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Why can you "cancel out" chemical species from a chemical equation that appear on both sides of a chemical equation?
Why is it acceptable in chemistry to "cancel out" chemical species from a chemical equation that appear on both sides of a chemical equation (as reactants and as products)?
On the one hand, it makes sense, since you could argue that if a set amount of a chemical species was used up (reactant) and the exact same amount of the same chemical species was also produced (product), it would seem as if the species was chemically inert and you could "ignore it" from the chemical reaction.
However, on the other hand, you could also argue that the reaction, might not happen if that said chemical species (that appears both as a reactant and a product in the equation) wasn't there.
For example, (this is from an application of Hess's Law exercise), the following reaction:
$$\ce{2CO2_(g)  + H_2 O(l) + 2C(s) +\frac{5}{2} O_2 (g) +H_2(g) -> C_2 H_2(g) + 2CO_2(g) + H_2O(l) +\frac{5}{2} O_2(g)}   \tag{1}$$
can also  be written as:
$\ce{2C(s) +H_2(g) -> C_2H_2(g)}
\tag{2}$
However, reaction (2), as it is written might not be able to happen, since maybe the chemical species that have been cancelled out, might be needed for the reaction to take place?

Reaction equations are fundamentally just mathematics, nothing but a system of linear equations written in a different way, purely to establish atom conservation. All the chemistry in reaction equations is tacked on after the fact. So with zero effort you can make mathematically consistent but chemically absurd reactions.

Related, possibly (near) duplicate: Why can we add/substract/cross out chemical equations for Hess law?

Does this answer your question? If a reactant appears on both sides of an elementary equation, can it be cancelled?

What you're talking about in the second part of your question, has to do with mechanisms of reaction, and within that particular subject the equations are written as to reflect the actual compounds/molecules formed in the reaction, even temporarily

@AlexVo I converted your answer into a comment. I agree with what you write but this needs to be expanded.

What is that water doing in the reaction in the first place? If it is a liquid and all other reagents are gases, is it a "spectator"? Or is the reaction occurring in the water (gaseous reagents diffuse in, products out)?

The nice thing about chemical equations is that they can be treated algebraically in many cases. You can add or subtract chemical equations. One can add something on both sides. For example in algebra, in 3+a+b = c+d+3, three cancels out leaving behind a+b=c+d. In the same way, if a chemical species is present on the left and the right, it simply would mean that particular entity did not participate in the chemical reaction and it can be cancelled.
On the other hand, there are chemical nuances regarding chemical equations as written. They do not contain all the information about the real reactions.
If we started with a mixture of carbon dioxide, water, carbon and hydrogen as suggested here, supplying heat would result in oxygen combining with hydrogen rather than carbon because of different reaction speeds. Hydrogen will combine with oxygen way to fast before carbon can begin to react.
2CO2(g)  + H2O(l) + 2C(s) +5/2 O2 (g) +H2(g) ---> C2H2(g) + 2CO2(g) + H2O(l) +5/2 O2(g)   ...(1)
In short, the context matters. When you are treating a system mathematically, chemical equations can behave like algebraic statements. However, when we are thinking in terms of actual chemical reactions, a written chemical equation may not describe the reality or actual products.

In hess's law, we are concerned with the overall energy changes happening due to the reaction that is final - initial and nothing in the intermediate steps. Species reappearing again on the product side do not change the overall energy as their contributions cancel out. Hence just to simplify and to shorten writing amount, we in a sense cancell them out. But this does not mean their presence is meaningless.

The Reactants that are also present on the product act like a catalyst. Or you can say that they actually are important for the reaction but don't involve the breaking and making of bonds. For example,
Photosynthesis can be written as:-   6CO2 + 6 H2O -----> C6H12O6 + 6O2
whereas what actually happens is:-   6CO2 + 12H2O -----> C6H12O6 + 6O2 + 12H2O
We remove the 6H2O in writing the equation. Although it's important for the reaction.

I think you made a mistake writing out that second equation?

Is that what really happens!! what about plants, chlorophyl, other functional compounds, sunlight etc.


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What is the hydrated diameter of a lithium ion？
When salts dissolve in water, the anions, cations, and strongly polar water molecules undergo hydration reactions, resulting in the formation of hydrated ions with a specific number of coordinated water molecules.
For example, from Ref 1:
Table S2. Diameter of hydrated ions.

Ions
Hydrated diameters/nm

$\ce{K+}$
0.66

$\ce{Cl-}$
0.66

$\ce{Na+}$
0.72

$\ce{Ca^{2+}}$
0.82

$\ce{Mg^{2+}}$
0.86

$\ce{Al^{3+}}$
0.95

$\ce{SO4^{2-}}$
1.00

BTW, please post the research paper that the data comes from.
References

Yixin Yang, Guorong Li, Dan Ouyang, Zongwei Cai, Zian Lin, Dual-activation interfacial polymerization based anionic covalent organic framework nanofiltration membrane for high-flux dye separation, Chemical Engineering Journal, Volume 456, 2023, 141008, ISSN 1385-8947, https://doi.org/10.1016/j.cej.2022.141008.

Hint: Alternatively, search for ion mobility. AFAIK hydrated ion radius cannot be (at least not easily) measured directly, but is derived from ion mobility and Stokes equation for laminar flow around sphere.

You may find useful reviewing of search results for
site:stackexchange.com OR site:libretexts.org OR site:wikipedia.org OR site:hyperphysics.phy-astr.gsu.edu  hydrated ion radius

The data you share (source of the table?) are poor; neither is there a reference what the author considers as hydrated cation -- because different coordination geometries require different volume, nor the method to determine the these dimensions.
In PAC, Persson for example compiles the following data (in Angstrom) from the literature about the distance between center/metal cation $\ce{M}$ and oxygen $\ce{O}$ of the water ligand:

aqua complex
$\ce{M-O}$ distance
$\ce{M^{n+}}$' ions radius/A
Shannon
Configuration

$\ce{Li(H2O)^+_4}$
1.95
0.61
0.59
Tetrahedron

$\ce{Li(H2O)^+_6}$
2.10
0.76
0.76
Octahedron

$\ce{Na(H2O)^+_6}$
2.43
1.09
1.02
Octahedron

$\ce{K(H2O)^+_8}$
2.48*
1.50*
1.51
Square antiprism

$\ce{Mg(H2O)^{2+}_6}$
2.10
0.76
0.72
Octahedron

$\ce{Ca(H2O)^{2+}_8}$
2.46
1.12
1.12
Square antiprism

$\ce{Al(H2O)^{3+}_6}$
1.89
0.55
0.535
Octahedron

(Persson2010, distances marked by an asterisk were determined in the solid state)
Recording the position of protons can be difficult (because the time scale of the proton exchanges vs the one of the analytic technique, very low diffraction power vs X-ray diffraction, etc.).
In a separate publication, Mähler and Perrson derive from Large Angle X-ray scattering (LAXS) and double difference infrared spectroscopy (DDIR)

It was found that the average Li−O distance in $\ce{Li(H2O)^+_4}$ was $\pu{1.942 Å}$ for 36 observed structures,and this number did not change noticeably if other neutral homo- or heteroleptic monodentate oxygen donor non-ether complexes were allowed. The Li−O distance of $\ce{Li(H2O)^+_6}$ is $\pu{2.134 Å}$ as an average for 10 observed structures.

X-ray diffraction indeed is one, but not the sole technique to determine the ions' dimension.  As pointed out in a comment by @poutnik, ion mobility can complement the insight.  Hayamizu et al. for instance monitored the diffusion of $\ce{Li(H2O)^+_x}$ by solution NMR spectroscopy, and determine an effective Stokes radius of $\ce{Li+}$ in diluted aqueous solutions of about $\pu{300 pm}$ ($ = \pu{3 Å}$).

The authors visualize the effective (Stokes) radii of a few ions with the following figure:

(Hayamizu et al.)
Persson, I. Hydrated Metal Ions in Aqueous Solution: How Regular Are Their Structures? Pure Appl. Chem. 2010, 82, 1901–1917. https://doi.org/10.1351/PAC-CON-09-10-22 (open access).
Mähler, J.; Persson, I. A Study of the Hydration of the Alkali Metal Ions in Aqueous Solution. Inorg. Chem. 2012, 51, 425–438. https://doi.org/10.1021/ic2018693 (open access).
Hayamizu, K.; Chiba, Y.; Haishi, T. Dynamic Ionic Radius of Alkali Metal Ions in Aqueous Solution: A Pulsed-Field Gradient NMR Study. RSC Adv. 2021, 11, 20252–20257. https://doi.org/10.1039/D1RA02301B (open access).
You might miss valuable literature references if you skip a keyword like ionic radii e.g.,
Marcus, Y. Ionic Radii in Aqueous Solutions. Chem. Rev. 1988, 88, 1475–1498. https://doi.org/10.1021/cr00090a003.
Conway, B. E.; Ayranci, E.  Effective Ionic Radii and Hydration Volumes for Evaluation of Solution Properties and Ionic Adsorption. J. Solution Chem. 1999, 28, 163–192. https://doi.org/10.1023/A:1021702230117.

The Hydrated Radius of Lithium ion is 76pm.
Thus the Hydrated Diameter should be 152pm (I think so).

Your answer doesn't state the origin of the data.

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Are trazodone and risperidone chemically related?
I am not a chemist and I didn't find any data on this at all but from a look over the two dimensional diagrams of the two molecules trazodone and risperidone I see a bit of resembelence, especially in the number and features of the rings. Besides the similarity in names (done suffix) they are both used by psychiatrists for various mood off-label uses.
Are trazodone and risperidone chemically related?
Update
In a well recieved comment it was written that they are related in that they both have two aromatic groups with a basic centre between them and they are both brain-penetrant.

Is this in Red the basic center and is it "basic" in the sense that it's not acidic?

What makes the two molecules brain-penetrant?

Trazodone:

Risperidone:

Chemically they are related in that they both have two aromatic groups with a basic centre between them and they are both brain-penetrant. They are different in their pharmacology - Trazadone's most important interactions are at 5HT2a and alpha 1, Risperidone is also highly active at 5HT2a but has significant activity at 5HT7 and  Dopamine receptors.

The carbon chains you've highlighted isn't what makes these molecules basic. It's those few nitrogen atoms. The two chains are also not really that similar. For most (= almost all) intents and purposes I would regard them as chemically different. They are probably similarly-ish in size, but that's basically it.

Generally and historically drugs ending with -odone are in the opiods family (see Oxycodone, Hydrocodone, ecc), but it's not a hard rule. In this case Trazodone is an antidepressant and Risperidone is an antipsycothic, but they are not opiods (so they do not derive/mimic natural opiods found in opium).
Being both neuroactive drugs they need to/tend to have some pretty common chemical properties for being capable of passing the brain-blood barrier.
They typically need to be water-soluble as they need to reach the brain via bloodstream (so they need to have some polar groups), but not too much ( so as to allow the drug to pass the brain-blood barrier. (see Lipinski's rule of five)
You can see that both molecules do not have strong polar groups (no alcohols -OH, no amines -NH2, ecc) and both have some basic centers (the central nitrogen atoms of both molecules). Generally, many drugs that act on the brain display this feature (basic nitrogen atoms); see alkaloids for more info.
The carbon chains you highlighted in red are simply some carbon linkers that bind the two "sections" of each molecule together AND gives the molecule some degree of flexibility, needed to better interact with the active sites present in the proteins in the brain.
Other than these pretty general characteristic of neuroactive drugs they are not particularly related.

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Bonding and antibinding orbital energy difference
Reading Claydens organic chemistry, in chapter 5 it states that Br-Br makes a good electrophile while the C-C in ethane doesnt because the energy difference between bromines bonding and anti bonding orbital is small and that of the C-C bond isnt. Is there a good way to deduce when a bond and anti-bond orbital energy difference for a molecule will be small? Why is it the case for bromine but not ethane?


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Difference in alkanes , alkenes and alkynes
Is there a way I can determine if a given hydrocarbon is an alkene, alkane or alkyne or even some other type. What are the KEY differences in them, like reactivity , ratio of elements and so on?

The NMR spectra a distinctly different

What even is the NMR spectra

If you really want to know start here: https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance_spectroscopy

Experimental determination can be done by using Baeyer's reagent or $\ce{Br₂ water}$. This, by far, is the qualitative analysis for it.

Alkanes have nearly no reactions with any chemicals, at room temperature. They only react with $\ce{Cl2}$ and $\ce{Br2}$ to produce a haloderivate and some $\ce{HCl}$ or $\ce{HBr}$.
Alkenes react by oxidative addition to dilute alkaline permanganate, chlorine, bromine, iodine in alcohol, and ozone. They also react by simple addition with $\ce{HCl}$ (in the presence of $\ce{AlCl3}$) and other acids, like nitrous, sulfuric, hydrobromic, hydriodic. Alkenes also react by polymerization.
Alkynes react in a similarly way with the previous reagents, but they can have two successive addition reactions. Furthermore, if they possess the ynil group ($\ce{·CCH}$), they are also able to make copper or silver derivates with $\ce{Cu(NH3)2^+}$ and $\ce{Ag(NH3)2^+}$ solutions, in reactions like $$\ce{RCCH  +  Ag(NH3)2^+ -> RCCAg  +  NH4^+ + NH3}$$ $$\ce{RCCH  +  Cu(NH3)2^+ -> RCCCu  +  NH4^+ + NH3}$$ These two reactions are specific properties of the ynil group.


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DIY hydrolysis of starch with copper and test them with polarizer sheet due to muta rotation
When I read about muta rotation — the interconversion of alpha to beta glucose — recently, I decided to do it with rice, because rice contain 66% of starch which is alpha glucose polymer. To hydrolyze the starch in rice, normally chemists use HCl-like acid as catalyst, but in my home there is nothing like that so I thought about copper, because I’ve seen that copper is used as catalyst in most cases. Will it work? I.e.,
(Cu) starch + H2O --->  alpha_D_glucose
Does the above reaction work? (We can see with polarizers due to chirality.)


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Can we use DFT to compute the potential energy of dynamic processes?
Suppose we want to determine the potential energy between a hydrogen atom and a surface, such as Cu(001), during hydrogen scattering on Cu(001) surface, assuming that the hydrogen atom does not change the surface structure.
I used DFT to performe single-point calculations to find the potential energy at various heights of the hydrogen atom above the surface. The potential energy was calculated using the equation:
$$
E_{potential} = E_{H+surface} - E_{H} - E_{surface}
$$
The results looked like

The question then arises: is the method outlined above reliable for computing the H-surface potential energy? Can DFT accurately calculate the potential energy in dynamical processes?

The real question is not about DFT but the validity of the hypothesis that the scattering process is slow compared to the copper atoms' dynamics.
DFT assumes the adiabatic approximation for the electronic dynamics. I.e., electrons follow the ionic dynamics instantaneously, always remaining in their instantaneous ground state. Considering the typical excitation energies and atomic scattering processes, I would take the validity of such approximation for granted in most cases.
A different story is related to the ionic time scales. The assumption that the hydrogen atom does not change the surface structure  does not guarantee that a sequence of static calculations where $Cu$ ionic positions are allowed to relax to their instantaneous equilibrium values accurately represents what is going on in the actual process. Indeed, even keeping the same structure, copper ions may move according to the normal modes of the surface slab, and depending on the energy of the hydrogen atom, the scattering process may interact with such modes.
The most direct way to assess the validity of a static calculation would be to perform a dynamic calculation, using ab initio molecular dynamics, with the parameters relevant to the experimental situation.


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How to decide between two lewis structures when formal charge is the same
I'm trying to draw the lewis structure of POCl3. At first, I drew it like this:

but after checking on google I saw that the correct structure is actually

My question is, are there any rules that allow us to decide between two structures such as these? I know that we want to minimize the formal charge, and if there is any formal charge we want to place it according to the atom's electronegativity (negative charge on most electronegative, positive charge on least electronegative). But in this case, both structures don't have any formal charge.
After looking at both structures, I noticed that in the correct one the electron geometry on the oxygen and the phosphorus seem better. In my incorrect structure, the P is AX3E and the O is AX2E2, while in the correct structure the P is AX4 and the O is AXE2. Hence the P has an atom instead of a lone pair and the O has one less atom around it, therefore there is less repulsion. Also there is a double bond instead of single bond, which releases more energy.
Are the electron repulsion and the double bond really why the second structure is preferred, or is there a different explanation? Are there any common rules to decide between lewis structures when there isn't a difference in the formal charge?

And if you learn a bit more, you'll find that having a pair of formal charges would be better here...


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Why doesn't sodium bicarbonate dissolved in water spontaneously effervesce when carbonic acid does?
It's well known that if you put carbonic acid $\ce{H2CO3}$ into water it will dehydrate into carbon dioxide $\ce{CO2}$ and water $\ce{H2O}$. Indeed this is how modern fizzy drinks generate their fizz. Taking this as a given what I don't understand is why doesn't sodium bicarbonate dissolved in water not slowly give off $\ce{CO2}$ via the following pathway:
$$ \ce{HCO3- +H+ <=> H2CO3} $$
$$ \ce{H2CO3 -> H2O + CO2} $$
This would lead to the solution of sodium bicarbonate slowly converting itself into a solution of sodium hydroxide. The first part of the reaction is reversible and is something that at equilibrium will have a non-zero amount of $\ce{H2CO3}$. In fact given that carbonic acid is a weak acid there will be a lot of $\ce{H2CO3}$ present at equilibrium. I also can't see how the presence of sodium ions in solution would inhibit the second reaction (since that is the only difference between this solution and a solution of pure carbonic acid which does effervesce) so according to my understanding we should be seeing a fizz .
However very clearly this bubbling does not happen in real life, as anyone can easily verify for themselves. So what inhibits it then?

Check all 4 parallel ongoing equilibrium processes and they eq. constants. CO2(g) <=> CO2(aq) <=> H2CO3(aq) <=> HCO3-(aq) <=> CO3^2-(aq).  All of them are reversible.

I agree with that. From my understanding these processes are the same in carbonic acid dissolved in water as they are with sodium bicarbonate dissolved in water. And in the former case for the level of partial pressure of CO2 in ambient air the equilibrium of this system lies very strongly towards CO2(g), however for the latter case it doesn't (as we don't see effervescence). My question can be rephrased as asking why there is a difference in equilibrium concentrations of CO2(g) between the two cases (as Na doesn't show up anywhere in the system you posted).

Conditions are very different. pH is the key.

Research the use of baking soda in baking, specifically soda bread; you will find at high temperatures bicarbonate does fizz. Aso the 400+ ppm of CO2 in the air raises the CO2 level in the water inhibiting the hydrolysis of HCO3- to give H2CO3 or CO2. Purging a bicarbonate solution with CO2 free air should eventually decompose all the bicarbonate.

It certainly effervesces in hot enough water

When sodium bicarbonate is dissolved in water, it produces $\ce{HCO3-}$ ions. These $\ce{HCO3-}$ ions can react with acidic ions in water, and produce some $\ce{CO2}$, as described in your question. But the available amount of acidic ions in water is rather low, about $\pu{10^{-7}}$ M. It has two consequences. First, the concentration of dissolved $\ce{CO2}$ is probably in the same order of magnitude. At such low concentrations, $\ce{CO2}$ will stay dissolved, and the solution will never fizz. To get a fizz, the $\ce{CO2}$ concentration in water must be greater than $0.04$ M, according to Henry's law. Second, if some $\ce{NaHCO3}$ produces  $\ce{CO2}$, it will also produce some $\ce{NaOH}$ in solution, and this $\ce{NaOH}$ will immediately react with all available $\ce{CO2}$ molecules around to produce $\ce{NaHCO3}$ backwards. This is why the $\ce{NaHCO3}$ solutions will never spontaneously produce any $\ce{CO2}$ bubbles.

HCO3- + H2O = H2CO3 + OH-; H2CO3 = H2O + CO2,  Standard hydrolysis equation. H2O is the acid in highest concentration in pure water. Atmospheric CO2 is also involved [or not].

... and as a kitchen experiment, adding a little vinegar to the water donates acidic ions, raising the concentration of $\ce{CO_2}$ enough to cause bubbling.

As a side-note, improperly stored samples of sodium hydroxide will react with moisture and carbon dioxide from the air to form the carbonate.

I was wondering how likely the direct decomposition $\ce{HCO3- <=>CO2 +OH-}$ in neutral water solution would be and ended up going down a quantum chemistry rabbit hole. It seems to be a topic of ongoing research, but AFAICT (from e.g. this recent paper) it may be a relevant route (maybe more than $\ce{HCO3- +H3O+<=>CO2 +2H2O}$ depending on pH), so it doesn't seem like water autodissociation is a necessary initial step. Of course, higher pH still drives the equilibrium left either way, so all this is really just mostly pointless quibbling.

(Also, AFAICT, apparently modern consensus is that $\ce{CO2 + H2O <=> H2CO3}$ doesn't happen in a single step in water, but goes through $\ce{HCO3-}$ first.)

@Ilmai Karonen. All you say is interesting and correct. But it does not change too much my answer. $\ce{CO2}$ may be produced by direct decomposition or by the reaction with acidic ions. Both reactions have been roughly described in my answer. Whatever the reaction, the amount of $\ce{CO2}$ would be rather poor, and no fizz will be produced. Apparently this is the most important part of the initial question.

authors: მამუკა ჯიბლაძე


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Does the E1cb elimination mechanism have a first-order or a pseudo-first order rate law?
Elimination reactions turning a single C-C bond into a double bond are classified as E1, E2, or E1cb, depending on the mechanism (concerted or with intermediate), the stereochemical outcome and their kinetics. The "1" stands for first-order reaction and the "2" for second-order reaction.
For E1cb, we first deprotonate to get a carbanion, then the leaving group leaves and the double bond forms:
$$\ce{RR'CX-CHR''R''' + B- <=> RR'CX-CR''R'''- + BH}\tag{1} $$
$$\ce{RR'CX-CR''R'''-  -> RR'C=CR''R''' +  X-}\tag{2} $$
If the base ($\ce{B-}$) is in excess and sufficiently strong, the intermediate forms stoichiometrically, and you have a first-order reaction for the second step, which is expected to be slower (acid/base chemistry vs. breaking and making bonds with carbon).
However, if there is an equilibrium between reactant and base $\ce{B-}$, and conjugate base of the reactant and $\ce{BH}$, this would be a fast pre-equilibrium, and the overall rate would depend on the concentration of base and of the reactant. In this case it is not first order (but if you keep the concentration of base in excess, it would be pseudo-first order).
What are the kinetics of E1cb, and what are some good references with original data to support the kinetic model?
There is a similar question already, but without answer: Kinetics for E1cb reactions


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Is light energy defined as work in thermodynamics?
Is Light/Radiant Energy considered Kinetic Energy or Potential Energy?
I have started studying Thermodynamics, and the concept of internal energy was introduced, and defined as the sum of kinetic and potential energies of a system.
Another important definition was that:
ΔU = q + w , where
q = heat absorbed/released by the system
w = work done on the system by surroundings/work done on surroundings by system
U = Internal Energy of a system
There are lots of chemical reactions, such as combustion reactions, where light is produced, but there aren't many examples in thermodynamics concepts that deal with light energy. Where does light fit into this concept of internal energy?
If light is produced by a chemical reaction, is that a form of work done by a system on the surroundings?
If light is absorbed, is that a form of work done on the system by the surroundings?

There is difference between thermal and non thermal radiation. The former is heat transferred by a radiative way. OTOH, non thermal radiation can go against temperature gradient.

related: https://chemistry.stackexchange.com/questions/32045/thermodynamics-of-photochemical-reactions

If you search the site for the terms "light thermodynamics" you'll find a number of related posts: https://chemistry.stackexchange.com/search?q=light+thermodynamics

Absorbing a photon will increase the potential energy of a molecule, and possibly a little  kinetic energy change if rotational and vibrational energy is changed in the process.  The translational kinetic energy is not changed immediately after absorption (there if no external force to do this) but can be imparted into products after reaction when bonds break.  Green photons have $\approx 25000\,\mathrm{cm^{-1}}$ vs thermal energy of $\approx 210\,\mathrm{cm^{-1}}$ so a lot of energy can be released on reaction. If light is emitted then this can simply heat up the surroundings by being absorbed by unreactive molecules or some work could be extracted if these photons are absorbed by, say, a photodiode/solar panel. (Excited molecules can also release energy by non-radiative means but this just heats up the surrounding molecules which energy then diffuses away into other molecules))

Thank you so much for your explanation. So light energy absorption or release can be categorised as both Heat and Work in the First Law of Thermodynamics equation? For example, when the absorption of photons increases the temperature of the system, would that be categorized as Heat? Or when a reaction like combustion produces photons, would those photons be categorised as Work?

heat and work clearly on emission, as to doing work on absorption , this seems to be more difficult, perhaps ionising a molecule producing electrons that could then be used to do work, or storing energy  in a reaction as in isomersation that could be released later to do work.

Electromagnetic radiation hereafter called EM, is energy. EM consists of individual photons or possibly of coupled photons. A photon has a specific energy, specific momentum and a spin or polarization. Bundled photons also have an entropy based on their frequency distribution, directionality, and source. Photons also have size in 3 dimensions and a questionable 4th dimension.
EM interacts with matter in at least three common methods. The First is Black Body Radiation, BBR. BBR is inextricably involved with the kinetic energy and momentum of molecules and of the photons. Molecular collisions with the repulsion and attractions of electrons induce the emission and absorption of photons. The energy distribution of the photons at thermal equilibrium is equal to the energy distribution of the molecular KE. When two substances have small differences in their thermal and BBR envelopes, heat energy will transfer from hot to cold by conduction, if in contact, and by radiation. When and where the energy envelopes do not overlap other things happen: chemical reactions, plasma formation, or even transparency. Example of the first: letting the water evaporate when cooking, the second: an atomic bomb, the third: outer space is transparent to infrared emitted by the Earth and the Earth's atmosphere is transparent to the visible light from the Sun. Further thought indicates that air at the same temperature as the ground is not transparent to emitted BBR. The reason BBR is lost to space is the continual lowering of atmospheric density with decreasing gravity; eventually molecular collisions cease, and the driving force is the entropy increase as the radiation is lost to space. In conclusion: radiation received or emitted by BBR raises or lowers the heat content of a system and is measured by the temperature change and the heat capacity. The energy transfers are quantum mechanical, however, the QM energy differences that involve molecular velocities and BBR are too miniscule to be resolved; they are described by continuous probability functions.
The second common interaction of radiation with matter is a Quantum Mechanical overlay over the black body radiation. These are the direct absorption and emission of photons into molecules resulting in transitions in the rotational, vibrational and electronic energy levels of the molecules. These transitions cause the "bumps and valleys" in the spectra of molecules. The function of measuring a background spectrum in spectroscopy is to remove the common BBR and isolate the overlay. The absorption or emission resulting in a QM transition does not cause an immediate change in temperature; it is not heat. It is a change in Chemical Potential Energy that can last for femtoseconds to hours or even days for phosphorescence [or hopefully forever in our computer storage]. The excited states can be converted into heat thru collisional deactivation, cause a chemical reaction with its change in heat and chemical potential energy, or reradiate with subtle consequences in molecular recoil [doppler effect] or interaction with other QM states. The initial interaction is work. The beauty of Thermodynamics is that it is mostly concerned with STATE Functions that are measured after everything is calmed down. Watching the fireworks can be exciting too, hence the study of kinetics.
The third common interaction is the variation on the Compton Effect. Photons can literally bounce off molecules in a process called scattering. There is an interchange in momentum with a small change in energy resulting in a change in the frequency of the photon and in the KE of the molecule. This is black body radiation that failed because the energy differences did not overlap. It appears to be an effect of momentum exchange and the size of the molecules and photons. This effect is weak. It results in the scattering of light, the blue sky; refraction of light, the reflection of light, the bouncing of some radio waves off the ionosphere, the Raman effect. The interchange of energy is small, much of the change is an entropy change, but energy interchange can be substantial if radiation is intense, just stick your hand in a laser beam [better yet don't!].
In conclusion: Black Body Radiation is heat where the energy envelopes overlap. Otherwise, there is everything from chemical reaction to chaos to transparency. Initial absorption of radiation resulting in changes in rotational, vibrational, or electronic energy levels is work but can change into heat, chemical potential energy or reemission with subtle changes in energy and entropy. Finally scattering results in entropy and energy changes in both the radiation and the substance depending on physical constraints. Thermodynamics studies the changes in the State Functions. Kinetics tries to study the actual processes or mechanisms.

Thank you so much for this answer! Even though the concepts are very complicated I appreciate all the details and explanations!


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Why mono and dioxides of germanium,tin and lead are amphoteric
I get that group 14 elements show +4 and +2 oxidation states so they can form both ionic and covalent compounds and so they are amphoteric but monoxide means that these elements have to show +2 oxidation state in these oxides and dioxide means that the oxidation state of these elements will be +4.
Monoxides of ge,sn and pb should be ionic because of fajan's rule and dioxides should be covalent then why are they amphoteric? I know that it is because they react with both alkali and acids but what is the reason for such behaviour?

Related: https://chemistry.stackexchange.com/questions/40651/what-is-an-acidic-oxide

Consider that elements in Group 1 are far less electronegative than than those amphoteric elements, and in Group 17, they're far more electronegative.
In solution , $\ce{Na+}$ readily donates an electron to tin, for example, form sodium stannate, $\ce{Na2(Sn(OH)6)}$.
On the other side, $\ce{Cl-}$ readily attracts an electron to form tin(II) chloride.


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Stability of alpha vs beta anomer in glucose
As per my knowledge, there seems to be 2 main factors influencing the stability of alpha vs beta anomers of glucose:

H - bonding

Considering this, beta anomer would be more stable due to the possible h bond between c1 - oh and c6 - oh

Anomeric effect

I havent been taught anything much about this, all i know is that more electronegative substituents(like oh) would prefer an axial arrangement which the alpha anomer has.

additionally i have read somewhere that hyperconjugation seems to play a role in making the alpha anomer more stable(no explaination was given) as well(I will try to mention the source as soon as possible, im unable to remember where i read that as of this moment)

My question is of two parts:

Ultimately, which anomer is more stable and why?
Does hyperconjugation actually play a role in differentiating their stabilities and how?


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Is This compound stable at room temperature?

This compound is planar conjugated and has 8 pi electrons on its periphery so it can be anti aromatic but if we polarize the pi bond at the bottom of the right ring the left ring will become aromatic making it stable. are there any theories or data indicating toward stability of this compound at room temperature? any kind of input is appreciated!

This is pentalene. It is indeed antiaromatic, and highly unstable tending to dimerize even at -100°C. With smaller rings the antiaromaticity and instability are much stronger than in the 8-pi electron system cyclooctatetraene.
There are two ways to stabilize this molecule. One is to introduce substituents that sterically hinder reactions such as tert-butyl, or that disperse the antiaromatic coupling such as cumulated benzene rings. The second method is to add two electrons, forming a dianion with the correct pi electron count to make an aromatic system.


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What is the size parameter of the direct red 80 molecule?
Basic information:
CAS number: 2610-10-08
Molecular formula: C45H26N10Na6O21S6
Molecular weight: 1373.07
Structural formula:

For example:
For congo red whose structural formula is demonstrated below, the molecule size can be expressed as ca. 2.56 × 0.73 × 1.13 nm3 (could be "length × width × height"?) (https://doi.org/10.1016/j.memsci.2021.119635)

I found a research article containing the information I need along with the parameters of Columbia blue, methyl blue, congo red, amaranth, and sunset yellow. They are in the supporting information.
https://doi.org/10.1007/s12598-022-02153-4
Hope it could help the ones in need.


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Help in balancing redox reaction
I've started balancing redox reactions, and one particular reaction has tripped me up.
$$\ce{PbS + O3 -> PbSO4 + O2}$$
$\Delta ON$ for $S =8$
$\Delta ON$ for $O =-2$
Here, $O_2$ acts as a spectator ion as its oxidation number remains $0$, and $PbSO_4$ is the RHS in both half reactions as its sulphur and oxygen get oxidised and reduced, respectively.
Writing the half reactions:
$$ \bigg( \,\, \ce{PbS -> PbSO4} \,\,\,\,\,\,\,\,\,\,\,\, \Delta ON  = 8 \,\, \bigg) \times 3$$
$$ \bigg( \,\, \ce{4O3 -> 3PbSO4} \,\,\,\,\,\,\,\,\,\,\,\, \Delta ON  = -12 \,\, \bigg) \times 2$$
Adding the two reactions:
$$\ce{3PbS + 8O3 -> 9PbSO4}$$
This is clearly wrong, as in this case, we would have to balance the $O_2$ spectator ion on the LHS, while it is clearly in the RHS in the original reaction, and upon further researching online, I found the actual balanced reaction:
$$\ce{PbS + 4O3 -> PbSO4 + 4O2}$$,
This indicating my half - reactions itself were wrong.
Upon doing some research online, it turns out this reaction utilises the n - factor concept, which I'm not very familiar with. If someone could point out where I'm going wrong in this concept, I'd be very grateful.

$\ce{O2}$ is not a spectator ion, as it is neither a ion nor a spectator. It is a reaction product.
You can formally separate the reaction to
$$\ce{4 O3 -> 4 O2 + 4 O}$$
with $\Delta \text{ON} = 0$ and
$$\ce{PbS + 4 O -> PbSO4}$$
with

$\Delta \text{ON}_\ce{S} = (+6)-(-2))= +8$
$\Delta \text{ON}_\ce{O} = 4 \cdot ((-2) - (0)) = 4 \cdot (-2) = -8$

You can effectively imagine the $\ce{O3}$ molecule as $\ce{O2}$ carrying an extra $\ce{O}$ as the oxidation agent.
Similar reaction, used for restoration of old painting where lead white ($\ce{2PbCO3·Pb(OH)2}$) was used is:
$$\ce{PbS + 4 H2O2 -> PbSO4 + 4 H2O}$$

Thank you for the answer. Could you please state the $\Delta ON$ for both individual half reactions?

This equation cannot be balanced unless one is given the exact stoichiometry and the mechanism is known. It might be reasonable to think that O2 could oxidize PbS:
PbS + 2O2 = PbSO4 and knowing 2O3 = 3O2  get PbS + 2O3 = PbSO4 + O2; Perfectly reasonable and balanced.
Were one told that one mole of PbS required 4 moles of O3 and produced 4 moles of O2 then the equation would simply be PbS + 4O3 = PbSO4 + 4O2. The reaction is not amenable to balancing by the half reaction method.
Should one remember high school chemistry that sulfide ores are roasted in air [O2] producing oxides of the metal and SO2 one might think that there was something special about ozone oxidizing sulfide to sulfate and jump directly to the second equation. There is more to chemistry than just balancing an equation. The equation is telling something!

Never thought of it that way. Thank you for your answer.


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van der Waals and the difference Cp - Cv
I would appreciate assistance with the following question please.
Show that for a gas obeying van der Waals' EOS, $$\left( p +\frac{a n^2}{V^2}\right)(V-nb) = nRT, $$
and whose internal energy is given by $$U = cT - \frac{a n^2}{V} + d,$$
that $C_p-C_v = nR(1 + (2n^2 a)/pV^2))$ where $a$, $b$, $c$, and $d$ are constants.
Clearly state any assumptions made.
My thoughts:
Expanding the vdW EOS gives $nRT = pv - nbp + an^2/V - abn^3/V^2$ and this final term can be assumed to be negligible for the purposes of this question.
From the expression for $U$,
$$
\begin{align}
C_v &= (\partial U/\partial T)_v = c\\
H &= U + pV\\
C_p &= (\partial H/\partial T)_p = (\partial U/\partial T)_p + (\partial (pv)/\partial T)_p
\end{align}$$
However, how are these results combined to give $C_p-C_v$ in the required form?

@Poutnik- yes I have put my apostrophe of possession in the wrong place in the OP, it should have been van de Waals' so thanks for pointing that out

Start with $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$ and $$dH=dU+PdV+VdP$$and show that $$C_p-C_v=T\left(\frac{\partial P}{\partial T}\right)_V\left(\frac{\partial V}{\partial T}\right)_P$$

Chet Miller, do you mind visiting this link, and answer my question? Thanks in advance


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Does nitroprusside test only work for ketones and not aldehydes?
Since the Brady test will be positive for both aldehydes and ketones, we need a way to distinguish between the two. The Tollen's test is only positive for aldehydes (and $\alpha$-hydroxyketones). I was taught in class that there's a nitroprusside test for ketones, but I'm not sure why it's specifically said to be "for ketones", and not aldehydes.
The mechanism I've been taught is this:
$$\ce{CH3COCH3 + OH- -> CH3COCH2- + H2O}$$ (the hydroxide ion is present because we add NaOH to the mixture)
$$\ce{CH3COCH2- + [Fe(CN)5NO]^2- -> [Fe(CN)5NO.CH3COCH2]^3-}$$
where the final complex is red (at least for the case of acetone shown above).
This abstraction of the acidic hydrogen using base doesn't seem unique to ketones. Does this test work for aldehydes too? If it doesn't, why not?

https://en.wikipedia.org/wiki/Urine_test_strip#Ketone_test

You know all these tests are lame and pretty much irrelevant as far as actual chemical practice goes?

Thanks for the link, but I'm not sure that it answers my question. As for the relevancy of this test to actual practice, that's a separate issue.


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Is it possible for sucrose to hydrolyse in water with a small amount of weak acid at room temperature?
Sucrose hydrolysis in the industry is done using HCl and heat. On the other hand I know that bee keepers add some of acetic  or citric acid to wet sugar to prepare emergency food supply for bees, what is called Sugar Board. After preparation the Sugar Board is kept for a month to "cure".
I'm wondering if a period of one month and weak acid are enough conditions for sucrose to hydrolyse in room temperature.
Beekeepers are doing that to achieve some results:

the acidic environment of the sugar mixture is similar to the acidity of natural nectar and honey
the acidic environment blocks the development of molds and bacterias in the hive
It is easier to create a plastic mixture when acid is added
glucose and fructose are more natural energy sources for bees.

First two points are clear, the third one comes from practice, but real question is the last one. Is it possible that crystal sucrose (the one used to sweet a tea) with weak acid will hydrolyse into glucose and fructose without heating?

Your heating only makes such reaction faster, so if you don't mind waiting a month...

I am willing to wait for a month, but can I be assured that this process will not take a year? Will hydrolysis work for crystal sugar that is not dissolved in water?


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What does it mean for a substance to be "sparingly soluble"?
I have trouble understanding what solubility means. Merriam-Webster defines it as follows:
1  the quality or state of being soluble
2 the amount of a substance that will dissolve in a given amount of another substance
For this discussion, the "other substance" of #2 will be water. I know that solubility can be represented in g/L (which can be converted in mol/L). Furthermore, when the concentrantion of a solute is equal to its solubility, the solution is said to be "saturated". Now, my first question is what it means for a substance to be sparingly soluble? From what I understand, the will always be a certain quantity of a solute, however small, such that when I put it in water it will be dissolved.

Such wording as a quantifier isn't very good. It's fuzzy and a bit open to interpretation. It is also a bit ambiguous. So I guess if you want to have a ballpark to play in: the highest concentration is somewhere around 0.1 mol/L or lower, but not as low as like nmol/L or so. But I guess different disciplines do different categories.

Highly dependent on the specific field. Few dopants in silicon are soluble much above 1 part per million (5E16/cc), and many non-dopants are soluble at even lower levels.

https://chemistry.stackexchange.com/questions/167106/why-do-we-call-salts-such-as-agcl-sparingly-soluble

https://hsm.stackexchange.com/questions/14694/when-was-the-term-sparingly-soluble-first-introduced-in-chemistry

https://chemistry.stackexchange.com/questions/70247/what-is-meant-by-fairly-when-used-as-a-quantifier-for-the-solubility-of-a-chem

Instant coffee is soluble in water and Ground coffee is sparingly soluble in water even though both solutions have about the same amount of coffee in them. The term seems to mean if one looks really hard a minute amount of material can be seen to dissolve. Another example is lead chloride PbCl2 A single crystal might just dissolve, definitely will if heated. Calcium and magnesium hydroxides are slightly or sparingly soluble. Pasta or rice are sparingly soluble in water; water is soluble in pasta and rice. It is a matter of degree and observation.


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Electrolysis of benzyl alcohol in methanol
I am conducting following electrochemical experiment on Au wires in organic medium:

I suppose the anodic reaction goes through a benzaldehyde intermediate:

The cathodic reaction is hydrogen evolution, so we are dealing with a disproportionation reaction. I suppose the hydrogen evolution comes from the methanol, so methanolate is regenerated in situ.
To confirm that this conversion is indeed possible, we wanted to take a NMR sample of out reaction solution in acetone-d6. However, as we conducted the experiment, the reaction volume turned yellow and cloudy, even in our second run of the experiment. Now our fear is that due to the high applied voltage also the phenyl rest of the benzaldehyde or the benzoic acid methyl ester gets reduced.
Does anyone have an idea what kind of side reaction could be responsible for this kind of discoloration and the reasons behind it and how it could be prevented?

Which voltage and current did you use?  On which basis did you choose your specific settings? How big is your electrode area? NMR would be a good start, also with reduced reaction time. Also from start material.

Voltage was 3V, current around 40 mA. Electrode area is pretty big, can’t calculate it rn. Gold surface oxide formation was around 1.4 V vs SHE and hydrogen evolution around -2V vs SHE on Au. That’s why the voltage was chosen at 3 V so not to form the gold oxide but maintaining high current

You are likely forcing your anode into reaction. Starting from 1.5 V oxygen mediated gold dissolution will take place. To make sure, you could weigh your electrodes before a test and then apply voltage just in the medium without starting material and see if you see the same precipitation. On any case you can compare the weight before and after. If it was become lighter, you know that it is not related to the intended reaction. In that case maybe Pt could be worth a shot. https://pubs.rsc.org/en/content/articlehtml/2013/ra/c3ra42684j#imgfig7

The source you mentioned performed this experiment in sulfuric acid though. I don’t think the gold oxide formation is responsible as if gold oxidation should happen, at the other electrode the counter hydrogen evolution needs to happen. We are in a two electrode setup, so the decomposition voltage of H+/H2//Au/AuO should be around 3.5 V, because in a CV the hydrogen evolution occurs in methanol at -2.5 V and Au oxidation at +1.4 vs SHE

Okay update, I believe the hemiacetal is formed with NaOMe [methoxy(phenyl)methanol], which remains deprotonated due to its high acidity (found on pubchem). This compound has a red color.

Your main concern is: "...the reaction volume turned yellow and cloudy.." Is this in NMR tube only or as usual in all experiments conducted?

In the experiments

The best alternate to see whether there is a benzaldehyde intermediate present in oxidation process is use benzaldehyde in place of benzyl alcohol and check the product (should have methyl benzoate as major product). Did you have done that?  If so, do you still get yellow cloudiness?

Yes I tried that, adding benzaldehyde to the methoxide, it immediately turned red and yellow upon dilution with more methanol, I guess due to acetal formation.

That’s why I didn’t pursue the electrolysis with benzaldehyde as starting material

According to the web search results, one possible side reaction that could cause the discoloration and cloudiness of your reaction solution is the oxidation of the gold anode by oxygen. This could result in the formation of gold oxide and gold chloride, which are yellow and brown compounds, respectively. These compounds could precipitate in the organic phase and make it turbid.
To prevent this side reaction, you could try to lower the applied voltage to avoid the oxygen evolution at the anode, or use a different anode material that is more resistant to oxidation, such as platinum. You could also weigh your electrodes before and after the experiment to check if there is any loss of mass due to the anode dissolution.
Reference:
(1) organic chemistry - Electrolysis of benzyl alcohol in methanol .... Electrolysis of benzyl alcohol in methanol.
(2) Selective oxidation of benzyl alcohol by two phase electrolysis using .... https://pubs.rsc.org/en/content/articlehtml/2012/cy/c2cy00424k.
(3) Ether synthesis by etherification (alkylation) - Organic Chemistry Portal. https://www.organic-chemistry.org/synthesis/C1O/ethers.shtm.
(4) Solvolysis of benzyl phenyl ether in high-temperature aqueous methanol .... https://pubs.rsc.org/en/content/articlelanding/2021/gc/d0gc04008h.
(5) en.wikipedia.org. https://en.wikipedia.org/wiki/Benzyl_alcohol.


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Question about the differences of white and grey NaH properties
We are having some problems with the next reaction in our lab. As you can see, is the sulfonylation of an indole in which we use NaH (60% dispersion in mineral oil) as a base. This reaction goes with good yields when the NaH is a white solid. But lately, we bought a new one (also 60% dispersion in mineral oil) but in this case its colour is grey. With this one, the reaction never completes and it goes with lower yield.
We bougth another one, because we thought that the other one came in bad conditions, and again, the product was grey.
We have also tried to wash the NaH with hexane and dry it carefully but nothing changed.
So my questions are: Do you know which is the difference between both of NaH? Where could we find the white one again? How can this happen? It's the same reactant but only with diferent colour.
Thank you so much for your help!!

I'm used to NaH (60% in oil) being grey and distrusted the material that was white - so not really any help.

I don't particularly see why this reaction should be problematic (though as raptorlane pointed out, NaH in DMF should be avoided), but you can always consider alternatives. For example, you could skip NaH altogether by performing an N-trimethylsilylation and reacting that with the sulfonyl chloride. It's two steps, but may end up being a cleaner and higher-yielding process overall.

Sodium hydride on mineral oil is grey. When it reacts with water, for example from air humidity it reacts to NaOH which leads to discolouration which is white. One can wash NaH  with hexane or THF to remove the oil and to increase reactivity, but unter practical consideration I did not have problems with using it directly. Pure NaH is available, but be careful when using it. Ideally this is handled only under protective atmosphere, because it is very reactive and will violently react to even traces of water. When the oil is removed, the neat hydride appears to be white. I honestly don't believe that the mineral oil is the problem for the described problem, because it is inert. More likely are:

The solvent was not dry. This reduced the amount of available NaH. Increase equivalents and or additionally dry your solvent. For example with mol sieve. You could try to use not 1 eq but 3 and see whats the result.
The atmosphere was not water free. Use nitrogen or argon with a schlenk line and dry your flasks beforehand reaction.
Did you account for the 40 Percent oil? One can easily forget this and than you have 0.6 eq instead of 1 eq in your reaction. Consider this in your calculations.
Try to increase the time for deprotonation and check if this makes a difference.
You could try to prove full deprotonation by a reaction that consumes the nucleophile fully. Check afterwards by tlc or lcms.
If possible use THF instead of DMF. I know, this is not always possible, but solvents like DMSO and DMF are not recommended with NaH, these mixtures can lead to a thermal runaway.

Explosion Hazards of Sodium Hydride in Dimethyl Sulfoxide, N,N-Dimethylformamide, and N,N-Dimethylacetamide
https://pubs.acs.org/doi/10.1021/acs.oprd.9b00276

Washing in dry pentane was my first thought, too (already visible from the freely available abstract of eEROS).
Regarding the determination of the concentration of the active reagent: in the chapter about the selection of bases, Loewenthal's Guide for the perplexed chemist recommends to determine the active concentration of $\ce{NaH}$ by titration.  So, rapidly weigh-in a sample of $\ce{NaH}$ to check in 2-propanol, intentionally add water to quench it and then titrate back with a standard solution of aqueous hydrochloric acid, all under with good swirling.
Else, if you have advanced access to a NMR spectrometer, Hoye et al. re-introduced No-D NMR spectroscopy to determine the active concentration of metal organic reagents including metal hydrides.  For greater accuracy, you might want to modify (usually increase) the $t_1$ and $t_2$ times used to collect a routine $\ce{^1H}$-NMR spectrum which sometimes works easier on elder instruments (e.g., Varian200) than e.g., Bruker's Avance DRX.  The specific times of course depend on the specific signal in the spectrum monitored.  With 16 to 32 fid collected (classical/standard probe head) you easily collect the data you need within 10 min, then integrate reference and sample signal, etc.
Hoye, T. R.; Eklov, B. M.; Ryba, T. D.; Voloshin, M.; Yao, L. J. No-D NMR (No-Deuterium Proton NMR) Spectroscopy: A Simple Yet Powerful Method for Analyzing Reaction and Reagent Solutions. Org. Lett. 2004, 6, 953–956. https://doi.org/10.1021/ol049979+.
Hoye, T. R.; Aspaas, A. W.; Eklov, B. M.; Ryba, T. D. Reaction Titration: A Convenient Method for Titering Reactive Hydride Agents (Red-Al, $\ce{LiAlH4}$, DIBALH, L-Selectride, NaH, and KH) by No-D NMR Spectroscopy. Org. Lett. 2005, 7, 2205–2208. https://doi.org/10.1021/ol0506011.
Creary, X. No-Deuterium Proton (No-D) NMR as a Convenient Method for Analysis of Organic Solvents. J. Org. Chem. 2023, 88, 11545–11551. https://doi.org/10.1021/acs.joc.3c00807 (open access).


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How does electric current flow in daniell cell?
The actual question in my mind is that ……..at anode zinc oxidises to form $\ce{Zn^{2+}}$ and $2$ electrons, but it should be the same case at cathode because copper should also oxidise to form $\ce{Cu^{2+}}$ and $2$ electrons. Then in this case the potential difference is zero hence there is no current flow……please clarify?

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

The Daniell cell is such a frequent topic on many sites involved in chemistry, including this one, that I have strong suspicion you have not done your searching homework before asking.

I have done it already but i coulnt find it

To search this site: https://chemistry.stackexchange.com/search?q=daniell+cell A sample of results: https://chemistry.stackexchange.com/questions/175183/simple-cell-vs-daniell-cell Look up also definition of cathode and anode.

I closed and then quickly reopened when I saw that my proposed duplicate is about a salt water cell with a different electrolyte. Similar principles hold though.

This is closer to what you are asking: https://chemistry.stackexchange.com/questions/126644/copper-electrode-in-a-daniel-cell

You may find useful reviewing of search results for
site:stackexchange.com OR site:libretexts.org OR site:wikipedia.org OR site:hyperphysics.phy-astr.gsu.edu  Daniell Cell schema

Physics student is wrong for two reasons. First, in the Daniell cell cathode, copper is not oxidized. It is the contrary. It is reduced from $\ce{Cu^{2+}}$ to $\ce{Cu}$, and it is reduced by the $2$ electrons produced by zinc at the anode. Second, the potential difference is not zero. It depends on the concentration of the ions in solution. But it is usually around $1$ Volt.

The copper electrode is at a different potential from the zinc electrode. Also is the Daniell cell not designed to give a well defined voltage  and its use restricted to a Wheatstone bridge environment of ZERO current draw, hence a constant voltage. This could be where the confusion is arising.


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What shold be the answer to the given reaction?

Q.75 The final product A, formed in the following reaction sequence is:
$$\ce{ Ph-CH=CH2 ->[(i) BH3 (ii) H2O2,OH- (iii) HBr][(iv) Mg, ether, then HCHO/H3O+] A }$$
Options

$\ce{Ph-CH2-CH2-CH2-OH}$
$\begin{align}\ce{Ph-&CH-CH3} \\&|
       \\ &\ce{CH3}\end{align}$
$\begin{align}\ce{Ph-&CH-CH3} \\&|
       \\ &\ce{CH2OH}\end{align}$
$\ce{Ph-CH2-CH2-CH3}$

I think 3 should be the answer as carbocation formed after HBr will rearrange to get resonance stablized.

If it would be 3) than what would be the point of hydroboration?

It is a reasonable argument to make, but the major product will be 1.
This paper details the experiment and records over 80% yields of 2-phenethyl bromide with 5-7% of the rearrangement product.


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Solubility of hydrogen
I'm interested in bacterial fermentation which produces molecular hydrogen as an end product. If bacteria produce molecular hydrogen in solution under anaerobic conditions, does this remain dissolved in the liquid media or is it likely to be released into the headspace? Would this be affected by oxygen exposure?

Molecular hydrogen has a boiling point of -253 °C. Volatile? It is a gas. And not very soluble in water.

If hydrogen $\ce{H2}$ is an end product in solution, it will soon produce gas bubbles, as the solubility of hydrogen is low : $\pu{1.5 10^{-4}}$ g in $100$ mL at room temperature. This is not affected by oxygen, as both gases are miscible at usual temperatures.

The idea of mixing O2 into H2 production is disconcerting.


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Estimation methods for temperature evolution of a high $\Delta H_{mix}$ dilution ($\ce{KOH}$ in water)
Question up front: Can temperature evolution in solutions be short-hand-estimated by simple state based calculations - with reasonable expectations of precision? (+/- 10° C)
I am trying to calculate the added hassle of ordering some 15 metric tonnes of potassium hydroxide at high concentration, so as to have a shorter lead time, less transportation and - surprisingly enough - pay less per unit of active ingredient. I can arrange slow dilution in a suitable tank, but I'd like to estimate the worst case scenario. We will be mixing $1$ m$^3$ potassium hydroxide into $12$ m$^3$ cold demineralized water, and then after a temperature reset repeat that 9-10 times with decreasing difference of concentration for each - I assume the first will be the worst. The enthalpy of mixing is worryingly large at  $-57$ $\frac{kJ}{mol}$ the need for double checking any estimation of the temperature evolution is acute.
I know there exists a complicated way of calculating a caloric estimation; if this was in a previous situation where I had access to such, I would have cheated and used the thermodynamic software. However I was wondering about the precision of doing a simple state based comparison.

state 1: given mass of $\ce{KOH}$ in a 46% concentration
state 2: given mass of $\ce{KOH}$ in a 25% concentration plus added water.

I believe I have available the $C_p$'s of the different concentrations.
$$C_{p,T1} \times m_{tot} \times T_1 \simeq C_{p,T2} \times m_{tot} \times T_2 + H_{i} \times m_{\ce{H2O,} added} + H_{mix} \times m_{\ce{KOH}}$$
With a known T1 this can then be solved for T2.
IIRC, I am now ignoring change in enthalpy based on temperature (small error). I believe I am also overforecasting due to the fact that I am not diluting to infinity, but stopping at 25%. I might also be slightly rusty in thermodynamics so my napkin-equation there might also be all wrong of course.


text:
What factors do heat of combustion of a organic compound depend on?
Let's say there are 3 organic compounds A, B, C and you know their structures.
How exactly could you go about deducing the order of their heat of combustions?
For an example,

Knowing the structures is of secondary importance since the compounds have different molecular formulas (C6H10, C5H8, and C4H6).

This problem has been thoroughly studied by Sydney W. Benson at the end of the $20$th century. He has developed a Theory of Increments for calculating the formation enthalpy of compounds of C, H and O. It states for example, that the increments are
-$42.2$ kJ/mol for $\ce{-CH3}$
-$392.1$ kJ/mol for $\ce{-COOH}$
-$312.6$ kJ/mol for $\ce{-COO-}$
-$131.8$  kJ/mol for ketones $\ce{-CO-}$.
The formation enthalpy of a given substance is the sum of its increments. The combustion enthalpies can then be calculated from the formation enthalpies with the usual method.
For acetic acid for example, its formation enthalpy is $-42.2 + (-392.1) = - 434.3$ kJ/mol.
For methyl acetate, the formation enthalpy is  $-42.2 + (-312.6) +(-42.2)$  = $-397.0$ kJ/mol.
These numerical values are taken from my copy of the original documents. Other values are given in the original publications, and also adding strains for cyclic compounds . Unfortunately, I don't have their values for rings like cyclopropane, and I have no more access to the original publications. If you can read these old documents, you will find them by checking Google, under the reference Sydney W. Benson Increments.


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Can a carbocation ever be more stable than a neutral molecule?

The question is about finding the most stable molecule or species. Now my thought process was that cyclopropenylidene(1) was a carbene so would be unstable even though it is aromatic. Cyclopenta-1,3-diene(4) is anti-aromatic so is very unstable. Between cyclopropenium(2) and cyclohexa-1,3-diene(3) though, I am confused, especially since my answer key has the former. I think (3) should be the most stable out of them all. Cyclopropenium is an especially stable carbocation, but it's still electron deficient. How could it possibly be more stable than (3) which is a neutral conjugated diene?
On another note, is it even possible to measure the stability of a cation with respect to a neutral molecule?

This is a meaningless question. Most stable with respect to what?

Reactivity I guess? Maybe the enthalpy of the species? This is a pretty common type of question in organic courses...

Unless it states what the criterion for stability is it's not even a bad question - as I say it's meaningless.

Even for a crappy jee/neet question this is quite bad. I suggest just skipping it. I know a lot of times they just ask which is more stable and leave it at that, but sometimes (co-ordination complexes for example) you can somewhat quantitatively go about things in terms of the stability constant. Here I really don't understand what they're going for as @IanBush said.

What this question is actually asking is simply "Which one of these is aromatic?", but it got redressed very badly.

Yeah it was probably about that, sometimes thinking about a question is a bad idea I guess. Thanks.

Yes this question is poorly stated, probably through no fault of the OP. (It looks like something from a textbook, in which case the authors of the book are to blame.) You can really compare only things that match up in mass, atomic composition and charge. They probably want the student to choose the cyclopropenyl cation, choice 2, because it is aromatic and the full symmetry of the ion (as opposed to choice 1 which is not fully symmetric) maximizes the effect of this aromaticity.
The cyclopropenyl cation offers a case where a salt containing the ion is more stable than a compositionally equivalent set of neutral molecules (which actually is a proper comparison); to wit, $\ce{[C3H3]+[SbCl6]^-}$ is more stable than the combination of neutral molecukes $\ce{C3H3Cl + SbCl5}$ ($\ce{C3H3Cl}$ = 3-chlorocyclopropene). When the latter two are combined at -20°C in carbon tetrachloride solvent, the salt spontaneously forms and precipitates[1]. This is considered evidence of the strong stabilization of the cyclopropenyl cation that would theoretically be predicted by the usual aromaticity rules.
Reference

Breslow, R.; Groves, J. T. (1970). "Cyclopropenyl Cation. Synthesis and Characterization". J. Am. Chem. Soc. 92 (4): 984–987. doi:10.1021/ja00707a040.

Thanks for the answer, and now I get the spirit of the question. Just a little thing I found weird though, isn't the stability of the cyclopropenyl cation a little overstated? Tropylium has stable bromides that dissolve in water with no reaction(though there is an adduct formed). It just seems weird to call that salt 'stable' when it's only stable in a non-polar solvent at -20°C. While hexa-1,3-diene is a liquid at room temperature...


text:
Hydrogen Peroxide 60% Dilution
Can anyone please help me? I have used various 'dilution' calculators' online, and done the equations on paper. However I DO NOT want to get this wrong and one  can be great at math, but help and guidance when applying is wise.
I currently have one 250ml bottle of 60% ( yes 60) Hydrogen Peroxide. I would like to make up 2ltr volume of 6% and 2ltr, 1ltr & 500ml of 3% dilution. I would be most grateful if anyone can tell me (not ratios) exactly how many mls of H2O2 to how many mls of water to achieve those dilutions and volumes.
These solutions are for household purposes and Chem was never my field at high school all the many years ago.
Also how do I safely store it in my home? (living through an NZ hot summer)
Thank you Science People

Send the peroxide for destruction!
You are clearly not equipped for this with either competence or actual equipment. You cannot store this safely in your home. It autodecomposes and the stronger and hotter the peroxide is the larger chance you have a runaway reaction that ends in  an explosion. It spontaneously ignites fires in wood and other household items. It violently decomposes if contaminated. It will cause severe injuries to skin or eyes if they are exposed to it. I would never handle hydrogen peroxide above 30% if avoiding it is an option - and I work as the site manager for a hydrogen peroxide factory! I am one of the two contact persons for accidents involving hydrogen peroxide for my country.
Furthermore, I don't know about NZ, but here in the EU hydrogen peroxide is a bomb precursor and illegal to posess without license under terrorism laws. If you were european, I would have reported this question to an antiterror hotline (in fact, I would be obliged to do so, as per agreement with the authorities). You do not want to be investigated by the police on terrorism charges.
If you cannot dispose of it in any safe way, if you somehow are reluctant to deliver it to the police; here is as safe as I can suggest with household items: Find an outside stove, fill a kettle with 5 liters of water. Bring it to a boil, and slowly add peroxide little by little to the kettle. This should not take less than an hour. It can still hurt you badly, and you are on your own as far as liabilites go.

Thank you. It was sold in a local grocer so  as a civil should be able to assume that they were legally allowed. Will investigate disposal and legality.

fire department called. and the grocer has been reported

@Constnace Must be a grocer doing very well then indeed, all those spurious fires, maimed clerks and detonations in the peroxide aisle must be costly - but it seems to still be in operation.


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Why is there a need to have H₂(g) gas even at the cathode of SHE (standard Hydrogen electrode) / Galvanic cell?
In almost all resources I have come across, I have noticed that there is an initial amount of $\ce{H2(g)}$ maintained at 1 bar, kept at the cathode of the standard hydrogen electrode, But what is the need to have this ? If the amount is 0, we get the log term of Nernst equation to be undefined. Wouldn't the reaction still occur as the reactant is still there?

As shown in this image, the $\ce{H2(g)}$ is sent to the cathode container. shouldn't it be released as a net whole?
I assume that the answer for the above would also answer why metal electrodes are used though they have no role in the reaction. (Eg: Cu and Zn solid metals are used as electrodes in the Daniell cell)

What do you mean by "cathode of SHE"? It makes impression a part of SHE is not a cathode. SHE is the cathode alongside the zinc anode. 1 atm H2 in SHE is the part of SHE definition. It plays no role elsewhere. Without it, it would be just HE at the best, or not even HE at all. // Why do you think metal electrodes play no role? Where do you think the voltage 0.76 V comes from?

Ok, please clarify this..In case we had no supply of H2 at cathode initially when we start, would the reaction still take place ? similarly in Daniel cell if we had a platinum electrode instead of Cu electrode at cathode with Cu2+ ions in cathode container, would the reaction still take place /cell Function as it usually does?

As the other answer suggests, the definition of the SHE requires operation with hydrogen gas at 1 bar pressure. In conformance with the Nernst equation, if the gas is not present at the electrode at that pressure, the cell potential will not have its standard value.
A more general form of the half-cell potential (at 298.15 K) of the hydrogen gas electrode
$$\ce{Pt,H2}(p_{\ce{H2}})|\ce{H+}(a_\ce{H+})$$
is
$$E=E^\circ -\pu{0.0591 V} \times (\frac12 \log\left(\frac{p_{\ce{H2}}}{p^\circ} \right)+\mathrm{p}H)$$
Here $p^\circ = \pu{1 bar}$ is defined as the standard pressure. The key word, in case it's not emphasized sufficiently, is standard. When $p_{\ce{H2}}=p^\circ$ and $\mathrm{p} H = 0$ (when $a_\ce{H+}=1$) the potential is $E=E^\circ_\mathrm{SHE}$. This potential is then defined as equal to zero, the standard value of the SHE ($E^\circ_\mathrm{SHE}=0$). It is important that the half-cell satisfy this condition when intended as a SHE since providing a specific half-cell voltage is central to its function as a reference electrode.
You don't have to operate a hydrogen electrode as an SHE. You can even pair up two hydrogen electrodes operating at different gas pressures and hydrogen ion activities in one cell, for instance:
$$\mathrm{SHE}||\ce{Pt,H2}(p))|\ce{H+}(a_\ce{H+})$$
where (in this example) one electrode is an SHE (by convention the left half-cell is the anode, therefore it is implied that the SHE is operated with hydrogen gas oxidized). The potential of the galvanic cell in this case is
$$\begin{align}E=&E_\mathrm{cathode}-E_\mathrm{anode}\\=&E^\circ-\pu{0.0591 V} \times (\frac12 \log\left(\frac{p_{\ce{H2}}}{p^\circ} \right)+\mathrm{p}H)-E^\circ\\=&-\pu{0.0591 V} \times (\frac12 \log\left(\frac{p_{\ce{H2}}}{p^\circ} \right)+\mathrm{p}H) \end{align}$$
The cell will operate as written (with the SHE operating as anode) when $E>0$. If we assume the hydrogen ion activity is the same in both half-cells (pH=0)  then this requires that the gas pressure in the cathode $p_{\ce{H2}}<p^\circ$ (the hydrogen gas pressure in the cathode should be less than in the anode).
Note that as written above the cathodic half-cell does not require hydrogen gas since it is a product of the reaction. However if you want the SHE to function as the cathode (swap electrodes) then
$$E=\pu{0.0591 V} \times (\frac12 \log\left(\frac{p_{\ce{H2}}}{p^\circ} \right)+\mathrm{p}H) $$
The SHE will serve as cathode if $p_{\ce{H2}}>p^\circ$ and gas must be present for anodic operation of the regular HE.
Reference
1 Skoog, West, and Holler. Fundamentals of Analytical Chemistry, 7th ed., Saunders College Publishing, 1996.

Note that SHE is intended to operate at practically currentless mode, safe the negligible input current of the voltmeter. So the reactions
$\ce{Zn(s) -> Zn^2+(aq) + 2 e-}$ on the zinc anode
$\ce{2 H+(aq) + 2 e- -> H2(aq,g)}$ in SHE
do occur, but with negligible amounts of involved substances. Therefore, if you do not have the $\pu{1 atm}$ hydrogen source for the SHE, you do not have SHE.
Considering the Daniell cell, where copper electrode is replaced by platinum one, the original expected half reactions still occur:
$$\ce{Zn(s) -> Zn^2+(aq) + 2 e-}$$
$$\ce{Cu^2+(aq) + 2 e-  -> Cu(s)}$$
with platinum getting covered by copper, effectively becoming the copper electrode with platinum core.
$\ce{H2}$ is not necessary for the reaction, but $\pu{1 atm}$ $\ce{H2}$ is necessary for SHE technical specification. As the potential of (general = non standard) hydrogen electrode depends on both pH and the partial hydrogen pressure, as Buck Thorn has written in his answer.

Thank you, got it! But a small query, so are we setting the path for the reaction to happen in a particular way by having that negligible amount of H₂. Is there any particular concept which governs this ?

I do not get what you exactly mean. Can you rephrase the question?

"Therefore, if you do not have the 1 atm
hydrogen source for the SHE, you do not have SHE."
Iam basically asking as to why that minimum amount of H2 is necessary for the reaction. And any associated factor which necessitates this condition.

Addressed in latest answer updates.

Another comment the drawing implies equal transference for the various ions, that is rarely the case. also NaCl is a poor choice for a salt bridge because of the different ion mobilities.


text:
Why can we calculate the molar mass of a non volatile solute using lowering in vapour pressure?
Apparently it is possible to calculate the molecular mass of a non volatile solute using relative lowering of vapour pressure using the equation:

Where P is the total vapour pressure of the solution, Po is the vapour pressure of pure solvent, W2 is the weight of the solute and M2 is it's molar mass.
My textbook derives this equation only using Raoult's law assuming we have a dilute solution.
But I wonder, relative lowering in vapour press is a colligative property so it should be independent of the identity of the solute yet we can calculate the identity of the solute (molar mass) using it. Maybe colligative properties are independent of the number of solute particles but they are dependent on the mass of the solute particles somehow? Is this correct?

Colligative properties depend on the ratio n(solute)/n(solvent), regardless of the solute. But n=m/M, so for given m(solute), n(solute) obviously depends on M(solute).


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My textbook writes hydronium ion as [H(H2O)]+, is this correct?

[...] Thus, it bonds to the oxygen atom of a solvent water
molecule to give trigonal pyramidal hydronium ion, $\ce{H3O^+}$ $\ce{\{[H (H2O)]^+\}}$ (see box).
In this chapter we shall use $\ce{H^+(aq)}$ and $\ce{H3O+(aq)}$ interchangeably to mean the same i.e., a
hydrated proton.

Isn’t this the way to write coordinate compounds? This may sound dumb but is the hydronium ion a complex compound?
The box that this passage refers to:

Hydronium and Hydroxyl Ions
Hydrogen ion by itself is a bare proton with very small size (~$\pu{{e-15} m}$ radius) and intense electric field, binds itself with the water molecule at one of the two available lone pairs on it giving $\ce{H3O+}$, This species has been detected in many compounds (e.g., $\ce{H3O+Cl-}$) in the solid state. In aqueous solution the hydronium ion is further hydrated to give species like $\ce{H5O2+}$, $\ce{H7O3+}$, $\ce{H9O4+}$. Similarly the hydroxyl ion is hydrated to give several ionic species like $\ce{H3O2-}$, $\ce{H5O3-}$, $\ce{H7O4-}$, etc.

Source
Chemistry Part 1, Textbook for Class XI
National Council of Educational Research and Training (https://ncert.nic.in), NCERT Publications
(version: October 2019 Ashwina 1941) https://ncert.nic.in/textbook/pdf/kech107.pdf

So... what is this question about? Notation? IUPAC rules? What is a coordination compound?

In water solution context, it is recommended and preferred to use $\ce{H+(aq)}$, as products of proton interaction with water are complex and still subject of research. $\ce{[H(H2O)]+}$ would be better to write as $\ce{[H(H2O)_n]+}$, what is nothing else than $\ce{H+(aq)}$.
An isolated, solvated hydronium is also possible in other protic solvents, where water is a minor/trace component. Hydronium is also abundant molecular ion in an interstellar space.
Solid structures with hydronium are known, like hydronium perchlorate [$\ce{H3O+}$][$\ce{ClO4-}$]. The corresponding bisulfate compound, $[\ce{H3O^+}][\ce{HSO4^-}]$, requires chilling to solidify, but even when melted the saltlike association of $\ce{H3O^+}$ and $\ce{HSO4^-}$ ions increases the viscosity and decreases the ionic conductivity of equimolar sulfuric acid-water solutions.
See hydronium and oxonium for more.

$\ce{H3O+}$ does not exist in aqueous solution outside text books
Due to the strong hydrogen bonds the species is better described as a mixture of $\ce{H5O2+}$, $\ce{H19O9+}$ or maybe $\ce{H37O18+}$ (though I just made those up). The point is that there exists a network of regular water around this lone proton that arranges itself according to the electrostatic potential of this proton. This solvation effect has very concrete implications.
To me the $\ce{[H(H2O)]+}$ notation more strongly conveys this message. There is a free proton in there, transiently coordinated by the solvent, water. The same thing happens in other hydrated chalcogenids and halogens. $\ce{HF}$ is similarily coordinated in aqueous solutions - and this shielding contributes to the weak apparent acidity of $\ce{HF}$.

"H3O+ does not exist outside of textbooks". That is beyond the pale. Salts with that ion are known (and so are other salts with additional water molecules incorporated into the cation). Only in solution do the arguments given here apply.

@OscarLanzi Fair point, the idea was in aqueous solution. I'll amend.

You'd think that the Grothuss mechanism would come naturally to an answer like this. And these "free" protons are far away from actually being free; they're just more mobile. I am trying to understand what you are saying, but I generally disagree with this interpretation to the point where I consider it wrong.


text:
Faraday's law of electrolysis and equivalent weight
I came across this question in an exam:

The mass of silver ($M(\ce{Ag}) = \pu{108 g mol-1}$) displaced by a
quantity of electricity which displaces $\pu{5600 mL}$ of $\ce{O2}$ at
STP will be _____ $\pu{g}$.

I know the calculation procedure for this type of question. My only doubt here is: can we really have one answer? Since a reaction for oxygen is not mentioned here, we cannot definitely say that its equivalent weight will be $32/4$. Oxygen could get converted to an oxide ($n = 4$), peroxide ($n = 2$), or even a superoxide ($n = 1$). So, according to me we could have $3$ possible answers to this problem.

You are overthinking it.  $\ce{2 H2O -> O2 + 4 H+ + 4 e-}$ //  I see only two options, involving water vapor or not.


text:
How can you find the Absorption of a specific compound in a solution with 2 other compounds using Uv spectrophotometer
I was wondering if there was a way to get the absorption of ascorbic acid in a solution with citric acid and disodium hydrogen phosphate?
I’m thinking of measuring the absorption of disodium hydrogen phosphate and citric acid solution and then subtracting the solution with the all three molecules to get the absorption of ascorbic acid by itself. Would this be fine? If not what can I do?


text:
How can I refine my method of determining iron content in spinach?
I have begun a chemistry experiment that attempts to determine how iron content in spinach is affected by the length of time it is cooked. Through research, I learned that iron content is often determined through a redox titration with potassium permanganate. I prepared my own spinach extract by mushing, dissolving in acid, and boiling, and titrated it, but I'm getting way larger iron levels than I should be (1 mg as opposed to 0.1). I know that this is because of other metals in the spinach which are also oxidized. How can I narrow this down to just iron levels? Is there some stoichiometry I can use to estimate it? How can I modify my experiment to produce a more reliable trend?

This does not work as oxidation by permanganate is not specific for iron(II), as many organic compounds from spinach can be oxidized too.

How should cooking change the iron content?

Note that spinach contains oxalates that bind to iron, making determination difficult. Likely, you'd want to calcine the sample, removing organic chemicals.

@Karl i plan to test the hypothesis that cooking spinach causes it to leach/lose its iron. cooking for longer periods of time may cause the spinach to lose its iron to the water, or, alternatively, it may begin to break down the oxalates and thus increase the iron content

@DrMoishePippik thanks for your advice. if calcining oxidizes the compound, how would this work with the titration? potassium permanganate is also an oxidizing agent, so if the iron is already oxidized from 2+ to 3+ by the calcification, i can't oxidize it again, right? like do i then have to re-reduce the iron somehow before titrating?

Potassium permanganate would also react with other chemicals in that mixture. See https://chemistry.stackexchange.com/questions/42866/how-to-determine-the-iron-content-in-spinach-from-a-titration-with-permanganate

You should only discard water by evaporating, during cooking. ;) Exeptions prove the rule (blanching, pasta water).

@DrMoishePippik i know that potassium permanganate reacts with organic molecules, but the experiment you linked did not use sulfuric acid. The idea in using sulfuric acid is that it dehydrates and degradates the organic molecules, freeing the trapped iron.


text:
Comparing Acidic and Basic Strength of organic compounds
If I find out the order of acidic strength of some organic compounds and simply flip them to get the order of basic strength, will it be correct?
Is there any case where this logic should be incorrect?

It depends how you apply this logic of yours, and such orders of acidic strength have serious limitations.

You'll rarely get asked about the acidity and basicity of the same compounds. See also: https://chemistry.stackexchange.com/questions/128120/strength-of-weak-acids-and-bases-is-methane-better-base-than-ammonia/128141#128141


text:
Confusion in Free Radical Reactions with alkenes
I was reading the free radical reactions, therein I saw two almost same(atleast that's what i think) but forming different products.
First, we have cyclohexene reacting with Br2 in presence of sunlight. The product it gives is 3-bromo cyclohexene. The C=C bond doesn't break homolytically
Second, We have Kharasch Effect (Anti Markovnikov Reaction), in which the C=C bond breaks homolytically .
I am kinda confused as to when we have to break the C=C bond and when we have to break C-H bond.

Wohl-Ziegler allylic bromination.


text:
Change in entropy with micelle formation
Micelles are formed only above the Kraft temperature.
We know ∆G = ∆H - T∆S (Gibbs energy relation).
For micelle formation to be spontaneous, ∆G has to be negative.
This implies ∆H - T∆S has to be negative. Since spontaneity increases with temperature, this must mean that ∆S is positive.
However this seems counter-intuitive because when micelles form, they come together. Shouldn't this actually decrease the entropy?

Very similar: https://chemistry.stackexchange.com/questions/7034/the-entropy-change-during-micellization?rq=1

You are forgetting that the micelles form in water which is polar and forms hydrogen bonds. The micelle formation is primarily entropically driven. For isolated surfactant molecules the polar head groups can stabilise water molecules around them, but the hydrocarbon chain is hydrophobic and so polar water has to bridge over this part of the molecule forming specific hydrogen bonds to do so. This reduces the water's entropy as these water molecules are more or less fixed in place. As more surfactant is added more water is forced into in this relatively low entropy state. At and above the CMC it is  entropically more favourable to form a micelle.  Lots of water molecules are now free up to take their normal behaviour, with increase in entropy, as the surfactant tails largely cluster together inside the micelle and the polar head groups are on the surface and interact with the water in just the same way as isolated surfactant molecules. The entropy decrease in forming  relatively few micelles is overwhelmed by its increase in releasing numerous water molecules.

Right. I got it, thanks a lot!


text:
Oxidation of Isopropanol by Permanganate in Basic Solution
so I've been designing an experiment where I will explore the effect of isopropanol concentration on the colour change of a solution containing permanganate, isopropanol, and NaOH.
I know that the isopropanol will be oxidized to propanone, but I'm not sure exactly what the permanganate will be reduced to?
From what I've read online, I've seen that it could either gain 1 electron and become a green manganate (VI), or it could be reduced to turn into a brown MnO2 precipitate. What exactly will the permanganate become? The sources are kind of conflicting for me.
Any help would be greatly appreciated!

It depends on pH.

Isopropanol has two successive reactions with the ion permanganate $\ce{MnO4-}$ in basic solution. First it reacts by gaining one electron and becoming manganate (VI) ion $\ce{MnO4^{2-}}$ in a reaction $$\ce{2 MnO4- + C3H7OH + 2 OH- -> 2 MnO4^{2-} + C3H6O + 2 H2O}$$ But a second (and even a third) reaction may immediately occur in solution ; $$\ce{MnO4^{2-} + C3H7OH ->  MnO2 + C3H6O + 2 OH-}$$ The third reaction is due to the instability of the ion manganate(VI), which is not chemically stable in solution (except in highly concentrated basic solution). Usually a disproportionation happens spontaneously according to : $$\ce{3 MnO4^{2-} + 2 H2O -> MnO2 + 2 MnO4-  + 4 OH-}$$ This equation can be added one, two, three or many more times to any of the two first equations cited above. It depends on the pH.
Finally there is no simple equation for explaining the reaction of an alcohol like isopropanol on permanganate ion. It depends on the importance of the third reaction, which itself depends on the pH.

I assume such oxidation is performed at very basic pH, possibly to expose RO- anions, where manganate was relatively stable.  I remember some manganometric back titration of organics and the intermediate solution with performed oxidation was  before acidification dark green.

I assume the reagent of the 2nd reaction is $\ce{C3H7OH}$ ( $\ce{(CH3)2CHOH}$ ) and not $\ce{C3H6O}$  ( $\ce{CH3COCH3}$ ). // Also see Pourbaix Mn diagram, suggesting $\ce{MnO4^2-}$ stability above pH 13.

@Poutnik. I am sorry. My second equation was wrong. I have now changed its writing and replaced propanone by isopropanol in the reagents. Fortunately, It does not change too much the whole mechanism.

The quick answer and rough rule when using potasium permanganate is that it changes oxidation number from +7 to +2 which is more favourable. It is reducing itself and acts like an oxidation agent (a very strong one). Most often, the side product would be black tar ($\ce{MnO2}$).
Anyway, in your case under basic conditions I think what Maurice said is right.


text:
How does increased polarity of O-H increase acidity? Shouldn't it reduce it as bond strength increases?
A polar bond will be stronger than a relatively less polar bond. So how does increasing polarity of O-H bond ever increase acidity. Shoudn't it make it harder to release H+ and become less acidic.

A polar bond is formed by the exchange of electrons between atoms to form a cation and an anion. Thus the bond between the cation and anion is due to electrostatic attraction. When such compounds(ionic) are dissolved in polar solvents(water, ammonia) these cations and anions exist as independent ions(hydrated ions to be precise). You can take the example of how hydrochloric acid releases protons in water. So here, as the polarity of the OH bond increases, there is a greater tendency for the oxygen and hydrogen atoms to form anion and cation respectively and hence when the compound is dissolved in water, they tend to ionize releasing protons into the solution.


text:
Is the charge spread throughout the molecule in polyatomic ions?
The following line was given in a chemistry textbook
"In case of polyatomic ions, the net charge is possessed by an ion as a whole and not by a particular atom."
I understand this in cases when that polyatomic ion has conjugated double bonds(nitrate, sulphate etc) allowing the charge to be spread throughout the molecule by means of resonance, but I fail to understand how charge could possibly be distributed in cases of ions like Ammonium(NH4+). Shouldn't the positive charge be on the Nitrogen atom alone?
This line was given under a topic discussing why formal charge isn't true charge.

N in NH3 significantly pulls electron density to itself, making H atoms partially positively charge. It happens even more in NH4+. N would not keep positive charge, if electrons can be pulled from N-H bonds.

Okay, but what about organic anions like acetate? The negative charge is shared only among the two oxygens, but not throughout the whole molecule.

Ion structures are different. The textbook notes were meant for more or less symmetric ions like ClO4-, NO3-, SO4^2- or NH4-. It obviously does not apply on big molecules with scattered charge groups like -OOC-(CH2)n-COO-,  proteins or ion exchanging resins.

Understood, Thanks!

You're not considering "inductive effects". They spread charge between neighbouring atoms. Mesomeric have longer range.


text:
Aromaticity of a fused ring compound

On a test, I marked this as aromatic, but the answer key says it's wrong.
My thought process:

It has 10pi electrons, so it follows Huckel's rule
All carbons are sp2 hybridized
It is cyclic
Every atom is conjugated

Everything points towards it being aromatic, is there something I'm missing?
Is there some sort of rule that says each individual ring must be aromatic for the whole molecule to be aromatic?
Any help would be greatly appreciated.

This is a carbene not carbanion, it changes a lot.

Because all of the hydrogens are implied at all other carbons not bearing the electron pair, it is fair to say that there is a hydrogen attached to the carbon in question. In which case the carbanion would be aromatic.

The electron pair on the right is meant to be outward pointing in the plane of the atoms, not part of the pi system. If you had the electrons in the pi system, then the carbon would be attached to a hydrogen atom and would have a negative formal charge; we would write this as $\ominus$ rather than as the electron pair.
So there are only eight conjugated pi electrons, and you can work from there.

Are there only eight electrons in the pie though? I don't think this is as easy as this. If there are 8 electrons in pi orbitals, this would make it anti-aromatic (in weird cases of thinking and extending Hückel's rules beyond the useful). But that couldn't be the ground state. So, there is probably a benzene ring with "intact aromaticity" and a fused second ring with a triplet carbene, which has some sort of delocalisation, 3c-3e bonding or so. There are means to look at this more thoroughly, but they don't fit a textbook narrative. It's a shame we don't have a citation for the exercise…

@Martin you may be right, but I am rolling with the punch. I am guessing the OP is supposed to be applying the 4n+2 rule for aromaticity however inaccurate that may be for a polycycle. In any event your model could also be called antiaromatic in the sense that it spontaneously localizes groups of pi electrons  rather than allowing a full delocalization.


text:
Alternative pH sensing methods?
I am looking for alternative methods to measure the pH of water (swimming pools, tap water, rivers,...) with (ideally) the following properties:

Durable: no maintenance or calibration required for 5-10 years.
Accurate: 0.1 pH or better.
In-line measurement.
Expected pH measurement range: 5 - 10

The 'durable' aspect rules out the traditional pH probe as it needs to be calibrated and replaced quite often.
So far most of the papers I have read use an electrochemical cell and a measurement technique such as cyclic voltammetry, square-wave voltammetry, electrochemical impedance spectroscopy,... which has the disadvantage of needing a reference electrode. A calomel electrode (which should be more durable) is not an option since it contains mercury and an AgCl electrode would also require calibration, maintenance or replacement I believe. For a working electrode they use all kinds of materials and coatings such as ITO (indium tin oxide), glassy carbon,...
I did find one commercially available product which seems to be promising: https://www.anbsensors.com/
They seem to use an entirely solid state sensor, using Nafion coated flavanone carbon
composite electrodes with square wave voltammetry. They claim no calibration is required and maintenance (abrasion) every 15.000 measurements.
I also looked into ISFET's (ion sensitive field effect transistor), they also seem to have some drawbacks such as drift, still needing a reference electrode, limited availability,...
Would determining the ratio between bicarbonate ions and carbonate ions / carbonic acid be a feasible option?
In short, does anyone have any pointers or recommendations on topics or measurement techniques to research further?
Sidenote: I have a background in electronics engineering and a basic understanding of electrochemistry.

You might face a great challenge in the durability requirement. There are many chemicals in pools, tap water and rivers that might eventually coat the electrodes (e.g., iron and manganese oxides), or destroy them (e.g., free chlorine). There are even bacteria and archaea that can build a biofilm on surfaces, often using the aforementioned metal ions for energy, and coating the surface with their oxides, too. Perhaps a mechanical device could be used to remove and wipe the electrodes between measurements.

@DrMoishePippik Yes, oxidation, biofilms and such are indeed a major concern, but there are also solutions to mitigate this problem such as a mechanical wiper, ultrasonic cleaning, a coating on the electrode surface etc.