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How is the notion that gases expand to the containing volume fit with the concept of molar volume?
I have a question about the concept of molar volume of a gas, which is defined as the volume occupied per mole of molecules of the gas.
I don't know details about what a gas is at a deep level, but from high school level knowledge, gases expand to fill the volume they are contained in.
If we have one mole of a gas in a balloon in a huge room at sea level at a location at some standard temperature (0 degrees C), and we release the gas from the balloon, what can we say about the volume the gas will occupy? Is it really the molar volume?
How does that go with the notion that a gas expands because molecules are moving very fast and chaotically at a relatively large distance from each other?
I can understand intuitively, from everyday experience, what a molar volume of a liquid would be (at specific conditions of temperature and atmospheric pressure), because I can see the liquid and how much volume it occupies.
Does the same thing occur with gases? If we have a gas that is red, will it have a fixed volume (the molar volume)?

For ideal gas: $$V_\mathrm{m} = \frac{V}{n} = \frac{RT}{p}.$$  Not to be confused with the own volume of $\pu{1 mol}$ of real gas molecules, with is very roughly 1/4 of $b$ parameter of van der Waals's real gas state  equation  (Less roughly for monoatomic gases, as the model assumes spherical molecules)

The confusion might come form comparing with molar mass (a constant for a given substance). Molar volume depends on the pressure and whether your gas is mixed with other gases. It is not a constant for a given substance. In my experience, complaining about downvotes leads to more downvotes. Improving the question, integrating what you learned from the comments, and asking for help improving the question are strategies that have been successful.

Related (thanks Mithoron): https://chemistry.stackexchange.com/q/105324

The gas laws define relationships under fixed conditions: molar volume is defined when all the other variables are fixed
The confusion you have seems to result from failing to consider the conditions under which molar volume is defined.
The ideal gas laws define the relationship between 4 things: the volume, the pressure, the temperature and the amount of molecules of the gas. Al sorts of combinations of those variables are possible. If the amount of gas molecules is constant (say 1 mole of the gas molecules) there are still 3 variables to play with. So there are plenty of combinations of volume, pressure and temperature are possible. Even if temperature is fixed, there are still a range of combinations of volume and pressure where the law will be satisfied.
But if the pressure is also fixed, there is only one possible volume where the law works. This is the molar volume (the volume occupied when everything else if fixed).
There is no contradiction between this definition and the fact that a gas whose volume is unconstrained will fill all the space available. But that case has too many free variable for a single volume to make sense (both the volume and the partial pressure can vary freely).
The case where molar volume does make sense is one where the pressure is held constant (imagine a vessel with a plunger that exerts a constant pressure in the vessel's contents. If that pressure is fixed, then the volume of a mole of gas released into the (empty) vessel will be the molar volume. And that volume will be the same for any gas (if far enough away from the point where it ceases to be an ideal gas).

For the air as mixture, one may consider the molar volume as the volume containing $\pu{1 mol}$ of any molecules $$V_\mathrm{M, air}=\frac{RT}{p_\mathrm{air}}.$$
For just the $\pu{1 mol}$ of the gas from the balloon, the molar volume is determined by the partial pressure of the gas $$V_\mathrm{M,gas}=\frac{RT}{p_\mathrm{gas}}.$$
And
$$p_\mathrm{gas} = \frac{n_\mathrm{gas}RT}{V_\mathrm{container}},$$
as the gas molar volume is equal to the container volume
In both cases, we can assume ideal gas behavior with negligible errors.

I am trying to get an intuitive idea of the concept of molar mass. The formulas are available online, but again, how do we reconcile the common but perhaps imprecise idea that a gas "expands" with the notion that it occupies a determined volume for a given amount of molecules?

Intuition and definition do not like each other. I do not see gas expansion in any controversy with definition of the molar mass.

No one said there was a contradiction. I don't know what you mean by "intuition and definition do not like each other", but my intuition tells me that that comment is very, very, very wrong. Intuitive notions can help one to understand definitions enormously. In that sense, I would say definition loves intuition. I can give you many examples in mathematics and physics. And the converse is true too, an intuitive notion loves to achieve the precise pinning down that a definition might offer.

Just look at the accepted answer. There are no formulas there, but rather an intuitive explanation of the formulas.

@evianpring Definitions are deliberately chosen relations between terms and their meanings. I would not describe verbose description as intuition. Intuition does not work with descriptions. Intuition does not need descriptions. Intuition just "knows"  that A is X.  But it cannot know if I have defined A is X or not.

What you are describing as intuition is called memorization, and it is not how understanding is obtained. Knowing is different from understanding. And understanding is helped by descriptions, didactical explanations, visuals/graphics, etc. Let me give you a very simple example: you can learn about matrix multiplication in high school and spend your entire life "knowing" how to do it but never having the insights about vector spaces contained in matrices. Therefore, you will never have intuition in all the places where matrices are applied.

Once you learn about vector spaces and linear combinations and someone explains to you that matrix columns and rows can be viewed as vectors, you look at a matrix and see a column space, a row space, a null space, etc. You can't unsee it anymore and so you have an intuitive understanding of what it means to do a matrix multiplication. The formulas don't matter anymore as much as your ability to recreate the formulas from your understanding of what is really happening. So, the original question was: what is really happening in this concept of molar volume? I know the formula. Is there insight?

You did not provide any of the latter. But the accepted answer did, and now I can see what was missing in my understanding initially, and my understanding has been augmented. The intuition is, as explained, that the ideal gas laws define a relationship between four things, and if we fix two of them (pressure and temperature) and then fix a third (1 mol), then because of the ideal gas law equation, volume is pinned down. From this point forward, my intuition about these variables will involve thinking in these terms.

@evianpring No, intuition as I understand it is nothing like memorization. It is guessing the right answer with some level of probability/confidence without exact knowledge how to get this answer. // I have just very recently, today or yesterday, in other question explained the ideal gas state equation is the system with 3 degrees of freedom, leading to 1 of 4 variables depending on the other 3.  // It is quite possible that we talk each about different kind of intuition. One is less than knowledge, the other more than knowledge.


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Constant in the molar conductivity equation
For the variation of conductivity with concentration for strong  electrolytes there is this equation relating molar conductivity and concentration:
$$\Lambda_\mathrm{M}=\Lambda^{\circ}_\mathrm{M} - A \sqrt{c}$$
Could someone explain what factors determine the value of the constant A in this expression?

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

Hello! I encourage you to read the Debye-Hückel-Onsager theory to undestand the two constants of that equation. It may or may not be an easy trip, depending on where are you right now at university. Kohlrausch's law is the name of the equation, but it was an empircal formulation, so there is no way we  can get a good grasp of what the two constants represent.

I'm currently in high school so it might be a little difficult for me but I'll check it out. Thanks!

Around each ion is a cloud of nonzero net charge density of the opposite charge, as ions statistically repulse the like and attract the opposite charges.
When the external potential gradient is applied, the a ion and its "cloud" move in the opposite directions and the "cloud" effectively decreases the gradient for the moving ion. This causes decreasing the ion mobility and molar conductivity.
Another factor, decreasing the effective ion mobility and molar conductivity, is partial forming ionic pairs that are indifferent to the potential gradient. It occurs mainly for higher concentrations and/or for electrolytes with higher charges (like magnesium sulphate in sea water).

Replying the follow up question:
Different electrolytes have generally different $A$ parameter for the Kohlrausch equation. As both complementing ions are contributors to the molar conductivity, the electrolytes with the common ion have generally more similar coefficient values, compared to ones without a common ion.
The effects contributing to the $A$ value via the affecting ion mobility:

Ions with higher mobility have bigger dynamic bias of their "anti ion cloud" leading to bigger $A$.
Ions with bigger charge have stronger "ion clouds" and ionic pair formation tendency, leading to bigger $A$.
Ions with bigger hydrated radius (note the opposite trend compared to naked ion radius) have weaker ion cloud", leading to bigger $A$. It correlates with the ion mobility.

The Kohlrausch law equation parameters may be found tabulated, or they can be determined experimentally from conductivity of solutions of different amount concentration.

Is the constant dependent on nature of the compound? Like for e.g, would KCl and NaCl have the same/ similar vales for the constant 'A' ?

See the question update.

Alright, thank you!

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What is the energy of activation of uncatalyzed ATP hydrolysis reaction?
If we ignore all the biological significans for a while, ATP hydrolysis, the reaction:
ATP + H2O -> ADP + Pi
is still a chemical reaction that must go through transition state to yield product. This is no doubt a nucleophillic substitution reaction that, according to all the paper I've found, occurs by SN2 mechanism. But what's totally missing is the energy of activation of the uncatalyzed reaction.
There are a lot of paper on the mechanism of ATP hydrolysis in the active site of this enzyme and that motor. There are a few paper on uncatalyzed reaction and various nucleophile. A few paper does estimate the activation energy of ATP hydrolysis by various enzyme and motor proteins. But I haven't found a single paper estimating the Eact of the uncatalyzed reaction.
My failure apart, what is the energy of activation of this reaciton without the aid of any enzyme or other catalyst? Did nobody make any estimation?

https://pubs.acs.org/doi/pdf/10.1021/ja01098a020

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How to calculate pH of a 0.1 M solution of ammonium bicarbonate considering simultaneous equilibrium along with hydrolysis?
We need to calculate the $\ce{[H+]}$ of 0.1M $\ce{(NH4+)(HCO3^-)}$ given the $\ce{k_b(NH4OH), k_{a_1}(H2CO3), k_{a_2}(HCO3^-)}$
The hint tells me to directly use the result $\ce{[H+]=\sqrt{k_{a_1}(\frac{k_w}{k_b}-k_{a_2})}}$
I'm trying to reach this result by systematic equilibrium analysis.
I considered the following simultaneous equilibrium:
$\ce{NH3 +H2O<=>NH4+ +OH^-:k_b = \frac{[NH4+][OH-]}{[NH3]}}$
$\ce{H2CO3<=>H+ +HCO3-:k_{a_1}=\frac{[H+][HCO3-]}{[H2CO3]}}$
$\ce{HCO3^-<=>H+ +CO3^{2-}:k_{a_2}=\frac{[H+][CO3^{2-}]}{[HCO3^-]}}$
$\ce{H2O(l)<=>H+ +OH-:k_w=[H+][OH-]}$
I'm not sure how I shall proceed further. I also tried to write the following charge balance equation:
$\ce{[NH4+] +[H+]=[HCO3-] +2[CO3^{2-}] +[OH-]}$
But in this case, wouldn't the $\ce{[NH4+],[HCO3-]}$ cancel out as they both are equal to 0.1M? Upon solving the charge balance I don't reach the given result. Where am I going wrong? How do I derive the formula?

It reduces to the generally applicable substitution method of solving the set of n nonlinear equations for n variables, with justified applying simplifications based of strong inequalities. The starting points are all equilibrium equations, the ion charge balance equation and total substance concentration equations. I highly recommend to perform formal symbol substitution by single-letter variables. BTW there is no NH4OH in water, it is decades old myth.

wouldn't the [NH4+],[HCO3−] cancel out as they both are equal to 0.1M?  They will hydrolyze and not in the same extent. [NH3] + [NH4+] = [H2CO3] + [HCO3-] + [CO3^2-]

@Poutnik I understand that it comes down to solving a system of equations, but I think either my method or the result given in the problem has an issue. If I can get the chemistry right, then the equations won't be a problem. About the $\ce{NH4OH}$, so should I edit that equilibrium to $\ce{NH3 +H2O<=>NH4+ +OH-}$? Would that be correct? But then how would I use the $\ce{k_b(NH4OH)}$?

the Kb, or pKb (about 4.75) remains it is just NH4OH(aq) versus NH3(aq).  N atom is unable to form 5 bonds, having only one 2s and three 2p orbitals. //  [NH3] + [NH4+] = 0.1 M   AND   [H2CO3] + [HCO3-] + [CO3^2-]  = 0.1 M

@Poutnik Okay, got it! What I did was I put the values in the given result to verify if it was correct and in the end I ended up getting that it'll only be true if $\ce{[HCO3-]=[NH4+]}$. Is the result wrong?

There is multiple "right" results, based on the chosen set of approximations. One of the possible is consideration $\ce{HCO3- + NH4+ <=> H2CO3 + NH3}$, with such ratios of respective forms that belong to the same pH, according to the respective dissociation constant. Then your suggested equation is approximately valid. But results of approximations should be checked how well they fit to exact equations.

@Poutnik We need to consider $\ce{k_b(NH3)=1.75 \times 10^{-5} M, k_{a_1}(H2CO3)=4.2 \times 10^{-7} M, k_{a_2}(HCO3^-)=4.2 \times 10^{-11} M}$, I've tried manipulating the seven equations and approximating and neglecting terms but it is becoming super messy and I can't quite figure out how to reach the formula they've used. I could get the correct answer just by brute-forcing the equations but my aim here is to reach the result mentioned in the problem's hint/solution. Considering the values I just mentioned, can I reach the result? Any hints would be great, thanks so much!


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Rate of reaction and rate constant confusion
If we had the reaction,
$A + B$ $\rightarrow$ $C$, and $A$ was second order, and $B$ was first order, we would say that $rate = k[A]^2[B]$, but this has confused me. When we say $rate$, do we mean the rate of change of the concentration of the products, or the reactants, and if it is reactants, why is it not negative (as the reactants are decreasing wrt time)
Thanks for the help in advance

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

coneventionally, we define rate of the reaction as the rate of the formation of products or in other words the rate will be positive if you are going from reactants to products.  As the reactants are getting consumed, the rate of change of the reactants IS going to be negative. But their rate of disappearance will be positive. So you are right in saying that the rate of change(not disappearance) of reactants will be negative.

But the reaction is still going forward (we have defined it that way, i.e, reactants to left, arrow towards right , and products at the right, thus forward direction) so even though change in concentration of reactants is negative the rate of the reaction will be positive.

If $A$ is a reactant, the rate $r$ of the reaction consuming $A$  is defined by :  $r$ = - ${d[A]/dt}$. Take care of the minus sign !
If the reaction is second-order in $A$ and first order in $B$ the rate is expressed by the following expression : $$r = - {d[A]/dt}= k [A]^2[B]$$

Alright, but say we kept $[B]$ constant, then we can say $r = k[A]^2$, where $k = k(rate) * [B]$ and then we plotted a conc time graph, we could find the rate of change of the concentration of $[A]$, would it then be acceptable to use this value to find $k (rate)$, if we knew the conc of $[B]$,

It is wrong to write that : $r = k[A]^2$ if $[B]$ is constant. No ! Whatever the values of $[A]$ and $[B]$ , the rate is always :  $ r = k[A][B]^2$

Thanks for the tip, the point I am making is that, is $r$ equal to the rate of change of the conc of the reactant, if there is one reactant, and what does it represent if the conc of one reactant is kept constant.

@nav if you keep conc of reactant constant then rate will also be constant

Only if B is in vast excess can it be considered constant (see pseudo-order reaction ),  in this case all A will be consumed and B remains effectively unchanged.


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Is ionization energy enthalpy change?
In physics, we learn how to calculate the ionization energy of a hydrogen atom using the Bohr model. The result is the following equation:
$$
E = - \frac{1}{n^2} \frac{Z^2e^2}{2a_0} = - \frac{Z^2 R_H}{n^2} = - \frac{Z^213.6\ \mathrm{eV}}{n^2}
$$
This equation shows that the ionization energy of a hydrogen atom is calculated as 13.6 eV.
On the other hand, in chemistry, ionization energy is treated as a type of enthalpy change, such as in the Born-Haber cycle.
When calculating ionization energy using the Bohr model, it appears that only the internal energy change ($\Delta U$) at 0K is considered and calculated. However, for the Born-Haber cycle, the enthalpy change ($\Delta H = \Delta U + p\Delta V$) at 25°C and 1 atm is required.
This raises the question: when defining "ionization energy," is the volume change of the gas due to the electron produced by the ionization of hydrogen--assuming such a volume change occurs--taken into account?
Or is "ionization energy" and "ionization enthalpy" distinct in concept but so close in value that the difference can be ignored? If so, why is this the case?

Ionization energy is the energy in any form that is required to excite electron from n to n=$\infty$. While ionization enthalpy specifically refers to the heat required (usually per mol) to do the same.


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Fine calcium carbonate powder from eggshells
My goal is to create a very fine calcium carbonate powder using only common materials and ingredients, no specialized chemicals, equipment or tools. I start by placing an egg in vinegar, and letting that sit until the eggshell is gone. Assuming there's not too much acetic acid, this should give me calcium acetate and water, right?
Next, I bake some baking soda at 200F for an hour or two, hopefully this produces sodium carbonate, which is more water-soluble than sodium bicarbonate. I dissolve this in water until it is saturated. (my goal here is to ensure that any powder produced is the target material, without precise measurements, so we combine two clear liquids, rather than adding a powder into a clear liquid)
Finally, I add the dissolved sodium carbonate to the calcium acetate solution until it looks like nothing is happening anymore, which should precipitate ~pure calcium carbonate, right?
It seems that calcium carbonate takes different forms, and I'd like to produce the variety that has certain spectral qualities (calcite?). Namely, reflects the majority of the solar spectrum (.3-2.5 μm) and emits in the atmospheric window spectrum (8–13 μm). Will this process work?

Please edit to explain what “reflects the solar spectrum and emits in the atmospheric window spectrum” mean.

@EdV I believe I'm trying to make micro/nano calcite crystals, as opposed to the other CaCO3 polymorphs like vaterite or aragonite. For the spectral properties, see line 154 of https://tinyurl.com/caco3-paint. So my main question is about the chemical composition of what I'm trying to produce, just to double check my logic, and the secondary question is about the specific variety of calcium carbonate produced, which might belong in its own question.

What are you looking for ? Adding a solution of $\ce{Na2CO3}$ to a solution of calcium acetate will produce a precipitate of $\ce{CaCO3}$.  Why do you want to know the crystallization system of CaCO3 ? This question was simply ignored in Lim's article. It does not seem to have any interest, as the used CaCO3 is a microcrystalline powder.

Ah, then I agree with @Maurice that your prep of calcium carbonate is feasible. Like most experimental procedures, it is subject to the old Yogi Berra saying: “In theory, theory and practice are the same; in practice, they ain’t”. So, good luck!

@Maurice I would like to know the polymorph because I believe calcite has the spectral properties I'm looking for (based on other sources)

@Reese, it will be difficult to know if the structure of the $\ce{CaCO3}$ precipitate is calcite or aragonite.


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Advice on Evaporation / Leakage of Marquis and Ehrlich's reagents
I had ordered both Marquis and Ehrlich's reagents from a manufacturer and they came in eye dropper like bottles inside of prescription pill like bottles. The perscription pill bottles also had cotton like strips inside them which I am assuming help absorb leakage. I put both these perscription pill bottles together inside two layers of Ziploc bags.
After a while, I noticed some condensation inside the Ziploc bags. I am worried because the safety data sheets of the reagents scared me and I had stored the reagents in a closet.
I am assuming that the condensation came from the HCl / propanol or ethanol in the Ehrlic's reagent but I'm not completely sure. I doubt that the sulfuric acid from the Marquis reagent would have evaporated and I know the formaldehyde is in such low quantity and so volatile that it would probably not remain in the clothes/items in the closet.
I am wondering if the evaporation could have escaped the plastic bags and got into the clothes / other items in the closet. I removed the reagents from the closet about 1.5 months ago so is it safe to assume that any evaporation has broken down and that there aren't chemicals in the clothes? Would washing these clothes neutralize these chemicals? Any advice would be appreciated.


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How do metal hydrides behave in a non-hydrogen closed system?
I am curious about the fundamentals of metal hydrides and how adsorption/desorption works in non-hydrogen systems. My understanding is that the atom ratio M/Hx of the metal hydride is a function of the hydrogen partial pressure and temperature. For a metal hydride in a closed system with hydrogen I can understand this relationship. In this case, the partial pressure and total pressure would be equivalent. However, if the metal hydride was placed in a closed system with another gas, presumably inert, I am unsure what would occur and how you could quantify the system.
If we assume that the metal hydride is moved directly from the hydrogen closed-system to an inert gas closed-system with the same temperature and total pressure, then initially the hydrogen partial pressure in the inert gas system would be zero. Under these assumptions, are we able to determine evolution of the system or at least the equilibrium hydrogen partial pressure and total pressure that would be reached? If not, what other information or assumptions would be necessary?

Creating metal hydride is not adsorption, but absorption. Most hydrides are just compounds, with hardly any equilibrium of decomposition in room temp.


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Silver Nitrate Test for Acetylene gas qualification; looking for an appropriate instruction
Phosphine and Hydrogen sulfide along with some other hazardous gases are of impurities of acetylene plants witch are based on the reaction between water and calcium carbide. One of the simple and fast tests for ensuring that these impurities are less than the limit amount is the "Silver Nitrate test". The gas is allowed to react with the Silver nitrate solution for a specific time. Silver Sulfide and Silver Phosphide that are dark in color will precipitate and a visual inspection will suggest that phosphine or hydrogen sulfide have passed the limit or not.
I have found two Standards for Dissolved Acetylene that explain how to measure Phosphine and Hydrogen sulfide by this method. We've been acquiring the necessary equipment for this test. But actually I think their explanations are not adequate and clear and we have problems understanding the procedure.
At first, their instructions are the same while they use two different dimensions for the funnels. why? Second, they haven't written anything about the internal diameter of the stem of the funnel and the required amount of pressure that has to be adjusted. I think these are the most critical points in the amount of gas flow through the filter paper and so affect the result of the test and must be considered.
What is your suggestion?
In these two standards only a maximum amount of 3 psi is specified.
Can a plastic funnel(polypropylene) be used instead of glass?
Do you know any better standard for "dissolved acetylene" or other safer tests than the Silver Nitrate test? (as it may form Silver acetylide which is explosive.)
You can observe the specific pages relating to our discussion of those two standard below.
The IS308 Standard:

The BB-A-106C Standard:

Thanks very much

The full document can be see at Archive.org.
However, the standard technique is specified in an arbitrary legal document, and a responsible official of the Indian government would have to answer any question on deviation from that standard measurement procedure if used for legal purposes, e.g., showing in court that the concentration of acetylene exceeds a legal limit. "[P]ublication of the Bureau of Indian Standards... MANAK BHAVAN, 9 BAHADUR SHAH ZAFAR MARG
NEW DELHI 11000... IS 308 (1988; Gases).
That said, for practical purposes, the pressure is limited to restrict flow rate and prevent damage to equipment. A plastic funnel might serve the purpose, if it could withstand the pressure and did not adsorb gases.
In short, if the purpose is to create a lab experiment or demonstration, any reasonable changes you make might be acceptable, but if this is to have any legal standing, the procedure should be followed verbatim.

Thanks for your recommendations and help.

Please accept the answer by clicking on the checkmark @RezaKoohpaee


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How did Joseph Proust measure mass composition of NATURALLY occurring compounds?
note: I am not a chemist, I am just interested in this
Proust famously compared the mass composition by element of natural and artificially prepared copper carbonate and concluded they were the same, but-

Since this is early atomic hypothesis days, how did he know to compare Cu, O and C and not Cu and CO2 or something ie. how did he know what the elements were?

I assume he determined mass composition of the artificial carbonate by measuring the elements he used to prepare them (a equation of what he did would be helpful), but how did he measure composition of the naturally occurring carbonates, don't the reactions for their breakdown leave compounds behind that are not easy to decompose to elements, and then he had to somehow separate and weigh them all

Any detailed account of his experiments would be helpful

:/ It didn't matter what he decomposed them into, just that the proportions of products were constant. It's not like weighing CuO or measuring volume of CO2 is difficult. What is the point of this question really?

So if you look at Maurice's answer below, you'll find that Cu, O, and C were elements to Proust...meaning Proust thought Cu, O, and C couldn't be decomposed further. Lavoisier's work from like...1789 actually shows how to decompose and synthesize acids, primarily through combustion or addition of a stronger acid. So Proust knew how to get out C, Cu, and O from existing techniques. He knew carbon dioxide was an acid (and hence not an element) from burning alcohol or burning charcoal.

For Proust and all chemists of this time, an element is a substance that nobody has been able to decompose into more fundamental constituents, whatever the method used (heating, electrolysis, etc.) For example, the copper carbonate that ThatApollo777 proposes, is not an element: it can be decomposed into two simpler substances, by heating: copper chalk (today copper oxide) and fixed air (today $\ce{CO2}$). Of course copper chalk is not yet an element, as it can be reduced (its mass is reduced) by using a reducing agent like flammable air (today Hydrogen $\ce{H2}$). This gives copper (which is an element) and water. Now water is not yet an element, as it can be decomposed by electrolysis into two substances that was shown to be elements: Hydrogen and Oxygen.
All these statements have been difficult to state and many times reexamined. This made endless discussions in the scientific world in the beginning $19$th century, which cannot be explained here in detail: it would last too long, and it is often boring.
Some generalities were sometimes discovered at Proust's times. For example, that metals are in general elements. The reason is that metals cannot be decomposed, whatever the attempts carried out to do it. On the other hand, some elements sometimes were not considered as such. For example, Chlorine $\ce{Cl}$ was for many years considered as an oxide (and not as an element), because all known salts in the years 1800 were made of at least three elements: one metal, one non-metal and oxygen. Of course, the main salt (today $\ce{NaCl}$) refused to be spliced into three parts. This failure was attributed to a lack of valid experimental techniques: chlorine $\ce{Cl2}$ was a long time considered as a compound of oxygen plus an unknown non-metal, to be discovered some day. Alas, it was never discovered. And chlorine stayed in the already long family of elements!
So as you see, there is no simple way of stating that a substance is an element or a compound, at Proust's time. Today the problem does not exist anymore, because atoms are known. Atoms were not really accepted at Proust's time. Their existence was only considered as an hypothesis. And there were other hypothesis in chemistry.

Thanks for the answer, although I was wondering if you can answer the 2nd part of the question, how he calculated mass composition of both the natural and artifical carbonate

Just as a note since history of chemistry interests me lately, this idea of elements as "anything which hasn't yet been decomposed" shows up in Lavoisier's Treatise on Elementary Chemistry, arguably the first or one of the first few works containing a table of elements. He talks about how to define elements pretty early on in that one.


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Does Warburg impedance predict zero DC current in an Electrode/Electroyte interface?
I'm trying to understand how/if the Warburg impedance, $Z_w$, of an electrode-electrolyte interface dominates the impedance at low frequencies. I cannot understand how this model can predict a DC current flow.
A commonly used electrical model of a Faradic electrode is from e.g. 1:

where $R_{ct}$ is the charge transfer resistance, $C_d$ is double layer capacitance and $R_b$ is the bulk resistance.
I found an approximation of $Z_w$ here.
$Z_w = \sigma/\omega^{1/2} - j\sigma/\omega^{1/2} \\
|Z_w| = \sqrt{2}\sigma/\omega^{1/2}$
According to this, the Warburg impedance approaches infinity as frequency approaches zero. This means that in the model in ref 1, there would be no DC current flow. Yet in ref 1 they say "At low frequencies, the effect of Warburg impedance can be neglected
and the element in series with the diffusion impedance is merely $R_{ct}$..."
This is where I get confused. I would be grateful for any help/guidance on this.
1 Comprehensive study of noise processes in electrode electrolyte interfaces
Arjang Hassibi,a) Reza Navid, Robert W. Dutton, and Thomas H. Lee, Journal of Applied Physics 96, 1074–1082 (2004)

On the face of it, there is the contradiction you have described: the DC limit of the Warburg impedance would be infinite and there would be no DC current flow. Presumably, the approximation of the Warburg impedance is simply not applicable in the neighborhood of zero Hz. You might be better off asking this at the electrical engineering stack exchange and providing more information about the circuit model(s) used for Warburg impedances. The EE folks are great at doing circuit modeling and can make short work of this if you meet them half way. Good luck!

I removed the link you provided to the equations as it currently leads to a strange page (surfing site?). The equations are available at the Wikipedia.

As Ed V mentioned, the Warburg approximation may be not applicable. This can be partially correct.
At least, it is worth to know that even for the Warburg element you should specify its type. You can find that there are 3 types of Warburg element:

Warburg 0.5-CPE: semi-infinite diffusion element; (I like to consider it as the constant phase element with an exponent n)
Warburg Short: finite-length diffusion element with transmissive boundary;
Warburg Open: finite-length diffusion element with reflective boundary.

https://en.wikipedia.org/wiki/Warburg_element
For the 2nd type (finite-length diffusion element with transmissive boundary), the impedance value is described by the equation
$$
{\displaystyle Z_{W_{\mathrm {S} }}={\frac {A_{\mathrm {W} }}{\sqrt {j\omega }}}\tanh \left(B{\sqrt {j\omega }}\right)}
$$
Specifically this type does have a typical 45° slope at high frequencies in Nyquist plot, but it gradually transforms into an arc. You can find an image describing it in the article Electrochemical Impedance Spectroscopy─A Tutorial, Alexandros Ch. Lazanas and Mamas I. Prodromidis (open access).
https://pubs.acs.org/doi/10.1021/acsmeasuresciau.2c00070
So, coming back to the equation, you can calculate that at low frequencies you get a real number:
$$
{\displaystyle \lim_{\omega \to 0} Z_{W_{\mathrm {S} }}=\lim_{\omega \to 0} {\frac {A_{\mathrm {W} }}{\sqrt {j\omega }}}\tanh \left(B{\sqrt {j\omega }}\right) = A_{\mathrm {W} }B }
$$
Therefore, in the scape of your concern, I prefer to consider that the authors assumed that $A_{\mathrm {W} }B \ll R_{ct}$, i. e., the effect of the Warburg impedance (Warburg Short element) is neglected at low frequencies.

(+1) Informative contribution! Welcome to the Chemistry Stack Exchange!

This answers my question. Thanks for the reference, I will study it.


text:
Why does KCl have FCC structure instead of BCC?
I have this table here.
\begin{array} {|r|r|}\hline \text{Radius Ratio}  & \text{Coordination number}    & \text{Type of Void} \\ \hline <0.155 & 2 & \text{Linear} \\ \hline 0.155-0.225 & 3 & \text{Triangular Planar} \\ \hline 0.225-0.414 & 4 & \text{Tetrahedral} \\ \hline 0.414-0.732 & 6 & \text{Octahedral} \\ \hline 0.732-1 & 8 & \text{Cubic} \\ \hline  \end{array}
I get the ionic radius for $\ce{Cl-}, r_{a} = \pu{181 pm}$
I get the ionic radius for $\ce{K+}, r_{c} = \pu{133 pm}$
For $\ce{KCl}$, Cation-anion radius ratio = $\frac{r_c}{r_a} = \frac{133}{181} \approx 0.735$
This value falls within the range of $0.732 \; \text{to}\; 1$, so I would expect $\ce{KCl}$ to have coordination number $8$, which corresponds to BCC structure similar to $\ce{CsCl}$. But in reality, it takes FCC structure like $\ce{NaCl}$.
Why is this so?

Note that the CsCl structure is not bcc. It's simple cubic because identical ions (such as the chloride ions) form a simple cubic array.

Ionic radii are not real things but an approximation to the truth - in reality all ions are not hard spheres but at least slightly "squishy" and close to the boundaries given by the radius ration rules there are exceptions. Also note KCl transforms to the CsCl structure at relatively low pressure - e.g. https://iopscience.iop.org/article/10.1088/0953-8984/14/44/377/meta Also as another example of the radius ratio rules not being golden note NaCl also transforms to CsCl under pressure https://pubs.aip.org/aip/jap/article/39/1/319/504587/Pressure-Induced-Phase-Transformation-in-NaCl

@OscarLanzi My apologies. After researching for a while, I discovered you're right. The material I'm reading from suggests that $\ce{CsCl}$ forms "BCC structure $\ce{Cl-}$ ions at the corners of a cube".

ChemLibreTexts describes the radius ratio rule as follows.

(We) consider that the anion is the packing atom in the crystal and the
smaller cation fills interstitial sites ("holes"). Cations will find
arrangements in which they can contact the largest number of anions.
If the cation can touch all of its nearest neighbor anions, as shown
at the right for a small cation in contact with larger anions, then
the fit is good. If the cation is too small for a given site, that
coordination number will be unstable and it will prefer a lower
coordination structure.

Note that this interpretation allows for the radius of cation to be slightly larger spacing the anions out. This is depicted below.

Say the left and right limits of the ratio for stability limit ratio or critical ratio corresponding to maximum packing efficiency are $r_\text{min}$ and $r_\text{max}$, respectively. The (more) accurate interpretation of the radius ratio rule is:
$$
\dfrac{r_\text{cation}}{r_\text{anion}} \in [r_\text{min}, r_\text{max}+\epsilon)
$$
where $\epsilon$ is a small number. Now consider the octahedral (FCC) range.
$$
\dfrac{r_\text{cation}}{r_\text{anion}} \in \pu{[0.414, 0.732+\epsilon)}
$$
Since the radius ratio of $\ce{KCl}\ \pu{(0.735)}$ only differs from the upper limit of the octahedral coordination range by $\pu{0.003}$; therefore, it can show both FCC and BCC structures.

The radius ratio rule is really just a geometric rule, and, that too, a rule of thumb in solid-state chemistry. It helps make a priori predictions regarding the structures, that seem (mostly) work while making a bunch of assumptions and not considering a lot of factors. At the end of the day, the structure is what it is.
Some Assumptions of Radius Ratio Rules

Ions are hard spheres.
Only Electrostatic attraction occurs.
Ions interact only to minimize electrostatic potential.
No Polarization or Covalent bond formation occurs.

The list goes on...
If the assumptions of the radius ratio rule were accurate, square planer geometry shouldn't exist at all, because every square planer geometry can also exist as an octahedral one while minimizing electrostatic potential.

Do note that FCC to BCC phase transitions$^1$ of $\ce{KCl}$ have also been observed under (not too) high pressures ($\pu{GPa}$).
References

Mashimo, Tsutomu & Nakamura, K & Tsumoto, K & Zhang, Y & Ando, Shinji & Tonda, H. (2002). Phase transition of KCl under shock compression. Journal of Physics: Condensed Matter. 14. 10783. 10.1088/0953-8984/14/44/377.


text:
Decomposition of ammonium nitrate
Consider the decomposition of $\ce{NH4NO3 -> N2O +2 H2O}$
How do I show this is a redox reaction. Within the $\ce{H2O}$ the oxidation state of $\ce{O}$ is $-2$ and the oxidation state of $\ce{H}$ is $+1$.
As ammonium nitrate is not a peroxide the oxidation state of $\ce{O}$ should also be $-2$. Hydrogen should be $+1$ again. It seems there is no transfer of electrons hear so somehow nitrogen must be both losing and gaining electrons. How is this possible?
Furthermore is this a disproportionation reaction as nitrogen both loses and gains electrons?

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As the oxidation state/number (ON) is generally a formal charge, oxidation/reduction is generally a formal electron exchange. Even if in many cases it is literal exchange.
Exact reaction mechanism is not probably known.
Formally, the nitrogen atom of ammonium is oxidized from $\text{ON}=-3$ to $\text{ON}=0$ by the nitrate nitrogen, which is reduced from  $\text{ON}=+5$ to $\text{ON}=+2$.
Depending on chosen formal reaction steps, oxidation of ammonium and reduction of nitrate can be done by formal homolytic break of bonds with hydrogen or oxygen, where formally free hydrogen and oxygen later recombine to water.

Disproportionation in Redox reaction context are two simultaneous redox reactions, leading from the same oxidation number to two different ones, like $\ce{2 Cu+(aq) -> Cu^2+(aq) + Cu(s)}$.
Comproportionation (synproportionation) is the opposite, like $\ce{SO2 + 2 H2S -> 2 H2O + 3 S}$.


text:
How can I approximate the actual HPLC gradient?
In High Performance Liquid Chromatography (HPLC), the actual gradient is a bit different from the true gradient programmed because of dwell volume and other effects. For example, in the figure added, we can see the programmed gradient in red and the actual gradient in blue.
I'm wondering if there are formulas I can use to approximate/estimate the actual gradient from my programmed gradient or gradient steps (given dwell volume, flow-rate, etc...)?
Approximations are ok, I don't expect this to be computable exactly. Any suggestions are welcome.

Figure just to illustrate the real gradient being different. source

There is no simple formula to calculate the actual gradients, as you can see yourself. The functional output is quite complicated (RC filter like output). I am sure if someone had a very detailed knowledge of fluid mechanics they could have easily predicted that. However there are empirical ways: Search this title. “Measure Your Gradient”: A New Way to Measure Gradients in High Performance Liquid Chromatography by Mass Spectrometric or Absorbance Detection

I assume the exponential response with transport delay is good enough.

Based on the comments of @AChem and @Poutnik the "exponential response with transport delay" seems to be ok for my use use-case.
The Paper
“Measure Your Gradient”: A New Way to Measure Gradients in High Performance Liquid Chromatography by Mass Spectrometric or Absorbance Detection, cited by @AChem. Summarizes it nicely:

The gradient produced by an HPLC is never the same as the one it is
programmed to produce. Gradient non-idealities are usually categorized
into three types:

Gradient delay time (dwell time)

Delay from the time the gradient is programmed to be produced to when it actually reaches the point where the sample is injected.

Gradient dispersion

Gradient dispersion is the rounding out of the gradient, resulting in more gradual changes in slope as if a low-pass filter
were applied to the gradient profile.

Solvent misproportioning

Any other gradient non-ideality that is not described by the former two categories

So I ended up modelling:

Gradient dispersion with exponential smoothing ("exponential response" mentioned by @Poutnik, "RC filter" mentioned by @AChem)
Gradient delay by adding a linear lag to my time domain ("transport delay" mentioned by @Poutnik)

it is good to know that you were able to model the gradient! Very nice work.


text:
Concrete curing
From what I've learned in school, mixing ingredients is generally used to speed up any chemical reaction.
However, people are mixing concrete for it not to cure.

How is that possible?
Is concrete curing a different kind of reaction? If yes, what is it called?
Are there other examples of reactions that are slowed down by mixing ingredients?

Concrete is a mixture of aggregate (stones of varying sizes, including sand), which is (almost) completely inert (usually), and cement and water. The cement is a mixture of various oxides and silicates which are somewhat reactive toward water, especially when the cement is finely ground.
And, yes, concrete setting and curing are different kinds of reactions in that the chemistry (the hydration of metallic oxides and silicates) and the physics (the development of compressive strength) have different time scales and different complexities. The first figure shows the typical strength gain of concrete for a month or so. (Ref 1)

The mixing processes which can delay the setting process take place in the first few hours of hydration. (Curing is a slower process of hydration which fills voids and provides strength.) The chemistry of hydration can be monitored most easily by heat evolution: it reaches a peak within a few minutes (from the aluminates), and then slowly falls over the next few hours and peaks again as the calcium silicates come out of dormancy; then it becomes difficult to track after a day or so. The next figure "Figure 4." shows the heat evolution during the first day or so of cement paste (cement + water, no aggregate) hydration. (Ref 2.) After about 6 hours or so, the concrete will not be remixable and will be gaining strength as a whole.

When the cement particles are first contacted by water, the calcium aluminates tend to react within minutes, and could stiffen the mix so much that it becomes unworkable. Calcium sulfate hemihydrate (added at the cement mill) diverts the aluminum hydroxide flocs to ettringite crystals. But these can also stiffen the mix because they are needles. Mixing can break these stiff needles and fluidify the mix, or at least keep it workable.
Even so, the calcium silicates will hydrate and stiffen the mix by adhering  particles together. Mixing the fresh (i.e., wet) concrete can break up large agglomerations of particles and prevent them from becoming larger. But eventually, the cement will become unmixable.
Vibrators can fluidify fresh concrete as if tiny stirring undoes some of the setting - but as soon as the vibration stops, the mix becomes viscous again.
The paradox here is that mixing/stirring might be expected to grind the cement particles, exposing more surface area, permitting more reaction and stiffening the mix even faster. In practice, the slow rotation of a concrete mix in a truck is not energetic enough to cause more rapid stiffening, but rather keeps cement particles from clumping together with aggregate to form large masses of stiff semi-set concrete.
Mixing can be a complicated process. A situation where mixing could slow down a process might be where there is a difference whether you pour A into B or B into A.
A longer description can be found at https://en.wikipedia.org/wiki/Concrete.
Ref 1. http://www.ce.memphis.edu/1101/notes/concrete/section_3_properties.html
Ref 2. http://matse1.matse.illinois.edu/concrete/prin.html


text:
Question about reaction in battery
I made a battery like this.
$\ce{(-)Al|HCl,CuSO4|Cu(+)}$
By piling up 6 battery of this(actually I made this using 10 yen and 1 yen coins:10yen coin is Cu and 1 yen coin is Al.) I could turn on an LED light.
The plausible reactions taking place at the cathode and the anode are as follows.
Anode: $\ce{Al ->Al^3+ + 3e^-}$
Cathode:$\ce{Cu^2+ +2e^- -> Cu}$  and  $\ce{2H^+ + 2e^- -> H2}$

I have an question here. Why doesn't all the electrons released from Al reduce $\ce{Cu^2+}$ or $\ce{H^+}$ immediately at the anode site?
https://cs.kus.hokkyodai.ac.jp/tancyou/vol.40/enn.htm
This url shows the battery on which mine based.
But I used paper wetted with HCl which contains CuSO4 , not NaCl saturated water like shown in the URL and picture above.

Be aware that attempts of theoretical evaluation of voltage and potentials of galvanic cells involving Al anodes is opening the Pandora's box .

Could you draw a diagram of your cell and battery (collection of cells)? Also "10 yes and 1 yen coins" is not clear. What do you mean by this?

And aluminum metal reacts with both hydrochloric acid and copper sulfate solution. Your cell notation does not show any salt bridge, so I am guessing you used the different coins and suitably wetted paper or cardboard dividers.

I assume 10 yen coins are from copper and 1 yen coins from aluminium. // Some electrons released by anodic Al reduce H+ to H2 even directly at the anode.

@Poutnik I am currently working as a tutor. Actually, when I showed this experiment to my student she said why doesn't all the electron react directly at the anode site. And I couldn't explain. I would appreciate it if you could explain the reason.

@ananta Thank you for your comment. I added a picture and a URL.

@EdV You are right. I used wetted paper like shown in the picture I added now.

Well, the aluminum actually does reduce both copper ions and hydrogen ions directly at the aluminum surface, so this is a cell that is not very good: it essentially has a parasitic short circuit. But this means it just is not working as well as a proper cell with a salt bridge. It works, but not as well as it could. And using hydrochloric acid is not good anyway. So this is kind of a lemon cell. You would be better off doing what Volta did around 1785: use pieces of zinc and copper, no aluminum, and use saltwater to wet the paper spacers. Have fun as a tutor!

Here I show that aluminum reacts with a solution of copper sulfate and sodium chloride: https://chemistry.stackexchange.com/a/128099/79678. Replacing the sodium chloride with hydrochloric acid would be even more favorable for reacting with aluminum. So other electrolytes should be used, i.e., not ones that spontaneously oxidize the aluminum.

I originally missed that there is a single electrolyte and cells have therefore wrong design, allowing Al to react with H+ and Cu^2+ directly, regardless being part of the galvanic cell settings.

@Poutnik Yes, it is just not a good cell design except for the copper! ;-)

@EdV Thank you for your explanation and URL. The experiment on the URL seems easy and interesting. . I understood that the cell I made loses much energy by parasitic short circuit. I will try to make a good cell with proper salt bridge.

Great! Have fun trying stuff: A. Volta’s “piles” design was used by Sir H. Davy more than 200 years ago in his discovery or isolation of several elements. Best of success!


text:
Effect of pressure on the vapour pressure of a liquid
In a textbook I am referring to, there is a section on the effect of external pressure affecting the vapour pressure of a liquid.

Source: A Textbook Of Physical Chemistry Volume 3 - KL Kapoor - Applications of Thermodynamics (Pg 26, 27).
(There is a small typing error in Eq 1.3.1, it is meant to be $\left(\frac{\partial \mu_{B(l)}}{\partial p} \right)_T = V_{m, B(l)}$)
They have provided a qualitative analysis as to why the vapour pressure changes with external pressure, however according to the coexistence curve, there should be a fixed vapour pressure at a fixed temperature and it should not vary with external pressure.
Also, at equilibrium, shouldn't the vapour pressure of the gas be equal to the external pressure?
What does the author imply here? From what I have learnt about vapour pressure, it should be true that vapour pressure depends only and only on temperature.
Clarification
I read about the Poynting effect @Chet Miller suggested. With reference to this answer, it seems like vapour pressure and saturated vapour pressure are different terms.
However I could not find the difference between the two. I think as the name implies, saturated vapour pressure is the one where pure fluid is only present in the system, however vapour pressure is a more general term, which is used (in this case) when there are other non-condensable gases in the system.
Further, would it make a difference if instead of non-condensable gases to create external pressure, we used, say, a piston. What would be the distinction between saturated vapour pressure and vapour pressure in that case?
Any help is greatly appreciated!

Can you provide correct reference of the book?

@MathewMahindaratne https://books.google.co.in/books/about/A_Textbook_of_Physical_Chemistry.html?id=Y-C-KWEs-eUC The photo I've given is from this book, pages 26, 27 cover section 1.7 I talked about

Note that this effect is used for medicinal compressed mixture N2O + O2 1:1, where pure N2O at pressure equal to its partial pressure in the mixture would be liquid, but the mixture is gaseous. The effect is significant due closiness of critical T near 35 Deg C.

Vapor pressure is sometimes used confusingly in 2 different meanings:  1/ Vapor saturated pressure (total or partial)  2/ Vapor partial pressure (saturated or not).  The usual case is the first meaning, but sometimes is better to be explicit.

when there are other incompressible gases in the system  - Being compressible is nature of all gasses.

Sorry, I meant non-condensable. @Poutnik So here we are talking about changing the vapor partial pressure by changing the external pressure, however the vapor saturated pressure is only dependent on the temperature?

@Omega At the first shot, just on temperature. At the second shot, on the total pressure too, due Poynting effect as Chet Miller suggested in his answer and so did I in my first comment.

Just for a final clarification, if there is pure liquid then vapour pressure is only dependent on temperature, however due to changes in external pressure we can change this vapour pressure of the pure liquid. So the statement that vapour pressure depends solely on temperature is incorrect?

@omega I will try to answer your question later by modifying my post, of course this is strange to you, because for a pure fluid the saturation pressure depends only on its temperature. Tomorrow I will post it.

@MetalStorm Sorry for bothering but can you please address the clarifications.

They have provided a qualitative analysis as to why the vapour pressure changes with external pressure, however according to the coexistence curve, there should be a fixed vapour pressure at a fixed temperature and it should not vary with external pressure.

This is correct, if you vary the pressure for a pure fluid in the coexistence curve, you end up at the liquid or gas phase.

Also, at equilibrium, shouldn't the vapour pressure of the gas be equal to the external pressure?

This is somewhat ok. Maybe a better formulation is that the gas and liquid phases are in mechanical equilibrium, and we call this pressure the vapor pressure or saturation pressure. However, the author likes the concept of "vapor pressure of a liquid" that may bring confusion.

What does the author imply here? From what I have learnt about vapour pressure, it should be true that vapour pressure depends only and only on temperature.

All you have said is fine, but the author is showing another thing, which I will try to explain. To help us we have the following figure:

Note that both phases are not in mechanical equilibrium. If you have seen the mathematical treatment of osmotic pressure, it is another case where the pressure of both phases are not equal.
Considering equal chemical potentials for a constant temperature in both phases yields
\begin{align}
  \require{cancel}
  \mu_\alpha(p,T) &= \mu_\beta(P,T) \\
  V_\alpha(p,T) dp - \cancel{S_\alpha(p,T) dT} &=
  V_\beta(P,T) dP - \cancel{S_\beta(P,T) dT} \\
  V_\alpha(p,T) dp &= V_\beta(P,T) dP \\
  \left(\frac{dp}{dP}\right)_T &= \frac{V_\beta(P,T)}{V_\alpha(p,T)} \tag{1}
\end{align}
Eq. (1) establishes the restriction in order to not have exchange of matter between phases. Both molar volumes are evaluated at different pressures. Moreover, for an evolution in both phases:

Since $\alpha$ is saturated liquid, we may consider its molar volume constant without considerable error.
If we approximate the behaviour of phase $\beta$ to that of an ideal gas.
Eq. (1) takes the form
\begin{align}
  \left(\frac{dp}{dP}\right)_T &= \left[\frac{RT}{V_\alpha(p,T)}\right]
  \frac{1}{P} \\
  \int_{p_1^\pu{sat}}^{p_2^\pu{sat}} \; dp &= \left[\frac{RT}{V_\alpha(p,T)}\right]
  \int_{P_1}^{P_2} \frac{dP}{P} \\
  p_2^\pu{sat} &= p_1^\pu{sat} + \left[\frac{RT}{V_\alpha(p,T)}\right]
  \ln\left(\frac{P_2}{P_1}\right) \tag{2}
\end{align}

If the pressure of phase $\beta$ increases, then the pressure of phase $\alpha$ must increase, or vice versa. Eq. (2) may be "useful" in order to obtain the final saturation pressure $p_2^\pu{sat}$ for an initial saturation pressure $p_1^\pu{sat}$, when the phase $\beta$ is compressed from $P_1$ to $P_2$.

Thanks for the reply! A few questions:
When we are considering phase $\beta$ as an ideal gas, shouldn't we write $V_{\beta} (P, T) = \frac{RT}{P}$?
The implications of the final equation also seem to be confusing... Does it mean on changing the external pressure, we can change the saturation pressure? (Without change in temperature)

Shouldn't it be $V_\alpha(p,T) dp - \cancel{S_\alpha(p,T) dT} =  V_\beta(P,T) dP - \cancel{S_\beta(P,T) dT}$?

@ananta Yes, wow. Thanks.

@omega Yes, I've made the changes in the final equation so it is more clear. Yes, you cannot leave unchaged both pressures or you would have transfer of matter from one to another.

They are talking about a pure liquid and and a gas composed of two or more chemical components, one of which the same vapor as the pure liquid and the rest of which are non-condensible.  So the total pressure is > the equilibrium vapor pressure of the pure liquid.  The corruption they are obtaining is the so-called Poynting correction.


text:
Can anhydrous magnesium sulfate decompose at temperatures lower than the expected temperature of 1,124°C; is this what happened?
Context
I am trying to synthesize and isolate boric acid from borax. I know that boric acid is less soluble in water than the NaCl that will be in the solution so it will precipitate out more readily, but I want to improve the yield by separating the additional dissolved boric acid from the solution. To do this I planned on completely drying the remaining solution and using (near) anhydrous isopropyl alcohol to dissolve the boric acid while leaving the NaCl. Because I didn't have any high concentration isopropyl alcohol around, I decided to use my anhydrous magnesium sulfate to concentrate it.
Afterwards I was going to dry it again which I did by first by lighting the alcohol on fire on a tray (which got it mostly dry) and then putting it in my oven (on broil for <10 min; I typically do it for longer at a lower temperature when going from the heptahydrate to the anhydrous (because I need to go slowly when there is that much water in it to not make a mess and to avoid caking).
That being said, I assume my oven does not reach temperatures anywhere near as high as 1,124°C (the decomposition temperature for anhydrous MgSO4) because the heating element is still red (not yellow like one would expect a metal to appear at around 1,000°C) aluminum foil doesn't melt in it, and information on the internet which corroborates this: https://genuineideas.com/ArticlesIndex/cookieBakingSimulation.html).
Despite this, when I took it out I could smell sulfur trioxide, and the appearance of the powder had changed from white to grey. I dissolved it into water (was exothermic as expected) and the solution was definitely acidic with suspended, now dark grey particles which I suspect to be magnesium metal particles covered in a layer of MgOH because of their grey color and apparent low density (as they remain suspended for a long time).
Question
All of this leaves me fairly confident that after the typical dehydration decomposition reaction chain, my anhydrous MgSO4 did decompose somehow, but leaves me wondering how it happened in my standard oven.
Does anyone have a better idea what is going on here?
Thanks in advance for your answers.

Decomposition of magnesium sulfate will give magnesium oxide, which forms a sparingly soluble hydroxide but is measurably basic. Not acidic.

OP's question:

[...] fairly confident that my $\ce{MgSO4}$ did decompose somehow, but leaves me wondering how it happened in my standard oven.

I'm sure OP's standard oven can heat up to $\pu{500 ^\circ C}$ even though OP didn't say how high it can go. However, according to Ref.1:

Magnesium sulfate hydrate will dehydrate below $\pu{500 ^\circ C}$ to form anhydrous magnesium sulfate. Different temperatures for dehydration are reported in the literature: $330$-$\pu{365 ^\circ C}$ (Ref.2), $400$-$\pu{500 ^\circ C}$ (Ref.3) and at least at $\pu{450 ^\circ C}$ (Ref.2). Magnesium sulfate will decompose into magnesium oxide and sulfur trioxide when the temperature is $\pu{700 ^\circ C}$ or higher, depending on the atmosphere under which the decomposition is carried out. Sulfur trioxide is not stable at temperatures in excess of $\pu{700 ^\circ C}$ and spontaneously forms sulfur dioxide and oxygen.
$$\ce{MgSO4 -> MgO + SO3(g)}$$
$$\ce{SO3(g) -> SO2(g) + 1/2O2(g)}$$

Also, keep in mind that the presence of some impurities such ad elemental $\ce{C}$ and gaseous $\ce{CO}$, which can be act as reducing agents will accelerate the decomposition process at lower temperatures (Ref.1):
$$\ce{MgSO4 + 1/2C -> MgO + SO2(g) + 1/2CO2}$$
$$\ce{MgSO4 + CO ->[\pu{600 ^\circ C}] MgO + SO2(g) + CO2}$$
References:

Madeleine N. Scheidema and Pekka Taskinen, "Decomposition Thermodynamics of Magnesium Sulfate," Ind. Eng. Chem. Res. 2011, 50(16), 9550–9556 (DOI: https://doi.org/10.1021/ie102554f).
H. -H. Emons, G. Ziegenbalg, R. Naumann, and F. Paulik, "Thermal decomposition of the magnesium sulphate hydrates under quasi-isothermal and quasi-isobaric conditions," Journal of thermal analysis 1990, 36(16), 1265–1279 (DOI: https://doi.org/10.1007/BF01914050).
M. Seeger, W. Otto, W. Flick, F. Bickelhaupt, and O. S.Akkerman, "Magnesium Compounds," In Ullmann’s Encyclopedia of Industrial Chemistry, Electronic Release; Wiley-VCH: Weinheim, Germany, 2005 (ISBN 13: 9783527310975).

Maybe I wasn't clear enough in that end question, but as suggested by the title and my context section, I am referring to decomposition of anhydrous magnesium sulfate. For the context section, tl;dr of what you seemed to miss is this: I describe getting a grey powder, which when I added water and did various basic tests, seems to be made of various (non-magnesium sulfate) magnesium compounds compounds and contained small amounts of sulfur trioxide (which formed a low concentration sulferic acid solution when the water was added).

And thank you for the section of your post on reducing agents, it certainly is relevant to my question. Do you know any reactions that would be plausible in my situation? I have updated my question to try to make it more clear.


text:
Why doesn't copper react with nitrate salts?
Copper reacts with nitric acid according to this equation:
$$\ce{Cu +  4 HNO3  ->  Cu(NO3)2 + 2 NO2 + 2 H2O}$$
This is a redox reaction with copper getting oxidized, and nitrogen getting reduced. If we think of the $\ce{H+}$ as a spectator ion (which I know is not exactly right, since it forms covalent bonds in water in the products), then you would have the following net ionic reaction:
$$\ce{Cu +  2 NO3-  ->  Cu^2+  +  2 NO2  + 2 O^2-}$$
I'm wondering why this reaction doesn't happen with solutions of nitrate salts (like $\ce{KNO3}$, for example). If nitrate is such a strong oxidizing agent, shouldn't the dissolved nitrate ions still be able to oxidize the copper, even without the $\ce{H+}$ around? It wouldn't be able to form water as a product, but should be able to form hydroxide ions ... something like this:
$$\ce{Cu + 2 KNO3 + 2 H2O -> 2 KOH + Cu(OH)2 + 2 NO2}$$
Is it just that forming water is necessary to make the reaction favorable, or is there something else that I'm missing about the process? (Maybe something in the mechanism where the H+ speeds it up?)

Without H+ it isn't strong anymore.

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O^2- is a very strong base, reacting with water as acid, forming OH-. NO2  reacts with OH-, forming NO3- and NO2-.

The simultaneous formation of $\ce{NO2}$ and hydroxides is impossible in chemistry. It will rather produce nitrites and nitrates.

Your $\ce{HNO3}$ acid acts as an acid but in your last reaction there is no acid did you try adding  some $\ce{H2SO4 }$ acid to dissolve that precipitate $\ce{Cu(OH)2}$. Which blocks the reaction. Not even pure $\ce{HNO3}$ acid will react continuously  with $\ce{Cu}$ because $\ce{Cu(NO3)2}$ is a precipitate. (I was refereeing to fuming nitric acid with no water). Normal $\ce{HNO3}$ acid has water and it dose react. Main cause is you are lacking an acidic medium.


text:
Could someone explain the ouzo/sambuca effect?
I know that ethanol is mixed with some ingredients as fennel which give to this its particular taste. Fennel and ethanol are mixed together. Ethanol is also mixed with water. The analogy is 60 percent water 40 percent ethanol.
When we add some water more the fennel gets free from ethanol and the drink takes its white color.
The question is why adding some water more changes the equilibrium between fennel and ethanol?

The ouzo/sambuca effect:

The ouzo effect, also known as the louche effect and spontaneous emulsification, is a milky oil-in-water emulsion that is formed when water is added to ouzo and other anise-flavored liqueurs and spirits, such as pastis, rakı, arak, sambuca and absinthe. Such emulsions occur with only minimal mixing and are highly stable.

Note the section about emulsion stabilization.

In a water-rich ouzo mixture the droplet coalescence is dramatically slowed without mechanical agitation, dispersing agents, or surfactants. It forms a stable homogeneous fluid dispersion by liquid–liquid nucleation. The size of the droplets has been measured by small-angle neutron scattering to be on the order of a micron.
Using dynamic light scattering, Sitnikova et al. showed that the droplets of oil in the emulsion grow by Ostwald ripening, and that droplets do not coalesce. The Ostwald ripening rate is observed to diminish with increasing ethanol concentrations until the droplets stabilize in size with an average diameter of 3 microns.
Based on thermodynamic considerations of the multi-component mixture, the emulsion derives its stability from trapping between the binodal and spinodal curves in the phase diagram. However, the microscopic mechanisms responsible for the observed slowing of Ostwald ripening rates at increasing ethanol concentrations appear not fully understood.

Fennel fruits/seeds contain various volatile  aromatic oils that are significantly soluble in ethanol, but much less soluble in water.
When the ethanolic extract is diluted by water, the oil solubility in the ethanol/water mixture dramatically decreases and dissolved oils eventually  separate, forming white emulsion.
This happens with many ethanolic plant extracts, used for medicinal or scented spirit purposes.
Quoting from the Fennel link:

It is a highly flavorful herb used in cooking and, along with the similar-tasting anise, is one of the primary ingredients of absinthe.

The aromatic character of fennel fruits derives from volatile oils imparting mixed aromas, including trans-anethole and estragole (resembling liquorice), fenchone (mint and camphor), limonene, 1-octen-3-ol (mushroom). Other phytochemicals found in fennel fruits include polyphenols, such as rosmarinic acid and luteolin, among others in minor content.

F. vulgare essential oil also has non-food uses. Pavela et al 2016 find the EO to be insecticidal.

I remember some empirical advice as natural repellent against ticks - carrying in pockets small fabric bags filled by fennel seeds.


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For radical polymerization reactions, azobis(isobutyronitrile) (AIBN) can also be used as a radical initiator
Draw the mechanism of radical formation if you know that N2 is formed as a by-product.

The solution they give us is

I know what radical polymerization is. The free electron of a molecule binds with another molecule and creates a link. The second molecule will then have a free electron that will again bind with a third molecule. And this reaction will repeat.
But I'm a bit confused by their solution. How can I guess that N2 will form as a byproduct if they didn't tell us directly ? Also, do I understand correctly that a third electron link will be created with the two electrons that were previously bonded to the carbon atom ? At the end, we have N2 and the molecule in the second picture with a free electron. Do I understand correctly that when a molecule has a free electron like this, it is now a radical initiator ? So this molecule can later be used to start a polymerization process ?
I hope my question is clear. I'm a bit confused by those terms and would really appreciate your help. Thank you so much !

"I hope my question is clear." - as usually such comment signifies that clarity of the post leaves much to be wanted. Title in particular is hardly OK.

@Mithoron How should the title be changed ? That’s the same sentence as in the exercise they give us

Your question rambles a bit. When AIBN thermally cleaves the two C-N bonds homolytically, two stabilized radicals are formed as you have indicated. The other "product" is (unpaired electron)-N=N-(unpaired electron). The unpaired electrons are actually paired forming a $\pi$-bond of a molecule of nitrogen.

One question at a time would make it easier to formulate a meaningful title.

@user55119 This exercise seemed confusing to me. I just wonder how we can foresee that the two nitrogens will pair up if they don’t tell us. Feel free to post your comment as an answer if you want, I will accept and upvote it.

@user55119 If you have time, I have one more question: how can you know that the radicals are now stabilized ? On Google it says a stable radical has a low energy level, how can we know that in this context ? Do we just assume that because the exercise says so ?

Question 1: Unpaired electrons on vicinal atoms do not exist. They constitute a -bond. Q2: 1,1-diphenyl-2-picryl hydrazyl is a stable (isolable) free radical. (CH3)2CCN is a delocalized radical given the presence of the cyano group. Delocalized radicals are less reactive than, say, a methyl radical, which has no resonance stabilization.

@user55119 Thank you very much ! It makes more sense now. Feel free to post an answer if you want


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Oxidation Mechanism of A Peroxide reacting with a sulfide to form Anthracene
I have the following reaction: A peroxide reacting with a sulfide to form anthracene, sulfoxide and water or sulfoxide and water depending on the reactant. The reaction is from a paper discussing the oxygen indused oxidation of bitumen (asphalt binder).

1

The paper cites another paper for the reaction mechanism which is of interest to me and is shown below:

2
In my opinion the two reactions differ greatly in their products and the mechanism doesn't make sense.
1. In the first step the why would the sulphur atom attack the negative oxygen atom?
2. Is the reaction even possible with hydrogen peroxide?

Edit: I have also found another paper which somewhat changes the initial from the sulphur attacking the oxygen to the oxygen attacking the sulphur. I assume either the oxygen is attacking or the sulphur is attacking. Who could be right?
J. Am. Chem. Soc. 2004, 126, 3, 900–908

1 Claine Petersen, J. “A Dual, Sequential Mechanism for the Oxidation of Petroleum Asphalts.” Petroleum Science and Technology, vol. 16, no. 9–10, 1998, pp. 1023–1059, https://doi.org/10.1080/10916469808949823.

2 Hargrave, R. “Oxidation of Organic Sulphides - VI. Interaction of Hydroperoxides with Unsaturated Sulphides.” Proceedings of the Royal Society of London. Series A. Mathematical and Physical Sciences, vol. 235, no. 1200, 1956, pp. 55–67, https://doi.org/10.1098/rspa.1956.0064.

The mechanism invokes the use of acid, H-X. R2S attacks the terminal oxygen to gives R2SO and an alcohol. Protonation of the hydroxyl group of the alcohol causes lose of water and the formation of anthracene. The mechanism shown does not elucidate the individual steps. Think of one of the peroxide oxygens as negative and the other one being positive, the site of nucleophilic attack. The reaction occurs with $\ce{H2O2}$ and ozonides.

@user55119 what could be possible individual steps? If another H atom would protonate the OH group it would form water but leave the R group positively charged.

I believe I told you the individual steps. What I think you are suggesting is to protonate the terminal oxygen,  lose water, and form anthrone, not anthracene. That doesn't get you where you want to go!

@user55119 "Protonation of the hydroxyl group of the alcohol causes lose of water" yes but also the formation of a carbocation instead of anthracene. Or is the step from carbocation to anthracene trivial?

Loss of water would give a positive charge on the oxygen that is still attached to the carbon framework. Loss of a proton would give anthrone. There would be no positive charge on carbon.

@user55119 how would the proton be lost in the last step?

Bobsburger: If I am understanding you correctly, you are proposing a reaction scheme wherein the hydroperoxide 1 forms anthracene via the loss of water. Pathway 1 $\rightarrow$ 2 $\rightarrow$ 3 $\rightarrow$ etc. does not provide anthracene but rather anthrone 4. There is no role for dimethyl sulfide (DMS) because hydroperoxide 1 and anthrone 4 are of the same oxidation level. This process is merely an acid-catalyzed rearrangement and not a redox reaction.
On the other hand and following the red arrows, attack of DMS on the terminal oxygen of hydroperoxide 1 affords an intermediates alkoxide 5 and the protonated form of dimethyl sulfoxide (DMSO) 6. A neutralization of these two species provides alcohol 7 and DMSO. This alcohol in acid medium forms carbocation 9 via the protonated alcohol 8. Loss of a proton from carbocation 9 provides anthracene 10. Accordingly, hydroperoxide 1 is reduced by two electrons and DMS is oxidized by two electrons.
Now, if this reaction were run in some sort of aqueous acid medium, then one might argue that protonation of the inner oxygen of 1 leads to a solvolysis to alcohol 7 via carbocation 8, both of which lead to anthracene 10. With or without water, hydrogen peroxide would also be formed. If this pathway were viable, then there is no need for the presence of DMS.

yes I was referring to reaction pathway 1-5-7-8-9-10 in your scheme. The last step, loss of a proton is still unclear to me. How does that happen?

9 is a strong acid. Anthracene is its conjugate base. Any convenient base such as water will effect the deprotonation.

I have found in a different paper that O attacks S in the initial attack instead of the other way around. what would be the explanation for this?

What is the paper? Reference? The O-O bond is electron deficient and S is a good nucleophile.

J. Am. Chem. Soc. 2004, 126, 3, 900–908. I’ve also put it in the question


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Why does Lewis dot structure of CO not have formal charges close to 0?
If we draw the Lewis dot structure of CO like this:

The formal charge on each atom must be 0 But in the actual structure, formal charge on carbon is $-1$ and formal charge on oxygen is $1$.

But I have studied that central atom may have an incomplete octet so that the condition of formal charges being close to $0$ is satisfied.
Then why does carbon form a coordinate bond with oxygen in order to complete it's octet making the formal charges far from 0?

Closely related: https://chemistry.stackexchange.com/questions/51262/how-to-rationalise-with-mo-theory-that-co-is-a-two-electron-donor-through-carbon    //  https://chemistry.stackexchange.com/questions/30797/how-can-the-dipole-moment-of-carbon-monoxide-be-rationalised-by-molecular-orbita

@Poutnik , can this be explained without Molecular Orbital Theory as I have not yet studied it?

@Poutnik This isn't even a question about CO, just about this lowest formal charge misconception. @ OP If there's a need and occasion for a dative bond to complete octet then it happens, end of story. Otherwise a concept of dative bond would be useless.

Note that CO has the highest known bond energy 1072 kJ/mol ( N2 has 942 kJ/mol), so forming the triple bond is energetically favourable. One may ask why C is not the donor, but its electron pair is in 2s orbital, not able to form a pi-bond.

@Mithoron thanks for clarifying.

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

check this paper, it specifically addresses the issue and is open-access: https://www.nature.com/articles/srep16307 . Long story short: yes, formal charge is negative for C, charge from population analysis is strongly positive on C and in reality C and O are strongly polarised and concept of atomic charge breaks down.

I think you are taking a view of the bonding which is very ionic in nature. Consider for a moment the bonding in water. Water has a sp3 oxygen which can be regarded as being a distorted tetrahedral atoms. The water has two sigma bonds between the oxygen atom and the hydrogens and a total of two lone pairs on the oxygen.
What you have in carbon monoxide is a sp oxygen and a sp carbon, you can think of carbon monoxide as being similar to acetylene and dinitrogen.
Between the carbon and the oxygen in CO we have a sigma bond, we also have two pi bonds which are formed from the p orbitals on the carbon and oxygen atoms. These pi bonds are at 90 degrees to each other. It is important to understand that the electrons in a covalent bond are shared between the atoms rather than kept by one atom alone.
It is hard to know what you mean in the biro drawing, do you mean that the oxygen has two lone pairs which are off the axis of the C-O bond or do you mean something else. If this is what you mean then the oxygen would be a sp2 hydridized atom.
If we consider where the lone pair is on a carbon monoxide then it will help to look at S.Niibayashi, K.Mitsui, K.Matsubara, H.Nagashima, Organometallics, 2003, 22, 4885, DOI: 10.1021/om0340701. In this paper they report the strucutre of a metal complex where the lone pair of the oxygen bonds to a titanium.
The Ti-O-C angle is 162 degrees which is only 18 degrees away from a perfect 180 degrees. I think it is likely to be packing effects which have distorted the bond away from 180 degrees. Here is a picture of the complex.

The differece in electronegativity of the carbon and the oxygen cause the covalent bonds to be polarized such that the oxygen gets more electron density than the carbon. As a result the oxygen end of the carbon monoxide is more negative than the carbon end.
So if anything the oxygen is more negative than the carbon.

Yes, in my drawing I meant that oxygen has 2 lone pairs(I am unable to understand what was the need of donating one of that lone pair, probably making it an exception to the "atoms should have formal charge close to 0" rule). Btw how can I view that paper? Can you please provide me the link? I copy pasted the name of the paper on my browser but I am not getting the exact paper.

Thanks for making it clear of what you were thinking, I will edit the answer shortly.


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Why does phosphinic acid not form pyroacids?
Why does $\ce{H3PO2}$ not form pyroacids? All I know is pyroacids are derived oxyacids obtained by removing one $\ce{H2O}$ molecule from two molecules of oxyacids. Correct me if this is wrong.

Look at the structure of the acid attentively, it should become obvious. Hint: the way you wrote it is misleading.

So what I could understand is that since H3PO2 is a monoprotic acid if we took two molecules of H3PO2 and removed two molecules of water, we would lose both the acidic hydrogen. Hence, we won't have an acid at all so no pyroacid either. correct me if I am wrong.

$\ce{H2(PO)-O-(PO)H2}$ is not an acid, like PH3 is not an acid. More exactly they are so weak acids we can forget about that. For similar reasons, there are no dinitrates, diperchlorates etc.

Browsing through Google and Wikipedia, it seems that $\ce{H3PO2}$ is sensitive to heat. When heated at $100$°C, it is disproportioned through :$$\ce{2 H3PO2 -> PH3 + H3PO4}$$ No pyro acids has the time to be formed.

I have forgotten to use the third conditional in the first comment.

It’s essentially both reasons mentioned in the comments.

“Pyro-” means “fire” or, more generally, “heat” (“pyrotechnics” = “fireworks”). Hypophosphorous acid, or more accurately phosphinic acid, is thermally not very stable decomposing at 110 °C. Thus there is no chance for condensation with water loss. See Wikipedia — Hypophosphorous acid.

Each monomer has one protic hydrogen. Condensing two of them with loss of water would mean both of the protic hydrogens area in the water instead of the dimer. So if a dimer were to form, it would not be a (Brønsted–Lowry) acid.

Phosphorus(III): more successful
With phosphorous/phosphonic acid, formation of the dimeric pyro acid can be accomplished by careful heating. Condensation to form the dimeric acid $\ce{H2(H2P2O5)}$ and possibly higher polymers were achieved at 120 °C, but disproportionation again to phosphine and phosphoric acid occurred at higher temperature. Several phosphite/phosphonate salts also gave generally impure dimeric pyro-salts upon heating.[1]
Reference

D. Grant, D.S. Payne, S. Skledar (1964).
"The pyrolysis of inorganic phosphites".
Journal of Inorganic and Nuclear Chemistry, 26, 12, 2103-2111.
ISSN 0022-1902,
https://doi.org/10.1016/0022-1902(64)80155-X.


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Electrolysis of Copper (II) Sulfate Solution with graphite electrodes
In the electrolysis of Copper (II) Sulfate solution with inert graphite electrodes, the electrons from the power pack will first go to cathode. I understand what happens at the cathode - the copper ions gets reduced and solidifies.
However, I'm unsure exactly what happens at the anode, specifically with it being made of graphite - as in the redox half equation. I know oxygen forms, but is it a hydroxide ion that gets oxidised, or is it water? Why?
The hydrogen cations then join with the sulfate to make sulfuric acid. From my understanding the sulfate cannot be oxidised because it cannot exist as a neutral atom.
Would anything change if it were a copper electrodes?
Thank you so much.

For more than what is in the Maurice's answer, you may try to search copper sulfate electrolysis, with and without site:stackexchange.com OR site:libretext.org. You should also include in questions a short conclusion from prior info searching, so others do not write things you already know, or should know if searched about it. If nothing was found, write what you have been searching for.

Sulfate can be oxidized on anodes at special conditions. Search for peroxysulfuric acid and peroxydisulfuric acid, resp. peroxysulfates and peroxydisulfates

In the electrolysis of $\ce{CuSO4}$, the following half-equation occurs at the graphite anode : $$\ce{2 H2O -> O2 + 4 H+ + 4 e-}$$ This production of electrons is due to $\ce{H2O}$ molecules and not to $\ce{OH-}$ ions, because $\ce{H2O}$ is more than $\pu{10^8}$ more frequent in solution than $\ce{OH-}$.
If the anode is made of copper, the production of electrons will be due to another half-equation, namely $$\ce{Cu -> Cu^{2+} + 2 e-}$$

CuSO4 solution has reportedly pH typically 3-4, so 10^11 - 10^12. SO4^2- does not get oxidized either (but at  high current density), as it has much higher redox potential than water oxidation.


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Redefinition of mole 2019
The mole was redefined in 2019 to $6.02214076\times 10^{23}$ elementary entities. I've learnt that this means that the mass of one mole of Carbon 12 is now no longer exactly 12g. However isn't Avagadros number or $6.02214076\times 10^{23}$ a measure of the number of atoms in 12g of Carbon 12? It seem this is no longer the case, so how exactly is Avagadros number now calculated?

The Avogadro constant $N_A$ was the approximate number (as far as measurement allowed to know) of carbon-12 atoms in $\pu{12 g}$ of pure carbon-12 isotope.
Now $N_A$ is   $\pu{6.02214076×10^23 mol-1}$ by definition, i.e. it is not calculated.

Therefore, $\pu{1 mol} = \pu{6.02214076×10^23}$ atoms of carbon-12 does not    have exactly the mass $\pu{12 g}$.

It relates the molar mass constant $M_\mathrm{u}$ and the atomic mass constant $m_\mathrm{u}$ currently $\pu{1.66053906660(50)E−27 kg}$:
$M_\mathrm{u} = m_\mathrm{u} N_\mathrm{A} = \pu{0.99999999965(30)E−3 kg⋅mol−1}$

Note that the element molar masses have uncertainty much bigger that the shift due Avogadro number redefinition, because of isotope abundance fluctuation.  The exception are elements with a single natural isotope, where it is comparable.


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Does I2 form coordinate covalent bond with PI3 in PI3.I2 (PI5)
As PI5 is uncertainly stated as the combination of PI3 and I2. There are suggestions that phosphorus donates its 3s electron pair to molecular orbital of I2 or I2 donates its electron to 3d orbital of phosphorus. Are these suggestions correct?

That's an adduct $\ce{PI3.I2}$ formed in solution, and $\ce{PI3}$ co-crystallizes with $\ce{I2}$. Related: Explanation of steric hindrance and existence of PI5

I mean an adduct like H3N->BF3 is formed between an acid lewis and base lewis. So specifically, in the case of PI3 and I2, which is acid lewis and base lewis?

Assuming the structure $\ce{I3P - I - I}$ exists at all (Wikipedia does not mention this isomer and the claim is unsourced in the question), the currently accepted molecular orbital theory posits a three-center four-electron bond, essentially a standard covalent bond that joins three atoms by being delocalized over two linkages:
$\ce{I3P:\space I - I\space <-> I3\overset{+}{P} - I\space :I^-}$
In this model the Lewis acid is actually the bridging iodine in the form of $\ce{I^+}$, which is partly bounded to each of two Lewis bases $\ce{I^-}$ (in the left structure) and $\ce{PI3}$ (in the right structure).


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How does sodium diacetate turn into two acetic acid molecules?
I'm trying to replace vinegar with sodium diacetate in a recipe, and I would like to understand better what is happening from a chemical point of view.
What I imagine happens when I add sodium diacetate to the pot:
$$\ce{NaH(C2H3O2)2 + H2O -> (C2H4O2)2 + NaOH}$$
But what happens to the sodium hydroxide? It's alkaline in a acidic solution so it should react somehow?
Also if I know that $\pu{100 ml}$ of vinegar contains $\pu{4 g}$ of acetic acid, how much sodium diacetate should I add to substitute the vinegar?

$\ce{NaH(C2H3O2)2(s) ->[H2O] CH3COOH(aq) + CH3COO-(aq) + Na+(aq)}$ with $\mathrm{pH} \approx 4.75$, forming pH buffer. / The solution will be just slightly acidic.

so the result will have positive and negative ions in solution? I find it hard to believe that they will not react (but I haven't touched chemistry in 15 years).
Are you telling me that sodium diacetate is a poor substitute for acetic acid?

Yes, it is a poor substitute. Sodium diacetate is in a way just a rearranged mixture of acetic acid and sodium acetate, with the latter being mildly alkalic. I would suggest using citrone juice instead of vinegar.

@Poutnik what if I add hydrocloric acid to the solution? I should end up with 2 acetic acid molecules and one sodium chloride right? or do I risk making something dangerous?
I'm already using a mixture of fumaric, malic, lactic and citric acid, but I see many professional chefs suggesting to use sodium diacetate in the mix.

What is bad with vinegar? Mixing equivalent amounts of hydrochloric acid and sodium diacetate would be the same as mixing vinegar with table salt. But it would be hard not to overdose the acid. I strongly advice against it.

@Poutnik:

You need to reduce the liquid, using a lot of heat and time, altering the food
it's more expensive
it's harder to make reproducible recipes
it's harder to handle, logistically
In sous vide cousine you try to avoid liquids, because they are a pain in the ass to manage

Let us continue this discussion in chat.

You can use citric acid with sodium diacetate:
$\ce{(CH3COO)2HNa + H3Citr -> 2 CH3COOH + NaH2Citr}$
You could use premixed or rather separate 2 solids. Premixed is more practical, but I would avoid its longer storage to prevent acetic acid creation.
It could help if a desiccant could be used, like silicagel balls in cups of effervescent tablets. But it should be tested. For myself, I would use empty containers from such tablets, recently used.

Note that there are 2 Acetic acid molecules and 1 Sodium Hydroxide molecule in the reaction so there will be an extra 1 Acetic acid molecule left after the complete reaction so, your reaction will always result an acidic medium.
To find how much sodium diacetate you should add to substitute the vinegar, you should first find how many moles of Acetic acid are in a 4g sample (divide the mass of sample from the molar mass of Acetic acid). Then according to your reaction, find how many moles of Sodium Diacetate are needed to get the same amount of Acetic acid you previously calculated using stochiometric ratios. Find the required mass of Sodium Diacetate by multiplying the amount of moles you calculated for Sodium Diacetate before with the molar mass of Sodium Diacetate.


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How to homogeneously mix a very small quantity of a salt in a very large quantity of another salt?
I'm trying to recreate super salt which is a mixture of:

90 parts NaCl
10 parts Monosodium Glutamate
1 part of disodium 5'-ribonucleotides (1 part disodium guanylate and 1 part disodium inosinate)

Anyhow this means that only 1% of final product is made of disodium 5'-ribonucleotides, which is usually a rounding error in my recipes.
How could I make sure that I have an homogeneous mixture of the 3 ingredients?
Solutions I thought:

Mix them in an empty bottle and shake for 30 minutes
Die the disodium 5'-ribonucleotides and visually check that an homogeneous solution has been reached
Create a oil / water based slush? (I think it's a bad idea)

Is there a better approach I could use to achieve my desired result, without buying any expensive machinery?

Don't mix all at once. Take your minor component, add 10 parts of the rest, mix, then add the remaining 90 parts.

As @IvanNeretin, use serial dilution, unless it's cereal dilution.


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Relative polarity - definition and meaning
I am struggling with understanding the parameter called "Relative Polarity" (RP). I know it is supposed to be an arbitrary scale, but I fail to grasp what the values represent. I understand that higher values mean "more polar", but nowhere have I found a proper definition.
As far as I know, the values are fixed between 0 and 1, 1 being the RP of water and 0 (I assume?) of some hydrocarbon (correct me please? I haven't found any compound for which RP = 0). In terms of physics I can think of several ways to describe polarity of a solvent:

Static dipole moment of the molecule
Spectral shift of the electronic spectral peaks of specified standards
(Arbitratily) normalized dielectric constant, i.e. how well does the solvent block electric field
Amount of specified standards that can be dissolved
etc.

Are any of the above the quantity by which RP is described?
Also, is the scale linear? Meaning, if I mix equal volumes of a solvent with RP 1 and a solvent with RP 0.5, will I get a mixture with RP 0.75? Or maybe it's not equal volumes, but weights/moles?
If there is a book where this is explained, feel free to direct me to it.

Provide the context of the Relative polarity. As polarity has several aspects and e.g. solvent A may be simultaneously more and less polar than solvent B according to the aspect that dominates.

https://en.wikipedia.org/wiki/Solvent#Solvent_classifications

I'm referring to the term "Relative Polarity" as can be found anywhere on google, which spits out tables with values between 0 and 1, e.g.: https://www.chem.rochester.edu/notvoodoo/pages/reagents.php?page=solvent_polarity
Yes, I am aware there are multiple scales, that is why I am asking specifically about the "relative polarity"...

Polarity itself is an ill-defined term, rather it is an umbrella term which includes several intermolecular interactions with a probe molecule. It is perhaps a futile exercise to assign a particular meaning to polarity or relative polarity without a full context. It is just like the common term popular among students which is "stability of a given compound", and my question is "stable with respect to what parameter?".
The table you referred to comes from Dr. Reichardt's work almost for all his life. The basic idea is based on solvatochromism, which means, that certain dyes change their color (hence the energy levels) depending on the solvent's polarity. From wavelength shift measures, one can calculate the energy of electronic transition using wavelength energy relationship. Now you assign a number to a chosen solvent, say water, and then use that to scale the rest of the energy of transitions of other solvents. Water then has a "relative polarity" of 1.00. With that one cannot have a relative polarity of zero, because the energy of electronic transition cannot be zero for a chosen dye.
Do not assign physics to these numbers. These are just convenient scales for measuring how polar a given solvent is.
See IUPAC's definition Definition

When applied to solvents, this rather ill-defined term covers their
overall solvation capability (solvation power) for solutes (i.e. in
chemical equilibria: reactants and products; in reaction rates:
reactants and activated complex; in light absorptions: ions or
molecules in the ground and excited state), which in turn depends on
the action of all possible, nonspecific and specific, intermolecular
interactions between solute ions or molecules and solvent molecules,
excluding such interactions leading to definite chemical alterations
of the ions or molecules of the solute. Occasionally, the term solvent
polarity is restricted to nonspecific solute/solvent interactions only
(i.e. to van der Waals forces).

You should read the article Solvatochromic Dyes as Solvent Polarity Indicators, Chem. Rev. 1994, 94, 8, 2319–2358, which will give you a much better understanding of the general topic of defining polarity. There's an obvious focus on Reichardt's own scales, but with a mention of numerous other ways it is done.
For the widely used Reichardt's normalized polarity scale $E\mathrm{^N_T}$ , the absolute transition energies $E\mathrm{_T}$ of a particular dye (and other closely-related analogues) in a chosen solvent are rescaled such that the value of tetramethylsilane (TMS) is defined as 0 and water is defined as 1. Taken directly from the article mentioned above:

So obviously this formula returns exactly 0 if you choose TMS as a solvent, and exactly 1 if you choose water as a solvent. But this zero is only as physically meaningful as zero degrees Celsius (which is to say, not at all) - you can go to negative values. In particular, you can measure the electronic transition of Reichardt's dye in a vacuum, and by the same normalization, one obtains a polarity of -0.111 for the vacuum. This makes sense, because no matter how poorly polarizable a solvent is, it will still be able to shield charges better than nothing at all. You can also go above 1, by the way. This can happen in ionic melts, and interestingly in the extremely strong hydrogen bond donor hexafluoroisopropanol (1.068). The 0 and 1 anchors of the normalized scale are arbitrary, but usefully chosen. If you want to extract physics out of these polarity values, then you have to go back to the absolute polarity scale $E\mathrm{_T}$, whose values actually have units of $\mathrm{kcal\ mol^{-1}}$ (for historical reasons) as opposed to the dimensionless values of the normalized scale.
As a last comment, the polarity of solvent mixtures (however you decide to define it) typically behaves in an extremely non-linear fashion if taken across the entire range of 100% solvent A to 100% solvent B, but for similar solvents and small intervals of composition change, you can interpolate approximate linear behaviour.


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Why does molar conductivity of a weak electrolyte exceed that of a strong electrolyte at low concentrations?
My textbook has this graph.

The only sort of explanation given is that conductivity increases on dilution because the salts ionise better. Which is incomplete because then the graph for KCl should be a flat line. But since strong electrolytes dissociate a 100%, you would expect it to set some sort of a limit, because that's as best as you can get in terms of dissociation, and that's the only difference between the two salts.

you should notice that is related to the mobility of the ions in solution. The molar conductivity of acetic acid can be expressed as $\Lambda_m^0(\ce{CH3OOH}) = \lambda(\ce{CH3OO^-}) + \lambda(\ce{H^+})$. If you check a table of limiting molar conductivites you would notice that the conductivity of \ce{H^+} and \ce{OH-} is much larger than other ions.

@PAEP tough day? You missed a carbon atom in acetic acid.

@PAEP okay, but what about weak electrolytes that do not have H+ or OH- ions? Do their conductivities not exceed those of regular ionic salts?

@Nikunj Mundhra I am not sure that I understand your question. In aqueous solution, the molar ionic conductivities of the ions are much smaller than those
of $\rm{H^+}$ or $\rm{OH^-}$ since their mobility is explained with a different mechanism than that other ions. As an educated guess I would not expect
to observe the kind of behaviour you illustrated for \ce{KCl} when compared to \ce{CH3COOH} molar conductivities.

You could find the Wikipedia page on molar conductivity (https://en.wikipedia.org/wiki/Molar_conductivity) useful. I would recommend you to check a Physical Chemistry (Electrochemistry) textbook.

@PAEP what i meant was if you have a weak electrolyte that does not have an H+ or OH- ions, their conductivities should not exceed those of strong electrolytes such as KCl. So the generalization of the graph breaks down. Also thanks, I will check the wiki out.

@Nikunj Mundhra  I would say that an electrolyte has \ce{H^+} or {OH^-} ions is not correct. I think that a more accurate description is that upon dissolution it generates \ce{H^+} or \ce{OH-} ions as in the case of \ce{CH3COOH}.  For electrolytes that do not generate those ions ... I guess you would expect a linear dependence with $\sqrt{c}$.

@PAEP Oh I'm sorry for the inaccuracy, I'm just a high schooler and that's how I had it in my head, thanks for correcting me. Also yeah that's what I'd been suspecting as well.