text:
Why do power requirement figures for water electrolysis on the web seem to be wrong?
I need to calculate the power needed to electrolyse 1 mole of water for a project.
We will either use Polymer Electrolyte Membrane Electrolysis or Alkaline Water Electrolysis, they operate on 1.75V-2.20V and 1.8V-2.4V respectively. And each has a maximum efficiency of about 69%. So, I have done further calculations assuming that the potential difference between the electrodes is 1.8V and the efficiency is 70%. As per my requirements, I have also considered that I have 8 minutes for conducting electrolysis.
Here are the calculations :
The reaction : 2H2O 2H2 + O2
Half Reactions:
Reduction at cathode: 2 H+(aq) + 2e−  H2(g)
Oxidation at anode: 2 H2O(l)  O2(g) + 4 H+(aq) + 4e−

1 mole of hydrogen gas will be produced from 1 mole of water ie 18 g of water or 18 cc of water.
Time available for electrolysis = 8 minutes ie 480 seconds
Number of moles of electrons = 2
Charge = 2 x 96,485 C/F = 192,970 C
Current = Charge / Time = 192,970 / 480 = 400 Amperes (approx.)

Considering 70% efficiency, Current required = 570 Amperes
Potential Difference = 1.8V (common to both PEM electrolysis and Alkaline water electrolysis)

So, Power Required = 1026 Watts ≈ 1 kW (for 8 minutes per mole)

I took reference of this webpage to learn how to calculate it.
However, then I came across this figure on the internet -
per mole water splitting requires at least 241.8 kJ of energy
Source: The first answer to this question.
So I am a little confused. I guess, this 241KJ energy figure came out because the person might be considering 1.23V potential difference (which is considered ideal), and 100% efficiency. I can't say anything because he has stated, that the values of enthalpy of formation (which he has used) are experimental, but hasn't provided the voltage and neither other conditions for the reaction.  He has just used the values of enthalpy of formation and not considered the efficiency of the method in use, in fact, that figure seems to be so general for all methods. As, PEM Electrolysis and Alkaline Water Electrolysis are the most efficient methods and both don't operate on that "ideal" 1.23V potential difference. Also, I think my reasoning is correct because if I redo the calculations considering 1.23V and 100% efficiency, the answer comes out to be the same as his.
Is my reasoning correct? and if not what is the reason for having different answers.

The essential topic for electrolytic efficiency is overpotential, especially overpotential of oxygen.

They said it right, "... requires at least 241.8 kJ ..." . You just found that one can simply calculate the molar formation enthalpy from the cell voltage, and vice versa, those two informations are redundant.
Of course you cannot run the electrolysis at the minimum voltage, because it will be extremely slow! You would just balance the electrochemical potential of the cell (created by the reverse reaction as soon as you make a small amount of hydogen and oxygen).
To make any actual electrolysis, you need a higher voltage, and the difference is converted into heat at the internal ohmic resistance of your cell. The minimum additional voltage is called the cell overvoltage, below that you don't get electrolysis. It is necessary to be able to break through the polarised electrolyte layers at the electrodes. So you cannot get towards 100% efficiency by running very slow.

Thanks. I had a question : you said that more voltage means speed of the reaction will increase. But according to what I have calculated (and according to that perdue web page: chem.purdue.edu/gchelp/howtosolveit/Electrochem/…) voltage is independent of the speed. So, could you please check my calculations and tell where have I gone wrong and why isnt my answer coming out to be the same as 241KJ.

Only to a very small extent: The diffusion speed of ions in your electrolyte is limited (by diffusion). So there is a small voltage range in which the elecrolysis rate increases, below it is practically zero, above the resistance of the cell simply increases, because diffusion limits the current.


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Why does gilman reagent react with acyl chlorides but not with other carbonyls?
According to HSAB (hard-soft acid-base) concept, a soft nucleophile preferentially reacts with soft electrophile and a hard nucleophile with a hard electrophile.
Since $\ce {R2CuLi}$ is a soft nucleophile it should react with ketones or aldehydes rather than acyl chlorides since acyl chloride should have higher positive charge density due to combined effect of chloride and oxygen.
I still don't understand HSAB completely as I don't know much about concept of orbitals. But I think acyl halides should have higher positive charge density at the carbonyl carbon.

You are making a mistake in applying HSAB. You have to compare the hardness and softness of the site of nucleophilic attack. Does it make sense now?

Here site of nucleophilic attack is the carbon attached with oxygen. Isn't the partial positive charge on the acyl carbon more than partial positive charge on carbonyl of ketone? Shouldn't then hardness of RCOCl>RCOR? Please reply to correct me.

HSAB is a concept, not a law. Use it with discretion.

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What is the black slag formed when sterling silver is melted with borax?
What is the black graphite looking slag that is formed when sterling silver is melted with borax?  Is there a way to remove it from the surface of the sterling silver?  It appears to be indestructible!
The melting is related to jewelry casting.  All melting was done in an electric Ney muffle furnace.  Here is a melting dish with some half-melted buttons of sterling silver.

I can't remember why I didn't completely melt this.  I assume the green is salts of copper from the sterling silver.  Notice how the black slag erodes the walls of the melting dish.
Here is a closeup of the erosion.

This erosion property ate through the bottom of my furnace! UGGG! It'll be $600 to replace the muffle.  I heated to 1,900 Fahrenheit for about one hour and this is what I saw when I opened the furnace to get the crucible with the molten sterling silver, but with glowing white-orange light.

Here I have the crucible laying on its side next to the hole.

Here is a close up of the hole and in one of the images I'm lighting it up with a flashlight.  I'd say it's about 1.5 inches in diameter and 1.5 inches deep.  There are 2.25 Ozt of sterling silver down in the hole somewhere because the eroding black slag ate holes through the wall of my crucible.

A closeup of the erosion in front of the main hole.  These two erosion spots were from the previous melt cycle with the same crucible.  I saw that the black slag was penetrating the crucible but I figure it was just minute capillary action and it'd be ok.  Notice that the black slag seems to be working like a catalytic reaction because the amount of borax in the crucible is only about the volume of one grain of white rice.  These two depressions were not depressions at the end of the previous melt cycle they were just flat black areas of the same size and shape.

Here is a view on the outside of the crucible.  I assume one of those holes is where the silver leaked out.

A view inside the crucible.  In person, the black slag has a metallic sheen to it.  It is very similar to the dark grey metallic sheen of graphite.

What is this black slag?

It is unnecessary and distracting to use edit statements. The history of the post is visible to everyone who is interested.

I have seen copper alloy penetrate a crucible like that when held at melting temperature for over an hour.

Hi Martin,
I rolled back your edits because you removed my joke at the end.  :)  Also, my edit statements are necessary because someone who had seen my original post would not know that I had made changes.  They would only know if they specifically looked to see if I had made changes.
James

blacksmith37, thanks for this thought.  I'll keep this in mind.  I still never found out what the black slag is. What I did was I made my own crucible out of stainless steel and inner refractory.  When the black slag eats through the inner refactory it is stopped by the stainless steel liner.  Then, when things cool down I fill in the erosion holes with more refractory.  The stainless steel liner would work by itself self but the molten sterling silver cools down too quickly without the thermal insulating property of the refractory, which is about 15 mm thick.

Please take the [tour] to learn how this site works. We do  care a little bit less about the few people who might have seen your post before, but more about the many people who might come to see it in the future. Your question should stand by itself anytime, without the history of edits being a requirement to know. This is a science site, and while some of us appreciate a joke, the questions is not a good place for it. For some casual fun visit [chat]. To notify people about your comment, put @ in front of their username.

"Indestructible" by what means exactly? How did you try to "destroy" it? Provided that sterling alloy is an alloy of silver and copper, the first guess is that your slag is primarily a copper (II) oxide. Depending on the actual composition of the alloy, it may contain other 3d-metal oxides. The formation of borates (not of silver, but of less noble metals) cannot be excluded, for borax is frequently used as a constituent of the high-temperature fluxes for spectroscopic and calorimetric applications. To what extent the borates are formed depends on the temperature and duration of your melting experiment.
You should be able to dissolve your slug in a boiling mineral acid of your choice. Personally, I'd try the $\ce{HCl}$ solution, or $\ce{HNO3}$ if you do not care that silver dissolves as well.
The following may be irrelevant to the question asked, but, based on my experience and some textbooks, people first try to purify silver and then melt the pure metal, not the other way around. The methods of purification are aplenty, but most of them are based on dissolving the silver-containing alloy in $\ce{HNO3}$. Then $\ce{AgNO3}$ can be transformed into $\ce{AgCl}$, which, in turn, can be reduced by metallic zinc in acidic environment or formaldehyde - in basic.

See images and new information concerning how the black slag has erosive properties to refractory materials.
Indestructible in that heating it to a very high degree with a propane-oxygen torch causes it just glow with a white light similar to how carbon behaves.  Also, soaking for hours in room temperature 10% solution hydrochloric acid or sodium bisulfate seems to only turn it to a powder conglomeration that needs to be brushed vigorously to remove it from the surface of the sterling silver.

I just remembered something that could be a clue.  In addition to the borax, I used this product, https://www.riogrande.com/product/Stop-Ox-II-Anti-Firescale-Coating-Spray-8-oz-Bottle/504071  Just a minute amount, maybe one or two sprays.  I now know that I didn't need to use this but at the time I thought it couldn't hurt.  Maybe it did hurt A LOT!  HAHA!  :D

The reference does not say anything about the composition of this product. But my experience tells me that it could be NH4Cl, or ammonium chloride. It protects metallic surfaces from oxidation during soldering.

Voffch mentioned that borax is a flux. Quite true; borax beads are used for determination of various metals, and copper or silver would give characteristic colors, like green and whatever silver gives. Fluxes are also solvents for metal oxides, like the refractory material of the crucible. While molten alloys should be containable by refractory crucibles (I'm surprised by blacksmith37's comment!), if the crucible is not properly chosen, I suppose it just might not do its job of containing the molten metal.
The borax was probably added to provide a cover for the molten alloy, but was too powerful a solvent for the crucible. There are many materials for crucibles, such as clay, alumina, graphite, and silicon carbide. There are many articles on the internet which address the specific issue of melting silver and sterling silver. Perhaps a better crucible would allow the use of borax (recommended for silver) or a different flux could be chosen.
The black slag has been called fire-burnt copper, an oxide which is very hard product when cooled; probably also combined with the borax.

I have experienced a situation that resembled yours when I got in a hurry and melted my silver nitrate right after filtering. So the silver was still saturated with nitric acid. My thought was it would just burn off before the Silver started to melt. Well the result looked like what you have shared.

When I was working with silver a back powder was often formed.
I have found out, that this black powder was silver oxide. Don't know if it is silver oxide in your case, but it likely is.

The black slag is copper oxide and silver oxide (both are black compounds). If you're trying to refine the sterling silver, i.e. remove the copper and other base metals, then you're approaching this incorrectly. You could theoretically remove the base metals via pyro, but you'll lose a ton of silver to the slag. I'd recommend a hydrometallurgical approach first.
Additionally, the crucible you're using is a fire assay crucible. It's made such that oxides will seep through it. Given the large amount of slag, it's no surprise it broke.

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Evaluate mass of salt needed to add to a buffer solution knowing only pH
This is an exercise taken from an old exam, I'm struggling with the resolution.

To a solution of a generic weak acid $\ce{HA}$ were added $\pu{2.40 g}$ of a potassium salt of the $\ce{KA}$. The solution has $\mathrm{pH} = 4.8$. Evaluate how many grams of salt are needed if we want to shift the solution to a $\mathrm{pH} = 5$.

Neither $K_\mathrm{a}$ nor $K_\mathrm{b}$ are given, nor anything. The only thing I got is just confusion.

I think there is all the data you need. $\mathrm{pH}$ of a buffer formed by a weak acid $\ce{HA}$ and its potassium salt $\ce{KA}$ can be found as
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{C(\ce{KA})}{C(\ce{HA})}}$$
On the other hand
$$C(\ce{KA}) = \frac{m(\ce{KA})}{M(\ce{KA})V}$$
where $m$ and $M$ are mass and molecular mass of $\ce{KA}$; $V$ is volume. In general, first equation can be rewritten as
$$
\begin{align}
\mathrm{pH}_i &= \mathrm{p}K_\mathrm{a} + \log{C_i(\ce{KA})} - \log{C(\ce{HA})} \\
&= \mathrm{p}K_\mathrm{a} + \log{m_i(\ce{KA})} -\log{M(\ce{KA})} - \log{V} - \log{C(\ce{HA})}
\end{align}
$$
Rearranging:
$$\mathrm{pH}_i - \log{m_i(\ce{KA})} = \mathrm{p}K_\mathrm{a} -\log{M(\ce{KA})} - \log{V} - \log{C(\ce{HA})} = \mathrm{const}$$
so that now we can equate conditions for both solutions and find the mass:
$$\mathrm{pH}_1 - \log{m_1(\ce{KA})} = \mathrm{pH}_2 - \log{m_2(\ce{KA})}$$
$$\log{\frac{m_2(\ce{KA})}{m_1(\ce{KA})}} = \mathrm{pH}_2 - \mathrm{pH}_1$$
$$m_2(\ce{KA}) = m_1(\ce{KA})\cdot10^{\mathrm{pH}_2 - \mathrm{pH}_1} = \pu{2.40 g}\cdot10^{5.0-4.8}\approx\pu{3.80 g}$$
So, in order to achieve $\mathrm{pH} = 5.0$, one has to add $\pu{3.80 g} - \pu{2.40 g} = \pu{1.40 g}$ of potassium salt.


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Does aniline react with diazonium ions at C or N?

In the azo coupling of aniline with benzenediazonium cation, I thought of two possible products 1 and 2 that could be formed. However, I can't work out which will be preferred. What would be the major product and why?

This is a really interesting question and the answer is that the reaction of benzenediazonium chloride with aniline is a bit different to most of the reactions of benzenediazonium salts in that the initial product is compound 1, diazoaminobenzene. It is possible to run the reaction to isolate diazoaminobenzene prep here.
These diazoaminobenzene compounds are unstable with respect to reversion to a diazonium salt + nucleophile, and so many references suggest that compound 2, 4-aminoazobenzene, is directly produced. The procedure for the transformation of diazoaminobenzene 1 to 4-aminoazobenzene 2 is by heating to 50 °C with aniline (prep here).

In general, for anilines reaction at nitrogen is kinetically faster than reaction at carbon. However, the C-substituted product is usually more stable than the N-substituted product, so can prevail under thermodynamic conditions.
As noted by Mayr et al. (the discussion below is taken from the same article):[1]

Similar regioselectivities are found in azo coupling reactions. It has long been known that anilines, as well as N-alkyl anilines, initially form triazenes in coupling reactions with benzenediazonium salts (N coupling), whereas C coupling takes place with tertiary aromatic amines. [...] It was reported that the reversible attack at the nitrogen atom is generally 20–25 times faster than the attack at the carbon atom. The mechanism (in the scheme below) shows that even diazonium ions prefer N attack of kinetic control. In the absence of base, the formation of the triazene is reversible and one observes the azo compound as the only reaction product.

The kinetic preference can be traced to several factors. Firstly, the charged intermediate 2a resulting from N-attack is more stable than the intermediate 2b resulting from C-attack. This can be explained by the fact that aromaticity is disrupted in 2b.
Secondly, there is a greater intrinsic barrier to a reaction at carbon, due to the large reorganisation energy involved in C-attack. To a first approximation, this can be understood in terms of atomic positions and electron density distribution: in both of these aspects 2a is closer to 1 than 2b is.
Both of these factors combined make the formation of 2a much faster than the formation of 2b (for more information, see: Marcus theory on Wikipedia). Hence, under kinetic control, the triazene 3a (E = PhN2) is formed from deprotonation of 2a.
If 2a cannot be deprotonated – this is the case if there is no base, or if the substrate is an N,N-dialkylaniline – the triazene 3a cannot be formed, and so the azobenzene 3b is the only product. Otherwise, it is necessary to use thermodynamic conditions to obtain the azobenzene 3b, which is more stable than 3a. This is consistent with the experimental procedures that Waylander has linked in his answer, where sodium acetate is used for the formation of the triazene 3a, and the HCl salt of aniline (without any added base) is used for the formation of azobenzene 3b.
Reference

Mayr, H.; Breugst, M.; Ofial, A. R. Farewell to the HSAB Treatment of Ambident Reactivity. Angew. Chem. Int. Ed. 2011, 50 (29), 6470–6505. DOI: 10.1002/anie.201007100.

Orthocresol Tertiary aromatic amines and primary aromatic amines are answered .How about secondary aromatic amines ?

“N-alkyl anilines” means the same thing as your “secondary aromatic amines”. Consider, e.g., N-methylaniline (= PhNHCH3).


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Hydrogen storage using activated carbons
On Wikipedia, a possibility of using activated carbons to store hydrogen is reported. According to the Wikipedia page, this solution can be used to store around 5 wt% by cooling down the hydrogen to -196$\;^\circ$C at 1 bar. I am wondering if the storage system has to be kept always at this temperature?

Short answer: yes, if you want to store the same amount of hydrogen and on condition that you do not mess with pressure.
Long answer: I'll skip some intermediate physicochemical derivations for brevity. The phenomenon we are dealing with is called adsorption. Perhaps, the simplest theoretical explanation of adsorption was given by Langmuir. Langmuir adsorption isotherm can be written as $$\theta = \frac{Kp}{1 + Kp},$$ where $\theta$ is a fraction of the surface of adsorbent (solid) covered by the adsorbate (gas) and $K$ is an equilibrium constant. It is seen that at a constant temperature the more pressure you apply to the system, the more surface is covered (i.e. the more gas is adsorbed).
But what about the temperature dependence? Since $K$ is an equilibrium constant, its temperature dependence is defined by the van't Hoff equation. Also, $K$ is directly proportional to the pressure required for a certain amount of gas to be adsorbed. In other words, when $\theta$ is constant, $\ln{K}+\ln{p}=\mathrm{const}$, transforming our van't Hoff into something like $$\left(\frac{\partial\ln{p}}{\partial(1/T)}\right)_\theta=\frac{\Delta_{ad}H^○}{R}.$$ Here, $R$ is the gas constant, and the sign of $\Delta_{ad}H^○$ - adsorption enthalpy - is negative, for adsorption is exothermic. Consequently, the slope of $\ln{p}$ vs $1/T$ line is negative. In plain English, the equation above reads like this: in order to obtain the same amount of adsorbate $\theta$ you should either decrease the temperature or increase the pressure.
If you're that interested in the topic, you may also want to check out how the sorption pumps work.

What would be the most efficient and repeatable way to store hydrogen when the size of the storage system is not as constrained as for the automotive industry? For high pressure storage the containers are quite expensive.


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Constructing potential energy surface of a periodic system, how to deal with atoms moving out of the supercell?
I'm using NVE ensemble for doing molecular dynamics (MD) simulations. The periodic system I'm working with contains $50$ atoms and the "calculator" for calculating potential energies and forces is a trained neural network. The platform I'm using for MD simulation is ase (atomic simulation environment).
After I ran the MD simulation for about $3$ps under $380$K, the atoms are moving out of the supercell. The ase module for MD simulation just faithfully updates the positions of atoms in the supercell (the supercell contains 50 atoms) based on the velocity. From its source code, it seems the module does not take periodicity of the system into account.
EDIT
@LonelyProf 's helpful comment reminds me periodic boundary conditions (PBC) should be implemented in the neural network calculator. The neural network calculator consists of two parts:

Generate Behler-Parrinello fingerprint given the Cartesian coordinates of atoms in the structure model. The fingerprint is actually a vector of symmetry functions that describes two-body and three-body interactions. Two-body symmetry functions are functions of atomic spacing, three-body symmetry functions are functions of both atomic spacing and angle.
In my original implementation (decorate the neural network potential to be a readable calculator by ase), there is not a module that adjusts the Cartesian coordinates of atoms "run wild" so as to move them back to the supercell.
Calculate the energy of a structure given its fingerprints. The calculated fingerprint of a structure in step one is treated as the neural network input to calculate the energy of the structure.

When constructing the potential energy surface of a periodic system whose structure model has a fixed supercell, should I always keep the atoms within the supercell?

It's not clear whether you suggesting that anything is wrong or not. My cursory reading of the ASE documentation indicates that the periodic boundary conditions are associated with the Atoms object, not with the MD algorithm itself. Provided that the Calculator takes the boundaries into account when computing energies, forces etc, there is no need to put the atoms back into the supercell. Can you clarify the nature of the problem?

@LonelyProf Thank you for your helpful comment! I should have taken boundaries into account when I constructed the calculator. I have edited the post to provide more detail, can you help me check it?

I think that it is best to consider the calculation of the potential energy (respecting the periodic boundary conditions) as a separate matter from keeping the atoms within the supercell. For simplicity I'm going to assume a cuboidal box with sides $(L_x,L_y,L_z)$, with periodicity in all three Cartesian directions.
Some people like to put the atoms back into the primary cell whenever they move out. This is done by replacing the atomic position vector $(x,y,z)$ by $(x,y,z)+(n_xL_x,n_yL_y,n_zL_z)$, where $(n_x,n_y,n_z)$ are integers, chosen to make the new coordinate values lie in the desired range, often $(\pm\frac{1}{2}L_x,\pm\frac{1}{2}L_y,\pm\frac{1}{2}L_z)$. If you do this every MD step, the integers will usually just need to take the values $0,\pm1$. Other people prefer not to apply this correction, and just keep track of the copy of an atom that diffuses into the periodic-image supercells.
Even if you do this correction to the atom positions, you still need to apply a similar correction every time you calculate a vector between atoms, for the purposes of computing two-body, three-body, potentials and forces. So, a properly written calculator for these quantities will not rely on the atoms lying within the same supercell. If the atomic coordinates are allowed to evolve freely, without periodic boundary corrections, then the integers $(n_x,n_y,n_z)$ will take whatever values are needed to bring the corrected interatomic vector into "minimum image" form, with components in the range $(\pm\frac{1}{2}L_x,\pm\frac{1}{2}L_y,\pm\frac{1}{2}L_z)$. There are standard prescriptions for doing this.
Often, the energy/force calculation is speeded up by a neighbour list calculation, which involves assigning each atom to a subcell of the original supercell. On the face of it, this seems to require that we put the atoms back into the supercell, before doing the assignment. But actually, a properly written function to identify nearest neighbours will have this mapping built in to it, and may just work with copies of the original coordinates. So, again, there should be no need to put the atoms back into the original supercell "by hand", if the neighbour identification is done this way, and if the vectors between neighbours are corrected as just described.
My impression, from a cursory look at ase, is that this mapping is indeed built in to the functions used to identify neighbours "within range". There is a simple Lennard-Jones calculator, for instance, which seems to rely on the neighbour list function for this purpose. The relevant lines are
        neighbors, offsets = self.nl.get_neighbors(a1)
        cells = np.dot(offsets, cell)
        d = positions[neighbors] + cells - positions[a1]

where I think that offsets contains the integers $(n_x,n_y,n_z)$ and I think that cell contains the cell lengths $(L_x,L_y,L_z)$ in the form of a $3\times3$ matrix (just the diagonal elements would be nonzero in this case).
If you are not using such a function to locate neighbours "within range" and pre-calculate the necessary minimum image corrections to the interatomic vectors, you will need to put it into your calculation of pair and three-body energies and forces, somewhere, yourself. As I mentioned, this is not the same as putting the atoms back into the supercell, when they move out.
Caveat! I am not an expert in ase, and if you want specific advice on that package, you should look at the package documentation, and at simple examples of calculators provided within it. Generally speaking, I believe that advice on debugging programs, and on specific packages, is considered off-topic for this site, so I think that's as far as I should go in any case.

Thank you! I never thought ase get_neighbors module will always find the neighbors within range. I tested this module on a structure where some atoms moved out of the supercell,  and this module identifies the same set of neighbors whether I mirrored those atoms back or not. You are right, to mirror the atoms back really is not a separate matter, what actually matters is the actual distance between two atoms. Thank you!


text:
Why does ligand substitution occur when OH- is a better ligand than NH3?
$$\ce{[Cr(H2O)3(OH)3](s) + 6NH3 -> [Cr(NH3)6]^3+ + 3H2O + 3OH-}$$
When excess $\ce{NH3}$ is added to triaquatrihydroxychromium(III), why does ligand substitution occur when $\ce{OH-}$ is a better ligand than $\ce{NH3}$?
Wouldn't the 3 $\ce{H2O}$ molecules be deprotonated to $\ce{OH-}$ and therefore form $\ce{[Cr(OH)6]^3-}$?
This also seems to occur when excess $\ce{NH3}$ is added to tetraaquadihydroxocobalt(II):
$$\ce{Co(H2O)4(OH)2(s) + 6NH3 -> [Co(NH3)6]^2+ + 4H2O + 2OH-}$$

"Excess" may mean that this occurs simply because the equilibrium is pushed to the right to form the hexaammine complex due to overwhelming conc. of NH3

Have a look at Jan's answer here: https://chemistry.stackexchange.com/q/68547/44877

This is an equilibrium reaction. Using e.g. LeChatelier's principle, you can figure you that if you increase the concentration of $\ce{NH3}$ sufficiently, you shift the equilibrium so far to the right that you get the hexamine as the main product. At neutral pH in water, the concentration of hydroxide is quite low (100 nanomolar). You would expect this reaction to be pH-dependent, and of course depend on the concentration of ammonia in solution.


text:
what is the cross aldol product of ethanal and acetone(major product)?
my teacher told that major product would be when acetone will form nucleophile because it will attack more positive carbon of ethanal(less +i than acetone) however i read in my book that since formation of carbanion is rds ethanal should be nucleophile (more stable due to less +i)

Remember the whole process is reversible

The anion of ethanal may form, and may react with acetone but it can just as easily break down reforming the starting materials if that is not the most energetically favourable pathway. Don't get hung up on which anion forms fastest, it is not the factor that determines the outcome.

ya i got it but why do we then in most cases decide major product using rds step?

Unclear what you mean by "most cases." I've got 15+ years of experience, and I'm not sure I've seen close to enough to count as having seen most cases.

As pointed out by Waylander, the question essentially wants students to go for the most stable product, or thermodynamic product. This is sort of like a well-understood thing for aldol product determination questions. Why? Well... Simply because there is usually only 1 thermodynamic product while there are several possible kinetic products. It is therefore hard to argue that the kinetic product drawn is really the major product under kinetic control. Under kinetic control, you may just get a mixture of products with nothing formed in significantly higher quantity.
Based on this understanding, we can go on to determine the thermodynamic product. This reaction is a very straightforward aldol reaction due to there being only two possible reaction pathways since one reactant is a symmetrical ketone while the other is an aldehyde. So there really are only two possible products...
One possible product comes from aldehyde as a the electrophile and acetone behaving as the nucleophile. Upon dehydration, we obtain A, which is a ketone and has a di-substituted conjugated $\ce {C=C}$ double bond. The other possible product comes from aldehyde as the nucleophile and acetone as the electrophile. Upon dehydration, we obtain B, which is still an aldehyde but it has a tri-subustituted $\ce {C=C}$ bond.

Which one is the thermodynamic product? Usually, we will just look at the substitution of the conjugated double bond formed. Very clearly, the kinetic product must be A since the more nucleophilic enolate from acetone attacks the more electrophilic aldehyde. However, is B really more stable? No, it isn't. The explanation presented below is based on qualitative molecular orbital theory.
Well... The key differences between A and B are:

A, being a ketone, has extra hyperconjugative interactions between the $\ce {C-H \sigma}$ MOs and the $\ce {C=O \pi^*}$ MO, compared to the B
B instead has extra hyperconjugative interactions between the $\ce {C-H \sigma}$ MOs and the $\ce {C=C \pi^*}$ MO, compared to A

The former hyperconjugative interactions are stronger since the $\ce {C=O \pi^*}$ MO is lower in energy and thus more energetically-accessible to the $\ce {C-H \sigma}$ MO. This is due to the higher electronegativity of the oxygen atom. Thus, we would expect the thermodynamic stability of A to be higher than that of B.
To provide even more concrete substantiation for the higher stability of A over B, Karsten Theis has kindly provided a source of data which indicates the Gibbs free energy of formation of the two compounds. As stated on chemeo, here and here, A has $\Delta G_f = \ce {-57.48 kJ/mol}$ while B has $\Delta G_f = \ce {-36.63 kJ/mol}$. This very clearly indicates that A has higher stability than B.
Thus, A is the answer.

On chemeo, the Gibbs energy of formation is tabulated as -57.48 kJ/mol for the ketone A and as -36.63 kJ/mol for the aldehyde B.

@KarstenTheis Thanks for your data. I will add it to my answer.

This is a question of thermodynamic vs kinetic control. For reactions that are irreversible, the kinetics will determine the product ratio. For reactions that are reversible, the product ratio will be determined by the stability of the products if you wait long enough before working up the reaction.
Highly reactive starting materials (e.g. Grignard reagents) typically are associated with kinetic control (you don't expect the Grignard reagent to reform).
Slow reactions that could go in either direction depending on conditions (e.g. esterification/saponification) typically are associated with thermodynamic control.
For some reactions, you can steer them in either direction. One example from molecular biology is taking single-stranded DNA and annealing it either by slow cooling (thermodynamic control, the best matches will be made) or fast cooling (kinetic control, the first matches made will be the products). This is used in the PCR process, where short primers will bind to long DNA because they are present at high concentration even though there is complementary long DNA in the mixture that would form a more stable product.


text:
How does the process of nucleation work for boiling liquids?
I have a somewhat clear picture how (on the molecular level) a gas turns into a liquids as it cools. When a gas has a certain temperature, its molecules on average have a high enough kinetic energy so that electromagnetic interactions cannot hold the molecules together and they are essentially free. When the gas cools, the kinetic energies are smaller and thus electromagnetic interactions between molecules start to pull the molecules together and the process of turning into a liquid begins. As more molecules lose kinetic energy, initially small numbers of molecules start to cluster up, forming "lumps" that continue growing as more molecules gather up into these clusters. I believe this process is called "nucleation".
I found this video that helps visualize the effect:
https://www.youtube.com/watch?v=wFT6G4CIL1o
But how does the reverse take place? Let's say I'm boiling water. I have a textbook that specifically states:

During nucleation, small droplets of liquid form in gases or gas bubbles form in water as it starts to boil.

As I heat water, the average kinetic energies of the molecules increase. How does nucleation create bubbles in this situation? I understand that a bubble is a "pocket" of gas inside a liquid. If nucleation works the same way in the process of liquid turning into a gas as it works in a gas turning into a liquid, by analogy there should be some "lumps" of gas forming inside the liquid. But what would hold these gas "lumps" together as the molecules in the gas have a higher kinetic energy as the molecules around them (that are still part of the liquid)? And how do these molecules with higher energy "find" each other to start forming clusters/bubbles? Or do I have a completely wrong visualization as to how bubbles form?

It is difficult for the initial bubbles to appear. If you put pure filtered water in the microwave (don't try this at home), you can make a super-heated liquid that will violently boil when you put a tea bag in it. In boiling organic solvent, you use boiling stones or capillaries to prevent formation of super-heated solutions.
Commonly, bubbles start at a surface or at an impurity (heteronucleation). To start a bubble in pure water, multiple water molecules have to lose all their hydrogen bonds (typically 4) at the same time. This costs energy, lowering the kinetic energy of the molecules or their neighbors. On the other hand, when the water molecule is adjacent to a wall or a piece of dust (hydrophobic), the hydrogen bond network is already disrupted, so it costs less to make a bubble.
On a stove top, bubbles form on the bottom of the pot because the temperature there is the highest. In a microwave, they can form anywhere (not sure whether the top of the liquid is actually hotter than the bottom, depends on the way the microwave oven is built).
Bubbles get bigger as they rise to the top for two reasons: the pressure decreases (not a big difference in a pot of water, you have to go 10 m deep to double the pressure) and the bubble picks up more water molecules if the water has a temperature above the boiling point.

what would hold these gas "lumps" together as the molecules in the gas have a higher kinetic energy as the molecules around them (that are still part of the liquid)? And how do these molecules with higher energy "find" each other to start forming clusters/bubbles?

It is the opposite. If you have - say - four water molecules with very high kinetic energy going in all directions, the space in between will be the newly-formed bubble. Once it is formed, more molecules in the liquid have to separate from it and go into the bubble. And gas molecules have to bounce off the liquid-gas interface to maintain the volume of the bubble.

Firstly, the average kinetic energy of the water has to be high enough (water obviously doesn't start to boil at room temp. and pressure the moment it is poured into a pan).
As the water gains more and more energy, small bubbles appear but don't fully rise to the surface as the energy of the gas particles is transferred to the surrounding water particles, eventually losing enough energy and becoming liquid again.
But if the energy of the gas particles is lost too slowly for this to occur they fully rise to the surface. This will only occur at a certain temp. difference i.e. the liquid has to be close to boiling temperature if not it will absorb too much energy and the bubble will become a liquid again.
A single bubble can increase in "size" (more molecules inside the bubble compared to when it started not talking about the diameter) as it rises, similar to the cluster effect you mentioned. This will only occur if the energy difference is high enough to vaporise the surrounding water molecules and not just increase their kinetic energy.
This can also not occur depending on how "violently" the water is being heated, or at which temperature the heating element is, therefore the temperature of the bubble being produced.

Thank you but you start with "..small bubbles appear..". I was interested to know why bubbles appear in the first place as a result of nucleation.

@S. Rotos it should be the same explanation. How do you select a limit to the size at which a bubble can be called small? Bubble form whenever molecules that are hot enough gather together. Otherwise it will take for ever to leave the liquid. I would have expected further Q on why objects or dirt can act as a bubble seed. I am sure I got or I was told the answer but at the moment I can't answer. Perhaps LR01 can tell us....


text:
Energy capture photochemistry with ionizing radiation?
Chlorophyll mediates a photochemical reaction: energy from a visible light photon is used to reduce CO2 to CHO, and the energy from that photon can later be recovered when CHO is oxidized back to CO2 in an organism.
Is there any analogous photochemical reaction where energy from shorter wavelength ionizing radiation (like Xrays or gamma rays) is captured in a molecule such that it can later be recovered chemically to do work?  I do not think any biochemistry happens along these lines but maybe synthetic corroles or porphyrins have been made that can accomplish this.

There is a field called betavoltaics, analogous to photovoltaics. However, my understanding is that organic materials would perform poorly in betavoltaic cells due to radiation damage.

The major effect x-rays have is to cause many energetic electrons to be produced from the solvent that then cascade down in energy ionising all in their path. It is a major headache in the x-ray crystallography of proteins.

IF the radiation is ionising it is usually too energetic to be useful for any chemical process. But near UV through to IR might, in principle and in some cases, be capturable by some chemical process.

The energy scales of x-ray and gamma radiation are too large to be chemically helpful in the way that the electromagnetic around visible wavelengths is.
X-rays, if absorbed, typically excite core electrons of atoms. As we know, core electrons are (to a first approximation) not affected by the details of chemical bonding in the valence, so any attempt to use a chemical bonding motif to capture x-rays (e.g. conjugation of porphyrins) is of the wrong energy scale. You could use the Auger Effect to eject a valence electron with x-ray scale energy. This does constitute radiationless conversion of the light into electronic energy, but it doesn't meet your qualifications for storing and extracting work.
Gamma rays are typically produced by the decay of one nuclear state to another, and are vastly different in energy scale than electronic states, so there's no way to do chemistry with gamma radiation, at least directly. The Mössbauer effect does allow us to use gamma rays to understand local structure in a crystal, but even then most of the energy is absorbed into exciting the nuclear state, which we cannot control chemically.


text:
1-Methyl-4-nitrobenzene proton NMR spectrum confusion

I wanted to analyse and suggest the peaks for the NMR spectrum and ran into a problem.
I knew that the methyl $\ce{CH3}$ produces a singlet peak with integration $3$ but when I ran into the benzene protons, I thought that the two equivalent protons (one on each side) would split into a doublet (considering if coupling occurs through 3 bonds or fewer only).
So this was my solution to the problem:

s, $\ce{3H}$, integrated $3$, ppm $2.0$ due to $\ce{CH3}$ group.
d, $\ce{3H}$, integrated $2$, ppm $7.2$ due to $\ce{H}$ on benzene furthest from $\ce{NO2}$.
d, $\ce{3H}$, integrated $2$, ppm $7.9$ due to $\ce{H}$ on benzene closest to $\ce{NO2}$.

However, the spectrum answer given by the mark scheme was as such:

Where have I made a mistake?

The 1H NMR spectrum of para-disubstituted benzenes is not as simple as the (n+1) rule predicts, because of complicated effects. But to a first approximation, a doublet is much more accurate than the supposed triplet given in the mark scheme. see e.g. https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-15-aabb.htm (Ctrl-F for "para-disub") or https://en.wikipedia.org/wiki/Magnetic_inequivalence#para-Disubstituted_benzenes You can see the NMR spectrum of this specific compound at https://www.sigmaaldrich.com/spectra/fnmr/FNMR011092.PDF

In the 1H NMR spectrum of p-disubstituted benzenes, the couplings to consider among protons in ortho, meta, and para relationship to each other.  Depending on the nature of your sample and the experimental conditions (e.g., resolution / how many data points per ppm the spectrometer records), the signals of the AA'XX' system might look different (simpler).  Cf. https://chemistry.stackexchange.com/questions/73434/1h-nmr-of-p-methoxy-phenol-32-possible-lines as follow-up of https://chemistry.stackexchange.com/questions/73328/what-is-the-proton-nmr-spectrum-of-p-methoxyphenol.


text:
Atomic carbon in rocket exhaust
Inspired by this article, I've been looking at using metallic hydrogen as a rocket fuel using CEARUN. I'm looking at diluting the atomic hydrogen with liquid methane to reduce the chamber temperature to something that could be handled using known materials. The reaction that powers the rocket would be $\ce{H^. + H^. -> H2}$. I used a mixture ratio of 4.5 mass units of liquid methane to 1 mass unit of atomic hydrogen. I got an ISP of ~760-780 seconds at reasonable chamber temperatures (3650 kelvin). The main exhaust products in terms of 'molar fractions' are as follows: $\ce{H2} = \pu{79.011}\%,$ $ \ce{C (gr)} = 20.949\%$, with a remaining $0.04\%$ being mainly $\ce{CH4}$ (methane). The exhaust temperature was 1505.2 kelvin.
Here's my main question; am I right in assuming that $\ce{C (gr)}$ means atomic carbon? Would atomic carbon at a temperature around 1500 degrees kelvin cause any major problems for a rocket engine? I'm assuming that at such high temperatures the carbon wouldn't solidify and cause any serious problems, correct?
Finally; can anyone suggest any other substances that could be used as a dilutent in the aforementioned reaction that would yield a higher specific impulse (anything that isn't liquid hydrogen)?
EDIT: So does anyone have any educated guesses on what would happen to the engine in this case? Thanks for all the responses by the way :)

C(gr) is most likely graphite (https://chemistry.stackexchange.com/questions/81209/physical-state-of-the-carbon-denoted-with-gr) and is definitely not atomic carbon.

Also: you are not lucky: in fact, carbon is the element with the highest melting point of all the periodic table (that's why, for instance, graphite vessels are used in high temperature devices like ETAAS).
The most common "form" (allotrope) of carbon, graphite, has a melting point around 3500°C

I think its worth noting that chamber temperature does not have to be reduced if you inject fuel along the wall of the combustion chamber to cool/insulate the materials.

Some known rocket fuels do produce a lot of carbon in their exhaust but theory is very poor at predicting their real performance. You should read the classic (and extremely well written) Ignition: an informal history of liquid rocket propellants by John D Clarke for some relevant evidence (and discussion of other extreme chemicals that have been tested like chlorine trifluoride.)

@A.K. : According to the authors even with active cooling measures such chamber temperatures could not be contained with any known materials.

@chuckstables Which is probably not the biggest problem. The biggest issue is that we do not know how to create metallic hydrogen in the first place and we can't even be absolutely sure it even exists.

@ matt_black : Yeah I'm totally aware of that. I was just sort of curious about a what if sort of scenario.

@chuckstables what about an ablative interior?


text:
What are the units of Kc and Kp?
They are both equilibrium constants as far as I know.
Kc is in terms of molarity and Kp is in terms of pressure. Also both of them are ratios of respective quantities [ ratio of molarity(s) in Kc and ratio of pressure(s) in Kp], so they should be dimensionless according to dimensional analysis. But in some places I have seen units mentioned along with both Kc and Kp.
It is quite confusing and I don't see any uniformity in this.
Anyone please help.

Try to express $K_\mathrm{c}$ or $K_\mathrm{p}$, say, for $\ce{N2O4 <=> 2NO2}$ and see how "dimensionless" they are. Equilibrium constant can only appear dimensionless when it so happens that the units are cancelled out.

But in this reaction    2(SO2)[g] + (O2)[g] <=> 2(SO3) [g]    a unit would appear in KC.

... and how it deviates from/contradicts with what I've written? Also, KC reads as potassium carbide.

The equilibrium constant $K_p$ is dimensionless as it is defined, ultimately,  in terms of the ratio of activities, which are themselves dimensionless. In practice we often use partial pressure instead of activities and then we use the numerical value by effectively dividing each partial pressure by 1 unit of pressure, say 1 atm.
In cases when a mole fraction (or concentration) is used then we find that  $K_p=K_x P^{\Delta n}$ where $\Delta n$  is the change in the number of moles, product minus reactant. However, $K_x$ is not a true equilibrium constant since its value must change as $P$ changes to keep $K_p$ constant.
If you are not sure you can check that $K_p$ must be dimensionless from the equation $\Delta G^\mathrm{o}=-RT\ln(K_p)$ as anything inside a function, such as a log, has to be dimensionless.
(Note that as the equilibrium constant $K_p$ is defined in terms of the free energy  of the standard states of the gaseous species at 1 atmosphere pressure it is independent of pressure.)

The equilibrium constant Kp is dimensionless as it is defined, ultimately, in terms of the ratio of activity coefficients, which are themselves dimensionless.

Excuse me? Won't $K_p$ have dimensions of pressure raised to some power? And what about $K_c$? I'm guessing $[L^{-3}]^a$ ... a is any number ... I think mol is dimensionless so I'm only considering the dimensions of litres.

@AnuragB. porhyrin is correct. There is a trap: $K_\mathrm{p}$ is indeed dimensionless as it deals not really with the pressures, but fugacity $f$, which, in turn is expressed as $p_i/p°$, where $p_i$ is partial pressure and $p°$ is standard state of pressure so that the units of pressure are cancelled out. Also, no sarcasm this time:)

The answer is OK-ish but there is one obvious error, which is now repeated in the comments. Actually, thermodynamic equilibrium constant is defined in terms of activities (fugacities) and not just activity coefficients! Fugacity, for example, is $f=\gamma \frac{p}{p^○}$ where $\gamma$ is an activity coefficient and $p^○$ is a standard pressure. Mind also that the latter is tricky: it was 1 atm in the past, but now IUPAC recommends it to be $10^5$ Pa, or 1 bar in non-SI units. So, you see that it is the activity (fugacity) which is dimensionless. Ask Atkins if you don't believe me.

@voffch, you are correct, I was careless and I have changed my answer  to be clearer, I was thinking of $K_x$ when I used 'coefficients'.

Different disciplines and different textbooks treat units of K differently, and you have to be very careful to figure out what their conventions are. Often, the conventions change within a textbook, and often, this is not state explicitly.
Examples where equilibrium constants are clearly treated as dimensionless are Kw = 10e-14, expressions involving ln(K) or -log(K), and once you make the switch from concentrations to activities, or from partial pressure to fugacity.
Examples where equilibrium constant are clearly treated as having dimensions is for the dissociation constant Kd used in biochemistry, or in converting Kc to Kp.
The way you make equilibrium constant dimensionless is to use dimensionless measures of concentrations in their definition. The best way is to use activities because then you get exact results. When that is not possible, you get dimensionless concentrations by dividing the concentration by the respective standard state (given in the same dimension) so that dimensions cancel out. The common standard state for solutes is a concentration of 1 M, whereas the common standard state for solvents is the pure liquid (that's why we usually leave out the solvent from an equilibrium expression). Be aware that unless you are using activities, all your calculations are estimates that become accurate only when working with infinitely diluted solutions.
As an aside, if you want to have a dimensionless version of converting Kp and Kc, you would use (RT M/atm) in the formula instead of (RT) if your agreed-upon standard states for Kc is 1 M and for Kp is 1 atm.

$K_c$ and and $K_p$ are not dimensionless quantity. Their dimensions change with the reaction you consider.
For example in the reaction you mentioned, clearly $K_c$ should have the units of $\mathrm{\dfrac{L}{mol}}$ which is evident of my statement.
For $K_p$, you can use the formula $K_p=K_c (RT)^{\Delta n_g}$ to find the dimensions. Remember that $R$ and $T$ also have their own dimensions.


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Bond enthalpy of hydroxide
Why does it require more energy to break the first O-H bond in water than is required to break the second O-H bond?

The Masterton Hurley (6th ed.) textbook has a bond dissociation energy of +499 kJ/mol for breaking water into a hydrogen atom and the hydroxyl radical OH. The bond dissociation energy for breaking the hydroxyl radical into another hydrogen atom and an oxygen atom is given as +428 kJ/mol.
You could make the argument that you are starting out with a situation where all electrons are paired and end up with two radicals in the first step, whereas in the second step you already start with unpaired electrons (the hydroxyl radical has seven valence electrons).
In the title of your question, you mention the hydroxide ion. This is not a relevant species because bond energies usually refer to homolytic bond cleavage, with one electron each remaining with the two fragments. It would require heterolytic bond cleavage to go from water to the hydroxide ion (water reacting as a Bronsted acid). This reaction merely requires +55 kJ/mol in aqueous solution.
Some sources contribute to the confusion between hydroxide ion and hydroxyl radical in this context. For example, Boundless states:

For instance, the HO-H bond in a water molecule requires 493 kJ/mol to break and generate the hydroxide ion (OH–). Breaking the O-H bond in the hydroxide ion requires an additional 424 kJ/mol. Therefore, the bond energy of the covalent O-H bonds in water is reported to be the average of the two values, or 458.9 kJ/mol. These energy values (493 and 424 kJ/mol) required to break successive O-H bonds in the water molecule are called ‘ bond dissociation energies,’ and they are different from the bond energy. The bond energy is the average of the bond dissociation energies in a molecule.

They incorrectly refer to the hydroxide ion (OH-) and should be referring to the hydroxyl radical (OH).


text:
Changes in pressure for an equilibrium reaction containing a solid/liquid on one side only
For a reaction where there is only one solid/liquid present, but gases present as both reactants and products (are there any examples of this type of reaction?):
$$\ce{A(s) + B(g) <=> C(g)},$$
if pressure is increased, it will favour the formation of the side not containing the solid/liquid (e.g. the forward reaction favoured), so as to decrease pressure of system – since one mole of a gas molecules has a considerably lower volume than a mole of any solid or liquid (far more spread out).
And the opposite if pressure decreased.
E.g. all that is considered is which side the solid/liquid is on, not the moles of gas present on either side. Is this true?

There are many examples of such reactions, for instance:
$$
\begin{align}
\ce{S (s) + O2 (g) &<=> SO2 (g)}\\
\ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\\
\ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
\end{align}
$$
However, your assumption

all that is considered is which side the solid/liquid is on, not the moles of gas present on either side

is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.

Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.

If you want to discuss using Ideal conditions, then pressure should have no effect on that equilibrium - the reaction produces and consumes an exact equal amount of gas - and with ideal conditions individual molecules have zero volume so the solid part doesn't take any. The assumption is fairly good, the volume of a mole of solid is completely negligible compared to a mole of gas of the same compound.
If you use Real conditions you have to evaluate the compressibility of the gas vs gas+liquid/solid along with some thermodynamic data about the compounds to get the shift of internal energy per atmosphere and compound. The magnitudes of these should show which way the equilibrium goes. And if you are really fond of doing long and complicated integrals and Legendre transforms and whatnot, you could also try to estimate where the new equilibrium would be by finding the minimum of the Helmholtz' free energy with varying the ratio of pressures of C and B but it does not sound like an interesting evening to me.

The equilibrium constant $K_p$ is defined in terms of the free energy  of the standard states of the gaseous species at 1 atmosphere pressure so is independent of pressure. This means that if the pressure is changed the degree of dissociation $\alpha$ may change and this is determined by working out the equilibrium in terms of $\alpha$ and the partial pressures.
If the partial pressures are expressed as mole fractions $x$ then $K_x =K_pP^{-\Delta n}$ where $\Delta n$ is the change in the number of moles, products minus reactants. Thus, unless $\Delta n =0 $, which is your case, the mole fractions of the components will vary if the total pressure changes even though $K_p$ does not.  If the partial pressures are expressed as concentrations then $K_c=K_P(RT)^{-\Delta n}$ and $K_c$  like $K_p$ is independent of pressure for perfect gasses.


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Isotopes of Elements
When we talk about the substance $\ce{H2O}$ with a molar mass of $\pu{18.0 g/mol}$, what hydrogen and oxygen isotopes are present in the $\ce{H2O}$ molecules? In other words, what isotopes do the symbols represent?

Most periodic tables quote the molar mass of an element as the average mass of the element given the natural abundance of the isotopes. Chlorine, for example, has a standard atomic weight of 35.45 which is a weighted average of ~75% 35Cl and ~25%  37Cl, the two common isotopes.
But isotopes of hydrogen and oxygen are rare. So the vast majority of water molecules will not contain them. So, at least to one decimal place, most water will have a molar mass of 18.0.
At it's natural abundance of about 0.25% 18O won't noticeably affect the molecular weight of water until you can measure it to at least 3 decimal places of precision. Slightly simplifying (assuming pure isotopic masses are exact integers, which they are not), water containing that isotope at its natural abundance will have an an average mass of 18.005 rather than 18.000.
When working with normal (unenriched) substances the best assumption is that they will contain the normal isotopic abundance of the component elements (which are well known and, for most elements, don't very much by source). But the effect on molecular mass will often be small and you will need a fair amount of precision to notice it. If you are not working to very precise masses, it is fair to assume natural isotopic abundance.

It's impossible to find isotopic composition of a compound with a molecular weight containing only one decimal point ($\pu{18.\underline{0} g mol-1}$). Typically, 6th decimal place is required in order to refine percentage of each isotope.
Symbols themselves (such as $\ce{H}$ and $\ce{O}$) don't carry any additional information except atomic number $Z$. For isotopes you need to use so-called AZE notation: $^A_Z\ce{E}$, for instance $\ce{^1_1H}$ and $\ce{^{17}_8O}$. Since $Z$ is already known from $\ce{E}$, simplified notation is often used instead: $\ce{^1H}$ and $\ce{^{17}O}$.

I'm just confused as to where the molar mass of a substance comes from. As in how do we know how much a mole of a substance weighs exactly? I could have 2 samples of 1 mole of water, and one could weigh more than the other because one may contain more heavy water (D2O) than the other.

Unless you took one water sample from the nuclear reactor and the other one from the river, isotopic distribution is going to be the same as elements have uniform isotopic distribution in nature. You probably want to read these topics: https://chemistry.stackexchange.com/questions/71942/what-is-the-standard-molar-weight-of-carbon-and-why and https://chemistry.stackexchange.com/questions/38082/what-is-the-difference-between-molecular-mass-average-atomic-mass-and-mola and any introductory chemistry textbook.

Hydrogen has 2 stable isotopes,H-1 and H-2, Oxygen has 3 stable isotopes H-16, H-17 and H-18.
There is very little of the minor isotopes so the molecular weight of water is 18.
However hydrogen and oxygen are light elements and differences in atomic weights have detectable differences in physical and chemical properties. For example H-2 compounds can have lower reaction rates than the H-1. Boiling points and freezing points of various isotopic forms of water differ slightly. Indeed it possible to distil water to separate the various isotopes.
This also happens in natural systems where minute variation in isotopic contents trapped in ancient  ice can be used to gauge the ocean temperature of that time


text:
how to measure the efficiency of carbon filter [or of activated carbon/charcoal]?
Fake HEPA filters are easy to spot using DIY PM sensor and DIY Air Purifier (we just need to turn the fan ON, place the sensor's tube in front of it, and read the data).
But what about carbon filters?
I bought a filter like this to clean the air, but I wonder if it really works (there are many fakes out there).
I'm looking for some tricks to know if it works, and (much better) to evaluate its efficiency (using Arduino sensors or any basic ways).

Well, the AliExpress link look very suspicious (instead of photos there is a bunch of pitch black rectangles). I think the classical method of testing would be determining the surface area of activated charcoal by Langmuir adsorption isotherm.

Thank you! I'll look at it. (I bought a HEPA filter from that brand, and it works ―amongst many others which did not work. So the seller might be honest).

You might try a something like a VOC sensor before and after your filter.
Just let some gasoline (or diesel, or nail polish, or whatever) evaporate in a fixed volume box (you could measure the amount by weight for instance). You can also have a fan in that box to ventilate the mixture, take the base reading with the VOC sensor. Add your filter to the fan an look what happens to the VOC reading.
It might be complicated to quantify the reduction rate professionally but could give a good impression of its effectiveness.
Edit: take 2 base readings, one of the box with "clean" air, one with contaminant. If you ever reach zero again you may even calculate absorption rates, you could also determine total absorption capacity of the filter, etc.

What sensor do you use (I tried few years ago, and it did not work at all)?

There are a few digital VOC sensors that seem to work relatively well (such as Sensirion SGP40 and modules based on it).

⚠ EDIT (after writing this answer). Be careful, when a carbon filter becomes "full", it can then release the molecules to which it has less affinity to capture the one to which it has more affinity.
The big question is: does it affect the test below?
Can iodine take the place of nasty stuff and thus give the impression that the active carbon isn't full? Anyone can answer?

1/ Use a TDS meter:

It's easy if you want to test the efficiency of a water filter:
 This Video from Project Farm use it to compare the efficiency of different water filters.

2/ The iodine method:

I found the answer on this video (from Cody's lab YouTube channel). Cody's show how to make activated carbon/charcoal and how to test its efficiency.
It's quite easy. The test below measures the absorption's efficiency of your charcoal filter (activated or not). (⚠ don't use it to test a water filter unless you throw it away after, because the smell/taste of iodine won't go away).
Quick facts:

Cody use iodine but it seems to work with any dye (it works with methylene blue and even red Fanta.
Non activated carbon (normal charcoal) does filter, but it's much less efficient.
To check if the carbon is activated or not just put it in some water, activated carbon is bubbling.

Tuto:

Take 1 gram of active charcoal (from your filter)
Put it in a test tube
Add 2ml hydrochloric acid (to make sure the charcoal is acidified, because otherwise, the alkaline ash in the charcoal will react with the iodine).
warms it a little to help the reaction.
Add 25ml of iodine (iodine with alcohol). Cody uses ≈ 4mg/ml.
Wait 24h, then look at the color of your solution: the lighter the color the more effective is the charcoal (the lighter, the more it has absorbed iodine which is dark). You could compare it with some active carbon you bought, or normal charcoal

To measure more precisely the efficiency of your charcoal:
The titration method:

Take 10ml of the solution you made above.
Add some thiosulfate to it (drop after drop) until your solution becomes clear.
Measure the quantity of thiosulfate that was needed.

See Cody's video for more details.

Someone in a comment says:
"When titrating iodine, it pays to use starch as an indicator. The change from dark blue to colorless makes the endpoint much clearer than the slow fading out of yellow you were dealing with. This would also be much clearer on camera."

3/ The Formaldehyde (or VOIC) method (requires a Formaldehyde Meter):

If you want to know if your activated carbon filter works, you can also test it using Thomas Talhelm method (he is the most famous air pollution DIY expert). But this technique requires a solution of formaldehyde, and a device to mesure the quantity of formaldehyde in the air.

Buy a formaldehyde solution.
warm it (e.g., using a rice cooker) in a closed room.
measure the level of formaldehyde
turn ON your activated carbon filters, and keep measuring.

If your filter is working you should see that:

With a fan only (red line), formaldehyde levels stayed high. But with
a carbon filter on the fan (blue line), formaldehyde levels went down.

What is the concentration of iodine $\ce{I2}$ used in the first Iodine method ?

@Maurice Cody uses ≈ 4mg/ml (*) but it only matters if you use the nitration method (to simply compare an unknown with a known/good active carbon it probably doesn't matter). *: https://youtu.be/GNKeps6pIao?t=833


text:
Is the adsorption of dissolved iodine onto activated carbon physical adsorption or chemical adsorption?
For an experiment, I am looking to see the effect of temperatures from 0–100 °C on the efficacy of activated carbon as an adsorbent. My adsorbate is iodine and I am confused as to whether the adsorption taking place below 100 °C is physical or chemical adsorption.
I have also read that physical adsorption decreases with an increase in temperature while chemical adsorption increases and then decreases with an increase in temperature. Additionally, I was wondering whether chemical adsorption takes place at all temperatures.

I am not familiar with chemical vs physical adsorption (and nobody is, given the number of answers). Could you give us some references?

The mechanism(s) of iodine adsorption into carbons are complex and strongly dependent on each carbon. That is, iodine can be physisorbed in the carbon micro/mesopores, but in carbons with a rich surface chamistry, it can also be chemically adsorbed through charge transfer interactions.
There are many papers about this, but this one shows clear evidence of both mechanisms. (Written by the carbon "connoisseur" Harry Marsh).
They clearly mention:

Thus, both classical physical adsorption theory and more specific chemical interactions have been proposed to model the adsorption of iodine by carbonaceous materials.

Strong evidence is provided in the paper itself.

What you are referring to are physisorption and chemisorption. They are differentiated mostly by the strength of the interaction between the sorbent and the sorbate. Skip to the highlighted text at the bottom for the punchline; otherwise, read on.
My Master's thesis - Desorption of High Explosives from Soil: Thermodynamics and the Role of Soil Organic Carbon (Minehardt, T. J, The University of Texas at Austin, August 1995) - was, until now, happily residing in a dusty part of my bookshelf, unopened for decades. Alas, it has not been digitized, and I cannot find it on the Internet (although at least one person has cited the DOE report that arose from the work I did), but there are nuggets of knowledge contained therein which I now impart to you and the masses.
First, I'll answer your question: you can, as I did, do experiments with solid-liquid mixtures at multiple temperatures and, from there, extract the enthalpies of adsorption and desorption. You need to fit data to isotherm models (Langmuir, Freundlich, linear, and BET, for example), find the best fit, and various parameters are then available to you. From these quantities, you will be able to determine the nature of the sorption interaction. Note that the best-fit isotherm type also is indicative of the type of interaction.
As for your queries about chemical sorption taking place at all temperatures, I would say that it takes place over a range of temperatures dictated by the system in question; similarly for the trend in chemical sorption - it depends. You can say, however, that physisorption does generally decrease as temperature increases and not run into too much trouble.
For my work, I used uncontaminated soils and solutions that were undersaturated with RDX and HMX (non-polar high explosives) and TNT (a polar high explosive). After a time which the mixtures were held at various temperatures, the supernatant and solid were both extracted and the concentrations of high explosives determined (via HPLC) - this is the adsorption part. For desorption, I used contaminated soils and DI or organic carbon-rich water (humic and fulvic acids), and repeated the experimental process. Coupled with v'ant Hoff plots, which yield enthalpies of solution (among other quantities), you can then make quantitative and qualitative statements about the nature of the interaction.
My primary reference was Voice, T. C. and W. J. Weber (1983). Sorption of hydrophobic compounds by sediments, soils and suspended solids I. Theory and background. Water Resources 17 (10): 1433-1441. I quote from my thesis below.

Physical sorption encompasses weak interactions of sorbent and sorbate, typically exhibiting heats of interaction in the 1 to 2 kcal/mol range. Bonds of this strength fall into the category of van der Waals forces, which is in turn composed of London dispersion forces and electrostatic forces.
Chemical sorption involves the formation of a much stronger bond between sorbent and sorbate, typically on the order of 15 to 50 kcal/mol. While there may be both physical and chemical sorption processes occurring simultaneously, one will commonly dominate. In fact, chemical sorption becomes more apparent at higher temperatures, when heat input into the system plays a more dominant role in the adsorption/desorption process than other factors, such as the thermodynamic gradients of hydrophobic reactions that characterize physisorption.
The electrostatic interaction of sorbent and sorbate is actually a component of physisorption. These forces are only readily apparent when the solute is polar enough to discount hydrophobic forces. A polar solute can interatct with other polar solutes, ionic species, and heterogeneous surfaces through dipole and quadrupole coupling.
The most commonly used tool in analyzing sorption equilibria is the isotherm. For the compound of interest, a plot of equilibrium solution concentration vs. solid-phase concentration (i.e., sorbed) is made at a given temperature. The resultant isotherm is best-fitted by one of a variety of sorption models. Various constants can them be evaluated from the equation that describes the fit to the data. These constants are then used to characterize the behavior of the compound of interest.


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Mechanisms of Birch and Bouveault-Blanc Reductions
The question is all about the sequence of a possible Birch- and Bouvault-Blanc-Reduction in this exemplary substrate. With which reduction the mechanism begin?

My first attempt was to classify the aromatic system as a nucleophilic center, whereas the ester group as an electrophilic center. But as which should the solvated, "free" electron be classified? Maybe, because of its charge, simple as a nucleophilic reagent, which therefore first attacks the nucleophilic aromatic ring? Is this simple explanation right? Because in the Bouvault-Blanc reduction the attack also takes places on the nucleophile, partial negatively charged O-Atom.
This question was inspired by the II,5th Supplementary problem of the book "The Art of Problem solving in Organic Chemistry" (Alonso-Amelot, Second edition). There, the Birch reduction takes place first, but the author delivers (for me) no neat explanation.

What makes you think everything's electro/nucleophilic? SET reactions are neither.

@Mithoron Since I use the Clayden as my primary educational resource, I was until know convinced that it is also common to classify radicals as electrophile and nucleophile. See p. 995 "Organic Chemistry" (Clayden, Warren, Greeves), Second Edition But yeah, it does not make real sense for plain metal radicals.

Molecules can act as  electro/nucleophiles, but electrons aren't classified as such and if electron is transferred not an atom it's not a "philic" reaction https://en.wikipedia.org/wiki/Outer_sphere_electron_transfer

@Mithoron Ok, thanks! Then the question remains: which reduction takes place first: Bouvault-Blanc or Birch?

Organic chemistry is an experimental science, at least historically. Do the reduction with a limited amount of sodium. Then analyze the product distribution.

@user55119 According to my resource (see link below, based on an experimental paper) the Birch reduction takes place first. And in my opinion, it is also common for an experimental science to interpret and explain these results (-;


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What is the displayed formula for methylpropene?
I am currently studying organic chemistry at school and I came across a question in my textbook which asked me to find the full displayed formula of methylpropene. I knew that there would be three carbons because of the 'prop' and that there would be a one-carbon side chain on the first carbon. However, in my chemistry textbook the one-carbon side chain is on the second carbon (the middle one). I don't know whether this is correct or not. Could someone please explain to me what the correct answer is?

There's only one place the methyl group can go for the compound to be a propene derivative i.e. the middle one. If you put it on the first carbon then it is a butene because you have a C4 chain; 1-Methyl-propene IS butene.

Related: Naming of 2-methylpropane. Nevertheless, the name "methylpropene" is not in accordance with IUPAC nomenclature; see also IUPAC rules on omission of locants in carboxylic acid

"Propene" is what defines the main chain of three atoms so that the maximum number of carbon atoms in any other chain cannot exceed this number. Hence your proposition that

there would be a one-carbon side chain on the first carbon

is wrong as in this case there would be a 4-membered chain of either 1-butene or 2‐butene:

In order to comply with the methylpropene name, one can only attach methyl group to the second carbon, and not to the terminal one:

To sum it up, your textbook is correct.


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What does [O] denote in organic redox reactions?
I've seen a few examples in the lectures we've been having and I wasn't able to find any sources talking about it. I think I remember a previous teacher saying it was a symbol for oxidation itself, but I doubt this is true.
Said examples from the lectures:

Could be, but I’m not sure. I’ll wait to see if an answer gets posted

Your memory is correct - it is a symbol for an oxidation

[O] is representing an oxidising agent


text:
Can an iodine clock reaction work without using a strong acid?
I'm going to do an iodine clock reaction for a project and we had to submit the materials and safety sheets for the experiment a while ago. I thought I could do one by using hydrogen peroxide, sodium thiosulfate, potassium iodide and starch, but online I'm only seeing experiments that use sulfuric acid or another strong acid of some sort which is worrying me. I talked to my teacher and it's too late to change/add materials, so I'm really hoping that it can work? I'd also like to know what the strong acid does in the reaction, and what would happen (or wouldn't happen) without sulfuric acid?
Here's a bit from wikipedia for reference about the reaction:
This reaction starts from a solution of hydrogen peroxide with sulfuric acid. To this is added a solution containing potassium iodide, sodium thiosulfate, and starch. There are two reactions occurring in the solution.
(note: I didn't try formatting anything so the numbers after the ^ are the charges)
In the first, slow reaction, iodine is produced:
H2O2  +  2I^−  +  2H^+  →  I2 + 2H2O
In the second, fast reaction, iodine is reconverted to 2 iodide ions by the thiosulfate:
2S2O3^2−  +  I2  →  S4O6^2−  +  2I^−
After some time the solution always changes color to a very dark blue, almost black.
When the solutions are mixed, the second reaction causes the triiodide ion to be consumed much faster than it is generated, and only a small amount of triiodide is present in the dynamic equilibrium. Once the thiosulfate ion has been exhausted, this reaction stops and the blue colour caused by the triiodide – starch complex appears.

You wrote the answer yourself:  the hydrogen peroxide consumes hydrogen ion according to the reaction
$\ce{H2O2 + 2 I^- + 2\color{blue}{H^+} -> I2 + 2 H2O}$
You need the acid to provide the hydrogen ions, especially since if you allow the solution to become basic then the iodine disproportionates (see "Chemistry and Compounds" section, "Iodine Oxides and Oxoacids" subsection).
In terms of pure chemistry, maybe you could have used a moderately strong acid such as phosphoric acid or a bisulfate salt.  But then the limited dissociation of such species might make the iodine formation reaction slower and thus slow down the "clock".
Looking at the Wikipedia article on this reaction, one alternative is to use a peroxydisulfate salt instead of hydrogen peroxide as your oxidant.  This reacts with iodide ion according to
$\ce{S2O8^{2-} + 2I^- -> 2 SO4^{2-} +I2}$
with thiosulfate again as the reducing agent to regenerate iodide ion.  Unlike hydrogen peroxide or the other oxidants, the above reaction does not consume hydrogen ions and is your best bet to run the reaction without a strong acid addition.  However, peroxydisulfates have their own hazards (one of which is requiring a strong sulfuric acid solution to make it in the first place), and if you make the common choice of potassium peroxydisulfate rather than the ammonium salt you could have solubility issues.

Would the experiment not work at all? Would anything happen in the reactions?

Not sure.  Wikipedia has several versions and most use a strong acid.  The only one that does not, appareny, is The one with potassium peroxyisulfate.  See the Wikipedia article.  When I get time I will edit the answer.


text:
ICHO Synthesis Problem 2018
I've several doubts about this ICHO synthesis problem, which was a part of the preparatory problem package of 2018 (so yes, it's not a part of an ongoing contest)

Before sharing the answer given, I'd like to share my thoughts and efforts towards the problem -

The first step is clearly an oxidation step, performed in presence of the peroxy acid above the arrow; which is subsequently reduced to the corresponding carboxylic acid. The oxidation product should be an amine oxide, but the question is, which one? (We have three nitrogens to play with)
For the sake of continuation, let's assume A is what they expect it to be (check the image below). Now, I'd expect more than just one hydrogen of the ring to be replaced by deuterium in the base catalysed tautomerism step. What's the next step (acid catalysed) for? Also, in basic medium (1) the initial ester would undergo saponification to acid.
B to C is a reducing step. Pd/C in presence of formic acid is known to be a reducing agent, but how do we know it reduces the amine oxide back to amine? (It'd be great if some sources could be provided for this, I couldn't find any)

C to the final product is pretty direct, the carboxylic acid group converts to acyl halide which undergoes nucleophilic addition by the ammonia derviative.

Thanks a lot.
P.S. Only a sound explanation for the A to B conversion remains - rest has been taken care of by the first posted answer. Please make additions/include relevant stuff for other conversions if you find any.

I would love to see a mechanism for the acid catalysed deuteriation of the N-Oxide adjacent carbon. I feel that the base is for saponification only and it doesnot catalyse deuteriation.

That's basically an enolate homologue. If you're reading this question, you probably know everything needed to understand that already. @EshaManideep

There's no reason why you can't form the N-oxide in both positions. The Pd hydrogenation should reduce both of them to the amine at the end.

Then, wouldn't deuterium substitution happen at the other ring too ? I guess n N-Oxide formation occurs at essentially one place @Zhe

@EshaManideep That's a good point. In light of that, the better answer is that the pyridine nitrogen is less sterically hindered than the pyrazole has an aryl substituent close to the N. This is somewhat easy to see if we draw out the hydrogens on the aryl rings explicitly.

You have asked quite a number of questions in your post so let's tackle it one by one.
Why does the oxidation take place at that particular nitrogen atom?
To understand this, we first need to acknowledge that peroxyacids $\ce {RCO3H}$ are electrophilic oxidising agents. Specifically, the oxygen atom attached to the $\ce {H}$ is the electrophilic atom. From the frontier MO perspective, the reactive LUMO of the peroxyacid is the $\ce {O-O \sigma^*}$. This implies that the most nucleophilic nitrogen atom is likely to be favourably oxidised.
In fact, the oxidation of pyridine, which is a chemical species containing a nucleophilic $\ce {N}$ atom, to form pyridine $\ce {N}$-oxide using peroxyacids is well-known. It is documented by Clayden, Warren & Greeves (2012) on p. 730 as a method of increasing the reactivity of pyridine towards both electrophilic and nucleophilic aromatic substitution reactions. Notice that the bottom $\ce {N}$-containing ring actually resembles a pyridine fused with another benzene ring. However, note that the lone pair on $\ce {N}$ is not delocalised into the ring(s). Thus, there is no significant decrease in nucleophilicity of the bottom $\ce {N}$ in this compound relative to pyridine's $\ce {N}$ atom, suggesting that oxidation of the peroxyacid can very likely take place there.
Of course, we must be thorough in our analysis and also take a look at the relative nucleophilicities of the other two $\ce {N}$ atoms. Immediately, we will notice that for the $\ce {N}$ atom that appears to be making 3 single bonds has a lone pair that is heavily delocalised, in both the benzene ring below and the 5-membered ring it is in. Thus, we can eliminate the possibility of the the oxidation taking place at that nitrogen atom.
For the final top-most $\ce {N}$ atom, the lone pair may appear to be of the same nature of that of pyridine, i.e. held in an $\ce {sp^2}$ hybrid orbital. However, note that the presence of the adjacent $\ce {N}$ atom may result in inductive withdrawal of electron density from that $\ce {N}$ atom, decreasing its nucleophilicity.
Ultimately, I believe this first part is trying to test students if they are able to relate the structure and reactivity of pyridine towards oxidation by peroxyacids to the bottom pyridine-like ring. If students can make the connection, then they would instinctively know that the oxidation has taken place at that $\ce {N}$ atom.
From A to B what happened?
Well... There are a number of things which happened here. The most obvious being the first step: Basic hydrolysis occurs, with the leaving of $\ce {EtO^-}$. The next step is an electrophilic aromatic substitution, with the acidic deuterium ions as the electrophile. For pyridine $\ce {N}$-oxide, the 2- and 4-positions are activated towards EAS reactions. This can be shown simply by drawing the resonance structures of the compound with an $\ce {N=O}$ double bond. Similarly, the EAS takes place at the 2-position, relative to the $\ce {N}$-oxide group. It is difficult to explain why it doesn't take place at the 4-position as well. In general, for this type of structure elucidation questions, we would need to predict structures by looking at the resultant structures as well, especially if they are given. In this case, we see that only the 2-position has deuterium substitution taking place and we should simply leave it as that. Of course, if anybody has a good explanation for why that is the case, please propose it. Also, the other positions on the other rings may not be activated enough for the substitution to take place. Thus, we do not see substitutions taking place at those positions. Again, I am not able to provide a rigorous explanation for why that is the case.
Finally, the third step with $\ce {H2O}$ effectively makes the pH of the medium neutral again and also changes the carboxylic acid group from $\ce {COOD}$ to $\ce {COOH}$. This is simply due to a concentration effect. The high concentration of $\ce {H2O}$ simply pushes the equilibrium towards species containing $\ce {COOH}$.
What about from B to C?
From Wikipedia, it seems that $\ce {B}$ to $\ce {C}$ is simply a reduction step. The ammonium formate decomposes to give $\ce {NH3}$, $\ce {H2}$ and $\ce {CO2}$ in the presence of $\ce {Pd/C}$. The hydrogen gas evolved then gets adsorbed onto the palladium catalyst, where it can reduce the $\ce {N}$-oxide to give back the original $\ce {N}$-heterocycle.
And the last step involving conversion of C to the final product is as you have proposed.
Reference
Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry (2nd ed.). Oxford University Press: New York, 2012.

Is there any source for B to C, which clearly says that amine oxides are reduced back to amines? Of course, the logic is sound but we can never really be sure about reduction reactions without a credible source.

Ammonium formate + Pd/C is a standard reaction for catalytic transfer hydrogenation. I've used it many times. If a reduction has been recorded using hydrogenation with H2 gas then catalytic transfer hydrogenation will do it.

This paper reports the deoxygenation of pyridine N-oxides by transfer hydrogenation with palladium catalysis  https://risweb.st-andrews.ac.uk/portal/en/researchoutput/deoxygenation-of-pyridine-noxides-by-palladiumcatalysed-transfer-oxidation-of-trialkylamines(c2f846af-97b4-4f9c-89a2-29efaad99187).html

"For the final top-most N atom, the lone pair may appear to be of the same nature of that of pyridine, i.e. held in an sp2 hybrid orbital. However, note that the presence of the adjacent N atom may result in inductive withdrawal of electron density from that N atom, decreasing its nucleophilicity." Please correct me if I'm wrong but - I think that, since electron density is lost by the top most N atom, shouldn't its nucleophilicity increase? Since it has less electron density, its a better place to attack for the electron-rich oxygen atom of m-CPBA, right?

@McSuperbX1 You are mistaken. Firstly, m-CPBA is an electrophilic oxidising agent. See here for greater clarity:  https://en.m.wikipedia.org/wiki/Prilezhaev_reaction The nitrogen atom donates a lone pair towards the electron-poor oxygen atom of m-CPBA. Secondly, the decrease in electron density of an atom would decrease its nucleophilicity in general. For example, consider ammonia and pyridine. Due to the lower electron density at N in pyridine due to delocalisation, pyridine is significantly less nucleophilic compared to ammonia.


text:
Why is excluded volume in van der Waals equation divided by 2?

Source: Physics Stack Exchange
The excluded volume calculated based on the diagram is 8 times the volume of a molecule. The volume is then divided by 2  to give the so-called excluded volume per molecule. Why divided by 2 only?
Obviously more than two molecules can come close together, and their assigned excluded volumes can overlap with each other, giving a smaller $b$. Is $b$ just an upper limit of excluded volume?

Collision is predominantly a bimolecular act. The probability of three gas molecules simultaneously colliding is incredibly low, therefore it quite accurate to say that the excluded molar volume $b$ is overcounted by a factor of 2.


text:
Formula of Oxone
A problem in my book asked the commercial name of sodium peroxide. The answer given was 'Oxone'. Even the website https://www.911metallurgist.com/oxone/ mentions oxone as sodium peroxide.
However, upon searching 'Oxone' on Wikipedia, gives another compound named potassium peroxymonosulfate.
So what exactly is Oxone?

The second one.

I believe if you search darknet for oxone, you are also going to find some new funky chemicals with the same name. Don't take everything posted on the internet for granted.

From [1, p. 828] (and many other organic chemistry textbooks), Oxone's formula is equally written as $\ce{KHSO5 * 0.5KHSO4 * 0.5K2SO4}$ or $\ce{2KHSO5 * KHSO4 * K2SO4}$, a potassium monopersulfate triple salt:

Oxone®
$(\ce{2KHSO5 * KHSO4 * K2SO4})$
A “triple salt”, providing a convenient source of potassium monoperoxysulfate (potassium hydrogen persulfate)
  [37222-66-51]
Commercially available
Notes: This reagent is a useful oxidizing agent.

Sodium peroxide is not mentioned as Oxone in any of the respectable literature sources I flipped through; rather, its names would be Solozone and Flocool.
References

Mundy, B. P.; Ellerd, M. G.; Favaloro, F. G. Name Reactions and Reagents in Organic Synthesis, 2nd ed.; Wiley: Hoboken, N.J.; 2005.


text:
What gas is released when heating sulphuric acid?
I'm doing a chemistry project and am really unsure about the gas released when heating sulphuric acid.

SO3 & H2O would be produced but if you continue heating, the reaction will goes on till you reach a constant boiling point (337 °C) in which an azeotrope is formed


text:
Other than ultracold lithium gas, are any ferromagnetic or magnetic gasses known or hypothesized?
I'm dealing with an interactive work of fiction right now, and one proposal is to combine two types of puzzles with a magnetic gas. I know that fermionic $\ce{^6Li}$, when cooled below $\pu{1 K}$, has shown verifiable ferromagnetic properties, but that is far too cold to be applicable to the work's setting.
(I'm only willing to suspend the user's belief so much.)
I know that on heating above the Curie temperature for a material, magnetic properties dissolve; but what that temperature is varies quite a bit. (I've seen them between $\pu{90 K}$ and $\pu{1400 K}$, which is pretty radical.) Has anyone yet discovered another molecule that remains magnetic in a gaseous phase? Are any of them at least hypothesized?

Gas, not really, but  plasma has magnetic properties and (non-thermal) may have reasonable temperature.

@Mithoron That's helpful, thank you! I was kind of hoping to fit it into a party balloon, though. It's something I can follow up on. Do you have any examples, off hand, of large-scale magnetic plasmas?

related https://chemistry.stackexchange.com/questions/27230/what-happens-if-you-make-a-gas-of-magnetic-particles https://chemistry.stackexchange.com/questions/33994/could-a-magnet-pull-oxygen-out-of-the-air?r=SearchResults&s=1|46.8673


text:
Can =NH2+ in Arginine's side chain form hydrogen bonding at physiological pH?
I know that at physiological pH, Arginine will have its $\ce{=NH}$ protonated to $\ce{=NH2+}$. I was wondering if this $\ce{=NH2+}$ can still form a hydrogen bond by being a donor? And does the positive charge on Nitrogen affect the hydrogen bonding's strength? My confusion stems from the definition of hydrogen bonding, which says that hydrogen's electron cloud needs to be withdrawn by a more electronegative atom like $\ce{F, O, N}$ to be partially positive enough to participate. Therefore, I believe that the positively charged N will withdraw electrons from hydrogens more strongly, leading to more partially positively charged hydrogen and eventually stronger H-bonding with another molecule. Could someone please verify this for me?
Moreover, I have seen $\ce{=O}$ and $\ce{=N}$ forming hydrogen bonding by being an acceptor. However, I tried search but did not find any example of $\ce{=NH3+}$ or $\ce{=NH2R+}$ or NHR2+ forming hydrogen bonding as a donor. So can a Nitrogen with a double bond still participate in hydrogen bonding as a donor?

Welcome to Chemistry.SE!  Please note that formulas can be better expressed with $\ce{ }$ for chemical formulas/equations, $ $ for math term/equations, and $\pu{ }$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies if you haven't already.

Yes, the arginine side chain is an excellent hydrogen bond donor. Charged N-H groups are even better hydrogen bond donors than the corresponding neutral N-H groups. Here is an article that gives examples of arginine side chains in hydrogen bonds. It showcases bidendate hydrogen bonds (i.e. two hydrogen bonds from the arginine side chain) with carboxylates and with nitrogenous bases of nucleic acids.
Specifically for proteins, if the hydrogen bond donor is positively charged (e.g. Arg+) and the hydrogen bond acceptor is negatively charged (e.g. Glu-), you have a combination of ionic and hydrogen bond interaction, which is particularly strong. It is often referred to as a salt bridge, such as in this example.

Both the glutamate and the arginine side chains are charged.


text:
How does Grignard reagent form?
So I've read that during the formation of the Grignard reagent, The R-X bond cleaves homolytically forming free radicals.
My question is how is this happeneing? Due to the high electronegativity difference between Carbon and Halide group shouldn't it result in R+ and X- ?
To be clear, I have no problem following through how the electrons move around. My only doubt is how exactly the Alkyl-Halide bond resulted in a free radical instead of an ion.

related https://chemistry.stackexchange.com/questions/67601/why-do-grignard-reagents-form

To be clear, I have no problem following through how the electrons move around. My only doubt is how exactly the Alkyl-Halide bond resulted in a free radical instead of an ion.

The electronegativity difference between carbon and the halogen used for Grignard reagents is actually not all that high.  Alkyl fluorides are generally un-reactive towards magnesium, and other halogen whose alkyl compounds are more reactive have electronegativity ranging from only a few tenths above carbon to a few tenths below carbon.
In any event the neutral alkyl halide molecule does not directly cleave into either radicals or ions.  According to the University of Texas, the first stage of the reaction involves a one-electron reduction of the alkyl halide to a radical anion; the magnesium is oxidized to a radical cation.  The alkyl halide anion, not the original molecule, cleaves off its halide ion to form an alkyl radical.  Thus, in effect, after the initial electron transfer you get "a little bit of both" -- the alkyl group is a radical but the halogen (even with its modest electronegativity) is an anion.  The alkyl radical, the magnesium radical cation, and the previously displaced halide ion then combine to get the Grignard reagent molecule (the combination process finishes off the oxidation of magnesium to the +2 state).  At no point in the given mechanism do you have a free radical halogen atom.


text:
How to find concentration of three species from given molar absorptivities
If I am given the molar absorptivities of three species at a given wavelength, and I only have the concentration of one of those species, how can I determine the concentrations of each of the remaining two species? The distance light travels, the molar absorptivities and the amount of light absorbed are given.

Welcome to Chemistry.SE!  Anyone is welcome to contribute answers but the aim of this site is quality and usefulness to future users. Please take a minute to look over the help center and tour page to better understand our guidelines and question policies. and please consider providing your thoughts in attempting to answer the question.

The equation you need would seem to be $\ce{A=\epsilon [C] L}$ where A is the amount of absorption (optical density), L the cell path length, your 'distance light travels' and [C] concentration. As you have 3 species absorbing at the same wavelength you have $\ce{A=(\epsilon_1[C_1] + \epsilon_2[C_2] + \epsilon_3[C_3])L}$. You know L and the $\epsilon$'s but only one of the three concentrations and therefore you do not have enough information to determine the two remaining ones.

Which, mathematically speaking, is "an underdetermined system" (https://en.wikipedia.org/wiki/Underdetermined_system).


text:
Reactivity of primary and tertiary alkyl halides
From USNCO 2004, Q58:

The reaction between which pair of reactants occurs the fastest for $[\ce{OH-}] = \pu{0.010 M}$?
(A) $\ce{CH3CH2CH2CH2Cl + OH-}$
(B) $\ce{(CH3)3CCl + OH-}$
(C) $\ce{CH3CH2CH2CH2Br + OH-}$
(D) $\ce{(CH3)3CBr + OH-}$

The following is my thought process:

This reaction occurs through one of the SN2, SN1, E1 or E2 pathways.
All pathways prefer better leaving groups. $\ce{Br-}$ is a better leaving group than $\ce{Cl-}$, hence (A) and (B) can be eliminated.
The nucleophile is strongly basic, so will favor E1 and E2 reactions.

However, I am unable to choose between E1 and E2, and option (D) or option (C), respectively.
Also, if there is any error in my reasoning, could it be pointed out?

in this question, I think, we are not talking about priority of SN2 or E2, but according to the Clayden, (D) would go through E2 and (C) would choose SN2.
But we know that reactivity of tertiary halides are much more than primary ones. So, I think (D) is the right choice, either.

(D), a tertiary halide, undergoes E2 much faster than (C), the primary halide.

How do I choose between E1 and E2 mechanisms? Does the low concentration of hydroxide rule out E2?

I don't think you can rule out a mechanism entirely, there's always a certain proportion of product that comes from the other mechanism.

As far as I understand, the reactivity of tertiary halides is greater than primary halides in both E1 and E2 reactions.