mixtures tags apply here.

With an appropriate heat of mixing, the solution might well be possible. Not sure if there is literature on this mixture, but if I have time I’ll give it a whirl...

@JonCuster I hope "give it a whirl" doesn't mean you're going to try to mix the two!

just trying to whip it into a nice froth for my coffee in the morning...

Is methane "heavier" (more dense) than $O_2$ etc at same low T? Why do I doubt this? This is not a primary source but suggests the opposite? Methane is certainly more polarizable but Mw is lower. My guess is the component with the higher melting point (methane) is perfectly soluble in the liquid. The entropy of mixing is probably enough to drive miscibility. As the answer suggests, they form ideal solutions at sufficiently low T.

There's a NASA report that looks into this: "ON THE SOLUBILITIES AND RATES OF SOLUTION OF GASES IN LIQUID METHANE", Hibbard and Evans, 1968 and concludes that such mixtures are possible.
Starting on page 8:

Figure 5(a) presents the curves for oxygen, argon, carbon monoxide,
  and nitrogen. Also shown are the two experimental values for nitrogen.
  Agreement is excellent at 99.83K and good at 110.9K. The curves for
  these gases show that solubility should decrease with increasing
  temperature and the nitrogen data confirm this. This figure shows the
  mole fraction solubility of oxygen to be 1.0 at 90K. This means that
  oxygen, which has a normal boiling temperature of 90.1K would
  continuously condense in, and be miscible in all proportions, with
  liquid methane at 90K. This is confirmed by reference 11 where, in a
  study of the solubility of methane in liquid oxygen, it was concluded
  that these formed a near-ideal solution at -297 F (90K)

(emphasis added) Reference 11 mentioned in there is "Hydrocarbon-Oxygen Systems Solubility", McKinley and Wang, 1960 (unfortunately paywalled) which also has interesting discussion of the stability (i.e. presence or absence of a tendency to explode) of various mixtures.
Figure 5 is reproduced below. Note how the solubility of oxygen rises rapidly as temperature drops.

Bingo! What a great find, thank you! I think you can (should) also post an answer at the linked question as well.

Pre-mixing fuel and oxygen looks like a Very Bad Idea. Any spark will ignite it, as opposed to fuel stored separately. How stable is such a mixture? How likely are external events which would ignite it, like sparks? Could it be managed technically, in a rocket; e.g. can one ensure that the combustion does not back-propagate through the pumps etc.? (Should I ask a question, perhaps in the space exploration department?)

@PeterA.Schneider have a look at all the answers to the question linked in the first sentence of the question above.

On the other side of the phase diagram is https://www.sciencedirect.com/science/article/pii/S0011227567800432/pdfft?md5=1976ed6238da8fe67a253815410e4d31&pid=1-s2.0-S0011227567800432-main.pdf (Hodges & Burch Cryogenics 7 112-3 (1967) where the looked at the distribution of methane between liquid and vapor phases of oxygen. A quote: "the very high solubility of methane in liquid oxygen precluded the possibility of methane precipitating from liquid oxygen". So, the would seem to mix quite nicely. Whether that is a good idea is left to the imagination (and hopefully left right there).

To add to the Bob's excellent answer (and expand a bit on my comment there), I've found two other potentially interesting papers to peruse.
The first is R.J. Hodges and R.J. Burch, Cryogenics 7 112-113 (1967), titled "The equilibrium distribution of methane between the liquid and vapour phases of oxygen". They note a "very high solubility of methane in liquid oxygen", with a equilibrium distribution coefficient at 93.15K (1.348 atm of O2) of 3.356 (so about 3x more methane in the liquid than in the gas phase). However, they kept the concentration of methane below the lower explosive limit so that it would not, well, explode.
Of perhaps more interest to the OP is R.L. Every and J.O. Thieme, Journal of Spacecraft and Rockets 2(5) 787-789 (1965) titled "Liquid oxygen and liquid methane mixtures as rocket monopropellants". In the introduction the authors note that previous work shows that the liquids are miscible in all proportions above 90K. Their tests show a specific impulse of almost 300 sec, and an exhaust velocity near 6000 fps.
In addition, since "shock sensitivities were reported" in the earlier work, they did some, well, vaguely disturbing experiments (hey, it was the 60's). "Tests were conducted to determine whether violent stirring or agitation, as found in an impeller-type pump, would detonate the mixture", although they didn't have any explosions there. They then went on to quantify the impact sensitivity, dropping a weight onto a stainless steel beaker of different mixtures from increasing heights until an explosion occurred. Impact sensitivities of 20-60 ft-lb were discovered. They also looked at adiabatic compression as a cause of explosion and found the liquid oxygen/methane mixtures were safer than some other things like nitromethane (which really isn't that comforting).
So, it would appear that the idea of using the liquid mixture as a monopropellant isn't totally off the wall, but please do it somewhere far from me...

Those were the days. ;-) This is really interesting info! You may want to also write an answer to Pre-mixing cryogenic fuels and using only one fuel tank while you're at it.

Just because it's off the wall doesn't mean the early rocket pioneers wouldn't do it.  Somewhere out there is a paper describing experimental tests of the feasibility of nitroglycerine as a monopropellant, and there was a general consensus that chlorine trifluoride would make an excellent oxidizer.

Shock sensitivities?  As in your rocket detonates when your ullage motors push the fuel to the bottom of the tank?


text:
Effect of spontaneous transmutation of atom in crystal
What happens to a compound when one of its atoms changes via spontaneous nuclear reaction to an atom with very different bonding? For example, beryllium-7 turns into lithium-7 with a half-life of 53 days. After 53 days half the divalent Beryllium in a crystal of BeF2 would be monovalent Lithium ions. Half of the flourine atome formerly bonded to the former Beryllium atoms would be "floating free", not something Fluorine is wont to do. What would actually happen? Would there be a fluorine ion trapped in the crystal? A flourine atom? Some kind of radical?
Plainly energies available from  the nuclear reaction are orders of magnitude larger than anything that would ever be available in chemical bond energies, so even crazy unstable species have plenty of energy to form.

Good question, this idea has bothered me for years. Still nobody ever answered it definitely.

Defects in crystals are quite well known. ‘Incredibly unstable’ isn’t quite what happens.

An extra unbonded fluorine in the middle of the crystal is actually pretty stable, much like a serial killer in a high-security prison cell.

@IvanNeretin - not quite the analogy that I would have used, but yes it quite likely finds a nice interstitial position.

@Chalo there is a brief discussion about this effect for Einsteinium compounds in section 12.6 here

For clarification: after 53 days, fully half of the crystal will contain these defects (half of the divalent beryllium will have turned to monovalent Lithium). This is not one serial killer in a cell. This is more like half the general population being serial killers.


text:
Stoichiometry of the redox reaction with crystallohydrate

Consider the following balanced equation:
$\ce{2 Al(s) + 3 CuCl2 . 2 H2O (aq) → 3 Cu (s) + 2 AlCl3 (aq) + 6 H2O (l)}$
If we began the experiment with 0.99 g of Al, according to the stoichiometry of the reaction how much $\ce{CuCl2.2 H2O}$ should be used to complete the reaction without either reactant being in excess?
Answer the following using the correct number of significant figures.

I cannot for the life of me solve this problem.  Should I be finding the mole of $\ce{Al}$ and $\ce{CuCl2.2H2O}$ then find the difference? Any help would be appreciated.

HINT - You need convert to convert grams into something that is proportional to the stoichiometric values.

Of all arithmetic operations, subtraction is the only one you won't be needing here.

Mass of $\ce{CuCl2.2H2O}$ needed can be found from its amount $n(\ce{CuCl2.2H2O})$ and molar mass $M(\ce{CuCl2.2H2O})$:
$$m(\ce{CuCl2.2H2O}) = n(\ce{CuCl2.2H2O})\cdot M(\ce{CuCl2.2H2O}) \label{eqn:1}\tag{1}$$
The unknown amount of copper(II) chloride dihydrate $n(\ce{CuCl2.2H2O})$ can be found from the reaction stoichiometry:
$$n(\ce{CuCl2.2H2O}) = \frac{3\cdot n(\ce{Al})}{2} = \frac{3\cdot m(\ce{Al})}{2\cdot M(\ce{Al})} \label{eqn:2}\tag{2}$$
Finally, plugging $\eqref{eqn:2}$ into $\eqref{eqn:1}$:
$$
\begin{align}
m(\ce{CuCl2.2H2O}) &= \frac{3\cdot M(\ce{CuCl2.2H2O})\cdot m(\ce{Al})}{2\cdot M(\ce{Al})} \\
  &= \frac{3\cdot\pu{170.5 g mol-1}\cdot\pu{0.99 g}}{2\cdot\pu{26.98 g mol-1}} \\
  &\approx \pu{9.40 g}
\end{align}
$$
Hence, you need $\pu{9.40 g}$ of $\ce{CuCl2.2H2O}$.

You made a bad rounding error. With a modern calculator you really should carry at least two extra digits in intermediate calculations and only round the final result to the appropriate number of significant figures.

@MaxW is right. What I would suggest is not to do any intermediate calculations at all, unless you are explicitly asked so. Solve the problem algebraically first, plug in the numbers at the very last step. Also, note that "number of moles" is a colloquial expression for the amount of substance and should be avoided. It's like replacing mass and voltage with  "number of kilograms" number "number of volts", respectively.

What "now"? There is still "No. of moles". Additionally to my last comment, note that algebraically the amount is usually denoted as $n(\ce{A_xB_y})$ — please don't write an essay in math formulas. If you want, I may try to rewrite your answer to better meet the standard notations.

I've rewritten your answer from ground up preserving your logic (which is correct), addressing the issues regarding the notations, units and significant figures. Simply put, this is how I'd answer this. If you are unhappy with this, feel free to revert my edit; otherwise have a closer look at the formatting and used notations so that you could improve your future posts:)

Thanks for your help folks.  I finally got the answer (now I can sleep at night).  The one problem I see with your solution is that your molar mass for CuCl_2*2H_20 is off.  I got 170.48 as the molar mass.  My answer ended up being 9.38 which I rounded to 9.4.


text:
Ionic equilibrium between multiple ions
This is a question from IIT-JEE 1993:

For the reaction
$$\ce{[Ag(CN)2]^- -> Ag^+ + 2CN^-}$$
the equilibrium constant at $\pu{298 K}$ is $\pu{4e-10}$. The silver ion concentration in a solution which was originally $\pu{0.10 M}$ in $\ce{KCN}$ and $\pu{0.03 M}$ in silver nitrate is?

This is what I've done:
$$
\begin{align}
[\ce{K^+}] &= 0.1 \\
[\ce{CN-}] &= 0.1 - 2x \\
[\ce{Ag+}] &= 0.03 - x \\
[\ce{NO3-}] &= 0.03
\end{align}
$$
$$\frac{(0.03-x)(0.1-2x)^2}{x} = 4\cdot 10^{-10}$$
But this gives $x = 0.03$, hence, $[\ce{Ag+}] \approx 0$.
The answer key says $[\ce{Ag+}] \approx 7.5\cdot 10^{-18}$. Where am I going wrong?

Wrong? To me, $10^{-18}\approx0$ looks about right.

In most of these type problems chemists look for ways to simply the mathematical expression by using approximations based on a knowledge of the chemistry. So $0 \approx 10^{-18} \ll 0.03$

@MaxW This was an MCQ question. There were 3 options in the order of $10^{-15} to 10^{-18}$. And the 4th was None of the Above. So, when I got 0, I went for the 4th one and was wrong

The gist is that because of the significant figures you assume that the reaction essentially goes to completion. So $\ce{Ag(CN)2^- \approx 0.0300}$ and $\ce{CN^- \approx 0.0400}$. Then you solve for $\ce{[Ag^+]}$.

That does indeed yield $7.5\times10^{-9}$

In most of these type problems chemists look for ways to simply the mathematical expression by using approximations based on a knowledge of the chemistry. The gist is that because of the significant figures you assume that the reaction essentially goes to completion. So $\ce{Ag(CN)_2^− \approx 0.0300}$ and $\ce{CN− \approx 0.0400}$. Then you solve for [Ag+]. So:
$4\times10^{-10} = \dfrac{\ce{[Ag^+][CN^-]^2}}{\ce{[Ag(CN)_2^-]}}$
or rearranging for $\ce{[Ag^+]}$
$\ce{Ag^+} = 4\times10^{-10} \dfrac{{\ce{[Ag(CN)_2^-]}}}{\ce{[CN^-]^2}} = 4\times10^{-10} \dfrac{0.03}{0.04^2} = 7.5\times10^{-9} $
Checking $7.5\times10^{-9} \ll 0.03$ so the assumption that the reaction essentially went to completion is valid. in other words:
$0.0300 - 7.5\times10^{-9} \approx 0.300$

The $\ce{[Ag+]}$ should be $7.5 \times 10^{-8}$, I believe.

@MathewMahindaratne - Thank you for the correction. There was a mistake. I had the equilibrium constant as $4\times 10^{-9}$ and it should have been  $4\times 10^{-10}$. With the correct equilibrium constant the value is $7.5\times10^{-9}$.

You are going wrong in your choice of $x$.
You should have recognized in advance that $\ce{CN-}$ is in excess, and hence $\ce{Ag+}$ would go to the complex almost completely, so the equilibrium $\ce{[Ag+]}$ is going to be awfully low, while the concentration of $\ce{[Ag(CN)2]-}$ (which you have unfortunately defined as $x$) is going to be really very close to 0.03 and even closer than that, like 0.29999 and more nines until your keyboard runs out of nines and stops working. No wonder that when you are doing your calculations with 3 (or 4, or any sensible number of) significant figures, it just reads as 0.03.
Throw everything away, define $x=\ce{[Ag+]}$ and start from the scratch.

I wouldn't say that the OP really setup $x$ wrong but rather the OP didn't understand the implication  of $x \approx 0.03$.

Well, it wouldn't matter what you define as $x$, if not for the strict adherence to the significant figures rules. It is the combination of the two that makes it wrong.

So, in the Kc equation, $\ce{[Ag^+]} = x$ & $\ce{CN^-} = 0.04 -2x$ & $\ce{[Ag(CN)2]^-} = 0.03 - x$ ? Doing that gives $7.5 * 10^{-9}$. What's causing the error in order of answer

It is $0.04\color{red}{+}2x$ instead of "-" (not that it matters much). Well, now I believe you are doing it right, and if the answer key says otherwise, there must be a mistake in it.


text:
molecular orbital theory
i read in my textbook that the energy diagram of moleculal orbits is different for atoms nitrogen and below them.
my question is which one of the energy diagrams to use in heteronuclear compounds like CO?

I'm assuming you only looked at homonuclear molecules ($\ce{O2}$, $\ce{H2}$) so far. The atomic orbitals are of rather different energies in $\ce{C}$ and $\ce{O}$ just because the nuclear charge is different (consider the ionization energies). That is why I don't expect anyone to produce a decent orbital diagram without performing some quantitative computational chemistry first.

(In heteronuclear diatomic molecules, atomic orbitals only mix when the electronegativity values are similar.)
(While MOs for homonuclear diatomic molecules contain equal contributions from each interacting atomic orbital, MOs for heteronuclear diatomics contain different atomic orbital contributions.)
(Orbital interactions to produce bonding or antibonding orbitals in heteronuclear diatomics occur if there is sufficient overlap between atomic orbitals, as determined by their symmetries and similarity in orbital energies.)

It depends on how many electrons your compound has. In $\ce{O2}$ , there are 16 electrons.
Likewise all compounds with more than 16 electrons follow the $"\ce{O2}"$ diagram and those with less than 16 follow the other diagram.


text:
Why neglecting OH Ions from charge balance?
I have a problem, in which the concentration of $\ce{Ca^{2+}}$ ions in water should be related to the partial pressure of $\ce{CO2}$ (assuming the water is an open system and in equilibrium with atmosphere).
All the approaches I saw so far, started by looking at the charge ballance of this system:
$$\ce{2 [Ca^{2+}] + [H+]} =\ce{2[CO3^{2-}] + [HCO3-] + [HO-]}$$
This is totally logical. But they continue with the assumption that the terms $\ce{[CO3^{2-}]}$,$\ce{[HO-]}$, $\ce{[H+]}$ are insignificantly small so that you can work with:
$$\ce{2 [Ca^{2+}]} =\ce{[HCO3-]}$$
That's the point I don't get.
Assuming at the beginning there is only $\ce{CaCO3}$ and Water. Both dissociate and let's say $\ce{CO3^{2-}}$ immediatly combines with $\ce{H+}$ to $\ce{HCO3-}$, then it is reasonable that the contribution of $\ce{H+}$ and $\ce{CO3^{2-}}$ can be ignored. But what about the remaining $\ce{OH-}$? Stupidly written, I would get something like:
$$\ce{H2O + CaCO3 } =\ce{ HCO3- + Ca^{2+} + OH-}$$
So why can I also neglect the charge caused by $\ce{OH-}$ Ions? For each bicarbonate/calcium-ion produced I would also get a hydronium ion. why can this still be ignored?

Is this pure water to start with, or something like ocean water that is buffered by other species already in solution? In other words, is the pH determined by $\ce{CaCO3}$ and $\ce{CO2}$ or set to some value?

Please state the problem that you're trying to solve exactly as it was given to you. Pointless to be guessing as to what the problem actually is.

@Karsten Theis : He mentioned water in equilibrium with the aerial CO2, therefore CaCO3 dissolution with CO2 support, forming carbonate hardness Ca(HCO3)2

This is not totally unreasonable.
Calcium carbonate as solubility product of $K_{\mathrm{sp}} = 3.3\times 10^{-9}$ (source).
This means that if you're dissolving the solid, you'd expect a maximum calcium ion concentration of $\ce{[Ca^{2+}]} = 5.7\times 10^{-5}\ \mathrm{M}$. This is a 2.5 orders of magnitude larger than either hydronium or hydroxide concentration in neutral solution. You'd expect the carbonic acid to be mostly dissociated (at least 50%), so the bicarbonate concentration will be similar to the calcium ion concentration (maybe factor of 2 smaller).
The moment you step away from a neutral solution though, this analysis totally falls apart. You'd need more details on the assumptions of the model in that case.

My official Android client says at 2 places "Undefined control sequence \pu" (While otherwise displaying chemical formatting correctly)

@orthocresol I think \pu is valid for us, but may not render correctly in general.

@Poutnik Official Android client is great for receiving notifications and reading Stack Overflow. Other Stack Exchange websites are poorly supported (e.g. tables and arrays will also produce the error) and I would recommend to use your web browser instead. See Undefined control sequence \pu in official SE Android app; MathJax in the Android App for more technical details.

I see. I will have it on mind. Otherwise, the client is very convenient for reading. Well written.

It looks like the Android browser support(Chrome,Brave, both mobile and PC page versions) for MathJax is worse than in the Client. So browser on PC, or Android client.

@Poutnik Everything looks fine to me, both \pu{...} and \ce{...} are rendered nicely (Chrome @Android 8, Moto G5). I suppose you have to troubleshoot at your side (try enabling/disabling hardware acceleration, turn desktop site version on/off, update your system etc.). App version is broken, as expected.

@andselisk Well, it looks fine on browsers now, after your change, but it destroyed rendering in the client, that was originally OK. :-). But as you say, that was expected. Thanks for update. The new version seems more difficult to edit. I use Android 9.

@Poutnik No prob, the syntax is indeed more complex, but unfortunately one must just bite the bullet. As I said earlier, some features as tables and arrays will always cause problems in mobile app anyway, not only the \pu{...} command, so better avoid it whenever possible IMO.

In short, $[\ce{H+}]$, $[\ce{OH-}]$ and $[\ce{CO3^2-}]$ are kept low by ion recombination, honouring the dissociation constants of water and involved compounds, mentioned below.
The reaction mentioned in the question
$$\ce{H2O + CaCO3 v -> HCO3−+Ca^2+ + OH−}$$
can be reformulated as
$$\ce{CO3^2- + H2O <=> HCO3- + OH- } $$
As there is abundance of $\ce{H+}$, created by the dissoved aerial $\ce{CO2}$,
$\ce{OH-}$ and $\ce{CO3^2-}$ quickly recombine with other ions.
$$\begin{align}
\ce{OH- + H+ &<=>> H2O} \\
\ce{OH- + HCO3- &<=>> CO3^2 + H2O} \\
\ce{CO3^2- + H+ &<=>> HCO3-} \\
\ce{HCO3- + H+ &<=> H2CO3 <=> CO2 + H2O} \\
\end{align}$$
Note that pure water saturated by the aerial $\ce{CO2}$ has $\mathrm{p}H=5.65$. At the beginning of $\ce{CaCO3}$ dissolution, $\ce{[OH−]} $ is not allowed to be > $\pu{4.10^-9 mol / l)}$.
The aerial $\ce{CO2}$ presence in water shifts the dissolution equilibrium. the calcium concentration is several orders higher than it would be in the pure, carbon dixode free water.
The major reversible summary equation below is basis of the calcite and aragonite cave phenomena:
$$\ce{CO2 + H2O + CaCO3 <=> Ca^2+ + 2 HCO3^-}$$
Typical $c_\ce{Ca^2+}$ and $c_\ce{HCO3-}$ in natural water are in order of units of $\pu{mmol/l}$.
$c_\ce{Ca^2+} = \pu{1 mmol/l}$ is equivalent to $\pu{5.6 dGH}$ (dGH on Wikipedia).
$[\ce{H+}]$ and $[\ce{OH-}]$ are limited by the water dissociation constant.
$$[\ce{H+}][\ce{OH-}] = K_\mathrm{w} = 10^{-14} \quad(\text{at}~\pu{25 °C})$$
If we consider water $\mathrm{pH}$ in $6-8$ range, $[\ce{H+}]$ and $[\ce{OH-}]$ are below $\pu{1 μmol/l}$, over 1000 times less than $[\ce{HCO3-}]$.
For $\mathrm{pH} = 7$
$$\begin{align}
\frac{[\ce{CO3^2-}]}{[\ce{HCO3-}]} &= 0.000469 \\
\frac{[\ce{HCO3-}]}{[\ce{CO2}]} &= 4.25
\end{align}$$
So again, $\ce{[CO3^2-]}$ is in range of $\pu{1 μmol/l}$
The $\ce{CO2}$ related equilibrium constants from Wikipedia:
$$
\begin{align}
\frac{[\ce{H+}][\ce{CO3^2-}]}{[\ce{HCO3-}]} &= K_\mathrm{a2} = \pu{4.69e-11} \\
\frac{[\ce{H+}][\ce{HCO3-}]}{[\ce{H2CO3}]} &= K_\mathrm{a1} = \pu{2.5e-4} \\
\frac{[\ce{H2CO3}]}{[\ce{CO2}]} &= K_\mathrm{h} = \pu{1.7e-3}
\end{align}
$$

I think downvoting without a comment or an edit suggestion is not much constructive. Note that the equilibrium concentration of Ca+2 and HCO3- is much higher than the other responder noted,, as CO2 presence shifts the dissolution equilibrium.

I will comment without downvoting: This answer is hard to read because you are not typesetting the chemical equations. The argument is difficult to understand because it is more of a data collection. What would be helpful would be the concentration of all species in a saturated solution of $\ce{CaCO3}$ that has a vapor pressure of $\ce{CO2}$ equal to the partial pressure present in the atmosphere today.

Thanks for the comment. It is interesting it could be hard to read,but perhaps I am too used to ASCII based Usenet of the old days of the Internet. I will try to improve it.

@andselisk While learning, I have noticed  I should not use \ce for variable subscripts, because of italic convention. It is a pity, as \ce is more compact than that mathxx string. :-)

@Poutnik \ce{...} is for chemical formulas only (you can remember it mnemonically as chemical expression), using it otherwise breaks readability and may cause weird (aka unexpected) MathJax behavior. As for italics, have a look at the quick guide Which symbols are written in roman (upright) font and which are italicized?. I made many mistakes at the beginning, but after numerous corrections from the Mod team I finally got used to it:)


text:
How to quickly diffuse a dye through still water
I'm doing an engineering experiment soon and would like to be able to quickly dye water. The water is still (i.e. not moving) and I would prefer not to physically disturb the water as best I can.
What techniques could I employ to quickly diffuse a dye through water? I was thinking that electrophoresis could be a possibility. If so, what dyes (or what could I add to a dye) to help speed up the process.
Thanks in advance.

Do you have some way of suppressing convection? How do you intend to prevent "disturbing the water"?

By definition diffusion is a passive process. You have to describe in detail what kind of distribution you want in the end. Electrophoresis will not give a uniform (homogeneous) distribution and can cause all sorts of convective currents due to Joule heating. You'll need to describe the experimental setup in more detail.

"Not moving" Molecules are moving. That is why diffusion takes place. If the water is not moving then certainly the water molecules can't move out of the way to allow for diffusion of a solute.

@NightWriter we have a slushy mixture of water and ice. I want to be able to distinguish between the two later. The idea is to quickly dye the water and the freeze the entire mixture. I would like to keep the shape of the composition. That is what I mean by not "disturbing the water".

I see, I suppose if you are concerned about breaking fragile crystal structures then stirring is not what you want.  Also, adding a dye in significant concentration will alter freezing point possibly or be incorporated as inclusion impurities into the crystals etc etc, so you don't want to add it before hand. Electrophoresis might significantly heat up your sample and take a long time, plus set up concentration gradients etc. You can initially add a high MW dye (polymeric) that is not incorporated into ice, at the price of some colligative effect which you can control for.

If you want a homogeneous distribution of dye, you will have to stir the solution.
If you want to use electrophoresis, you need a charged dye.
Azo dyes used for hair coloring have colour and charge, for instance Basic Red 51 and Basic Yellow 87:

Of course, heat will increase the diffusion rate but will also disturb the water.

Thank you for the response. Homogeneity is not that important. I just want to get some dye into all of the water. As I responded above, we have a slushy mixture of water and ice. I want to be able to distinguish between the two later. The idea is to quickly dye the water and the freeze the entire mixture. I would like to keep the shape of the composition. I will look into the two dyes you mentioned.

You are welcome. If it has been useful to you, you might consider up voting and/or accepting the answer.


text:
Copper powder amount dissoved in water
How much copper powder dispersed into 2 liters of water will result in a 10 ppm concentration? I am not a chemist.

Adding copper powder to water won't result in a homogeneous solution and there is no concentration to speak of since copper solubility in water is negligible.

However if you manage to disperse it, it amounts to 0.00002 g. It is the term "dissolved" that isn't correct. Or it can be in ionic form. For the calculation, at this high dilution we consider the density of solution equal to that of water,  so calculate ppm is straightforward.

If you want 10 ppm copper in 2 liters of water, the best way to make it would be to start with some soluble copper salt, like $\ce{CuSO4.5H2O}$, make up a convenient concentration (like 1%), then dilute it to 10 ppm.
As was mentioned in the comments, copper metal is not noticeably soluble in water. When rainwater washing thru copper downspouts leaves a blue stain on concrete, the copper metal is dissolving because of tiny solubility of copper oxides or carbonates. It might be 10 ppm, or 1 ppm in a lot more water.


text:
Finding all Chemical Reactions Given Products and Reactants
Suppose only a set of chemical formulas is given. How can you find all mathematically possible chemical equations whose educts and products are only from this set?
Take e.g. the set $\{ \ce{C, H2, O2, N2, CO, CO2, H2O, NH3, NO, NO2, NO3}\}$.
Or consider e.g. the example in MathStackexchange: Finding all chemical equations (Linear Algebra).
To find all combinations and partial reactions between such given species by hand, is possibly incredibly tedious or near impossible for more complex chemical reaction systems.
I guess these methods are topic of Mathematical Chemistry, Computer Chemistry or Chemoinformatics.

Are you interested in the chemistry or the math of this? If it is the math, try to ask a question on StackExchange math that can be answered without knowing any chemistry. If you are interested in chemical reactions that actually happen, improve your current question to make clear what you are asking.

@KarstenTheis I am interested in the math more than the chemistry, yes. However, Floatzel98 asked the same question on StackExchange maths and their post was also put on hold because it was deemed not mathematical enough. If this can't be answer in chemistry or maths forums, where can it be answered? Thank you

@IV_ just because it was closed as unclear and clarified, does not mean it is in line with our guidelines. Unclear was the reason that got the most votes. This question is still too broad.

It isn't that broad of a question, it is just that everyone assumes stoichiometry is simple and thinks of reactions only in terms of the equilibrium constant and as a result, doesn't know the great breadth of math behind it. You should be reading this book by Smith & Missen. And so should everyone who thinks this question is too broad. If I remember correctly, Chapter 9 will be of great service to you.

The purely combinatorial methods are a first step in modeling chemical reactions. They alone however cannot consider the possibility of the chemical reactions and the chemical stability of the reaction products.
$\ $
1.)
You can build all combinations of at least two formulas, all from the original set of given chemical formulas, and treat this as your set of given formulas. But you can also treat the whole original set of given chemical formulas.
I demonstrate the method with the first example given in the question:
$$\{ \ce{C, H2, N2, O2, CO, CO2, H2O, NH3, NO, NO2, NO3}\}$$
We set up one chemical equation with all chemical formulas that are in the given set. Let the stoichiometric factors in the chemical equation be denoted by $\nu$. Because we don't know which of the given substances will be educt and which of them will be product, we write the chemical equation without reaction arrow. Instead, a negative stoichiometric factor will later mark an educt, a positive a product:
$$\nu_1\ce{C}+\nu_2\ce{H2}+\nu_3\ce{N2}+\nu_4\ce{O2}+\nu_5\ce{CO}+\nu_6\ce{CO2}+\nu_7\ce{H2O}+\nu_8\ce{NH3}+\nu_9\ce{NO}+\nu_{10}\ce{NO2}+\nu_{11}\ce{NO3}$$
We have to obey one of the first fundamental laws of chemistry: the Law of multiple proportions. To fulfill this law, the total sum of each chemical element in our chemical equation has to be $0$. And if ions are among our given chemical formulas, the total sum of all electric charges has to be $0$ also. But no electric charges are involved in our given example.
We will now build a mathematical model of our chemical problem.
For each of the given chemical elements, we have to set up its balance equation from our chemical equation above:
$\ce{C}:\ \ \ \nu_1+\nu_5+\nu_6=0$
$\ce{H}:\ \ \ 2\nu_2+2\nu_7+3\nu_8=0$
$\ce{N}:\ \ \ 2\nu_3+\nu_8+\nu_9+\nu_{10}+\nu_{11}=0$
$\ce{O}:\ \ \ 2\nu_4+\nu_5+2\nu_6+\nu_7+\nu_9+2\nu_{10}+3\nu_{11}=0$
This is a linear equation system. In the general case, too, a linear equation system results. It consists of the coefficients (the numbers above) and the stoichiometric factors $\nu_i$, which are sought. In our example, the equation system has 4 equations (the number of different chemical elements and the electric charge) and 11 unknowns (the number of chemical formulas in the given set).
Linear Algebra says how a linear equation system can be handled. We use here the matrix presentation.
Each of the 11 given chemical formulas is presented by a column vector which contains the frequency of occurence of each chemical element in the chemical formula in a prescribed order. Each of the 4 given chemical elements is presented by a row vector which contains the frequency of occurence of the chemical element in the chemical formulas in a prescribed order.
All 11 column vectors or all 4 raw vectors of the equation system are combined to give the coefficient matrix $A$. Our 11 wanted stoichiometric coefficients build the solution vector $x$, which is sought. The matrix representation of our linear equation system is than:
$$A\cdot x=\emptyset,$$
wherein $\emptyset$ is the zero column vector with 4 rows. It is written out:
$$\left(
\begin{array}{}
1&0&0&0&1&1&0&0&0&0&0\\
0&2&0&0&0&0&2&3&0&0&0\\
0&0&2&0&0&0&0&1&1&1&1\\
0&0&0&2&1&2&1&0&1&2&3
\end{array}
\right)
\cdot
\left(
\begin{array}{}
\nu_1\\\nu_2\\\nu_3\\\nu_4\\\nu_5\\\nu_6\\\nu_7\\\nu_8\\\nu_9\\\nu_{10}\\\nu_{11}
\end{array}
\right)
=\left(\begin{array}{}0\\0\\0\\0\end{array}\right)$$
The solution vector $x$ can be found by solving the linear equation system by methods of Linear Algebra.
The solution vector of our example is:
$$x=\left(
\begin{array}{c}
2\nu_4+\nu_6+\nu_7+\nu_9+2\nu_{10}+3\nu_{11}\\-\nu_7+3\nu_3+\frac{3}{2}\nu_9+\frac{3}{2}\nu_{10}+\frac{3}{2}\nu_{11}\\\nu_3\\\nu_4\\-2\nu_4-2\nu_6-\nu_7-\nu_9-2\nu_{10}-3\nu_{11}\\\nu_6\\\nu_7\\-2\nu_3-\nu_9-\nu_{10}-\nu_{11}\\\nu_9\\\nu_{10}\\\nu_{11}
\end{array}
\right)$$
The different combinatorially possible chemical equations are obtained by choosing suitable values for the free variables $\nu_3,\nu_4,\nu_6,\nu_7,\nu_9,\nu_{10},\nu_{11}$.
$\ $
2.)
Further investigations could be made by searching the solutions that don't contain any other solution. That are the minimal solutions, that means the solutions that are linearly independent of the other solutions.
I found 663 chemical equations and 83 minimal solutions for the example above.
Generating only the minimal solutions in polynomial time needs particular algorithms:
Schay, G.; Pethö, Á.: Über die mathematischen Grundlagen der Stöchiometrie. Acta Chim. Acad. Sci. Hung. 32 (1962) 59-67
Pethö, Á.: Zur Theorie der Stöchiometrie Chemischer Reaktionssysteme. Wissenschaftl. Zeitschr. 6 (1964) 13-15
Pethö, Á.: Algebraic treatment of a class of chemical reactions in stoichiometry. Acta Chim. Acad. Sci. Hung. 54 (1967) 107-117
Pethö, Á.: On a class of solutions of algebraic homogeneous linear equations. Acta Math. Acad. Sci. Hung. 18 (1967) 19-23
Pethö, Á.: Kémiai reakciók egy osztályának algebrai elemzése. Magyar Kémiai Folyóirat 74 (1968) 488-491
Kumar, S.; Pethö, Á: Note on a combinatorial problem for the stoichiometry of chemical reactions. Int. Chem. Eng. 25 (1985) 767-769
Pethö, Á.: The linear relationship between stoichiometry and dimensional analysis. Chem. Eng. Technol. 13 (1990) 328-332
Szalkai, I.: Generating minimal reactions in stoichiometry using linear algebra. Hung. J. Ind. Chem. 19 (1991) 289-292
with a software code at the end of the article
Pethö, Á.: Mathematical discussion of the application of Hess‘s law. Hung. J. Ind. Chem. 21 (1993) 35-38
Pethö, Á.: Further remarks on the linear relationship between stoichiometry and dimensional analysis. Chem. Eng. Techn. Chem. Eng. Tech. 17 (1994) (1) 47-49
Pethö, Á.: Further remarks on the analogy between stoichiometry and dimensional analysis: The valuation operation. Hung. J. Ind. Chem. 23 (1995) 229-231
Laflamme, C.; Szalkai, I.: Counting simplexes in $R^n$. Hung. J. Ind. Chem. 23 (1995) 237-240
Laflamme, C.; and I. Szalkai, I.: Counting simplexes in $R^3$ Electron. J. Combin. 5 (1998) (1) #R40 11. Printed version in: J. Combin. 5 (1998) 597-607
Szalkai, I.: Handling multicomponent systems in $R^n$ I Theoretical results. J. Math. Chem. 25 (1999) 31-46
Szalkai, I.: On valuation operators in stoichiometry and in reaction syntheses. J. Math. Chem. 27 (2000) 377-386
Szalkai, I.: A new general algorithmic method in reaction syntheses using linear algebra. J. Math. Chem. 28 (2000) 1-34
Szalkai, B.; Szalkai, I.: Counting minimal reactions with specific conditions in $R^4$ J. Math. Chem. 49 (2011) 1071-1085
Szalkai, I.; Dósa, G; Tuza, Z.; Szalkai, B.: On minimal solutions of systems of linear equations with applications. Miskolc Mathematical Notes 13 (2012) (2) 529-541
Szalkai, B.; Szalkai, I.: Simplexes and their applications - A short survey. Miskolc Math. Notes 14 (2013) (1) 279-290
Szalkai, I.; Tuza, Z.: Minimum Number of Affine Simplexes of Given Dimension. Discr. Appl. Math. 180 (2015) 141-149
Szalkai, I.: Reakciómechanizmusok algoritmikus és matematikai vizsgálata (Algorithmic and Mathematical Examination of Reaction Mechanisms). PhD thesis, University of Pannonia, Veszprém, Hungary, 2014
Szalkai, I: An algorithmical and mathematical investigation of reactions. English extract of the author's PhD thesis, 2014
See the examples and references therein.
Tóth , J.: Reaction Kinetics: Exercises, Programs and Theorems. Springer, 2018
$ $
3.)
The answer above treats overall reactions. The underlying elementary reaction steps (elementary reactions) can be generated by working with the Ugi-Dugundji model: The chemical formulas are presented as graphs together with their graph matrices (Bond-Electron matrices (BE-matrices) and Reaction matrices (R-matrices)). The single elementary reaction steps will be generated successively by stepwise shifts of electrons or bonds in the set of given chemical formulas by stepwise shifting of single edges in the reaction matrix.
Ugi, I.; Gillespie, P.; Gillespie, C.: Chemistry, a finite metric topology - synthetic planning, an exercise in algebra. Trans. New York Acad. Sci. II 34 (1972) 416-432
Dugundji, J.; Ugi, I.: An algebraic model of constitutional chemistry as a basis for chemical computer programs. Top. Curr. Chem. 39 (1973) 19-64
Dugundji, J.; Gillespie, P. D.; Marquarding, D.; Ugi, I.; Ramirez, F.: Metric spaces and graphs representing the logical structure of chemistry. in: Chemical applications of graph theory. Balaban, A. T. (Ed.), Academic Press, London 1976, 107-174
Brandt, J.; Friedrich, J.; Gasteiger, J.; Jochum, C.; Wolfgang Schubert, W.; Ugi, I.: Computer programs for the deductive solution of chemical problems on the basis of a mathematical model of chemistry. in: Wipke, W. T.; Howe, W. J.: Computer-Assisted Organic Synthesis. American Chemical Society 1977, chapter 2, 33-59
Ugi, I.; Brandt, J.; Friedrich, J.; Gasteiger, J.; Jochum, C.; Lemmen, P.; Wolfgang Schubert, W.: The deductive solution of chemical problems by computer programs on the basis of a mathematical model of chemistry. Pure Appl. Chem. 50 (1978) 1303-1318
Ugi, I.; Brandt, J.; Friedrich, J.; Gasteiger, J.; Jochum, C.; Lemmen, P.; Schubert, W.: Computer Programs for the Deductive Solution of Chemical Problems on the Basis of a Mathematical Model of Chemistry. Pure Appl. Chem. 50 (1978) (11-12) 1303-1318
Behnke, C.; Bargon, J.: Computer-assisted topological analysis and completion of chemical reactions. J. Chem. Inf. Comput. Sci. 30 (1990) (3) 228-237
Ugi, I; Dengler, A.: The algebraic and graph theoretical completion of truncated reaction equations. J. Math. Chem. 9 (1992) (1) 1-10

If all the elements are part of the set of species, the solution is trivial: Write down how the compounds are made from the elements, giving $n$ balanced equations for $n$ compounds. Any linear combination of these will also be a valid balanced chemical equation (where you can cancel anything that appears both as reactant and product to arrive at e.g. $\ce{CO2 + NO -> CO + NO2}$).

Here is the same solution vector with free variables matching the compounds, and coefficients for the elements showing the stoichiometry of the compounds: $$x=\left( \begin{array}{c} \nu_5 + \nu_6\ \frac{1}{2}(2 \nu_7 + 3 \nu_8)\ \frac{1}{2}(\nu_8 + \nu_9 + \nu_{10} + \nu_{11})\ \frac{1}{2}(\nu_5 + 2\nu_6 + \nu_7 + \nu_9 + 2\nu_{10} + 3\nu_{11})\ -\nu_5\  -\nu_6\  -\nu_7 \  -\nu_8 \  -\nu_9 \ -\nu_{10}\  -\nu_{11} \end{array} \right)$$

Only the chemical formulas of products and reactants of a chemical system are given. What mathematical methods can be used to determine all combinatorially possible chemical reactions between species of this reaction system?

Let's take the example mentioned in the question, plus dihydrogen to make a point:
$$\ce{CO2, C7H10N, H2O, O2, H2 and NO2}$$
We can represent product and reactant specides by vectors where the components of the vector represent the number of H, C, N and O in each. Thus,
(0,1,0,2) is $\ce{CO2}$,
(10,7,1,0) is $\ce{C7H10N}$,
(2,0,0,1) is $\ce{H2O}$,
(0,0,0,2) is $\ce{O2}$, and
(0,0,1,2) is $\ce{NO2}$.
A balanced chemical equation is a linear combination of these vectors with a sum of zero (negative coefficients indicate reactants, positive coefficients products). How many different balanced chemical equations exist for this set of molecules?
In a set of species that could potentially result in a balanced equation, every element has to appear at least twice. If I start with $\ce{C7H10N}$, it is clear I also need $\ce{CO2}$ (for C), $\ce{NO2}$ (for N) and either $\ce{H2O or H2}$ for H. Because we have to use $\ce{NO2 and CO2}$, we also need something with O. The two possible sets are:
$$\ce{C7H10N, CO2, NO2, H2O, and O2}$$ and
$$\ce{C7H10N, CO2, NO2, H2, and O2}.$$
Also, you can use the set of
$$\ce{H2O, H2, and O2}.$$
Here are the three balanced equations:
$$ \ce{C7H10N + 10.5 O2-> 7CO2 + NO2 + 5H2O}$$
$$ \ce{C7H10N + 8O2 -> 7CO2 + NO2 + 5H2}$$
$$ \ce{H2O -> H2 + 0.5 O2}$$
Using reaction 3, you can turn reaction 2 into reaction 1. You would have to refine the question a bit to specify as how many reactions this counts (probably 2 rather than 3). Obviously, multiplying all coefficients by a constant should not count as a new equation.

And how do these mathematical methods work?

I would probably start with writing a program that does the work using an algorithm like the one I have outlined. I would guess a hundred different species would be doable on a single CPU machine. The combinations will grow astronomically, but there are some good criteria to exclude many of the combinations. It does depend on how different the species are in their atomic composition. If they are all alkanes plus elemental carbon and elemental hydrogen, it looks like a huge number of reactions, but they all boil down to making each alkane from the elements, and then having linear combinations of the already balanced equations (again, not sure how you would count unique reactions).


text:
Are the dsp3 hybrid orbitals degenerate?
I learned in general chemistry that hybridization of n orbitals produces n hybrid orbitals with the same energy (degenerate).
However, in Housecroft and Sharpe inorganic chemistry, I read the following:

The mixing of $\mathrm{s}$, $\mathrm{p}_x$, $\mathrm{p}_y$, $\mathrm{p}_z$, and $\mathrm{d}_{z^2}$ atomic orbitals gives a set of five $\mathrm{sp^3d}$ hybrid orbitals, the mutual orientations of which correspond to a trigonal bipyramidal arrangement. The five $\mathrm{sp^3d}$ hybrid orbitals are not equivalent and divide into sets of two axial and three equatorial orbitals; the axial orbital lobes lie along the $z$ axis.

The above excerpt seems to imply that the 5 resulting $\mathrm{sp^3d}$ hybrid orbitals are not degenerate and have different energies. However, this is contradictory to what I learned in general chemistry. Why is this so?

They never say a word about energy. The orbitals are geometrically different, that's the point.

Also notice here that the energy of the orbitals is for the unbonded atom in free space. When you bond the five orbitals then they are not necessary equivalent anymore because of the geometric arrangement of the substituents.


text:
Iodoform on acetone gives one product?
During the Iodoform reaction of acetone why does the second iodine addition happen on the carbon that already has undergone substitution instead of the other carbon?

There are no both. As soon as your first group succumbs to iodine and falls off as iodoform, you no longer have acetone, so why would the other group react?

I think a better way to frame the OP's question is to ask: why does the second iodine addition happen on the carbon that already has undergone substitution instead of the other carbon?

@ Ivan Neretin it does not form iodoform at once . According to the mechanism halogenation happens step by step. Meaning that alpha hydrogens are replaced one by one by halogen atoms


text:
What exact color does ozone gas have?
This answer to a now closed question says that ozone gas has a "slight blue" color. But there are lots of blue colors: sky blue, ultramarine, phthalo blue, many others — what is closer to the color of ozone?
This page shows photos of ozone solutions in water, like the one below, but this may not be representative of the color of the gas.
 (source)
So, what exactly color does ozone gas have? Is it a single hue, or does it change depending on concentration/amount of ozone?

For some reasons, the ozonated water bottle looks odd. I searched Google and elsewhere nobody ever mentions that ozonated water has a bluish tinge except the link shared by the author. Are there any other references which mention that ozonated water is blue?

@M.Farooq I think you've not seen this anywhere else because the water in the photos has quite high concentration of ozone. As is told in the blog entry linked in the OP, "The blue color in the water starts to become visually evident above 50 ppm.".

I was asking out of curiosity as to why nobody mentions the color of ozonated water anywhere. Ozonation is a very old technology but how come every one is silent on the color of ozone dissolved in water. The key question is if we have 50 ppm of O3 gas in air in a glass tube, will be see blue color?

Due to Chappuis absorption, ozone does have a bluish color. To determine exactly what kind of blue it is, let's first look at the spectrum of absorption in the Chappuis band. The following plot was done using these data for 293K.

This is spectral cross-section of absorption. To determine color from this spectrum, we need to choose some parameters:

Number density of ozone molecules,
Thickness of ozone layer we're trying to visualize,
Illuminant.

If we denote spectral radiance of our illuminant as $L(\lambda)$, thickness of ozone layer as $d$, ozone molecule number density as $\rho$, and absorption cross-section as $\sigma(\lambda)$, then we'll get the following expression for spectral radiance transmitted through the layer:
$$L_T(\lambda)=L(\lambda)\exp\big(-\sigma(\lambda)\rho d\big).$$
The most sensible illuminant to choose for showing color of a material on the web is the CIE illuminant D65, whose color is the white point of the sRGB color space. Its spectrum can be found e.g. here.
We can find the color in XYZ space using CIE 1931 color matching functions (can be found e.g. here). The expression is
$$c_X=\int_{300}^{830}L_T(\lambda)\bar x(\lambda)\,\mathrm d\lambda,$$
and similarly for $Y$ and $Z$ coordinates. Then these can be transformed to sRGB using linear transformation matrix $\mathrm{XYZ}\to\mathrm{sRGB}$ given e.g. here and gamma-correcting to $\gamma=1/2.2$ to yield final sRGB values.
Then, for ozone molecule number density $\rho=10^{25} \frac{\mathrm{molecule}}{\mathrm{m}^3}$ we'll get the following colors for different layer thicknesses:

For comparison, typical ozone column in the atmospheric ozone layer is about 300 Dobson units, which is equivalent to $2.687\times10^{20}\frac{\mathrm{molecule}}{\mathrm m^2}$; with our $\rho$ chosen above this corresponds to $d=8\,\mathrm{mm}$. So for daylight, ozone column has negligible effect on the sky color (unlike the evening — see the history of Chappuis absorption!).
As can be seen in the above plot, hue does change with increasing layer thickness. If we normalize the RGB values to see the hues of the thick layers (this would correspond to increasing illuminant power to compensate for absorption), we'll get the following hues:

Note that the violet hues in the above plot aren't accurate: they can't be accurately represented on sRGB monitors, so the plot only approximates them. They should be more saturated. Here's how the chromaticity changes from the white point to the most violet with increasing layer thickness (dashed triangle denotes the sRGB gamut):

Nice answer, but why did you ask the question? ;-)

@Karl Own Q-A duets are perfectly fine :)

@Karl because the other question where I could have posted the answer is closed, and unlikely to be reopened given how long ago it happened. But since I think the question is still relevant (in the form "what color" instead of "does it have a color"), not easily (if at all) answered by googling, and interesting, I made this Q&A.

I just wanted to say that I've always loved sunsets, but until now I would never have guessed ozone was part of the reason; those deep blue-purples, contrasting with the orange-reds from Rayleigh scattering. You learn something new every day! Thanks so much for this contribution!

That passes pretty close to "true" sRGB blue. Any relation? Or just coincidence?

@TLW I don't think there's any relation. sRGB was based on capabilities of CRTs of its time. Colors in CRTs are generated by luminescence, not by absorption, and blue phosphor is usually ZnS:Ag (P11 phosphor), which has no relation to ozone.

For some reasons, the ozonated water bottle looks odd. I searched Google and elsewhere nobody ever mentions that ozonated water has a bluish tinge except the link shared by the author. Are there any other references which mention that ozonated water is blue?

@M.Farooq - you're likely thinking of much lower concentrations than that solution. From the link "The blue color in the water starts to become visually evident above 50 ppm.", whereas e.g. https://msu.edu/~brook/publications/aeis/aeis655.htm suggests "Ozone is effective and safe at concentrations 3 to 8 ppm in water" (although that is for produce washwater.)

@TLW, Actually I was doing a general search and couldn't find anything. What is the color of ozone saturated water? Is it really that blue? So far there is only one website which shows a picture of ozonated water with 50 ppm concentration. If we have 50 ppm ozone in a gas phase, can we see it's color at ambient condition?

@M.Farooq at 20°C and 101.3 kPa 1-liter cube of ozone gas is visible. See a plot of colors in these conditions


text:
How to describe the enthalpy of acid dissociation in aqueous solutions?
How to describe the enthalpy of acid dissociation in aqueous solutions?
If we had a weak acid "PB" that dissociates to give a proton and a conjugate base. Classically it would be described like the following:

Unfortunately, things aren't that simple in aqueous solutions. Each of the dissociated molecules will have a charge which will change its polarity and water hydration (solvation).
a better way would be to describe dissociation in gas phase then the solvation of each:

Is this description accurate? Is the enthalpy of protonation measured by calorimetry or calculated by the Van't Hoff analysis the "dissociation" or the "Ionization"?

https://en.wikipedia.org/wiki/Acid_dissociation_constant This link may help.


text:
Does the Location of Line-Dash-Wedge Notations Matter?
In Line-Dash-Wedge notation, when illustrating that an atom or bond is directed towards or away from the viewer (solid triangle for coming forward, dotted lines for going back), does the location of these lines matter when doing illustrations?
For example, a THC molecule.

The two hydrogen atoms there. Could they be placed inside the carbon rings instead of appearing outside, like below?

I understand that there's a whole host of issues with that otherwise, but just asking about the triangle and dashes in particular.

Yes they could be placed inside, it doesn't matter.

Yes, both inner and outer placements are possible with some nuances.
From Graphical representation of stereochemical configuration (IUPAC Recommendations 2006) [1, p. 1926]:

ST-1.3.2: Stereogenic centers at ring fusion atoms
Stereogenic centers at ring fusion atoms should be drawn with hashed wedged or solid wedged bonds to the exocyclic substituent at the fusion atom whenever possible. If necessary, an implicit hydrogen should be made explicit in order to provide an exocyclic substituent to bear the hashed wedged or solid wedged bond. When one of the fusion bonds is oriented vertically and the exocyclic substituent is graphically small (such as a hydrogen or a methyl group, or even such physically larger substituents as a phenyl group represented with a graphically small “Ph” group label), the exocyclic substituent is also
  preferentially oriented vertically and opposite to the vertical fusion bond. This orientation is preferred even if it results in the substituent being placed within the ring system, and is particularly common in the depiction of steroids and other natural products.

From Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008) [2, p. 354–357]:

GR-4.2.2 Fusion atoms with one substituent drawn
Ring fusion atoms already have three bonds within the ring system. For external ring fusion atoms, the substituent should preferentially be positioned outside the ring system and oriented so that it bisects the angle between the adjacent bonds.

[...]
When two substituted fusion atoms are adjacent, placing both substituents outside the ring system will typically cause them to overlap. In those cases, it is better to draw one of the substituents within the ring system. A substituent drawn in this way should preferentially be drawn exactly vertically or horizontally, whichever direction would minimize overlap with other atoms and bonds.

If a substituted fusion atom is adjacent to an unsubstituted fusion atom, there typically is enough room to orient the substituent either inside or outside the ring system. That substituent should be oriented outside the ring system if all substituents can be oriented outwards, but should preferably be drawn exactly vertically or horizontally if any other substituent is placed within the ring system to avoid
  conflict with an adjacent substituent.

Small substituents on the C-8 and C-9 atoms of steroids should be drawn within the ring system in this fashion even if there are no substituents on the adjacent fusion atoms C-10 and C-14. Certain other natural products [23] also have preferred orientations that place substituents within the ring system.

A substituent should not be oriented within a ring system when attached to a fusion atom that lacks any vertical or horizontal fusion bond. The substituent should be oriented outside the ring system, minimizing overlap as best as is possible. Sometimes, it might be possible to reorient the ring system so that the fusion bonds are vertical; the substituents should certainly be placed vertically if so.

It is not acceptable to orient two substituents within the same ring if the ring has fewer than eight atoms. One or both of the substituents should be oriented outside the ring system, minimizing overlap as best as is possible.

Substituents on interior fusion atoms have no option other than being drawn within the ring system. As above, such substituents should preferentially be drawn exactly vertically or horizontally, according to the direction that would minimize overlap with other atoms and bonds.

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)
Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). Pure and Applied Chemistry 2009, 80 (2), 277–410. https://doi.org/10.1351/pac200880020277. (Free Access)

I have issues with some of the structures labelled "not acceptable." Yeah, some of them are not acceptable given overlap, but others are quite valid from first principles. Most particularly, in the section starting with "Small substituents on the C-8 and C-9," it seems extremely arbitrary to call the right structure "not acceptable."

@Zhe The way I interpret that section is that for particular classes of compounds (steroids for one) there's conventions on how you place those hydrogens, such that they will match across a diverse compound series (and with reader expectations). It's sort of like the conventions on how to orient certain molecules (e.g. for steroids the 5-member ring is in the upper right) - other ways of placing them (e.g. 5-member ring on the bottom or on the lower left) aren't technically incorrect, but are so against convention that you shouldn't do it.

That sounds 100% reasonable. Though I would strongly prefer the language of "unconventional" or "dispreferred" instead of "not acceptable."


text:
Is it stupid to put used acid into a pool after cleaning the chlorinator?
I have a salt pool, and when the chlorinator salts up with calcium, I clean it with a diluted acidic solution.
What happens to the acidic solution after the reaction happens? I.e., does the pH change?
Would it be stupid to put the used solution into the pool afterwards? Am I correct in thinking the following sentence? "We are trying to get the calcium out of the pool, but putting it back in just means the chlorinator has to get it out again." Or, it is stupid for a different reason, e.g. the acidic solution is too weak after the reaction anyway?

You put more calcium to the pool. If this reflects on the interval between cleaning up the chlorinator isn't sure.  The real concentration in the pool should not be really affected, nor the pH

The bicarbonate buffer capacity of the pool  water probably would deal with the remaining acid, but the pH would IMHO get lower.
But returning calcium to pool is not wise, as it will cause calcium accumulation and progressively shortening periods of calcium precipitation.


text:
I conducted some experiment and I want to know the resulting redish black thing
I placed two electrodes in a jar
and poured the sparking water with lemon flavor.
After about 5 hours, the water turned to blue and
the negative probe had the redish black thing.
I'd like to know what it is. Please, reply for me.

Materials : 9V battery, battery holder&wire, jar, copper wire and trevi (sparkling
water)

Connecting battery

After about 5 hours

Result : redish and black thing.

Can you please list the content (ion composition) of the sparkling water? 2. In your text you wrote that you're added lemon flavour. What exactly does this mean? Fresh lemon juice? Something you've bought? If so, does it have a composition list? BTW, this lemon flavour is missing in your list of materials.

I guess the composition of the sparkling water is irrelevant. It is obviously copper that is responsible for both phenomena (blue solution and reddish-black dirt).

1-2. The sparking water is just one I bought. "Composition List : 정제수(distilled water), 이산화탄소(CO2), 천연향료(레몬향, natural flavor; lemon) 0.02%". I'm sorry for lack of english skill.

@IvanNeretin well, knowing the water composition allows us to rule out

Solvated copper(II) ions for the blue color, copper(II) oxide for the black (as hinted by IvanNeretin). To prove this hypothesis, one could dissolve the copper oxide in sulphuric acid, neutralize the solution and add ammonia. The deep blue color indicates the $\ce{[Cu(NH3)4(H2O)2]^{2+}}$ complex. The latter can also be done for the blue solution currently present.

...and metallic copper for the red.

Thank you for answering to me. You mean that (1) the positive probe : Cu -> Cu(ll) + 2e    (2) the negative probe : Cu(ll) + 2e + H2O -> CuO + H2, am I right?