text:
What is the difference between a single displacement reaction and a two competing reaction system?
For example, consider the single-displacement reaction
$$\ce{AB + C <=> BC + A}$$
My question is then if writing this reaction as the system
\begin{gather}
\ce{AB <=> A + B} \\
\ce{B + C <=> CB}
\end{gather}
is equivalent (surely, adding the reactions up gives the correct stoichiometric coefficients).
The thought behind this question is confusion regarding the notion of a "displacement" reaction. I am confused why in textbooks the displacement reaction is presented as if C is reacting with AB directly; wouldn't C instead be reacting with the free B present in the solution (i.e., the reaction is written more fundamentally as the two reaction system)?
The only time I can think of when this might not be true is if B being bound to A induces some electronic structure change localized on B that makes C able to react with B (and free B is unable to react with C).

Either is possible, but if AB is a normally stable molecule (in absence of C) then your second scheme seems unlikely, but only experimental data can sort this out.

Just read about SN1 and SN2.

As @porphyrin suggested, you can not make a general guess in absence of experimental data or chemical knowledge for a
concrete reaction.
Just a simple example of the first case, an elementary isotope exchange gas phase reaction
$$\ce{H_2 + D <=> HD + H}$$

Upon thinking about this most displacement reactions follow your scenario.
If the two reactions are independent then A is not necessary to the production of BC and is not in the equation for its production. This is a common fallacy in describing consecutive or concurrent reactions especially redox reactions that are often multistep and only accidently stoichiometric because the kinetics are too fast to see what happens. This is what is behind the idea of the net ionic equation for reactions. In solution NaCl +AgNO3 = AgCl + NaNO3 This happens regardless of the Na+ ion [but possibly not regardless of the nitrate because there are few soluble silver salts]. It is correct to write Ag+ + Cl- = AgCl not NaCl + Ag+ = AgCl + Na+ unless it were done in a solid-state reaction. This example meets your requirements because it is two separate independent reactions: the dissolution of the two salts and the forming of the precipitate. Let's dissolve some HCl gas in water HCl + H2O = Cl- + H3O+ Now the H and the Cl cannot be separated in a twostep reaction. Hopefully you woke some of us up a bit. Equations are a synopsis of a process. They might relate somewhat to the actual mechanism but mechanisms are usually more complicated than the equation. Keep that in mind and look behind the scene.


text:
Why silicon doesn't have the electronic configuration [Ne] 3s¹ 3p³?
I have read that half or fully filled orbitals provide more stability to the element due to symmetry and exchange energy.It is the reason for electronic configuration of Cr-[Ar]4s¹3d⁵.
Then, why doesn't silicon have the electronic configuration [Ne] 3s¹ 3p³? Doesn't that provide extra stability?
If this is not the case when do half and fully filled orbitals provide extra stability?

Because having the $s$ level happy is the biggest win here.

It depends what requires higher energy - the pairing electrons in the current orbital group, or filling the next orbital group instead. The energy difference 3s-3p is higher than for 4s-3d, with the latter even swapping the levels.

The ground state electronic configuration of silicon is $\ce{[Ne] 3s^2 3p^2}$. It seems that the atom would have gained more stability if it were $\ce{[Ne] 3s^1 3p^3}$ because half filled orbitals are very stable. But here is a s orbital which is actually fully filled in the former configuration. Since fully filled orbitals are also stable, this provides some stability.
But here comes another factor; since two electrons of opposite spin are paired, energy increased and stability decreases. But we also see that s orbital is closer to nucleus which means the positive energy of inter electronic repulsion is actually compensated by negative potential energy of attraction from the nucleus.
So these four factors are underlying below the configuration of silicon being $\ce{[Ne] 3s^2 3p^2}$.

The energy difference between the s and p orbitals in multielectron atoms is much much bigger than the difference between s and d orbitals when talking about transition elements.

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text:
Shouldn't the negative charge be on carbon in this scenario according to molecular orbital theory
I am sorry in advance for posting an image in this question. I know it makes searchability difficult, but I don't see any other way

The above screenshot is from Claydens organic chemistry. We can see clearly from the screenshot that the pi electrons jump to the antibonding pi* molecular orbital. As far as my memory goes oxygen is more electronegative than carbon, which in molecular orbitals would mean that the energy of the atomic orbital of oxygen is lower i.e. the pi orbitals of oxygen would contribute more to the sigma molecualr orbital as there is less difference in energy.
We would also see that the pi orbitals of carbon would be make more contribution to antibonding pi molecular orbital*
So my question is since the energy difference between the pi orbital of carbon and the anitbonding pi* molecular orbital is lesser in comparison with pi orbital of oxygen and the antibonding pi* molecular orbital,
Shouldn't the negative charge be on carbon by virtue of its energy being comprable to (pi*) orbital?
PS- I am reffering to the molecular bond formation between oxygen and carbon. If my questions sounds unclear please let me know, I will try to frame it more clearly
Thanks in advance to the answerer

Too much MO theory getting in the way. how would you draw the product with a negative charge on carbon?

I know carbon doesn't want to have a negative charge on it, but the MO theory suggests just that. Maybe I am wrong somewhere

When you form an additional bond, there is no more pi orbital. Or look at it this way. You add a hydride, two electrons and a proton. Two electrons (formally, and only for the purpose of visualising it) go to the pi* orbital, which would give 2- to the whole molecule (1- at oxygen and 1- at carbon). The proton then goes to the carbon and forms the bond. Where is the remaining charge?

Maybe just repeating what others have said, but in a different way - when determining the charge distribution of the product, you forget about the $\pi*$ orbital, since it's no longer there. Start fresh with an MO diagram of an alkoxide (ie the product). What does that look like and where does the charge end up?

It would seem the mechanism is a straight forward nucleophilic addition of hydride to the carbonyl group.  O- would be the most stable intermediate.  The reaction is a reduction of the carbonyl to an alcohol.
With carbonyl groups, the carbon is slightly more positive, the hydride adds there.

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text:
Is it possible to physically stretch US paper currency?
This is not a facetious question.
I am curious how one can physically stretch US paper currency, which is allegedly 75% cotton and 25% linen. Unlike cotton or linen fabric, paper currency does not appear to stretch when wet, hot, etc.
Is there a chemical process that can be applied to paper currency that will allow it to be stretched, but that may also keep the ink on it?
Edit: It should be noted that this is for personal use only -- I have no intent to distribute or circulate modified currency.

Tampering with currency may not be a good idea.  Find out if linen stretches.  Cotton does.

I would be doing this for personal use only. It does appear that linen does not stretch -- however I'm wondering if there is a way to modify the chemisty or material properties of the bills to allow them to stretch.

Anything can be stretched within limits. I think there is a law prohibiting the defacement or modification of money. Bills run thru the washer and dryer [one time] come out nice and crisp and shrunk so there is some empty space in the formula. Since paper making usually involves pressing the wet pulp into sheets and currency is rather dense I would think that there is not much leeway in further expansion. You could try an Instron or try soaking and pressing with a hot iron between cloth. The common way to enlarge things nowadays is to take a high resolution color accurate photo and enlarge.

@jimchmst - I am aware of that law, however this is for personal usage only and not for redistribution. I'm also unaware of this law (for bills) ever being enforced. Printing and rescaling is not an option here, I'd like to retain the original physical origin.

If you can't do it with a bill, try a coin in a press.

Another way to stretch currency  is to spend it judiciously. ;-)

Fibers hardly stretch, i.e. they have a very high modulus in the direction of the fiber. Woven fabric does stretch, and differently depending on the direction, because the fibers are oriented in a regular (rectangular) pattern. One direction has straight fibers and is very stiff, perpendicular they are woven and more elastic, and at a 45° angle, you can also distort the pattern, so the modulus is even lower.
Perhaps US dollar bills contain a nonwoven of some kind. Or several layers of fabric oriented in an an angle to each other.


text:
Problem with standard addition
I am working on the problem 8-23 of Skoog's textbook. In brief, a standard addition protocol is followed to determine the concentration of nitrite. In order to estimate the concentration, a colorimetric reaction is done using 1 mL of an unknown solution to which 24 mL of reagent is added before measuring the absorbance (A = 0.300). To 50.00 mL of the original solution is added 1.00 mL of a nitrite standard solution at 1 mM. The same color-forming procedure was followed and a new absorbance of 0.530 was obtained. In my opinion, the step involving the dilution of the solution with reagent (1 + 24 mL) should be ignored since it is repeated for both measurement. However, in the solution manual, the concentration 0,025 mM is corrected by multiplying by 25, so the answer given is 0.623 mM. I think the answer is wrong. What is your opinion?

Can you post the exact question in the text? Skoog has plenty of editions.

The description is too brief. If the calculation determines the concentration of added standard in the diluted solution and from that the concentration of analyte in the diluted solution, you do have to multiply by 25. You are not showing that part of the calculation, so there is no way to answer this.

The exact question is "The method of standard additions was used to determine nitrite in a soil sample. A 1.00-mL portion of the sample was mixed with 24.00 mL of a colorimetric reagent, and the nitrite was converted to a colored product that produced a blank-corrected absorbance of 0.300. To 50.00 mL of the original sample, 1.00 mL of a standard solution of 1.00E-3 M nitrite was added. The same color-forming procedure was followed, and the new absorbance was 0.530. What was the concentration of nitrite
in the original undiluted sample?" My answer is 2.49E-5 M, Skoog answer is 6.23E-4 M.

You correctly measured the concentration in the diluted sample. imagine had you used a different dilution.


text:
Does the acidity constant depend on the solvent?
Today I came across $pK_a$ values that depend on the solvent. For the hypothetical reaction $HA \rightarrow H^+ + A^-$, I had in mind that the acidity constant is defined simply as $K_a = \frac{\{H^+\}\{A^-\}}{\{HA\}}$. Therefore, no reference to the solvent seems to appear.
Of course, you wouldn't find protons by themselves, but would it still be a "pure" acidity constant if you used $H_3O^+$ and $H_2O$? In this case you would then have $HA + H_2O\rightarrow H_3O^+ + A^-$. As water is a liquid $\{H_2O\} = 1$. Thus, for the constants to be equal $\{H^+\} = \{H_3O^+\}$. But that should not hold for all solvents since different values come up, right?

The solvent acts here as an implicit Broensted-Lawry base. A weaker solvent base means lower acid dissociation. Acid dissociation is neutralization reaction HA + B <=> A- + BH+, where BH+/B is conjugate acid/base pair of the solvent. If as a solvent is used acetic acid, bases get much stronger and acids much weaker. And vice versa, if some suitable organic base is used as a solvent. This can be used for titration of very weak acids or bases. If the solvent is a mixture of more solvents (e.g. 95/5 vol% of ethanol/water), parallel equilibrii are established with all solvent components.

Right! The acidity is an equilibrium with the solvent so each solvent will have its set of equilibrium conditions and the appropriate equilibrium constant.

The hypothetical reaction you're referencing is an over-simplified version of the actual reaction that is taking place. In strict terms, the hydronium ion does no exist as "$[H^+]$", but rather as "$[H_3O^+]$". The complete reaction (if the solvent is water) would be:
$$\ce{HA(aq) + H_2O(l) <=> H_3O^+(aq) + A^-(aq)}$$
And the equilibrium constant for this reaction would be:
$$K_{eq}=\frac{[H_3O^+][A^-]}{[HA][H_2O]}$$
However, because the solvent (in this case, water) is in excess, its concentration is considered a constant value. This means $[H_2O]$ can be grouped up with $K_{eq}$ (another constant value):
$$[H_2O]*K_{eq}=\frac{[H_3O^+][A^-]}{[HA]}$$
This product is defined as:
$$K_a=[H_2O]*K_{eq}$$
So we have:
$$K_a=\frac{[H_3O^+][A^-]}{[HA]}$$
Even though using $[H_2O]=1$ leads to $K_a = K_{eq}$, that's not the case because $K_a ≠ K_{eq}$
The actual value of $[H_2O]$ at 25°C is:
$$[H_2O]=\frac{\rho_{H_2O}}{M_{H2O}}=\frac{1000 g/L}{18 g/mol}=55.56 mol/L$$
So, if we repeat the reaction using a different solvent "$S$":
$$\ce{HA(aq) + S(l) <=> H_3O^+(aq) + A^-(aq)}$$
$$K_{eq}=\frac{[H_3O^+][A^-]}{[HA][S]}$$
$$[S]*K_{eq}=\frac{[H_3O^+][A^-]}{[HA]}$$
$$K_a=[S]*K_{eq}$$
$$[S]=\frac{\rho_{S}}{M_{S}}$$
In other words, $K_a$ is a function of $[S]$
Since "S" in this case isn't water, it follows that:
$$[S]≠[H_2O]$$
In conclusion:
(1) For the same reaction at the same temperature, $K_a$ calculated with water as a solvent is different than $K_a$ calculated with a different solvent.
(2) $K_a$ depends on the solvent used.


text:
Doubt on the process to determine the amount of precipitate of a salt, product of a reaction between two solutions
I am attempting to determine whether there is going to be the formation of a $\ce{AgCl}$ salt precipitate if the following two solutions were to be added together:

$50mL$ of $\ce{NaCl}$, $0.04M$,
$50mL$ of $\ce{AgNO3}$, $0.05M$.

Knowing the $K_{sp}$ of $\ce{AgCl}$ is $1.6 \times 10^{-10}$.
My thought process is the following:

Determine the reaction that takes place: $\ce{NaCl + AgNO3 -> AgCl + NaNO3}$ (is this an equilibrium? How could I certainly say?),
Determine the amount of $\ce{AgCl}$ that forms at the end of the reaction - noticing that $\ce{NaCl}$ limits the reaction and that, thus, there are going to be at most $n$ moles of $\ce{Cl-}$. I would say the amount of $\ce{AgCl}$ that forms, at most, is $50mL \cdot 0.04M = 0.002mol$,

And at this point I'm stuck. If the solubility product, $K_{sp}$ of $\ce{AgCl}$, expresses the amount of ions that can be dissolved in the solution at the equilibrium, I would be able to say that, since the amount of silver chloride that forms ($0.02M$, in $100mL$) is greater than the value of $s$:
$$s = \sqrt{K_{sp}} = 1.265 \times 10^{-5}M = \ce{[Ag+]_{eq} = [Cl-]_{eq} = [AgCl]_{eq}}$$
from $K_{sp} = s \cdot s$, since there are going to be equal amounts of $\ce{Ag+}$ and $\ce{Cl-}$ ions dissolved in the solution, then I would assume that some $\ce{AgCl}$ would precipitate out: $\ce{AgCl}_{precipitate} = 0.002mol - 1.265 \times 10^{-5}mol = 0,0019987$, or $(107.9u + 35.45u) \cdot 0,0019987 = 0.286g$.
Are my assumptions on the solubility product correct?
Also, if I wanted to calculate the molar concentration of the ionic species ($\ce{Ag+}$ and $\ce{Cl-}$) in the saturated solution... wouldn't I take the previously found value of $s$ and divide it by the total volume of the solution ($100mL$)? That would be $1.265 \times 10^{-5}M$ for both $\ce{Ag+}$ and $\ce{Cl-}$?
Thank you!

Instead of $50mL$ ( $50mL$ ), write $\pu{ 50 mL}$ ( $\pu{ 50 mL}$ }. As typographically, there is space between the value and unit, with the latter never in italic. Similarly for M and mol.

Here is a summary of the steps I would use:

Determine the concentration of silver and chloride ions in the mixture, assuming that no silver chloride precipitates.
Compare the solubility product to the product of these concentrations.
If point 2. suggests precipitation, determine the equilibrium concentration of ions if requested.

The concentrations of silver and chloride ions are $\pu{0.025 mol L-1}$ and $\pu{0.02 mol L-1}$, respectively. Dropping the unit $\pu{mol L-1}$ and multiplying the values gives a product larger than the solubility product, so the solution is supersaturated and silver chloride will precipitate as the system attains equilibrium. At equilibrium, the following holds:
$$(0.025 - x)(0.02 - x) = \pu{1.6E−10} = K_{sp},$$
where $(0.025 - x) \pu{mol L-1}$ is the remaining concentration of silver ions, and $(0.020 - x) \pu{mol L-1}$ the remaining concentration of chloride ions. The silver ions are in excess by $\pu{0.005 mol L-1}$; because they form a 1:1 salt, this excess will remain as silver chloride precipitates.
If you try to solve this quadratic equation the usual way, you might run into numerical problems. Instead, we can substitute
$$ y = 0.02 - x$$ and solve the equation below instead:
$$(0.005 + y)(y) = \pu{1.6E−10} = K_{sp},$$
where $(0.005 + y) \pu{mol L-1}$ is the remaining concentration of silver ions, and $y\ \pu{mol L-1}$ the remaining concentration of chloride ions. To avoid the quadratic equation, estimating $(0.005 + y)$ as 0.005 gives a solution of:
$$y \approx \frac{\pu{1.6E−10}}{0.005} = \pu{3.2E-8},$$
very close to the physically sensible solution of the quadratic equation.
So the remaining concentration of chloride ions is about $\pu{3.2E-8 mol L-1}$ and the remaining concentration of silver ions is about $\pu{0.005 mol L-1}$.
This calculation assumed that the given values are exact, and no corrections for non-ideality (i.e. activity coefficients different from 1, volumes of solutions not additive, etc.) were attempted.

First let's calculate the initial amounts of both reactants, $\ce{Ag+}$ and $\ce{Cl^-}$. They are :
$\ce{n(Cl^-) = c·V = 0.05 L·0.04 M = 2·10^{-3} mol}$
$\ce{n(Ag+) = c·V = 0.05 L · 0.05 M = 2.5·10^{-3} mol}$
As a consequence, the ions $\ce{Ag^+}$ are in excess (and the excess is n($\ce{Ag+) = 0.5·10^{-3}}$ mol). So the amount of $\ce{AgCl}$ is probably equal to the initial amount of $\ce{Cl-}$, and it is $2· 10^{-3}$ mol. But it is necessary to know the level of the error on n$\ce{(AgCl})$ before going further away, and taking the solubility product into account.
Of course, n($\ce{AgCl}$) has a yet unknown uncertainty. But let's start from the beginning and assume that the minimum uncertainty on the initial concentration of $\ce{NaCl}$ is $0.04$ M ± $0.001$ M. This makes sense as no figures are given after the figure $4$ included in the value  $0.04$ M. This absolute error (or uncertainty) corresponds to a relative uncertainty on the initial n$\ce{(NaCl})$ and on the final n($\ce{AgCl})$ which is equal or bigger than $\pu{\frac{0.001}{0.04} = 2.5}$%. As a consequence, the absolute uncertainty on the initial amount of $\ce{NaCl}$ (and on the final amount of $\ce{AgCl}$) is equal or greater than $\ce{0.025·2·10^{-3} mol = 5·10^{-5}}$ mol. So the final amount  of $\ce{AgCl}$ is $\ce{n(AgCl) = (2 ± 0.05)·10^{-3}}$ mol.
Please note that this value is provisory, because it has been obtained without taking the $\ce{AgCl}$ solubility product into account. Let's do this calculation now.
As the solubility product of $\ce{AgCl}$ is $1.6· 10^{-10}$ $\ce{M^2}$, the remaining concentrations in the final solution are :
$\ce{[Ag^+] = n(excess)/V = \frac{0.5·10^{-3} mol}{0.1 L}= 5 ·10^{-3}}$ M
$\ce{[Cl^-] = \frac{K_s}{[Ag^+]} = \frac{1.6·10^{-10}}{5·10^{-3}} = 3.2·10^{-8} }$ M
The corresponding amounts of $\ce{Ag+}$ and $\ce{Cl-}$ remaining in the final solution are then :
n$(\ce{Ag^+) = [Ag^+]·V = 5·10^{-3} M·0.1 L = 5·10^{-4}}$mol
n($\ce{Cl^-) = [Cl-]·V = 3.2 10^{-8} M·0.1 L = 3.2·10^{-9}}$ mol
As this amount of $\ce{Cl^-}$ is remaining in solution, it means that $3.2·10^{-9}$ mole of $\ce{AgCl}$ should be removed from the amount of $\ce{AgCl}$ supposed to be precipitated previously : $(2 ± 0.05)·10^{-3}$ mol. So the exact final amount of $\ce{AgCl}$ is :
$\ce{n(AgCl) = (2 ± 0.05)·10^{-3} \mathrm{mol} - 3.2·10^{-9}}$ mol = $\ce{(2.00 ± 0.05)·10^{-3}}$ mol.
This long demonstration shows that the effect of the solubility product is smaller than the final errors. So it is negligible.

@Karsten. Thank you for the message. I have edited the correction.


text:
Why doesn't contact adhesive stick well to cured modified silane polymer?
A contact adhesive(dissolved neoprene rubber) sticks well
to rubbery smooth materials like rubber, so why
doesn't contact adhesive stick well to a cured modified silane  polymer (caulk like hybrifix super 7)?
Is there any adhesive that can stick to a cured piece of ms polymer?
(besides the ms polymer itself)
Thanks.
EDIT.
Ms polymer caulk does not stick well to all the materials,
so if you bond two objects (different materials) with ms polymer and
after it cured, one of the object gets loose, what type of adhesive can
you use in order to bond the loosen object to the cured ms polymer?
And why doesn't contact cement stick better to ms polymer?

Sorry for the downvotes, but what you may have is a more polar siloxane sealer, which would tend to repel "organic" carbon/hydrogen compounds.
The repulsion is similar to fluropolymers such as Teflon.
It may be possible to find a more compatible match for your contact adhesive by finding a sealant polymer with side groups that are more chemically matching.


text:
Standard Enthalpy of Formation
I have hard time understanding why in the definition of standard enthalpy of formation (SEF), the reactants and the products are required to have fixed temperature and pressure. Specifically, LibreTexts define SEF as follows: "SEF is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298K) is formed from its pure elements under the same conditions" (link here).
My question is the following: consider the case where we want to get 1 mole of O, given 1/2 mole of O_2. So, we start with a container with 1/2 moles of $O_2$ in 1atm, 298K. Then, we add heat to break the covalent bonds. Suppose we managed to have only isolated O atoms in our gas. However, the temperature and possibly the pressure are higher than the desired values (1atm, 298K). By adjusting the volume and the temperature, we can then reach the desired values. So now, ideally, we have 1 mole of isolated O atoms in 1atm, 298K. However, this is an unstable configuration, so we will not achieve it. Am I missing something here?

SEF does not say what path is to be taken to get from elements at STD conditions to the compound at STD conditions, as it does not depend on the path. It may have many formal and real steps.

Please explain : What is your question, Plemath ?

My question: how is it possible to end up with 1 mole of isolated oxygen atoms in 1atm, 298K? Such a configuration is unstable, and O_2 molecules will start forming.

Look in your table of SEF's.  Does it give a value for water vapor at 1 atm and 298 K?  How is this possible?  The answer to your question is the same as the answer to this question.


text:
How does one achieve a non-stick layer over a copper plate?
We normally perform the process of coating of aluminum on BOPP under vacuum.
In this process, we use a copper material holder to electrically evaporate the aluminum. With this, the balance aluminum vapors get cooled down and sticks to the holder.
Due to this sticking, we need to remove the aluminum by scraping it off which consumes a lot of time. Is there a way to make the holder non-stick to eliminate this process from the overall process?

I assume "scrapping" meant scraping the metal off. Feel free to correct if it was wrong.


text:
Osmotic pressure for non-electrical neutral region of a solution
For calculation of the osmotic pressure, we can use Van't Hoff equation as below:
π=CRT
If we put the solution in an electrical field, the cations and anions move towards negative and positive pols, respectively. To calculate the osmotic pressure near the poles in which the ions are not neutral (for example we have 3 times more Na+ than Cl- near the negative pole), Should we calculate the osmotic pressure by summing the molarity (C in Van't Hoff equation) of each ion and use the Van't Hoff equation?
π=(CNa++Ccl-)RT

3x more  cations than anions or vice versa would cause the discharge blowing solution from the vessel and possibly yourself from a room.

But, electrodialysis is working on this basis; the anions couldn't pass through CEMs and vice versa. It causes accumulation of ions near the membrane if they couldn't pass through it (for example accumulating of anions on the surface of CEM).

Could it be 1.5 times higher?

A very slight charge disbalance creates a strong potential gradient acting against external potential gradient.You can calculate it, using estimated solution permitivity.

Another thing is, if the excess is being eliminated by electrode reactions during electrolysis, with providing or withdrawing of electrons.


text:
Bond symmetry in AX2 type triatomic molecules
Will triatomic molecules of the AX2, AX2E, AX2E2 etc. (e.g. H2O or CO2), where the central molecule is bonded to two identical atoms, always have symmetric bond lengths in the ground state? If so, is there a simple explanation for why this symmetry should result in the lowest possible energy?

Now that's a good question.

The hydrogen atoms in carboxylic acid dimers are bound to two oxygen atoms, but the distances are different: https://commons.wikimedia.org/wiki/File:Carboxylic_acid_dimers.png. For $\ce{[FHF]-}$, on the other hand, either symmetrical or slightly asymmetrical structures are reported, doi:10.3390/cryst6010003

@Karsten I don't think these are quite what the OP is asking about. In the dimers though the element is the same on each side of the H, oxygen, they are in chemically different environments so one would expect asymmetry, and in the bifluoride case, while I need to read it in detail, the asymmetric cases are in crystals where the asymmetric field due to the other ions could be expected to distort the ion. I think what is being asked is is there a case for an isolated AX2 type molecule where the two AX bonds are of different lengths.

If I understand you correctly to my understanding I would expect it to always be symmetric in an isolated triatomic molecule unless there is an underlying asymmetry such as orbital occupation, which immediately made me think of Jahn-Teller - CO2 has a doubly degenerate HOMO, does anybody know the geometry of isolated CO2+, BO2 or similar?

Damn - I forgot Jahn-Teller only applies to non-linear molecules so forget that idea for CO2, and I think the HOMOs for both H2O and SO2 are non-degenerate

@IanBush The carboxylic acid dimers are in the gas phase. It is not clear whether the chemical environment is different because you can write to equivalent Lewis structures just like you would for benzene.

@IanBush As for the hydrogen bifluoride anion, the crystal environment could be symmetric or asymmetric, depending on the type of crystal symmetry. If the environment is symmetric but the anion is not, you might get disorder. If the environment is asymmetric but the anion is symmetric (in the gas phase), you might see a distortion. In any case, it is possible to discuss the symmetry of the anion (and tackle the question computationally if experiments are confusing).

When the central atom is bonded to equal atoms, the bonds does not need to be symmetrical. For example
NO2. The order of the bond is 1.5, and the order of the other bond is 2. So they are not the same longer.

You should read about resonance...


text:
How to remove a strong, (silicone?) adhesive that holds felt onto a tennis ball?
I want to put new felt on tennis balls. But first, how do I get the old felt off?
I tried heating up the felt to 305C using a soldering gun. I started at lower temperatures before that. The adhesive may have started to melt, but the felt started to burn. This wouldn’t be a problem in some situations, but smoke is a problem in this situation.
I placed one overnight in acetone, thinking that could break down the adhesive. Some of the yellow dye came off the tennis ball, but no adhesive.
I placed one in 5%HCl as well. Again, no luck.
Somewhat useful information can be found in a 3 minute How it’s Made video on Youtube. https://youtu.be/ovalwd7VZLU
These are tennis tennis balls, not dog tennis balls, not practice balls. I have to clarify because some cheaper balls use a cheaper adhesive.
It’s possible this is a silicone adhesive as opposed to some other kind of adhesive, which is part of the problem!

You try scoring it and just pulling it off?

I'm trying to remove the felt but keep the tennis ball in a "ball" shape. I'm trying to remove the felt from the silicone ball, not cut the ball in half.

Try hydrogen peroxide. There are lots of answers out there to remove glue.

Try using NMP or diglycolamine. NMP eats through all resins and diglycolamine amazingly eats through silicone vacuum grease as well


text:
Separate aluminium from brass threaded bushing without affecting the brass
I wanted to separate aluminium from brass threaded bushing without affecting the brass.
I also suppose that brass isn't the brass but it's a phosphor bronze.
Does brass and phosphor bronze amalgamate with mercury without scratching?
Is it possible to dissolve aluminium without affecting the brass?
I've also read that aluminium can corrode in the presence of brass, but I suppose brass is isolated from bushing or pressed really tight to prevent corrosion.

As brass contains zinc, it would probably get corroded, because aluminium and zinc have similar dissolution behaviour.

Try NaOH. Hot NaOH is needed for Zinc but aluminium is good with room temperature NaOH

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.

Aluminum in contact with any copper alloy will be anodic and corrode rapidly if any conductive liquid is present ( eg. water).

So hydrogen peroxide would be the best option? Corroding aluminium would not damage copper alloy?

@MalumWolfram, where did you the idea to use H2O2? The comments do not suggest it.

@DrMoishePippik I've read that $H_{2}O_{2}$ accelerates corrosion.


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In a liquid-in-glass thermometer, how does the gas pressure over the liquid affect the linearity of measurement?
In a common ethanol thermometer, the space above the alcohol column is filled with nitrogen. It also contains varying amounts of the alcohol in the gas phase.

Source: http://www.aashtoresource.org/university/newsletters/newsletters/2016/08/02/the-anatomy-of-a-liquid-in-glass-thermometer
At high temperatures, the gas phase volume decreases because the liquid expands. As a consequence, the pressure rises (lower volume, higher temperature, more ethanol in the gas phase because of the increased vapor pressure). Does this pressure increase contribute to the non-linearity of the measurement, i.e. position of the meniscus vs. temperature?
Even without a possible pressure effect, the rise of the liquid is non-linear with temperature because of the non-linear expansion coefficient of ethanol. I am wondering whether the pressure effect significantly contributes to the non-linearity. I am also wondering whether the expansion chamber (at the tip of the thermometer) helps to prevent excessively high pressures.

Considering high values of Young modulus for liquids, I doubt there would be significant compression effect.

@Poutnik I just saw the claim that liquids don't have a Young modulus because they are isotropic. They do have different compressibility, and this is much lower than that of a gas.

You are right. Ethanol has the bulk modulus 1.06 GPa and thermal expansion 750 ppm/K, what make 1 K temperature difference equivalent about 8 atm of pressure difference.

Here is a bit on how thermometers are built, filled, and given a scale: http://www.madehow.com/Volume-1/Thermometer.html, and a historic movie showing manufacturing of mercury thermometers.

[OP] Does this pressure increase contribute to the non-linearity of the measurement, i.e. position of the meniscus vs. temperature?

The main non-linearity is due to non-linear expansion of the body of the thermometer, and of the liquid in the thermometer (both expand with temperature, but the liquid expands more, leading to the rise of the liquid column). The graduations on a thermometer are equidistant, and the inside is built to have constant cross-sectional area. Therefor, you expect a non-linear response. This effect is smallest for a small temperature range, and largest for a large temperature range. If you look at the specifications of ASTM-certified thermometers, the temperature is supposed to be accurate to 0.01 °C for the smallest range thermometers (e.g. 18.9 to 25.1 °C, ASTM type 116C), up to 1 °C for the largest range thermometers (e.g. −6 to +400°C , ASTM type 11C).
Two other factors influencing the accuracy of the reading are the type of liquid (mercury, while highly toxic, has the advantage of not wetting the glass) and whether the thermometer is total immersion (more accurate) or not (less accurate because the temperature of the liquid above immersion depends on the ambient temperature).
So what about the gas pressure above the column? According to ASTM document E1-14,

[ASTM:] An expansion chamber is provided for relief of gas pressure to avoid
distortion of the bulb at higher temperatures.

So the pressure is indeed a potential problem, but it affects the shape of the bulb (i.e. bends the glass) more than the volume of the liquid.


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Why does carbon dioxide forms simple molecular structure, but silicon dioxide forms giant covalent structure?
I was wondering about this in class, drawing the structures of carbon dioxide and silicon dioxide.
Carbon and silicon are both in Group 4/14, but coming up with oxygen, one can only form a simple molecular structure while the other can form a giant covalent structure. I found no satisfactory in my chemistry textbook and online. What difference between carbon and silicon is this caused by?

Silicon is larger than Carbon, so Silicon's electron cloud is more diffused. Thus, if a hypothetical  bond was formed between Silicon and Oxygen, the p - orbitals would not be able to come close enough to overlap enough to form a sufficiently strong  bond. The  bond would be too weak, and not enough energy would be released in this hypothetical bond formation, so this is energetically unfavourable.

Compare it to the analogical case $\ce{N≡N}$ versus tetrahedral $\ce{P4}$ with 3 single bonds per atom.
Carbon forms in $\ce{CO2}$ or $\ce{CS2}$  2 double bonds, while silicon in $\ce{SiO2}$ or $\ce{SiS2}$ 4 single bonds.
Generally, elements from the 2nd period (C, N, O) have a stronger ( C,N much stronger) tendency to form double/triple bonds than their counterparts in the 3rd period ( Si, P, S), mostly because of better overlapping of p orbitals at forming $\pi$ bonds.
If there is bad orbital overlapping, eventual $\pi$ bonds are weak and multiple single bonds are energetically preferred to 1 multiple bond. By other words, forming hypothetical $\ce{O=Si=O}$ would release much less energy than forming polymeric $\ce{(SiO2)x}$

What about the analogous $\ce{CS2}$ and $\ce{SiS2}$? Is there something special about 2nd period elements making multiple bonds with each other, or is it sufficient to have one 2nd period bonding partner for a tendency to form multiple bonds.

CS2 is molecular, SiS2 is polymeric, even if differently to SiO2.

CS2 is also notoriously unstable to combustion.


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Can an element be a single atom or a molecule made up of atoms of the same element?
Some online websites and some books as well suggest that elements are either atoms (e.g. Ne) or molecules (e.g. $\ce{H2}$, $\ce{O2}$).

Original source: Tro, Chemistry: A Molecular Approach (2008)

Other sources say that only Fe and Cu are elements and $\ce{H2}$ and $\ce{O2}$ are not.

Source: https://slidetodoc.com/matter-matter-anything-that-has-mass-and-takes-5/

Can anyone explain and provide a reputable reference to the correct definition?

Refer to IUPAC Gold book that is taken in context of chemistry as the authoritative reference for term definitions.

I have tried to look up there but that didn't provide examples so I was not able to grasp what exactly it wants to say. I searched element , it shows look for chemical elements, I looked for that and it gives 2 definitions , the second one seems like it includes molecules as well. Can you confirm it?

://doi.org/10.1351/goldbook.C01022.
1-A species of atoms; all atoms with the same number of protons in the atomic nucleus.
2-A pure chemical substance composed of atoms with the same number of protons in the atomic nucleus. Sometimes this concept is called the elementary substance as distinct from the chemical element as defined under 1, but mostly the term chemical element is used for both concepts.

Generally, there is a frequent case that a term does not mean either this(1) either this(2), but there is defined meaning term(1)=this(1) and term(2)=this(2). E.g. an orbital has 3 distinguished meanings (a wave function, a quantum state, a 3D region)

chemical element(2)=A pure chemical substance
composed of atoms with the same number of protons in the atomic nucleus  O, O2 and O3 are all composed of atoms with the same number of protons in the atomic nucleus, therefore are chemical element(2)
.

https://chemistry.stackexchange.com/questions/2879/what-is-the-definition-of-of-compound-mixture-element-and-molecule

I read that old post beforehand but I couldn't find uniformity in the answers. It was still confusing and also I wanted to ask specific difference so I posted this one.

Here is one possible definition:

each of more than one hundred substances that cannot be chemically interconverted or broken down into simpler substances and are primary constituents of matter. Each element is distinguished by its atomic number, i.e. the number of protons in the nuclei of its atoms.
Source: Oxford dictionary

This starts with the classical definition (can not broken down into simpler substances) and then gives the modern picture of subatomic particles.
Here is the more technical definition:

1-A species of atoms; all atoms with the same number of protons in the atomic nucleus. 2-A pure chemical substance composed of atoms with the same number of protons in the atomic nucleus. Sometimes this concept is called the elementary substance as distinct from the chemical element as defined under 1, but mostly the term chemical element is used for both concepts.
Source: IUPAC Gold book, doi.org/10.1351/goldbook.C01022 as cited by the OP in the comments

So samples containing pure iron, pure argon, pure dioxygen ($\ce{O2}$) or ozone ($\ce{O3}$), or a perfect diamond are all elements. The bonding is different in all of these (or absent, as in argon), but all are considered elements. Among the examples, only dioxygen and ozone form molecules.
The first IUPAC definition (1-A) implies that you can e.g. say water is made up of the elements hydrogen and oxygen, even though water itself is a compound and not an element. The IUPAC definition also makes clear that you can use the term element either at the macroscopic level ("This sample is pure iron, an element") or at the particular level ("The helium atoms are diffusing through the membrane. Other elements, with larger atomic radii, do not diffuse readily").

[OP] Can an element be a single atom or a molecule of same atom?

Yes, element can refer to single atoms, to a molecule or, at the macroscopic level, to a sample made up of atoms grouped as molecules or not. The only requirement is that all the atoms in your set have the "same number of protons in the atomic nucleus".

I believe that same elements should have same chemical properties. If O2 and O3 are both elements then why do they differ in properties?

The only property that has to be the same is the number of protons in the nucleus. Isotopes have different mass, graphite and diamond have different market value, red phosphorus and white phosphorus have different color, liquid nitrogen and common nitrogen have different temperature etc. All they mean to say is don't rip apart the nucleus, then you will no longer have the same element.

Good question, there is an entire book dedicated to your question: What Is A Chemical Element? A Collection of Essays by Chemists, Philosophers, Historians, and Educators, 2020, published by the Oxford University Press.

The concept of a chemical element is foundational within the field of
chemistry, but there is wide disagreement over its definition. Even
the International Union for Pure and Applied Chemistry (IUPAC) claims
two distinct definitions: a species of atoms versus one which
identifies chemical elements with the simple substances bearing their
names. The double definition of elements proposed by the International
Union for Pure and Applied Chemistry contrasts an abstract meaning and
an operational one.

You may wish to get hold of this book!
At one time, element was a substance that resisted further analysis, i.e., it was impossible to break it down into simpler substances. This implied classical chemical analysis. For example, if we chemically try to break down the Fe atom by heating in an ordinary laboratory, it will be an unsuccessful attempt. The substance after heating at laboratory temperatures followed by cooling will still behave the same.
IUPAC modernized this definition, by stating that A species of atoms; all atoms with the same number of protons in the atomic nucleus. It is a collective and a comparative label for a group of atoms.
As an example, if we isolated an argon atom from a bottle of argon gas, we could not conclude that it was an element without counting and comparing its protons with those from other atoms in the bottle. In that bottle, if we find that all atoms have 18 protons, we can label it as an argon gas, since each atom has 18 protons.
Now coming to your main question: What about Fe vs. $\ce{O2}$. Iron is clearly an element by elementary definitions. You can say that $\ce{O2}$ is a molecule formed by two identical chemical elements, called oxygen.
However, now in everyday parlance, we can say that many elements exist as molecules in nature but not as free atoms like Ne, Ar, Kr, Xe. For example the element sulfur exists as $\ce{S8}$, the element called oxygen exists as a diatomic molecule $\ce{O2}$. The element are oxygen and sulfur atoms but in nature they exist as molecules. If someone gives a solid yellow piece of sulfur in your hands, colloquially, we might say that we are holding the element sulfur, but in effect the actual referal is to individual sulfur atoms, regardless of the fact that the atoms are chemically bonded to each other.

Level1: Some of the properties of elements are related to their molecular structures including the fact that they can form different molecules or different crystal structures or different compounds with other elements and even different nuclear reactions for the different isotopes. This is something to learn about but not to have beliefs about.

@AChem and jimchmst thank you both of you for providing further information.


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Why is the bond angle of SO2 less than 120 degrees?
I've learnt about VSEPR theory and have a general idea of how lone electron pairs affect the molecular geometry by repulsions. But my teacher told me that when the angle between bond pairs or lone pairs is 120 degrees, there is no repulsion. $\ce {SO2}$ has $sp^2$ hybridization and thus has trigonal planar electron geometry. Thus, angle between the lone pair and each of the $\ce {S-O}$ bonds is 120 degrees. This means there should be no repulsions and the bond angle between the two $\ce {S-O}$ bonds should remain 120, instead of becoming less than that. So then why is it, in fact, less than 120?
Edit: The possible duplicate to my question is answered by saying that (a) VSEPR isn't perfect and (b) in this case it actually does do a pretty good job of explaining the bond angle.

...the results give a bond angle ∠(O−S−O)≈119∘. This is indeed very close to 120∘ so you can say that the VSEPR method gives a good approximation of the SO2 bond angle.

The essence of my question revolves around the fact that lone and/or bond pairs separated by an angle of 120 degrees do not create repulsions, so considering that, why is the $\ce{O-S-O}$ bond angle compressed down to 119?

It is written not as $\ce SO_2$  ($\ce SO_2$), but as   $\ce{SO2}$  ($\ce{SO2}$) // Note that CH SE prefers plain text in titles.

@Mithoron I've edited my question to explain why that is not a duplicate.

Under the VSEPR model, rationalization of bond angle compression to less than 120$^\circ$ can be made by examining how diffuse a lone-pair is relative to a bond. In a bond, the electrons are confined to the band of space between the two atoms—the presence of the positive nuclei on each end attracts electron density into a tube, which makes up the bond(s). A lone-pair does not have this same restriction; instead, a lone-pair is only bound to a single atom (by definition) so further from the nucleus, the electrons can diffuse across a larger space. There is also less positive charge as you move away from the nucleus, and so the negative charge of electron density over there is more potent. This has the net effect of a lone-pair being more repulsive than a chemical bond, which causes the $\ce{O—S—O}$ bond angle to compress.
In reality, most bond angles are a slurry of effects from various lone-pairs and bonds influencing what angles are most stable. Addressing the nature and impact of these factors on chemical reactivity is called sterics.

I understand why there is repulsion in general, but shouldn't the 120 angle between lone pair and $\ce{S-O}$ bond mean there shouldn't be repulsion at all, and thus no compression of bond angle? (See edited question)

A lone pair is more cumbersome than a covalent bond. It repels more strongly the covalent bonds.

@AVS probably best not to think about it as "no repulsion at all" and think of 120 degrees as "where 3 equally repelling things will be relative to each other because that is the farthest they can get from each other"

Oh, so then since the 3 "things" in this case are not equally repelling (two bond pairs and one lone pair) the repulsion still takes place. Have I understood this right?

@AVS mostly—i would still say when the three "things" are equal, there is repulsion. it's just the least repulsion possible or "minimal repulsion" which you could consider as a definition for no repulsion. if you add all the force vectors, there is no net repulsion, but the force vectors are still there.

Oh, this nicely propagates the myth of the "lone pairs need more space" argument. It is a terrible post-rationalisation for a concept that failed. Sorry, but this is an explanation from the last millennium, hence my down-vote.

@Martin-マーチン this is much simpler to frame in the context of the VSEPR model than discussing the linear algebra necessary for justifying slight changes in hybridized orbital p-character and whatnot. the normalized electron density of a lone pair orbital still must come to two electrons, but the orbital certainly occupies a WIDER space (more spherical electron density) which induces more charge and Pauli repulsion. the explanation retains the core of what’s going on—VSEPR theory isnt bad, it’s just a model with limited uses. do you use general relativity when trying to model basketball throws?


text:
GLP mass to volume conversion problem (Rotary Meter)
this is my first post here on this forum.
I would like to share a question that is bothering me:
I have a gas powered system.
I would like to measure the total gas consumption in kg.
The problem is that the meter I use gives me the value in m³.
I use the g10 DN40 threaded delta meter from the link below.
It is a rotary principle meter.
https://www.itron.com/-/media/feature/products/documents/brochure/delta-en-web.pdf
Currently I use the factor of 2.5kg/m³ for the conversion.
The system is periodically supplied with lpg gas and the supplier gives me an invoice with the total value of gas supplied in kg.
When adding up the total supply carried out in the cylinders within a period of a few months and comparing with the consumption measured by the meter, in the same period, I always obtain lower consumption values. Could it be that the company is supplying me with less gas than it says.
Note: The gas meter is not right after the cylinders.
he is after a few meters of pipes and after a pressure reducer.
Do I need to do a temperature and pressure compensation to get another conversion factor from kg to m³?

Maybe - but you're looking at conservation of mass, so my inclination is that your conversion factor is not quite right because you certainly don't have the value for pure propane or butane. Have a look here for conversion factors and report back.

@ToddMinehardt
The conversion factor that is on the website is a lower value (1.8315 kg/m³ ) than the conversion factor I use, it would make the measurement worse.


text:
What is the nature of container when talking about Boiling Point of a Liquid [ Vapour Pressure related ]?
From what I understand, whenever talking about calculating the value of Vapour Pressure we need to take a closed container as there needs to be an equilibrium established {R evaporation = R condensation} for us to be able to find it's value.

My question is, whenever we talk about finding the boiling point of a liquid -
Boiling point of a Liquid is the temperature at which the Vapour Pressure of the Liquid becomes equal to the External / Atmospheric pressure.
So the container taken here need to be rigid or non-rigid ?
Both cases kinda made sense to me, but are equally confusing because I don't think both can be right at the same time.
Because if we take a non-rigid container so that the atmospheric pressure can be felt inside the container, but in real life examples we mostly deal with rigid containers.

For a calculation, you just assume some idealised conditions. No need to take into account the container wall properties.

@Karl then how would the increasing vapour pressure know that it has become equal to the External pressure on it and then the liquid starts boiling, because in a rigid container, the atmospheric pressure cannot be felt right ? [Now obviously the vapour pressure doesn't have a conscience, but please try to understand what I'm trying to say]

Boiling point as normally measured implies an open container as, by definition, it is at atmospheric pressure. Measuring vapour pressure implies an equilibrium has been established but that can be approximated well even in open containers if the setup is right.

Measuring of the equilibrium vapour pressure and measuring of the liquid boiling point are operations with very different settings.
Vapour pressure measurement is based on measuring of equilibrium pressure in a container, containing the liquid and pure liquid vapour at the given temperature.
Boiling is non-equilibrium process. Measuring of boiling point is done at constant pressure of open container.

Vapour pressure dependence on temperature can be measured alternatively as boiling point dependence on chosen external pressure.

Near the boiling point, there are being continuously created nano bubbles by random local processes (mostly local overheating), filled by saturated vapor. Hydrostatic pressure (external pressure included) of surrounding liquid is either high enough to make them disappear back to the liquid, either is too low, bubbles rapidly grow and leave the liquid.
The container is filled by vapour, that has the same pressure as is the external pressure.

That actually makes sense, but let's suppose another scenario we are boiling a liquid in a closed container (whatever may be the nature of the container), then how would the increasing vapour pressure know that it has become equal to the External pressure on it and then the liquid starts boiling, because in a rigid container, the atmospheric pressure cannot be felt right ? [Now obviously the vapour pressure doesn't have a conscience, but try to understand what I'm trying to say]

If you heat liquid in a  closed container where is some gas already that the liquid would never boil, as the pressure would be ALWAYS higher than vapour pressure (assuming slow heating, without significant local overheating ).

@ Paras Gupta: You may need to read your physics notes to understand pressure calculations. In here we are talking about equilibrium pressure closed to the liquid surface. Thus, it doesn't matter liquid is in rigid or non-rigid vessel. It should be open (see the answer).


text:
Help with chemical bond associated with 1500cm-1 and 1600cm-1
I have IR spectra of polystyrene with characteristic peaks at 1450cm-1. 1500cm-1 and 1600cm-1. What chemical bond is associated with 1600cm-1 and 1500cm-1?

As shown here (Figure 2), the lines at 1600 and 1492 (~1500) cm$^{-1}$ correspond to aromatic ring modes.
From my doctoral dissertation (Classical and Quantum Dynamics of Vibrational Energy Flow in Benzene: the CH($\nu = 2$) Overtone. Minehardt, T. J. The University of Texas at Austin. May, 1999):
the specific modes correspond to E$_{1u}$ (CC stretch and in-plane CH wag) at 1494 cm$^{-1}$; and E$_{2g}$ (CC stretch, CCC bend, and in-plane CH wag) at 1607 cm$^{-1}$.
Thus, the corresponding bonds are carbon-carbon (stretch), carbon-carbon-carbon (bend), and carbon-carbon-hydrogen (in-plane wag).


text:
Relation between activation and threshold energies
Activation energy $E_\mathrm{a}$ and threshold energy $E_0$ appear to be equivalent quantities:

In particle physics, the threshold energy for production of a particle is the minimum kinetic energy a pair of traveling particles must have when they collide.

In chemistry and physics, activation energy is the minimum amount of energy that must be provided for compounds to result in a chemical reaction.

How are activation energy and threshold energy defined, exactly? Why do both terms exist? Is there a context where both terms are used, and they refer to different aspects of the same phenomenon?

@Karsten Sometimes, the context is clear and OP does not ask for the solution to be done. He implicitly asks to do figurative homework of searching for him. It is sometimes clear no searching was done, as using his own post keywords lead to an answer.

@Karsten I use multiple canned comments, one of them is:  The site wants to prevent others to provide what users already know or can easily find. Therefore, it expects  questions to contain explicit compact summary of prior effort to answer the question, based on user  knowledge and on searching for existing related  info or answers.

The activation energy $E_a$ of a reaction is the amount of energy required to take reactants to products across some reaction coordinate. The threshold energy $E_0$ is related, but defined as the minimum kinetic energy of a particle in order to bring about collisions where atom-exchange is possible. The terms are mostly used in different contexts (activation energy is common in discussions of organic synthesis, but kinetic energy of particles isn't) but they also are not fully congruent. Specifically, reactions that don't require much collision for reaction progress may not have them congruent.
For example, the activation energy of thymine-thymine dimerization in DNA strands is large—this energetic barrier is surpassed when ultraviolet light excites the appropriate electrons and initiates the dimerization. This process is ambivalent to kinetic energy (inside the DNA strand, the nucleic acid residues do not move very much.)
Regardless, the concepts are very close and mostly a matter of context. Threshold energy can more easily be extrapolated to other contexts, like particle physics. Chemical reactions are a subset of particle collisions, and the term is more relevant there.

Arrhenius activation energy $E_\mathrm{A}$ is one more closely related quantity that needs to be distinguished from activation and threshhold energies. The difference between the quantities is briefly outlined in IUPAC Green Book [1, pp. 64, 67]:

The term activation energy is also used for an energy threshold appearing in the electronic potential (the height of the electronic energy barrier).
For this “activation energy” the symbol $E_0$ and the term threshold energy is preferred, but $E_\mathrm{a}$ is also commonly used.
Furthermore, $E_0$ may or may not include a correction for zero point energies of reactants and the transition state.
It is thus recommended to specify in any given context exactly which activation energy is meant and to reserve (Arrhenius) activation energy only and exactly for the quantity deﬁned in the table.
$E_\mathrm{A}$ depends on temperature and may be written $E_\mathrm{A}(T).$

$E_0$ is the threshold energy for reaction, below which no reaction is assumed to occur.
In transition state theory, it is the difference of the zero-point level of the transition state and the zero-point level of reactants.

Reference

IUPAC. Quantities, Units and Symbols in Physical Chemistry (the “Green Book”), 3rd ed.; Cohen, R. E., Cvitaš, T., Frey, J. G., Holmström, B., Kuchitsu, K., Marquardt, R., Mills, I., Pavese, F., Quack, M., Stohner, J., Strauss, H. L., Takami, M., Thor, A. J., Eds.; RSC Publishing: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.


text:
What is the flash point of acetylene?
On episode s10e11 of the show Criminal Minds, a confrontation occurs in a freezer full of acetylene gas. In this confrontation, the main character confidently fires her gun because the freezer was "18 below" which is below the flashpoint of acetylene.
This fact seemed ludicrous to me because if the temperature really was below the flashpoint of acetylene, then my knowledge of how flashpoints work tells me that the acetylene should be condensing out of gas phase into either solid or liquid, and acetylene sublimes at -84°C.

So what really is the flash point of acetylene?
PubChem says it's -18°C citing an AirGas MSDS. I've tried to search for a primary source, but I feel like all the MSDSs are just citing each other.
The flash point must be below the sublimation/boiling point right? (Which is -84°C for acetylene)

I believe your reasoning is correct. Perhaps the Airgas MSDS refers to acetylene dissolved in acetone? We can't know for sure until we find the original source!

Considering acetylene dissolved in acetone (liquid pressurised acetylene would be powerful unpredictable explosive), they would rather measure the flashpoint of acetone, which is -20 Deg C.

The SDS on this gas states -18C or -.29F as flashpoint. In the show, that would be referencing -18F which would tranlaste to almost -30C, in which case she is right, and they are well below the flashpoint stated on the official Safety Data Sheet.

My comment is getting cited by Google, so here's my best attempt at calculating the flash point. I'd still love for someone to give a primary source for the -18°C number though.
In 1915, Burrell and Robertson did quite a bit of research on acetylene. They published one paper1 showing that acetylene has a range of flammability behaviors between 2.4 and 3%, but concluded that the Lower Explosive Limit for acetylene is 2.53%. In another paper2, they calculated the vapor pressure of acetylene all the way down to -143°C, but they flipped a sign in their equation and so I don't think modern sites like NIST picked it up. (It should be + 0.001511).

Using their corrected equation along with the 2.53% Lower Explosive Limit, I calculate -124°C (149 K) as the flash point of acetylene.
1Burrell, G. A., & Oberfell, G. G. (1915). The Explosibility of Acetylene (Vol. 112). US Government Printing Office.
2Burrell, G. A., & Robertson, I. W. (1915). THE VAPOR PRESSURES OF ACETYLENE, AMMONIA AND ISOBUTANE AT TEMPERATURES BELOW THEIR NORMAL BOILING POINTS. Journal of the American Chemical Society, 37(11), 2482-2486.

The UK National Fire Chief Council reports the flash point of acetylene with $\pu{-17 ^\circ{}C}$ (reference); while the US American (Californian?) firefightinsider states $\pu{-0.7 ^\circ{}F}$ or $\pu{-18.15 ^\circ{}C}$.  It didn't take long to find these values (open access) on the internet.  I speculate average Jane and Joe (regardless if they are [gas] welding, or not) don't mind the small difference but treat the gas with the respect it deserves.

Well so, you've identified the crux of the issue again here. Everyone reports values of -18C, but where did they get that number from? Acetylene boils at -84C, so presumably it's flash point should be around -100C or something. Already at -84C, acetylene is capable of reaching 100% air concentrations.

There still is the possibility to get in touch with the suppliers in the field (e.g. AirLiquide) for a literature reference to cite. Though, for safety reasons (e.g., to prevent decomposition when closing the valves) their gas cylinders typically contain it as solution in acetone on solid support …


text:
Why does DIBAL-H not reduce esters to alcohols directly?
This source says that DIBAL-H reduces esters to aldehydes. Further, it says that, like NaBH4, it reduces aldehydes too.
Why in reduction of ester, the aldehyde so formed is not converted to alcohol, if it can reduce both ?

It depends on the reduction conditions.
1 eq of Dibal-H takes esters to aldehydes at low temperature. Use multiple eqs at a higher temperature and you get further reduction. This is because of the formation of a hemi-acetal aluminium complex intermediate that is stable at low temperature and resistant to further reduction.

DIBAL is an “electrophilic” reductant. That is, the first step in the reaction is coordination of a lone pair from the carbonyl oxygen (a nucleophile) to the aluminum (electrophile). It is only after coordinating to its carbonyl host that DIBAL delivers its hydride to the carbonyl carbon, resulting in formation of a neutral hemiacetal intermediate that is stable at low temperatures.

source here

Thanks sir but when we say 1 eq. , we theoretically make product that would be formed if only one molecule react. But in the solvent, there are many molecules and practically we should get more reduction products even if 1 equivalent is used?

@Answer_Me  if you control the temperature of the reaction the intermediate (top right of the diagram) is stable and does not react with more reagent.  Of course, if you add reactant too fast and get local heating then it can react and reduce your yield. In practice, run at -75C and with slow addition, this is a good high yielding reaction.


text:
Conditions for Pd/C reduction of alkene
I am trying to do a $\ce{H2}$ reduction of cyclic alkene (otherwise simple hydrocarbon with 1 methyl group along with the chain about the weakly-strained cycle) using  $\ce{Pd/C}$, the cycloalkene neat, and some glacial acetic acid.  I am starting with a  $\ce{N2}$ atmosphere, and charging the vessel (Parr) with high pressures of  $\ce{H2}$ $(\gt \pu{400 psi})$ and from ambient to elevated temperatures $(\pu{150 ^\circ C})$.
But the reaction isn't working (GC data only)... What am I overlooking? Or doing wrong?  (I'm a chemist, but have never done this).
Could I be reducing the  $\ce{N2}$, producing  $\ce{NH3}$ and poisoning the  $\ce{Pd}$?  Am I sintering the  $\ce{Pd}$ by adding too much  $\ce{H2}$ too quickly or heating too high? Simply a slow reaction and I'm been impatient (running $\gt \pu{100 h}$ though)?

This is a simple reaction that should work easily at much lower pressures. Have you tried using a different batch of catalyst? Have other researchers used this batch successfully recently?

I agree, the few hydrogenations I have done were at low pressure and ambient T. The reaction was smooth and fast, monitored by the lowering of H2 pressure. Research some methods in the library not the internet

You started with a nitrogen atmosphere. Did you purge with hydrogen? Parr apparatus should be in a hood.

Catalytic hydrogenation is a fundamental reaction, specifically in organic chemistry (Ref.1). Yet, the reduction of multiple bonds using hydrogen gas $(\ce{H2})$ and a metal catalyst (e.g., $\ce{Pd/C}$) is a reaction familiar to all organic chemists. That is the same approach taken by OP in the reduction of cycloalkene to corresponding alkane in this question. However, OP has not given detailed reduction condition(s) so that we cannot answer OP's specific questions (For example, no GC traces have shown, so we don't know whether some side reactions happened). Thus, I'd attempt to give a different approach, which does not require $\ce{H2}$ gas thus, no need to control pressure and can avoid hazardous conditions such as flammability.
For this method called catalytic hydrogen transfer reduction of carbon-carbon double bonds, a plethora of examples has been published Ref.2). It is important to note that variety of solvents (e.g., methanol, ethanol, isopropanol, etc.) can be used as hydrogen-donor in this method. In theory, any organic compound whose oxidation potential is sufficiently low can donate hydrogen in this method, of cause, in the presence of catalyst. Thus, the choice of donor is generally determined by the ease of reaction and availability (Ref.2).
Recently, in a study of catalytic transfer-hydrogenations of olefins, glycerol has been used as the hydrogen donor and solvent (Ref.3-5). A typical reaction is illustrated in following image:

In general procedure is given in Ref.3:

In a typical procedure, $\pu{0.50 g}$ of cyclohexene and  $\pu{0.03 g}$ of (5 wt.%) $\ce{Pd/C}$ were added to a vial with  $\pu{5 g}$ of glycerol (all purchased from Aldrich). The mixture was placed in a preheated oil bath and heated to  $\pu{70 ^\circ C}$ after which it was magnetically stirred for  $\pu{5 h}$. At the end of the reaction, the reaction mixture was cooled and extracted with  $\pu{2 mL}$ of diethyl ether. Finally, the ether phase was analyzed by GC analysis using an HP-5 column ($\pu{30 m} \times \pu{0.25 mm}$, $\pu{0.25 \mu m}$ thick) to determine the conversion.

References:

S. Nishimura, In Handbook of Heterogeneous Catalytic Hydrogenation for Organic Synthesis; John Wiley & Sons: New York, NY, 2001.
Gottfried Brieger and Terry J. Nestrick, "Catalytic transfer hydrogenation," Chem. Rev. 1974, 74(5), 567-580 (ODI: https://doi.org/10.1021/cr60291a003).
Dorith Tavor, Sergay Popov, Christina Dlugy, and Adi Wolfson, "Catalytic transfer-hydrogenations of olefins in glycerol," Org. Commun. 2010, 3(4), 70-75 (ODI: Has not given).
Robert H. Crabtree, "Transfer Hydrogenation with Glycerol as H‑Donor: Catalyst
Activation, Deactivation and Homogeneity," ACS Sustainable Chem. Eng. 2019, 7(19), 15845–15853 (ODI: https://doi.org/10.1021/acssuschemeng.9b00228).
Adi Wolfson, Christina Dlugy, Yoram Shotland, and Dorith Tavor, "Glycerol as solvent and hydrogen donor in transfer hydrogenation–dehydrogenation reactions," Tetrahedron Letters 2009, 50(43), 5951-5953 (ODI: https://doi.org/10.1016/j.tetlet.2009.08.035).


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Is there a way to neutralize aluminum hydroxide in pool water?
Pool water sat unfiltered for 12 months with heavy leaf debris. Once manual removal was completed a variety of chemicals were introduced. In particular alum powder. This alum powder has decomposed into sulfuric acid H2SO4 and aluminum hydroxide Al(OH)3. Aluminum hydroxide is a colloid that remains in suspension, producing a cloudy effect.
Is there a way to neutralize this effect so that the suspended particles drop and water clarity is regained?

Dissolution would probably mean it is not pool water anymore. // Mechanical filtration // water replacement

@Poutnik something about the way the problem is formulated suggests this is homework and has a tidy answer involving chemistry. Could be wrong though.

@BuckThorn It is a followup question to this, Not sure about being a HW, seems like a desperate pool keeper throwing in water whatever was sitting on the shelf.

Use a good pool filter to remove the suspended $\ce{Al(OH)3}$. As you state, they're fine particles, and should glom onto (technical word for "adsorb to") the diatomaceous earth... eventually.
Aluminum hydroxide is quite stable, and dries out to aluminum oxide, $\ce{Al2O3}$, which is very stable. $\ce{Al2O3}$ (sapphire and ruby, i.e., corundum) lasts millions of years, and any chemical that could break down $\ce{Al(OH)3}$ or $\ce{Al2O3}$ is not something in which you want to swim!


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Transform a star's carbon and oxygen into elemental hydrogen
Forgive me if this a really dumb Question, and yesterday Star Trek: Deep Space 9 suggested a terraforming specialist might transform a star's carbon and oxygen into elemental hydrogen, no details of how or why being given.
Ignoring the logistics, is that not chemically an impossible mission?

Well, that would be a weird idea, to put it mildly, as stars are literally made of hydrogen. Even white dwarfs have hydrogen in atmospheres.

@Mithoron Quite. That was my fear, exactly.

I had higher opinion about Star Trek until now. It is a very dumb idea. Chemically impossible. Using nuclear processes of breaking C and O nuclei, theoretically possible, but technologically very demanding, using particle accelerators. Energetically would need  many orders more energy than classical chemical ways. It would be like a fussion power plant going in reverse mode, but even worse.

@Poutnik Again, quite. That was my fear.

It is an impossible mission, because the usual Carbon nucleus is made of 6 protons and 6 neutrons. And its mass is a little bit less than 6 times the mass of one free proton, plus 6 times the mass of one free neutron. For lack of mass, the splitting of the carbon nucleus into separated protons and neutrons is impossible.

Maurice, that sounds wholly  reasonable but why did you go that far? My problem is that you did not just dismiss the whole idea out of hand… which suggests it is not wholly unreasonable.

I don't understand the mass argument - a proton is a proton. Once you separate them, they'll necessarily have the correct mass. The reason carbon is less than the sum of the free masses is E=mc^2

A free proton has not the same mass when included in a nucleus. Of course, the difference is due to the formula E = mc^2. It means that to remove one proton from a nuclues, this energy must be provided to the nucleus. And this energy is rather huge. This is why a proton cannot be removed from a C nucleus. No experiment can give this huge amount of energy to the proton.

Chemically, it's definitely impossible; hydrogen is another element than carbon or oxygen, and chemical reactions don't change elements into other types. You'd need nuclear fission for that (because hydrogen is smaller than the source elements) and a lot of energy (think Large Hadron Collider - note that that works on the scale of a few atoms, not an entire star).

And yet as we speak carbon-14 is cycling into nitrogen-14. The scenario is far-fetched but if one were to take the proton and electron being moved around here and liberally (somehow) fuse it into hydrogen, it might pass muster. And there's plenty of similar wiggle room for other beta decaying chains as well as alpha decay.

Thanks, Glorfindel.
@Todd Minehardt what is that '(somehow) fuse it into hydrogen, it might pass muster'? That looks like a useful Answer and what does it really explain?

@ToddMinehardt - but a nucleus finding that converting one of the protons into a neutron, or vice versa, is a very different thing than the nucleus deciding it is better to completely fall apart into a bunch of protons. Both 14C and 14N are more stable than being separated into protons+neutrons, its just that 14N is more stable than 14C.

@JonCuster I stand edified. I'm not entirely in my comfort zone here and advise the OP to heed Jon's comment.

Are you sure they were not talking about making some C, O, N etc to terraform a planet with too much H.  Setup a super fusion reactor and come back in 30 years!!


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Is eta hexachloro cyclohexane optically active compound?
Quoting from Wikipedia,

Hexachlorocyclohexane (HCH), $\ce{C6H6Cl6}$, is any of several polyhalogenated organic compounds consisting of a six-carbon ring with one chlorine and one hydrogen attached to each carbon. This structure has nine stereoisomers (eight diastereomers, one of which has two enantiomers), which differ by the stereochemistry of the individual chlorine substituents on the cyclohexane.

According to Wikipedia, only $\alpha-$hexachlorocyclohexane can exhibit optical isomersim. Here is an image of $\eta-$hexachlorocyclohexane,
Clearly it doesn't have any plane or centre of symmetries. So why it isn't an optically active compound?

Make a mirror image of this molecule and try to superimpose the structure you have here.

There is a mirror plane of symmetry normal to the plane of the ring.

@ToddMinehardt I am unable to see the plane of symmetry.

@AChem I tried superimposing the images.  I found that they are non-superimposable.

The plane is between the 12 o'clock to 2 o'clock and 6 o'clock and 8 o'clock.

Your compound is related to allo-inositol where OH groups replace Cl. All nine stereoisomers are discussed here.

@ToddMinehardt that is not a plane of symmetry when considering the 3D conformation

by 2/m i’m interpreting that as C2h symmetry, is that what you mean by that? because if so, i hard disagree with you

OP's comment: Clearly it doesn't have any plane or center of symmetries. So why it isn't an optically active compound?

It has a plane of symmetry:

Thus it is not optically active compound (a meso-compound):

If we consider the chair conformer of cyclohexane I don't observe any plane of symmetry.

this plane of symmetry is only a plane of symmetry in this representation of cyclohexane—a chair representation of cyclohexane does not have this plane of symmetry and thus it cannot be a meso-compound by this reason

i have left another response that i believe is more accurate—in re Matthew's answer, note that the mirror plane splits two sets of adjacent chlorines: to point in the same direction, one side of the mirror must have the axial orientation, and the other HAS to be equatorial.

i am not saying it is optically active, i’m saying “it has a mirror plane of symmetry” is not the reason it’s achiral

@ToddMinehardt I don't quite get your comment; you can't show that a conformer is optically active by the presence of a symmetry element, only the absence. The conformer doesn't have any improper rotation axes, thus it is chiral. The lack of optical activity in a real sample arises not because of the presence of a symmetry element, but because of conformational interconversion (between the two enantiomeric chair conformers).

@orthocresol Your summary is likely what I was attempting to communicate (poorly).

This molecule does not have any symmetries—however, we can show that its enantiomers are equivalent with some visualization and demonstrate thus that it is not optically active. We'll start by drawing out each enantiomer, like so:

One property of substituted cyclohexanes is that it is isomerically equivalent with the isomer where the axial and equatorial substituents are swapped. This conformational isomerism is known as a ring-flip—after a ring-flip, all upward facing carbons face downward and vice versa; every axial substituent becomes equatorial; and every equatorial substituent becomes axial. Let's now perform a ring-flip on the left enantiomer of $\eta$-hexachloro-cyclohexane:

Remember: a ring-flip is a conformational isomerism—no exchange of bonds needs to occur for this isomerism to happen. This means that these enantiomers are equivalent to each other and so the molecule is not optically active. They are not enantiomers because they can "freely" (ignoring the kinetics and thermodynamics of the transition states/intermediates) move between each other. There does not need to be a plane of symmetry in this case. In fact, this molecule does not have any standard symmetries—it belongs to the $C_1$ point group, as it lacks a symmetry across any mirror plane, axis of rotation, and inversion center.
You can prove mathematically via cyclic permutations as well that there won't be superimposability between the isomers, but they will be isomeric across a ring-flip. Let's represent a conformation $\eta$-hexachlorocyclohexane according to its sequence of axial ($A$) and equatorial ($E$) chlorines:
$$ AAEEAE $$
Rotation of the molecule about the z-axis can be represented as a cyclic permutation of this sequence. A 60$^\circ$ rotation clockwise would look like this:
$$ AAEEAE \longrightarrow EAAEEA $$
A reflection across a mirror plane looks like reflecting the sequence, like so:
$$ AAEEAE \longrightarrow EAEEAA $$
We do not need to note the orientation (up or down) of each individual chlorine, because we know that relative to each other there will always be four chlorines in one direction, and two going the other. However, rotating the molecule by any axis other than the one previously described 180$^\circ$ reorients the molecule and swaps ups for down and downs for up. This is isomorphic to a binary sign-flip operation, so we will represent all 180$^\circ$ rotations with a sign flip, as such:
$$ (+) AAEEAE \longrightarrow (-) EAEEAA $$
This would change our mirror reflection notation to:
$$ (+) AAEEAE \longrightarrow (+) EAEEAA $$
The question is: is this above transformation possible using only the rotations discussed? This is the mirror reflection; if rotations bring us to the same sequence as a mirror reflection, then our compounds are superimposable. If you play around with this for a while, it isn't possible. Any 180$^\circ$ rotation to get the sequence in the right orientation will change the sign such that they can't be lined up anymore.
However, we can prove that a ring-flip will bring us to the same sequence as the mirror reflection sequence. A ring-flip takes all axial to equatorial and all equatorial to axial, so we can represent a ring-flip like so:
$$ (+) AAEEAE \longrightarrow (+) EEAAEA $$
A ring-flip does not change the direction the chlorines are facing—only their orientation—so the sign is maintained. The following sequence incorporating the ring-flip brings us to the enantiomer sequence:
$$ (+) AAEEAE \longrightarrow (+) EEAAEA \longrightarrow (+) EAEEAA $$
The first operation was the ring-flip, and the second operation was a 120$^\circ$ clockwise rotation. This proves that even though there is no symmetry in this molecule it is conformationally isomeric with its enantiomer via the ring-flip isomerization.
It turns out that if the skeletal representation of a cyclohexane ring has an internal plane of symmetry, that molecule will be achiral—this is a convenient shortcut for assessing the achirality of a compound, but the mirror plane in the skeletal representation should not be mistaken for a mirror plane in the chair conformation.

@toddminehardt to have a mirror plane of symmetry is not the only reason something can be achiral—that is a shortcut for looking at the flat representation of a molecule and seeing if it is achiral. the molecule itself does not have a true mirror plane, ONLY its skeletal representation does. the reason the literal molecule itself is achiral is because the enantiomers are equivalent through a ring-flip isomerization—achirality comes from conformational equivalence of two enantiomers, which we typically think of as rotating but can also be through changes in internal degrees of freedom

@ToddMinehardt does cis-1,2-dibromocyclohexane have a mirror plane of symmetry? no, one of the bromines is equatorial and the other is axial—it’s literally impossible. but its skeletal representation does. the reason it is achiral is via a ring-flip isomerization.

@ToddMinehardt i also—just to make sure i wasn’t crazy—made the molecule in GaussView and tried to superimpose it with its reflection (couldn’t figure it out after toying for like 10 minutes) and the only point group GaussView would tolerate was C1, even with the highest tolerance for RMSD. so i’m pretty confident it’s not superimposable with its enantiomer without a ring-flip.

cf. my and others' answers on cis-1,2-dimethylcyclohexane

Yes, ring flips are the reasons it isn't optically active. But the fact that the flat configuration of the ring has a mirror plane is a good indicator that this will happen when cyclohexanes are known to interconvert between common configurations easily at room temperature. There is little point in considering the chirality of an individual chair form when the molecule is not frozen and the various forms interconvert rapidly.


text:
Pressure vs. Size of Particles
If two containers with different sized particles are held at the same temperature, which gas will exert more pressure?

Depends on the initial pressure in each, don't it?

It is a very poorly conditioned question.

The ideal gas law assumes that gas molecules are point particles, so regardless of how big the molecule actually is, if you have the same number of molecules at the same temperature contained within the same volume, the pressure will be the same.
If you incorporate the modifications that the van der Waals equation introduces, where molecular size and intermolecular forces are taken into account, then the pressure goes up as the size of the molecule increases. We can see this by rearranging the van der Waals equation to solve for the pressure $P$:
$$
P = \frac{RT}{V/n - b} - a\left(\frac{n}{V}\right)^2
$$
where $T$ is the temperature, $V$ is the volume, $n$ is the number of molecules, $R$ is the gas constant, and $a$ and $b$ depend on the identity of the gas where $a$ accounts for intermolecular forces and $b$ accounts for the molecular size. Thus, we care mostly about $b$, though of course the size of the molecule changes $a$ as well. From the equation, we can see that, all else held equal, if $b$ increases (i.e. the molecule size increases) then the pressure goes up.

@Poutnik yes, you're right, oops... I edited my comment, thanks!


text:
How long does it take for zinc oxide powder and heinz vinegar to fully react into zinc acetate at a 1:2 ratio of zinc:heinz vinegar?
How long should I leave a solution of zinc oxide powder and heinz vinegar in order for it to react as completely as possible? The solution is 1(zinc oxide powder):2(heinz vinegar) in a glass flask. How much is this process increased by heat (say 50 deg C)? Without heat, should I cover the flask to prevent the acetic acid from evaporation while it reacts?

Heinz vinegar is not a chemical substance to determine the ratio. If the vinegar contained 60 g acetic acid/L (about 6%), than you need minimally 2 L of vinegar for about 81 g of zinc oxide, but little excess would be needed.

So the short answer to the title question is: It would never fully react.

@Poutnik thank you so much for this response! How long would it take for the reaction to occur at room temp vs. gentle heat/stovetop?

And also just for future reference, what kind of calculation is done to determine how long these kind of reactions take? I now know you can use stoichiometry to figure out the ratios (lol), so what's the kind for duration as a function of temp?

Heterogenous reactions have too many unknown variables, so nothing beats empirical data. As a rule of thumb, reaction rates increase typically 2-4 times per 10 °C increase.

You might be far better off to react vinegar with zinc carbonate, yielding zinc acetate and CO2.
Zinc carbonate is much more soluble in water, no need for heating, just stir a bit.

Thanks for this I will definitely do that next time - but zinc oxide is what I have on hand....

@tomatopies you can try it with ZnO.  Just leave it overnight.  Beware acetic acid fumes if you reflux with heat (do it in a hood).  If no luck consider going to ZnCl with HCl to get it into solution, then make the carbonate with baking soda.


text:
Are Cp and Cv functions of temperature?
I have read that Cp and Cv are independent of temperature for the case of a perfect gas whereas for an ideal gas it will vary with temperature. The variation in Cp and Cv with temperature is attributed to the activation of the additional DOFs at high temperature. Although over the normal range of temperature it can be considered almost independent of temperature.
My question is

Do Cp and Cv vary at the same rate with respect to temperature at high temperatures (I have read somewhere that the variation of Cp and Cv are different but it was long back ago and I can't find the reference).
If Cp and Cv vary with temperature at different rates, then the Meyer's relation will be applicable only to certain limited cases like at low temperatures for ideal gas.

Update :

For monoatomic gases there are no rotational DOFs. I don't know about vibrational modes. So I think the Meyer's relation will hold true at all temperatures for monoatomic gas.

Link regarding the convergence of Cp and Cv value at absolute zero temperature.

If we take for 1 mol of an ideal gas the definition of enthalpy:
$$H = U + pV$$
then differentiation according to T leads to:
$$\frac{\mathrm{d}H}{\mathrm{d}T} = \frac{\mathrm{d}U}{\mathrm{d}T}  + \frac{\mathrm{d}(pV)}{\mathrm{d}T}$$
As $\frac{\mathrm{d}H}{\mathrm{d}T} = C_p$, $\frac{\mathrm{d}U}{\mathrm{d}T} = C_V$, $pV=RT$,
it directly leads for ideal gases to;
$$ C_p(T) = C_V(T) + R$$

For some real diatomic gases, check the chart at Wikipedia. There is easily seen that:

Rotational modes for light molecules like  $\ce{H2}$ are at room temperature still discriminated  by quantization.
Heavy molecules of halogens have their vibrations already active, with various discrimination degree.

I still couldn't get my head around the answer to first question. In the case of ideal gas does Cp and Cv vary at different rate with respect to temperature? I have edited the question with a new link to a page. It says that the Cp and Cv converge to 0 at absolute zero temperature. So clearly when Cp and Cv have different value at same temperature and if they change at same rate with respect to temperature ,then there is no way they could converge at 0 at absolute zero temperature (assuming extrapolation was done linear).

Yes, heat capacities do converge to 0 for T->0, but not for ideal gas, that is always gaseous. Remember that ideal gas model fails miserably to approximate gas behaviours near 0 K. It is unusable extrapolation.

What gas behavior near zero K? An ideal gas is simply a mathematical figment not a gas. Heat capacity is a discontinuous function at absolute zero IF it is actually zero that implies if 0 K is ever reached it will be permanent, No Exit! The behavior near absolute zero could be related to possible quantization of molecular motion.

@jimchmst No reality models of science are reality. We speak here in context of ideal gas model of real gas behaviour, not about real gases. The heat capacity of real solid matter near T=0 K is proportional to T^3, as there is no translation nor rotation, and vibration modes of solid lattice are heavily quantized.


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Magnesium vs calcium vs barium ions from these 3 tests
How do I pick up the difference between barium, calcium, and magnesium ions? I had to do a quantitative analysis practical exam: I was given an aqueous solution of a salt, and when adding separately dilute sulphuric acid, magnesium chloride, and sodium carbonate to the solution, a white precipitate formed in each tube. The notes say that all three ions make insoluble precipitates, so I don’t know how to get the difference.

The unfortunate thing is that salts of these ions are white and can't be differentiated by just looking. But you can take solubility to your advantage.
The solubility of the sulfates of alkaline earth metal decreases down the group. So, you can form their sulfates and differentiate based on their solubility.
\begin{array}{c|c}
\mathbf{Salt} & \mathbf{Solubility~at~r.t.}  & \mathbf{Inference}\\\hline
\text{Magnesium sulfate} & \text{$\ce{35.1 g/100 ml}$} & \text{Very Soluble} \\
\text{Calcium sulfate} & \text{$\ce{0.26 g/100ml}$} & \text{$\ce{Slightly Soluble}$}\\
\text{Strontium sulfate} & \text{$\ce{97 mg/l}$} & \text{$\ce{Almost Insoluble but not negligible}$}\\
\text{Barium sulfate} & \text{$\ce{2.5 mg/l}$}& \text{Insoluble}\end{array}
You can make oxalates of ions and then dissolve in hot acetic acid. Barium oxalate is readily dissolved in acetic acid while calcium and magnesium oxalate are practically insoluble in acetic acid (calcium and magnesium oxalates can still be separated as described here)
Their chromates can be considered but I would not recommend as chromates are toxic and carcinogenic. Barium chromate is practically insoluble in water and acetic acid while calcium and magnesium chromate are soluble in water and acetic acid (the latter is more soluble).
But the most trustworthy test is flame test. Barium - pale green ; Calcium - Brick red ; Magnesium - Salts are colorless due to formation of oxide layer
Previous discussions:

How to separate and analyse a sample of cations of alkaline and earth alkaline cations and ammonium?
In qualitative analysis for basic radicals, why are group 2 cations not precipitated along with group 4?
How to test for magnesium and calcium oxide? (how to separate calcium and magnesium salts -- easiest way is turning them to their respective oxide and check its solubility in water)

Magnesium sulfate $\ce{MgSO4}$ is soluble in water. So when adding some dilute sulphuric acid $\ce{H2SO4}$ to the solution of magnesium sulphate, no precipitate should occur. It should occur with barium ions $\ce{Ba^{2+}}$ for sure, and with calcium $\ce{Ca^{2+}}$ if the solution is concentrated enough. So sulphuric acid helps deciding if the solution contains magnesium or not.
Solutions of sodium carbonate $\ce{Na2CO3}$ and magnesium chloride $\ce{MgCl2}$ are no use in your exam. Adding sodium carbonate will produce a precipitate with all ions. Adding magnesium chloride will never produce any precipitate.
To distinguish between calcium $\ce{Ca^{2+}}$ and baryum $\ce{Ba^{2+}}$ ions, the traditional method is to add some drops of potassium dichromate $\ce{K2Cr2O7}$ plus $2$ mL sodium acetate $\ce{CH3COONa}$ as a pH buffer. If there is barium in solution, it will produce a precipitate $\ce{BaCrO4}$. Calcium and magnesium do not produce a precipitate in this mixture.

Magnesium chloride could form a precipitate from some calcium or barium bearing solutions, especially if the solution is basic.

@ Oscar Lanzi. Yes, if the solution is basic. Is it basic ?

Unknown from the information given.


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Why is hexacyanoferrate(II) coloured whereas tetrachlorocadmiate(II) is not, even though both lack unpaired electrons?
$\ce{K4[Fe(CN)6]}$ is coloured whereas $\ce{Na2[CdCl4]}$ is not. Even Laporte selection rule would suggest the tetrahedral cadmium complex to have a higher intensity of colour than the octahedral iron complex. Could the difference in color be explained on the basis of only d-d transition, or would you have to invoke charge transfer?

Related posts: https://chemistry.stackexchange.com/questions/44165/color-of-anhydrous-transition-metal-salt and https://chemistry.stackexchange.com/questions/42802/predicting-color-of-precipitate?noredirect=1&lq=1


text:
What matters for ion exchange: net or local charge?
I am interested in separating two small organic molecules by ion exchange chromatography.
Following the example set by several old papers for these molecules, I want to use a cation exchanger. Both molecules have guanidino groups which bind very well to the resin. Older papers used dilute ammonia to neutralize the guanidinos and elute the molecules but I found a rather poor separation with this approach.
There is a range of pH where both molecules would have different net charges, but in that range of conditions, the guanidinos would still be charged on both molecules, and the net charge changes are different mostly because of the number of negatively charges group elsewhere on the molecules (which, in either state, charged vs. uncharged, are not expected to interact much with the resin).
My question is the following: for predicting whether a molecule is expected to elute, what matters more: the overall net charge, or the "local" presence of the ad hoc resin-binding group?

Ion chromatography or ion-exchange chromatography (sadly) does not totally rely on net/local charge. Imagine the separation of standard simple inorganic ions such as fluoride, chloride, bromide and nitrate on ion-exchange resins. All of them have a single minus charge yet they have significant retention time difference as fluoride, elutes first, always, and nitrate in the end. In simple cases, one can say that the hydration radius of each ion is different and that is why they interact differently with the stationary phase (recall Coulomb's law)
The problem starts with organic ionizable molecules, here all simple predictions fail. Recall that the ion-exchange's resin's bulk is an organic polymer. You will have not only coulombic interactions, but also hydrophobic interactions. Also not all ion-exchange resins are created equal. So just saying a cation exchanger is not enough. The polymer backbone of resins interacts strongly with organic molecules.
The word "poor separation" does not mean anything to a separation scientist. It can mean broad but separated peaks, or tailing or asymmetric peak shapes, or just chromatographic resolution < 1.5! You characterize chromatograms with resolution, peak efficiency, and peak asymmetry.
Coming to your secondary query whether it is the net charge or the local charge. In my opinion, it is the local charge that matters. Retention of organic compounds is not well understood. This is an example from anion-exchange resin from a nice paper (Environ. Sci. Technol. 2019, 53, 16, 9734–9743), but the same ideas apply to cation exchange resins:


text:
Can the number of moles be a ratio of mass percentage and molecular mass?
I recently started studying mole concept and came across this formula,
$$ n=\frac{\text{mass percentage}}{\text{molecular mass}}$$
I don't know how it came into existence. How does this formula work?

Never saw this before. What is the context?

Aside of easy dimension check, do not confuse molar and molecular mass. // We do not say number of kilograms either.

The exact formula is :  $n = \ce{\frac{mass}{molar mass}}$. It is not : $n = \ce{\frac{mass percentage}{molar mass}}$, as written in your text. "My" formula can be checked by dimension analysis. The mass is expressed in grams. The molar mass is expressed in gram per mole. The unit of ratio $n$ by "my" formula is expressed in $\ce{\frac{gram}{g/mol} = mol}$. And, of course, the mole is the unit of amount of matter (what you call "number of moles")

Actually, the base SI unit for the mass is kilogramme. But that is (all in all) irrelevant to this answer.


text:
Why does Bachmann reaction differ in batch characteristics as opposed to flow characteristics
I have studied this reaction for over a year and it makes zero sense. When synthesizing HMX from Bachmann reaction, an important intermediate is DPT. DPT is formed by reaction of nitric and hexamine in acetic acid and acetic anhydride with a 60-65% yield. 63% to be precise according to literature. But, there's a caveat. Everyone who runs this reaction adds them simultaneously. All 3 reactants. I added them in steps like 2 together, 1 in a mixture of 2, all 3 together and in a flow setup as well. Adding them not altogether always decreases the yields. Similarly, decreasing acetic acid and acetic anhydride also reduces yield. One would assume Acetic Acid is the solvent and acetic anhydride the water remover in the reaction but even in a flow setup, if you reduce their ratios, the yields go down. Solvents are not supposed to behave in that manner. IF anyone has any experience with this reaction, can they just explain to me what goes on in it?

https://chempedia.info/info/bachmann_process/

This is an overview. I have studied this reaction for over a year. There's nothing in those papers that explains my conundrum which is why I am asking here.

And it would be nice if you linked it yourself. Also, I wouldn't be surprised if it still can provide some insight.

I have read each and every single word in all those papers. There's one mention to the effect I am encountering in the 1953 paper by bachmann himself where he cites private communication on observing the effect I observed. We also got similar yields. 18% mine 20% theirs. I do not have that much insight into flow chemistry and nor do I suspect that they had it so I am asking here.

The illustrations you use show that different conditions for the reaction give worse yields not that flow conditions give worse yields. The actual reaction might be more complex than you think in terms of what is a solvent and what isn't. So are you just asking why you can't further optimise a reaction that has already been optimised?

@matt_black I think the OP is trying to ask what's the difference between adding the reactants all at once or separately (of course it could be a totally different reason), but the person who answers should address that

Exactly @DialFrost Its clear in the isotopic study of the reaction that the whole ring opens and then somehow instead of forming the more stable 6 membered ring, it forms the 8 membered parent compound. Also, if you break it in parts, it goes along without breaking though there's only DFT study of that not isotopic. I am trying to understand what's going on in this reaction? If one of the I guess 4 reactants is in excess, results are completely changed. Why? Why is that?