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How much chlorine is there in 1 g of sodium dichloroisocyanurate?
According to Wikipedia, molar mass of anhydrous sodium dichloroisocyanurate is approximately 220 g per mole, and molar mass of Cl2 is 71 g per mole. Therefore there is 323 mg of chlorine in 1 g of NaDCC. But according to this page, 33 mg of NaDCC releases 20 mg of chlorine, and so there is 606 mg of chlorine in 1 g of NaDCC. Where have I made a mistake?

You're wrongfully assuming the amount of chlorine released is the same as the amount of chlorine contained in NADCC, but it does not dissociate completely.

By my calculations 33 mg of NaDCC would release 10.65 mg of chlorine, which is almost two times less than what that page claims.

In your page, "chlorine" isn't referring to $\ce{Cl2}$ but rather to $\ce{HOCl}$, which is released when added to water.

Assuming that all of the chlorine goes from NaDCC to HOCl, 33 mg of NaDCC would result in 15.75 mg of HOCl, which is still less than what is claimed on that page.

$\ce{HOCl}$ also partially dissociates into $\ce{OCl^-}$ ions, which are also accounted for in the 20 mg of chlorine produced.

@Sam202 OP is saying in that comment that if all the chlorine is turned into $\ce{HOCl}$, then that would be the mass. If you consider dissociation, the mass of the 'considered' active chlorine species would only decrease.

@M.L the addition of NaDCC to water is complex and involves around 10 equilibrium systems which all partially contribute to the total mass of chlorine disolved in water. Therefore, if only $\ce{NaDCC}$ dissociation is considered, there will be a deficit in amount of $\ce{Cl_2}$ produced.

@Sam202 Well the dissociation process might involve many steps, but ultimately, you should end up with active chlorine in the form of bleach.

I believe the discrepancy comes with the $\ce{Na+}$. Each NaDCC has that positive ion. Since there is one $\ce{Na+}$ and two $\ce{Cl}$ per NaDCC. What I suspect happens is that there is that the page derives its values from calculating the abundance of two species: $\ce{NaClO}$ rather than $\ce{ClO-}$ and $\ce{HClO}$. The exact ratio between the two may not be 1:1, but I believe that the mass of all the $\ce{Na+}$ was included in the calculation of the amount of active chlorine.
Revising the calculation, you'd get about $\pu{19.0g}$ of chlorine species including $\ce{NaOCl}$ and $\ce{HOCl}$ with the following expression below (assuming the two species are in 1:1 ratio):
$$\frac{\pu{33 mg}}{\pu{220 g/mol}}\cdot MM(\ce{NaOCl)} + \frac{\pu{33 mg}}{\pu{220 g/mol}}\cdot MM(\ce{HOCl)} = \pu{19.0 g/mol}$$
This is much closer to what was reported. Note that the ratio of the two species is likely not 1:1. However, this would give the largest possible value as all the $\ce{Na+}$ is used up from NaDCC. I am not too sure how got and marketed the value of $\pu{20g}$, but I am guessing it has to do with what I mentioned above and maybe some slight bias in rounding.


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Resorcinol turned brown
I stored resorcinol in a transparent plastic container with an airtight lid. There was some air captured inside the container too. After some days, it turned brown. please tell what happened to it?

Essentially, this is a almost a duplicate of a similar question on phenols. Resorcinol is sensitive to light, if impure, and also, that reference states it's readily oxidized in the presence of water, especially in alkaline conditions.
See Nilay Gosh's fine answer for a more detailed explanation.

source: Stack Exchange

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

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How to use the "equilibrium favors reactants" symbol in Microsoft Word?
Is there a Unicode symbol or any other way to insert the symbols "equilibrium favors reactants" and "equilibrium favors products" as text?
I have been looking online for a while and found nothing useful.

Turning my comments into an answer: I assume you're referring to the symbols $\ce{<=>>}$ and $\ce{<<=>}$. I don't believe these are included in unicode, but there are some similar ones that I believe are IUPAC acceptable replacements, which use full arrows instead of harpoons. These are
unicode 2942: ⥂
unicode 2943: ⥃
unicode 2944: ⥄
An alternative is to use a LaTeX package such as mhchem.

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


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What is the main difference in between PGNAA and PINS?
I understand that both use a neutron emitter to trigger the material being analyzed to emit gamma radiation which is then analyzed to determine the material, but what is the exact difference between these methods? Is it only that PINS is portable, but PGNAA is not?
PGNAS (Prompt Gamma Neutron Activation Analysis) and
PINS (Portable Isotopic Neutron Spectroscopy)

Is there a difference?


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Is the given figure correct? Does the electron really flow out as depicted by the arrow?

Someone up here questioned about the figure and that the arrow is pointless and doesnt make sense? Or is it?

See also CH SE META post about policy for "Am I right?" questions

Both arrows are correct: conventional current flow is defined as from more positive to less positive. It is the opposite of electron flow. Blame Ben Franklin for this convention.

It would be better if the legend showed arrows as “net movement of negative charge”, and there were another arrow in the salt bridge. The transport mechanism in the wire is distinct from that in solution, yet there is a frequent misconception that free electrons move through the solution.

The direction of electron flow as shown in the figure is correct, however, as the reduction potential of Zn is -0.76 V and that of Cu is 0.34 V the resultant cell potential obtained by
E cell = E cathode - E anode.
E cell = 0.34-(-0.76) = 0.34+0.76 = 1.1 V.
For this to happen the Zn(s) electrode must get oxidized to Zn2+ ions giving up electron to Cu electrode as denoted by the arrow. On the other hand the Cu2+ ions present in the cathodic compartment take up the electrons coming through wire in the copper electrode and get reduced to Cu(s). However, in the figure the arrows drawn in the solution part of the figure are not correct.
The arrow for anode compartment(containing Zn electrode) should show movement of Zn atoms from electrode to electrolyte solution as Zn2+ after oxidation.
The arrow for Cathode compartment (containing Cu electrode) should show movement of Cu2+ ions from electrolyte solution to the electrode as deposited Cu metal after reduction.

The arrows are correct: see my main comment.

It sounds like you are assuming that the arrows at the electrodes are intended to show the direction of flow of ions, but the figure label clearly indicates those arrows represent electron flow, for which the indicated direction is correct.

I also find that the yellowish arrows in solution are confusing. The electrons originate in the Zn and Zn ions shed into solution and electrons are 'consumed' in the Cu metal by combining with Cu ions. The electrons do not flow through the solution as is suggested by the arrows in the figure.

in solution there are no electrond, conduction takes place because of ions. Therefore using arrows to show electron movement in solution is technically incorrect


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in the reaction of aldehydes with NaHSO3 why will sulphur's lone pair act as nucleophile instead of O-
aren't negatively charged species like oxygen anion more nucleophilic than neutral atom like S
why cant O- which is bonded to na act as nucleophile

The product shown in your mechanism has more structures stabilized by resonance than the product that would be formed in your other scenario.

Sam 202 is correct- if you consider the mechanism if oxygen serves as the nucleophile, then there are only 2 different resonance structures possible, but the structure that you have drawn 3 different resonance structures possible.
The dissociation of the ions is not something that needs to be considered in the mechanism you have offered, but if you considered a mechanism with the O- as the nucleophile, then when it bonds to the carbon, it no longer has a negative charge, and so the sodium ion dissociates.
This is significant because this oxygen has formed two bonds, and thus cannot be assigned a negative charge in a theoretical resonance structure. Hence, although the sulfur atom is still bonded to 3 oxygens, only two of them can be assigned a negative charge, whereas in the final structure in your mechanism, all 3 oxygens can be assigned the negative charge.
Theoretical mechanism of oxygen as the nucleophile below:

As you can see, here only 2 distinct resonance structures are possible, less than the 3 in your final product.
It follows that the product of your mechanism is more stable, and hence more thermodynamically favourable, so it’s production, and a mechanism which leads to its production will be favoured and occur over one which uses O- as a nucleophile, which would produce a less stable product.


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MgSO4 is soluble while BaSO4 isn't, what's the role of Lattice energy here?
I came across a question asking the reason behind

MgSO4 is soluble while BaSO4 is insoluble in H2O

Now, I know

Solubility has inverse relation with Lattice energy. While Lattice energy is more when ionic size is less.

So, how come BaSO4,with less Lattice energy be insoluble?

To determine if a salt is soluble you need to work out the change in Gibbs energy, which combines enthalpy (internal energy) and entropy (disorder) into a single value. Only if the solubility process causes an overall reduction in Gibbs energy, then the salt is soluble (Here it's explained for NaCl https://www.linkedin.com/pulse/solid-liquid-solubility-how-does-work-nikhilesh-mukherjee/).
∆G = ∆H - T∆S [ G is Gibbs free energy, H enthalpy, T is temperature and S is entropy]
Lattice energy: energy required to convert the crystal into infinitely separated gaseous ions in vacuum.
Hydration energy: the amount of energy released when one mole of ions undergoes hydration
Lattice energy BaSO4= +2474 kj/mol
Hydration energy BaSO4 = -2368 kj / mol
T = 298 k, ∆S = 102 j/k-mol
∆H = Lattice energy + Hydration energy = 2474 + [- 2368] = +106 kj/ mol
∆G = [106 – 298 x 0.102] = +75.6 kj / mol (positive therefore BaSO4 is insoluble!)
Now, let's see MgSO4:
Lattice energy MgSO4= +2833 kj/mol
Hydration energy MgSO4 = -2927 kj / mol
∆H = Lattice energy + Hydration energy = 2833 + [-2927] = -94 kj/ mol
The enthalphy change associated with the solubility of MgSO4 is negative (-94 kj/ mol), given the entropy contribution is always negative when salts dissolve, then you can immediately conclude that the change in Gibbs energy is negative, this means MgSO4 is soluble!
https://chemistryklipz.files.wordpress.com/2016/12/born-haber-cycles-and-enthalpy-of-solution.pdf
http://chemistry-reference.com/reaction.asp?rxnnum=88


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Computing the molar volume for a 2D-simulation of phase separation
I am a mathematician and I want to simulate phase separation that occurs in a sodium silicate glass ($12.5\,\text{Na}_2\text{O}\cdot 87.5\,\text{SiO}_2$) at $T = 923\ \text{K}$ as it was done, for example, in Kim and Sanders (2020).
In particular, I want to do this simulation in two-dimensions. However, certain parameters that are implicitly required for the Cahn-Hilliard equation are only given in units that concern 3D simulations. For example, the molar volume for the mixture above can be computed to be
\begin{equation}
V_m = 25.13 \cdot 10^{-6}\ \frac{\text{m}^3}{\text{mol}}.
\end{equation}
Question: How do I have to deal with that in a 2D simulation? From a geometrical point of view, I just would like to consider
\begin{equation}
(25.13 \cdot 10^{-6})^{\frac{2}{3}}
\end{equation}
as a corresponding quantity for a two-dimensional simulation (unit: $\frac{\text{m}^2}{\text{mol}^{(2/3)}}$ ?). The quantity mol refers basically to a  3D framework, isn't it?

Define relation between molar volume and molar area. [m2/mol]

My proposal simply comes from the geometric average but actually does not fit toe the notion of a mol. I still think that the notion of a mol is meat for 3D. Therefore, I have no idea how to deal with that notion in 2D.

At least to me, "molar area" is not that common. And I have also no idea how to compute this quantity (without the corresponding mass density).

Computing such things in 2D is not common either. You have to use 2D variants of 3D quantities. Similarly, density in kg/m3 would have 2D variant area density in kg/m2.

Actually, in most cases simulations (especially simulating phase separations) start with 2D simulations. One paper is cited in the questions. Here is another one. And I was wondering where I can find the corresponding 2D parameter variants or how these 2D variants are computed.

I am not saying 2D models are unusual. I have just said by other words, that 2D volume is area. Twice molar volume means twice molar area. The only problem is the proportionality constant.

Since one really wants to have the quantity (molar area) in the unit area per mole (and not per $\text{mol}^{\frac{2}{3}}$), the guessed solution $(25.13 \cdot 10^{-6})^{\frac{2}{3}}$ above is in fact not what we are looking for. Instead, it seems to be common to compute
\begin{equation}
A_m = \left(\frac{V_m}{N_A}\right)^{\frac{2}{3}} \cdot N_A = V_m^{\frac{2}{3}} \cdot N_A^{\frac{1}{3}},
\end{equation}
where $N_A = 6.02 \cdot 10^{23}$ is the Avogadro constant.


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Is hard tap water useful for "salting out" a liquid organic solute?
We use methylene iodide ("MI", CH2I2) as a heavy liquid for sink-swim mineral separation. We use acetone to wash the MI off of the minerals and glassware after we're done. MI is pretty expensive, so we save that MI/acetone mix to later recover the MI. This is done by adding lots of water to the mix. The solution starts off very cloudy, but with enough water (probably something like a 10:1 water-acetone ratio), the solution become clear, and a well-defined blob of MI is present at the bottom of the container.
My question is if it is better to use deionized water for this, or if it's better to use hard tap water that's high in calcium bicarbonate? I thought the hard tap water might actually be better for this, because of all the dissolved ions interacting with the water, limiting the space for MI to stay dissolved to the water and acetone, similar to a "salting out" precipitation reaction.

It may depend on the composition of the organic and of the inorganic phases. What are the metallic ions in solution ? And in which concentration ? Are they complexed by EDTA or similar agents ?

Why don't you just vacuum distill it? Like why add contamination to it? Pull a vacuum on it, the acetone is pretty cheap so even if it gets wasted yeah who cares and then just distill the Di iodo methane?

@PulkitSharma You mean just forego the water all together? Just try to evaporate out the acetone, leaving the MI behind? I hadn't thought of this because I was always told that MI was actually fairly volatile, despite it being a heavy liquid. https://haz-map.com/Agents/8547 That says its vapor pressure is only 1.2 mm Hg, though.

Somewhat irrelevant note: sea water was used for photo processing aboard ship in WWII: https://www.pbagalleries.com/m/lot-details/index/catalog/542/lot/177698/Archive-of-World-War-II-Photo-Technical-School-Papers , IF#52-7 Sensitized Materials - Sea Water Processing. 1943 . However, this was useful because it removed unexposed silver halides, rather than for salting out ;-)

No I mean distill iut the MI. pull a vacuum of 5mm and purify that fucker before next use too


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Reduction formula for infinite group order?
In group theory, what do we do when we have infinite group order, like in D∞h or C∞v but we want to mathematically determine the SALCs (symmetry adapted linear combinations) using the projection formula.
If we did, we would get a function that tends towards 0 which isn't that useful. I asked my lecturer, and he suggested that it's probably because it's obvious with linear molecules so we wouldn't need to mathematically determine the SALCs. Is there some way of doing it though?

This is an interesting question, but I think it'd benefit from having a specific, concrete example. I also think it'd help if you explicitly wrote out this function which "tends towards 0" and how you obtained it.

Generally when working SALCs for $D_{\infty h}$ or $C_{\infty v}$, you use a subgroup like $D_{2h}$ or $C_{2v}$ using an equivalence table or "descent in symmetry" table between the irreducible representations. For example the Jacobs University tables for $D_{\infty h}$ indicate the A / E representations as well as $\Sigma$, $\Pi$, $\Delta$ etc.

Since 9 months have gone by with no answers, I think it's fair to let you know about Matter Modeling Stack Exchange which also answers group theory questions, but make sure to delete the question from here if you're cross-posting there (or to at least put a note at the top of your question, saying that it's cross-posted).


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Which is really water soluble, amylose or amylopectin？
In my textbook it was written that amylose is more water soluble than amylopectin. But as I searched for the reason I found a totally opposite information.
https://pubs.acs.org/doi/pdf/10.1021/ed052p729
Do you know which is really true?
Or this problem is more complex than I imagine?

Check quite detailed respective  Wikipedia pages, eventually their references. Amylopectin reportedly seams more soluble, being loosely packed with less hydrogen bond linking, in spite of having larger and branched molecules.

Thank you for your comment. It seems many textbooks are making mistakes.

Always search for more sources when 2 different statements clash.

Both amylose and amylopectin are high molecular weight glucose polymers found in plants.
Both are insoluble in cold water, but can be made soluble through a process known as starch gelatinization, which is essentially a cooking process.
Some of the confusion may arise from the fact that amylose is straight chained and amylopectin is more branched.  The tight helical structure of amylose may require more extreme treatment to fully hydrate it, but then it has a higher degree of solubility than amylopectin.
As gelatinization techniques vary, further study of potential applications of both types of starch may help clarify which one is more desirable.

Thank you for your answer sir. Could you explain the definition of solubility in your answer?

@satorukurita that really is the crux of the issue.  Very large polymers are not the same as simple salts.  These "gels" exist somewhere in between, depending on conditions such as temperature and whether or not they have been hydrolyzed.  They are used as thickeners in cooking.


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Why is a ring carbon C-R of a substituted ferrocene more downshifted on 13C-NMR?

Upon assigning $\ce{^{13}C}$-NMR for 1,1'-bis(α-methylmethanol)ferrocene, we discovered the substituted FcC-CH peak (95.14 ppm) was more downshifted than the CH-OH peak (65.59 ppm) as was confirmed by HSQC. The other four FcCH peaks were found at 67.77, 67.67, 66.12 and 66.07 ppm.
I can't rationalise why the FcC-CH peak is so downshifted as the lone pairs of oxygen can't donate in via resonance and if it was purely inductive, surely the CH-OH peak would be further downshifted.
Upon transformation to 1,1'-bis(α-methoxyethyl)ferrocene and then 1,1'-bis((α-methyl)methylpropanoate)ferrocene, the FcC-CH peak was still seen to be further downshifted than I expected.
If anyone can offer an explanation, I'd be very grateful!

First of all, electronic effect on $\ce{^{13}C}$-chemical shifts of 1,3-cyclopentadiene $(\pu{132.2 ppm}, \pu{132.8 ppm})$ is well evident due to the iron metal in ferrocene $(\pu{67.7 ppm})$. Yet, when ferrocene is substituted with any group, the $\ce{^{13}C}$-chemical shift of the carbon attached to the substituent shows a blue shift. I think this seemingly large blue shift in $\ce{C}$1 of the substituted ferrocenes is due to the sheilding-deshielding effects due to steric hindrance (similar to chemical shift difference shown in cis- and trans-alkenes). The same effect we have usually shown in substituted benzenes as well. I also think this chemical shift difference in $\ce{C}$1 of substituted ferrocenes is not too much of a difference than that shown in substituted benzenes:

For example, when $\ce{R = CH3}$ (see above structures), $\ce{^{13}C}$-chemical shift of $\ce{C}$1 of ferrocene is $\pu{83.4 ppm}$ $(\Delta\delta = \pu{15.7 ppm})$ while that of substituted benzene is $\pu{137.4 ppm}$ $(\Delta\delta = \pu{9.3 ppm})$ where $\Delta\delta =$ $\ce{C}$1 of unsubstituted - $\ce{C}$1 of substituted (Ref.1).
Similarly, when stereo bulk is increased, say $\ce{R = C(CH3)3}$, then $\ce{^{13}C}$-chemical shift of $\ce{C}$1 of ferrocene is increased to $\pu{101.7 ppm}$ $(\Delta\delta = \pu{34.0 ppm})$ while that of substituted benzene is $\pu{150.5 ppm}$ $(\Delta\delta = \pu{22.4 ppm})$ (Ref.1).
Also, in a compound related to OP's mentioned compound where $\ce{R = CH2OH}$, the $\ce{^{13}C}$-chemical shift of $\ce{C}$1 of ferrocene is given as $\pu{87.7 ppm}$ $(\Delta\delta = \pu{20.0 ppm})$ while that of the substituted benzene is $\pu{141.1 ppm}$ $(\Delta\delta = \pu{13.0 ppm})$ (Ref.1).
There are few other examples in Ref.1 suggests that the difference in this chemical shift is due to the stereo hindrance. Also keep in mind that extra large $\Delta\delta$ shown in ferrocenes are due to the electronic influence from the iron metal.
References:

A. A. Koridze, P. V. Petrovskii, A. I. Mokhov, A. I. Lutsenko, "Electronic effects in the cyclopentadienyl ring. The $\ce{^{13}C}$ NMR spectra of monosubstituted ferrocenes," Journal of Organometallic Chemistry 1977, 136(1), 57-63 (DOI: https://doi.org/10.1016/S0022-328X(00)87967-X).


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What is the difference between technical-grade and food-grade tripotassium phosphate?
There is an article about tripotassium phosphate that states the following:

Consumers may have health concerns about why this cleaning agent can
be used in food, but that is the technical grade, not the food grade.
When used as a food additive, it almost has no side effects and its
safety has been approved by the FDA...

There is an answer here that nicely addresses the difference between lab-grade and food-grade purities but does that answer my question? My question specifically is what is the difference between "technical-grade" and "food-grade" tripotassium phosphate? Is it just a matter of impurities? At the atomic level, K3PO4 is K3PO4, whether it's being used as a commercial degreaser, a stain remover, or a food additive, no?

Food grade must have acceptably low levels of impurities, such as heavy metals, as dictated by applicable authorities. Technical grade does not necessarily have to meet this requirement because it is not intended for consumption.

It is just a question of purity. Usually the manufacturer of chemicals prints their composition on the label. For example, the purest form of potassium triphosphate produced by VWR is
Minimum purity : > $98$ %;
Free alkali : < $1.0$ %;
Dipotassium hydrogen phosphate : < $1.0$ %;
Sodium : < $0.5$ %;
Chloride < $0.003$ %;
Iron : < $0.001$ %;
Lead : < $0.002$ %;
Total nitrogen : < $0.001$ %
I don't have the result of the analysis of the same product, in a  technical-grade quality. But I am sure that the same values are significantly higher. Technical products are cheaper. But they contain more impurities.

Correct my lay understanding here. When a chemical like tripotassium phosphate is made, compounds are concocted (in this case, to neutralize phosphoric acid) that, by the nature of the concoction, yield a chemical with impurities. This chemical then undergoes purification. When the purified chemical is tested for human safety, as a food additive, do the researchers simply look for the presence of side effects or do they try to determine if the side effects are caused by the chemical itself (in its purest form) or caused by its impurities? And can we even test this chemical with 0% impurities?

Potassium triphosphate is made by adding enough $\ce{KOH}$ to neutralize phosphoric acid. But both these substances ($\ce{KOH}$ and $\ce{H3PO4}$) are themselves not 100% pure. They contain some impurities, that usually remain in the potassium triphosphate. Of course, these impurities can be removed by dissolution and reevaporation. But this treatment costs and is often not necessary, depending on the use of the product. If a phosphate is used as a fertilizer, the purity is not important. If a phosphate is used as a food additive, the product must be tested for side effects of impurities.

So when the phosphate is tested for human safety as a food additive no distinction is made between side effects caused by the chemical compound itself (in its purest form) versus its impurities?

@trndjc if it was at some point tested for safety as a food additive, an even purer form may have been used.  Individual batches aren't tested for safety, though quite possibly are for purity, or at least for the presence of known contaminants.  Certain impurities may be totally benign while others are harmful; we in food the concern is far more about the latter.  The other potassium phosphates, for example, are generally recognised as safe meaning they're so well established that modern tests aren't needed.


text:
Rate of Reaction and half life
During the course of chemical kinetics while studying zero order or first order they consider rate of reaction to be dependent only on single Reactant
For ex:- A--> product,Rate of Reaction=k[A]^0 or [A]^1 and then define half life of such reaction But it can be [A]^1[B]^-1 for zero order and for first order it can be something like [A]^3/2[B]^-1/2 So why we don't consider these cases?can we define half life of these cases?and pls tell what is the significance of half life of a Reaction?

See e.g. wikipedia.org :Half-life and generally do at least basic prior search withing available offline and online resources.

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

The big issue for a half-life for reactions that are not simple first order reactions is that the half-life varies both with time and initial state. For simple first order reactions, half-life is a constant that does not depend on time or the starting amount of material. So it is pretty meaningless for more complex reactions.

Consider a reaction that follows the power law at constant volume, temperature, and pressure of the form:
$$\ce{aA +bB->cC +dD}$$
The rate of this reaction would be:
$$-r_\pu{A}=-\frac{\mathrm{d}C_\pu{A}}{\mathrm{d}t}=k\;C_\pu{A}^{\alpha}\;C_\pu{B}^{\beta}$$
We can express both concentrations in terms of conversion $X$:
$$C_\pu{A}=C_{\pu{Ao}}\;(1-X)$$
$$C_\pu{B}=C_{\pu{Ao}}\left(\frac{C_{\pu{Bo}}}{C_{\pu{Ao}}}-\frac{b}{a}X\right)$$
Substituting into the rate law:
$$-\frac{\mathrm{d}[C_{\pu{Ao}}\;(1-X)]}{\mathrm{d}t}=k\;C_{\pu{Ao}}^\alpha\;(1-X)^{\alpha}\;C_{\pu{Ao}}^\beta\left(\frac{C_{\pu{Bo}}}{C_{\pu{Ao}}}-\frac{b}{a}X\right)^{\beta}$$
Grouping up terms and re-arranging:
$$\pu{C_{Ao}}\frac{\mathrm{d}X}{\mathrm{d}t}=k\;\pu{C_{Ao}}^{\alpha+\beta}\;(1-X)^\alpha\;\left(\frac{\pu{C_{Bo}}}{\pu{C_{Ao}}}-\frac{b}{a}X\right)^\beta$$
$$\frac{\mathrm{d}X}{(1-X)^\alpha\;\left(\frac{\pu{C_{Bo}}}{\pu{C_{Ao}}}-\frac{b}{a}X\right)^\beta}=k\;\pu{C_{Ao}}^{\alpha+\beta-1}\;\mathrm{d}t$$
Half-life is defined as the time it takes for the concentration of limiting reagent $\pu{A}$ to reach half of its initial value:
$$t=t_{1/2}\implies\pu{C_A}=0.5\;\pu{C_{Ao}}$$
And conversion is also defined as:
$$X=\frac{\pu{C_{Ao}}-\pu{C_{A}}}{\pu{C_{Ao}}}=\frac{\pu{C_{Ao}}-0.5\;\pu{C_{Ao}}}{\pu{C_{Ao}}}=0.5$$
For reaction half-life to make sense when 2 or more reactants are involved, two additional conditions must be met:
$$\pu{C_{Ao}}=\pu{C_{Bo}}$$
$$a=b$$
This guarantees that both concentrations of $\pu{A}$ and $\pu{B}$ start at the same initial values and get consumed at the same rate, since it follows that:
$$\pu{C_A=C_B}$$
$$\pu{-r_A=-r_B}$$
So if you want to find the half-life for a reaction that meets all conditions mentioned, you would have to integrate:
$$\int_0^{0.5} \frac{\mathrm{d}X}{(1-X)^{\alpha+\beta}}=\int_0^{t_{1/2}}k\;\pu{C_{Ao}}^{\alpha+\beta-1}\;\mathrm{d}t$$
You will obtain an equation that you can solve for to calculate half-life $t_{1/2}$.
If the overall reaction order is not 1:
$$\alpha+\beta≠1\implies\pu{C_{Ao}}^{\alpha+\beta-1}≠1\implies t_{1/2}=f(\pu{C_{Ao}})$$
So half-life would depend on initial concentration of A.
Conversely, when the overall reaction order is 1:
$$\alpha+\beta=1\implies\pu{C_{Ao}}^{\alpha+\beta-1}=1\implies t_{1/2}≠f(\pu{C_{Ao}})$$
So half-life would be independent of initial concentration of A.

But does it make sense to define half life of a reaction ?half life of a species is understable for A-->Product. But for a reaction for a half life to be defined both reactants need to be halved at the same time. Which is usually not the case? Can you pls explain this

In my book it was also written that for a reaction involving two or more reactants,half life of reaction is defined only when the initial concentration of reactants were taken in their stoichiometric ratio. So does that mean we can't define half of these reaction always???

One of the reactants is usually the limiting reagent, which means it will be totally consumed first before the others get a chance to. Conversion $X$ is defined for such limiting reagent $A$, as shown above.

http://goldbook.iupac.org/terms/view/H02716 this is link of ipuac book in which it is written that half life can only be defined when reactants are in their stoichiometric coefficient

If such condition is necessary, then: $\frac{C_{Bo}}{C_{Ao}}=\frac{b}{a}$, which means $C_A=C_B$, and both concentrations will experience the same consumption rate. In other words, stoichiometric initial concentrations guarantees equal change in both reactant $A$ and $B$ after half-life $t_{1/2}$ elapsed.

@Ronit Even if reactants are in their stoichiometric ratio, the half-life of other then the 1st order reactions is concentration dependent. While it can be formally defined and practically used, e.g. for controlling of the synthesis, it is not a constant, but scenario specific parameter.


text:
Would a container made of sugar eventually dissolve if filled with a sugar-saturated solution of water?
Suppose you had a bottle made of sugar glass, and filled it with water that was already saturated with sugar.
I know that with a fully saturated solution of sugar in water, no additional sugar can dissolve in the water. However, just like how a reaction in an equilibrium state simply has equal amounts of transformations between the two substances in both directions, could sugar from the water randomly precipitate and sugar from the walls of the container at those points in time dissolve into the liquid? If this process continued for long enough, the water could, hypothetically, dissolve its container and break free. My question is simply whether this could happen.
When thinking about this as a form of equilibrium, I’m reminded of the triple point of a substance—that, too, is an equilibrium, but even in equilibrium, the substance shifts between all of the states. I know these are not directly related in the sense in which I’m talking about, but it highlights what is possible for an equilibrium.
I would assume that whatever principle governs the answer to this question also applies to any substance dissolved to full saturation in any liquid, which is made into a container into which that liquid is placed.

Is is similar case as for halogen incandescent bulbs. If tungsten vapours  were returned to the places from which they evaporated, it could shine eternally. But they are returned elsewhere and they eventually get the filament broken. It could take very long time for the sugar case.

Remember also recrystalization of near unsoluble precipitates in gravimetry, like BaSO4. It has initially very small particles, partially passing paper filters. the procedure includes heating the suspension, accelerating dissolving smaller ones and growth bigger ones, what is thermodynamically preferred. Not exactly the same case, as there are very very tiny differences in solubility, but almost there.

Arguably, with sugar it won't take as long as with tungsten.

@IvanNeretin I would rather agree. But hard to make  time estimation and hard to test. They seldom sell  bottles from sugar glass these days. "Very long" was meant absolutely, not relatively wrt a halogen bulb.

You picked a bad example, I think. In lieu of such speculation, you better read about supersaturation.

Consider that you can buy sugar-syrup sweets with a hard sugar shell (try saying that five times fast!). See this patent for how they're made, https://patents.google.com/patent/US4614658A/en, or this example product: https://www.walmart.com/ip/Arcor-Assorted-Filled-Fruit-BonBons-Bulk-Hard-Candy-assorted-fruit-candies-4-pounds/756758946 . My guess is shelf life is decades, if not centuries.

The sugar will be slowly dissolving and recrystallizing That does not mean the crystallization takes place where the dissolving did. It takes time A similar process happens in limestone caverns or in the old way of making apple jack by freezing a barrel of hard cider [in places like Minnesota and the Adirondack area of New York].


text:
Self Consistent Field method and LCAO
I am reading about the Self Consistent Field Method and Linear Combination of Atomic Orbitals.
Suppose we have one electron and one nucleus, then we can solve the Schrodinger equation explicitly.
If we have two electron and two nuclei, then if we neglect the inter-electronic repulsion, then the individual hamiltonian becomes
$h_i=-\frac{1}{2}\nabla_i^2-\sum_\alpha\frac{Z_{alpha}e^2}{r_{i\alpha}}$
So, $H=h_1+h_2$
We can write the total wavefunction as product of individual wavefunction.
$\psi(x_1,x_2)=\psi_i(x_1)\psi_j(x_2)$
Now we can expand the individual wavefunctions in basis of hydrogenic wavefunctions.
$\psi_i(x_1)=\sum_{n=1}^kc_{n1}\phi_n(x_1)$
Similarly, $\psi_j(x_1)=\sum_{n=1}^kc_{n2}\phi_n(x_2)$
So, $E=E_1+E_2$
$\displaystyle\langle E_1\rangle=\frac{\sum_{n,n'}a_{n'}a_n\int \phi_{n'}\hat H\phi_ndx}{\sum_{n,n'}a_{n'}a_n\int \phi_{n'}\phi_ndx}=\frac{\sum_{n,n'}a_{n'}a_n H_{n',n}}{\sum_{n,n'}a_{n'}a_nS_{n',n}}$
Now, minimizing this $\langle E_1\rangle$ we get $k$ linear equations in $k$ variables.
$\begin{pmatrix} H_{11}-E_1S_{11} & H_{12}-E_1S_{12} & ... & H_{1k}-E_1S_{1k} \\
H_{21}-E_1S_{21} & H_{22}-E_1S_{22} & ... & H_{2k}-E_1S_{2k} \\
...  \\
H_{k1}-E_1S_{k1} & H_{k2}-E_1S_{k2} & ... & H_{kk}-E_1S_{kk}\end{pmatrix}\begin{pmatrix}c_{11}\\c_{21}\\...\\c_{k1}\end{pmatrix}=\begin{pmatrix}0\\0\\...\\0\end{pmatrix}$
For the non-trivial solution, we have to set the determinant of the above matrix to be 0, and as a result we get $k$ values of $E_1$. The lowest value of $E_1$ will be the good approximate to the ground state energy. Corresponding to this we get set of $c_{n1}s$
So, we get $\psi_i(x_1)$ and $E_1$ by the variational principle.
Similarly we can find the $\psi_j(x_2)$
But we have to take into account the spin also. So there are four possible states.
$\chi_1(x_1)=\psi_1(x_1)\alpha(\omega)$
$\chi_2(x_1)=\psi_1(x_1)\beta(\omega)$
$\chi_3(x_2)=\psi_1(x_1)\alpha(\omega)$
$\chi_4(x_2)=\psi_1(x_1)\beta(\omega)$
$\alpha$ corresponds to spin up and $\beta$ corresponds to spin down.
So, total wavefunction $\bar\psi(x_1,x_2)=\text{slater determinant}(\chi_i(x_1)\chi_j(x_2))$
where $i$ can take value $1$ or $2$ and $j$ take $3$ or $4$.
Doubts
i) Is my above analysis correct? While forming combine wavefunction $(\bar\psi)$ how can we determine which combination of $\chi$ will be the lowest in energy? Or is this arbitrary?
ii) In $H_2$ molecule, I have seen that the HOMO is $\sigma_{1s}$ which is formed by the linear combination of $1s$ orbital of first and second $H$ atom. How is this possible?
In the above analysis, we are expanding the $\psi_i(x_1)$ as the linear combination of $1s$, $2s$, etc orbitals of first hydrogen atom. Similarly, $\psi_j(x_2)$ is expressed as the linear combination of $1s$, $2s$, etc orbitals of second hydrogen atom. What is the point of expressing the $\psi_i(x_1)$ as the linear combination of the atomic orbitals of both the hydrogen atoms?
Am I missing something?
Please clarify the doubt.

Do you mean this for helium? Because for an $\ce{H2}$ molecule, the hamiltonian would be different because you have to consider the repulsion between nuclei.

@M.L No the above is fine. The energy for the total energy of the system will include the inter-electron repulsion, but the effective one electron Hamiltonian, which is being considered here, doesn't include such a term. The confusion I think is about what MOs are, and how they are expressed in basis functions - the OP seems to think pis_1 only include basis functions from the 1st atom, when it should be expressed as a linear combination of all the basis functions, it represent a MOLECULAR orbital.

@M.L, by Born-Oppenheimer approximation, we can assume the nuclei to be at rest. So, the- internuclear repulsion will be just constant.

But this is with $\ce{H2}$, so it would be two separate nuclei, no?

Both the nuclei will be considered to be at rest, so the separation between them will also be constant and thus the electrostatic potential energy. So, neglecting internuclear repulsion won't affect the analysis because it is just constant.

Well, isn't the LUMO of an $\ce{H2}$ $\sigma *$. And the formation of molecular orbitals from atomic orbitals shouldn't be the 1s and 2s of one atom. It's simply the 1s from both H atoms forming two molecular orbitals, one bonding and one antibonding.

Yes, it is HOMO. Sorry for typo.

I will provide an answer in multiple parts, and most of the information I describe here comes from source [2].
For a $\ce{H2}$ molecule, the Hamiltonian must consider the interactions between both electrons with both nuclei. So the condensed version would be as described in source [1] and applying the Born-Oppenheimer approximation:
$$\hat{H} = -\frac{1}{2}(\nabla^2_1 + \nabla^2_2) - \frac{1}{r_{1A}} - \frac{1}{r_{1B}} - \frac{1}{r_{2A}} - \frac{1}{r_{2B}} + \frac{1}{r_{12}} + \frac{1}{R}$$
Here, $r_{iA}$ represents how far electron $i$ is from atom A and atom B; $i=1,2$. $r_{12}$ accounts for the inter-electronic interaction and $R$ represents the distance between nuclei. While $R$ can be fixed, the energies are typically plotted against the internuclear distance like a function $E(R)$, hence where you get the Morse potential from.
History and Methods
We will focus on the spatial part for now. In 1927, Heitler and London proposed a method to analyze $\ce{H2}$ in particular called the valence-bond method. The point here is that we'll be constructing a trial function to approximate the actual wave function. There are several different approaches, one of which involves the linear combination of atomic orbitals with molecular-orbital method, but I'll get into that later. First, let's consider a trial wavefunction between two hydrogen atoms that are infinitely apart:
$$\psi_1 = 1s_A(1)1s_B(2)$$
Where $1s_A(1)$ represents electron 1 on the 1s orbital of atom A and $1s_B(2)$ represents electron 2 on the 1s orbital of atom B. One thing we have to consider here is that when the atoms are put closer together, the electrons are indistinguishable. That means we must also consider
$$\psi_2 = 1s_A(2)1s_B(1)$$
For example, electron 2 may be on $1s_A$ and electron 1 on $1s_B$ from the viewpoint of atomic orbitals. This means that the overall wave function can be represented as a linear combination of the two
$$\Psi_{VB} = c_1\psi_1 + c_2\psi_2$$
However, we should also include an ionic term. For example, if two electrons are on one side of the molecule, you'll have a positively and negatively charged molecules on either sides.
$$\Psi_{ionic} = 1s_A(1)1s_A(2) + 1s_B(1)1s_B(2)$$
We will come back to this, but we need to first talk about $\ce{H2+}$ and molecular orbitals.
$\ce{H2+}$ Molecular Orbitals
In molecular-orbital method, we essentially derive the molecular wave functions as a product of molecular orbitals (which are single-electron molecular wave function, see source [4]).
So, to understand how we came to the trial wave function for $\ce{H2}$, we need to first look at $\ce{H2+}$. The Hamiltonian is
$$\hat{H} = -\frac{1}{2}\nabla^2 - \frac{1}{r_A} - \frac{1}{r_A} + \frac{1}{R}$$
Where we have two nuclei A and B and one electron. While it is possible to solve this system exactly, we will solve it approximately to construct the molecular wave function for $\ce{H2}$ eventually. So take:
$$\psi_{\ce{H+}} = c_11s_A + c_21s_B$$
This is a linear combination of atomic orbitals. Again, the point is to create a trial wave function that can approximate the actual wave function of $\ce{H2+}$ well. Taking a linear combination of atomic orbitals has shown success, as we will later see in the treatment of $\ce{H2}$, and that is why it is widely used. The reason why we take the linear combination from both atoms is because they both contribute to the overall wave function.
For this trial wave function, the associated secular determinant would be
$$\begin{vmatrix}
H_{AA}-E\cdot1 & H_{AB} - RS\\
H_{AB}-ES & H_{BB}-E\cdot1\\
\end{vmatrix}$$
When we solve this, though it is somewhat complex and I don't want to type it all out, we will get two solutions, corresponding to the molecular orbitals, bonding and antibonding, formed with an $\ce{H2+}$ molecule. They are
$$\psi_+ = \frac{1}{\sqrt{2(1+S)}}(1s_A + 1s_B)$$
$$\psi_- = \frac{1}{\sqrt{2(1+S)}}(1s_A - 1s_B)$$
where $S = \int dr1s_A1s_B$. Source [2] provides the pathway to the solutions. Here, we've developed a set of molecular orbitals that are analogous to $\ce{H2}$, and we can use them to tackle it.
$\ce{H2}$ Analysis
What the molecular-orbital method wants us to consider is constructing a molecular wave function by combining molecular orbitals. With $\ce{H2+}$, we found the molecular orbitals using a linear combination of atomic orbitals as a trial function. Since $\psi_+$ is the molecular orbital with the ground state energy, we can basically describe $\ce{H2}$ by filling that bonding molecular orbital with another electron with opposite spin. This can be represented by a Slater determinant
$$\Psi = \begin{vmatrix}
\psi_+\alpha(1) & \psi_+\beta(1)\\
\psi_+\alpha(2) & \psi_+\beta(2)\\
\end{vmatrix}$$
$$= \psi_+\cdot\psi_+[\frac{1}{\sqrt{2}}[\alpha(1)\beta(2)-\alpha(2)\beta(1)]]$$
Considering only the spatial part, we get
$$\Psi_{MO} = \frac{1}{2(1+S)}[1s_A(1) + 1s_B(1)][1s_A(2) + 1s_B(2)]$$
$$ = 1s_A(1)1s_B(2) + 1s_A(2)1s_B(1) + 1s_A(1)1s_A(2) + 1s_B(1)1s_B(2)$$
Interestingly, if you do the math, this is almost just like combining the valance-bond method with the ionic terms but without the coefficients. This method is called the LCAO-MO method. This essentially explains how we got to molecular orbital theory, why we express molecular orbitals as a linear combination (as in the section about $\ce{H2+}$), and how we use that to analyze $\ce{H2}$.
Although this is the basis of molecular orbital theory, it is the simple version. We should technically extend molecular-orbital theory as the in our earlier equation with $\Psi_{VB}$ and $\Psi_{ionic}$, each term isn't contributing equally, so we should add coefficients to each term in our trial wave function and solve variationally. This is done by considering the excited antibonding state $\psi_-$, and, amazingly, when we do, we get precisely the form $\Psi_{CI} = c_1\Psi_{VB} + c_2\Psi_{ionic}$. The subscript in $\Psi_{CI}$ represents the extension of simple molecular-orbital theory by including excited-state configurations, known as Configuration Interaction.
In all, we could have used $\Psi_{VB}$, $\Psi_{MO}$ (simple version), and $\Psi_{CI}$ (extended version), or any other more complicated trial wave function to analyze $\ce{H2}$. It's just that the molecular-orbital method with configuration interactions considered is widely used. Source [2] uses this extended form to analyze $\ce{H2}$, but it's rather complicated.
Note that we are not taking into account the $\ce{2s}$ orbital in this analysis as are considering the minimal basis set which leads to two molecular orbitals. Even in molecular orbital diagrams, you will see that the $\ce{H2}$ molecular orbitals arise from a linear combination of the 1s orbitals. We could choose a different trial function to approximate the wavefunction with for example: $\Psi = c_11s_A + c_21s_B + c_32s_A + c_42s_B$. What we find is that when variationally computed, the ground-state molecular orbital becomes $\Psi = 0.7071(1s_A + 1s_B) + 0.00145(2s_A + 2s_B)$. The corresponding energy basically has no improvement compared to solely analyzing the linear combinations of $\ce{1s}$ orbitals with the valance-bond method (see source [2]). So, we are consider solely $\ce{1s}$ here.
Your Analysis
This derivation is in ways similar and different than yours. You basically have the form
$$\Psi_T = [c_11s_A(1) + c_21s_B(2)][c_31s_B(1) + c_41s_B(2)]$$
Sure, you could go about solving this by partitioning the Hamiltonian and ignoring the electron-electron repusion term, but all you really get is the $\ce{H2+}$ wavefunction for each electron. Also remember that the energies depend on the internuclear distance, and that is considered in the analysis of $\ce{H2+}$. So if we also don't include that in the Hamiltonian, then we aren't really analyzing a molecule anymore. Plus, the hard part is not constructing the secular determinant, the hard part is try to mathematically analyze the integrals inside the secular determinant.
As for your comment on spin, for this investigation, we say that the Hamiltonian does not depend on the spin, so the energy is determined by the spatial elements of the wave function.
Sources
[1] http://websites.umich.edu/~chem461/QMChap10.pdf
[2] Quantum Chemistry - Donald A. McQuarrie
[3] https://aip.scitation.org/doi/pdf/10.1063/1.1701472?casa_token=If9tKkNIyIQAAAAA:xYrx4IoQExf2NFTRSNA-ElywJRrhTmbsRrYZIlSPx_vZS9e9DI5EHcMKyU2jJ0AqrIU2d821bbE
[4]http://lampx.tugraz.at/~hadley/ss1/molecules/mo.php#:~:text=In%20the%20simplest%20approximation%2C%20the,r%E2%88%92%E2%83%97ra%7C.
Other Sources
[5] https://www.sciencedirect.com/topics/mathematics/london-theory#:~:text=The%20Heitler%E2%80%93London%20(Ref.,as%20shown%20in%20Figure%2013.2.
[6] https://dasher.wustl.edu/chem478/reading/frontier-orbitals-anh.pdf
[7] http://websites.umich.edu/~chem461/QMChap10.pdf
[8]https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/5%3A_The_Chemical_Bond/5.2%3A_Valence_Bond_Theory#:~:text=along%20the%20axis.-,The%20Valence%20Bond%20Wavefunction,Equation%205.2.
[9] https://www.sciencedirect.com/topics/chemistry/molecular-orbital-method
[10] https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1951.0240


text:
Linking galvanic cell and power supply for electrolysis of water

P- Power supply
Circle - Water electrolyzer
- and + are the terminals
The salt bridge is soaked in potassium nitrate
Electrolytes - magnesium nitrate and zinc nitrate

The water electrolyzer is made of distilled water and sodium bicarbonate ($\pu{1.5 L}$ of distilled water and $\pu{60 g}$ of sodium bicarbonate)
They are connected in series. The power supply and the galvanic cell should add, but it is not working out.
Is the problem the galvanic cell?

Sorry what do you mean like so it is not possible to have 2 electrolytic cells? (I am new to this topic sorry)

This is like a project of mine and I wanted to understand why isn't it possible could you explain to me?

Well like I tried to use the (Galvanic cell and power supply) to electrolyze water. I did this becuase my galvanic cell didn't provide enough to electroylze water, so I needed a power supply. When I performed it with the powers supply alone it worked, but not with galvnic cell.

I tried to also do that and it made no difference I had some help of like some teachers also and they couldn't also make it work

Yeah the current was like very low and it was like 0 amps

Like when I used the power supplt alone for the electroylysis there was 0 amps but the water was being electroylzed I saw the bubbles forming.

Comments reformatted and posted as an answer, my comments then deleted.

Why such a strange scenario ? Why would one want to add a galvanic cell to an electrolysis circuit? // Your have polarity of the galvanic cell reversed, having 2 electrolytic cells.

Sorry what do you mean like so it is not possible to have 2 electrolytic cells? (I am new to this topic sorry). This is like a project of mine and I wanted to understand why isn't it possible could you explain to me?

It is possible to have 2 electrolytic cells, just guessing it is not what you have wanted. From description, it looks you want a galvanic cell and an electrolytic cell.

Well like I tried to use the (Galvanic cell and power supply) to electrolyze water. I did this becuase my galvanic cell didn't provide enough to electroylze water, so I needed a power supply. When I performed it with the powers supply alone it worked, but not with galvnic cell.

Partly because you have the cell orientation reversed, partly possibly due cell internal resistence, depending on the cell capability.
Key:

Electrolytic cell mode: + on +, - on -
Galvanic cell mode: + on -, - on +
Mg electrode is the - one.
You have $\ce{Power- -> ^{-}elyt.  cell_1^{+}  -> ^{-}elyt.  cell_1^{+}  -> ^{+}Power}$
You need $\ce{Power- -> ^{+}galv.  cell_1^{-}  -> ^{-}elyt.  cell_1^{+}  -> ^{+}Power}$

I tried to also do that and it made no difference I had some help of like some teachers also and they couldn't also make it work –

Then the too high cell resistance comes to the game, not able to pass high enough current. Did you test the cell for current provisioning?

Yeah the current was like very low and it was like 0 amps.

If you connect 2 water pipes (power and cell) and one is very narrow(cell), it does not matter if the other(power) is very wide.

Like when I used the power supplt alone for the electroylysis there was 0 amps but the water was being electroylzed I saw the bubbles forming. –

If it was bubbling , the current was > 0.


text:
What chemicals can remove or greatly reduce the smell of burning diesel fuel?
I am not a chemist myself but I am trying to solve a practical problem I am experiencing.
I have a nicely working diesel stove but it produces some smell.
I've read that adding urea to the exhaust may reduce the diesel specific smell? Is that true?
Also what other common (easy to find) chemicals can supress the specific smell of burning diesel?
What about potassium permanganate for example? Is urea the best chemical to supress these smells or there are others which can do better?

depends on the specific setup of the burner and the causes of the smell. Urea is used in cars to reduce NOx in the exhaust but isn't just added to the diesel but injected into the post catalytic converter exhaust in a controlled way.

A vent with sufficient draw. Many diesel-fuel stoves have 12 VDC vent fan to help.

Chemicals generally don't reduce each other's smell. At best, they stun your senses so that you stop feeling the other smell, much like shooting yourself in the leg makes you forget about your sore tooth. A dubious remedy, though.

First off try low sulfur diesel, or better yet: kerosene.
Next try and find a catalytic converter, perhaps off an old Colman heater or maybe at your local wood burning stove store.  Cats do the same wonders in stoves as they have done in automobiles.
NOx should not be too much of an issue if the stove if properly vented.
Try kerosene first.


text:
Neutralizing/ removing the mothball/naphthalene smell from textbooks
I was recently given some excellent (and very expensive) academic textbooks. Unfortunately they must have been immersed in mothballs for a long time. When I take them out of the box, the smell of naphthalene actually gives me a headache within minutes. I have tried airing them, covering them in sodium bicarbonate, but nothing has worked. Any ideas please?

Slowly in, slowly out. It is diffusion driven. Long term heating could work, but also may shorten the book life. Or storing next to well spread active carbon, which would be visual danger for the books.

If the weather is good and sunny, keep the opened books outside during daytime. Repeat until the smell goes away. This may take 1-2 weeks. Baking soda won't do anything.

Check out these solutions from Book Repair Supply (not affiliated in any way).


text:
d orbitals overlapping
I have two questions.

Why can't dx2-y2 orbitals only make sigma bonds? If there is another dx2-y2 coming through z axis, can't they possibly make delta bonds since 4 overlappings seem possible?

Is there a reason why  there is no dz2-y2 or dz2-x2 orbitals except experimental results?

You only have sigma bonds in this pic. but pi and delta certainly exist.

d orbitals can make different types of bonds depending on the orientation and the other orbital. All cases are possible.
the other orbitals are possible but they can combine giving an orbital with S-symmetry. You can have only 5 orbitals with l = 2 so there is an extra combination which is not linearly independent. Cartesian d orbitals in certain electronic structure programs come in groups of six.


text:
Why are free radicals not considered as positive or negative even though they usually have 1 excess electron?
Same as the question, I looked up radicals etc, and could see that methyl radical or phenyl radical had 1 extra electron that would have been a bonding electron normally. However it is not considered a negatively charged compound. Why is it so?
Also do positive free radicals exist? How stable are they?

"Extra electron" of radicals is misleading expression . It is not extra in sense giving negative net charge, but rather extra as an odd electron, not paired with another is a lone electron pair or in a chemical bond. Radical ions exist too, like superoxide anion $\ce{^{.}O2^{-}}$. E.g. a hydrogen atom is a radical, It stops being a radical with no electron as a proton, or with 2 electrons as a hydride anion.

The formal charge of a species is defined as:
$$C=G-N-B$$
Where:
$C$ is the charge of the species.
$G$ is the group number the species belongs to.
$N$ is the total number of non-bonded electrons the species has.
$B$ is the total number of bonds the species has with other atoms.
For example, the formal charge of the carbon atom in a methyl radical would be:
$$C=4-1-3=0$$
A chlorine radical would also be electrically neutral:
$$C=7-7-0=0$$
As would a hydrogen radical:
$$C=1-1-0=0$$
As others have pointed out, there are some cases where the radical can also be negative, like the superoxide radical.
The charge of the oxygen atom that carries the free radical is:
$$C=6-5-1=0$$
While the charge of the other oxygen atom is:
$$C=6-6-1=-1$$
Which gives a net charge of $-1$.

Imagine a neutral molecule. Cut the molecule in half along any single bond. Each piece gets one electron from that bond. Now each half has a single unpair electron, but each piece is also still neutral based on the charge of the original molecule by conservation of charge. Most radicals that we look at fall into this category.
However, you can also have cationic radicals or anionic radicals, where the species with the unpaired electron is respectively positive or negative.
A good example of this is when a sodium-ammonia (Birch) reduction of benzene. The benzene molecule picks up an extra electron, which must be unpaired since all of the other electrons are paired. But since we added an electron, the radical must also be anionic.


text:
What are the relevant pK values on HEPES?
HEPES is 2-[4-(2-Hydroxyethyl)piperazin-1-yl]ethane-1-sulfonic acid. I have read (MA Al-Ghobashy, Bull of Faculty of Pharmacy, Cairo University 2014; 52: 71-78) that the sulfonic acid group is completely dissociated "over almost the whole pH range" and there are two basic nitrogens N1 and N2 stated to have pKa 3 and 7.5 in the mentioned paper and in Wikipedia article. But what is the pKa of the sulfonic acid group? And, in the opposite end we have an OH group - is this inert or does it ever give off the proton?

Both of these values are quite extreme, not really relevant in aq. solutions. Also they probably weren't investigated. Still these are probably typical values for sulfonic acids and alcohols. Did you search for what is usual for them?

@Sam202 No, it doesn't. These are pKa values for both nitrogens. To protonate sulfonic group, you'd need to first have them both protonated, and then go a lot further than that.

@Mithoron This source claims otherwise.

@Sam202 It's pretty impressive they make such error even in paper! This paper isn't about this compound's acidity, though, so could happen. If there weren't two nitrogens, it would be true.

@Mithoron You're right. However, I'm finding conflicting information with respect to those pKa values belonging to the nitrogen atoms. According to this source the pKa of HEPES is 7.55 (at 25°C), and its conjugate base referenced there is the deprotonated form of HEPES with respect to the sulfonic group.

Also, the two nitrogen atoms in HEPES don't have any hydrogen atoms bonded to them, so they should have pKaH values but not pKa values.

HEAP is a zwitterionic sulfonic acid, which should be in neutral form in aqueous solutions like amino acids. Therefore, pKa of 7.55 sounds correct.

@Sam202 Yes, these are both $p\mathrm{K_{BH^+}}$ values. Below pH=8 it's a simple anion. Between about 3-7 it's a zwitterion like a usual amino acid and at higher pH its second nitrogen is also protonated.

Sulfonic acid strength is typically estimated as 10^6 stronger than the respective carboxylic acid. R-OH aciity will be comparable to ethanol.


text:
property of carbon in pi bond that determine a reaction mechanism

This is a mechanism provided in a problem set in my ochem course.
As far as I know, double bond is electron rich, which is nucleophilic and attacks electrophilic species. In this case, why does one double bond attack another carbon in double bond?
IMO, wouldn't a SN2 reaction make more sense?

You need to consider the heteroatoms in the two molecules

SN2 and SN2' (SN2 prime) reaction of this allylic substrate leads to the formation of the same product.The attack of nucleophile to the double bond is explained using frontier orbitals.The attack of nucleophile to the double bond is called SN2 prime reaction.

Several points about the question itself. First, the enamine of n-butyraldehyde is unquestionably of the (E)-configuration and its preparation does not require an acid catalyst. Mannich[1] prepared 1,1-dipiperidinobutane (1,1'-(butane-1,1-diyl)dipiperidine) from piperidine and n-butyraldehyde in the presence of potassium carbonate.  Elimination to the enamine and piperidine occurred upon distillation. Secondly, the enamine employed in the allylation is that of propionaldehyde.
Whether or not $\ce{S_N2 or S_N2'}$ allylation occurs is not straightforward. N-Allylation may be kinetically favored over direct C-allylation. Stork, et al.[2] have noted that Cope rearrangement of the N-allylated product leads to a C-allylation aldehyde which would, if C-allylation were kinetically favored, lead to allylation at the more hindered end of the allylic electrophile. The following is from the Stork paper:

C. Mannich and H. Davidsen, Chem. Ber., 1936, 69, 2106.
G. Stork, A. Brizzolara, H. Landesman, J. Szmuszkovicz, and R. Terrell, J. Am. Chem. Soc. 1963, 85, 207–222.

+1 Nice answer. BTW, is there any reason you chose to screenshot the relevant text from the paper instead of manually writing it?

@Nilay Ghosh: If I were able to type rather than "hunt and peck", I would have considered that option.


text:
What carbonaceous form would result from pyrolysing HMW hydrocarbons and powdered biochar together?
I understand that charcoal/biochar are mostly "amorphous carbon," and that the production of graphite via pyrolysis requires using a <60% C feedstock that will exists in liquid phase during at least a portion of the pyrolysis heating.
However, if a stable (mechanically), mostly-solid piece of carbon were the only goal of the process, could a liquefiable feedstock be used to "wet out," and bind together powdered charcoal, or bio-char in a second pyrolysis/carbonization cyle? And if so, are there any resources where I could find information about the physical properties the resulting final product would display?


text:
Why do we take active mass of water 1 but while calculating pKa of water as 55.345?
First, I should mention that the question was already asked here:Why is active mass of a pure solid or liquid always taken as unity?
But while calculating $\mathrm{p}K_\mathrm{a}$ of pure water we take active mass of pure at $\pu{25 ^{\circ}C}$ as 55.345 and therefore get the $\mathrm{p}K_\mathrm{a} =  15.74$.
$$K_\mathrm{a}=\frac{[\ce{OH−}][\ce{H+}]}{[\ce{H2O}]}=\frac{\pu{e−14}}{55.345}=\pu{1.807E−16}=10^{−15.74}$$
Shouldn't we take active mass of water as one the do the calculation as
$$\mathrm{p}K_\mathrm{a}=−\log{([\ce{H+}][\ce{OH−}])}$$
For $\pu{25 ^{\circ}C}$, $[\ce{H+}][\ce{OH−}]=\pu{e−14}=_\mathrm{w}$
Thus follows $\mathrm{p}K_\mathrm{a}=14$
Why do we use different active masses in the two situation?Any help would be appreciated.
A similar answer is given but why we take active mass of water as 55.345 is not discussed - What is the real pKa of water?

but we consider active mass of pure liquids as 1 so here why don't we consider [H2O] = 1?

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Check also results of Google:  site:chemistry.stackexchange.com pKa water

What makes you think that pure water has one molecules of water per liter?

55 isn't any "active mass", just how many moles of water are in a liter.

The activity of the solvent is not 1. but is just reasonably constant and the concentration is included in such equilibrium constants such as Kw, Ksp, and Ka and Kb. This works because these constants are useful in dilute solutions and the concentration of solvent does not change much. The changes are also accommodated by the incorporation of activity coefficients of ions as a function of concentration and ionic strength. The only case that I can think of where pure liquid and solid have the same activity each defined as 1. is the freezing-melting point at standard pressure or possibly the three phases at the triple point.
One of the reasons for arguments about the acidity of water, methanol, ethanol etc. is inability to consider the different solvent concentrations.

If the standard chemical potential is considered for the pure liquid then the activity of the pure liquid is 1. $\mu = \mu_0 +RT\ln{\frac{a}{a_0}}$

Just to confirm, activity is defined as the effective concentration right? And effective concentration is said to be one for pure liquids... So activity IS nearly 1 right? And not just any constant


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What does it mean stir at room temperature overnight?
It was mentioned in synthetic procedure to "stir at room temperature overnight".
What does this mean?  (I want the exact duration for how long should I keep stirring.)
How is it possible to carry this out during day time, due to my limitations?

Do you have method for following the progress of your reaction? The procedure means that the reaction is not time sensitive once it has gone to completion, it can be left running. It may be that it takes 12+ hours to go to completion. Yes, it can be worked up before that at the risk of a reduced yield.

It can be calculated. the needed start and stop of stirring  depends on the geographic latitude, longitude, date and if there is meant astronomical night, nautical night, citizen night or Sun (sunset-sunrise) night. // Now seriously, overnight intentionally means the exact duration is not critical. Unless there is light sensitivity, it can be done over working hours too. doing so overnight means more handy waiting while sleeping.

If the instructions say that then "exact" time is not an issue. Probably anything around 12±6hr will work.

There may be magnetic stirring bars available for the lab or even your local high school. Simply get one(or ask prof where it is), put it in the beaker and put the beaker in a rotary. Then go to sleep.

"Stir at room temperature overnight" is lazy researcher's way of saying the next morning that s/he forgot to write down when the stirring started. You are correct. Duration is important.

If you are talking about a solution, you may use a magnetic stirrer to stir your solution overnight, preferably kept in a fume hood or well ventilated area. Depending on the volatility of your solution you might want to cover it to prevent evaporation and weight loss.

or run it under a nitrogen blanket

What is the point of stirring a solution? Force of habit perhaps. A two phase mixture, certainly.

@user55119 diffusion is very slow for centimer-scale lengths.  https://www.youtube.com/watch?v=MMAYOvlzD6w


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How to understand the acid-base properties of a molecule with many functional groups: example biotinyl-5'-AMP
Background
I'm interested in the acid-base properties of a metabolite, biotinyl 5' AMP in the cell. Biotinyl-5'-AMP can react with protein lysine side chains at the anhydride linkage to form biotinyl-lysine. This immediate reaction is not, however, the focus of this question. I'm struggling to understand for sure whether, in its soluble state, biotinyl-5'-AMP is likely to adsorb or donate (further) protons to its immediate chemical neighbours in the cell, particularly weak bases or acids.
Question
How can I intuitively understand the acid-base properties of an organic molecule in solution with multiple functional groups? Biotinyl-5'-AMP is deprotonated in solution, as far as I am aware. Would this deprotonated species then have further ability to donate or adsorb protons? I have found database information about this molecule suggesting the pKa of the strongest acid and strongest basic groups (0.83 and 4.99 respectively), and a line that the molecule is likely to be a 'strong basic compound'.
Given those database figures and the structure/functional groups (figure below): How can I intuitively understand the acid-base properties of this compound in solution? How should I interpret and rationalise those database pKa values? How should I then predict the propensity of biotinyl-5'-AMP to influence the pKa of immediately-adjacent weak acids and bases in the cell?

It's essentially just another amino acid. It's a zwitterion around pH=5. This pKa about 1 is for cation (its protonated form).

How can I intuitively understand the acid-base properties of an organic molecule in solution with multiple functional groups?

You should view the molecule as a set of different acids and bases that just happen to be connected in the same molecule.  Some of those acidic groups like the phosphate anhydride might have low pKas; some of those basic groups like the exocyclic amine might have high pKas.  If you understand how mixtures of strong and weak acids and bases work you understand how this molecule works.  The only additional constraint is that because all the groups are connected in the same molecule, all the different acid/base species here have the same concentration, instead of independent concentrations.

Biotinyl-5'-AMP is deprotonated in solution...

Well that depends entirely on the pH!  (Although at low pH where the phosphate is less likely to be protonated I would expect the rate of degradation via hydrolysis to be fairly significant...but that's just an technicality of this particular molecule that's not relevant to acid/base chemistry of "big" molecules in general...)

How should I interpret and rationalise those database pKa values?

With extreme caution.  The folks who compute these pKas aren't always right, and even when they are right it's not always clear what acid/base equilibrium they are characterizing with their prediction.

How should I then predict the propensity of biotinyl-5'-AMP to influence the pKa of immediately-adjacent weak acids and bases in the cell?

The same way any other mixture of acids and bases would...if you are trying to understand how this molecule affects a single molecule to which it is bound, it will be more helpful to think of H-bond donating and accepting.  "Acidic" and "basic" make a bit more sense as bulk solution ideas

Re: 'deprotonated in solution' : As a biochemist I momentarily forgot that pHs actually exist outside of the 7.2--7.8 range :D Thanks for the correction.

This answer is very useful and explains the landscape I was trying to understand very clearly. Regarding your last paragraph, I think that I more trying to ask a question along the lines of, "would a high local concentration of this molecule induce an increase or decrease in pH at a point within a cell" (starting pH ~ 7.4). Can you include a briefly reasoned approximation based on the prominent ionisable groups which you see? Or, is it too complex to be able to approximate? It would be useful to know the degree to which one can intuitively answer this problem without a rigorous calculation.

The diffusion time of protons is very fast.  For cells that are 10 microns big, the diffusion time is approximately ~5 milliseconds.  So whatever perturbation this molecule has on a local area in the cell, it isn't going to last long.  Or, if you like, treat the relevant length scale as 10 nm.  This is probably (ballpark) the hydrodynamic radius of biotinyl-ATP.  The diffusion time over 10 nm for protons is like 5 nanoseconds.  So whatever "local concentration" of this metabolite exists at a point might perturb the pH, but not for very long.


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Pressure Washing with sodium hypochlorite (bleach) but I want to use coated sodium percarbonate
I am in the pressure washing industry. I don't want to use sodium hypochlorite (bleach) to clean exteriors of properties because it is so corrosive to my equipment and people's homes and businesses. I want to start using coated sodium percarbonate powder (peroxide). How much SP do I need to add to a gallon of water to equal 12.5% SH cleaning power. Also, it seems to only last about 6 hours. Is there another chemical I can add to it to make it last much longer?

Why don't you first dilute your bleach solution, before tying to change its composition ?

Define SH cleaning power, please? Bleach is indeed corrosive.

I'm afraid this might be a violation of Chemistry.SE's homework policy.

@AChem Coming from a non-chemist my definition of SH cleaning power is how much SP per gallon do I add to one gallon of water to equal the same cleaning power of 12.5% SH.

@John, You never defined what SH is ? After that you introduced a new abbreviation SP. Please write full words.

@AChem I believe OP is abbreviating sodium hypochlorite and sodium percarbonate but I agree with you to write full words.

I agree but he still does not define what cleaning power is. There is no measure of "cleaning"...hence there is no answer

While I applaud you're attempt to avoid more-toxic chlorine compounds, I don't know that there is a simple comparison of cleaning efficacy between hypochlorites and peroxides. Perhaps the manufacturers have a guide? Also, would not the coating on coated sodium percarbonate interfere with the pressure washer? See https://propowerwash.com/board/upload/threads/premade-sodium-percarbonate-siding-cleaning-mix.30867/


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Pourbaix Diagram and Galvanic Cell
I am trying to understand how a galvanic cell where copper and mild steel are linked occurs. It seems that on most galvanic series, Fe and Cu are far enough apart (>.2V difference) that Fe would preferentially ionize to Fe+2, while copper is "plated" onto the copper electrode from solution.
I understand how to calculate the voltage difference given by the galvanic cell (roughly .45 V difference). What I do not understand is why the iron electrode even corrodes at all. On Pourbaix diagrams of Fe around pH of 7, it seems that either the cell potential of -0.45V or the galvanic series potential of -1.1V for iron would both put it near/within the stable Fe area of the Pourbaix diagram.
My questions is:

Which potential/voltage do I use to determine stability on a Pourbaix: overall cell potential or galvanic series potential?
If the pH and potential on a Pourbaix put it in the stable metal range, does that mean that no reaction will occur (since Cu needs to "get" e- from Fe turning into Fe+2)?

I believe you might not be completely sure on how to read a Pourbaix diagram.
The potential you use in the Pourbaix diagram is the potential at your electrode, this might be the mixed potential at your electrode, applied potential, potential formed when you connect the electrochemical circuit, etc.
A Pourbaix diagram is constructed using the Nernst potential, which is not a potential difference, but an equilibrium potential (relative to some reference). Your system will not be thermodynamically stable so long there is a net driving force, in your case until the concentrations of Fe2+ ions and Cu2+ ions are so that the cell potential (difference in Nernst potentials) is zero.
If the Pourbaix diagram indicates that you have the Fe species, then Fe2+ ions will be plated on your electrode until the cell potential is 0. If there are no Fe (or Cu) ions anymore you might notice that the Nernst equation blows to infinity. In reality no reaction will indeed occur, or other species (for example the solvent) might start donating/accepting electrons.
In short, current will always flow (although possibly negligible amounts) in the case that there is a non-zero cell potential.


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is it possible to make crystals of multiple organic compounds that form a homogenous mass under normal conditions
I was thinking about crystals of organic compounds recently after I had to clean beeswax off a pot. it's an extremely tedious process to remove. you have to boil water in the pot and then continually remove the liquid wax while hoping it doesn't adhere to the side. so I looked up what beeswax is soluble in, it's soluble in most notably acetone. I grow crystals just for fun mostly because they look cool. anyway, beeswax is composed of multiple organic compounds and im wondering if anybody has tried to grow crystals of beeswax or similar organic compounds. lastly is it even possible to make a crystal with multiple primary constituents? I can't really find a lot of information on this so I decided to ask here

You've surely seen some books on organic chemistry, right? Seen their breadth? Now that's how far are some organic compounds from some other organic compounds (and farther than that, too). Some would form nice sparkling crystals readily, and some wouldn't until you persuade them really, really hard, or maybe ever at all. Beeswax is on the bad side.

nope I have not. I would consider myself a hobby chemist really.

Well, then the short answer is: no, you won't crystallize beeswax or any of its components. Try something nicer, like the examples from DrMoishe Pippik's answer.

Some chemicals, organic or not, naturally separate when crystallized. For example, mixed dextro- and levo- forms of tartaric acid ("racemic acid") were separated by Louis Pasteur by the arduous process of picking out right-handed from left-handed crystals.
Some chemicals form mixed crystals, such as chrome alum, $\ce{KCr(SO4)2}$ and ammonium alum, $\ce{(NH4)Al(SO4)2}$. Since these are both octahedral, with similar layout, crystallization would not serve to separate the components of a mixed solution; in fact, one can grow crystals ranging from colorless to deep purple by varying the proportions. This is an example of miscible solids.
As you state, beeswax is a mixture of many compounds. You might try to grow crystals of one component using various solvents, leaving most components undissolved, but crystallizing the mix is not feasible. BTW, honey naturally crystallizes since it's mostly glucose and fructose, already supersaturated. Emulate Pasteur, and pick apart some glucose and fructose crystals from crystallized honey. Eat the rest.

thanks this was very informative. I now believe that it is unlikely that more than one crystal compound would form because as soon as one crystallized the solution would likely drop below the necessary saturation to form crystals of any of the remaining compounds.

this is a continuation: perhaps by this procedure, it would be possible to get crystals of varying composition in the same container without removing the solution. essentially 1. make a supersaturated solution of beeswax in acetone and let one of the compounds condense into a crystal. 2. make more batches of solution and repeat step one with those. 3. take the liquid solution from a batch then concentrate probably via evaporation to below a supersaturation point where crystal formation is possible again(not sure what this concentration would be).

continuation: 4. add the concentrated solution from step 3 to the original solution while the formed crystals in that solution are still present. this should probably be done while the solution is heated from above slightly to increase the saturation to the point where it can form crystals while avoiding disturbing the liquid. lastly, a cooling system at the bottom of the vessel could help in forming crystals by increasing the saturation in lower layers, creating a temperature gradient resulting in convection, and lowering its ability to carry the organic compound next in line for formation.

continuation: 5. repeat steps 1 through 4 each time removing from the solution to be added to the primary vessel the last crystallized compound in the primary vessel till you have in the primary vessel all the constituent compounds of beeswax in crystal form(if this is possible at all)

it's also possible a centrifuge could expedite the process if the densities of the constituents of beeswax are not too similar in density. if visual separation is off issue perhaps you could find a process to bind dyes to various compounds put the dyed beeswax through the centrifuge and separate based on color. looking over this whole thing it's not super worth it to satisfy my love for shiny or geometrically interesting objects.

it is possible that since beeswax is a homogenous liquid when melted that it could readily form cocrystals(I just learned this is a thing so it adds some reasonability to the idea thanks mostly to your comment about alum crystals). it's also reasonable that beeswax could form a non-visually evident crystal structure dissolved in itself similar to how a plastic water bottle can be below it's freezing point and in a state of supersaturation in itself until a nucleation point is provided by hitting the water bottle freeing plastic particles.

Great ideas...  but beware that some pure compounds, and many mixtures, never crystallize. "[M]any organic compounds continue to resist attempts to crystallize them," https://www.ncbi.nlm.nih.gov/pmc/articles/PMC8239258/

thanks for the paper. also, I have recently learned that the wax part of beeswax forms crystals called spherulites so it is most certainly possible to crystallize large scale the waxy polymer compounds of the wax can by crystalized.


text:
Doubt in heterogenous equilibria
Let us take the following equilibrium:
$$\ce{NH4HS(s) <=> NH3(g) + H2S(g)}$$
I have been taught that since $\ce{NH4HS}$ is in the solid state, its concentration cant be taken and hence we write $$K_{\mathrm{eq}}=[\ce{NH3}][\ce{H2S}]$$
This means that according to Le-Chatelier's Principle, alteration in $[\ce{NH3}],[\ce{H2S}]$ results in equilibrium is disturbed. However, alteration in $[\ce{NH4HS}]$ won't disturb the equilibrium. This seems a bit confusing to me.
Logically, if we had large quantity of $\ce{NH4HS}$, large quantities of $\ce{NH3}$,$\ce{H2S}$ would be produced. However, the equation tells us that $[\ce{NH4HS}]$ won't matter at all. Please help resolve my confusion.

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The change in $\ce{[NH3]}$ and $\ce{[H2S]}$ when $[\ce{NH4HS}]$ is added is negligible compared to the change in $[\ce{NH4HS}]$ when $\ce{[NH3]}$ and $\ce{[H2S]}$ are added.

Large quantity of [X] is nonsense. Either large quantity of X, either high [X], either high concentration of X.

The amount of gas produced only depends on the free (gas) volume of the vessel. Obviously, because otherwise any large enough vessel with only a little free space would burst!

For WHat reason my question is being downvoted? I have asked for a simple physical chemistry concept along with my undertsanding.

If no downvote reason is provided, there can be often applied the default one, revealed at the downvote bottom ( The question does not show any research effort; it is unclear or not useful. ). // Always try to find answers on your own in offline and online resources, before asking.  It will give the user more than if one just waits to receive the answer.  Questions that look like it has not been done are frown upon on StackExchange site network.

Equilibrium constants deal with ratios of reactants and products (under a given set of conditions), not quantities.  Sure, if you had more NH4HS, you would expect more NH3 and H2S.  Also, consider rate of reaction, or how much time it takes to reach equilibrium.

Short answer:

A different amount of solid substance does not shift the equilibrium.
A different volume ( but the same partial pressures) of gaseous phase does not shift the equilibrium.

Because these are extensive properties of systems.
An equilibrium is based on intensive properties, like concentration, activity, partial pressure, molar fraction or chemical potential.

However, alteration in [$\ce{NH4HS}$] won't disturb the equilibrium.

The point is, adding $\ce{NH4HS(s)}$ does not mean alteration in [$\ce{NH4HS(s)}$], similarly as more of the same gaseous phase does not mean alteration in [$\ce{NH3(g)}$] nor [$\ce{H2S(g)}$].

At the reaction equilibrium, there is minimum of the Gibbs energy
$$G=H-TS=U + pV - TS,\tag{1}$$
therefore $$\mathrm{d}G = 0 \tag{2}$$.
There is defined the quantity chemical potential $\mu$ of a substance as the partial derivative of Gibbs energy of the system per molar amount of such a substance.
$$\mu_i = \left(\frac{\partial G}{\partial n_i} \right)_{\text{const }T,p,n_j; j \ne i}\tag{3}$$
At constant temperature and pressure
$$\mathrm{d}G = \sum_i{\mu_i \cdot \mathrm{d}n_i } \tag{4}$$
$$\mathrm{d}G = \mu_{\ce{NH4HS(s)}}\cdot \mathrm{d}n_{\ce{NH4HS(s)}}  + \mu_{\ce{NH3(g)}}\cdot \mathrm{d}n_{\ce{NH3(g)}} + \mu_{\ce{H2S(g)}}\cdot \mathrm{d}n_{\ce{H2S(g)}}\\
=\left(\mu_{\ce{NH4HS(s)}}  - \mu_{\ce{NH3(g)}} - \mu_{\ce{H2S(g)}}\right)\mathrm{d}n_{\ce{NH4HS(s)}}\tag{5}$$
and due (2):
$$\mu_{\ce{NH4HS(s)}}=  \mu_{\ce{NH3(g)}} + \mu_{\ce{H2S(g)}}  \tag{6}$$
Now, $\mu$ of solid substances is practically independent on their amounts, if we neglect changes of pressure of otherwise gaseous systems.
$$\frac{{\mathrm{d}\mu_{\ce{NH4HS(s)}}} }{ {\mathrm{d}n_{\ce{NH4HS(s)}}}}  \approx 0 \tag{7}$$
Therefore, $\mu$ of $\ce{NH3(g)}$ and $\ce{H2S(g)}$ and therefore their partial pressures or concentrations are (practically) independent on the amount of $\ce{NH4HS(s)}$, as for ideal gases,
$$\mu = \mu^{\circ} + RT\ln{\frac{p}{p_0}}\tag{8}$$
Therefore, the amount of the solid component does not have (in the first approximation) impact on the equilibrium.
Note that (molar) concentrations are not much used for gases. The formal relation of concentration and partial pressure is:
$$n_i=\frac{p_iV}{RT} \implies \frac{n_i}{V}=\frac{p_i}{RT} \implies c_i = \frac{p_i}{RT} \tag{9}$$

Feedback response:
For a pure solid substance, its $\mu$ is a parameter characteristic for the substance itself, not for its amount. By other words, $\ce{NH4HS(s)}$ does not have bigger thermodynamic tendency to decompose if there is a bigger amount of it.
It is analogical as having bigger volume of the gaseous phase, but with the same partial pressures of $\ce{H2S(g)}$ and $\ce{NH3(g)}$, therefore having their respective $\mu$ independent on the gas phase volume.
There would not be created more of $\ce{NH4HS(s)}$ if there is doubled the gas phase volume, keeping partial and total pressures. OTOH, if there is increased their partial pressure, their $\mu$ would increase and the there would be formed more $\ce{NH4HS(s)}$, until $\mu$ of substances get balanced again.

So our chemical potential of a compound is analogous to something like density which is also independant of initial mass/concentration. Am I right?

See the answer update.

So, adding more $\ce{NH4HS}$(s) will not produce more $\ce{NH3}$(g) and more $\ce{H2S}$(g)

@Maurice Assuming the system is already in equilibrium and addition of solid causes zero (p=const) or negligible shift of partial pressures, it will not. Different initial amounts will lead to the same equilibrium partial pressures. The key is Eq. no (6).

@Poutnik. So you agree with my interpretation.

@Maurice. I have thought it was obvious, I have just elaborated it. Perhaps not to your profit. :-)

I have been taught that since $\ce{NH4HS}$ is in the solid state, its concentration cant be taken and hence we write $$K_{\mathrm{eq}}=[\ce{NH3}][\ce{H2S}]$$

The other way to think about it is that the concentrations in the equilibrium constant expressions are relative to a standard state (so they are dimensionless, without units). The standard state for a solid sample is the pure solid sample, so it is always 1.

This means that according to Le-Chatelier's Principle, alteration in $[\ce{NH3}],[\ce{H2S}]$ results in equilibrium is disturbed. However, alteration in $[\ce{NH4HS}]$ won't disturb the equilibrium. This seems a bit confusing to me.

Here is a home chemistry example. If you have a saturated salt solution, will adding more solid pure salt change the concentration in solution? No, it will not. However, if you lower the concentration of salt in the solution (by adding more water), the equilibrium will be disturbed, and more solid salt will dissolve.

Logically, if we had large quantity of $\ce{NH4HS}$, large quantities of $\ce{NH3}$,$\ce{H2S}$ would be produced. However, the equation tells us that $[\ce{NH4HS}]$ won't matter at all. Please help resolve my confusion.

Quantity and concentration are two different things (the first is extensive, and the second is not). Let's say you have some $\ce{NH4HS}$ in a closed bottle. It makes a certain amount of ammonia and dihydrogen sulfide, according to the equilibrium expression. If you want to make more of these gases, you could set up more bottles. This shows you that you can scale this up. The concentration in each bottle would still be the same, but you would have a larger amount.
Alternatively and less complicated, you could switch to a larger bottle, or you could continously pump out the gases from your bottle to make more (higher amount of substance) of the gas species. However, the concentration of the gases would still be the same (if it were higher, you would produce some $\ce{NH4HS}$ again.

A PURE solid or a PURE liquid has a constant chemical composition and constant chemical activity.  There is a dependence on crystal or droplet size when either become very small and most of the molecules are exposed and not in contact with others hence the interest in nanochemistry. The amount does have an effect on kinetics, the time it takes to reach equilibrium, and there must be enough to establish equilibrium. You seem to think that its concentration, activity, is not considered. It is definitely considered! It is a constant and therefore can be defined; it is convenient to define it as 1.. If the solid or liquid is impure its activity must be redefined, usually as the mole fraction or as molality [moles per constant mass of solvent].
Instead of worrying about solid state chemical reactions take a step back and investigate simple phase changes such as water-ice where both activities are 1. or water-water vapor where the activity of the water can be considered 1. and the activity of the vapor is its vapor pressure. [The second case is more complicated since the activity of the water is a function of temperature because of the higher vapor pressure with higher T but you can redefine it as 1. at each temperature.] Once understood then try some simple reactions such as NH4HS decomposition


text:
Why has acetaminophen only two different NMR chemical shifts related to phenyl?
I learned that, by using NMR, you could tell how many different hydrogen environments an molecule has. For example, benzene has only one.
Since acetaminophen has an amide bond, and the bond has planarity, I thought the phenyl group of acetaminophen has four different hydrogen environments.
But the textbook said only two. What is wrong with my thinking?

Nomenclature: It's not phenyl, which is -C5H6, and in NMR, what you called "wavelenght" is termed "chemical shift".

@Karl typo maybe? Phenyl = C6H5.

@OscarLanzi Oops. :)

The 'amide bond being planar' refers to restricted rotation about the C(O)–N bond, not the N-aryl bond.

The restricted rotation about the C(O)–N bond does lead to inequivalence in NMR, depending on the compound: a good example of this is N,N-dimethylformamide, where the two methyl groups have different chemical shifts.*
However, in your compound, this restricted rotation is irrelevant. It is rotation about the N–aryl bond which makes the pair of ring protons H-2 and H-6 chemically equivalent (and H-3 and H-5).†

* At room temperature, at least. If you heat it up enough the two peaks coalesce because there is enough energy to overcome the barrier to rotation.
† They are not magnetically equivalent, see e.g. NMR magnetically equivalent protons for a 1,4-disubstituted benzene ring, although that is a separate matter from what you're asking about.


text:
Which substrates will undergo an E2 reaction to give a (Z) alkene?
What I do know: E2 reactions are concerted and have an anti-coplanar transition state, where the leaving group and the hydrogen being abstracted have a dihedral angle of 180 degrees.
What I don't understand: A does not have a halogen that acts as a leaving group. B has two halogens (bromine and fluorine). Thus, how do I determine the substrate(s), if any, that will undergo E2?

A has a mesylate that is every bit as good a leaving group as a halogen. In B is there a proton with a dihedral angle of 180 to a halogen?

I would say that it's possible thath both of them could undergo an E2 step. A has the mesylate group that acts like an halogen. B in one of its conformation (may be not the stablest one) has an hydrogen atom in anti-coplanar.
To choose between F and Br: Br is a better LG than F. To remove an F atom you need very strong conditions

Be careful with the CIP rules in determining the E/Z assignment. While Br is the better leaving group, configurationally, does it matter whether HBr or HF is lost?


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Are aromatic methoxy-groups more para-directing than aromatic TBDPS-ethers?
I'm searching a comprehensive explanation for the following issue. Even though I knew before, that a regioselective bromination of compound 1 could be problematic, I wondered that the undesired 6-brominated product 3 was always the main product under various conditions. I assumed that the sterical hinderance of the bulky TBDPS-group could drive the reaction towards the desired 2-brominated product 2.
However, as this does not seem to be the case and as both protecting groups are electron donating-groups (+M) I am somewhat lacking an explanation for this issue. I am familiar with the alpha and beta effect of silicon and think they don't play any role in this case. Is there anything I overlooked? Thank You very much for helping me out with this :)

I suspect the observed outcome is a combination of two major factors.
First, it is indeed expected that siloxy substituents are poorer electron donors via mesomerism, compared to alkoxy substituents. One indication comes from the comparison of the pKas of the parent alcohols and silanols. For example, trimethylsilanol has an aqueous acidity constant approximately eight orders of magnitude larger than the carbon analogue, tert-butanol (pKa 11 vs. 19, respectively). This suggests the conjugate base is being stabilized by the silicon, and pulling the electron density away from the oxygen atom. In the past this was attributed to partial $\pi$ bonding between the 3d orbitals of silicon and 2p orbitals of oxygen (a type of hypervalency), but this is now not thought to be an accurate explanation. Instead, it comes down to hyperconjugative effects, where the oxygen lone pairs donate into the Si-C $\sigma^*$ orbitals.
Secondly, one could argue that the tert-butyldiphenylsiloxy substituent is less sterically demanding than may be expected from intuition. Organic chemists spend a lot of time dealing with p-block elements from the second row of the periodic table, which are anomalously small due to the lack of radial nodes in the valence 2p orbitals. Most atoms are rather larger. Typical Si-O and Si-C bond lengths are ~40-50% longer than the C-O and C-C equivalents. This means that the apparently enormous amount of steric bulk from the simultaneous tert-butyl and diphenyl substituents on silicon is actually spread out of a considerably larger volume, weakening their steric hindrance effects. A very crude attempt at visualizing this difference is shown in the structures below. The left structure has default bond lengths between all atoms, whereas the structure on the right has had the Si-X bond lengths increased by 40%, to more accurately reflect the increased size of the silicon atom. Note the increased space surrounding the position opposite to the methoxy substituent, even in the most crowded conformation. Of course, this is a 2D representation of a 3D structure, so in reality there is an extra spatial dimension further diluting the steric bulk. Roughly speaking, the silicon atom causes the bulky substituents to be diluted in approximately $1.4^3=2.74$ times as much volume.

I completely agree with the hyperconjugative effects apparent in the Si-O bond. However, I thought this effects only attribute to the inductive (-I) character of the whole TBDPS-protecting group and thus only play a minor role in the directing activity. Thank you for the explanation regarding the sterics. I know that these bonds are longer, however, I never heard of this explanation when it comes to compare sterics of protecting groups. Generally, the TBDPS group is denoted as a much more sterically hinderinggroup compared to the methoxy-group.

To make the long story short: Maybe I will try a trityl-protecting group. It would have no electron-withdrawing hyperconjugation effects on the oxygen (3 aromatic cores in the second plane should have no significant impact) and would also be located much closer to the ortho-positions.

A trityl is an interesting option, though I have no experience with it as a protecting group. It sounds like it should work in principle, but it might be worth checking SciFinder/Reaxys for halogenations in the presence of a ArOTr group. Best of luck!

Would iodination rather than bromination be an option if you are trying to uses steric effects to direct substitution? As it is larger it will be more subject to hindrance.

Thanks for your input. The next step would be a lithium-halogen exchange reaction. I think Iodine would work also for this reaction, however, in the literature there are comparably more bromine substituents used for that purpose and even more information about regioselective procedures to introduce the Br...


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Chemical formula for this potential denitrification inhibitor
I am trying to track down a supplier of this compound (a potential fungal denitrification inhibitor to reduce N20 emissions from pastureland), but all I have is its structure from the computational chemistry paper where it was identified. The supplier wants the formula at the very least. I am not a chemist, but hoped someone here could translate the structure into a formula (or a name.)

Too short for an answer, but a search using Chemspider (you can draw the structure) gives : http://www.chemspider.com/Chemical-Structure.1553572.html?rid=b828c58e-9722-42e5-9a00-478cede7fcd7 - so "2-{[4-(1-Piperidinyl)phenyl]amino}-6-propyl-4(1H)-pyrimidinone"

If you got your answer, feel free to post a self answer. Self answers are appreciated and it helps future readers.

Thanks to Gilleain's comment, the formula for that structure is:
2-{[4-(1-Piperidinyl)phenyl]amino}-6-propyl-4(1H)-pyrimidinone
Thanks to Chemspider's links, I was able to find a supplier. (But unfortunately, the octanol/water partition coefficient is high (3.6) - which means it's considered bio-accumulative under REACH and very unlikely to be approved for field-spreading.)

The preferred IUPAC name is: 2-[4-(piperidin-1-yl)anilino]-6-propylpyrimidin-4(1H)-one


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Reproducing Hydrogen Molecule Hamiltonian in OpenFermion
I am learning quantum chemistry at the moment and I'm trying to understand the Hamiltonian generated by the OpenFermion package. I'm now stuck at understanding how openfermion calculates the coefficients in second quantization Hamiltonians.
Take the for Hydrogen molecule for example. My understanding is that the coefficients

$h_{00}$, represents electron 1 in 1s orbit with spin up.
$h_{11}$, represents electron 1 in 1s orbit with spin down.
$h_{22}$, represents electron 2 in 1s orbit with spin up.
$h_{33}$, represents electron 2 in 1s orbit with spin down.

As the wavefunction of 1s orbits takes the form
$$\psi_{1s}(r) = \frac{1}{\sqrt{\pi}}e^{-r}$$
The $h_{ii}$ should be
\begin{equation}
2\pi \int_0^{\infty}\int_0^{\pi}\psi_{1s}(r)\left(-\frac{1}{2}\nabla^2-\frac{1}{r}-\frac{1}{\sqrt{r^2+R^2-2r R\cos\theta}}\right)\psi_{1s}(r)r^2 \sin\theta d\theta dr
= -1.07123
\end{equation}
when $R = 0.74$ Bohr radius, which does not match with openfermion's
\begin{align}
h_{00} &= h_{11} = -1.2524635735648988\\
h_{22} &= h_{33} = -0.47594871522096416
\end{align}
Can anyone tell me why I am wrong?

Is there spin down in $h_{22}$ ? Is it not spin up ?

yeah it should be spin up. I just fixed the typo

Isn't this just basically considering two separate H atoms, not a hydrogen molecule? I'm quite inexperienced in this, but do we not have to consider the attraction of two electrons to both separate nuclei? Also, why is $h_{11}$ not equal to $h_{33}$ as both have the same spin and the assignment of which electron is electron 1 or 2 is arbitrary.