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Simple Cell vs. Daniell Cell
In pretty much all introductory chemistry courses, the galvanic cell (i.e., battery) is introduced as a system of two half-cells with a salt bridge converting chemical energy to electrical energy. From what I understand, this is what is called a Daniell cell. For example, $\text{Zn}_{(\text{s})}|\text{Zn}^{2+}_{(\text{aq})}||\text{Cu}^{2+}_{(\text{aq})}|\text{Cu}_{(\text{s})}|$ with an $\text{NaCl}_{(\text{aq})}$ salt bridge.
However, we can design a much simpler battery using only one cell: $\text{Zn}_{(\text{s})}$ and $\text{Cu}_{(\text{s})}$ electrodes both dipped in the same $\text{H}_2\text{SO}_{4\text{(aq)}}$ solution (similar to a lemon battery). Both designs would produce the same voltage (~$1.10 \text{V}$). But since the latter design is simpler/more practical/cheaper/etc., why do most resources exclusively talk about the Daniell cell? Are there flaws in the single-celled system?
I tried looking for an answer in Why can't a galvanic cell be a single cell? and Why can't an electrochemical cell be a single cell?. They both point towards the idea that single-celled batteries would not do useful work on the external circuit, but I'm not convinced. In the single-celled design, when $\text{Zn}_{(\text{s})}$ dissolves to $\text{Zn}^{2+}_{(\text{aq})}$, it won't react with $\text{Cu}_{(\text{s})}$. Also, free electrons can't flow in an aqueous solution. Hence the only path for electrons is through the wire. Am I missing something?

Read about the history of the battery: https://en.wikipedia.org/wiki/History_of_the_battery. The single reservoir of sulfuric acid works, but only on short time scales because the zinc reacts with the sulfuric acid. The Grove cell and Volta’s pile were better and the Daniell cell, of course.

Zn + Cu in H2SO4 would not produce 1.1 V, but about 0.77 V or less, being a zinc–hydrogen cell.

It seems you confuse a cell like a compartment (that may contain  a half-cell) versus a cell as a complete galvanic system, consisting of 2 half-cells, that may or may not be physically separated.

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

Yes, you're missing something. The overall reaction in a Daniel cell is:
$$\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu }$$
The salt bridge isolates the $\ce{Cu^{2+}}$ ions so that they do not react directly on the zinc electrode. If the $\ce{Cu^{2+}}$ ions do react directly on the zinc electrode then two counterproductive things happen. First of all no electrons flow through the external electrical circuit since the copper is reacting directly at the zinc electrode not the copper electrode. Secondly the zinc electrode itself gets coated with copper which eventually stops the reaction altogether.
Now the other concept to grasp is that in a working battery electrons flow through the external electrical circuit and ions migrate in the battery itself. Both half cells in the battery however must remain charge neutral. So for the Daniel cell cations from the salt bridge have to flow into the copper half cell and anions from the salt bridge have to flow into the zinc half cell.

Thanks, your answer justifies why we need two cells (two beakers) for a battery with two electrolytes (e.g., Zn and Cu dipped in ZnSO4 and CuSO4 respectively). But, regarding the single-celled system with an H2SO4 electrolyte, there won't be any Cu2+ ions in solution, so we won't run into the two problems you outlined. There will only be Zn2+ ions, which won't react directly on the Cu electrode. So, it seems like the single-celled system works just as fine as a Daniell cell, right? Or is there a flaw in the single-celled battery that explains why most textbooks only discuss the Daniell cell?

I can't conceptualize how a single cell for Cu/Zn would work. You can't just have solid Cu and Zn electrodes in an electrolyte. You need $\ce{Cu^{2+}}$ ions in solution which can be reduced.


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Crystal Field Splitting for a d0 metal ion
Its well-known that crystal field splitting occurs because the electrons in d orbitals repel incoming ligands and the d orbitals are no longer degenerate.
My question is now: is the repulsion (with the incoming ligands) between the electrons in the d orbitals or the orbitals itself? If its the former,then does Sc3+ not undergo d orbital splitting ie it exists as 5 degenerate 3d orbitals?

It is between the electrons and the ligands. "empty" orbitals per se don't exist as they more or less only dictate the probability distribution of an electron around a nucleus. However, if the electrons of the sublevels are excited to the d-states, it will be excited into ligand-split d-orbitals, that is correct.

An orbital has 3 major related meanings: 1/ a mathematical wave function being a solution of the wave equation  2/ An electron quantum state belonging to such a function 3/ 3D region of probable occurrence of such an electron.

The answer to this question is quite obvious, an orbital has no physical meaning, it's a mathematical object, an electron is a physical charge particle.  An occupied orbital is therefore the state of the electron, an empty orbital is a virtual state.  Virtual orbitals are useless, have no meaning and are used for convenience to understand things like transitions, the available states and so on. Sc$^{3+}$ has only unoccupied virtual $d$ orbitals, which cannot be split by a crystal field.

so im guessing that this image in https://docbrown.info/page07/transition01Sc.htm is wrong? since it shows that Sc3+ has d-orbital splitting?

@whitingsheep Well, energies the electrons would achieve IF reaching quantum states belonging to empty 3d orbitals MAY split, e.g. in magnetic field - Zeeman effect.

"useless" does not mean "don't exist", and if they have "no meaning" how are they "used to understand things"? Surely the things they are used to understand give them meaning

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Temperature and electrolysis
Based on my understanding, metals conduct electricity better at lower temperatures than at higher temperatures as the resistance to electron flow is lower.
Then, what about ionic compounds? Do they conduct electricity better at higher temperatures? The electrical conductivity of ionic compounds is tested through the set-up of electrolysis. So I mean, is there a larger current flow in electrolysis when the aqueous solution/ molten liquid of ionic compounds is at higher temperatures?

I expect them to behave similar to metals and different from semiconductors.

Based on my understanding, metals conduct electricity better at lower temperatures than at higher temperatures as the resistance to electron flow is lower.

This statement is correct, the conductivity of a metal decreases with temperature, this is due to the collision rate, the Drude model explains it very well.

Then, what about ionic compounds? Do they conduct electricity better at higher temperatures?

The temperature increases a random velocity of ions and the electric field creates a drift velocity, the effects combined give a better conductivity. The ions can also be scattered, the rate depends on the size of the ions.

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In cubic system how many atoms arrangement exist in nature?
All websites say the answer should be 3 , however shouldn't it be four ?

Primitive
BCC
CCP
End Centered

What's end centered? If you mean base centered, then it is not cubic.

Could you edit the question to give some context? The question asks about the number of atoms (or arrangements of atoms, not clear), but then then it asks to choose one of four different centering operations. It would be nice to see the original question or some more background. Googling I learned that polonium crystallizes in a primitive cubic structure, which is otherwise rare for simple structures.

CS, FCC, BCC : https://en.wikipedia.org/wiki/Cubic_crystal_system

End-centered or base-centered cubic actually defaults to face-centered cubic.
Start with a cubic arrangement having eight points at the vertices. To make the end-centered or base-centered cube you add two points at the centers of a pair of opposing faces. But then cubic symmetry, by definition, requires retaining a threefold symmetry axis around a body diagonal of the cube. With the additional pair of points on the faces you identified as bases, this is possible only if you also place points at the centers of the other two opposing face-pairs, and then your array has become face-centered cubic.

Or you could say it does not exist. Once you have centering on one face, you also have it on the others.

The idea behind my logic is to explain why a separare end-centered cubic lattice fails to hold up.

In cubic system how many atoms arrangement exist in nature?

There is no limit to the number of atoms. Below is an example of a plant protein (PDB 1eav) crystallizing in space group $\ce{P 2_1 3}$. It is a primitive setting, and there are lots of atoms (some are disordered water molecules, not shown) in a huge unit cell:

The image is a cross-eyed stereo diagram. If you look at it at a typical viewing distance and then focus on a object (like your index finger) about 10 cm - 15 cm away, the two pictures overlap. Some people are able to focus on the figure while keeping the eyes crossed, achieving a 3D effect. Others need special glasses and the images reversed (for wall-eyed viewing).
Stereo or not, you can see the 3-fold axis along the space-diagonal of the cube, demonstrating that the crystal is cubic. As there is no centering, it is primitive cubic, but centered cubic space groups exist as well.


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Fluorescence and resulting vibrational state
I was taking a look at Jablonski schemes and had some doubts regarding it. Now assume a regular molecule which follows this Jablonski scheme:

We can see that when molecules absorb certain quantized wavelengths, they get promoted to vibrationally excited, excited electronic states. But when they emit the photon again, they do that from the vibrationally lowest, lowest excited electronic state (vibrational ground state of S1 (or T1 for phosphorescence)), this is a result from Kasha's rule.
However, when they emit photons, does it happen to any of the vibrationally excited states of the electronic ground state (although the probability into which vibrational mode of S0 is given by the Franck-Condon-principle) or into a specific one? Is there such a rule?
Edit: As I am currently working on lanthanides, investigating the antenna effect. I was wondering which states the excited electronic atomic states (see the diagram) decay to; the picture shows all kinds of terms can be emitted to, but I am not sure in what proportion. Also, one usually only detects one emitted wavelength in these kinds of complexes (especially of Eu3+, since I worked with it), but I don't understand why because there should be many emission lines:

The de-excitation is into the manifold of vibrational levels of the ground electronic state, just as the cartoon figure shows. You can see this by looking at the fluorescence emission spectrum of, e.g., anthracene. See this excellent answer by @porphyrin: https://chemistry.stackexchange.com/a/77449/79678.

By the way, azulene is a rare exception to Kasha’s rule: https://en.wikipedia.org/wiki/Azulene.

Thats a really good answer, thanks; on another note: since atoms themselves can't rotate or vibrate, Kasha's rule shouldn't apply to atoms right? For example, since atoms can't relax vibronically, we should, in theory, see all emission bands from other electronically excited states? So for example, S1-S0, but also S2-S1 and S2-S0 etc.

(execpt intersystem crossing, which should be possible)

Atoms in the gas phase (or in plasmas) are different: they have energy levels. See, for example, the sodium emission 2D spectrum here: https://chemistry.stackexchange.com/a/164168/79678.

For atoms, we have Grotrian diagrams, as shown in this excellent answer: https://chemistry.stackexchange.com/a/154055/79678. Off topic, but I had the honor of meeting Kasha in 1986. Amazing scientist and guitar maker.

In the condensed phase (your figure) emission is usually from the lowest vibrational level and can occur to any vibrational level in the ground state according the the FC principle. (no rule other than this)  In the vapour phase emission may occur from the vibrational level initially excited to the ground state and is also controlled by FC factors, or, if collisions occur , emission can be from any number of other vibrational levels in the excited state. It just depends on rate of collisions vs fluorescence decay rate constant.

Okay Grotrian diagrams I knew, but never knew that they were called like this, very interesting. I was asking because I am currently working in lanthanide chemistry which (for the most part) behave like isolated 3+ ions, but I was wondering to which states the electronic transitions could be attributed to. Look at the edit if you have the time.

It gets complicated. For example, neodymium ion fluorescence is the basis of the various solid state neodymium lasers: https://chemistry.stackexchange.com/a/158765/79678. Even europium (III) oxide strongly fluoresces in the red. There are plenty of papers you can find, so I think you have good search options.


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Which xylenes, if any, are immiscible with DMSO?
"Xylene" is reported to be immiscible with dimethyl sulfoxide (DMSO), according to a few solvent miscibility tables found online (chart 1, chart 2, chart 3). I have tried to locate some experimental data on this system to verify and quantify the miscibility gap, but have not located anything promising so far.
Seeing as DMSO is reported to be miscible with benzene and toluene, I could expect miscibility with one or more xylenes to be right near the limit. Absood et al. (J. Chem. Eng. Data, 21, 304 (1976)) reported that mesitylene (1,3,5-trimethylbenzene) shows some immiscibility with DMSO below 21.6°C, but also specifically stated that p-xylene (1,4-dimethylbenzene) shows no phase separation in mixture with DMSO. Aralaguppi, et al. (J. Chem. Eng. Data, 37, 298 (1992)) reported various measurements for binary DMSO mixtures with m-xylene (1,3-dimethylbenzene) and other aromatic compounds, at and above 25°C, and report no liquid-liquid separation. Unfortunately the actual literature data I have found so far are fragmentary and not fully useful for constructing a thermodynamic understanding of these mixtures.
So are any pure xylene isomers immiscible with DMSO? Are mixed xylenes immiscible with DMSO? Is there any more experimental data on these mixtures available in the literature?

It's possible to check the solubility theoretically and numerically quite accurately if there is no available experimental data.

@M06-2x without giving a few hints as to the equations or approximation methods involved, your comment comes off as kind of useless. Solubility is well-known to be like a completely foreign language to today’s physics and chemistry.

Modeling interactions between two condensed phases, especially near the "marginal miscibility" range we likely have in this system is going to be troublesome. Activity models, molecular simulations, and even DFT should refer to some experimental data to be based in reality. I am searching for that basis.


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Compounds with fluorescence and/or phosphorescence, what can do both?
What else (besides benzophenone) can both do phosphorescence but not fluorescence, and what else can do both phosphorescence and fluorescence? For example, does pentacene can do both?
I'm reading that there are more compounds that can only do 1, than do both. But is there any inherent contradiction that if it had the properties, the properties will contradict?
For example, if a compound has properties A and B to do fluorescence, and properties C and D to do phosphorescence, then a compound that has A, B, C, and D, should do both fluorescence and phosphorescence, and will not have any contraindicating properties?
And what do we have for fluorescence/phosphorescence in the visible-light range, i.e., absorb violet and emit red.
So far I only have organic compounds examples, what about for inorganic compounds examples? As well as for organometallics.

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An excited state has initially five possible fates, (1) internal conversion to a state of the same spin, (2) intersystem crossing to a state of different spin (e.g. singlet to triplet) (3) emission to a state of the same or different spin multiplicity, i.e. fluorescence, phosphorescence or in general luminescence,(4) chemical reaction such as bond breaking, isomerisation, electron or proton transfer or other photochemistry. Not all molecules exhibit chemical reaction (there may not be enough energy in an excited state for this to happen) but I add it for completeness. (5) quenching by other species in solution. This could be energy or electron transfer or 'heavy atom' effect which is the same as quenching by paramagnetic species such as dissolved oxygen. Clearly this pathway cannot occur with isolated molecules in the gas phase.
The probability of any one process occurring depends on the nature of the excited state and its closeness in energy to other states, (and the nature of any quencher) but any process always occur in competition with all the others, its just a matter of the yield of each process. If the rate constants are $k_i$ then the yield (probability) of event $i$, say fluorescence, is $\varphi_i=k_i/\sum k_i$ and always $0<\varphi_i <1$ and of course the sum of all different yields is unity.

(+1) Nice answer! Back in the early 1980s, I used to do laser-excited two photon photoionization of PAHs in solvents such as n-hexane. I used a pulsed nitrogen laser at about 337 nm. So sometimes excited molecules can suffer another fate: get hit again and ionize. ;-)

Anyways, a compound I hear that does phosphorescence but not fluorescence (or poorly fluorescence), is benzophenone, I believe, at room temperature.

@NealConroy The Hope diamond is blue and famously phosphoresces a deep red color. Phosphoresce is rare compared to fluorescence and especially so in liquid solutions at room temperature and in the presence of oxygen.

It is true that Benzophenone has almost 100% intersystem crossing forming triplets and these are used to sensitise reactions. Phosphorescence is observed in O2 free solutions (Parker & Joyce, Chem. Comm, 1968, p749)

@ed V thanks, I had forgotten about ionisation.

@porphyrin Ed V I had only recently found out that the mechanism behind glow-in-the-dark toys, is not phosphorescence, but called persistent luminescence. Which is something in its Wikipedia article said the mechanism is not fully understood, and it is commonly mistaken as phosphorescence. According to an 2009 article written by a chem professor "the challenge of finding an efficient red emitting persistent phosphor is still waiting." So those 3, and chemiluminescence, are like 4 categories of its own. What do you guys call the study of this? I guess either photochemistry or photophysics.

I was describing what happens in discrete molecules only, in materials such as semiconductors electron - hole recombination can lead to luminescence.

@porphyrin another compound that does phosphorescence but not fluorescence, is pentacene? For same reasons or different reasons as benzophenone?

@Neal Conroy Pentacene is highly fluorescent, did you mean fluorescence but not phosphorescence. I can se no reason why it does not fluoresce, as do naphthalene and anthracene.

@porphyrin oh, so pentacene is a good example for something that does both? It's high molecular weight and rigid, so that's in line for good phosphorescence?


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Risk of spontaneous combustion with cedarwood oil
Is there a risk of spontaneous combustion of cedarwood oil used with rattan diffuser reeds as in the photograph at the bottom?
An NFPA safety sheet talks about the risk of spontaneous combustion of "oily rags wet with flammable or combustible liquid," but does not specify which liquids are and are not a risk.
According to Popular Mechanics, it is the "drying oils" specifically that are a problem, e.g. linseed oil. Wikipedia describes drying oils as "consist[ing] of glycerol triesters of fatty acides...characterized by high levels of polyunsaturated fatty acids, especially alpha-linolenic acid."
This NIH study on the human health risks of cedarwood oil lists the typical composition of three different types of cedarwood oil. These have varying concentrations of thujopsene, cedrol, cedrenes, widdrol, methyl thujate, thujic acid, and thujaplicins. "The ISO standard specifies an alcohol content (expressed as cedrol) of 35-48% with a minimum cedrol content of 20% for Texas cedarwood oil" and a maximum cedrol content of 14% for Virginia oil.
The chemical names in the Popular Mechanics and NIH articles are different, but I lack the expertise to know whether that is meaningful either way. Can anyone with more expertise weigh in?

The risk of spontaneous combustion is greatest when a drying oil (i.e., one readily oxidized) is exposed to air but is thermally insulated, so that heat builds up. The greater the surface area exposed to oxygen (e.g., when spread thinly over a rag), and the more effective the insulation (e.g., layers of surrounding cloth), the more likely it would ignite.
Though I could not find out how easily it is oxidized, it is certainly flammable. Since the diffuser is not insulated, heat buildup is less likely (though where the sticks cross inside the jar could be an issue). Keeping the diffuser in a large nonflammable container and away from any source of ignition seems like a good idea.

The sticks in hat jar are not a risk. These things that stink up flats are sold everywhere. You need both the thermal insulation and the large surface of the rag before anything can happen.

Apparently thujopsene, cedrol, and cedrene are all sesquiterpenes. Sesquiterpenes are a class of terpenes, which according to Wikipedia are unsaturated hydrocarbons. According to the Handbook of Industrial Hydrocarbon Processes, unsaturated hydrocarbons can spontaneously combust. So the answer would seem to be yes, theoretically it could, and therefore it should be treated as a spontaneous combustion risk. I suspect that determining whether or not it is a significant risk would require empirical testing (e.g., here or here).


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Pyrazole iodination
I'm now working on synthesizing 4-Iodo-1,3-dimethyl-1H-pyrazole starting from 1,3-Dimethylpyrazole, iodine, and hydrogen peroxide. I think the mechanism is just a simple aromatic nucleophilic substitution reaction, but I'm not sure if the mechanism is same because of 2 nitrogen atoms in aromatic ring. The reagents are pretty much same as from the textbook.


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What is the probability the bonds broken with sunlight reforming back?
The maximum wavelength $\lambda$ capable of breaking a chemical bond can be estimated as $\lambda = hc/E.$
UV-A $(\pu{380 nm})$ can break bonds with the energy up to $\pu{315 kJ mol^-1},$ UV-B $(\pu{320 nm})$ can break $\pu{374 kJ mol^-1}$ bonds, etc.
So, the sunlight (at UV) can break both the $\ce{C-C}$ and the $\ce{C-O}$ bonds. Sunlight has enough energy to break up vitamin C molecules in juice. Supposedly plants have some excision repair.
But on the chemistry side, when UV breaks a bond, what's the probability the bond can just reform back?
I'm going to assume there are multiple factors. What are factors where 50% of the time, the bond can just reform, as well as factors where ~90% and ~10% of the time, the broken bond can just reform back?

You are asking about geminate recombination (cage effect) if in solution, in the species can recombine rapidly depending on solvent viscosity or much more slowly (as 1/sqrt(time)) at longer times, again depending on solvent properties, so there is no answer as such as time for $50$% it depends on species etc. In the gas phase an isolated molecule will have no geminate recombination if it dissociates.

@porphyrin by species, can be water? I generally have 2 questions for this, the 1st can just be vitamin C in water.

If the bond was instantly re-generated, how would you know it was ever split? In other words, what are you trying to find out here? With a significant probability for that process, it would just look like the interaction cross section of that compound with light of a given frequency would be somewhat smaller. Smaller than a theoretical value that is very hard to predict.

@Karl, we can see bonds breaking under certain circumstances by using femtosecond resolved spectroscopy (pump-probe method) . A bond cannot break in less than half a vibrational period and it is possible to time resolve this. There is a famous example in NaI molecules by A. Zewail where the bond breaks in a step-wise manner, see fig 15 of his Nobel Prize lecture.

@porphyrin what kind of vibrational period?

What do you mean, 'type of .. period' ?

Oh @porphyrin so that didn't include rotational or translational I see.

Does anyone here have clout with Reddit chempros section? Apparently I got permanently banned from there today for posting this same question. They don't want easier questions made for /chemistry as well as no homework help question, which someone accused this question of being. Sigh.

Hey @porphyrin here's an answer I got "2 that come to mind are solvent cage and spin state effects. The solvent can, keep the radicals close together, increasing their likelihood of recombination through the cage effect. Upon excitation, you form initially an excited singlet state. However, if intersystem crossing is facile (or you use a triplet photosensitizer), you can form a triplet radical pair, which can’t recombine as their electron spins are the same (so their unpaired electrons can’t occupy the same bonding orbital)."
Would you say that after sunlight to juice, radicals form some time

UV light creates an excited electronic state, on this electronic state the nuclei are no longer in an equilibrium position and begin to move, i.e. start to move towards breaking the bond. This is the simple semi-classical picture where we view nuclei as classical point particles. More correctly we should speak about the dynamics of the nuclear wavefunction on the excited electronic state potential. But the idea is the same. The initial nuclear wavefunction is no eigenfunction of the excited electronic state and thus "moves".
It is very difficult to describe the dynamics of this process as it involves nuclei and electrons at the same time. A bond can be broken if the wavepacket moves along a vibrational coordinate, elongating the bond. Another aspect that is the exchange of energy between electrons and nuclei. At some point the electronis typically relax via internal conversion to the electronic ground state. Energy is conserved during this step and the decrease of potential energy of the electrons is transferred to the nuclei as kinetic energy.
This excess energy can lead to bond breaking in the groundstate, since the nuclei are now in an excited vibrational state that may lie across a potential barrier that kept the initial bond stable.
This answer is also useful if you want to get a better understanding how bonds are broken.
Lastly we also need to consider exchange of energy with the environment, which allows the molecule to cool down by exchanging vibrational energy with other molecules. This will make the whole process irreversible and stop the dynamics eventually.
The question for a simple probability to reform a bond is thus very difficult and  has no simple universal answer. You need to look at all the steps that I have described, which dependent on the molecule in question and the conditions, to work out an answer. You may be able to measure these processes but then again it will depend on the molecule and the conditions and there is no universal answer.

You mentioned sunlight hitting the nucleus, but yet for covalent bonds like C-C and C-O, the UV must hit the electrons. Perhaps you can divide it into sunlight hitting the electrons vs. hitting the nucleus? Also, another set of category is when sunlight hits ionic bonds.

@NealConroy I don't think I mention "sunlight hitting the nucleus". The very first sentence "UV light creates an excited electronic state" means that electrons are excited. This excitation of the electrons can also cause the nuclei to move.  Bond breaking usually means that the nuclei forming the bond also move apart, or at least an increase in bond length. I attempted to describe this process from a physical/theoretical chemists point of view.

Okay, you think bond reformation also happens more likely in liquids than in gas, as what porphyrin was suggesting?


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What is the meaning of "*" in this compound notation?
I am reading a book on hydrogen energy (handbook of hydrogen energy), and came across the following compound on the Cu-Cl thermochemical cycle: $$\ce{CuO\text{*}CuCl_2(s)}$$
What exactly is the role of the "*" symbol here, how can I better understand this compound?

How is this a duplicate of the linked question? They are about different (admittedly equivalent) notations, and the linked answer and none of the replies to it mention use of a "*" at all.

Although the asterisk is this context denote adduct, it can also mean other things. It can denote electronically excited states. In organic chemistry, it can denote chiral centers: https://chem.libretexts.org/Courses/University_of_Missouri/MU%3A__1330H_(Keller)/25%3A_Chemistry_of_Life%3A_Organic_and_Biological_Chemistry/25.07%3A_Chirality_in_Organic_Chemistry

A similar post for my above point: https://physics.stackexchange.com/questions/443235/nuclear-reaction-what-is-the-asterisk-an-indication-of

${CuO*CuCl_2}$ is  another way of writing $\ce{CuO*CuCl2}$. This substance is a stoichiometric mixture of $\ce{CuO}$ and $\ce{CuCl2}$. The symbol asterisk (*) is often used in chemistry instead of the point or dot (·) , especially when using machine languages like LaTeX.

I'd argue that using correct typography is one of the strengths of latex. I've never seen anyone deliberately using an asterisk instead of a center dot.


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What is the oxidation state of atoms in interstitial transition compounds?
My teacher said that even though interstitial compounds are neither ionic nor covalent, we still define an oxidation state for them. The example she gave was $\ce{TiC}$. She said the carbon is in $-4$ oxidation state. My thought was that there is no chemical bonding in interstitial compounds, and oxidation state is defined as the hypothetical charge on an atom if all its bonds were ionic, and since there are no bonds here, all atoms would be in their elemental state, and oxidation states would be zero. Which one is it?

In my opinion nobody can determine the oxidation state of such interstitial compounds.

" there is no chemical bonding"? Of course there is! It can be considered metallic, but that's irrelevant. C atoms don't bind with each other here, so if you reaally want to ascribe ox. states they are just like you said

From a different perspective perhaps, if the Gibbs free energy changes (decreases) upon introduction of the interstitials then there is 'binding' of some kind going on. This is certainly the case for TiC, and way too many to list other transition metal carbides, oxides, nitrides, borides, ...

Also, interstitial compounds are sometimes non-stoichiometric. So, it's hard to determine their oxidation state.

In the case of titanium carbide $\ce{TiC}$, there is a slight reaction with water to form methane, consistent with an oxidation state assignment $\ce{Ti^{IV}C^{-IV}}$, and even Group 5 carbides with the 1:1 stoichiometry (like niobium carbide $\ce{NbC}$) give methane as well.
From Avgustinik et al[1]:

Contrary to the generally held view that carbides of the transition metals of groups IV and V are inert with respect to water, the authors show on the example of titanium carbide that the carbides are decomposed by water, although the extent of attack is not more than 5–30 atomic layers, i. e., not more than 100–150 Å.
The hydrolysis of TiC proceeds according to the reaction: $\ce{TiC + xH2O -> CH4 + TiO2·xH2O}.$
[sic; the coefficient on water should be $x+2$]
Methane is the principal constituent of the gaseous hydrolysis products, and consequently it is possible to regard carbides of the transition metals of groups IV and V as methane derivatives. Hydrogen evolution in both carbides of group V and nonstoichiometric carbides of group IV is believed to take place as a result of the presence of free electrons in the carbides.

The "free electrons" in the reference apparently refer to the possibility that stoichiometric Group V carbides may contain metal(V) and free electrons (e.g. $\ce{Nb^{V}C^{-IV}e^-}$) or a Group 4 carbide might be carbon deficient and thus not have all valence electrons bound to carbon (e.g. $\ce{Ti^{IV}C_{1-x}^{-IV}(e^-)_{4x}}$). The overall story is that at least for methanides, the distinction between "ionic/saline" and "metallic/interstitial" carbides is actually not sharp.
Reference

A. I. Avgustinik; G. V. Drozdetskaya; S. S. Ordan'yan (1967). "Reaction of titanium carbide with water". Powder Metallurgy and Metal Ceramics. 6 (6): 470–473. doi:10.1007/BF00780135. S2CID 134209836

Although $\ce{TiC}$  cannot be strictly described only by covalent or ionic bonds, it does include both combined.  The hardness of $\ce{TiC}$ is much greater than that of Ti and C in the same configuration, which simply means that they interact to form bonds, contrary to the argument where "there is no chemical bonding" that suggests a process like an intercalation of Li between two graphene layers in Li-ion batteries, where there are no such bonds (covalent or ionic).
Impurities tend to localize orbitals, as this affects the translational symmetry, the ionic effect is more effective. But this explanation is not intuitive and satisfying. There are articles on the subject giving an explanation in the case of $\ce{TiC}$ as a combination of a metallic, a partial ionic and covalent bonds. As they interact, the charge fluctuates and there is an oxidation state.

Oxidation state doesn't actually have anything to do with charge, or electronegativity. It's an arbitrary parameter. Also, even in things like intercalated Li, there is a lot of chemical bonding - more of a metallic kind again.

@Mithoron This is a very clumsy criticism, if you confirm that you have a very good understanding of electronic structure and quantum mechanics, I will expand my answer or respond to your comment.


text:
Which part of a soap molecule is the hydrophilic part?
Is the hydrophilic part of the soap both the cation ($\ce{Na+/K+}$), just the $\ce{COO-}$, or both of those sections together like the picture shown below? Do the cation and the anion dissociate from each other instead, leaving only the negative charge on the soap; therefore, intermolecular forces are due to the negative charge attracted to the partial positive charge on water's hydrogens?

For table salt NaCl, is hydrophilic Na+, Cl- or both?

Both? But only when they're dissociated, I think?

And when they are not? There are no NaCl molecules on NaCl crystals. Similarly there are no soap molecules in classical soap but sodium and carboxylate ions.

It's not hydrophilic? So the soap must be dissociated in the water in order to be hydrophilic?

In that case, the Na+ doesn't really belong to the soap molecule when you explain its hydrophilic head and hydrophobic tail?

All ions are hydrophilic.

Related: https://chemistry.stackexchange.com/q/122562

Do the cation and the anion dissociate from each other instead, leaving only the negative charge on the soap

At low concentration in water, cation and anion would dissociate, and they would be solvated by water. As you remove the water (e.g. in a soap bubble on a dry day), the ions would form pairs again. If your solution contains other cations, the soap (technically not a molecule but an anion) might pair with these other cations.
The details of this are quite complicated, as glancing at the abstract and figures of e.g. this paper illustrates.

Which part of a soap molecule is the hydrophilic part?

The carboxylate is commonly called hydrophilic. Calling the counter-ions $\ce{Na+}$ or $\ce{K+}$ hydrophilic seems strange to me. They are not attached to the soap, although they will be enriched near the negatively charged carboxylate groups e.g. when the soap forms a micelle to maintain local electroneutrality.

Soap is made of equal amounts of cations $\ce{Na+}$ and anions $\ce{RCOO-}$, where R is usually a chain of $n$ carbon atoms generally included in $\ce{C_{n}H_{2n+1}}$. This anionic chain is hydrophobic, but the charged $\ce{COO^-}$ terminal group is hydrophilic. In rather dilute solutions,  the ions are separated from one another by water molecules. But their attraction remains, so that $\ce{Na+}$ cations stay not too far away from the hydrophilic end of the $\ce{RCOO-}$ anions. If the soap concentration increases, colloidal micelles start to be formed : the long hydrophobic carbon chains of the anions are joining one another by Van der Wals attractions. This  makes a small bubble called a micelle whose outer shell contains the hydrophilic $\ce{-COO^-}$ groups.

Soaps belong to a class of electrolytes called colloidal electrolytes. The term "colloidal electrolyte" was defined by McBain, J. W. as "salts in which an ion has
been replaced by a heavily hydrated polyvalent micelle
that carries an equivalent sum-total of electrical
charges and conducts electricity just as well or even
better than the simple ion it replaces."$^\text{1}$

The distinguishing feature of colloidal
electrolytes is the combination of low osmotic or colligative effects with good electrical conductiviy.$^\text{1}$

The properties of colloidal electrolytes are different than general inorganic electrolytes: even below critical micelle concentration (CMC), solutions of colloidal electrolytes show different properties; consider, for example, surface properties.$^\text{2}$
The hydrophobic-hydrophilic nomenclature applies to the nature of soap molecule arrangement in a micelle or at the surface. For hydrocolloids, hydrophilic ends form the surface of the micelle or are submerged in the water at the surface. Nevertheless, being weak electrolytes, soap molecules do not dissociate to an appreciable extent, meaning that the hydrophilic surface of the micelle consists of both carboxylate ends and sodium carboxylate ends. Since polarity itself leads to hydrophilicity, stating that either the cationic or anionic part of the molecule is hydrophilic is incorrect.
The polar end of the molecule is hydrophilic, while the nonpolar end is hydrophobic.
References

Merrill, R. C. (1950). Colloidal Electrolytes. J. Chem. Educ., 27, 6, 312. 10.1021/ed027p312
McBain, J. W. (1940). Colloidal Electrolytes. Nature,
145, 702–703. 10.1038/145702a0

Depending on the pH, weak acids (such as fatty acids) will dissociate fully. Traditional ("anionic") soap is already the deprotonated salt of the fatty acid, and will remain ionized unless you add acid to lower the pH and make the solution acidic.


text:
which is more acidic thiols or carboxylic acid?
According to me as the key atoms are different so when the h+ is released and each of them gets a negative charge as sulphur is a bigger atom compared to oxygen charge is more dispersed , hence the conjugate base formed by thiol is more stable.(I am not considering the equivalent resonance structure of carboxylic acid as i have learnt that only if the atom on which negative charge is present  is same then only can we use effects like mesomeric inductive etc.)

The clue is in the name - carboxylic ACID

Some $pK_a$ values:
$\ce{H3C–SH}$: 10.4
$\ce{H3C–C(O)–OH}$: 4.8
This comparison clearly indicates that the carboxylic acid group imparts stronger acidity than the thiol group.
The determinimg factor is not the greater electronegativity of oxygen versus sulfur. Rather it's the stabilization of the conjugate-base anion by pi-bond delocalization. If we equalize the pi-bond delocalization factor, the $\ce{S–H}$ hydrogen is actually more acidic than the the $\ce{O–H}$ hydrogen due to the weaker hydrogen-sulfur bonding versus hydrogen-oxygen bonding. Thioacetic acid, $\ce{H3C–C(O)–SH}$, with delocalized pi bonding and hydrogen bonded to sulfur, beats acetic acid with a $pK_a$ of $3.4$.


text:
Enthalpy of reaction and work
Hello I am currently struggling with enthalpy of reaction and its use cases.
Let's say I am conducting the following reaction in a piston as a closed system:
$$\ce{C2H4(g) + 2 O2(g) -> 2 CO(g) + 2 H2O(g)}$$
This reaction has an enthalpy of reaction of $\pu{-1560 kJ/mol}$. As far as I know, this refers to the heat released under constant pressure. So for example, if I let my reaction run in a piston with outside pressure of 1 bar, the reaction will heat the gases up by $\pu{-1560 kJ/mol}$, when the system is in pressure equilibrium with the outside pressure again (after heating).
But what about the work the gas does to its surroundings (in my case to the atmosphere)? Do I need to subtract the work done by the gas from the enthalpy of reaction to get the real value? Because for sure, the gas will expand against the outside pressure and lose some of the heat of reaction?

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
// Optionally, here are formatting guides for texts and formulas/equations/expressions.

Ok will look into it

If you treat the mixture as a mixture of ideal gases, then the enthalpy is independent of pressure.  But only at constant pressure is the heat added equal to the enthalpy change.

You appear to be conflating two distinct concepts: internal energy and enthalpy.
From my understanding, the term 'enthalpy' is applicable exclusively under constant pressure conditions. As such, its relevance extends only to reactions occurring at constant pressure. In cases like constant volume, enthalpy becomes a mere mathematical value devoid of any practical significance.
At constant pressure, enthalpy signifies the net heat energy exchanged between a system and its surroundings. Notably, this measurement excludes any involvement of pressure-volume work.
On the other hand, internal energy accounts for both heat and expansion work.

To initiate, let's compute the expansion work carried out by the system. For this calculation, let's assume a constant temperature of 298K and a system perpetually in equilibrium with its surroundings, implying the reaction progresses in infinitesimal steps. This assumption enables us to replace the external pressure of the surroundings with the internal pressure of the system.
Expansion work done by gas Wexp = -∫Pext.dv = -∫Pgas.dv = -(∆ng)RT = (4-3)×8.314×298 ≈ -2477j ≈ -2.48kj
(∆ng) : change in no of moles of gaseous state in reaction
It's essential to note that the energy of approximately 2.48 kJ relinquished by the system due to expansion is unrelated to enthalpy changes.

But what about the work the gas does to its surroundings (in my case to the atmosphere)? Do I need to substract the work done by the gas from the enthalpy of reaction to get the real value? Because for sure, the gas will expand against the outside pressure and lose some of the heat of reaction?

Yes you must add or subtract, depending upon process, expansion  work to get change in internal energy not in enthalpy.
Think of it this way: The system released more heat energy than was needed to break the bonds of the reactants. This resulted in a net release of 1560 kJ of heat energy. Additionally, the system lost 2.48 kJ of energy in the form of work. Therefore, the change in internal energy of the system is the sum of the heat released and the work done, which equals -1562.48 kJ.
I hope I have addressed your queries satisfactorily.

Enthalpy as a state function is applicable anytime. But only in isobaric cases it is really useful as it's change is equal to sum of exchanged heat and non expansion work. Expansion work is implicitly contained in Enthalpy definition here.

@Poutnik why are you not active in chat discussion https://chat.stackexchange.com/rooms/147631/discussion-between-poutnik-and-bharat-prajapat

Partly because I was not notified as you have not used explicit @username syntax in chat, partly I am not going to get involved in extensive interactive education sessions. That would be resource demanding. There are more suited free study resources available to you.

For isobaric systems, enthalpy change differs to internal energy change by the factor of the volume work done by the system,
This volume change related work mutually transforms with the extra heat provided to/obtained from isobaric system, compared to isochoric conditions.
$\Delta H = \Delta U + p \Delta V = Q_V - W_{pV} = Q_p \gt Q_V$
with the sign convention of positive work done on a system.

So pV work is already included I guess?

So if I want to get internal energy change (and therefore the temperature change) I can just subtract the work ?

Questions should not be answered if the other side could get them on their own easily and quickly, as this way makes answers much more valuable. As the understanding records itself more permanently this way.

@Mäßige "So if I want to get internal energy change (and therefore the temperature change) ? " Note you can only find temperature change by just knowing internal energy change if there is only ideal gas which is not undergoing  any reaction , by the expression ∆U = nCv∆T.  Even for a real gas , not undergoing any reaction, this expression doesn't hold. Similarly for a reaction you can't calculate temperature change looking only at internal energy change. In my answer I assumed reaction occuring at constant temperature at 298 , but there is a change in internal energy by -1562.48 kj of energy.

I think I got it; at constant pressure, the enthalpy of reaction already includes the internal energy change + work done by the gas. To only get the internal energy change, one needs to add the work done to the (negative) enthalpy of reaction. @Poutnik, However, in one comment you stated that “enthalpy is change is equal to sum of exchanged heat and non expansion work.” Why non expansion work? dH=dU+pdV

@Mäßige ΔH=ΔU + pΔV = Q_p + W_nonV = Q_v + pΔV  + W_nonV.  // The volume work is not an independent parameter, but is fully determined by Q_p.

Now I get it. We usually disregard other types of work but I get your point

@Mäßige Consider the classical Joule's experiment about work-heat equivalence.

Yes but isn’t the Joule experiment only for real gases (because they have interactions), therefore shouldn’t have an influence when we assume ideal gases?

@Mäßige No, it was transformation of mechanical work to heat, using a propeller on a crankshaft, submersed in water.

Yeah sorry I was confusing Joule experiment with Joule-Thomson experiment. Link is dead for me (maybe because I am from Europe). What were you getting at?

I am from Europe too, but it works for me. It should be possible found other link, it is quite well known.

Coming back to this question, I still don't know what Joule's expiment has to do with W_nonV. Why are we incorporating it

See also wikipedia  Mechanical_equivalent_of_heat


text:
Why does brandy have a higher methanol-to-ethanol ratio than beer, despite the distillation?
From my googling, beer tends to contain around 16mg/L of methanol. While for brandy, typical methanol content is around 1000mg/L (give or take a factor of 3, depending) The distillation that brandy goes through attempts to concentrate ethanol and remove methanol, yet we end up with an 8x worse methanol-to-ethanol ratio.

If they did distillation to decrease the ratio, the brandy would have decreased ratio, compared to the distilled fermented solution. Note that different sources have different ratios after fermentation.

There is a lot of water in beer.

Different raw materials yield different amounts of methanol upon fermentation, see e.g. https://pubs.acs.org/doi/10.1021/acsfoodscitech.1c00025

@Karsten Very interesting! It's the pectin...

Brandy is made from wine or other fermented fruits. You can also distill beer (German chat), which usually has less methanol unless a home-brewer messes up.

You state, "The distillation that brandy goes through attempts to concentrate ethanol and remove methanol." AFAIK, the intent of distillation is to concentrate ethanol only.
However, the statistics you quote are misleading for a few reasons.

Brandy is not generally distilled beer, but rather from wine. Your comparing apples and oranges... well, beer and brandy. Would it not be better to compare alcohols in wines and brandies? Also, pick the same type of wine and brandy, e.g., plum wine vs. slivovitz.

Water is removed by distillation, leaving a higher percentage of ethanol, so the ratios you quote are misleading: comparing beer, about 4%-6% ethanol with brandy, about 40%-60% ethanol, one should measure concentration based on the amount of alcohol in the beverage, rather than on the total volume. That would make the beer about 0.4 g/L methanol:ethanol, and the brandy about 2 g/L, based on the statistics you gave for the amount of methanol. What is the source of those statistics?

both alcohols are lowerboiling than water distillation at a low 50-60 proof is designed to add water to the distillate not remove it. Distillation removes alcohol from the wine together with some flavoring chemicals probably by steam distillation. Since this necessitates an inefficient distillation process discarding the first fractions would probably remove more flavorings than methanol.


text:
Calculate the pH of a buffered solution after adding strong acid
Calculate the pH of a solution formed by the addition of $0.0020$ moles of $\ce{HCl}$ to $\pu{100.0 mL}$ of $\pu{0.200 M}$ $\ce{CH3NH2}$/ $\ce{0.100 M CH3NH3Cl}$.
$K_\mathrm{b}(\ce{CH3NH2})=4.4\times 10^{-4}$
Any hints will be much appreciated. :)

We will not help you solve this problem if you don't send us what sort of trials and calculations  you have done previously.

I know this is a buffer solution that resists a change in pH. I can calculate the pH of the buffer by itself, but how do I proceed after the HCl is added? I just want a nudge in the right direction.

there is a general method outlined here https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566#95566

Adding 0.2 millimole HCl to the buffer solution is consuming 0.2 millimole CH3NH2 and creating 0.2 millimole CH3NH3Cl. Go on !

@Maurice Thanks! Can you take a quick look at my solution?

It should be widely known that strong acids react with weak bases, forming respective salts as weak acids themselves, what changes conjugate acid/base pair ratio of buffers.

Okay, here goes nothing.
$$\ce{CH3NH3+ <=> H+ + CH3NH2}$$
$$K_a(\ce{CH3NH3+})=\frac{[\ce{H+}][\ce{CH3NH2}]}{[\ce{CH3NH3+}]}\tag{1}$$
$\displaystyle K_a(\ce{CH3NH3+})=\frac{10^{-14}}{K_b(\ce{CH3NH2})}=\frac{10^{-14}}{4.4\times 10^{-4}}=2.27\times 10^{-11}$
Initial moles of $\ce{CH3NH3+}=0.100\times 0.100=0.010$
Initial moles of $\ce{CH3NH2}=0.100\times 0.200=0.020$
$$\ce{H+ + CH3NH2 -> CH3NH3+}$$
$0.0020$ moles of $\ce{H+}$ react with $0.0020$ moles of $\ce{CH3NH2}$ to form $0.0020$ moles of $\ce{CH3NH3+}$.
Final moles of $\ce{CH3NH3+}=0.010+0.0020=0.012$
Final moles of $\ce{CH3NH2}=0.020-0.0020=0.018$
$V=100.0\text{ mL}=0.1000\text{ L}$
$\displaystyle[\ce{CH3NH3+}]_f=\frac{0.012}{0.1000}=0.120$
$\displaystyle[\ce{CH3NH2}]_f=\frac{0.018}{0.1000}=0.180$
Substituting these values in equation $(1)$, we get
$\displaystyle[\ce{H+}]=\frac{2.27\times 10^{-11}\times[\ce{CH3NH3+}]}{[\ce{CH3NH2}]}=\frac{2.27\times 10^{-11}\times 0.120}{0.180}=1.51\times 10^{-11}$
pH$=-\log[\ce{H+}]=10.82$
Does this look good?

What was the information you needed when you posted the question? Which comment did it? What would happen if you added 0.03 moles of strong acid instead?

@Karsten. Why do you ask this question ? Why transform the initial question ? What does it help ?

@Maurice I was surprised by the self-answer and was wondering what the missing step was initially.  My 3rd question in the comment goes after limits of the recipe shown here.

@Karsten. OK. I see. Thank you for the information. I was also surprised by the self-answer.


text:
Alternative synthesis for LK-99 proposed room-temperature superconductor
This is probably not the best place to post this, but I'm posting in case it might reach the right people who could give this a try.
The current synthesis for the proposed LK-99 superconductor is a multi-step process that goes through a sulfate compound, while the end product is a phosphate compound [1, 2]:
$\ce{PbO + Pb(SO4) -> Pb2(SO4)O}$
$\ce{3 Cu + P -> Cu3P}$
$\ce{Pb2(SO4)O + Cu3P -> Pb_{10-x}Cu_x(PO4)6O + S(g)}$, where $(0.9 < x < 1.1)$
It may produce several byproducts when the equation is balanced:
$\ce{5 Pb2SO4O + 6 Cu3P → Pb9Cu(PO4)6O + 5 Cu2S + Pb + 7 Cu}$

Since the current synthesis is already based on $\ce{PbO}$, I propose the following alternative route:
$\ce{9 PbO + CuO + 6 H3PO4 -> Pb9Cu(PO4)6O + 9 H2O(g)}$
Single pot, anywhere above $\ce{100C}$, to let the water evaporate. Since there were some suggestions that it needs some extra doping with oxygen or sulfur [3], then probably do this is in an open flask, or mix in a bit of sulfur in there.
This is a simple acid-base reaction, which might even occur in the solid phase. Or maybe it can be done in $\ce{H3PO4(aq)}$, letting the product precipitate. Then dry it in a vacuum furnace.
What are some possible complications with this reaction?

https://en.wikipedia.org/wiki/LK-99#Synthesis
https://arxiv.org/abs/2307.12008
https://arxiv.org/abs/2308.00676

Not working at all would be pretty big complication ;) See, you can't just add reagents and hope it's gonna work like you want. Also this whole "LK-99" thing is highly dubious - there's likely no superconductor  like that to be made, at least not some magically room temp. stuff.

There's plenty of reason to expect that it won't be superconducting, or won't even have the right crystallographic structure. But, is there a reason to expect that it won't give the expected compound (chemical formula)? Or that there won't be a reaction at all? This is a simple acid-base ion exchange. But I was asking more towards e.g. it reacting too violently (or not at all, as you suggest), or giving off toxic fumes, etc.

Well, if acid was conc. enough, it should dissolve it, but products would be different. I don't see an option for mixed oxide, just some mixture of (hydrogen)phosphates.

The target material is an apatite, so it is a phosphate material. The source oxides are both amphoteric AFAIK, so should dissolve in mineral acids. But whether or not phosphoric acid is strong enough, I don't know. A less direct route is to dissolve the oxides in HCl and react with sodium phosphate.

(Though lead chloride is only mildly soluble.)

My original thoughts were to react with anhydrous $\ce{H3PO4}$. Either in the solid phase and dry the product later, or if done above $\ce{100C}$ to let the water evaporate, then the phosphoric acid would melt, and it becomes a mixed phase reaction. Mixed oxide powder with liquid phosphoric acid.


text:
Synthesis of Cyclohexenon from Anisole
(I'll write organic compounds as SMILES notation)
I am learning about carbonyl compounds these days, and I have a problem.
One way of cyclohex-2-ene-1-one synthesis is using anisole following the reaction.
\begin{aligned}
\mathtt{c1ccccc1OC>Li.NH_3}&\mathtt{>C1C=CCC=C1OC}\\
\mathtt{>H^+}&\mathtt{>C1C=CCC=C1O}\\
\mathtt{>H^+}&\mathtt{>C1C=CCCC1=O}&(\mathtt{cyclohexｰ3ｰeneｰ1ｰone})\\
\mathtt{>H^+}&\mathtt{>C1=CCCCC1=O}&(\mathtt{cyclohexｰ2ｰeneｰ1ｰone})
\end{aligned}
But I cannot understand the last reaction. What kind of reaction is taking place here, it would be helpful if you could indicate the reaction mechanism.
(This is the structure formula of the above reaction)

It's two tautomerisations, like penultimate reaction going backward and again forward.

https://chemistry.stackexchange.com/questions/164480/acid-hydrolysis-of-birch-reduced-anisole
Mechanism of this rearrangement is explained here


text:
Consequences of Nitrate fertilisers
Consider the following multiple choice.

What is the environmental consequences of the uncontrolled use of nitrate fertilisers?

The choices are.
A. Acid rain
B. Low oxygen levels in streams
C. Ozone depletion
D. Greenhouse effect
The answer is B which is obvious why this must be an effect. However, I have read that using nitrate fertilisers also release nitrous oxide which is a greenhouse gas. So shouldn't D also be a consequence?

To me the question is poorly worded, It is not always the case that nitrates cause excessive algae growth, and subsequent death, removing oxygen, and yes, nitrates in air (e.g., wind-blown soil) could increase acidity of rain, and some nitrous oxide does come from fertilizer: https://news.stanford.edu/2020/10/07/laughing-gas-growing-climate-problem/ . However, the author of that question has a definite answer in mind. sigh If you can speak with the author, you might mention that citation. or not

The answer should be "All of these".
Increased use of nitrogenous fertilizers in agriculture can significantly alter the global N-cycle because they release nitrogenous gases especially $\ce{NO_x}$. The emission of nitrous oxide ($\ce{N2O}$) contributes to the global greenhouse gas accumulation and that is causing stratospheric ozone depletion.
Nitrogenous gas can react with moisture and water vapor forming nitric acid which is a key component of acid rain. Also, as you said, these fertilizers contains nutrients which accelerate the growth of plants and crops. However, if they flow in rivers/ponds, it can lead to uncontrollable growth of algal blooms and aquatic plants leading to oxygen content in water. The process being known as "Eutrophication".
There are many papers online which has discussed on this topic but for starters, you can check below articles which has explained in simple language:

https://www.ndsu.edu/agriculture/extension/publications/environmental-implications-excess-fertilizer-and-manure-water-quality
https://www.epa.gov/sciencematters/understanding-impacts-synthetic-nitrogen-air-and-water-quality-using-integrated


text:
Concentration units for kinetic mechanism of combustion
I am reading Tamás Turányi & Alison S. Tomlin's Analysis of Kinetic Reaction Mechanisms (2015). On page 10-11, the authors state that for a mechanism with $N_R$ reactions involving $N_S$ species, where $\nu_{ij}^L$ denotes the stoichiometric coefficients on the left-hand side,

[...] the rates of elementary reactions can be calculated by assuming the rule of mass action kinetics. According to the chemical kinetic law of mass action
$$
r_i = k_i \prod_j^{N_S} Y_j^{\nu_{ij}^L}, \tag{2.5}
$$
where $r_i$ and $k_i$ are the rate and the rate coefficient, respectively, of reaction step $i$, and $Y_j$ is the molar concentration of species $j$.
[...]
The kinetic system of ordinary differential equations (ODEs) defines the relationship between the production rates of the species and rates of the reaction steps $r_i$:
\begin{align}\label{eq26}
\frac{\mathrm{d} Y_j}{\mathrm{d}t} &= \sum_i^{N_R} \nu_{ij}r_i;& j&=1,2,\dots,N_S. \tag{2.6}
\end{align}
[...]
An analogous equation to Eq. (2.6) can be written when other concentration
units are used, e.g. mass fractions or mole fractions [see, e.g. Warnatz et al. (2006)],
but Eq. (2.5) is applicable only when the “amount of matter divided by volume”
concentration units are used. The amount of matter can be defined, e.g., in moles or
molecules, whilst volume is usually defined in $\pu{dm3}$ or $\pu{cm3}$ units.

Can anyone help me understand why the two bolded sentences are true? I looked in the book of Warnatz et al. (2006) and did not find the answer.

Hello! Are you sure of how you've written Eq. (2.5)? Maybe $\nu_{ij}^L$ is the exponent, i.e., $Y_j^{\nu_{ij}^L}$?

In addition, maybe Eq. (2.6) is a sum and not a product?
$$ \frac{\mathrm{d}Y_j}{\mathrm{d}t} = \sum_{i}^{N_\mathrm{R}} \nu_{ij} r_i $$

[AUTHOR] The kinetic system of ordinary differential equations (ODEs) defines the relationship between the production rates of the species and rates of the reaction steps $r_i$:
$$ \frac{\mathrm{d}Y_j}{\mathrm{d}t} = \sum_i^{N_R} \nu_{ij}r_i \quad
   j=1,2,\dots,N_\mathrm{S} \tag{2.6} $$

This equation can be obtained by performing a mole balance of chemical species $j$ on an element of volume.
The rate of change of species $j$ takes place due to $N_\mathrm{R}$ reactions. Let me call this magnitude just $r_j$
$$ r_j := \sum_{i = 1}^{N_\mathrm{_R}} \nu_{i,j}r_{i} \tag{1} $$
The mole balance in this case is simplified, because in our reactor there are no species flowing into or out of it, so the first two brackets are zero
\begin{align}
  \begin{bmatrix}
    \text{Flow} \\
    \text{of species $j$} \\
    \text{INTO the} \\
    \text{system} \\
    \text{[mol/s]}
  \end{bmatrix} +
  \begin{bmatrix}
    \text{Flow} \\
    \text{of species $j$ } \\
    \text{OUT OF the} \\
    \text{system} \\
    \text{[mol/s]}
  \end{bmatrix} +
  \begin{bmatrix}
    \text{Generation} \\
    \text{of species $j$} \\
    \text{within the} \\
    \text{system} \\
    \text{[mol/s]}
  \end{bmatrix} &=
  \begin{bmatrix}
    \text{Accumulation} \\
    \text{of species $j$} \\
    \text{within the} \\
    \text{system} \\
    \text{[mol/s]}
  \end{bmatrix} \\
  \begin{bmatrix}
    \text{Generation} \\
    \text{of species $j$} \\
    \text{within the} \\
    \text{system} \\
    \text{[mol/s]}
  \end{bmatrix} &=
  \begin{bmatrix}
    \text{Accumulation} \\
    \text{of species $j$} \\
    \text{within the} \\
    \text{system} \\
    \text{[mol/s]}
  \end{bmatrix} \\
  G_j &= \frac{\mathrm{d}N_j}{\mathrm{d}t} \tag{2}
\end{align}
where $G_j$ is the rate of generation and $N_j$ is the amount of species $j$. The rate of generation is a function of the space coordinates in the volume $V$. So, imagine that we break up this volume in a huge number of elements of volume $\Delta V_1$, $\Delta V_2$, and so on. These elements of volume are so small, that we assume that the rate is constant in those tiny volumes. The picture is the following:

The rate of generation $G_j$ can be calculated by summing all the rate of reactions evaluated in their respected tiny volumes, multiplied by this same tiny volume. If we take the limit and use the definition of a Riemannian integral
\begin{align}
  G_j &= \lim_{M \to \infty} \sum_{k = 1}^M r_j(\Delta V_k)\Delta V_k \\
  G_j &= \int_{0}^V r_j(V) \, \mathrm{d}V \tag{3}
\end{align}
combining Eqs. (2) and (3) and relating the amount with the concentration via $N_j = C_jV$
$$ \frac{\mathrm{d}(C_jV)}{\mathrm{d}t} = \int_{0}^V r_j(V) \, \mathrm{d}V \tag{4} $$
Eq. (4) is super general. Now comes the two most important assumptions:

In the case that the reactor is perfectly mixed, the rate of reaction may be taken out of the integral because it is not a function of position. This mathematical step is done in real life when you use a magnetic stirrer in the laboratory to ensure a homogeneous behavior of the variables in the beaker, e.g., during a titration. There is no mistery.
The volume $V$ is constant. This is an excellent approximation in liquid phase reactions, because it is generally assumed incompressible. Strictly, the molar volume of the mixture is going to change in time, because species are being formed and consumed during the reaction (pure $A$ will not have the same molar volume than pure $B$ in the academic reaction $A \to B$).

Thus, Eq. (4) becomes
\begin{align}
  \require{cancel}
  \frac{\mathrm{d}(C_jV)}{\mathrm{d}t} &= \int_{0}^V r_j \, \mathrm{d}V \\
  V\frac{\mathrm{d}C_j}{\mathrm{d}t} &= r_j \int_{0}^V \, \mathrm{d}V \\
  \cancel{V}\frac{\mathrm{d}C_j}{\mathrm{d}t} &= r_j \cancel{V} \\
  \frac{\mathrm{d}C_j}{\mathrm{d}t} &= r_j \qquad \text{(Use Eq. (1))} \rightarrow
  \boxed{\frac{\mathrm{d}C_j}{\mathrm{d}t} =
  \sum_{i = 1}^{N_\mathrm{_R}} \nu_{i,j}r_{i}} \tag{5}
\end{align}
Eq. (5) is the one that is presented in the book.
1. Eq. (5) in terms of the mass fraction The relation between the amount and the mass fraction of species $j$ is
\begin{align}
  N_j           &= \frac{m_j}{M_j} \\
  N_j           &= \frac{\omega_j m}{M_j} \\
  \frac{\mathrm{d}N_j}{\mathrm{d}t} &= \left(\frac{m}{M_j}\right)
  \frac{\mathrm{d}\omega_j}{\mathrm{d}t} \tag{6}
\end{align}
since the mass $m$ and the molar mass $M_j$ are constant over time. Combining Eq. (4) and (6), under the assumption that $r_j$ is not a function of position, yields
\begin{align}
  \frac{\mathrm{d}N_j}{\mathrm{d}t} &= r_jV \\
  \left(\frac{m}{M_j}\right)\frac{\mathrm{d}\omega_j}{\mathrm{d}t} &=
  r_jV \\
  \left(\frac{m}{M_j}\right)\frac{\mathrm{d}\omega_j}{\mathrm{d}t} &=
  \sum_{i = 1}^{N_\mathrm{R}} \nu_{i,j}r_{i}V \rightarrow
  \boxed{\frac{\mathrm{d}\omega_j}{\mathrm{d}t} =
  \left(\frac{M_jV}{m}\right)\sum_{i = 1}^{N_\mathrm{R}} \nu_{i,j}r_{i}} \tag{7}
\end{align}
The mass $m$, which will not change because we are dealing with chemical reactions, can be calculated by the initial conditions of the differential equations. Knowing the concentrations $C_1, C_2, ..., C_\mathrm{S}$ and the volume $V$, we can obtain the amounts $N_1, N_2, ..., N_\mathrm{S}$ and then with the molar mass of each species get the individual masses.
2. Eq. (5) in terms of the mole fraction With Eq. (4) under the assumption that $r_j$ is not a function of position, and that $N_j = x_jN$
\begin{align}
  \frac{\mathrm{d}N_j}{\mathrm{d}t} &= r_jV \tag{8} \\
  \frac{\mathrm{d}(x_jN)}{\mathrm{d}t} &= r_jV \\
  x_j\frac{\mathrm{d}N}{\mathrm{d}t} + N\frac{\mathrm{d}x_j}{\mathrm{d}t}
  &= r_jV \\
  \frac{\mathrm{d}x_j}{\mathrm{d}t} &=
  \frac{1}{N}\left(r_jV - x_j\frac{\mathrm{d}N}{\mathrm{d}t}\right) \tag{9}
\end{align}
to obtain $\mathrm{d}N/\mathrm{d}t$ we use its definition, the total amount is equal to the sum of all the amounts (the presence of inert species won't change the final result, because the time derivative will kill them)
\begin{align}
  N &= \sum_{j = 1}^{N_\mathrm{S}} N_j \\
  \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{j = 1}^{N_\mathrm{S}}
  \frac{\mathrm{d}N_j}{\mathrm{d}t} \qquad \text{(Use Eq. (8))} \\
  \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{j = 1}^{N_\mathrm{S}}r_jV \\
  \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{j = 1}^{N_\mathrm{S}}
  \sum_{i = 1}^{N_\mathrm{R}}\nu_{i,j}r_iV \\
  \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{i = 1}^{N_\mathrm{R}}
  \sum_{j = 1}^{N_\mathrm{S}} \nu_{i,j}r_iV \\
  \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{i = 1}^{N_\mathrm{R}} r_iV
  \sum_{j = 1}^{N_\mathrm{S}} \nu_{i,j} \\
  \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{i = 1}^{N_\mathrm{R}}\nu_ir_iV
  \tag{10} \\
\end{align}
where we defined $\nu_i := \sum_{j = 1}^{N_\mathrm{S}} \nu_{i,j}$. This number is the sum of the stoichiometric coefficients of reaction $i$. For example, for the reaction scheme
\begin{align}
  \ce{A + B  &->  C + 2D} \qquad \nu_1 = (1 + 2) - (1 + 1) = 1 \\
  \ce{B + 2C &-> 4C + 5E} \qquad \nu_2 = (4 + 5) - (1 + 2) = 6
\end{align}
or you could use matrix form if you like. Note that in the case of a zero change of amounts, then $\nu_i = 0$, as expected.
Combining Eqs. (9) and (10) yields the final result
\begin{align}
  \frac{\mathrm{d}x_j}{\mathrm{d}t} &=
  \frac{1}{N}\left(\sum_{i = 1}^{N_\mathrm{R}}\nu_{i,j}r_iV -
  x_j\sum_{i = 1}^{N_\mathrm{R}}\nu_ir_iV\right) \\
  \frac{\mathrm{d}x_j}{\mathrm{d}t} &=
  \frac{V}{N}\left(\sum_{i = 1}^{N_\mathrm{R}}\nu_{i,j}r_i -
  x_j\sum_{i = 1}^{N_\mathrm{R}}\nu_ir_i\right)
  \rightarrow
  \boxed{\frac{\mathrm{d}x_j}{\mathrm{d}t} =
  \frac{V}{N}\sum_{i = 1}^{N_\mathrm{R}}(\nu_{i,j} - x_j\nu_i)r_i}
   \tag{9}
\end{align}

So in your explanation could C_j be the mole fraction or the mass fraction of species j? @MetalStorm

@math_lover0105 There I added the results when you want the ODE in terms of the mass fraction or the mole fraction. Tell me if it is clear afterwards.


text:
How does bond strength affect catenation?
I am wondering why C with single bond strength 347 Kj/mol is able to form much longer linear bond chain, while S with single bond strength 266 Kj/mol is able to do so with difficulty, although not even comparable.
It might appear, and even obvious, that the longer the chain, the greater is the chance to break any segment of chain, and thus unstable.
So my question is, does the bond energy in any other way affect the catenation property.
For example, consider a chain of 3 carbons. Does the bond between C1 and C2 affect the bond between C2 and C3. Molecular Orbital Theory might say, yes. But I am not able to understand how?

There is no stable linear chain of carbon atoms with single bonds, you should consider the sp hybridization.

Difficulty? You melt sulfur, you heat it a little more, maybe to $200\ ^\circ C$, pour it in the cold water, and here you have them, the long chains of S.


text:
What is the difference between Charge transfer and Electron transfer?
This problem was given to me yesterday that What is the difference between Charge transfer and Electron transfer? I was taking it thesame though the professor have given me some hint that charge transfer is of two types which are Metal to ligand charge transfer (MLCT) and Ligand to Metal Charge transfer (LMCT) and he defined electron transfer as the movement of electrons from one molecular species (donor) to another molecular species (acceptor),I need the actual  difference between the two. please help I have exam tomorrow

In a polar molecule of hydrogen chloride, there is a charge transfer from hydrogen to chlorine atom, but not an electron transfer. There is an electron transfer between them during hydration and dissociation.
The charge transfer is a fractional transfer of a charge, based on the shift of the electron density, caused by electronegativity difference or by orbital configuration rearrangement.
The electron transfer is a transfer of an electron as a charged particle, making the left molecular entity more positive and making the entity the electron lands on more negative.


text:
Why would my pH meter have changed its voltage setpoints?
I'm using a Thermo Scientific Orion Dual Star pH/ISE Dual Channel Benchtop Meter.
The manual says that a $\mathrm{pH}~7$ buffer should read about $\pu{0 mV},$ and a $\mathrm{pH}~4$ buffer should read about $\pu{177 mV}.$
We use a two-point calibration with the $\mathrm{pH}~4.00$ and $\mathrm{pH}~7.02$ buffers at $\pu{20 °C}.$
I'm getting $\pu{-41 mV}$ for $\mathrm{pH}~7$ and $\pu{133 mV}$ for $\mathrm{pH}~4,$ i.e. my $\pu{mV}$ readings seem to be offset by about $\pu{-40 mV}.$
The slope is approx. $99\,\%,$ which is fine.
Our internal methods require a voltage range of $\pu{\pm30 mV}$ for the $\mathrm{pH}~7$ buffer as validity criteria.
When I calibrate as is, I can test a $\mathrm{pH}~6.00$ buffer I use for verification and get $\mathrm{pH}~6.00$ consistently.
The $\mathrm{pH}~6.00$ buffer is not part of the calibration, it's an independent test to see if the calibration is valid.
The buffers are purchased from a certified supplier. The CoA says they are made of $\ce{KH2PO4},$ $\ce{Na2HPO4},$ $\ce{NaHCO3},$ $\ce{Na2CO3}$ in DI $\ce{H2O}.$
Why would this happen? Can I recalibrate the voltage or enter an offset somehow?

I am sure you will have another electrode from Thermo. Replace the electrode and see if the values change. Also it does not make sense that you are so off and the slope is 99% of the theoretical?

Electronic components and electrodes are aging and their parameters drifting. Or the pH electrode is off because of its bad conditions.
As pH meters are usually also mV meters, check its voltage reading by independent electronic device.
Check the pH electrode by using other one, if both their reading on the same device match.
That will help to determine the cause.

As temporary workaround, one can technically always use fallback solution of measuring the voltage and recalculation to pH. Sure, if the standard operation procedure does not allow to continue, it will not help.

The voltage measured is the potential difference between the glass electrode and the reference electrode. This depends on the construction of both and is set to zero by the calibration control of the meter. My procedure in using a meter was to record mv readings at pH 7 and 4 daily, monitoring the trend. If there was a change, determine the reason. Generally, there was a slow progression in voltages over months with little change in slope. This is with double junction flowing reference electrodes. An offset of 41 mv seems large and suggests a reference electrode effect, Check it out and either match the electrodes or simply monitor the voltages daily to see if everything is constant. Does the meter read to 0.1 mv?
The criterium for 30 mv: what is that based on? I have found that the voltage is constant to fractions of a mv with a slow drift over time if an electrode system is working properly.  A side note I have found Fisher Scientific color-coded pH buffers to be extremely precise bottle to bottle. Different brands were generally agreeing within fractions of a mv. Also a better check is a pH 10 buffer that will stress the system a bit more. [Do not set the slope at 10 though]


text:
Kinetic energy change in reaction
If we consider a reaction occurring at constant pressure and temperature , we know that enthalpy represent the net heat released or absorbed by the system and internal energy accounts for both heat and expansion work. ( Further reading )
suppose we perform a reaction H2(g) + O2(g) —> H2O(g) at constant temperature 25°C and pressure 1 atm
• Bond energy of H2 ≈ 436 kj/mol
• Bond energy of O2 ≈ 498 kj/mol
• Bond energy of O-H bond in H2O(g) ≈ 464 kj /mol
Calculation
• ∆H = Energy required to break the bonds of H2 and O2 - Energy released in formation of water molecule -
= (436 + 498/2) - (2 × 464)  = -243 kj
• Expansion work = Wexp = ∆(ng)RT ≈ +1.2 kj/mol
• ∆U = ∆H - ∆PV = ∆H - (∆ng)RT = -241.8 kj
Data is taken from : Question out of syllabus
As we can see here enthalpy is evaluated just by calculating difference between bond energies and kinetic energy of reactants and products are not included i.e. enthalpy is not related to kinetic energy.
Now I want to know that system absorbed some expansion work energy but in what form does it store it within itself i.e. kinetic energy or potential energy.
In the book Chemical Physics Electrons and Excitations By Sven Larsson · 2012 page 146 Mr Sven writes that we the term pv ( which accounts for expansion work) takes care of kinetic energy i.e. expansion work is equal to change in kinetic energy during reaction.

But problem is that the expression  of kinetic energy and expansion work although looks like similar but have a difference of factor 3/2 as kinetic energy change at constant temperature ( According to kinetic gas theory ) is equal to 3/2(∆ng)RT but expansion work is equal to (∆ng)RT.
Could someone prove that change in kinetic energy in a reaction ( occurring at constant pressure and temperature) is equal to Expansion work ?

Why do not you use enthalpies of formation instead of bond energies?

Standard enthalpy of formation of H2 = 0.         Standard enthalpy of formation of O2 = 0.          Standand enthalpy of formation of H2O = −241.818

3/2nRT is ideal gas thermal energy(=translational kinetic energy in this case) only for monoatomic gases or for very low T of polyatomic ones. For low enough T it is 5/2nRT for biatomic  and linear triatomic ones, 6/2nRT for other triatomic ones, due 2 or 3 axis of rotation. For higher T, there are being progressively added energies of vibration modes, up to nRT each.

But reaction described here is at low temperature 25°C

that is enough for rotation,not for vibration. H2 and O2 5/2RT, H2O 6/2RT. (All not exactly,but approximately, you can search for tables or charts of molar heat capacities = f(T) ) // See also https://en.wikipedia.org/wiki/Molar_heat_capacity // Expansion work is equal to the difference between system heat capacities at constant pressure and constant volume.

Select wisely what you ask about, as the last resort choice. Frequent asking is demanding on resources of others and some users could get tired and annoyed by it. Most such things have been already described by many ways on many places, few keywords and mouse clicks away. Bring rather just particular details from such resources you do not understand even after failed effort.

Sorry, I had no intention to bother respected  community members but will remember your suggestion for future.

While doing some more research and collecting data , I found ( Quantum Bits and Quantum Secrets page 128 )  that average kinetic  energy of a particle at room temperature is equal to 4 × 10 ^ (-21) joule so now if we further calculate energy of one of gas at room temperature that will be equal to 4 × 6.023 × 100 j ≈ 2.4 kj.

Net kinetic energy change of reaction ≈ Change in no of moles × energy of 1 mole ≈ (1/2) × 2.4 ≈ 1.2 kj
To my wonder this is exactly same as volume work done by surrounding to system.
Comment if there is anything wrong!

Taking 1mol of He, being heated isobarically by 1 K,  the provided heat is 5/2R J = 3/2 R J to increase He thermal energy + R J of heat is converted to volume work.

@Poutnik last question, according to kinetic theory of gas kinetic energy of a molecule is 3/2kt whereas as I described above that kinetic energy of a atom at room temperature is kt , which is correct ?

Neither is correct. // The mean translational kinetic energy of gaseous molecules or atoms is 3/2kT. But in case of molecules, it is not the only form of their kinetic energy. They have another 2/2kT (linear ones) or 3/2kT of rotational energy (except of very low T where
they progressively stop rotate due rotation quantization) and at higher T also progressively (due vibration quantization) 2/2kT for each vibration. At very high T comes also excitation and ionization. Generally, in classical thermodynamics, each degree of freedom is covered by 1/2kT of energy contribution.


text:
Do covalent networks experience dipole-dipole forces?
After my research, I have learnt that dipole-dipole forces only exist between polar covalent molecules.
I am aware that covalent networks are not molecules, thus they obviously cannot be classified as a polar covalent molecule. However, covalent networks such as SiC are overall rendered polar, and I have read that SiC does experience dipole-dipole forces between its compounds.
I don’t understand how SiC can experience dipole-dipole forces as it is not made up of molecules - or am I mistaken here? Are covalent networks made up of molecules?
I would be grateful for any help in clearing up my confusion on the overarching question of: Do covalent networks such as SiC experience dipole-dipole forces, and if so, does this force exist between silicon carbide molecules?

I object to this very question. Silicon carbide molecules are not a thing, except maybe in the gas phase which itself is not much of a thing, and the forces in a covalent network do not matter, because the network is going to stay regardless of those.

Dipole - dipole forces occur any time where there are 2 or more pairs of dislocated but connected charges. Be aware they are nothing else but ordinary coulombic forces, just reorganized in their evaluation, as their connection limits the degrees of their freedom.
Therefore, if there is some covalent chain
$$\ce{\bond{-}A^{(+)}-B^{(-)}-A^{(+)}-B^{(-)}\bond{-}}$$
then the sequence of A and B atoms with partial alternating charges can be formally evaluated as the sequence of connected and oriented dipoles
$$\ce{A^{(+)}-B^{(-)}},$$
regardless of if such an approach is useful.
The same applies for dipoles in different molecules or different branches of the same molecule (e.g. in proteins).


text:
Can we mix non polar and polar molecules by providing heat?
Polar molecules don't dissolve with non polar molecules because of difference in forces of attraction between them. But if we provide enough heat will they mix?

It is not like do not nor do, but rather do little or do more.

Entropy of mixing increases with temperature (TΔS term increases), so at elevated temperatures, mixing can become possible.

This really should use proper jargon. Substances dissolve, not molecules. Everything can "mix", but issue is what would in achieve: colloid, true solution, suspension? If the question is supposed mean "Does solubility rise with temperature" then it better say it...

Can you run faster than a man? Well, the answer is exactly that: it depends. The men are different, and so (or more so) are the molecules.

Sure Simply vaporize both and they will mix as a gas. Polar water vapor mixes well with nonpolar N2 and O2.  In the condensed phases mixing and dissolution are complicated ask anyone in the cosmetics business how to make the perfect lipstick or hand cream, or even ice cream for that matter, here cold is important.

Your assumption (at least substituting "substances" for "molecules") is wrong. There is no binary distinction between polar and non-polar substances or their relative solubility. Non-polar hydrocarbons happily dissolve in ethanol (which is fairly polar) but not in water. But water happily dissolves in ethanol.

Sure. When cooking my cranberry-orange jam I boil sugar solution containing the cranberries and some cut oranges, and the hydrocarbon that gives the orange peel its flavor mixes in. Yum!


text:
In collision theory why we multiply collision frequency by N/2?
In kinetic theory of gases we know that the average number of collisions, $N_\mathrm{col}$, is given by
$$N_\mathrm{col}  = \sqrt{2\pi\sigma^2 \overline{v}_\mathrm{rel}}\ N $$
where $N$ is the number of particles, and $\overline{v}_\mathrm{rel}$ its average relative velocity.
If we consider bimolecular collisions we should have divided the total number of collisions $N_\mathrm{col}$ by $2$. But instead we multiply it by $N/2$. Why?

I guess when you refer to bimolecular collisions you are really meaning a collision between like atoms or molecules (A + A) instead of disimilar species (A + B). You have to take into account that you have N molecules of each kind and ieach of this molecules can collide with the rest of the molecules in the sample.

2 objects can only collide 1 way: a-b. 3 objects can collide 3 ways: a-b, a-c and b-c. 4 objects can collide 6 ways: a-b, a-c, a-d, b-c, b-d, and c-d. The more objects, the more ways they can collide, and the more collisions, so N is a factor.

@paep I'm not sure I get the meaning of the angle signs. In MathJax They're set as relational operators. I'm confident that's not what you wanted to achieve.

@DrMoishePippik So this means that using the factor 'N' along with 1/2 gives us an accurate measurement of the number of collisions

Thanks for the tip @Martin-マーチン. I meant to write $\overline{v}_{rel}$.

I'm not sure if this answers your question, but in the special case where a collision/reaction combines two identical molecules $\mathrm{A+A} $ the number of indistinguishable combinations, i.e. the number of distinct A-A pairs of the two species must be calculated and this makes the bimolecular rate $\displaystyle \sim \frac{N_A(N_A - 1)}{2!}$ collisions. Normally, in dealing with chemical kinetics, $N_A$ is so vast that we can use $\displaystyle N_A^2/2$ without any error.

This carries similarities to a single-rounded round-Robin tournament with $n$ participants, and $\frac{n(n-1)}{2}$ games (every participant plays only once with every other participant than him/herself).  Note, the denominator I used here is $2$, not $2!$ (though indeed $2! = 2$ is true).

Or the simple puzzle how many handshakes would occur if N friends met and all did handshaking with all others.

@Buttonwood only for interest, in a termolecular reaction the chance is $n(n-1)(n-2)/3!$, which is needed if a Monte Carlo method are used to calculate populations during reactions, rather than $n^3/6$, not that termolecular reactions actually ever occur except in a computer :)

@porphyrin Then (I now speculate) $2!$ is the more general, easier extendable approach to describe the situation with mathematics.

@Buttonwood, yes I guess it is.


text:
How might humidity inhibit the drying or hardening chemical oxidation process of linseed-based oil paint?
I understand that linseed-based oil paint essentially dries or hardens by the chemical process of oxidation and not by evaporation. If it is possible that humidity somehow inhibits the chemical drying process in a related physical way, how might that occur?

Has anybody ever shown that humidity inhibits the chemical oxidation of linseed-based oil paint ? Never heard about this process.

I’m guessing conservation studies tend to examine humidity's effect on linseed oil painting after being painted. I believe actual curing takes months. I have seen one or two artists’ websites which suggest humidity may slow drying, for example.

see "The temperature of your room" subheading under https://finearttutorials.com/guide/how-long-does-oil-paint-take-to-dry/#How_does_oil_paint_dry.
The conclusion of the conservation article below notes humidity can inhibit cross-linking, but I'm not sure about how humidity would slow drying.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6445089/


text:
Effect of presence of competitive inhibitor on observed reaction constant
Suppose there is a solution of enzyme, its substrate, a competitive inhibitor, and a suicide inhibitor. The reaction rate constant for the reaction of the suicide inhibitor with the enzyme is known. The Ki for the competitive inhibitor is also known. The occupation of the enzyme's active site by the competitive inhibitor prevents deactivation of the enzyme by the suicide inhibitor. The initial concentrations of the inhibitors are also known. How would the presence of the competitive inhibitor affect the observed reaction rate constant for the inactivation of the enzyme by the suicide inhibitor in the following cases:

When the enzyme has been inactivated by the suicide inhibitor, the competitive inhibitor cannot enter the enzyme's active site.
When the enzyme has been inactivated by the suicide inhibitor, a molecule of the competitive inhibitor can still enter the enzyme's active.
A molecule of the competitive inhibitor can still enter an inactivated enzyme's active site, but at a lower probability compared to that of the active enzyme.

Is there an equation to determine the observed inhibition reaction rate constant based on the actual rate constant, the Ki of the competitive inhibitor, and the concentrations of the enzyme and the inhibitors.

Hello! Still interested in an answer, @user73910 ?

@MetalStorm yes

Yes, there are many such models you could make.
A general, difficult approach
You could make a very general model that has lots and lots of unknown parameters, and try to numerically fit observed kinetics data to this "full" model.  But you'll probably need a lot of data and you'll need an accurate model.  Fitting large (>=3 or 4) parameters by numerical integration of ODEs is not fun; you need to carefully consider which of your parameters are independent.  You'd probably want to non-dimensionalize your ODE model to be sure the parameterization you have chosen is actually fully identifiable from your data.
A specific approach, with lots more assumptions
You could also make much simpler models by making some simplifying assumptions.
$$\ce{E + S <=>[K_d] (ES)}$$
$$\ce{(ES) ->[k_f] E + P}$$
$$\ce{E + I <=>[K_i] (EI)}$$
$$\ce{E + C ->[k_{cov}] C}$$
One of the simplest sets of assumptions might be:

Covalent inhibition by covalent inhibitor $\ce{C}$ is slow relative to the equilibration of free enzyme with substrate and with non-covalent, reversible inhibitor, and also slower than the reaction of (ES) to form product.

Non-covalent reversible inhibition by $\ce{I}$ happens only to free enzyme.  Once an enzyme is inactivated by $\ce{C}$, nothing else can happen to it.  (This violates your third bullet point in your question, but the goal here is simplicity.)

Usual quasi-equilibrium assumptions about $\ce{ES}$.  These can often be replaced by less stringent steady-state approximations about $\ce{ES}$, but at the expense of harder math.

You are using a large excess of covalent inhibitor relative to the initial total amount of enzyme.

Usual assumptions about a large excess of substrate relative to initial total enzyme.

I think with these assumptions, you can write a rate law for the formation of product like:
$\frac{d P}{d t} = k_f E_{tot} \frac{1}{(1 +\frac{I}{K_i})} \frac{S}{(K_M + S)}$
This is the usual equation for competitive inhibition, but now $E_{tot}$ is a time-dependent variable that has first-order depletion kinetics.  So you could substitute in:
$\frac{d P}{d t} = k_f E_0 e^{-k_{cov}t} \frac{S}{(\frac{I}{K_i} + 1)K_d + S}$
This equation is itself already complicated enough!  But it is a starting point.  By going through the math of competitive inhibition you could see if you could relax some of the assumptions I've made.  The hardest one to relax will be #1; if covalent inhibition is not slow relative to all other reaction steps, everything gets considerably more complicated.


text:
Finding ΔH and ΔS given Gibbs free energy at two temperatures
Question: Gibbs free energy of a reaction at $\ce{300 K}$ and $\pu{310 K}$ are $\pu{-28 kcal}$ and $\pu{–28.5 kcal}$
respectively. Determine $\Delta H$ and $\Delta S$ at $\pu{300 K}$.

I tried forming equations using the given data. If I assume the enthalpy to be a constant irrespective of temperature, I will get a system of two simultaneous equations in $\Delta H$ and $\Delta S$ which can be solved. Using: (1) $\Delta G = \Delta H - T\Delta S$, and (2) replacing $\Delta G$ for the mentioned values, I obtain
\begin{align}
  \pu{-28 kcal}   &= \Delta H - (\pu{300 K}) \Delta S \tag{1} \\
  \pu{-28.5 kcal} &= \Delta H - (\pu{310 K}) \Delta S \tag{2}
\end{align}
However, I think this approach is incorrect. Enthalpy is a temperature dependant quantity and $\Delta H$ at $\pu{300 K}$ won't be equal to $\Delta H$ at $\pu{310 K}$.
I think the question lacks sufficient data to be solved.

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
// Optionally, here are formatting guides for texts and formulas/equations/expressions.

You overcomplicate it.  You are right that ΔH but even ΔS are temperature dependent, based on the difference of heat capacity of reagents and products. But that simple exercise neglects it for simplicity. Give them what they expect with a relevant note.

Note that various equations like the van't Hoff one or Clausius-Clapeiron(?) one imply constant ΔH, assuming they will not be used over wide T range.

The parameters ΔH and ΔS have units, and the numerical values have uncertainties. Which ones ? If the two values of ΔG are known to ± 0.1 unit(?) and if the temperatures are known to ± 0.1 K,  it means that the calculated value of ΔS is known with an uncertainty of at least ± 60%. Does it make sense ?

@Maurice You are true. But I guess you go behind the expected complexity level of the task. I think they do not expect statistical evaluation.

Over such a small temperature range it is safe to assume changes in H and S are nil . Thus you have two equations and two unknowns.


text:
Why the water in an lead-acid battery isn't self-electrolysed (at any significant rate)?
I'm reading Detchko Pavlov's Lead-Acid Batteries: Science and Technology (maybe an old version than the link) and in the first chapter the book mentioned "it is currently unknown why electrolysis of water happens so slowly".
If we think about it, isn't that strange? We have a pair of electrodes in a pot of acidic water, with a potential of 2+ volts, but nothing happens? And I do remember asking the same question when I was in middle school but really didn't get a convincing answer.
Do we know the answer now?

Here's the text of the book. Mine is a Chinese translation version so I will try to translate it back.

As is shown by the E/pH diagram of Figure 2.1, an lead-acid battery in open-circuit is thermal-dynamically unstable. The self-discharge reaction between the electrodes will electrolyse water into $\ce{H2}$ and $\ce{O2}$. At the same time, the following reactions happens on the electrode plates:
On the positive plate:
$$
\ce{PbO2 + H2SO4 -> PbSO4 + H2O + \frac{1}{2} O2}
$$
On the negative plate:
$$
\ce{Pb + H2SO4 -> PbSO4 + H2}
$$
If the battery is constructed with high-purity lead and sulfuric acid, the above reactions happens at a very slow rate. To this day, we have not fully figured out what basic reactions so strongly inhibits the precipitation of $\ce{H2}$ and $\ce{O2}$.

Also in this document

What is the exact shape of this outgassing-rate/voltage curve and how is it derived?

In the body of questions, of answers, and of comments, chemistry.se allows you to use mhchem; an easier approach to provide chemical equations.  Because it is a technique not all web browsers process equally well, do not use mhchem in the title of questions, though.

For electrolysis to happen, you need a (sufficiently large) potential difference between electrodes, and a closed electrical circuit with connection of anode and cathode.
Electrolysis of the electrolyte certainly is/was a (safety) issue with the simpler/older design of car batteries when it is/was up to the user to refill electrolyte

(image source)
and use of traditional battery chargers:  it was possible to «overcharge» the accumulators which lead to the generation of hydrogen gas.  Thus, supervised charging (monitoring the density of the electrolyte), closing the individual cells with perforated screw caps (to allow venting) in an aerated place was recommended.
Today, there are car batteries either capture the gases (to some extend) or/and use an electrolyte which is trapped in a gel; then, manual refill of acid usually is not necessary anymore.  Equally, contemporary chargers now check the status of the battery to prevent overcharge / stop before the electrolyte decomposes (in electrochemical terms: before you leave the intended electrochemical window*).  Or (the still more expensive way), the lead battery is replaced altogether by a different type (one of the variants using lithium, or a molten salt).
* A term used e.g. in cyclic voltammetry (a freely accessible primer).

So how much is the over potential needed for the H2 and O2 to bubble out? What's the underlying mechanism that determines the bubbling threshold? The way it is stated in the book seem to indicate the threshold is already reached in the battery.

Also in this document: https://mesa-tec.com/wp-content/uploads/2014/06/Hydrogen-Gas-Management-For-Flooded-Lead-Acid-Batteries.pdf   "Flooded batteries vent continuously, under all states"

Overpotential may depend on multiple factors (concentration, or better activities of the reagents [it isn't a dilute aqueous solution], said temperature and pressure, current densities) and design of the electrodes. Because it can alter a lot (e.g. instead of Pt wire, use of a platinized Pt electrode; or electrolysis of aqueous NaCl typically does not yield hydrogen and oxygen, but hydrogen and chlorine), a true electrochemist likely can answer this better.

For multi-electron transfer reactions you typically need a higher overvoltage because you need to (usually) make an intermediate which is highly energetic. Look at proton reduction, in the case your electrode does not interact with your reagent you need to generate a hydrogen radical as an intermediate. You can imagine the energy needed to do so is incredibly high. Then why if we use the right electrode does it work without a high overvoltage? Well the right catalyst can stabilize the intermediate, for example Pt forms adsorbed hydrogen. No need for a hydrogen radical as intermediate.

Hmm... I'm old enough to have used some of the lead acid car batteries that had screw caps and needed filling. I never used anything but water. As I remember the batteries were shipped dry then filled with the sulfuric acid electrolyte before installation.

@MaxW The picture likely depicts only the small refill with aqua dest (no visible yellow skin coloration of the epidermis [bare toes] which can happen after accidental though repeated/extended exposure to sulfuric acid) which flows with greater ease/has lower viscosity than battery acid.

This town banned cars (except tiny electric ones), a video by Tom Scott, published on August 7th, 2023 about Zermat in Switzerland, a community which went fully electric for their mode of automobile transportation.

Very minor addition, but evolution of hydrogen in large lead acid batteries on submarines is a known issue and is controlled by sensors, ventilation, alarms, etc.  It does not happen precipitously, but could endanger the ship (especially while charging) if these safety systems did not exist.  It is definitely measurable.

I remember in Das Boot the movie, the crew was constantly monitoring the hydrogen level when they are submerged (i.e. not charging). I guess during WW2 the self discharge of the battery was high but the hydrogen purging device wasn't invented yet, so hydrogen accumulation in a sealed submarine was quite a big problem at the time and could be the limiting factor on how long a submarine could stay submerged.

Catalytic converters to recombine hydrogen and oxygen appear to be common on lead-based accumulators of greater capacity than typically used in cars (add by a Polish company, a review by a British author).