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CCSD and CISD size consistency problem
I have performed a Gaussian16 calculation to explore the size-consistency problem for the CISD and CCSD methods. The system I am studying is the water dimer and the results are not what I expected at first. I have computed the energy of the dimer for both models (single point calculation) where the two monomers are 100 Å away and the energy of the and the energy of the monomer.
I wanted to compute the difference between those energies for both methods as two times the energy of the monomer and the energy of the dimer. If a method is size-consistent, the difference should be zero. These are my results:
$$
\begin{array}{lrr}
\hline
\text{Quantity}  & \text{CCSD} & \text{CISD} \\
\hline
\text{Monomer energy}/\pu{Ha}  & -75.012  & -75.012  \\
\text{Dimer energy}/\pu{Ha}    & -150.030 & -150.026 \\
\text{Abs. difference}/\pu{Ha} & 0.004    & 0.002    \\
\hline
\end{array}
$$
I have been reading many articles and papers and what I expected is that CISD is not size consistent for a system of more than two electrons while CCSD is size-consistent.
However, the value of the difference is larger in the case of CCSD, what makes me think that it is telling me that CCSD is less size-consistent for my problem than CISD. Am I wrong? What do these values mean?

When you write monomer energy, is it the true monomer energy, or did you use the "dimer" at 100 Angstrom distance?

It is the monomer energy, while the distance of the dimer is the energy of the dimer where the two monomers are 100 A away

I don't have the answer to your question, but you could opt to migrate this question to matter modelling SE. That site is specifically for computational modelling and there may be a better chance of an answer there.

Thank you for the suggestion. I have already posted the question and can be found here: https://mattermodeling.stackexchange.com/questions/4693/cssd-and-cisd-self-consistency-problem


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Schrödinger equation and degeneracy of atomic orbitals
How does the application of the Schrödinger equation to model a system, such as a particle in a box, help us understand the origin of the degeneracy of atomic orbitals?

It doesn't. Unless we apply it to a model system which is a hydrogen atom, that is.

I would argue it does if you consdier a 2D (or 3D etc.) box and consider under what conditions you can get degeneracies. But I also would argue this isn't the world's greatest ever exam question.

I'd say the simplest model possessing degeneracy is the particle on a ring, which is quasi 1D for fixed radius.

Degeneracy refers to states having the same energy.
If you have two or more coordinates that can be related by a symmetry operation then the system will contain degenerate states because swapping the coordinates associated with those degrees will result in a Hamiltonian with the same solutions.
This is the case in both the hydrogen atom and in the particle in a cube.

Non accidental Degeneracy implies symmetry but symmetry does not always imply degeneracy. The 1D particle in a box system does have inversion symmetry if you construct it with the center at the origin but it has no degenerate states. A good discussion is found here, https://physics.stackexchange.com/questions/319096/what-is-the-relationship-between-symmetry-and-degeneracy-in-quantum-mechanics

I know what you mean and your answer is ok. My comment is intended for questionener's/readers that may want are more in depth view at symmetry and degeneracy. It's easy to jump to the wrong conclusion that symmetry always implies degeneracy so i thought it wouldn't hurt to mention it in a comment.

@HansWurst You're right, and I repost my comment because I was unclear. I meant symmetric coordinates ie rotate x to y. If you have the same Hamiltonian after a symmetry operation that swaps coordinates then degeneracy follows. That's all I meant.

This is not right. Consider the Hamiltoninan $H=-\partial_x^2-\partial_y^2+(x-y)^2$ with Dirichlet boundaries at $x,y=\pm1$. It's invariant with respect to reflection $x\leftrightarrow y$, but the symmetry group is abelian, so you don't get degeneracy (except maybe accidental).

@Ruslan undoubtedly you are correct. However I don't think referring to Abelian groups will enlighten the OP (or me for that matter, who cannot explain it other than saying abelian group elements commute). Maybe I should add something hand-wavy about particular symmetry conditions that must be satisfied. I am looking for an explanation that is accessible to someone who does not know of group theory.

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How much energy is required to liquify oxygen?
Does anyone know where I can find some documentation of a real-world liquification process, and the energy costs?
I can calculate the heat that must be removed from O2 gas to get to liquid, but how would this translate to an electricity cost for cooling equipment?

What is the efficiency of your chiller?

Can I assume some arbitrary value just to see how the calculation works?

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Confusion in solubility expressions in Henry's Law
While reading about Henry's law and solubility I frequently come across two relations:

$C = k_{h}P$ (c = concentration of a dissolved gas)
$P = k_{h}x$ (x = solubility/mole_fraction)

What is the difference in these two expressions, do they seem to contradict each other? I am not able to decide whether a gas with higher $k_{h}$ would be more soluble or less soluble.
Please help me out with the same I am new to this topic

Please define all variables. Note that asterisk is not a proper multiplication sign and should never be used as such outside code listings. A meaningful title also wouldn't hurt.

kh in 1. is not the kh from the Henry law, but rather a solubility constant at given partial pressure. There should be used a different symbol for that. Note that the Henry constant uses the capital K.

@Poutnik google search of henry's law constant gives me that expression. Also I think I've read it elsewhere too. But as mentioned earlier I am not sure about this topic so it would be good if you could correct me wherever you think I am going wrong

It cannot be proportional to p ( in lower letter ) for the same molar fractions and at the same time reciprocal to p for the same concentrations. As  concentrations are not reciprocal to molar fractions. See https://en.wikipedia.org/wiki/Henry%27s_law for different used symbols.

@Poutnik Wikipedia mentions $H = \frac{x}{p}$ where x is the molar mixing ratio. I assume that they mean that x is the mole fraction. But isn't this formula different that what is used otherwise like in here https://chemistry.stackexchange.com/a/29118/106010

@Poutnik Also incase that the symbols are different it would great if you could explain me the use case of both the expressions ? Like in what cases do we use expression 1 and in what expression 2.

There is more H constants there, with explicit superscript indexes, aside of K_H. The question is not, when to use which equation, but when to use which symbol and constant.

Okay I understand that's probably the reason for my confusion. But then in a nutshell if someone was to ask me the relation of solubility with henry's constant what would my answer be

Please see my answer. The notation used in your equations is obviously going to lead to a contradiction.

Sander (Ref. 1) has compiled a useful review of Henry's law constants in water that includes an introduction showing notation and conversions.
There are two types of Henry's law constants:

Solubility constants convert from pressure to concentration in solution (solubility):

$$c=Hp$$

Volatility constants convert from concentration in solution (solubility) to pressure:

$$p=Kc$$
Usually it is possible to determine which constant is reported by inspecting the units.
References

R. Sander. Compilation of Henry’s law constants (version 4.0) for water as solvent. Atmos. Chem. Phys., 15, 4399–4981, 2015.
www.atmos-chem-phys.net/15/4399/2015/doi:10.5194/acp-15-4399-2015


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What is left over when water and sodium hypochorate decompose, i.e. household bleach?
I am having some confusion in understanding the decompostion of household bleach. One website says bleach decomposes to  salt and sodium chlorate, simple enough.
Orther wedsites say bleach decomposes to salt and nothing else. I called Clorox customer service and their vote is it converts to salt only. I have 3 votes for salt only and 2 votes for salt and sodium chlorate thus far.

@Nilay Ghosh.  Yes my motivation for my question was a safely concern as you guessed. Thank you for the link. It is very helpful

It depends on temperature. At higher temperatures, hypochlorite has tendency to disproportionate: $$\ce{3 ClO- -> 2 Cl- + ClO3-}$$ Otherwise it mostly releases atomic oxygen : $$\ce{ClO- -> Cl- + O}$$
In reality, it would be combination of both in various ratio, depending on relative reaction rates,  all ending as mixture of sodium hydroxide, carbonate, chloride and chlorate.


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Finding the concentration from a dilution
This must be a very basic question but I have gotten confused over it.
How would I find the concentration of each component of the reaction for each experiment from this data? I am guessing it can’t be the concentrations given, is that for some stock solution?

Solving one out of the four would be good enough to understand the concept behind this. Therefore, I shall explain the first experiment and the concentrations of the components in the solution produced.
In experiment 1 all four are added in equal quantities. Finding the total volume, we get the total volume
$$V_\text{tot} = \pu{4 ml}.$$
Assuming the amount of $i$th solute to be denoted by $n_i,$ its molarity $c_i$ can be found as
$$c_i = \frac{n_i}{V_\mathrm{tot}}.$$
Now, we add $\pu{1 ml}$ of $\pu{2.0 M}$ acetone. This is equivalent to saying we add $1/1000$ of $\pu{2 mol}$ of acetone. Therefore we add $\pu{2 mmol}$ (millimoles) of acetone.
This means that we have $\pu{2 mmol}$ of acetone in $\pu{4 mL}$ of water. Using the formula for molarity we get that acetone is in $\pu{0.5 M}$ in solution.
Hope you can follow on for the other parts as well, the concept remains the same.


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Invisible spin allowed transitions in Tanabe-Sugano diagrams
In class we are looking at Tanabe-Sugano diagrams, more concretely at transitions for a d2 complex.
I understand that the only spin allowed transitions are those that have a triplet state, therefore that there are 3 spin allowed transitions
3T1g (P) <- 3T1g (F)
3A2g (F) <- 3T1g (F)
3T2g (F) <- 3T1g (F)
These can be seen in the diagram. In addition, were are given the orbital occupancies for each energetic state some of which I have attached for clarity.

From the  occupancy image, it can be seen that the 3A2g state corresponds to both d electrons being in the high energy eg orbitals. Hence my question is what process can lead to both electrons being in such an excited state? Is it possible for a photon to excite both electrons at the time? Which I had never heard of before, or is it that in order to achieve the 3A2g excited state, 2 photons need to be absorbed in rapid succession, going through an intermediate excited state?
If two photons need to be absorbed does that effectively make the 3A2g state "impossible" to reach and thus the absorption band isn't seen?
Thanks for your help.

It is better to imagine as the energy is absorbed by the whole system not by individual electrons. Two-electron excitations exist (simultaneous, not just one by one).

Two photon absorption is well known using short pulsed lasers as high flux is needed, but only on electron is promoted. (Different selection rules also apply) To get two electrons excited will need two absorptions one after the other, probably at different energies, so v. v. hard to achieve in practice  as excited species always decay rapidly back to the ground state. In a thought experiment, of course, you may be able to excite an excited state again to reach the state you want.


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Why is atomic mass unit/unified atomic mass defined against Carbon 12?
I have searched the web all over this question and I have not really found an answer to my satisfaction. From my understanding of the topic so far, I understand that reason why concept of unified atomic mass works better than providing absolute mass of protons/neutrons (in grams or kilograms) is because "u" provides between clearer presentation of relative masses among elements in whole numbers than the negative exponents would in regards to absolute mass. If this premise is true than why not define atomic mass unit in terms of multiple of protons and neutron? Like since proton and neutron are roughly equal in mass ~ 1.66* 10^-24g. Why not say 1u = mass of one proton and 1 neutron. Why does the definition of amu need to include 1/12 of carbon 12, because mass of 1/12 of carbon gives me mass of 1 proton 1 neutrons and this seems redundant to me?
But now if we accept the definition of amu as 1/12 of carbon 12? Why not include other elements like amu is 1/1 of Hydrogen atom,  1/4 of helium atom, 1/16 of oxygen atom, because all this would would give me same value of 1u= 1.66*10^-24g ? If anyone can give their input on this, that would be grea.

Saying "roughly equal" doesn't cut it, it needs to be really, really, really, really, really precise.

Why 12C? — see Why is the definition of the mole as it is?

This will be a very long answer but the good news is that it has been addressed in detail in the International Journal of Mass Spectrometry and Ion Processes, 86 (1988) 1-19. EVOLUTION OF THE UNIFIED SCALE OF ATOMIC MASS, ‘*C =12u *.

Why is/was the standard for a kilogram a metal cylinder under triple bell jars in Paris?

Hi. So is absolute mass ever known in nature? Or is it always relative against some already known entity? Thanks.

Be aware that the mass of nucleus does not depend only on the nucleon count, but also on total bounding energy. E.g. the mass of 4He is 0.7% less than of 2p+2n. I rather wonder, why was not chosen an element with a single isotope. Probably because new values would be more different from those based on 16O.

@TLo Evaluating the value of an absolute quantity by referring to a chosen unit of this quantity does not make this quantity relative quantity.


text:
Potassium permanganate structurre
So I'm learning about Lewis structures for university. I was reading a book written by some teachers from my university and in one exercise they ask us to draw the Lewis structure for potassium permanganate ($KMnO4$). I attempted to do it, but when I compared my answer to that of the book I realized what I did was wrong.

This is the book's structure. However, when calculating the formal charge of the central manganese atom, I got $7-6-0 = 1$, and as far I understand, the best structure has a negative charge on one of the oxygen atoms, since they are more electronegative
So I looked up the structure online and I found that both my structure and that of the book were wrong.

This is the most common structure I found. Here, the formal charge for Mn is $7-7-0=0$, for the upper oxygen atom $6-1-6=-1$ and for the other oxygen atoms $6-2-4=0$, which makes more sense according to what I wrote before about oxygen being more electronegative than manganese.

However, I also found this another possible structure, this time on Wikipedia, where the formal charge of the central Mn atom is $7-8-0=-1$, and that of the oxygen atoms is $6-2-4=0$. So according to what I mentioned before regarding the formal charge, this structure is wrong as well.
I think the only right structure is the second one. Am I right?
Thank you in advance! :)

I won't go so far as to say the second is right, but I would say it is the least incorrect of the 3 you show :) If you said that it was number 2 but that a better picture had all 4 Oxygens equivalent with a formal charge of -0.25 I would be happier, but maybe you don't know enough Chemistry to appreciate that yet - and I suspect that is what the last picture is trying to say, but it isn't very clear. The first picture is horrible, well done on rejecting that! Potassium won't ever form a covalent bond to Oxygen, it will always be ionic.

Formal charges are but a man-made concept.

In MnO4- all the Mn-O bonds are equivalent and they have the same length. This type of structure is difficult to represent with a single Lewis structure, and formal charges only make sense for a Lewis structure.

Thank you all for reading my post and answering to my question! @Ian Bush, you're right. I'm still grasping the concept of formal charge. I can't believe I used to draw Lewis structures without knowing about it!

And @Shoubhik R Maiti, that makes sense. I thought about that after I found two different possible structures online.

Yes. And you can get an intuitive sense of this by analogy with benzene. The usual benzene structure has three double bonds and three single bonds, but obviously all 6 C-C bonds are same. (which is why sometimes benzene is drawn without the double bonds, but with a circle inside) In spite of the fact that the Lewis structure is not completely right, we still use it. (partly because it helps us draw mechanisms better, and partly because people like to draw octet complete Lewis structures I guess)


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Are acid anhydride and anhydride the same thing?
In my organic chemistry course, I have been made familiar with the term acid anhydride and anhydride.
I know that anhydride denotes the structure :

But I am confused about whether acid anhydride means the same thing.
Please provide some clarification on this.

This only acid anhydride.

The thing is, my instructors have repeatedly used the term 'anhydride'. So is it just a shorter way of saying 'acid anhydride'?

@AniruddhMukherjee if you are learning Organic Chemistry and especially about carboxylic acids, anhydrides would refer to acid anhydrides. It depends on the context.

An anhydride is anything that has a water removed from it, how would you remove water with two acid molecules of the form $\ce{R^1 -COOH}$ and $\ce{R^2 -COOH}$ respectively.

Acid anhydrides are of 2 types: Organic acid anhydride $\ce{R(CO)O(CO)R'}$ and inorganic acid anhydride which refers to acidic oxide.

I would point out that many compounds, that formally fit the general definition of an anhydride are not referred as anhydrides, or just seldom. The only anhydrides regularly referred so are anhydrides of organic acids. Carbon suboxide C3O2 is anhydride of malonic acid in general sense, aside of the narrow sense "true" malonic anhydride C3H2O3. Or, as a very extreme absurd example, carbon is an anhydride of "carbs", which has origin in the German term Kohlenhydrate = hydrate of carbon. Similarly, acid-forming oxides of nonmetals are seldom referred as anhydrides, unless this aspect is stressed.

@Poutnik This is what I understood, correct me if I am wrong: Acid anhydrides are obtained by the removal of water from carboxylic acids, and hence the 'acid' in 'acid anhydride', and they fall under the set of anhydrides which may or may not be obtained from the dehydration of acids. And, in organic chemistry, we use the terms 'acid anhydrides' and 'anhydrides' interchangeably for the sake of simplicity.

I would say so. But it is formal path. Real obtaining of anhydrides may used different reactions. Some acids may decarboxylate or otherwise deteriorate instead of forming anhydride. Neither you can get AlCl3 by heating AlCl3 . 6 H2O.

Defining the concept of anhydride
An ”anhydride” is a chemical entity that has been obtained by dehydrating a substrate (containing water either as water of crystallization, or bonded with the substrate). Thus, all acid anhydrides are anhydrides, but not all anhydrides are acid anhydrides.
Here is a more professional definition, from the Merriam-Webster dictionary:

Definition of anhydride
: a compound derived from another (such as an acid) by removal of the elements of water

A quite usual example from the field of inorganic chemistry
For example, copper (II) sulfate pentahydrate is, well, the hydrated form of the copper-based salt. Water is present there as water of crystallization, the water molecule is embedded within the crystal structure. Let's see the differences between the hydrate crystal and the anhydrous form (that is, the anhydride)

This inorgani salt is different from the acid anhydrides so commonly encountered in organic chemistry! However, anhydrous copper (II) sulfate is still an anhydride. And so are the acid anhydrides you have mentioned.


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Does the position of a functional group in an organic compound have any effect on the compound's heat of combustion?
Is there a link between where the functional group is located in an organic compound and the compound's heat of combustion? It has been mentioned before that it is difficult to accurately get the enthalpy of a reaction experimentally; thus, enthalpies of formation are used instead when calculating it.This is why I looked up the enthalpy of formation of liquid at STP data for two compounds, 1-Heptanol and 2-Heptanol, which are $\pu{403.4 kJ mol^{-1}}$ and $\pu{416.9 kJ mol^{-1}}$ respectively.

You can find some calculation here.

Sure it has an effect. Why it should not be?

@MathewMahindaratne Thank you for the notice, I updated the post.

@Alchimista Why would there be an effect? I don't think I understand it.

Because two different molecules are different in basically every property, except composition in isomers like in this case. Even the heats of combustion of conformers - imagine to burn them separately - are different!

@Alchimista Then the explanation is "because everything else is different". I don't get it. Is it because the bonds are stronger that the enthalpy is higher in 2-Heptanol? For example, I also checked out their boiling poins and 1-Heptanol has a higher boiling point: 175°C versus 160°C; which makes me think that because a 1-Heptanol can approach another one's functional group to form a bond from a range of angles broader than that of 2-Heptanol, its boiling point is higher. I can't find an explanation like this one for the enthalpies though.

Deciding the order of heat of combustion of isomeric alkanes; Heats of combustion and stability of rings; What causes the difference in heat energy released by combustion for the butanol isomers?

@abtoiew see the link posted below. For the aspect of boiling points, in the specific case the reason is as you sketched. In general it will depend on shape and size.

@MathewMahindaratne The source you linked actually did help a lot! Do you have any other sources where I can find heat of combustion values for other organic compounds like Aliphatic Aldehydes with more than 7 Carbons.

The exhaustive combustion of an organic compound consisting of carbon and hydrogen only will yield carbon dioxide and water.  Thus, if the heat of formation for isomers differ from each other, then the heat of combustion will differ, too.
This entry in the English Wikipedia states, for example
$\pu{−146.9 kJ mol^{-1}}$ for n-pentane, $\pu{−154.4 kJ mol^{-1}}$ about isopentane (methylbutane), and $\pu{−167.8 kJ mol^{-1}}$ for neopentane / dimethylpropane; all of same number of atoms $\ce{C5H12}$ (pentane), but with different connection of methyl groups to a backbone in common.  The very table continues for other branched isomers of alkanes, too.

Possibly there is a misunderstanding between the computation of heat of formations from paper (or, in a computer program), on one hand, and the experimental determination of these properties, too.
For the former, the approach typically taught in schools to determine the enthalpy of formation is

to imagine the reaction of $\ce{C_xH_y}$ with $\ce{O2}$ to yield $\ce{CO2}$ and $\ce{H2O}$
to look up the standard heat of formation of $\ce{CO2}$ and $\ce{H2O}$, hopefully with a discern between water as gas and water as liquid
to apply Hess's law to compute $\Delta_fH^\circ$ for $\ce{C_xH_y}$ with the convention that $\Delta_fH^\circ$ for elements in their most stable form (here, oxygen) is set zero

This approach however is an approximation which does not account how the atoms are connected with each other and their neighbours, or bond order; as if e.g., a carbon in acetylene does not differ from the carbon in methane.
If one leaves the imaginary reaction to $\ce{CO2}$ and $\ce{H2O}$, there are empirical increment systems based on experimentation, set to improve this approximation when computing this on paper.$^*$  With the advent of quantum chemistry, it is up to you to select a theory suitable enough to compute this in silico, too (example by MOPAC).
Among the classic approaches to determine experimentally the heat of formation via the heat of reaction (again, Hess' law is applied) is bomb calorimetry.  Here, exhaustive combustion a known quantity of your compound takes place in an atmosphere of pure high-pressure oxygen.  It is one of the experiments students may encounter early when studying physical chemistry.
$^*$  Publications like 1993ChemRev2419, 2011JPhysChemA10576, or 2012JPhysChemA7196 review / present extensions, or are basis for more recent work, too (cited by below the landing page of the publication) in addition to the increment system above mentioned Wikipedia entry currently presents.  Of course, one equally may  adjust existing / establish a new increment systems, e.g., as combination of own work (bomb calorimetry) and using data already recorded by others and presented in primary references (articles) and tables (e.g., CRC Handbook of Chemistry and Physics, NIST Webbook Chemistry)

The increment systems you linked work for compounds that contain C and H only. Though, what would I do if I needed to improve the approximation when calculating the enthalpy of formation of a compound that has C, H, and O like Aldehydes?

Either a) identify other increment systems including elements other than C and H in the literature (examples 1993ChemRev2419, 2011JPhysChemA10576, 2012JPhysChemA7196 review / present extensions; and are basis for more recent work, too [cited by]).  Or b) establish a new increment system, maybe as combination of own work (bomb calorimetry) and using data already recorded by others and presented in primary references (articles) and tables (e.g., CRC Handbook of Chemistry and Physics)


text:
Trans junctions in fused cyclopentane rings
Is it possible for the ring junction to be trans in a molecule containing 2 fused cyclopentane rings? I am aware it is not possible for fused cyclobutane rings but am not sure about the 5-membered counterpart. In case the question isn't clear, the ring junction I am referring to is the one that has stereochemistry drawn in the decalin molecules below.

Are you asking whether trans-octahydropentalene exists?

@andselisk Yes, that is what I was asking, thank you for the insight. A lot of (probably bad) sources I have found have argued that it was impossible for some reason...

Maybe you want to answer the question yourself?


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How to avoid explosion when there is a leakage of LPG gas and coal gas/town gas?
How to avoid explosion when there is a leakage of LPG gas and coal gas/town gas? What are the conditions of that happening? The related underlying chemical reactions are hydrogen and butane combustion in oxygen, right? What is the reaction mechanism? We should avoid using any electrical appliances, electric light, electric door bells and smartphone , just anything that could produce sparks and turn off any flames, right? So I should turn off the electrical appliances immediately after noticing a leakage or should I just keep them on?

Aside of open flame and operating electronic devices, another danger hard to eliminate is static electricity, mostly due synthetic material of clothes, or rubber soles of shoes.
The safest and the most available prevention choice at emergency scenario is intense ventilation. All others are rather hope it would be enough, as it is hard to eliminate or not to forget all possible ignition sources.
Be aware that switching devices off may be more dangerous than keeping them on, as interrupting high voltage  AC circuit usually creates sparks, unless this is addressed by parallel capacitors. This is typical for devices transforming AC to  low voltage, like most low power electronics.

Coal gas is of little use these days, with major fuel components  carbon monoxide and hydrogen.
Much more common is natural gas, based on methane, or LPG, based on butane (typically 80%) and propane.
The oxidant is in all cases aerial oxygen.

the chemical element(s)/compound(s) (reactants) involving are hydrogen and oxygen in the case of coal gas and alkane and oxygen in the case of LPG gas, right?

See the answer update.


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Do Hartree-Fock (or other model Hamiltonian) electron densities fullfill the Kato theorem?
I have done Hartree-Fock calculations on a single He atom and now I tried to check numerically if the electron density fulfills the Kato theorem. It apparently doesn't. Instead I obtain a cusp corresponding to an effective charge of about $1.76$. That looks suspiciously like an effect due to the effective shielding exerted from the second electron. The Slater rules would suggest to expect an effective shielding of $0.3$ which is not that far away.
Also by theory one would expect such an effect, as HF is a "mean-field theory" treating the electrons seeing the mean potentials from the other electrons. Possibly one could even derive the exact deviation?


text:
Adjusting redox reactions
$$\ce{KCl + KMnO4 + H2SO4 -> K2SO4 + MnSO4 + H2O + Cl2}$$
Attempt. I want to adjust this redox reaction, I've figured out that Chlorine and Manganese are the 2 elements that change thier oxidation number. But I've encountered the problem that if I do:
$$\ce{KCl -> 1/2 Cl2}$$
I don't have any ideas on how to adjust the K atoms, similar stuff with the K and S on the manganese half reaction. Is this impossible to do? Beforehand, I've been doing redox reactions where none of this issue ocurred and I could adjust O and H by adding $\ce{H2O}$ or $\ce{H+}$ but what do I do here?

Be careful in confusing 0 and O in formulas. Most of chemists still think 0 means zero. For eventual writing and formatting of chemical or mathematical formulas or equations, see how to use  MathJax with  mhchem

If each child wants just 1/2 of an apple, how many children you can serve with 1 apple ? Basically, all the enumeration of chemical equations can be reduced to counting until 20 and other basic stuff of elementary school arithmetics.

..what are you talking about? that doesn't help me at all

Try to apply it. It is very easy, if you forget it is chemistry. If apples fit in 3 rows with 5 columns, how many rows are needed if there are just 3 columns ?

Cl2 had 2 chlorine atoms. KCl has 1. How many KCl you need for 1 Cl2 ? Do you see ? Just children, apples, rows and columns. Many people lose elementary school skills when it comes to chemistry.

Start with the change of oxidation numbers. The respective coefficients of these formulas will have the reciprocal ratio. Rows and columns. Multiply all to get integer Cl2 count. Count cations and supply enough H2SO4 to match them.

I'm not understanding you at all, I know you have to put a 2 in front of KCl to balance the atoms of Cl, but that does not have anything to do with what my question is

do i just add K to the other side to balance K atoms?

Let us continue this discussion in chat.

Apparently the discussion does not go much ahead. So I will try to speed it up and show how it goes. First try to establish the half-reaction with ions, not with molecules and formula units. We will go back to neutral compounds later on and get rid of the positive and negative charges at the end.
Here, in the first half-reaction, the ion $\ce{MnO4-}$ is reduced to $\ce{Mn^{2+}}$, and the $4$ lost oxygen atoms are transformed into $\ce{H2O}$ molecules, here $\ce{4H2O}$ to be added on the other side of the arrow. These $4$ water molecules introduce unwanted $\ce{H}$ atoms, that must appear on the other side of the arrow as $\ce{8 H+}$ ions In the second half-reaction, the ion $\ce{Cl-}$ is oxidized to $\ce{Cl2}$. Hopefully you know how to build these half-equations. If you don't, tell us, and we will explain it in more details. Here are these half-equations :
$$\ce{MnO4- + 8 H+ + 5 e- -> Mn^{2+} + 4 H2O} \\ \ce{2 Cl- -> Cl2 + 2 e-}$$ Now you have to add these two half-equations after multiplication of both in order to make the electrons disappear. This can be done by multiplying the first half-equation by $2$ and the second by $5$. Look. $$\ce{2 MnO4- + 16 H+ + 10 e- -> 2 Mn^{2+} + 8 H2O} \\ \ce{10 Cl- -> 5 Cl2 + 10 e-}$$
This produces $10$ electrons on both sides, so that the electrons can be skipped after summing. $$\ce{2 MnO4- + 10 Cl- + 16 H+ -> 2 Mn^{2+} + 5 Cl2 + 8 H2O}$$ The final equation is obtained. But now you would like to have the ions disappear, so that only neutral species appear in the equation. So you must add $2+10 = 12$ ions $\ce{K+}$ ions on the left hand side to compensate the anions and form neutral potassium salts. You also must add $8$ ions $\ce{SO4^{2-}}$ to compensate the $\ce{H+}$ ions. Afterwards you must take care of these new ions and make some neutral salts on the right hand side. So the two $\ce{Mn^{2+}}$ ions recombine with $2$ of the $\ce{8 SO4^{2-}}$ ions, forming $\ce{8 MnSO4}$. The remaining ions produce $\ce{6 K2SO4}$ . This gives :  $$\ce{2 KMnO4 + 10 KCl + 8 H2SO4 -> 2 MnSO4 + 5 Cl2 + 6 K2SO4 + 8 H2O}$$ Is it OK ?


text:
What kinetic model does the hydrogenation reaction of dibenzyltoluene follow?
So during the course of my research, I've read countless papers related to the hydrogenation of dibenzyltoluenen and the dehydrogenation of perhydro-dibenzyltoluene. Now what I've noticed, is the fact that I could not find any data concerning the kinetic models linked to this hydrogenation process (I did find kinetic data concerning dehydrogenation of perhydro-dibenzyltoluene).
Is there a specific reason for this?
Many thanks in advance.

What kind of hydrogenation?


text:
How to carry out basis set fitting and extrapolation
I am playing around with basis set extrapolations, and i would like to perform some simple tests by doing some parameter fitting and checking the errors.
The functional form of the fitted curve is this
$$
E(X) = E_\infty + Ae^{-\alpha X}
$$
where $X$ is the cardinal number of the basis set used, and $E$ is the energy. This function makes sense because the rate of decay empirically fits the convergence (of correlation concistent basis sets) and the exponential term vanishes as $x\rightarrow\infty$.
We want to use energies from two low-cardinal numbers, and extrapolate to the $E_\infty$ as given by the function and the fitted parameters.
Assume that for the cardinal numbers $X$ and $Y$ we have the corresponding energies $E_X$ and $E_Y$. We then have the relationship
$$
E_X = E_\infty + Ae^{-\alpha X} \\
E_Y = E_\infty + Ae^{-\alpha Y}
$$
By using that $A = \frac{E_X - E_\infty}{e^{-\alpha X}}$, we can solve for $E_\infty$ to obtain
$$
E_\infty = \frac{E_Ye^{-\alpha X} - E_Xe^{-\alpha Y}}{e^{-\alpha X} - e^{-\alpha Y}}
$$
and solve for $A$ to obtain
$$
A = \frac{E_Y - E_X}{e^{-\alpha Y} - e^{-\alpha X}}
$$
From what I can see, if all I want is to obtain the extrapolated energy, then I do not need take $A$ into any consideration. I just need the cardinal numbers, the energies, and $\alpha$. In Python syntax this would be something like this
def extrapolate(X, Y, E_X, E_Y, alpha):
    gamma = np.exp(-alpha*X)
    omega = np.exp(-alpha*Y)
    return (E_Y * gamma - E_X * omega) / (gamma - omega)

Is the above approach the normal one for curve fitting and extrapolation?
In some tests I find that the extrapolated energy is virtually identical to the energy with the highest cardinal number used in the extrapolation. So either there is no benefit of my extrapolation, or I am doing something wrong.
(Note that I am not using Gaussian basis sets, but a MultiWavelet basis, and that the cardinal numbers are not really cardinal numbers, but a defined precision relative to the CBS. Still the approach should be the same regardless - even if the function I use for the fitting perhaps not is the perfect choice.)

Hm, I would prefer 3 equations for 3 unknown parameters ( E_inf, A, Alfa) and therefore 3 degrees of freedom. Otherwise 1 parameter would be left for arbitrary choice and the other 2 would depend on its choice, still fitting both equations.

So given $X$, $Y$, and $Z$, and corresponding energies, I would derive expression for E and A in terms of these?

How are you determining the alpha?

Alpha is fitted to multiple training cases. I guess I take the average value from these.

Why not just fit the three points to the curve with scipy curve_fit ? Taking alpha from training means you are introducing an empirical parameters.

@Yoda If alpha is given, you can obviously use just 2 points. But it may not reflect reality, if real alpha is different. And, for measurement error minimalization, I would suggest rather least square approximation through multiple values by linearized function.

I wanted to follow the standard procedure in basis set extrapolation, but I got confused by the literature. If you know this procedure, perhaps you can outline it in an answer?

If I well understand your question you have a set of experimental points $(E_1,X_1), (E_2,X_2), ... , (E_n,X_n)$. You want to fit to this set of points the function :
$$
E(X) = E_\infty + Ae^{-\alpha X}
$$
So, you want to approximately determine the parameters $E_\infty$ , $A$ and $\alpha$.
This requires a non-linear regression calculus. For example see https://mathworld.wolfram.com/NonlinearLeastSquaresFitting.html
This calculus is iterative and requires to set some "guessed" values of the parameters to start the iterative process.
Another method (much simpler but for approximative result only) is not iterative and doesn't require "guessed" values is explained in : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales
You could try it in folowing the numerical example below with $\quad y(x)=E(X)$ , $a=E_\infty$ , $b=A$ , $c=-\alpha$

Note : This simplified method is not reliable if the number of points is too low.

+1 This is a good answer with maths, but I suspect OP was looking for a python based solution.

@Shoubhik R Maiti. Thank you for the judicious comment. In fact I was expecting that the OP could implement on pyton this very simple matrix calculus and gives a comparison of results as feeback. This method has been applied many times  in other contexts but not in Chemistry as far as I know.

Another thing I would point out is that in general it would be difficult to get more than 4-5 points for basis set fitting, because as x (which is determined by basis set size) increases the computational cost increases rapidly, so its not practicaly possible. Does the non-iterative method work for that? I am a chemist mainly, so all of this math stuff is difficult for me :-)

I cannot answer without having an experimental data of 4 or 5 points. But a-priori I am pessimistic in this case. This depends also on the scatter. It is possible that the method be not convenient in this Chemistry context. In any case this would be interesting for me to see what happens and to look why.

I would add that even if the non-iterative method doesn't give good results in some cases (not enough points and too high scatter) generaly the values obtained for the parameters $a,b,c$ are at least of the correct magnitude. So they can be used as starting values for a classical iterative non-linear regression. This was used in a few cases instead of "guessed" initial values.

The common procedure in basis set extrapolation is to use just two empirical data points, and then extrapolate a curve previously optimized on multiple data points. Basis set convergence is very systematic, and so such two-point procedures are very common, since one can get much better accuracy at the fraction of the cost. I wanted to use some reference data and fit a curve on, say, 4 data points, and then use just two data points from my test data to extrapolate the previously fitted line. Perhaps approach this is very limited to basis set extrapolation studies.


text:
What is the proof of Kohlrausch's Law of Independent Migration of Ions?
What is the proof of Kohlrausch's Law of Independent Migration of Ions? How did Kohlrausch calculate the limiting molar conductance of a weak electrolyte? I am asking this in context of electrochemistry and limiting molar conductance of a weak electrolyte.

There is a simple demonstration of Kohlrausch's law. Start with four electrolytes with common ions, say AX, AY, BX and BY. The law says that the difference in limiting molar conductivities of two salts with a common ion is independent of that ion, for instance: $$\Lambda_m^\circ (\ce{AX})-\Lambda_m^\circ (\ce{AY})=\Lambda_m^\circ (\ce{BX})-\Lambda_m^\circ (\ce{BY})$$
The reason for this is that A contributes the same amount to the conductivities of solutions of AX or AY, and similarly B contributes the same amount to the conductivities of solutions of BX or BY (provided molar concentrations are the same of course).
If you want a more fundamental explanation you can use Coulomb's law to compute the energy of a pair of charges at a distance r from each other. Apply it to the pair of counterions formed when a simple 1:1 salt dissociates in solution:
$$U = \frac{e^2}{4\pi \epsilon\epsilon_0 r}$$
where $\epsilon$ is the dielectric constant of the medium (assumed constant, approx 80 for water).
Consider now how the energy of the interaction changes with distance. At $\pu{r= 1 Å}$ the energy is $\pu{17.4 kJmol-1}$. Repeat the computation for other distances and tabulate the results:
$$\begin{array}{c|c|c}\hline r(Å) & U (\pu{kJmol-1}) & \pu{C(M)}\\\hline 1 & 17.4 & 481\\ 10 & 1.74 &0.481\\  100 & 0.174& \pu{4.81e-4}\\ 1000 & 0.0174&\pu{4.81e-7} \\\hline \end{array}$$
Compare the energies to kT, a measure of the thermal energy available to the molecules causing random motion in the solution. If kT is greater than the interaction energy, then the ions are unlikely to be close to each other. At a temperature of 300K, $RT=\pu{2.5 kJmol-1}$. This means that if two ions are at a distance of more than a few nm they are unlikely to have any significant effect on each other's position. In other words, they will act independently. This is embodied by the corollary of Kohlrausch's law that conductivity is linearly dependent on concentration at very low concentrations, because in that regime ions act independently.
You can also estimate how the mean distance between ions changes with concentration (using some rough back-of-the envelope calculations), this is shown in the last column of the table. It suggests that if you keep to ~mM concentrations of simple salts, the behavior of the electrolyte should approximate the zero concentration (limiting) prediction of Kohlrausch's law.
As for the second question, the limiting molar conductivity $\Lambda_m^\circ$ is a value you obtain by extrapolating experimental data (molar conductivity versus concentration) to the limit of infinite dilution (zero concentration). For a weak 1:1 electrolyte you can perform a linear fit using Ostwald's dilution law
$$\frac{1}{\Lambda_m} = \frac{1}{\Lambda_m^\circ} + \frac{\Lambda_m c}{K_a (\Lambda_m^\circ )^2}$$
and determine the intercept.
In general $\Lambda_m^\circ$ can be calculated by fitting the experimental data (molar conductivity versus concentration) to a polynomial in concentration and computing the value expected at zero concentration (the intercept).

What is 'k' and 'T' in 'kT = 2.5 kilojoule per mole'?

That's Boltzmann's constant...

@BuckThorn Then it should be RT. Aside of the value wrong by N_A, kT has dimension of energy. (As is known R=k . N_A.)

And for some unknown reasons, energy in the post uses exponent -1. ( $kT= \pu{2.5 kJ-1 mol-1}$)

@Poutnik thanks for bringing that to my attention!


text:
How to use Sodium Thiosulfate for fixing dicopper (I) tetraiodomercurate (II)?
DISCLAIMER
DO NOT TRY TO DEMERCURATE USING DESCRIBED PROCEDURE. CONSULT YOUR EMERGENCY SERVICE.

I found a discrepancy between a government-approved procedure of living quarters demercuration and the patent RU 2081198 C1 the procedure is based on.
The last phase of cleaning involves fixing remaining mercury salts $\ce{HgI2}$, $\ce{Hg2I2}$, $\ce{Cu2Hg2I2}$ (I didn't find any mentions of $\ce{Cu2Hg2I2}$ in literature), and $\ce{Cu2[HgI4]}$ using $\ce{Na2S2O3}$. According to the patent one should use acidified $\ce{Na2S2O3}$:
$$\ce{Na2S2O3} + 2\ce{H+} \rightarrow \ce{H2SO3} + \ce{S} + 2\ce{Na+}$$
$$\ce{mercury salts + S -> \ce{HgS} + other products}$$
However, the government-approved procedure suggests using not acidified $\ce{Na2S2O3}$ or $\ce{HOOC-COOH}$. I am wondering if it is a mistake. I found that $\ce{HgI2}$ is soluble in $\ce{Na2S2O3}$ in the Russian book on analytical mercury chemistry. So using $\ce{Na2S2O3}$ for $\ce{HgI2}$ removal makes sense.
My question
Is $\ce{Cu2[HgI4]}$ soluble in presence of $\ce{Na2S2O3}$ or $\ce{HOOC-COOH}$? Or are there other reactions that make it less dangerous or make $\ce{Cu2[HgI4]}$ easy to remove?
Reactions involved in demercuration according to the patent RU 2081198 C1
$$2\ce{CuSO4} + 4\ce{KI} \rightarrow \ce{Cu2I2} + \ce{I2} + 2\ce{K2SO4}$$
$$\ce{Hg} + \ce{I2} \rightarrow \ce{HgI2}$$
$$\ce{Hg} + 2\ce{KI} + \ce{I2} \rightarrow \ce{K2[HgI4]}$$
$$\ce{Hg} + \ce{Cu2I2} + \ce{I2} \rightarrow \ce{Cu2[HgI4]}$$
$$\ce{HgI2} + 2\ce{KI} \rightarrow \ce{K2[HgI4]}$$
$$\ce{K2[HgI4]} + \ce{Cu2I2} \rightarrow \ce{Cu2[HgI4]} + 2 \ce{KI}$$
Binding mercury wapors???
$$\ce{Hg} + \ce{Cu2I2} \rightarrow \ce{Cu2HgI2}$$
The method of demercuration from the patent RU 2081198 C1
Materials

$\ce{CuSO4} \bullet \ce{H2O}$ $10\%$ aqueous solution
$\ce{KI}$ $10\%$ aqueous solution
$\ce{Na2S2O3}$ $25\%$ aqueous solution
$\ce{HOOC-COOH}$ $5\%$ aqueous solution
sawdust (if a floor in demercurated room is made of material with many cracks, e.g. parquet)

Procedure

Remove visible mercury droplets
Spray copper sulfate until the floor is moistened
Spray double volume of potassium iodide over the floor
Wait for 10 days. Mercury droplets are located as red spots on the white layer of salts
Remove located mercury and salts from the floor
If the floor is made of material with many cracks, the floor is covered by a layer of sawdust impregnated by the mixture of $\ce{CuSO4}$ and $\ce{KI}$ and exposed for another 10 days. There may be several repeats of sawdust mercury absorption.
Fix remaining salts of mercury by the spraying of $\ce{Na2SO3}$ and then spraying of $\ce{HOOC-COOH}$.
Wait for 1 day
Wet cleaning.

The method of demercuration according to government-approved procedure
Materials

$\ce{CuSO4} \bullet \ce{H2O}$ $10\%$ aqueous solution
$\ce{KI}$ $10\%$ aqueous solution
$\ce{Na2S2O3}$ $25\%$ aqueous solution or $\ce{HOOC-COOH}$ $3-5\%$ aqueous solution

Procedure

Remove visible mercury droplets
Spray copper sulfate $0.1 - 0.15 L/m^2$
Wait for 2-3 minutes
Spray potassium iodide $0.2 - 0.3 L/m^2$
Wait for 1-3 days exposition for spilled mercury or 5-10 days if mercury is removed from the air
Remove mercury and salts
In rooms with the floor made of material that is not penetrable for mercury one turn of $\ce{CuSO4}$ and $\ce{KI}$ cleaning is enough
If the floor is made of parquet mercury droplets are removed from cracks using a flat knife
In living quarters, schools, and kindergartens remaining mercury salts are fixed using $\ce{Na2S2O3}$ or $\ce{HOOC-COOH}$ $0.15 - 0.2 L/m^2$.
Wait for 1 day
Wet cleaning

My question
Is $\ce{Cu2[HgI4]}$ soluble in presence of $\ce{Na2S2O3}$ or $\ce{HOOC-COOH}$? Or are there other reactions that make it less dangerous or make $\ce{Cu2[HgI4]}$ easy to remove?
Related questions:

about chemical properties of $\ce{Cu2[HgI4]}$,
about toxicity of $\ce{HgI}$ and $\ce{Cu2[HgI4]}$.

Could you add the actual question at the end? I am finding it hard to understand what it is that you are asking for..

@SafdarFaisal Done. I have 4 questions.
Should I make them separate posts?

Separate questions would be great. I was just about to ask if you could do that.

@SafdarFaisal I left 1 question is this post and created a new https://chemistry.stackexchange.com/questions/149855/chemical-properties-of-dicopper-i-tetraiodomercurate-ii
I'll create a question about $\ce{Cu2[HgI2]}$ and $\ce{HgI2}$ toxicity in 30 mins.

@SafdarFaisal question about toxicity https://chemistry.stackexchange.com/questions/149860/toxicity-of-mercury-ii-iodide-and-dicopper-i-tetraiodomercurate-ii


text:
Is there a way to convert aniline to phenol without diazotizing it?
I am a high school student and while reading about reactions on amines, I was trying to figure out a way to convert aniline $(\ce{C6H5NH2})$ to phenol $(\ce{C6H5OH})$ without diazotizing it with sodium nitrite $(\ce{NaNO2})$ and dilute hydrochloric acid $(\ce{HCl})$ in temperature conditions of $\pu{273 K}$ to $\pu{278 K}$ and then reacting it with $\ce{H2O},$ i.e. anything that does not involve the following reaction:
$$\ce{RNH_2 + NaNO_2 + HCl ->[$273–\pu{278 K}$] [RN2^+Cl^-] ->[H2O] ROH}\tag{R1}$$
I tried converting it to an isocyanide via the carbylamine reaction, but that seemed to make things more complicated. The closest I found to solving this was Tilden's reagent $(\ce{NOCl}),$ which reacts with primary amines in the following way:
$$\ce{RNH2 + NOCl -> RCl + N2 + H2O}\tag{R2}$$
However, this is limited to primary amines by definition.
Is there any way to convert aromatic amines to aromatic alcohols without diazotizing the compound? More specifically, is there a way to convert aniline to phenol without converting it to a benzene diazonium ion?

Names of compounds aren't supposed to be capitalised.

There's plenty of ways, but reaction itself is pretty much exercise in futility - pointless in practice.

Benzenediazonium salt is so versatile that it can be used to make many compounds. That's why, diazotization reaction is so popular and useful. The reaction conditions is not that harsh and is easily achievable. Coming to the reaction, diazotization is a straight one-way procedure to achieve aniline to phenol conversion. If you talk about other ways of converting aniline to phenol, the reaction conditions for such ways are generally harsh and hence not done practically to obtain phenol. In addition, there are so many steps, that the conversion is practically not feasible.
One of the ways of converting aniline to phenol without diazotizing is to convert aniline to acetanilide and then hydrogenolyze it to make benzene which can be then used to make phenol.

Step 1

Step 2

The reaction is taken from a paper1. The paper discussed the use of $\ce{NiMo}$ based catalyst along with novel neo-microporous composite material HSM/ZSM-5 as support. The reaction result is compared with other supports like $\ce{\gamma-Al2O3}$ and lone ZSM-5 and HSM system.

Step 3

Reference

Sajedeh Jafarian, Ahmad Tavasoli, Hasan Nikkhah, "Catalytic hydrotreating of pyro-oil derived from green microalgae spirulina the (Arthrospira) plantensis over $\ce{NiMo}$ catalysts impregnated over a novel hybrid support," International Journal of Hydrogen Energy 2019, 44(36), 19855-19867 (https://doi.org/10.1016/j.ijhydene.2019.05.182) 44. 10.1016/j.ijhydene.2019.05.182(PDF).

Step 2 does not guarantee hydrogenation would stop at benzene, it may continue to give cyclohexane under the conditions. It is also possibility that N-ethylaniline may converted to cyclohexylethylamine first, which then hydrogenalize to cyclohexane. I attached the PDF in the reference for your information.


text:
Why does hydrogen burn with a pale blue flame while its emission spectral lines are red in colour?
I was studying oxidising flames and realized that sodium burns with a bright yellow flame the wavelength of which is around 588 nm. Then I searched for emission spectra of sodium and found that the lines around 588 nm are the most intense ones, so it appears yellow. The same happened with potassium.
But when I arrived at hydrogen, I knew it burned with a pale blue flame, but its emission spectral lines were most intense in the red region. So why does it still gives a blue flame?

As you see, it emits blue light too, but the red one is the most intense and so electrical discharge of hydrogen emits red light. But its oxidising flame is pale blue.
Would it burn with a red flame at higher temperature or something like that or are spectral lines not related to flame colours?

@Buttonwood This is the page of libretexts, https://chem.libretexts.org/Courses/Solano_Community_College/Chem_160/Chapter_07%3A_Atomic_Structure_and_Periodicity/7.03_The_Atomic_Spectrum_of_Hydrogen

The emission max is not mostly in the red.

It is a very interesting question, but comparing a combustion spectrum with an atomic emission one is like comparing apples and oranges. A flame is a luminous gas phase chemical reaction where the hydrogen atoms are combining with oxygen atoms. It is a chemiluminescence phenomenon. A discharge tube emission is an atomic emission.
You posted a picture of a hydrogen spectrum from a discharge tube which consists of discrete lines in the UV, visible, and IR regions. In the same way, the hydrogen-air flame has an emission in the ultraviolet, visible and in the IR. The blue flame is not a line spectrum, as you might speculate. It is a continuum (broad band), which indicates that this is due to molecular emission in the flames, not from hydrogen atoms! This is in contrast with the alkali metals whose compounds easily atomize in the flame.
Also note that what may appear to us as a single color, like blue, is rarely a single wavelength.
There is a beautiful article by R.W. Schefer , W.D. Kulatilaka, B.D. Patterson, and T.B. Settersten on Visible emission of hydrogen flames, Combustion and Flame 156 (2009) 1234–1241. Go to Google Scholar and find it; it is open access on Google.
Look at their experiment: They view the "blue" hydrogen flames by various optical filters which block or pass certain wavelengths. See many colors are present in the "blue" flame of hydrogen. In fact most of the emission is in the ultraviolet region.

Direct flame luminosity photographs in a laminar, diffusion H2-jet flame. (a) Unfiltered, f /2.4 aperture; (b) short-wavelength pass filter with 550 nm cutoff wavelength,
f /2.4 aperture; (c) long-wavelength pass filter with 530 nm cutoff wavelength. Images taken at f /2.4 aperture with 90-ms exposure time. Fuel jet exit velocity is 47 m/s
with a coflow air velocity of 0.57 m/s. Reynolds number = 837.
Most of the emission is in the UV and infrared as you can see from the spike near 300 nm and intense band after 700 nm. Most people cannot see beyond 700 nm. The infrared emission is due to excited water molecule or perhaps its fragments.

Note: The above spectrum consists of two parts, UV on the left, and visible extending from $\pu{350 nm}$ to $\pu{850 nm}$ on the right hand side, recorded separately. For this figure, the later was amplified by a factor of 6.5 with respect to the OH in the ultraviolet.

I think the subsequent figure (https://ars.els-cdn.com/content/image/1-s2.0-S0010218009000170-gr006.sml) is actually the one you want to demonstrate why the flame is blue. While interesting, the UV and IR spectra have little to say about human color perception.

Are you sure it is open access? Doesn't it just point to sciencedirect.com where only the abstract is available ("Purchase PDF" and "Access through your institution")?

If you search Google with this title, the first result should be the pdf of this article. It is hosted by zenodo.org  "Zenodo is a general-purpose open-access repository developed under the European OpenAIRE program and operated by CERN." I think this is a legit website as per Wiki information.

@PeterMortensen If you use Firefox with its add-on unpaywall, then Elsevier's landing page of this publication is marked on the right hand side by an open padlock on green background.  While this then links directly to the pdf, just remove the two last layers of this string https://zenodo.org/record/1258847/files/article.pdf to reach zenodo's  landing page including the copyright note (here: Creative Commons Zero v1.0 Universal) for this work.


text:
Can the guanidinium ion give stable N-nitrosamines under mild metalworking conditions?

Can the guanidinium ion react with inorganic nitrite under mild basic metalworking conditions ($\mathrm{pH\ 8 - 10}$ and temperature $\lt \pu{70^\circ C}$) to give stable N-nitrosamines?

Usually, the carcinogenic nitrosamines are generated from secondary amines. But guanidinium ion is not a secondary amine. Guanidinium ion is a mesomerism-stabilized system, and the nitroso group could be also stabilized by mesomerism therefore. So I'm afraid that it can build nitrosamines readily. But is this possible under basic conditions as mentioned above?

What do you mean by this?  "Give" how?  What reactants would the guanidine be reacting with to "give" the N-nitroso-substituted derivatives?  Just itself and oxygen from the air?  With some other NO donor?

Nitrosoguanidine may be precipatated from cold water and dried at 40C according to this ref  https://prepchem.com/synthesis-of-nitrosoguanidine/

By definition, N-nitrosamines are aliphatic or aromatic derivatives of secondary amines, which have a nitroso group ($\ce{–NO}$) attached to nitrogen. In general, N-nitroso compounds are usually formed by the interaction of a nitrosating agent ($\ce{NOX}$) derived from either nitrite salts or nitrogen oxides with an amino substance such as secondary and tertiary amines, secondary and tertiary amides, peptides, substituted ureas, guanidines, and urethanes. Some important N-nitroso compounds are shown below:

As structures indicated, resonance in the framework has nothing to do with $\ce{-N-N=O}$ function (e.g., 1-Methyl-3-nitro-1-nitrosoguanidine is stable at room temperature and commercially available). Nonetheless, guanidine derivatives interact with $\ce{NOX}$ to give stable nitroso-guanidine derivatives (which are type of N-nitrosamides). With $\ce{NOX}$ derived from nitrite salts, the formation of N-nitroso derivatives of guanidine rarely occurs above $\mathrm{pH \ 4}$, but increases steadily with increasing acidity i.e. decreasing $\mathrm{pH}$, a maximum at  $\mathrm{pH \ 2.5-3.5}$ where the yield is inversely proportional to the basicity of the amino substance  (Ref.1).

On the other hand, guanidines do not react with $\ce{NOX}$ derived from nitrogen oxides in aqueous media (Ref.2,3). Note that $\ce{NOX}$ derived from nitrogen oxides worked best with strongly basic amino substances (e.g., aliphatic amines and heterocyclic amines) under non-aqueous or neutral and alkaline aqueous conditions (Ref.2,3).

References:

Sidney S. Mirvish, "Formation of N-nitroso compounds: Chemistry, kinetics, and in vivo occurrence," Toxicology and Applied Pharmacology 1975, 31(3), 325-351 (DOI: https://doi.org/10.1016/0041-008X(75)90255-0).
B. C. Challis, "Nutrition and nitrosamine formation," Proceedings of the Nutrition Society 1985, 44(1), 95-100 (DOI: https://doi.org/10.1079/PNS19850015) (PDF).
Brian C. Challis,  Soterios A. Kyrtopoulos, "The chemistry of nitroso-compounds. Part 11. Nitrosation of amines by the two-phase interaction of amines in solution with gaseous oxides of nitrogen," J. Chem. Soc., Perkin Trans. 1 1979, 299-304 (DOI: https://doi.org/10.1039/P19790000299).


text:
Can oxidizing agents be used to increase the combustion efficiency of waste oil?
Waste oil is atomized and combined with air to produce a very high temperature flame capable of melting iron. I was wondering if it's possible to further increase its combustion efficiency by mixing it with an oxidizing agent such as Sodium Perchlorate, Hydrogen Peroxide or some other oxidizer in mixable form.
Conceivably, it could enhance the combustion since it adds more oxygen to the fuel, especially at higher temperatures. At the same time, it could also inhibit the combustion since the additive may absorb energy from the reaction than it produces, since it's already an oxide.

It is safer to keep the process as simple as possible without the addition of perchlorates or peroxides to the oil.  Perchlorates are salts with low solubility in oil; hydrogen peroxide equally is a polar compound with little affinity to form a solution with oil.  Thus, the salts could clog the fuel delivery, or in case of $\ce{H2O2}$ simply yield a biphasic liquid mixture if stored in the same container as the fuel.  It equally would increase the potential hazard to run such a combustion engine; perchlorates are heat sensitive and could explode when you don't want them to do so. In addition, over time residues of salts and salt residues facilitate corrosion (already a problem for the transportation of crude oil); here, the elevated temperatures could amplify the problem e.g., at the burner's nozzle, or as deposits in the fire box.
Thus, stay with the delivery of liquid oil and air.  If your are lucky in terms of viscosity of the oil, you still may obtain an aerosol but not the fine mist of air and fuel as characteristic for Otto engines, slow motion at 2:56 min onward).  More often, however, similar to oil driven heaters, you will rely on special burners running which bring pressurized air via fans into contact with a fine film of oil:

(credit)

If additives are acceptable (and especially if the viscosity of the waste oil is high), a low viscosity hydrocarbon, richer in hydrogen, could boost the flame temperature. This could be adjusted so that the waste oil is reduced until economics is optimized. And, of course, the best low viscosity hydrocarbon would be methane.


text:
Given initial moles of weak acid, find moles of conjugate base to be added to get a final pH (volume unknown)
I am preparing notes on buffers and I came across this problem on a website:

Given 10.0 mmol of a weak acid HA with $\textrm{p}K_a =5$, how many mmol of NaA needs to be added to have a final $\textrm{p}H = 5$?

The solution method shown used the Henderson-Hasselbalch equation
$$\textrm{p}H = \textrm{p}K_a + \log_{10} \frac{ [\ce{A-}] }{ [\ce{HA}] } \hspace{1in} (\textrm{Eqn 0})$$
Plugging we get
$$[\ce{A-}] = [\ce{HA}] \qquad \iff \qquad n(\ce{A-}) = n(\ce{HA})$$
Thus, at equilibrium the moles of $\ce{A-}$ needs to equal the moles of $\ce{HA}$. Next, the author assumes that this implies the initial moles will also equal
$$n(\ce{A-}) = n(\ce{HA}) \implies n_\circ (\ce{A-}) \approx n_\circ (\ce{HA})$$
Since $n_\circ (\ce{HA}) = 10$ mmol this means $n_\circ(\ce{A-}) = 10$ mmol which means the amount of NaA needed is also 10 mmol.
I wanted to solve the problem without making that assumption.
Mass Balance At equilibrium the "atoms" $\ce{A}$ appear only in the HA and $\ce{A-}$ and they only come from the initial $\ce{HA}$ of 10 mmol or from the added moles of $\ce{NaA}$ which equals the moles of $\ce{Na+}$
$$n(\ce{HA}) + n(\ce{A-}) = 10 + n(\ce{Na+}) \hspace{1in} (\textrm{Eqn 1})$$
Charge Balance
$$n(\ce{H+}) + n(\ce{Na+}) = n(\ce{A-}) + n(\ce{OH-})\hspace{1in} (\textrm{Eqn 2})$$
Equilibrium for the weak acid
$$ K_a = \frac{ [\ce{H+}] \cdot [\ce{A-}] }{ [\ce{HA}] } = [\ce{H+}] \,\frac{n(\ce{A-})}{n(\ce{HA})} \hspace{1in} (\textrm{Eqn 3})$$
Equilibrium for the water
$$
      K_w = [\ce{H+}] \cdot [\ce{OH-}] \hspace{1in} (\textrm{Eqn 4})$$
In the above we know $K_a = 10^{-5}$, $K_w= 10^{-14}$, and from the p$H$ we know $[\ce{H+}] = 10^{-5}$ mmol/mL.  The fourth equation gives $[\ce{OH-}] = 10^{-9}$ mmol/mL.
However, the final volume of the solution $V$ is not given. No volumes are given. So $n(\ce{OH-})$ can not be found from $[\ce{OH-}]$.  This means the first three equations can not be used to get $n(\ce{HA})$, $n(\ce{A-})$, and $n(\ce{Na+})$. Notice dividing the first equation by $V$ does not help since it 10 mmol would become $10/V$.
Is this correct?  Without knowing the volume I can not find a value for $n(\ce{Na+})$ unless I make the assumption given above?
Clarification Notes Added: Yes, Eqn 0 is just the negative logarithm of Eqn 3 (i.e., HH is no more than an algebraic rearrangement of the Law of Mass Action). Therefore, it is not surprising that Eqn 3 also gives $n(\ce{A-}) = n(\ce{HA})$ (after plugging in $\textrm{p}K_a=\textrm{p}H$).
The point is that I do not want to then assume that equivalence of the equilibrium mole amounts implies the same in the initial mole amounts.
In other acid-base problems this may be avoided by including an additional mass balance, charge balance, and auto-ionization of water.  Here it appears to be insufficient.  I am wondering if this is because of missing data (e.g., volume) or is there something more profound going on?

I don't see why it is necessary to do those extra calculations. Notice eq. 3 is actually henderson hasselbalch equation when you take -log of both sides. However, in this case, simply Ka = [H+] so [A-] = [HA] at equilibrium. By adding the exact same moles of NaA, it would not shift the equilibrium at all.

Yes, I know HH is just the log of eqn 3.  I am trying not to then assume that the final equilibrium concentrations may be replaced by the initial concentrations.

I think that is always the case because if you add equimolar amounts of NaA and HA initially, the equilibrium would not change and initial concentrations should equal equilibrium concentrations. Just make an ICE table and you will find that x has to equal 0. It could just be me misunderstanding the question and if that is the case, sorry.

The equilibrium ratio always somewhat shifts wrt initial ratio, because H+ ions are made from HA. But as the task is set, it is assumed negligible, with assumption [HA]>>[H+]. Otherwise, more input data would be needed, like volume or concentration.

@Poutnik  Thanks for the confirmation.  That is exactly what I was wondering.  I know it shifts and granted usually (as here) [HA]>>[H+].  It is really obvious that volume (or concentration) is needed?  That is, I am wondering is there no exact way in which to get the answer?

@ML I think you are misunderstanding and it is surely my fault.  There is a shift in the molar amounts of both HA and A- due to the production of H+.  While typically small, this will make the final equilibrium values appearing in both HH and my Eqn 3 no longer quite equal to the initial values.  An ICE table is in effect what my Eqn 1 and Eqn 3 are.  That is, I do not think x is zero.

The degree of ionization is concentration dependent. To keep the same pH at 10 times higher dilution, 10 times more of acid has to be ionized. so you need higher initial HA/A ratio for more diluted solutions. Or, the other way, the same initial ratio would lead to high pH for more diluted solutions.

@Poutnik  I agree.  So I think your conclusion is mine: that there is no way to rigorously get an answer with the data given?

Without assumptions, no, there is no rigorous way. But I think you overcomplicate it, I would take the assumption and was done with it.

Well I am not trying to solve this problem.  I am more interested in this class of problem.  I am curious when these assumptions fall apart.  Thanks for the help.


text:
Do phosphoryl thiocholines, their uncharged analogs, and other organophosphates, interact differently with Acetylcholinesterase?
As asked in the title, would the various salts -such as hydrochloride or methyl iodide- of V-Agents interact any differently with acetylcholinesterase compared to the neutral compound? For instance, would the interactions between 33-SN and AChE differ in any way from those between 33-SN+ and AChE? Also, would the phosphoryl-thiocholine analog of Soman have different interactions with AChE compared to Soman itself? In short, would a set of organophosphate AChE inhibitors which only differ in whether the leaving group during the process of AChE inhibition is a fluorine, dialkylaminothiolate, or thiocholine, interact differently with AChE??

As a matter, of fact, yes, uncharged phosphoryl thiocholines do seem to interact with acetylcholinesterase differently compared to their charged analogs. For example, it has been reported that

Sundwall (1961) reported that P2S was considerably more effective as
an antagonist of the lethal action of 37 SN
+(isopropoxy-2-trimethylammonio-tiomethylphosphino oxide) in mice than it was of those of Sarin and 37 S-N
(isopropoxy-2-dimethylaminothiomethylphosphine oxide).

P2S refers to the methyl sulfonate salt of pralidoxime. This would seem to suggest that, at least in the case of 37-SN+, the adduct formed by the charged inhibitor is more susceptible to reactivation by oxime therapy as opposed to that formed by the neutral compound†.
Regarding the second half of the question, the answer is also yes. According to Boskovic, Acetylcholinesterase that has been inhibited by O-pinacolyl S-2-diethylmethylammonioethyl methyl phosphonothioate methyl sulfate undergoes aging two orders of magnitude more slowly compared to AChE that has been inhibited by Soman, and is also more susceptible to reactivation. The explanation proposed by the author is that the interaction between a phosphoryl-thiocholine and AChE results in the formation of an adduct in which the phosphoryl group is oriented differently compared to the adduct formed as a result of inhibition by an uncharged organophosphate. The author also points out that AChE which had been inhibited by the compound Ro 3-0411 is more easily reactivated compared to AChE inhibited by diisopropyl fluorophosphate.
In conclusion, it can be said that, for a set of organophosphorus compounds which only differ in whether the leaving group during AChE inhibition is a fluorine atom, a thiocholine, or dialkylaminoethylthiolate (And in which the other side chains are the same), each of these compounds would interact with AChE differently, which influences, probably among other things, their susceptibility to reactivation by oximes and the rate at which the inhibited enzyme undergoes aging.
†It should be noted that, under physiological conditions, some amount of these neutral compounds becomes protonated, although I do not know to exactly to what extent and how and if that affects their pharmacology.
References:
BOSKOVIC, B. (1981). The treatment of soman poisoning and its perspectives. Fundamental and Applied Toxicology, 1(2), 203–213. doi:10.1016/s0272-0590(81)80059-0
BošKović, B., Maksimović, M., & Minić, D. (1968). Ageing and reactivation of acetylcholinesterase inhibited with Soman and its thiocholine-like analogue. Biochemical Pharmacology, 17(8), 1738–1741. doi:10.1016/0006-2952(68)90237-2


text:
Choosing hyperconjugation when resonance doesn't give a reason for stability

Which of the following structures would be least stable?

I initially approached it like this:
As both are allylic carbocations, and that both are in fact resonating structures of the same molecule, so resonance can't be used to judge them. So I went for hyperconjugation. The first molecule has four alpha hydrogens and the second has only one alpha hydrogen.
Since there is more hyperconjugation in the first, it should be more stable. But I also have an intuition that there might be something more that we should have a look upon but I'm not able to figure it out. Should we always choose hyperconjugation as our second option when we can't make judgements using resonance?

To me, the reasoning sounds fair enough.

Well, the alkenes have the same degree of substitution, resonance can be ruled out, all that's left is, indeed, hyperconjugation. And you have compared the two correctly. That should be it.

In addition to the hyperconjugation (resonance) effect there is also an inductive (sigma) effect that provides stabilization.  See this earlier  answer for an explanation.

Will we only be checking for hyperconjugation stability of the cation or should we also check for stability of double bond. I think in general 2-ene are more stable than 1-ene cause of more alpha hydrogen, but here we have a cation too so not sure which one to choose

@Ashish yes! that's what we're doing when we are accounting for hyperconjugation here. More the number of alpha hydrogens, more would be the stability

@PrajwalTiwari but we're doing that for cation at the cost of alpha hydrogens of the double bond. So can we say that cation stabilization has more dominance here ?

My teacher says that while comparing the stability of the species concerned, we should remember that stability provided by Resonance>hyper conjugation>inductive effect. Of course there are exceptions(like in the case of chloro benzene where according to resonance the ring should be activated but inductive effect dominates it and the ring gets deactivated) to these but I think we can follow it in majority of these cases and there is no reason to not be confident about it.So likewise in your case resonance is not enough so you look upon the next one i.e hyper conjugation

suggestions regarding anything is appreciated


text:
Thermodynamic properties of enantiomers
Pairs of enantiomers have the same chemical and physical properties (except rotation of plane polarized light) within an achiral environment. However, I am curious about the thermodynamic properties of pairs of enantiomers: say a molecule has a certain $\Delta H_f$, $\Delta S$, and $\Delta G_f$; will its enantiomer have the same exact values?

When Isolated, yes. If adsorbed or solvated by something optically active not, or perhaps not depending on the interaction. So strictly the enthalpy of formation is the same. Thermochemistry might differ.

All the normal thermodynamic properties of a pair of enantiomers are the same.
Unless there is some interaction with a chiral environment. The only way any properties will differ is if the environment in which they are measured is chiral.
For example, the heat of combustion of D glucose is the same as for L glucose (as combustion doesn't care about the stereochemistry). But only D glucose is usable by higher organisms in metabolism as all the enzymes involved are chiral (despite the thermodynamics of the overall reaction being exactly the same).


text:
According to Markovnikov rule and stability of carbocation which is most likely to form if HI "excess" is not given?
Problem

Answer                          >
I thought "I" can stabilizes carbocation formed intially by resonance so there can be a shift of (+) to the carbon close to "I" and hence forming the product 1,1-diiodopropane but given is 1,2 product , whats wrong in this ? Or is that H+ and +I combined effect of CH3 is slightly greater than compared to resonance of "I"?

yes! the inductive effect of halogens is a special case. That it overweighs their +R

No exception? In any case?

not that I know of, but I believe there must be some exception as in almost all concepts, we get to see at least one exception in chemistry. Someone who has more idea about this might be able to help

Do you know two -OH groups on a single carbon are unstable? And here we have I atoms! In fact, how would you explain the formation of 2-iodopropane when excess HI is taken?

What the OP is asking is that could the carbocation shift to the carbon to which the I is initially attached, then we could get a 1,1 di iodo product

@WayBig, Is it HI in excess?

@PrajwalTiwari https://chemistry.stackexchange.com/questions/72156/mechanism-of-the-reaction-of-an-alkene-with-excess-hi-in-presence-of-ccl4

@Safdar Faisal Not excess pls read the title of the question

Yeah ik @Light Yagami , i was just asking one step before that elimination thing happens which is the cause for 2 iodopropane .  And i was asking in case someone who dont know about the excess product and was just given this as problem then what would he be guessing nothing more

Exactly @Prajwal Tiwari  and i see i was just asking if u have encountered any exceptions

There are three reasons as far as I can see as to why the 1,2 product is favored. There may be more.

Steric hindrance
The iodine atom is large in size compared to the carbon atom. Having two iodine atoms on successive carbon atoms in a hydrocarbon is quite unstable. In fact, should two successive carbon atoms possess iodine substituents, they usually undergo elimination reactions. And here the 1,1 product would have two iodine groups on the same carbon atom!

Inductive effect of halogens
While it is true that halogen atoms have a positive mesomeric effect, this effect is almost always dominated by their negative inductive effect. If you see a halogen atom, you can usually neglect it's +M effects in favor of it's −I effects.

Carbocation stability
The carbocation formed in the process of formation of the 1,2 product is more stable (due to hyperconjugation) than the one formed in the process of formation of the 1,1 product (again, due to the −I effect of iodine).

Either way,  1,2-diiodopropane is more stable than  1,1-diiodopropane.

Ty i got your pt one more thing wanted to ask any exceptions uk when that -I of halogens fails over +M ?

@WayBig I have not heard on any exception.

I see , Can u check this link out too its regarding same reaction type ? https://chemistry.stackexchange.com/q/149835/106350


text:
What is the maximum number of water molecules that has been recorded till date to present as water of crystallization in a compound?
Context: I was casually looking through some old questions when I found this question. The OP wanted a compound that has all the hydrates ranging from monohydrate to decahydrate. I found chromium(III) sulfate [$\ce{Cr2(SO4)3.$x$(H2O)}$ where $x=0,\ldots,18$] to be suitable. But, the actual formula of the compound is $\ce{[Cr(H2O)6]2(SO4)3·6H2O}$, so 6 of the 18 molecules are present as water of crystallization. So, I wondered, what is the maximum number of water molecules that can present in a compound as water of crystallization?
I can agree to what @BuckThorn said in his answer. It might be protein crystals. Wikipedia also supports the fact:

Compared to inorganic salts, proteins crystallize with large amounts
of water in the crystal lattice. A water content of 50% is not
uncommon for proteins.

Now, the question is what protein has the maximum number of water molecules as water of crystallization?

I did some cursory searches and looked for inorganic hydrates (see edit history: 36-hydrate is maximum I could find). I am also looking for such compounds which is in inorganic domain.

There are gas hydrates which are pretty much mostly ice, but I guess they don't count since technically they are clathrates, not hydrates.

Would it make more sense to do wt percent instead of number of water molecules so you aren’t biased to larger unit cells?

I edited tags to reflect that your interest is in inorganic hydrates. I sort of apologize for my answer. I realized proteins weren't perhaps  what you were looking for, but I'll leave it up anyway as food for thought. Hint: look for a really big and spherical inorganic molecule....

Maybe a duplicate?: https://chemistry.stackexchange.com/questions/128679/what-is-the-largest-noncrystalline-non-polymer-inorganic-molecule

@BuckThorn No. I am not interested in inorganic hydrates (I was initially) but I edited the question to reflect your POV (+ a quote from Wikipedia) and that I am interested in any compound(organic or inorganic) that has the maximum number of water molecules in its water of crystallization (which happens to be proteins). I will accept your previous answer once the bounty period is over (I was waiting for other answers). Regarding that link, you happened to find the answer (it would have been a duplicate if I hadn't edited the question). +1 to both of your answers.

However, I am removing the 'inorganic chemistry' tag to show that I am not bounded under the 'inorganic domain. I am just looking for a canonical answer which describes a compound of inorganic domain.

It might be a protein crystal. Proteins are large and the interstitial space is filled to a large extent with water. The volume of a protein crystal occupied by the solvent can be 50% or so. However the fraction of waters observed in crystals (relatively fixed) is often low, on the order of 10%.
For instance, from a report on the crystal structure of crambin (chosen arbitrarily, see Ref. 1):

The volume of solvent (32%) is low but not atypical for proteins [values of 30-78% solvent are found.] Crambin is small (Mr 4720), and most residues are solvent accessible. Although crystals were grown from 60% ethanol, only 7%(vol/vol) ethanol has been located. Of the 85 water molecules expected (based on the density and the presence of 2 ethanol molecules), the model includes 64 [...]

It's not clear whether the 85 expected or the 64 found water molecules would be considered crambin's water of crystallization (presumably all count, even if not observed).
As another example, the model crystal structure of BPTI contains 63 waters (observable waters, see PDB structure 5PTI).
While I was looking for the protein crystal structure with greatest number of waters in its model, Karsten Theis suggested as an example the pyruvate-bound crystal structures of ADP–glucose pyrophosphorylase from the bacterium Agrobacterium tumefaciens (pdb 5W5T, see the file for a description of the solvent model), which has 3752 waters in the crystallographic model (59.12% of the volume).
Many waters may have irregular positions, and some may form clathrate-like solvation structures (as in gas hydrates, as andselisk comments), and some can be placed at regular locations.
References

M. M. Teeter. Water structure of a hydrophobic protein at atomic resolution: Pentagon rings of water molecules in crystals of crambin. Proc Natl Acad Sci U S A. 1984 Oct; 81(19): 6014–6018. doi: 10.1073/pnas.81.19.6014

You might also take a look at https://pubs.rsc.org/en/content/articlehtml/2016/sc/c5sc03447g which is an ab initio study of crambin, and considers the water associated with the crystallised protein

At room temperature, most of the water in typical protein crystals is in the liquid rather than the solid state. So it is not clear if you want to count those, or only the ones that are fairly stationary.

@KarstenTheis I've been looking around for the protein crystal structure with the largest reported (static?) number of interstitial waters. You have more experience and may know?

I know that photosystem I crystals have 90% water with a huge unit cell. There are currently 8 structures in the PDB that have more than 10,000 water molecules in the model. One of these structures is 5W5T, if you are interested in an example.

Some papers for ref.: http://scripts.iucr.org/cgi-bin/paper?S0907444993013368 ... https://pubmed.ncbi.nlm.nih.gov/1592185/ ... https://www.jstor.org/stable/2417785?seq=1 ... https://pubs.rsc.org/en/Content/ArticleLanding/2020/RA/C9RA09601A#!divAbstract

This feels like cheating, since andselisk did all the hard work already and could have provided you with the answer, but I would think
$$\ce{Na48[H_xMo368O1032(H2O)240(SO4)48] · \text{ca.}~1000 H2O}$$
is a good candidate for the largest inorganic hydrate. Note the "circa" 1000 waters. Unfortunately I don't have access to the Angewandte Chemie article and could not confirm whether any waters were observable spectroscopically or by scattering, clarifying how many waters are in fact crystalline.
The reasoning why this would be the largest hydrated inorganic compound follows from my other answer: the larger and more spherical the inorganic compound the more interstitial space is available; more interstitial space means more room is available for interstitial water. Crystallization however usually requires regular specific interactions with the inorganic core, at least in order to observe coherent scattering (water may instead be crystalline but irregularly distributed).
Please see the answer to "What is the largest, noncrystalline, non-polymer inorganic molecule?" for more on the compound.

The crystal-water content has been determined thermogravimetrically. Also, some oxygen atoms from the water molecules have been allocated on the electron-density map as seen with the deposited structure (CSD-412283). As for posting the answer, I didn't think it actually counts as a record since I figured OP is asking about normalized water content per metal center, which, being calculated for $\ce{Mo386}$ cluster, is not that impressive.


text:
Finding order of acidic strength for given compounds

The decreasing order of acid strength of the following compounds:
The given compounds are
Water ($\ce{H2O}$)
Ammonia ($\ce{NH3}$)
Acetylene ($\ce{C2H2}$)
Triphenyl system ($\ce{Ph3CH}$)
a) $\ce{NH3}$ > $\ce{Ph3CH}$ > $\ce{C2H2}$ > $\ce{H2O}$
b) $\ce{H2O}$ > $\ce{C2H2}$ > $\ce{Ph3CH}$ > $\ce{NH3}$
c) $\ce{C2H2}$ > $\ce{H2O}$ > $\ce{Ph3CH}$  > $\ce{NH3}$
d) $\ce{Ph3CH}$ > $\ce{C2H2}$ > $\ce{H2O}$ > $\ce{NH3}$

I gave the answer to be $\ce{Ph3CH}$ > Acetylene > Water > $\ce{NH3}$. Here I have marked the triphenyl system as the most acidic because it has resonance stabilization involved and $\ce{NH3}$ as the least acidic because it has basic nature. Using these two facts I have marked the answer as above however it was wrong. What is the correct order and why?

The real, not the thought order is to be explained. To create the real order by comparing quantities, it is very useful to know them. C2H2 pKa is very different to H2O one, guess which direction ? Even more for (CH3)3CH.

Think like this: Which is the stronger base? $\ce{NH2-}$ or $\ce{OH-}$? Thus order is opposite for their conjugate acid ($\ce{NH3}$ and $\ce{H2O}$ ) strengths.

Here, this was a simple question of checking the most acidic compound out of the four and this would have given you the answer. $\ce{H2O}$ will be most acidic as the negative charge will be formed on an oxygen compared to nitrogen and carbon in the other two cases.
However, the interesting question here is the part in the middle, what about $\ce{C2H2}$ and $\ce{Ph3CH}$? Which is more acidic?
If you take a look at the $\mathrm p K_\mathrm a$ values of the given compounds, you would notice that the values would come out to be:
\begin{array}{l|c}
\hline
\text{Compound} &\mathrm {p}K_\mathrm {a} \\
\hline
\ce{H2O} & 15.7 \\
\ce{NH3} & 38\\
\ce{C2H2} & 24\\
\ce{Ph3CH} & 31.5 \\
\hline
\end{array}
Now, this is the definitive set of values which states the order should be  $\ce{H2O > C2H2 > Ph3CH > NH3}$.

Note: Next part is speculative as I do not have a solid reason for $\mathrm pK_\mathrm a$ values being so for acetylene and the triphenyl system.
A question that may arise at this point is why is this so? Why would a triphenyl system with $3$ phenyl rings be less acidic than acetylene? What about the three rings providing resonance?
This is because in reality, only one phenyl ring provides resonance at a time. Due to the size of phenyl groups, the structure is not fully planar. Another reasoning is that the negative charge lies on an $\mathrm{sp^3}$ carbon whereas in acetylene it is on an $\mathrm{sp}$ carbon.

Also, lone pair in the carboanion is not exactly in a $\mathrm{p}$-orbital.

It is in a $\mathrm{sp^3}$ orbital, I believe. So $\pi$-system of phenyl groups does bot participate. You already have edited and added that to the post.

Why is this a Community Wiki?

@user1271772 I have asked a question in the process of making this question that was answered by the community. However, if you notice, I didn't make it a community wiki until I got an answer from the site itself.


text:
HOBt and DCC in the synthesis of peptides
DCC is a well known reagent used for the synthesis of peptides, aiding the formation of the amide bond by forming an intermediate with the carboxylic acid that contains a leaving group. During my research on the reagent HOBt, or hydroxybenzotriazole, I came across a scheme which showed DCC and HOBt being used in conjunction as a method for peptide synthesis (picture shown below). What is the point of the HOBt being used here? Doesn't the intermediate formed by the DCC molecule already contain a sufficient leaving group?

HOBt is a good nucleophile and the HOBt ester is a better leaving group than the DCC adduct. This simply gives better yields than using DCC alone


text:
Reason for triphenyl methane being less acidic than acetylene
This question came through answering a question that talked about the acidic order of four compounds.
Upon researching the actual $\mathrm p K_\mathrm a$ values I noticed that even though there was a resonance destabilization of electrons (albeit partial) in triphenyl methane, the $\mathrm p K_\mathrm a$ values of acetylene were less than that of triphenyl methane? I was unable to find a solid reason for why this is the case.
Does anyone in the community have the reason?

Basically you are seeking confirmation for your answer there, which is unusual. Anyway, I just upvoted it because is it at least in the right direction.

@Alchimista yeah. I would convert my answer there into a community post if I can get a proper explanation for this one. Since the first part answers the actual question but not the question I pose.

Triphenyl methane is some kind of propeller shaped, so it has less conjugation, making it very less acidic than what we except.

As mentioned by @orthocresol, just comparing methane and acetylene shows a difference of $\approx 20$ $\mathrm pK_\mathrm a$ units and the amount of resonance required to overcome this difference is very high.
Comparing $\mathrm p K_\mathrm a$ of phenol($10.0$) and methanol($15.5$), we see that the presence of one resonating ring only produces a 5.5 $\mathrm p K_\mathrm a$ difference. Resonance in triphenyl system is effectively only one phenyl group at a time due to the steric effects due to the bulkiness of the phenyl groups.


text:
Determining whether a compound is ionic or molecular in nature

$\ce{HCl}$ is a molecular compound because there is a covalent bond between $\ce{H+}$(a proton) and $\ce{Cl-}$ (a chloride ion)
$\ce{NH4NO3}$ is an ionic compound because there is an ionic bond between $\ce{NH4+}$ (an ammonium ion) and $\ce{NO3-}$ (a nitrate ion) even though all the atoms comprising the molecule are non-metal.
Then, which of the following is true?
$\ce{H3C-COOH}$ is an ionic compound because there is an ionic bond between $\ce{C2H3O-}$ (an acetate ion) and $\ce{H+}$ (a proton)? OR
$\ce{H3C-COOH}$ is a molecular compound because there is a covalent bond between $\ce{C2H3O2-}$ (an acetate ion) and $\ce{H+}$ (a proton)?

How do you tell if a bond between the two ions are molecular or ionic? (regardless of whether they are polyatomic or monoatomic ions)

A bond between 2 ions is always ionic, otherwise there would be no ions. HCl(g) has covalent bond, as there are no ions. OTOH, HCl(aq) means separated ions H+(aq) ( or H3O+(aq)) + Cl-(aq).

what ...? i don't get it. I heard someone said acetic acid is an ionic compound, and i thought it was molecular, we had a debate, the other guy said it dissociates in water to give 2 ions and so the bond would be ionic, and i gave him the case of HCl,  and polyatomic ions. but couldn't be sure of the answer; if I had a place to ask this question, i probably have asked it to my teacher; which obviously isn't the case;

There is a big difference between what compounds do in solution and how they behave as pure compounds. Some compounds are ionic as pure compounds (like sodium chloride and ammonium nitrate; some are only ionic in solution like HCl or strong organic acids).

What's the point of us giving you the answer, if you already know it, but you can't be sure? Well, if this helps: you were right to mention HCl, which is covalent in pure form and ionic in a solution; so is CH3COOH. Generally speaking, no pure compound with H+ is ionic.

In principle, when studying a structure like HCl, CH3COOH, or any other compound, the first thing to do is drawing the symbol of the atoms in a reasonable geometry on a sheet of paper. Then consider the outer electron layers : put enough points around each symbol to get a reasonable Lewis structure. For example, one point near $\ce{H}$ and six points around $\ce{O}$. Then try to join atoms with their neighbors to make covalencies (or doublets). When all atoms are joined to neighbors, count the number of electrons around each atom, taking into account that the two electrons included in a covalence between A and B  are at the periphery of both atoms A and B. Check all atoms. If an atom respects the octet law, don't modify it : its structure is covalent. If an atom A does not respect octet law, try to remove one electron from its outer layer, and send this electron to the B atom, which becomes negatively charged. The covalency disappears from the region between A and B, and the bond AB becomes ionic with a positive charge on the A atom, which becomes $\ce{A+}$, and a negative charge on $\ce{B}$ which becomes $\ce{B-}$.
Using this method, you will see that $\ce{HCl}$ and $\ce{CH3COOH}$ are both made of covalencies, because all their atoms are respecting the octet rule. You will also see that $\ce{NaCl}$ may first be drawn in a covalent structure. But then the sodium atom does not fit the octet rule. It will if an electron is removed from the sodium atom, to be given to the chlorine atom, producing then $\ce{Na+}$ and $\ce{Cl-}$ ions.
Hydroxides contains two sorts of bonds. Hydroxides like $\ce{NaOH}$ are made of covalent bonds between $\ce{O}$ and $\ce{H}$ atoms, and of ionic bonds between $\ce{Na^+}$ and $\ce{O-}$.
Oxygenated acids are a special case. They may get two sorts of bonds between $\ce{A}$ and $\ce{B}$. For example sulfuric acid $\ce{H2SO4}$ can be drawn with this method to give first a covalent structure with two bonds $\ce{H-O}$, two bonds $\ce{O-S}$, and two double bonds $\ce{S=O}$. This is OK for all atoms except the central Sulfur atom, which is surrounded by six doublets or six covalences. At this point, some chemists are thinking that it is a case of hyper valence. Some think that the octet rule may be obtained by removing the second bond from each double bond from the sulfur atom, and sending it to the outer Oxygen atom. This makes a central sulfur atom charged $2+$ (respecting the octet rule) and two oxygen atoms charged $1-$. Both theories can be discussed.
The only problem with this reasoning is the fact that it does not work properly  for atoms of the third column. For example, $\ce{BF3}$ should be ionic making $\ce{B^{3+}}$ and $\ce{F-}$ ions. Indeed $\ce{BF3}$ remains covalent, and it is hard to justify. Maybe, on may admit that Boron atom prefers to stay covalent rather than to loose three electrons. Three is probably too much.


text:
Does a geminal or vicinal diiodide form upon addition of excess HI to but-2-yne?
Reaction: $\ce{CH3-C#C-CH3 + HI (excess) -> ?}$
Product : I thought of different carbons by the reasoning from this post earlier I posted. But it is given that both iodine atoms exist on the same carbon.
Is this an exception or is the answer given incorrect?

https://chemistry.stackexchange.com/questions/138882/mechanism-for-the-addition-of-hydrogen-iodide-to-3-3-dimethylbut-1-yne
Does this help?

I think the reason might be since this reaction follows a termolecular mechanism whereas the reaction you had posted earlier involved us to check the stability of the carbocation being formed and the final product being formed.