text:
Mass Spectrometry Resolving Power Normalization to Average Mass
The resolving power in MS is often defined as $m / \Delta m$, or the reciprocal of the smallest discernible mass difference $\Delta m$ normalized to average mass $m$.
I wonder why the normalization to the average mass $m$ makes sense. Is there a physical reason, or a mathematical or sociologic reason? Is 1 at 100 really same as 2 at 200, does this resolving power scale like that or is that just copied from optics? In optics I would accept that resolution depends on wavelength, but how so for mass?

Not likely sociologic, whatever that means. This is not my field but isn't the resolution just $\Delta m$? Guess not!: https://en.wikipedia.org/wiki/Resolution_(mass_spectrometry)

Related: https://chemistry.stackexchange.com/questions/314/resolution-in-mass-spectrometry?rq=1

$m / \Delta m$, or the reciprocal of the smallest discernible mass
difference $\Delta m$ normalized to average mass $m$.

Good question because the resolution in mass spectrometry and chromatography have been misused and there is an element of inadequacy if we look deeply. I don't know where you got this definition, but whoever wrote it has the notion of $m$ wrong. It is not normalized to average mass. It is the single mass of your analyte or molecule of interest. There is an entire 2022 article in a leading journal dedicated to your query.
Resolution and Resolving Power in Mass Spectrometry
Kermit K. Murray in J. Am. Soc. Mass Spectrom. 2022, 33, 12
The proper definition is as follows, in case there is a problem in accessing it.

For a single peak made up of singly charged ions at mass m in a mass spectrum, the resolution may be expressed as m/Δm, where Δm is the width of the peak at a height which is a specified fraction of the maximum peak height. It is recommended that one of three values 50%, 5%, or 0.5% should always be used. For an isolated symmetrical peak recorded with a system which is linear in the range between 5% and 10% levels of the peak, the 5% peak width definition is technically equivalent to the 10% valley definition. A common standard is the definition of resolution based upon Δm being Full Width of the peak at Half its Maximum height, sometimes abbreviated "fwhm". This acronym should preferably be defined the first time it is used.

Nowhere it says, average mass! $m$ is a single mass of your interest, and its width is $\Delta m$.

Thanks so much! However, the main question is how the normalization makes sense.. How would you interpret the choice of the definition? I understand why a resolution would use the reciprocal of the smallest discernible difference. But why divide by the mass?

Welcome, that is just a human definition of resolution. There is no right or wrong.


text:
Does the electric field of a battery exist before a wire is connected outside?
Imagine a galavanic cell.
What I am wondering is, what causes the inital oxidation of the less nobel metal to occur initally. Is the electric field there before you connect a wire, which would allow electrons to move through the outside wire?
If not, what does the wire change on the physical level?

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

Yes, there is an electric field around a disconnected cell -- an electrostatic field, just as if you ran a comb through your hair and p;laced it near an electroscope.
In fact, one of the longest-running experiment is the Clarendon Bell at Oxford University, running since 1840, relying on that field.

Electricity can flow even if there is no direct current path, through capacitance. When a charged object is placed near another object, electrostatic force induces a charge in the other object.
Think of it as a verrrrry low frequency electromagnetic wave. If one takes ten seconds to move the objects, the fundamental frequency is 0.1 Hz. Note that such rdiation can extend to X-rays and gamma rays, as when a fast-moving electron is stopped abruptly: bremsstrahlung.

What I write here may be opinion based, please be patient.
Inside the bulk metal, Ohm's law always applies, involving disconnected cell.
Ohm's law is expressed as $\mathbf{J}=\sigma\mathbf{E}$, where $\mathbf{J}$ is the electric current density, $\sigma$ is the electric conductivity and $\mathbf{E}$ is the electric field.
Inside the bulk metal of the disconnected cell, no electric current density is experimentally observed; this means $\mathbf{J}=0$.
Since $\sigma$ is positive, the total electric field must be zero ($\mathbf{E}=0$) in this disconnected cell case.
When the circuit is connected and closed, we observe non-zero electric current denisty, which means that the electric field is non-zero for such a closed circuit.
This discussion does not mean that the electric field caused by the electric scalar potential ($\mathbf{E}=-\nabla\phi$)
is zero in the case of an open circuit. One source of the electric field is the scalar potential $\phi$ and another source of the electric field comes from the time variation of the vector potential $\mathbf{A}$. That is, in the mathematical expression $\mathbf{E}=-\nabla\phi-\frac{d}{dt}\mathbf{A}$.
Unfortunately, the role of the electric field coming from the vector potential is not clarified even in 2024.

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text:
Do electrons really hang around the electrode while ions go take a swim?
This answer to this post addresses "what happens if you dip a single zinc electrode into some electrolyte solution." It explains that

However, though the zinc ion can diffuse through the solution, there is nowhere for the two electrons to go, so they are trapped within the electrode.

This seems very counter-intuitive to me. The word "can" seems important. Perhaps it is a point of minor importance that the author did not think through. In the absence of an applied external potential can or does immersing metallic zinc in water result in the surface retaining electrons while ions diffuse into the water? Why are neutral zinc metal atoms insoluble in water, or are they soluble, just very reactive? The questions are partly addressed in comments below the answer and links to articles are provided, but this seems worthy of a separate post (it may have been asked already I realize).
What actually happens to zinc when immersed in pure water?
Edit: assume the initial conditions as anaerobic pure water (no electrolyte), room temperature and pressure.

Have a look at Gouy-Chapman-Stern (and related) theories/methods for interface at metals/solutions, in short there is a structured area which is diffusion-limited for ions near the surface and then a continuum as some distance thereafter. Shells or planes can be defined, I haven't time to elaborate but your answer might be found with some targeted Googling.

Thanks @ToddMinehardt There's a lot happening from the moment you dip the metal in water. My question is perhaps like asking what  happened seconds after the Big Bang.

This study describes what happens in the presence of air and electrolyte over time: https://www.sciencedirect.com/science/article/pii/S2452199X18300240

I have a related question about self-diffusion of metal atoms/ions at this interface: Let's say we have the natural abundance of Zinc isotopes in the solution in the form of a soluble salt. The metal however is 99% enriched with the 68Zn isotope, which usually has a natural abundance of 4%. If we let the metal rest in the solution for a long time, are we going to see an exchange of the metal 68Zinc atoms and solution Zinc ions? So basically the self-diffusion of neutral 68Zinc and ionized Zinc in the solution? Or should there be a huge energy barrier for this kind of self-diffusion?

@Timo If you have a question please create a new post rather than post the question as an answer to an existing question. The site does not function as a forum. Each page corresponds to one question and potentially multiple answers to the same question.

If you have access to chat on the site then you can instead visit the common Periodic Table room and discuss this there. All the same: I think the barrier will be too high for a significant alteration in composition of the bulk metal; the Zn will be consumed into solvated ions and I doubt the reverse reaction will gain traction until very close to equilibrium. See one of the answers for a computation of how much is expected to be consumed. But you propose an interesting approach to study the system.

Zn in solution will all be ionized, I expect. You could monitor the concentration of the different isotopes to follow the rate of consumption of the metal.

We can expect one of three things to happen:

The electrons also diffuse out from the electrode, becoming solvated electrons.

If the electrons cannot diffuse, their negative charge creates an electric field that opposes the diffusion of the zinc ions, so the latter are reined in.

The electrons react with the water to evolve hydrogen and leave behind hydroxide ions, which then do (1) or (2) plus maybe undergoing secondary reactions with the zinc ions (precipitation, complexation).

With zinc in water we don't see evidence of (1) or (3), but (2) requires no visible change and thus is an experimentally plausible option. Magnesium just barely, and alkali and heavier alkaline earth metals more prominently, give (3). Alkali metals in ammonia solvent readily give (1) combined slowly with (3).

Thanks, this begins to address the mechanistic problem. Do you have suggestions for supporting references? After spending some time googling it seems that eventually water will react with Zn to form passivating oxide and release H2. I suspect Zn never leaves the electrodes.

If a zinc plate is dipped into pure water, maybe a small number of zinc atoms produce $\ce{Zn^{2+}}$ ions while the corresponding electrons are staying back in the metal. But this is only valid for the first few zinc atoms. When additional zinc atoms will do the same, they are prevented by the presence of positive $\ce{Zn^{2+}}$ in water and the negative charges in the metal. So additional zinc atoms will not be ionized and dissolved. So few ions are generated that the concentration of zinc ions in solution is totally negligible. But it is sufficient to prevent further ionization of zinc atoms plus dissolution of the corresponding $\ce{Zn^{2+}}$ ions in the future.

So is it about the interionic forces in water?

2 electrons would not make so big difference. Consider the capacitance of the electrode and the equation dE = dq/C. Very roughly some 10 millions of electrons just  for Delta E = 1 V for C = 1 pF. More at a much higher capacitance. Finally, potential gets so low electrons start to get hydrated and quickly reducing water or  H+(aq) to hydrogen.

@Poutnik I guess that explains the non-reactivity of Zn towards cold/hot water and reactivity only towards steam?

Golden flat surface electrode reportedly exhibits capacitance $\pu{10  \mu F cm-2}$.

@Poutnik. I expected your answer. I perfectly know that one and only one atom cannot be ionized and dissolved. But it was an image. I think instead of one, I should have said "the first millions". But this amount is not known with accuracy. And it would start a long series of remarks about its real values, which is not important for the present problem. It is better to choose a simple number like "one", even if it is probably not one.

But it is severely misleading, and without much justification. "Many" is enough.

@Maurice The intent might have been to be vague, but "the second zinc atom" bespeaks a certain accuracy. I've taken the liberty of making edits to generalize and "unspecify" your answer. Hopefully the edits align with your intent.

Outsider trying to understand: is the gist that, as more zinc atom(s) ionize, the consequent increasing charge on the electrode makes it more difficult for subsequent zinc atoms to ionize and dissolve (where an equilibrium is reached with very few atoms having ionized, relative to the quantity of Zinc and pure water)?

Yes, Doktor J. What you say is exactly what I have tried to explain my self. without too much success, alas. Thank you.

This a question that offers an answer with electrochemical equilibrium. Our experiment is as follows:

We have a typical beaker of volume $V = \pu{200 mL}$ of water. On top of it we have a volume $V_\mathrm{a} = \pu{1 dm3}$ of air, at ambient conditions of $p = \pu{1 bar}$ and $T = \pu{25 °C}$.
We add a mass $m = \pu{1 g}$ of zinc.
Determine the final state after a long period of time.

First attempt
One of the most important issues with these type of problems is to determine which types and how many reactions will take place. Let's consider for simplicity that only the oxidation of zinc takes place:
$$\ce{Zn(s) -> Zn^2+(aq) + 2e^- \tag{1}}$$
However, if only this happens we have the following problems:

The electrolyte solution becomes positively charged with $\ce{Zn^2+}$ without any counterbalance of anions. This is a violation of charge conservation taking place at the macroscopic scale in the liquid phase.
The solid zinc becomes negatively charge with $\ce{e^-}$. This is a violation of charge conservation taking place at the macroscopic scale in the electronic phase.

Thus, we reject that only Eq. (1) takes place.
A more rapid way to show that this is impossible is the known phrase taught at at general chemistry: 'an oxidation needs a reduction or vice versa', so we cannot have a single oxidation.

Second attempt
We know that self-ionisation of pure water brings negligible but some amount of protons and hydroxides to the solution. This raises the interesting question: can the two electrons from Eq. (1) be grabbed by the hydrogen ions, and produce hydrogen gas? This situation will not violate any law of charge conservation, so we see whether this is possible by thermodynamics.
The system has two electrochemical reactions and one chemical reaction
\begin{align}
  \ce{Zn(s)          &-> Zn^2+(aq) + 2e^-}  \tag2 \\
  \ce{2H+(aq) + 2e^- &-> H2(g)}             \tag3 \\
  \ce{H2O(l)         &-> H+(aq) + OH^-(aq)} \tag4 \\
\end{align}
Eqs. (2) and (3) lead to two electrochemical equilibrium, while Eq. (4) to a chemical equilibrium
\begin{align}
  E &= E^\circ(\ce{Zn^2+/Zn}) - \frac{RT}{2F}\ln ([\ce{Zn^2+}]) \tag5 \\
  E &= E^\circ(\ce{H+/H2}) - \frac{RT}{2F}\ln
  \left(\frac{p_\ce{H2}}{[\ce{H+}]^2}\right)
  \tag6 \\
  K_\mathrm{w} &= [\ce{H+}][\ce{OH-}] \tag7
\end{align}
and we count 5 unknowns: $E$, $[\ce{Zn^2+}]$, $p_\ce{H2}$, $[\ce{H+}]$, and $[\ce{OH-}]$. We need two more equations. The first one is charge conservation
\begin{align}
  [\ce{H+}] + 2[\ce{Zn^2+}] &= [\ce{OH^-}] \tag8 \\
\end{align}
The other one is a relation between the partial pressure of hydrogen and other species. Let $\xi_2$ and $\xi_3$ be the extents of reactions 2 and 3. Then, the amount of hydrogen, protons, and hydroxides in terms of these extents are
\begin{align}
  n_\ce{H2}  &= \xi_2 \tag{9}  \\
  n_\ce{H+}  &= n_\ce{H+}^0 - 2\xi_2 + \xi_3 \tag{10} \\
  n_\ce{OH-} &= n_\ce{OH-}^0 + \xi_3 \tag{11}
\end{align}
By Eq. (11) we can write $\xi_3 = n_\ce{OH-} - n_\ce{OH-}^0$. Introducing this expression and Eq. (9) in Eq. (10) gives
\begin{align}
  n_\ce{H+}  &= n_\ce{H+}^0 + 2n_\ce{H2} + n_\ce{OH-} - n_\ce{OH-}^0 \\
  2n_\ce{H2} &= n_\ce{H+}^0 -  n_\ce{H+} + n_\ce{OH-} - n_\ce{OH-}^0 \\
  2n_\ce{H2} &= ([\ce{H+}]_0 - [\ce{H+}] + [\ce{OH-}] - [\ce{OH-}]_0)V \\
  p_\ce{H2} &= ([\ce{H+}]_0 - [\ce{H+}] + [\ce{OH-}] - [\ce{OH-}]_0)
  V\left(\frac{RT}{2V_\mathrm{a}}\right) \tag{12} \\
\end{align}
and we note that the volume $V$ refers to the species in the liquid phase, which is not the same as the volume $V_\mathrm{a}$ for the gas phase.
And we have a set of 5 non-linear equations for the system, which we repeat again
\begin{align}
  E &= E^\circ(\ce{Zn^2+/Zn}) - \frac{RT}{2F}\ln ([\ce{Zn^2+}]) \tag{13} \\
  E &= E^\circ(\ce{H+/H2}) - \frac{RT}{2F}\ln
  \left(\frac{p_\ce{H2}}{[\ce{H+}]^2}\right) \tag{14} \\
  K_\mathrm{w} &= [\ce{H+}][\ce{OH-}] \tag{15} \\
  [\ce{H+}] + 2[\ce{Zn^2+}] &= [\ce{OH^-}] \tag{16} \\
  p_\ce{H2} &= ([\ce{H+}]_0 - [\ce{H+}] + [\ce{OH-}] - [\ce{OH-}]_0)
  V\left(\frac{RT}{2V_\mathrm{a}}\right) \tag{17} \\
\end{align}
It is possible to combine this equations to a single unknown, but this will unecessary extend the answer. We tabulate the solutions:
\begin{array}{|c|c|}
  [\ce{H+}] \; \pu{(mol dm^-3)} & \pu{9.0632E-13} \\
  [\ce{OH-}] \; \pu{(mol dm^-3)} & \pu{5.5168E-2} \\
  [\ce{Zn^2+}] \; \pu{(mol dm^-3)} & \pu{1.1034E-3} \\
  p_\ce{H2} \; \pu{(bar)} & \pu{0.27352} \\
  E \; (\pu{V}) & \pu{-0.69598} \\
\end{array}
The results indicate:

Indeed, the hydrogen ions were consumed and a bit of zinc entered the solution.
The electrode potential $E = \phi^\mathrm{M} - \phi^\mathrm{S} < 0$, thus the electric potential of the liquid phase became more negative.

By mass conservation, we can finally see how much the mass of zinc solution changed (remember we put $\pu{1 g}$ initially)
\begin{equation}
  m_\ce{Zn} = m_\ce{Zn}^0 - [\ce{Zn^2+}]VM(\ce{Zn}) \to
  \boxed{m_\ce{Zn} = \pu{0.99278 g}}
\end{equation}
where $M(\ce{Zn}) = \pu{65.41 g mol^-1}$ is the molar mass of zinc.

It's good that you kept oxygen out of the analysis. I should have been clearer and stated the initial conditions as anaerobic, RT and pressure.

@BuckThorn The system can get more complicated as you wish, as more reactions keep being added... Oxygen may enter the liquid phase and react with water or $\ce{H+}$, which I did not consider...

Yes, and oxide can eventually form, at least according to a reference I placed in comments under the question. But that describes an experiment in the presence of electrolyte exposed to air.

I will read it later. However, you were interested in the case of introducing zinc in pure water, right?

Yes, pure water (no salt). I could not find an article that addresses that problem.

Observe a galvanized garbage can exposed for several years the zinc definitely reacts. Pure water is much slower.

@jimchmst Hi. For sure, I just wanted to so some calculations to arrive at a rough number of what would happen.

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The sign of the lobes in an atomic orbital
In atomic orbital diagram of an atomic orbital, the lobes are represented as either positive or negative. It is said that the wave function is 'positive' where the lobe is depicted positive and 'negative' where it is depicted negative. But what I am confused about is that when electrons are paired in an atomic orbital, is one the electrons in the positive region and the other in the negative? And if this is the case, what is the exact configuration of the electron pair in the 1s orbital where there isn't a negative lobe? And what about the case where there are multiple positive and negative lobes like the d-orbitals, do the 'positive electrons' move between the two positive lobes and vice versa for 'negative electron'?

Forgive me for my poor terminology as I am not used to treating electrons as waves and I do not know the mathematics of the Schrodinger equation.

The sign of the wave function has no relation to the probable location of a particular electron in the orbital.


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What is the full mechanism for the controlled reduction of an alkyne using LiAlH₄?

Was going through my notes and realised I didn't completely fill them in. The first few steps I think are quite self-explanatory but I'm really confused about the last step where I go from the five-membered ring to the final product by adding water.
I'd be really grateful if someone could clear this up or provide any suggestions!

Does this related reduction help?


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Where are the electrons located when two atomic orbitals overlap? (VBT)
According to the valence bond theory(VBT), a covalent bond is formed by the overlapping of atomic orbitals containing electrons of opposite spin which pair together. However what is the physical meaning of overlapping here? Does it mean that the electrons of either orbital are now constrained to move in the region where the orbitals overlap?
Note: I know that in the case of MO theory the atomic orbitals just combine linearly(atleast approximately) to give molecular orbitals but I wonder what is the explanation that VBT provides.

Orbitals have fuzzy boundaries so the extent of their overlapping is a matter of taste in the first place.
In the context of this linear combination approach electrons are free to use the whole combined "volume" of the molecular orbital.

No, you misunderstand my question. I am asking for the explanation which the Valence Bond Theory provides for the electronic behavior in an orbital overlap, not the molecular orbital theory. Besides in MO theory, the orbitals don't overlap, they combine.

Well, then your understanding is OK - while the electrons are not "constrained" anywhere, their density is highest where the orbitals overlap.


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TMSBr and 2,4,6 Collidin in Acetonitrile leads to white precipitate
I want to hydrolyse a phosphonic acid  ester with TMSBr.

I think this reaction works only with the addition of base. Without base, the product in my specific case is prone to deterioation due to acid sensitivity. In the literature I found, that Collidin is used here, which is the reason why I used this base as well. I used ACN as solvent. During the addition of base and than TMSBr to the educt I immediately observed the precipitation of a white solid. UPLC indicated, that at that time the product was not yet formed. I am wondering if white precipitate might indicate the presence of water due to formation of the Collidin-Br salt? Or is there any other explanation? The ACN was dried before over mol sieve.

Did UPLC show the starting material gone?

Nope, still intact. When it is the Collidin br salt, than the Acetonitrile must have been very wet, but it has been dried over mol sieve....

Waylander BTW This is the acid we were discussing already for some time. It is easier to to do it this way than attaching a benzyl group which did not work well, the product was not formed in big quantities. I did not check however the silayting approach yet, but I would prefer this route, because the diethylester is formed before the monofluorphosphonic acid.

Did you dry your collidine? It is known to be hygroscopic

No, used straight from the bottle (without sept). But it is fairly new batch, just couple of months old.

It is worth putting some activated sieves in the collidine and adding a septum

I tried mixing Collidin and TMSBr in ACN and chloroform - no precipitation. It has to be related to my compound. I actually saw my wished product in lcms and NMR but contaminated. This is a progress however, because now I know for what mass signals I need to look for. The problem in this methodology is, traces of acids during the workup may lead to product deterioation, I observed this yesterday when I evaporated the excess TMSBr and the addition of Water led to acidic pH unfortunately. Also, during reaction controll I see my mass, but the signal seems to be accompanied by a lot of others.


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Can we replace surface stabilizers of CuS nanoparticles with a solid matrix?
Is there a method to simultaneously remove surface stabilizer molecules from $\ce{CuS}$ (copper monosulfide) nanoparticles and form inclusions of the nanoparticles into a solid crystal growing from a liquid suspension (aqueous?) of the particles in a saturated solution of a material forming the matrix as it precipitates?
If anybody happens to know several methods, the one that achieves the highest number density of inclusions without clumping of $\ce{CuS}$ particles is preferable.


text:
Why does solution turn yellow at higher temperatures in potassium iodate-sodium metabisulfite variation?
When I did the potassium iodide-sodium thiosulfate experiment, the reaction mixture did not turn yellow at higher temperatures. For example, it was blue-black at 50 degrees celsius, however, for iodate variation, it turned yellow at higher temperatures, like 50 degree. I do not understand why this happens.
According to what I understand starch must decompose, however, why doesn't it happen for iodide-thiosulfate?
To clarify the doubts
This is iodine clock reaction and both of them are variations of this experiment. Firstly, diluted potassium iodide reacts with acidified sodium thiosulfate with starch (starch acts as an indicator).  Solution turns blue-black when reaction is complete. For the iodate variation, diluted potassium iodate reacts with acidifed sodium metabisulfate with starch (which acts as an indicator). And, I am trying to understand why it turns yellow instead of blue-black for the iodate variation at higher temperatures?

Please give us more information. What is this potassium iodide-sodium thiosulfate experiment ? Iodide should not react with thiosulfate. What is the composition of the reaction mixture ? What do you mean when stating that "starch must decompose" ? Is there starch in this mixture ? Any other component ?  Why do you say that there is a "iodate variation" ? Iodate variation ? So it is not an experiment iodide + thiosulfate as informed in the first sentence.  It is a mixture of four instead of two components : iodide + iodate + starch + thiosulfate ? Is the mixture acidic or basic ? Explain !

@Maurice There is no space before question, or exclamation, mark... BTW why should OP explain every tidbit if iodine clock reaction is well known?

This appears to be a duplicate of https://chemistry.stackexchange.com/questions/181480/why-does-iodate-and-acidified-metabisulfite-solution-turn-yellow-instead-of-blue, and one should be deleted.

Hello Maurice! Both of them are separate experiments. I have edited the question and I hope that it is more clear now :)


text:
Removing carbon matter from molten beeswax
In my candle manufacturing process, I melt down used candle stubs, filter the molten wax (heated to ~ $\pu{75^\circ C}$) and use this to create new candles (in addition to fresh beeswax). I'm currently using oil filter paper to gravity filtrate the molten wax, this does a good job of removing the larger carbon particles but doesn't remove all carbon matter.
Is there a chemical process that I can use to perhaps make the carbon elements heavier, to allow them to sink to the bottom, and harvest the cleaner molten wax from the top?
Edit:
http://www.dave-cushman.net/bee/waxdecont.html talks about melting wax in water just below boiling and adding hydrogen peroxide. Another filtering option mentioned in the article is dissolving beeswax in carbon tetrachloride (CCl4) which is meant to result in the liquid flowing more readily through a filter. If these are realistic options, guidance would be appreciated on the approach / amounts to use with these approaches.

I have tried coffee filter paper, besides it taking much longer to filter, it made no notable difference.
I have experimented with leaving the molten wax for longer periods of time undisturbed, and scooping from the top. This worked well if the stubs were left to melt undisturbed until molten, and scooped from the top. It made no difference if the wick/soot was disturbed during the melting process or when processing the 'dregs' at the bottom that were full of soot. See photo of the difference. Although I'm able to process a significant amount of stubs without disturbing the soot, I still need a process to clean this soot stained beeswax.

Both samples originate from the same shade of candles. I can confirm this is soot stained beeswax (on the left), and not overheated beeswax (which discolors when exposed to high temperatures). Color of the beeswax on the right is of satisfactory shade, and very close to the original color of both samples.

Try finer filter paper. Coffee filter paper is inexpensive, and might do a better job of trapping particles. However, colored candles would still have dissolved dyes in the wax that cannot be eliminated easily.

Thanks @DrMoishePippik - I gave coffee filter paper a go, no notable difference (besides taking much longer to filter). Thank you for the suggestion!

You should check the T.V. program "Backyard Science" to get ideas of the project, It might help you.

Hydrogen peroxide as suggested will possible just bleach the particles, not remove them. I think also, that the miscibility with an aqueous solution and wax may be a problem here. There are probably other bleaching agents that could be used. Be aware that this agent may lead to oxidation of metal pots. CCl4 (carbon tetrachloride) is cancerogenic and not advisable here.
From my point of view, melting the wax and let it flow through a filter medium may not be the optimal solution, because the fluidity will at some point decrease and the wax may have problems passing through the filter when it goes back to solid form.
However, pillow cloths may work well as a filter media when you have a sufficient big amount that will not solidify before going through the filter (bigger amount equals a bigger heat capacity).
https://m.youtube.com/watch?v=-1pc8lMrsYM&pp
In addition I found a guide, were the wax is put in a mesh, and this mesh is heated in water. The wax will liquidity and the solid contaminations will stay in the cloth. You could perform this cleaning multiple times with finer meshes to achieve better cleaning.

https://backyardbeekeeping.iamcountryside.com/honey-beeswax/steps-for-filtering-beeswax/
If you have access to a centrifuge you could try to melt the bee wax and centrifuge the molten wax. The particles should sink to the bottom if they have higher density.
Alternatively one could try to use a "not-so toxic" solvent like acetone, ethyl acetate or ether (all of them are flammable, ether is very volatile so proceed with caution and avoid open flame) to dissolve the bee wax and then filter the liquid. The solution would have less viscosity and will flow much readily through a filter membrane or a fine glas frit, or alternatively a filter bed filed with fine sand, sealite etc. Using this technique you are probably able to filter with a much smaller pore size. I would advice against CCl4, benzene or dichloromethane, because they are cancerogenic and toxic and not safely handled outside a lab setting with proper fumehoods and air exchange system.

centrifuge the molten wax - Ohh, I'll see if I can do a small scale test of this!

I'm not sure how to induce agglomeration of the burnt particles to speed up their sedimentation.
I'd try one or both of the following:

Maintain the wax in its molten state longer without stirring, then scrape off the particulate-rich layer.
Melt the wax in hot water, hoping the particles selectively disperse into the water rather than staying in the wax.

The carbon particles should be more dense, a possibility would be to set up a vertical zone melting setup to slowly melt a portion of the wax and allow the carbon particles to remain in the melt.

The dark color in beeswax is likely due to two distinct materials: #1. soot (i.e., particulate matter, including partially burnt wicks) and #2. dissolved matter, essentially dark colored molecules (of undetermined, but very large molecular weight).
Removal of the larger particulates by filtration may not improve the color sufficiently. Removal of the residual very small particles and high molecular weight materials could be improved by using a filter bed of granular carbon/charcoal to adsorb them.
Selection of the filter material will be a time-consuming experimental journey. There are many carbons on the internet. Some things to consider:

Activated carbon for aquariums is likely to be tiny-particle-free, and colorless

You could wash the carbon to remove color by using water, alcohol, or petroleum solvents. Dry the carbon before using it with the beeswax.

Large particles could be crushed to give more surface area for adsorbing the beeswax impurities, but might produce tiny particles that go thru the filter

A long tube of coarse particles could be equivalent to a fine filter paper, but would need only a coarse screen to hold the carbon (i.e., a chromatography column or column filtration).

The chromatography column shown would allow you to discard the darkened filter material, but if you are using black carbon to remove dark colored material (ironic, isn't it?), this will not be possible. Just keep using the same column till the output beeswax isn't light enough, then toss the column and continue with a new batch of carbon filter material.

I made a comment about zone purification that seems to have been ignored. If the fine particles are actually denser and do settle with gravity a vertical zone purification setup might hasten the separation. It could be tested on a small scale using a glass or plastic cylinder and a heating tape to see if the particles settled and were aided by the progressive melting and solidification of the wax. It is explained exhaustingly in Wikipedia https://en.wikipedia.org/wiki/Zone_melting#High-resistivity_devices

Why are suggestions , however inane, down voted, Someone gives you an idea perhaps there is some good in it just say thanks but I can't or don not want to do that. I Thnk zone purification might work by putting the molten wax in a tall cylinder and cooling ir slowly from the top down. Put it in hot water to met the wax and slowly remove the cylinder from the heat allowing the wax to crystallize. Easy to try is this idea worth a downvote?

Hey there @jimchmst - thank you for the answer. Just to be super clear, I haven't personally downvoted any answers

This could actually work, with possible troubles with scaling up.

The first thought was to add $\ce{BaTiO3}$ powder, melt, and mix. My understanding is, carbon is a polarizable solid and $\ce{BaTiO3}$ is ferroelectric and is also much heavier, so it might bind to carbon particles and sink to the bottom. Later I would reuse the $\ce{BaTiO3}$ powder if carbon can be removed by burning it.
Care must be taken not to breathe the dust.
I never tried it myself, I don't even have any beeswax, just an idea.

Not so crazy Carbon particles usually have some free radicals about so might be sightly magnetic [maybe I am mixing concepts?]

@jimchmst Did you mistake 'ferroelectric' for 'ferromagnetic' ?

I think I have a test for zone purification. Fill a narrow cylindrical glass jar [similar to what salad olives come in] with molten wax. Place the covered jar in water heated just above the melting point, insulate the jar bottom so the jar bottom is not touching the bottom of the pot. Allow the water to evaporate slowly so that the wax in the immersed jar is molten and the wax in the exposed jar crystallizes. This will require some effort. Eventually the front will move down the jar leaving pure wax on top if it works. This relies on Brownian movement and gravity both free with no moving parts. It works for the Arctic Ocean ice that freezes from the top down. If it does not work too bad not much lost. If it does work I see one problem it might also separate wax molecules by weight a type of size exclusion. I see no problem with scaleing up: a taller pot, a taller tube and some more time.

I gave this a go and it wasn't successful - thank you for the suggestion though!


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Food safe Green flame
I'm a master bartender that is trying to create a green flame that is food safe and can be burned indoors safely. I'd be interested in any other colors as well.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

A pinch of food grade borax should work.

@Mithoron, your comment was tongue-in-cheek, I assume (no pun intended!). Borax is not considered safe to ingest (https://www.foodauthority.nsw.gov.au/consumer/keeping-food-safe/other-topics/borax-and-boric-acid: "The Food Standards Code does not permit borax to be used in foods.") In the USA, the Pure Food and Drug Act was written because of borates in food (among other kissues): https://guides.loc.gov/chronicling-america-pure-food-drug-act-1906 Ummm... would you suggest copper chloride, for a nice green flame???

@DrMoishePippik You assume wrong. If borax wasn't used in food, there would be no need for mentioning it in such documents. On the other hand copper chloride isn't mentioned, because nobody uses it - it's much more dangerous. Borax got LD50 about as high as table salt, and is used similarly in a few Asian countries.

Borax is a sodium borate. Will the D lines interfere with the desired green coloration? Can we use boric acid?

There is none. Your only options are yellow, red and some in-betweens, all other flames are unsafe indoor and require strong ventilation.

@Mithoron, yet borax is banned in food in USA, Australia, and likely other countries... perhaps it has to do with life expectancy -- if they don't expect one to live long enough to accumulate a significant amount of boron, it doesn't matter. As for copper chloride, that definitely was meant tongue in cheek! Take that statement with a grain of salt... but not CuCl2

@OscarLanzi Not too much, but it changes the shade. https://en.wikipedia.org/wiki/Colored_fire

Trimethylborate burns green is volatile but it reacts with water . Fill a butane lighter with a mix and use it to light the alcohol. It's been a while since the Coconut Grove https://en.wikipedia.org/wiki/Cocoanut_Grove_fire so we probably need some more fire in bars.

I'm trying to make different colored flames on cocktails. It has to be food safe and safe to burn inside.


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Why does iodate and acidified metabisulfite solution turn yellow instead of blue at high temperatures (above 50)
We are doing a kinetics investigation at school where we are trying to find the activation energy of the reaction between potassium iodate and sodium metabisulfite. Solution A contains 2g of potassium iodate per dm^3 of water and solution B, which is acidified, contains 0.25g of sodium metabisulfite per dm^3 of water and some starch. Solution A was diluted by five times and the ration used in the experiment was 6cm^3 solution A, 4cm^3 solution B and 10cm^3 water. At room temperature, when A and B were mixed, the solution turned blue which is what is to be expected, however, at temperatures greater than 40 degrees celsius, the solution turns yellow. Is there an explanation as to why this is?

https://chemistry.stackexchange.com/questions/181473/why-does-solution-turn-yellow-at-higher-temperatures-in-potassium-iodate-sodium

Iodine in water, as you've noticed, is yellowish to brown, depending on concentration. Starch contains amylose, which combines with iodine to make a blue-colored colloid.
You might simplify the experiment:

Start with a solution of "metallic" iodine in water, which needs some iodide (e.g., $\ce{KI}$) to form the brown complex triiodide ion, i.e., Lugol's iodine solution.
Add some of the starch suspension to get the blue amylose-iodine complex.
Heat it to 40°C and see if the blue color reverts to yellow-brown.
Allow it to cool to room temperature to see if the blue color returns, and if it is the same shade as when the experiment started.

Does the color return?
Here are two hypotheses:

The iodine is driven out of the amylose by heating.
The amylose is destroyed by heating.

What do the results of the experiment show?


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Question Involving Bonding (VSEPR AND VBT)
So basically any chemistry book will tell you about the extent of overlapping in different orbitals.
For formation of sigma bonds, head on overlapping happens. Chemistry textbooks always state that $p-p$ overlap is more effective than $s-s$ overlap. The explanation given is that the s-orbital is spherically symmetric. So it is non directional, while the p-orbital is definitely directional.
However, if we think of hybridised orbitals, $sp$ orbitals are known to form stronger bonds than $sp^2$ and $sp^3$ hybrid orbitals. The explanation given is that the higher percentage of the s-character makes a bond stronger. If that is the case, then shouldn't an $s-s$ overlapping be stronger than a $p-p$ one?
And if $p-p$ overlapping is better than $s-s$ one, shouldn't the hybrid orbitals with higher p character form stronger bonds?
I am kind of confused. I would appreciate an explanation.

I can't think of a lot of bonds where you expect s-s overlap (excluding hybridization). That makes a comparison of strengths  of bonds involving p-p vs s-s overlap difficult. It's always good to support statements by citing a source rather than writing " any chemistry book". In any case I suspect most "proofs" are theoretical in nature.


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On the relativistic contraction of orbitals
In Platinum $\text{Pt}$, why do $\text{s}$ and $\text{p}$ orbitals contract when electrons start travelling at relativistic speeds $|v|\sim c$, while $\text{d}$ orbital expands?
The way I reasoned about this is that due to relativistic effects, relativistic mass of an electron increases:
$$|v|\sim c\rightarrow m_c = \frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}\uparrow$$
Which means that they possess higher kinetic energy and Bohr radius shrinks. However, due to  $\text{d}$ orbital suffering from shielding effect, it expands while $\text{s}$ and $\text{p}$ orbitals contract in size.

The $d$ and $f$ orbitals do not expand due to the speed (or more accurately, the ratio of energy drawn from the electrostatic field to the rest mass-energy) of the innermost electrons in those orbitals ($3d,4f$). They expand because the shielding effect changes.
In the case of $s$ orbitals, only for the innermost $1s$ subshell do the electrons actually draw an amount of energy from the electrostatic field that's comparable with the rest-mass energy. That orbital then contracts relativistically. But now the $2s$ orbital, seeking the lowest spherically symmetric state orthogonal to the $1s$ orbital, shrinks in response even though the relativistic effect is smaller. This indirect effect cascades to all the $s$ orbitals as they respectively seek the lowest available orthogonal spherically symmetric states.
This cascading shrinkage is greatly diminished for $p$ and higher angular momentum orbitals because their wavefunctions have nodes passing through the nucleus and thus much less probability density in the strongly electrostatically-attracted region. Instead the shrunken $s$ orbitals now shield the higher-angular-momentum orbitals more effectively, causing the latter orbitals to expand under the decreased unshielded nuclear charge. The net expansion is greatest for $d,f,$ etc. orbitals because their multiple nodes through the nucleus decreases the probability of hitting the strongly attracted region more than is the case for $p$ orbitals.
The net result of all this is that for the heaviest elements, the "inert pair" effect is accompanied by a "reverse inert pair" effect in which higher-angular-momentum electrons are activated.


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Interpretation of a Phase Diagram
I am a bit confused about the correct way to interpret a phase diagram. I was told that the line separating the liquid and gas phases gives the vapor pressure of the liquid substance as a function of temperature. I am not fully sure why this must be the case as in my mind, that line represents the boiling point of the liquid as a function of pressure and I don't see clearly how this connects with vapor pressure.
Furthermore, I know that the line separating the solid and liquid phases gives the melting point as a function of temperature. I am wondering if there is an alternative interpretation of this line involving an analog of vapor pressure between the solid and liquid phases (i.e. I know that there is a vapor pressure between the solid and liquid/solid and gas phases. I am wondering if there is some analogous equilibrium pressure between the solid and liquid phases. If so, how would this be represented on the diagram and if not, why is the solid/gas, liquid/gas boundary special?)

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

Liquids and solids have vapor pressure, even though well below their boiling point. For example, the vapor pressure of solid $\ce{CO2}$ is high enough that it sublimes, i.e., goes directly from solid to gas.
In the diagram below, adapted from Wikipedia, notice that

Water boils at 100°C, i.e., the vapor pressure is equal to atmospheric pressure, and causes bubbles to form and burst out of a container.
At 50°C, e.g., a very hot day in Death Valley, AZ, the vapor pressure of water is ~10 kPa (about 1/10 atmospheric pressure). This pressure is still high enough to make water evaporate rapidly, though it is not visibly boiling.
At about 180°C, perhaps the temperature of an old steam-engine boiler, the pressure is 1 MPa (about ten times atmospheric pressure). If the boiler were to burst, all the water would flash to steam. Yet, in a deep-sea hydrothermal vent, that would still be liquid water at that temperature.

Thank you! That makes sense. However, if even at low temperature, some pressure of the gas exists, why can't the same be said for solids and liquids. For example, why is it not true that for a solid, there is some amount of liquid in equilibrium with the solid? Also, while I understand that vapor pressure exists, I don't quite see why the phase boundary between liquids and gases, which describes the temperature at which liquids boil at a given pressure also gives the vapor pressure at a given temperature. It's more the specific interpretation of those aspects of the diagram that confuse me

@JohnnySmith, there certainly can be. Some materials do not have a clearly defined melting/freezing point, such as window glass or non-eutectic alloys (e.g., 50% Sn/50% Pb solder). Also note that some substances have incredibly low vapor pressure, e.g., gallium. Though liquid at body temperature, Ga does not reach boiling point at 1 atm until 2403°C!

that makes sense. I am still not sure how to rigorously argue that the boiling point curve of T vs P is the same as the vapor pressure curve of P vs. T. For context, I am a physics student taking an upper level pchem class so while I have never been exposed to these specific topics, I am looking for a rigorous argument. Thank you so much!

@JohnnySmith, think of "boiling point" as the temperature where the molecules of liquid can push back against the vapor and, so to speak, "violently" escape. Below that pressure, some molecules have sufficient energy to escape (as Orwell never stated, "some molecule's energy are more equal than others."), but at the b.p., the pressure is the same as that above the liquid... and it bubbles.

That phase diagram is incorrect the liquid- solid-vapor lines are disjunct at the triple point. My post explains it.

@jimchmst, you might make any correction at Wikipedia, from whence the diagram was adapted (cut down, and 2 temps added). No mention in my answer was made of the triple point, which, as you state in your answer, has no degree of freedom. It's unique.

It is not my job to correct Wikipedia so you can be prevented from posting incorrect information.

The phase diagram for a pure substance has only one component, the substance in its phases, and describes the equilibria among the phases. The Gibbs Phase Rule: Freedoms = Components - Phases + 2; F = C- P + 2, describes the diagram. For one phase present there are 2 freedoms, T and P are independent; this is an area in the diagram.
If two phases are present there is one freedom, T and P are entangled, for each value of one there is one value of the other. This is a line in the diagram. This line is continuous and monotonic until a new phase is encountered. There are no boiling points or freezing points just continuing equilibrium. The line does not separate the phases, it is the Equilibrium between the phases. Remember boiling points and freezing points are defined at a SPECIFIC temperature and pressure; the pressure is the open atmosphere and is variable. Addition of a second component, the atmosphere adds an additional freedom, the atmospheric pressure. [Even more when the composition of the atmosphere is of concern.]
A point, the intersection of 3 lines, [This is interesting because this is nonEuclidian, the lines are not straight or continuous but end at the triple point] is invariant with no degrees of freedom; there is a distinct T and VP. Three lines are necessary because the solid and liquid have the same VP but different latent heats and volume changes. When an inert gas, an atmosphere, is introduced the change in pressure has a different effect on the liquid and solid phases changing their vapor pressures differently. The difference in vapor pressures causes a temperature change to equalize the vapor pressures and results in a new triple point at every inert gas pressure. This gives a 4 dimensional aspect to the phase diagram with a continuous overlay of triple points and the corresponding phase diagrams. This is why melting points and boiling points in an [inert] atmosphere must be defined by a temperature and an applied pressure while a one component triple point has invariant T and VP.

What do you mean by "non-Euclidean"? Isn't that just confusing the subject? If I interpret you correctly, what you suggest is that the S-G and L-G coexistence lines are discontinuous at the triple point (there is a discontinuity in the derivative dp/dT).

There is a definite discontinuity. The heats of reaction that effect the slope are different for the 3 equilibria and the volume changes differ affecting the slope. I am not a good enough mathematician but it seems that the triple point is really not a point in 2 dimensions but a cusp in 3 dimensions and the addition of an additional component, the atmosphere, adds a fourth dimension to the math.

Well, I agree with you, just not sure bringing up non-Euclidean geometry is at all necessary, it makes it sound more complicated than it is.

Everyone is showing excellent understanding of the subject . Be it me to confuse all. J. Willard Gibbs explained it nicely with his phase rule. It is very simple starting with first principles very complicated if they are glossed over. If you agree with me and my explanation is correct upvote me so it is read. If I am wrong correct me and give a correct answer. I hope to get people to think about these simple things

Boiling is a special mode of liquid substance evaporation (transition to its gaseous phase). In this mode, the evaporation occurs not only at the liquid surface, but also inside the liquid, where are being formed bubbles of the gaseous phase of the substance.
The curve between liquid and gaseous phases on phase diagrams have two equivalent meanings:

The liquid vapor pressure at the given temperature at equilibrium
The liquid boiling point at the given pressure

As at temperature of the boiling point, the liquid vapor pressure is equal to external pressure and is able to push the liquid away, against the external pressure. The formed space is filled with the vapor.
The relation of temperature and vapor pressure relates to equilibrium conditions of a closed container containing just the liquid and its vapor, reaching the particular pressure at given temperature.
The boiling is a dynamic process with ongoing on (usually) open system, that is far from equilibrium.
There are two small deviations:

Forming of a bubble needs to overcome the liquid surface tension, so there is needed vapor pressure little higher than external pressure.
In presence of gravity, boiling point raises with the depth due contribution of hydrostatic pressure.

The curve between the solid and liquid phase gives primarily the dependance of the substance melting point on pressure.
There are few hidden thermodynamic relations to this (s)-(l) curve.
Chemical potentials $\mu_{i}=\left( \dfrac {\partial G}{\partial n_i}\right)_{p, T, n_j, j \ne i}$ of solid and liquid phase are equal at the melting point.
If there is 2 component system with an inert gas, coexistence of 3 phases is not limited to the triple point with given temperature and pressure, but there is still one degree of freedom = 1 parameter we can deliberately choose, usually temperature. If we keep such a system at temperature of melting point, partial vapor pressures over both condensed phases are equal and $$\mu_text{s}=\mu_text{l}=\mu_text{g}$.


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Replacing calcium acetate with sodium polyacrylate for homemade Sterno?
I was researching making canned heat using denatured(ethyl) alcohol, water, and calcium acetate from scratch.
However, it occured to me that sodium polyacrylate also absorbs water and alcohol and is pretty easy to retrieve by taking apart diapers.
Would this substitution give the same effect as calcium acetate does?

Technically it may work, but I would be concerned by burning the sodium polyacrylate.


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Why cant H2O be linear?
My understanding is that a H2O molecule cannot be linear due to repulsions between the electrons of the non-bonding domains and the hydrogen:

However i was wondering why the structure could not take this form where the repulsions cancel out, resulting in a linear molecule:

Any answers would be helpful!

sp3 orbitals seem to bring more electrons closer to the nuclei than the p orbitals for the unshared pairs. It is a balance between attraction and repulsion.

There is a detailed answer here https://chemistry.stackexchange.com/questions/14981/why-does-bond-angle-decrease-in-the-order-h2o-h2s-h2se/57973#57973


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Does styrene really react with KMnO4?

A benzene derivative $ \ce{C_8H_x} $  did not produce white precipitate with the
ammonical silver nitrate solution but decolorised the cold dilute
alkaline $\ce{KMnO4}$ solution. The compound is having _____ number of
hydrogen atoms.

The absence of white precipitate indicates it is not an alkyne.
The benzene ring must have an alkene group attached with it such that it decolourises $\ce{KMnO4}$

Now for the question, the answer key tells it must be styrene, however I argue that the answer should be one homologous higher than styrene, because styrene's resonance doesn't allow it to react with $\ce{KMnO4}$.
Is my reasoning right?

No, it would react just fine. Also, the next one in the homologous series would be C9, not C8.

Oh my, I think I overlooked that part about $C_8$. Due apologies. But isn't the double bond in conjugation with the benzene? Shouldn't the electron delocalization prevent further attack?

It is in conjugation, but that's not enough to prevent it from most attacks.

Hmm makes sense. Thanks for helping out!

Styrene has been reported to react with potassium permanganate to yield 1-phenyl-1,2-ethanediol (at room temperature, albeit promoted by either ultrasound or hydrogen peroxide), phenylglyoxylic acid (albeit at 70 °C), and benzaldehyde (at room temperature, albeit in the presence of alumina). None of the above-mentioned methods strictly adhere to the "cold dilute alkaline KMnO4 solution" mentioned. Yet, I argue that styrene is more than happy to react with KMnO4 if given a little nudge. Meanwhile, phenylacetylene also reacts with KMnO4, yet again requiring a little heat (refluxing DMC).
Nonetheless, as @jimchmst pointed out, phenylacetylene couldn't be the right answer because it reacts with silver nitrate.This leads us to conclude that the correct answer is styrene.
References:

https://doi.org/10.1016/S0040-4039(98)01671-2
https://doi.org/10.1039/C5RA01646K
https://doi.org/10.1021/ja01265a511
https://doi.org/10.1055/s-2001-16760
https://doi.org/10.1055/s-1978-24783
https://orgsyn.org/demo.aspx?prep=CV5P0517


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How does the following cyclization reaction work?

The following is a reaction from a school provided lab manual. However, I am unable to find anything similar to the cyclization reaction taking place. All of the reactions I have found online make use of either aldehydes or carboxylic acids for the ring closure. It is obvious from the product that the carbon connected to the 2 OMe substituents is the carbon that participates in the reaction. I just can't find anything that shows the possible mechanism that results in the product. Any help would be appreciated in how I should tackle this problem.
Edit:

I've taken Waylander's advice and come up with the above mechanism.
However, before I read his comment I also managed to come up with this mechanism:

Would either mechanism work? Or would one be more correct than the other?

Look up Bradsher reaction. For your particular reaction, https://doi.org/10.1055/s-1974-23372 and Wikipedia — Bradsher-Cyclisierung (in German) appear to be relevant.

Hello, thank you for the reply! Could I trouble you for further clarification? I took a look at the resources you linked and they use involve HBr. In this case would the H2SO4 act in the same way HBr does? The journal article you linked has compounds that seem to undergo the same reaction (2) --> (3), but there is no suggested mechanism so I am finding it hard to apply to my particular problem.

The key step here is the protonation of one of the -OMe groups. This gives a cation which is  which is stabilised by the other OMe group and captured by the aromatic ring. Further protonation/elimination of the remaining -OMe and the -OH gives aromatisation of system. It should be noted that there will most likely be a second isomer formed by cyclisation into the ortho position

I think your mechanism in pic 2 (apart from the unfortunate wandering alcohol group) is more correct. The site of first protonation is imho one of the acetal oxyens

Well, you do have there a protected aldehyde. BTW you'd do better not putting every proton exchange into separate step.

Thank you all for the input! Mithoron, so do you have a similar opinion to buttonwood that the reaction should involve the deprotection of the aldehyde group which is then "attacked" by the benzene ring pi bond?

The stated reaction conditions of cH2SO4 in acetic will not transform the acetal to the aldehyde, just swap the -OMe groups for -OAc. The cyclisation reaction does not need the aldehyde as intermediate.

Oh, could you explain why the conditions would not allow for the transformation of the acetal to an aldehyde? As I understand, the transformation only requires acidic conditions.

There is no free water present to displace -OMe, the only available nucleophile is acetate

Terminal to the alkyl chain, you have the motif of an acetal.  In presence of aqueous acid, it can pass to hemiacetal and free aldehyde.

From there, the aldehyde possibly is activated further by protonation.  The arene (a pi-bond nucleophile) then reacts with the carbonyl-C (the electrophile) to establish the new ring.* The intermediate carbocation rearomatizes.  In principle the subsequent dehydrations of the the alcohols pass equilibria, but the eventual gain of aromaticity (again) drives the reaction forward. (Maybe the sequence of which alcohol is dehydrated first, a tertiary vs. benzylic alcohol is a different one, than depicted; the result were the same.)

* At this stage, the pi-bond nucleophile can be recognized easier by moving the electron sextet, see the second structure.  (We all know that neither one of the two mesomeric formulae represents the true state fully.)

I don't think you need to postulate cleavage of the dimethyl acetal to aldehyde. Simple protonation of the acetal is enough.

@Waylander Possibly influenced by ortho esters' hydrolysis in Claisen reactions, does the dehydration of a tertiary alcohol (as in the starting material) typically outrun the (re)generation of an aldehyde from an acetal? The example of a Bodroux Chichibabin reaction in Organic Synthesis couples this with steam distillation to harvest n-hexaldehyde and (hence) does not state the reaction time.


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Why is some racemisation possible in Sn2 reaction?
Partial racemisation is known to occur in Sn1 reactions, but why does Sn2 reaction also show little racemisation?

Because it's not really a binary thing - real processes are somewhere in between $\mathrm{S_N1}$ and $\mathrm{S_N2}$.

Your question is a potentially good one, but the way of asking it can be many times better. Please include supporting resources, your research work and the source from where you read it. As such, your question is too vague to be answered.

If this is a question you've been asked to answer, it is a poor one. SN2  reactions (as they are generally understood) don't show racemisation at all, but some reactions are a mixture of SN1 and SN2 as outlined by Mithoron.
That said, I recently read this paper which blew my mind a little:
Backside versus Frontside SN2 Reactions of Alkyl Triflates and Alcohols
It leads to a more general point: should we think of SN2 reactions in terms of kinetics (i.e. second order nucleophilic substitutions) or mechanistically (i.e. backside attack into a low-lying σ*)? It so happens that the two definitions usually correspond, but I suppose this isn't necessarily the case.
But I'm certain the answer they were looking for was "because it does SN1 too, a bit".


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Absolute entropy of a compound at a given temperature is S0→T=∫Cp(T)Tdt + ΔH(trans)/T(trans).Is there an assumption of constant pressure here?
Change in entropy of system is given by dQrev/T, Qrev=nCpdT for ideal gas only at constant pressure,so when we find absolute entropy of a gas by integrating nCpdT from 0K, is constnat pressure maintained? It doesn't seem practical that a solid at 0K and when you take it to a gas, that gas and that solid have the same pressure which is needed to apply the mentioned formula.

Yes, it's possible for p to be constant. You can change T to go from condensed to gas phase.

Note also your equations contain errors. Should be $dq_{rev}=nC_p dT$, for instance.


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4 nitro groups on benzene
I was just wondering if there exists a phenol having 4 nitro groups. I tried searching the internet but couldn't find anything. If it exists what all features should it possess, and will it be more acidic than a trinitrophenol?

Even 5 is possible.

PubChem's records about tetranitrophenol, and the one about pentanitrophenol.

@Buttonwood Thank you sir.Got a great web to check all compounds which come to mind but I don't come across in my book

I would not like to try to isolate it

They "exist" but not much more then that. And are weaker acids, because of SIR.

You might find Derek Lowe's article, Can't Stop the Nitro Groups, https://www.science.org/content/blog-post/can-t-stop-nitro-groups, enlightening (and entertaining) -- you might get a real bang out of it.


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Why atomic isotopes cause a different melting/boiling point?
Different experiments show that replacing atoms with isotopes can change intermolecular forces. For example, heavy water has a higher melting point than normal water, and D2 has a higher boiling point than H2. But the trend is not always "the heavier the higher", as C13-methanol and ethanol, have a "lower" boiling point than C12 ones.
I've already read this topic, in which it is mentioned that vibrational energy is the cause of it. heavier atoms vibrate slower, so they have a higher melting point. (which again, the methanol case can't be explained by it).
My argument is that the "temperature" is defined as the average kinetic energy due to vibrations. So when D2O and H2O reach 0. they both have the same average vibrational energy, regardless of how much energy they hold. But with the same kinetic energy, intermolecular bonds in H2O are broken, and in D2O are not. Can someone explain why it happens? or persuade me that D2O and H20 can have the same temperature with different vibrations?

Why should they cause the same melting/boiling points?

You change something, so it has to be different. It should rather be "how" than "why".

The different masses directly indicate that the vibrations will be different (consider the classic mass on a spring from freshman physics). In any event, by changing how the different parts move relative to each other you make the interactions slightly different, meaning they respond to adding energy differently.

The argument about temperature and vibrations ignores big quantization steps for bond vibration energies, that spoil the simple proportionality rule of classical thermodynamics.

@Mithoron I have read this post. My question is, how the vibration is different at the same temperature? temperature is basically an average of vibrations.

@Poutnik can you explain more? also, here we are talking about intermolecular forces, not intramolecular bonds.

Vibrations are not limited to intramolecular forces. If they were, there would be no liquids nor solids - only gases.(excluding giant molecules like diamond structure or some polymers).

Quick search C-13 methanol has slightly higher BP slightly lower MP. Different sources? Temperature is not vibrations it is relative molecular velocities resulting in average KE. Vibrational states in molecules are populated by temperature. Kinetic isotope effects result from changing of vibrational states in the transition state making zeropoint energy difference between the isotopes important.  The effect on total energy affecting MP and BP is much more subtle. The differences are profound for H2, D2, H2O, D2O, He[3] and He[4], less so, but real, for others needing careful analysis.

You’re confusing ‘vibrations’ with ‘translations’. Temperature is more related to the movement of molecules in space (translation), as opposed to the movement of atoms within a molecule (vibration).


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Deviations in trend for boiling points of straight chain alkanols
I was given a set of data about the boiling points of straight chain alkanols from methanol to dodecan-1-ol and noticed that the trend line seemed to be in the shape of a polynomial 4. I have also noticed a similar pattern in a data set for boiling points of alkanoic acids ranging from methanoic acid to decanoic acid. Can anyone explain why this oscillation occurs?

https://chemistry.stackexchange.com/questions/160008/continuous-variable-for-graph-of-boiling-point-of-different-chain-lengths

https://chemistry.stackexchange.com/questions/50047/elevated-boiling-points-in-alkanes-with-even-number-of-carbons


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Cellulose acid hydrolysis
Recently i tried to hydrolyse some 10g of cellulose with 50ml of 75% H2SO4. Right after adding the acid, the entire solution turned blue black. With no heat applied.


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Why does hydrogen peroxide decompose during the oxidation of an aryl amine to and nitroarene (rt) and what are the possible overoxidation products?
Small scale:
I attempted the oxidation of an aryl amine (2-amino-6-bromonaphtalene) to the nitroarene using hydrogen peroxide (ArNH2, H2O2, MeCN (15 mL as solvent). After trying it on small scale (0.1 mmol), I figured out I needed much more hydrogen peroxide than in the reference paper to observe my product. Nothing special was observed during the reaction, even after using more than 10 equ of hydrogen peroxide.
Larger scale:
I eventually repeated the reaction on a 4 mmol scale using 80 mmol of hydrogen peroxide (20 equ.). After 5 min of stirring at room temperature, the reaction bubbled up. The use of a very large flask and being lucky allowed me to stop the stirring and contain the reaction. Even after stopping the magnetic stir bar, the reaction continued to foam for another 2 or 3 minutes.
After pouring into cold water followed by a very short silica plug using ethyl acetate I obtained the pure product (confirmed by NMR) but with only 15% yield. The TLC from the reaction before purification showed two spots. One that corresponded to my product, and one that would not come off the baseline (hence it could be multiple) even in pure DCM.
Mechanism
A possible mechanism pf the oxidation reaction is shown here:

Possible explanation
To make things easier, I only considered the formation of 3.
To explain this I proceeded to do a thermodynamic cycle and ended up finding that to go from the starting material to (3), around 900 KJ/mol was released. ( enthalpy energies found here )
Total broken bonds: 2OH + 2xC≡N + C=N + O-O + N-O + C-N + N-H
Bonds made: 2CO + 3*NH + O=O + N=O + C=N + N-O + C-N
deltaH= - (2CO + 2NH + O=O + N=O) + (2OH + 2xC≡N + C=N + O-O)
= - (2358 + 2391 +(495-146) + 405) + ( 2463 +2*(891-615-293) + (615-293) + 146) )
= - 892 kj/mol
This number is on the same order of magnitude as combustion reactions for methane (-890 kj/mol). Based on the reaction mechanism, only 4 mmol should react (theoretically), hence producing 3600 kj.
Using Q = m * Cp * dT with Cp = 2.09 J/g·°C for acetonitrile and rearranging leads to an increase in temperature (15 mL of MeCN) of 135 degrees celsius, above the decomposition temperature for hydrogen peroxide. This thus may explain the decomposition I am observing.
This reasoning suggests that the stochiometric addition of hydrogen peroxide to this reaction will produce enough heat to drastically increase the temperature of the flask, hence decomposing any excess hydrogen peroxide.
Regarding the very low yield of my reaction, I am quite confused on what could be happening to my starting material to produce something so much more polar than the nitro arene. Following the mechanism, it would feel natural to suggest an over-nitration.
Questions
Is my reasoning above correct to explain why I am observing the decomposition of hydrogen peroxide?
Since we do not always have access to the mechanisms, are there ways to predict a reaction to be very exothermic to manipulate safely?
Any advice or suggestions on what could be happening to my starting material for it to be totally consumed into extremely polar products? Could the decomposition of hydrogen peroxide affect the starting material?
Notes:
My starting material is quite stable under ambient conditions.
Regarding the TLC mentioned at the start, I am aware that DCM is not that polar, so I am planning to use methanol:dcm if I redo the reaction. Nevertheless, going from a bromo naphtalene to a product that does not elute in pure dcm seems rather unusual. My product elutes to 0.5 rf in 50% DCM:Hex.
My apologies for such a long description. I do hope this may be useful to inexperienced chemists if manipulating hydrogen peroxide.
Ref:
Scheme:
Capperucci, A.; Tanini, D. Synthesis of Nitroarenes by Oxidation of Aryl Amines. Chemistry 2022, 4, 77-97. https://doi.org/10.3390/chemistry4010007
Errika Voutyritsa, Alexis Theodorou, Maroula G. Kokotou and Christoforos G. Kokotos. Organocatalytic oxidation of substituted anilinesto azoxybenzenes and nitro compounds:mechanistic studies excluding the involvement of a dioxirane intermediate. GreenChem., 2017,19,1291. DOI: 10.1039/c6gc03174a
Original reaction
Gupta, S., Ansari, A., & Sashidhara, K. v. (2019). Base promoted peroxide systems for the efficient synthesis of nitroarenes and benzamides. Tetrahedron Letters, 60(39), 151076. https://doi.org/10.1016/J.TETLET.2019.151076

H2O2 is no good unless you use some catalyst. It doesn't need much of a reason to decompose.

Bit of Ag and $\ce{H2O2}$ and your project could really take off! https://www.youtube.com/watch?v=iNUppNTT9RI


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Are CAS numbers considered proprietary?
I’m interested in working with a large dataset consisting of CAS numbers and each corresponding molecule/chemical unit’s name. From my understanding, CAS Registry Numbers are assigned and stored by a division of the American Chemical Society.
However, it's not entirely clear to me how these numbers are controlled outside CAS's own database. Does ACS “own” these numbers?

What do you mean by controlled? You can't assign these numbers yourself, but once assigned I don't think there is much legally that prohibits how you use the number. I'll await an answer though. In the CAS faqs it states: "A CAS Registry Number license is required anytime an organization will “publish” CAS Registry Numbers to the public or use them to support features of a platform that is publicly or commercially available."

Basically I think CAS sells licenses to a service that checks that the numbers you are using are correct.

When you have the numbers, as check the last digit in the CAS number is a check digit, obtained by multiplying each preceding digit by its position in the number; taken in reverse order, starting by multiplying the last digit by $1$, the second to last by $2$, etc. then adding the result and calculating modulo $10$ of the number produced. For naphthalene $(91 - 20 -3)$; the $3$ is the check digit. The  sum is , $0 \times 1 + 2 \times 2 + 1 \times 3 + 9 \times 4 = 43 $, finally, $43 \text{ mod } 10 = 3$, where the $3$ is the remainder of a whole number of divisions of $10$ into $43$.

CAS numbers itself don't contain chemical information, and initially set to help ACS' Chemical Abstract Service to organize their records.  It is the only body to create and curate (both assign and withdraw) them.(1)  The rules behind the format of a CAS number are public.
You have to pay for a look-up in their databases (i.e., Scifinder).  Depending on your local law, the mere assignment of a catalogue number to a chemical(2) might not represent something creative (or: creative enough) to grant a granting copyright.  A couple of years ago, this was a point of discussion between Wikipedia and its property boxes stating CAS numbers, on one hand, and CAS on the other (see, for instance a blog post by Peter Murray-Rust in 2008 here about the curation of these public entries(3)).
If you want to be on the safe side regardless the jurisdiction, the set of about 500k entries of common chemistry is a larger public database curated by CAS under the permissible Creative Commons CC BY-NC 4.0 license.  A search by either (English) chemical name, SMILES, InChI, or CAS RN yields basic entries like chemical name, CAS RN, a 2D .sdf file and other identifiers.  Below a screen photo about the open chain form of glucose (note the use of registry marks)

(screen photo of glucose in commenchemistry.cas.org)
There is no guarantee larger public chemistry-relevant databases as for instance PubChem, or ChemSpider have for every entry on their records the corresponding CAS registry number(s).  Like vendors of chemicals, they are not obliged to use them (under the presumption CAS assigned at least one), or follow an other scheme.

(1) One of the motivations for InChI was to provide a non-proprietary identifier, an other that the string would relate to (at least some) chemistry of the record.
(2) There are caveats like 1317-80-2 about rutile, and 1370-70-0 about anatase -- both $\ce{TiO2}$ polymorphs -- vs. 13463-67-7 titania.  Have a look on the many entries of the later «Deleted or Replaced CAS Registry Numbers», bottom right.
(3) The English edition of Wikipedia contains about 21k chemicals searchable by the structure explorer.


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Azo dyes with different colours than Yellow, Orange and Red
Why are there just Azo dyes with colours of Yellow Orange and Red. Do the molecules become to be unstable with a greater pi-electron System? Every dye or indicator with other colours are'nt Azo dyes.

There are also blue azo dyes, e.g. trypan blue. You can have most any colour you want, all that is needed is the C-N=N-C linkage.

More complex azo dyes can be other colors, as with trypan blue. With its more complex structure comes more extended conjugation, which shifts visible light absorption from the shorter violet and blue wavelengths (which gives the more common yellow to red colors as the complementary color) towards absorbing yellow to red wavelengths (thus giving a complementary purple to blue color).
Structure of trypan blue:
Source file
Typical case of staining an organic specimen with trypan blue:
Source file

Wikipedia article Category:Azo dyes list you all the azo dyes in the
whole color spectrum, with colors explicitly indicated (yellow, orange, brown, red, blue, black) in the names of some of the dyes. So, the premise of your question that azo dyes ranges in only red to yellow-orange is false.
Also see: What is the origin of the colour of azo dyes?

In addition to this, the presence of the diazo motif doesn't hinder the molecule to bind to metals; this opens additional variation including the colors perceived (example).


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What is the major product of reaction of phenol with HNO3 in the presence of H2SO4?
In many articles I have seen they say the reaction proceeds to picric acid but the amount produced is very small. But the major product is formed by ring cleavage (oxidatively). Can anyone mention the major product and if possible provide the mechanism?

Without initial sulfonation step, phenol oxidation has many steps and branches..

Why do you guys keep asking about it? There's no specific product or mechanism. You just get some gunk. Still, it's unlikely to be "ring cleavage" - it's rather oxidative coupling.

If you use dilute nitric acid, you will get mononitrophenol in various proportions depending on phenol/HNO3 molar ratio. Please find yield here: https://www.researchgate.net/figure/Composition-of-the-products-of-phenol-nitration-with-nitric-acid-in-aqueous-dioxane-in_tbl1_294735990
If you use concentrated nitric acid, the reaction proceeds towards picric acid. See previous discussions:

Does nitration of phenol with nitric acid yields 2,6-dinitrophenol?
Preparation of picric acid


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Help with ppm calculation
Recently, me and my class had a lab about spectrophotometry, where we try to determine the concentration in ppm of an unknown sample contaminated with manganese.
In the process of calculating this concentration, we have to make a linear regression using the absorbance data from known concentrations in ppm. Me, my classmates and the teacher have conflicting answers about how to calculate the ppm concentration and I wish to confirm if my method is right or wrong. It goes as follows:
Vial 1 has 2 mL of a 0,0008 mol/L solution of manganese, diluted to 100 mL with water. Since 1 ppm = 1 mg/L, the calculation goes as follows:
There are 0,0008 x 0,002 = 0,0000016 moles of manganese in Vial 1, which by the molar mass of manganese (54,938 g/mol) is 0,0000879 grams, which is equivalent to 0,0879008 mg. Since 100 mL = 0,1 L, the ppm concentration is then 0,0879008 mg/0,1 L = 0,879008 ppm. Is this correct?
My teacher and classmates used 1 L instead of 0,1 L in the calculation for reasons I do not know, which makes it 0,0879008 ppm instead.

Be aware that using of ppm relative unit is strongly discouraged by IUPAC. as it often brings confusion.  Saying 1 ppm means 1 mg/L makes 2 implicit assumptions. 1/ It is implying 1 ppm is related to parts per million in units of mass (can be volume or amount)  2/ It is implying the density of the fluid is 1 kg/L. // The recommended way is, instead of 1 ppm,  explicitly writing 1 mg/L, if it is the case.

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


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BS Chemistry Thesis Topic Suggestions in the Areas of Natural Products, Synthetic Inorganic and Organometallic Chemistry, Physical Chemistry
I am a 3rd year Chemistry student. In partial fulfillment of the requirements for the degree Bachelor of Science in Chemistry, we are to present an undergraduate thesis. We were highly encouraged to choose a topic that interests us, however, I find it difficult to find a topic because my interests are too broad. I also want my undergraduate thesis to be publishable internationally but I don’t know what topics would top that. I’ve had a difficult time finding topics that are relevant, unique, cost-effective, and feasible for an undergraduate that interests me. Can you suggest thesis topics in the areas of natural products, synthetic inorganic and organometallic synthesis, or physical chemistry? We are given 10 months to propose, experiment, and defend our thesis. Thank you.

Huh, while you may be encouraged to do that, I don't see how it's a good idea. Try finding a good supervisor and talk what the topic could be. More importantly you question isn't suitable for the site, as it's got no specific answer.

Here is one problem that could be studied in an undergraduate thesis : the kinetics of copper dissolution. I explain. If you put always the same copper cylinder in always the same volume of different concentrations of sulfuric acid, and if you introduce a rubber tubing producing constant bubbling of air into this solution, you will observe that the solution gets slowly more blue. You could measure the concentration of copper by a colorimeter. The strange thing is that there is no corrosion of copper in highly diluted solutions, and in high concentrations of sulfuric acid. Why ? I don't know. The corrosion is maximum for a given intermediate concentration. Which one ? I don't know, You may study many different solutions simultaneously by using as many beakers and as many pumps as needed. Another strange phenomena : when the solution gets saturated in copper sulfate, the corrosion does not stop. It continues, and well crystallized copper sulfate is slowly developing on the copper cylinder. You may weigh this amount of cristal versus time. Try ! It is easy. In my opinion, it has not been studied so far. I may be wrong,

Maybe you want to post this as a question instead?

Why do people decide to disqualify the strange facts I am mentioning ? I will not  post this strange behavior as a question now, as suggested by Buck Thorn. I have not enough measurements. It would be better to repeat this experiment, with a lot to different concentrations. I have done it with only four solutions.  It should be repeated and the kinetics seriously determined. It could be a nice subject fo an undergraduate thesis.

As you are probably aware by now, questions such as in this post tend to be considered off-topic as "opinion-based" (well, here is was voted "unfocused"), and answers are sometimes discouraged with downvotes.


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Why cool a reaction down when adding reagents if you're only going to reflux subsequently?
I've seen several procedures using thionyl chloride (SOCl2) to make an ester, where the first step is to add the thionyl chloride to the reagent/solvent on ice or subzero temperature, followed immediately by raising the temperature to reflux.
A labmate asked me a question that stumped me: what is the point of subzero cooling during addition, if you only plan to elevate the temperature immediately after for reflux?
My "real" answer is that I don't want deviate from procedure in case that is detrimental for some unspecified reason. Perhaps a better answer is that on initial addition you don't want a runaway reaction while hand is over the RBF or a strong spike in temperature. Given that the substrate/rxn mix survives reflux, I'm not sure if the latter explanation is sufficient or whether there is actually a good rationale for cooling during reagent addition prior to heating/reflux.

Sometimes adding thionylchloride will lead to gas development, cooling down the mixture will lead to a decreased speed in the reaction. Also, sometimes you can avoid side reactions by adding the reactants in a slow and controlled fashion.

It is all about control of the reaction. Thionyl chloride, as noted, can readily evolve gas with a great deal of bubbling and a rapid exotherm - very easy to lose half your reaction mixture out of the reaction vessel if it takes off.

Great answers. Just to clarify to fully spell it out, if gas evolution is problem, why is refluxing ok? Just to slow the rate?

Thionyl chloride reacts in a 2-step mechanism to first form a chlorosulfite ester that internally reacts to give the chloride or possibly can react with a second alcohol to give a dialkyl sulfite. For some reason the thionyl chloride is usually added to the alcohol in a manner to quickly give an excess; cooling controls the reaction which is then allowed to warm. I have run several of these with yields in the 90%+ range but there was always a small amount of high melting solid that spectroscopically seemed to be a sulfite ester. Possibly the alcohol should be added to SOCl2?

Adding a catalytic amount of DMF generally improves the reaction and reduces sulfite ester formation. You don't reflux until the initial gas evolution is complete.

When the side product is a problem, you could also try Oxalylchloride for the same purpose. As mentioned, DMF will form the Vilsmeier haack reagent as an intermediate. A very small amount e.g. 1-5 drops from a syringe are enough (30-150 microliter) and will simplyfy the workup.


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How can equilibrium constant be derived by irreversible reaction?
I am solving the example from the book [Essentials of Chemical Reaction Engineering, Chapter 3, P3-12B] and finding out a contradictive point.
"The rate law for the A->B reaction was obtained at low temperatures. The reaction is highly exothermic and, therefore, reversible at high temperatures. Suggest a rate law for the reaction at high temperature."
The reaction A->B is irreversible at low temperatures and the rate law is
$-r_A=k C_A^2$
The answer given in the book is:

However, if I look back at the definition of the equilibrium constant

The answer should be $K_c=\frac{C_B}{C_A}$.
If I use the irreversible rxn to derive the answer, it will be $K_c=\frac{C_B^2}{C_A^2}$
I am confused by two different answers.

Only elementary reactions satisfy the condition that the order of a species is equal to its molar coefficient: $\alpha=a$; $\beta=b$. If the reaction is not elementary, $\alpha\neq a$; $\beta\neq b$.

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

No matter what the kinetics, the equilibrium constant for the reaction $\ce{A <=> B}$ is $$K = \frac{[\mathrm{B}]}{[\mathrm{A}]}$$
On the other hand, the equilibrium constant for the reaction $\ce{2A <=> 2B}$ is $$K = \frac{[\mathrm{B}]^2}{[\mathrm{A}]^2}$$
The rate law of the forward step does not determine the equilibrium constant expression. At equilibrium, all steps in the reaction are at equilibrium. The elementary steps have the same rate laws as they have before reaching equilibrium, but the overall rate law is different than the rate law you might figure out starting with reactants only (no intermediates or product yet).
For the relationship of rate laws and equilibrium constant expression, see this answer I wrote, and the questions and answers linked to it.


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Why do we assume the total number of valence electron in a compound are the sum of its atoms' electrons in ground state?
Take for example H2O, when drawing the lewis structure for this, we will assume, by adding up the valence electrons of 2H and O in their ground states, the final compound has a total of 2+6=8 valence electrons. We also do a similar practice of assuming charge is preserved in writing balanced chemical reactions. But why do we assume so? Is there any law saying that the charge of a compound's reactants is preserved as they form the compound itself?
Edit - I am not questioning the balance of electrons, I am questioning that why a molecule of H2O has to be composed of its ground state electrons, why can it not be split up into a oxygen ion and 2 hydrogen atoms. I have no reason to believe that the charge of the oxygen, has anything to do with H2O, which is a new entity, in which all I know is it is composed of one 2 hydrogens and 1 oxygen.

There is a very basic, fundamental law of nature, which says that if you've put 10$$$ in your pocket, you'll get 10$$$ out of your pocket. Same thing with electrons.

Charge is preserved, it has been never observed the opposite. But charge can be interchanged between several molecular entities.

"are the sum of it's atom's electrons in ground state?" I think you mean "are the sum of its atom's valence electrons in ground state?"

It's unclear what electrons you are talking about, for instance when you state "why can it not be split up into a oxygen ion and 2 hydrogen atoms". An oxygen ion would be missing or have gained electrons, which, given the principle of conservation of charge implies the H together must have gained or lost the same number of electrons. Maybe it would help to include an image.

@BuckThorn I am talking specifically about the valence electrons. I am confused why in a chemical reaction, the charge has to be the same on both sides. One way to think justify it could be that we start out and end up with the same total number of protons and electrons, so a deficiency (or excess) of electrons will cause the them resultant molecule to be charged aswell. But I feel this is oversimplistic, it would not be too far fetched to assume bonds could alter the charge

Are you aware that electrons have a charge of -1 and protons +1? I assume as much. Chemical reactions don't destroy or create electrons or protons, therefore the total charge (sum of protons minus electrons) has to be constant. Where electrons end up is another question. That can get awfully complicated and is one of the essential question chemistry asks ("what is a bond?")

I think your question might become clearer with a schematic illustrating an example of what you mean. How would a bond alter the charge? You write: "the charge has to be the same on both sides." Yes. Chemical equations are balanced. No fundamental particles (protons, electrons) are created or destroyed. However one thing that can happen is for atoms to gain or lose charge, but because protons and electrons are the charge carriers, the total charge has to be constant.

@BuckThorn Thanks for following up. I realised my question is more about net charge itself than the conservation of it. In polar molecules like HF there is no net charge, yet they can attract electrons - they behave similarly to ions. So I guess my question is what does a net charge (excess or deficiency of electrons) mean physically. It seems like chemical objects do not need to have a non-neutral net charge in order to attract other charged particles. For another example, take halogens, despite being electrically neutral they attract electrons.

I created a seperate post for my follow up question, I figured it was too physics related - https://physics.stackexchange.com/questions/808415/do-neutral-atoms-have-a-electric-field

@KrypticCoconut Ok, I understand, I had a suspicion that that was your question. The internal distribution of charge is essentially a quantum mechanical phenomenon. That is a hand-wavy way of saying that it is weird. Electrons are particles and yet they can "spread" themselves about a molecule. You can still try to edit your question to fit this site.

Your opening question at physics, "If a atom is neutral, does it mean it does not have an net electric field?", is easy: the total field about a neutral particle (atom or molecules) is zero. There are field lines emanating and ending at the particle, but the number emanating and ending is the same.

@BuckThorn Thanks for sticking out with my question. I'm confused by what you mean the total field is zero in a neutral atom. There will still exist points of negative electric field that repulse electrons outside the neutral atom right?

The number of electric field lines emanating from positive and negative charges is the same, just in opposite directions. If you have an electron and a proton enclosed in a volume and sum the lines crossing the surface they add to zero (where you add lines heading out, subtract those heading in). This is Gauss's law. Yes, there will be lines no matter how far you go, whether neutral or not. But the lines from plus and minus charges cancel so the net field will become weaker faster as you move away from a neutral molecule.


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Is there intramolecular H-bonding in 2-hydroxybenzonitrile?
Is there an intramolecular hydrogen bonding in 2-hydroxybenzonitrile? If so, then the $\mathrm pK_\mathrm a$ order should be meta > ortho > para but the actual order is meta > para > ortho.
I think since the $\ce{-CN}$ group is linear in shape hence the distance between $\ce{-OH}$ group and $\ce{-CN}$ is large so there should not be any hydrogen bonding between them.

You think right.


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Separation of dried blood using solvent
What is the best solvent to separate the metabolites from dried blood?

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

What metabolites? Short of knowing that, your question sounds a bit vague.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

According to the procedure that follows from Ref. [1] a mixture of water and methanol is an appropriate solvent:

Metabolites from DBS samples (3.2 mm diameter punch) were extracted onto 96-well plates in 100 µL 80% methanol. The supernatants (75 µL) were transferred onto new plates, dried under nitrogen, reconstituted in 75 µL 2.5% methanol, and transferred (65 µL) onto the final plates before injection.

Blood metabolites, unless associated with a carrier such as liposomes or BSA, are significantly water-soluble (salts or polar compounds). Methanol is thoroughly miscible with water but lowers the polarity of the mixture helping solubilize less polar compounds such as lipids, and methanol is also easily removed by evaporation.
Reference

Olarini, A.; Ernst, M.; Gürdeniz, G.; Kim, M.; Brustad, N.; Bønnelykke, K.; Cohen, A.; Hougaard, D.; Lasky-Su, J.; Bisgaard, H.; et al. Vertical Transfer of Metabolites Detectable from Newborn’s Dried Blood Spot Samples Using UPLC-MS: A Chemometric Study. Metabolites 2022, 12, 94. https://doi.org/10.3390/metabo12020094


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How to design a lab experiment showing differences between reversible and irreversible work?
Some of my students keep asking for more information about reversible and irreversible work. I've been trying to think of a demonstration/experiment that might highlight the broad concepts but haven't gotten far. Does anyone know of or have any experience on such demonstrations or experiments?

As inspiration, see this simulated experiment LearnChemE.com - simulations - thermodynamics - reversible-irreversible-expansion-compression-work

Make a list of processes that involve reversible work and a second list of processes that involve irreversible work.

Pull a weight on a rubber band, and let it snap back. Pull a weight on a rubber band until the band breaks. Where did that work go?

In a rapid irreversible compression, do you think that the gas satisfies the ideal gas law?

It probably does All the processes are the result of random molecular motion.

Each of them is a living example of a series of almost equilibrium reactions that extract work from a process. Compare eating a cookie, digesting it, getting energy to move etc., compared to just putting the cookie in a fire. The end products are the same: CO2 + H2O + energy. Processes are irreversible because they have the combination of negative deltaH and positive deltaS and cannot reach equilibrium. The trick is to control the entropy change.

@jimchmst
Thanks for your reviewing my post. Perhaps I wasn't clear in my original post, but by reversible and irreversible work I'm referring to expansion work (there's more on this in Atkin's Physical Chemistry). So, in essence I'd like a somewhat intuitive experiment or demonstration to show how the maximum expansion work for instance is the maximum work possible ...


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Which one has a greater dipole moment: benzamide or benzoic acid?
My guess is that, since the relative electronegativities are similar, therefore the difference would arise only due to magnitudes of difference in electronegativities. And hence the strength should be more for benzoic acid since it has more electronegative oxygen which boosts the individual dipole strengths.
But the answer in my textbook is the opposite given without a reason.
What is the fault in my guess, is the lone pair that makes a difference?

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Both are -R groups, check with the order and decide.

If you compare formic acid (1.415 D) and formamide (3.71 D) you note that the amide has the larger dipole moment, ie there appears to be a trend. If you add a ring it has a secondary (minor) effect of enhancing the dipole moments, more for formic acid (by about 0.3 D) than formamide (by about 0.05D), but we are motivated to ignore the ring and focus on the amide and carboxylic acid groups.
The planar structure of an amide can be explained by invoking resonance between two forms, one neutral with double bonded C and O, another with double bonded C and N and resulting partial charges on O and N. That separation of charges accounts for the large dipole moment of amides.

In this case, the lone pairs of both Nitrogen(Benzamide) and Oxygen(Benzoic acid) are in conjugation. Hence, when delocalization of electrons take place, a partial positive charge develops on both of them. As you stated, oxygen is more electronegative. Hence, relative to the Nitrogen of Benzamide, its ability to handle that positive charge is lesser. So, the tendency of delocalization of electrons is greater due to which it is actually feasible for it to exist with all of that partial charges as compared to the Oxygen of Benzoic acid. Then again, this is all an assumption since dipole moment is mostly calculated experimentally

Partial positive charge develops on both of them? Not at all. It should be clear from the second resonance structure of amide group.

@Mithoron second resonance structure shows positive charge on Nitrogen, what are you trying to say?

See Buck's answer.

no i was talking about the nitrogen of amide and oxygen of benzoic acid @Mithoron