text:
Why is Sulfur Trioxide has SP² hybridization?
When I google what is the type of hybridization in SO₃ it gives me "SP² hybridization". The problem I have is why does it has the above hybridization when it has 3 double bonds with the oxygen atoms.
For example, take CO₂. It has just 2 double bonds with oxygen, and it has SP hybridization. it has two π bonds that are not hybridized, and 1 P orbital hybridized with the S orbital. However, SO₃ has 3 double bonds and if it has SP² hybridization (which means only S and P orbitals participated in making hybrid orbitals) how does it has 6 bonds altogether (there is only 1 S orbital and only 3 P orbitals). SO₃ has three π bonds but, and it still has SP² hybridization (how did it get more than 3 P orbitals which are free or hybridized)
From my understanding hybridization occurs to minimalize the repulsion in electrons in orbitals. So, these S and P (D orbitals sometimes) orbitals make an intermediate orbital so that they can create bonds. Why and how does Sulfur Trioxide has SP² hybridization when it contradicts both of these behaviors I mentioned above?
Sorry in advance if my English is not that good.

https://chemistry.stackexchange.com/questions/13483/is-the-s-o-bond-strength-in-sulfur-trioxide-or-sulfite-anion-larger

@Mithoron I read the answer. From what I understand it is better to determine the hybridization using the shape of the molecule, than considering the way the bonds of these molecules have bonded (considering the number of π and σ bonds). Is my assumption correct?

@Mithoron Also is it okay to call the double bonds in SO₃ as π bonds?

Your "hybridization" is irrelevant. Geometry of a molecule is important. There's no double bond in SO3 - all of them have limited π character.


text:
Entropy of the system is different for irreversible and reversible adiabatic process?
Entropy change of a system is a state function...
But if we consider reversible adiabatic process, the entropy change of the system is zero, whereas the entropy of the system is greater than zero in irreversible adiabatic process. Shouldn't the entropy of the system in both cases must be same?
This answer: How would one calculate the entropy change for an adiabatic irreversible process? deals with how to find the the entropy of the system but it doesn't clear how can we say entropy is state function in case of irreversible adiabatic process.

When a system goes from state A to B  once reversibly, the other time irreversibly, the system entropy change is the same. What is not the same is the state change and entropy change of the neighborhood.

It is imposible to design s reversible Adiabatic Process between the same two end states as an irreversible Adiabatic Process.  The reversible path cannot be adiabatic.
To learn how to determine the change in entropy for an irreversible Adiabatic Process (or any other irreversible process), see the following link:  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

By other words, endpoints of both adiabatic process paths have to be connected by another process, typically isobaric or isochoric one.

@Poutnik Those would be two possible options for the reversible path..

That would be a different case. I mean  irreversible adiabatic compression would lead to higher T and p at the final volume. Then followed by isochoric reversible cooling would lead to the final state of reversible adiabatic compression. As the result, extra work of neighborhood would be converted to thermal energy of neighborhood.

@ChetMiller, so entropy change of system in case of irreversible system is non-zero. But what about entropy change in reversible adiabatic process?

@Natasha There is no reversible adibatic path between the same two end states as  for an irreversible adibatic process.

@Poutnik You can always inset reversible segments into a reversible path, including even Carnot cycles.  There are an infinite number of reversible paths between two end states; all have the same entropy change.

That is clear, but not my case. I mean joining both endpoints of reversible path and partially irreversible path. Both with the same entropy change for the system, but not the same for neighborhood.

@Poutnik I don’t follow.

Path1:  Reversible adiabatic compresion p1,T1,V1->p2,T2,V2.  //   Path2:    Irreversible adiabatic compression(p_ext>p_int or friction or..) p1,T1,V1->p3,T3,V2 where p3,T3 > p2, V2  + reversible isochoric cooling p3,T3,V2->p2,T2,V2.  Entropy change for the system is the same for both paths.  Entropy change for the neighborhood is the same for both paths.

I agree with this, but don’t understand what “entropy change for neighborhood" means.

Rather "of neighborhood". E.g. that isochoric cooling passes heat to neighbourhood. //   $\Delta S_\text{irrev,tot} = \Delta S_\text{irrev,sys} + \Delta S_\text{irrev,nbhd} \gt \Delta S_\text{rev,sys} + \Delta S_\text{rev,nbhd} = \Delta S_\text{rev,tot}$, while $\Delta S_\text{rev,sys} = \Delta S_\text{irrev,sys}$ and  $\Delta S_\text{irrev,nbhd} \gt \Delta S_\text{rev,nbhd}$

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text:
Should I remove inhibitor from methyl acrylate?
I need to run multiple aza-Michael additions of methyl acrylate with various amines (dimethylamine, benzylamine, phenylethylamine, aniline, etc.) in order to get the bis-ester. I was wondering if the inhibitor (MEHQ,hydroquinone, others)  that is in the acrylate as-purchased will cause me any problems/prevent the intended reaction? I asked my coworkers and no one can agree on if I should remove it or not.
Oh, and I’m using methanol as my solvent.
Thank you!!

It is common for the hydroquinone inhibitor to be added to aclates to prevent polymerization during storage. It is possible that MEHQ could react with the amine and hinder the aza-Michael addition. To minimize the potential interference, try removing the inhibitor first to prevent side reactions.
You mentioned that you are planning to use methanol as your solvent. In aza-Michael reactions, the amine acts as a nucleophile and attacks the electrophilic carbon of the Michael acceptor. Methanol can act as a proton donor and protonate the amine, making it less nucleophilic. Have you considered using THF or DMSO instead?

Methanol protonate the amine? Fat chance of that.  Even MeHQ would do that only slightly.

I would not worry that the typical polymerization inhibitors added to acrylic esters will interfere with the aza-Michael reaction. I guess you will add the amine in excess, and the relatively small amount of the inhibitors, if they react at all (which I doubt), will do no harm. Depending on your reaction conditions, you may even want inhibition of the radical polymerization, for example at elevated temperatures.
I also assume you will have effective workup procedures to isolate your products in pure form, so you may want to pay attention to remove the inhibitors then so that they will not disturb whatever you plan later on. After the double bond is lost due to the reaction, the inhibitors are not needed any more.
Something else to keep in mind: Your reaction products are $\beta$-aminoesters which are much more prone to hydrolysis or alcoholysis than their counterparts without the amino group.
I have used various acrylic esters successfully in the past in Michael-type addition reactions and never bothered with removing inhibitors first.

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text:
Evaluating the density of a gas equilibrium mixture
I'm a beginner in chemistry (first year uni course) and I have this exercise that I need some help with. I don't really know where to start so for now. I've tried random approaches to see what works but nothing does.
This is the exercise:

For the reaction $$\ce{N2O4(g) <=> 2NO2(g)}$$
the $K_c$ is $0.060$, at $\pu{80 ^\circ C}$. Evaluate the density of the mixture once the equilibrium has been reached (at equilibrium, $T = \pu{80 ^\circ C}$ and $\pu{494 mmHg}$).

What I've tried to do is:
I have found $K_p$ using the $K_c$ and $K_p$ relation
$$K_c = K_p \left(\frac{1}{RT}\right)^N,$$
but now I can't figure out how to proceed.
If receiving the full solution is not an option is it possible to get a step by step guide on what I have to do?

What are the random approaches you tried? Do you know how to calculate the density of pure $\ce{NO2}$ under these conditions? Do you know how to calculate the fraction of reactant and product under these conditions?

My best attempt involved using Kp and and I.C.E table to try and find the partial pressures of the substances and then i would use the found partial pressures (The part i couldn't manage to do) to find the mole percentages and then find the molar mass of the substence after it was balanced and then finally tweaking the formula PV=nRT to find the density.

The ICE table is less useful here. You have two unknowns, and you know the total pressure from the ideal gas law. The first equation is that the partial pressures add up to th total pressure, and the second is a quadratic from the equilibrium constant expression. Solve those for the partial pressures (or the mole percentage) first.

In principle the constant $K_c$ should have a unit. Is it in mol/L ?

Kc should have a unit but in our exercises we are told to use it only as a simple unitless value.

The approach of trying to solve the 2 unknowns is the one i ended up with and while the values are close to the ones in the exercise result sheet they are still a decent bit off the mark.

You can check your answer by substituting the partial pressures into the equilibrium constant expression. If your answer is off by a percent or so, it might have to do with rounding intermediate values instead of keeping at least a guard digit.

Deriving $\rho_{\text{mix}}$
Density of mixture
$$
\rho_{\text{mix}}= \frac{w_{\ce{N2O4}}+w_{\ce{NO2}}}{V}
$$
Let total number of mol at equilibrium be $n$ and $n=w/M$
$$
\rho_{\text{mix}}= \frac{n}{V}(M_{\ce{N2O4}}\frac{n_{\ce{N2O4}}}{n}+M_{\ce{NO2}}\frac{n_{\ce{NO2}}}{n})
$$
From ideal gas equation (the subscript 'eq' is dropped),
$$
\frac{n}{V}=\frac{P}{RT}
$$
and mole fraction $x_j=n_j/n$,
$$
\rho_{\text{mix}}= \frac{P}{RT}(M_{\ce{N2O4}}x_{\ce{N2O4}}+M_{\ce{NO2}}x_{\ce{NO2}}) \tag{1}
$$
Assuming initially 1 mol of $\ce{N2O4}$ is present and degree of dissociation is $\alpha$,

Time
$\ce{N2O4}$
$\ce{NO2}$

$t=0$
$1$
$0$

At equilibrium
$1-\alpha$
$2\alpha$

$$
n=1-\alpha+2\alpha=1+\alpha \\
x_{\ce{N2O4}}=\frac{1-\alpha}{1+\alpha} \\
x_{\ce{NO2}}=\frac{2\alpha}{1+\alpha} \tag{2}
$$
We also have $M_{\ce{N2O4}}=2M_{\ce{NO2}}$ then from $(1)$ and $(2)$
$$
\rho_{\text{mix}}=\frac{2PM_{\ce{NO2}}}{RT(1+\alpha)}
$$
Calculating $\alpha$
For the equilibrium
$$
K_p=\frac{p_{\ce{NO2}}^2}{p_{\ce{N2O4}}}=\frac{x_{\ce{NO2}}^2}{x_{\ce{N2O4}}}~P
$$
where partial pressure $p_j=x_j~P$
In terms of $\alpha$,
$$
K_p=\frac{4\alpha^2}{1-\alpha^2}~P
$$
Solving for $\alpha$
$$
\alpha=\sqrt{\frac{K_p}{K_p+4P}}
$$
Now $K_p=K_c~(RT)^{\Delta n}$ and $\Delta n=2-1=1$. Therefore,
$$
K_p=RT~K_c
$$
Substituting the above in $\alpha$
$$
\alpha=\sqrt{\frac{RTK_c}{RTK_c+4P}} \\
\begin{align*}
& P=\pu{494 mmHg}=\pu{0.66 bar}\\
& T=\pu{353.15 K} \\
& R=\pu{0.08314 bar-\ell~K^{-1} mol^{-1}} \\
& K_c=\pu{0.06 M}
\end{align*}
$$
From the given values $\alpha=0.63$
Calculating $\rho_{\text{mix}}$
Note: Directly substituting the values will give the density in $\pu{g/\ell}$
$$
\begin{align*}
& \alpha=0.63 \\
& M_{\ce{NO2}}=\pu{46.00 g mol^{-1}}
\end{align*}
$$
Therefore,
$$
\rho_{\text{mix}}=\pu{1.27 g/\ell}=\pu{1.27E-3 g/m\ell}
$$

The assumption of a closed vessel is taken in the solution

Try chemistry markup with mhchem's \ce{…}. If you want to know more, please have a look here and here.

This is a more direct calculation (or sketch of a calculation) compared to megamonster68's correct answer:
The partial pressures add up to the given pressure $P$, and are related as part of the equilibrium constant expression:
$$p_{\mathrm{\ce{N2O4}}}= P - p_{\mathrm{\ce{NO2}}}$$
so
$$K_{\mathrm{p}} = \dfrac{{p_{\mathrm{\ce{NO2}}}}^{2}}{P - p_{\mathrm{\ce{NO2}}}}$$
Solving for the partial pressures, you get (as the physically possible solution to the quadratic equation):
$$p_{\mathrm{\ce{NO2}}} =0.504\  \mathrm{atm}$$
$$p_{\mathrm{\ce{N2O4}}} =0.146\  \mathrm{atm}$$
You can calculate the densities separately and add them up:
$$ρ= ρ_\ce{N2O4} +  ρ_\ce{NO2}= \frac{1}{R T} (p_\ce{N2O4} M_\ce{N2O4}+  p_\ce{NO2}  M_\ce{NO2}) = 1.26\  \frac{\mathrm{g}}{\mathrm{L}} $$
I used three significant figures throughout, even though two of the quantities are given with only two significant figures. In other words, differences in the third digit of the answer are not significant. Surprisingly, the only time the stoichiometric factor 2 appears is in the equilibrium constant expression.
Here is a link to the full calculation.

Adding a second alternative method for arriving at the correct answer.
Let:
$$A = {\ce{N2O4}}$$
$$B = {\ce{NO2}}$$
We can calculate the total concentration at equilibrium $C$ with the ideal gas law by using the given temperature $T$ and total pressure $P$ at equilibrium:
$$C=\frac{P}{RT}=\frac{\pu{494mmHg}}{\left(\pu{62.36\frac{mmHg\;L}{mol\;K}}\right)\pu{353K}}=\pu{0.02243mol/L}$$
Using the relationship between $K_X$ and $K_C$, and noting that $\Delta n =1$, we can calculate the former:
$$K_C=K_X\;C^{\Delta n}\implies K_X=\frac{K_C}{C}=\frac{\pu{0.06mol/L}}{\pu{0.02243mol/L}}=2.675$$
Next, we can use the expression for $K_X$ and its value to obtain an equation involving the molar fractions of both species. Considering we're dealing with a binary mixture: $X_A+X_B=1$, and:
$$K_X=\frac{X_B^2}{X_A}=2.675\implies \frac{X_B^2}{1-X_B}=2.675\implies X_B=0.775,\;X_A=0.225$$
Next, we can use the molar fractions to calculate the molar mass of the mixture at equilibrium:
$$M=X_AM_A+X_BM_B=(0.225)(\pu{92g/mol})+(0.775)(\pu{46g/mol})=\pu{56.35g/mol}$$
Finally, we can calculate the density of the mixture at equilibrium using the total concentration and molar mass of the mixture:
$$\rho=CM=(\pu{0.02243mol/L})(\pu{56.35g/mol})=\pu{1.263g/L}$$


text:
Why isn't the unit of rate constant for any reaction always per second?
It is formally agreed upon that $K_{eq}=\prod_{j}a_j^{\nu_j}$ where $\nu_j$ is stoichiometric coefficient of species $j$ (negative for reactants and positive for products). This equation results in equilibrium constant not having any units. Experimentally, of course $K_c$ and $K_p$ are measured (and are of practical application). But everyone agrees that equilibrium constant has no units.
Similarly, why do we not formally define rate constant as
$$
\frac{d}{dt}a_P=k\ \prod_{j}a_j^{\alpha_j}
$$
where $\alpha_j$ is order of reaction relative to species $j$ and $a_P$ is the activity of the product.
This would mean that $k$ always had the unit $\textrm{s}^{-1}$ and, depending on the approximations used, the "unit" of rate constant may depend (analogous to $K_c$ and $K_p$). Why do chemists differentiate between the constants when it comes to units?
Another thing to remember would be that $K_{eq}=k_1/k_{-1}$, so it definitely makes sense that rate constants have same units so as to treat $K_{eq}$ as a ratio.

Rate equation hardly has to be in such simple form.

You ask a lot at once, and seem to confuse some things and the reader along the way. Units and dimensions are not the same thing. And one can, and might be particularly inclined and justified to, write a rate equation like you suggest, using activities instead of concentrations (or pressures). How "formal" this is I can't say.


text:
Why does fluorine form only one hydrogen bond, but oxygen forms two?
Oxygen forms two hydrogen bonds as it has two lone pairs. So, why does fluorine only forms one hydrogen bond having three lone pairs?
Both oxygen and fluorine have nearly the same size, so steric hindrance should not be an issue.
I'm aware of the following facts, so please do not use them as an explanation:

hydrogen forms only one hydrogen bond;
fluorine forms one hydrogen bond in hydrogen fluoride $\ce{HF}$;
oxygen forms two hydrogen bonds in water $\ce{H2O}.$

Are there any compounds in which fluorine forms three hydrogen bonds?

The number of lone pairs has no importance in the calculation of the bonds of an atom. Only the number of electrons non included in lone pairs is important. Oxygen has $2$ such electrons. It can make $2$ bonds with $\ce{H}$ atoms. Fluor has only one such electron. So it can make only one bond with $\ce{H}$  atoms.

Be aware of possible confusion of "fluorine H-bond" as 1/ an ordinary polar covalent bond H-F as in hydrogen fluoride. 2/ Hydrogen bond as e.g. between $\ce{HF}$ molecules as $\ce{H-F\bond{...}H-F}$.

Who said F only forms one H-bond, and in which contetxt?

Fluorine in hydrogen fluoride can form only a limited amount of hydrogen bonds because there is only one (protic) hydrogen atom per fluorine. Ammonium fluoride has enough protic hydrogens to form hydrogen bonds with all four electron pairs on each fluorine — and so they do, in a wurtzite-type arrangement of the ions.


text:
Potential Energy Function for Two Atoms

The potential energy between two atoms, in a molecule, is given by
$$U(x)=\frac{a}{x^{12}} -\frac{b}{x^6}$$ where $a$ and $b$ are positive
constants and $x$ is the distance between the atoms. The atom is in
stable equilibrium when: (NEET 1995)

Differentiating and equating to zero gives a single solution, which must then be the stable equilbrium:
$$\frac{dU}{dx}=0\Rightarrow x=(\frac{2a}{b})^\frac{1}{6}$$
Question: Is the potential energy function here an actual one for quantities like bond length, or just made up for the sake of this question?

Does this answer your question? Lennard-Jones Potential repulsion by nucleus nucleus or Pauli repulsion?

I wanted to identify the function (which the comments have done for me), not ask about the properties of that function. In general, I do not think the other questions/answers answer my question.

I would think the Lennard-Jones potential is an empirical potential but based on solid theoretical grounds. Basically, the $C_{12}/r^{12}$ term describes the short range Pauli repulsion and the $C_6/r^6$ term is a long-range attractive potential that you can associate to dipole-dipole interactions. I think originally was developed to describe the interatomic interactions in gases but its used extended to many fields.

Although there are better interatomic potential functions, it gives you a good approximation of the isotropic part of the potential of Van der Walls complexes and in that constext an idea of their stability and bond length. You would find a more nuanced description in the interatomic potencial chpater of a Physical Chemistry textbook (e.g. Atkins, Engel, ...).

The empirical potential
$$U(x)=\frac{a}{x^{12}} -\frac{b}{x^6}$$
is known as the Lennard-Jones (or 12-6) potential. It is ubiquitous in theoretical chemistry as an approximation of real interatomic potentials. It's "made up" but captures essential features of interatomic potentials, has been particularly useful because of its mathematical simplicity, as explained in another answer, and continues to be broadly implemented, for instance in MD simulation packages.

Is the potential energy function here an actual one for quantities like bond length, or just made up for the sake of this question?

Note that the real potential is "known" only to the interacting matter. As the number of interacting particles (nuclei, electrons) increases the number and complexity of interactions grows staggeringly fast. We do our best to approximate the interactions to facilitate predictions. Even QM simulations are slow to perform on all but the smallest systems. An L-J potential is a pair-wise mean-field potential and as such brushes under many of the complexities of real interactions. You might also find useful information on uses of the L-J potential in our sister site Matter Modeling.


text:
PH - Free Radical - O2 confusion
some context 1st, this is my understanding:
I understand PH as the balance of protons vs electrons 7 being neutral, 1 being too many protons and 9 too many electrons.
A free radical in the body as a particle with excess protons (thus acid?)
O2 takes electrons from other molecules (I don’t know other mechanisms than having a + charge or excess of protons). Reason why the process of plucking electrons is called (misleadingly) oxidation, and why elements that donate electrons are called anti-oxidants.
Confusion/contractions:

How then O2 “alkalinize the body”?
How ClO2 (Chlorine Dioxide) is an oxidant but donates 5 e-? (Donating is what antioxidants do?).

pH is a measure of the hydrogen ion (or hydronium in water) activity ($p\mathrm{H}=-\log(a_{\ce{H+}}$)),where activity approx equal concentration in dilute solutions.

There are some concepts you seem to have misunderstood. Your definitions of pH balance and free radicals for one are not correct.

Your understanding is often wrong.
$1$) The pH value is NOT a balance of protons vs. electrons. It is not a balance at all. It is the logarithm (or rather minus the logarithm) of the concentration of $\ce{H+}$ ions in solution : pH = - log [H+]
$2$) When pH = $1$, it does not mean that there are too many protons. There are never TOO MANY protons, whatever the solution. It only means that the concentration of the $\ce{H+}$ ions is $0.1$ M, since the logarithm of $0.1$ is $-1$.
$3$) When pH = $9$, it does not mean that there are too many electrons. There are never too many electrons, whatever the solution. And why choose pH = $9$ ? This value has no particular interest. pH = $9$ is obtained with a solution where the concentration of the ion $\ce{H+}$ is $\pu{10^{-9}}$ M, which also means that $\ce{[OH-]}$ = $\pu{10^{-5}}$ M.
$4$) A free radical is NOT a particle with excess protons. It is not an acid. A free radical has no excess at all, whatever the particle involved. A free radical in a group of atoms where a bond going to an outer atom has been broken in two equal parts, leaving one lonely electron on the atom belonging to the broken bond.
$5$) The oxygen molecule does not have an excess of protons. And elements that donate electrons are not called anti-oxidants. They are called reducers.
$6$) Oxygen does not alkalinize the body. Why the body ? $\ce{O2}$ never alkalinizes any substance, whatever this substance.
$7$) Why do you think that $\ce{ClO2}$ donates $5$ electrons ? And why $5$ ? $\ce{ClO2}$ does not donate electrons, as it is an oxidant.

Thanks for the clarifications, I think I was clear enough in that I had a few misunderstandings, and where they generated contradictions. Not sure why all the negatives, If I knew chemistry I wouldn't need to come here to ask in the 1st place.
I'll leave this question undeleted (as suggested) because I don't care about me rep here, and I'm sure these are things many people are confused about, in the health space.
On (6) you are right, it's the removal of CO2 that does the trick.

(7) chlorine atom in chlorine dioxide has an oxidation number of +4. For this reason chlorine dioxide accepts 5 electrons when it is reduced to chloride.
I had it backwards. reducing/accepting | oxidizing/donating are confusing terms.


text:
Schlenk line simultaneously opening vacuum and gas
I’ve been working in the lab for some time now, but I’ve found conflicting information when asking this question, so maybe you know the answer.
When working in a Schlenk line with double oblique stopcocks, imagine the following situation. One Schlenk flask needs to be under vaccum, and at the same time another one needs to be filled with argon gas.
Now some colleagues told me that you can’t simultaneously open the argon gas and vacuum even though the schlenk flasks are connected with a different tube and the argon manifold and the vacuum manifold are separate from each other. I’m referring to two different double oblique stopcocks.
My guess would be that it doesn’t matter as long both manifolds are seperate i.e independent from each other. Why should it be a problem? The setup is like this:

You can't open a single hose to both as the taps switching the lines won't let you. They either open to one manifold or the other and can't do both at the same time. A vessel attached to two hoses, though, could be open to gas on one line and vacuum on the other. There is no restriction on what you can do with separate hoses/valves/vessels. I suspect your colleagues advice was garbled in translation.

Yeah I edited my question to clarify; opening two hoses to two different Schlenk flasks. Not a single one

in which case the answer is there is no problem at all. I've certainly run multiple different operations on a single Schlenk line without any problems.

I've learned it as a precaution. Before drawing vacuum on one line, you close all others. If you open the valve the wrong way, the worst that could happen is you ruin one product. Also, depending on what you are doing, the sudden pressure increase (and drop) might have funtastic outcomes for all the other hoses. But then I became a computational chemist, because I have thumbs for fingers... So my advice is probably not the best.

Okay @matt_black just as I figured, thanks.

There are no general reasons to not use vacuum and gas at the same time but there are some subtleties in practice
The design of the valves in Schlenk lines prevent the (obviously bad) idea of opening a single vessel to vacuum and gas at the same time. But should a complete Schlenk manifold only be used for one operation on one vessel at a time?
There is no solid reason why not. The vacuum systems are independent of the gas systems and the use of one does not, in principle, interact with the use of the other. Why, after all, do typical manifolds have multiple valves?
But, in practice, doing too many things at once sometimes can be ill-advised. For example, exposing a vessel to the vacuum side of the manifold is used for a variety of purposes. Sometimes it is to ensure all the oxygen is removed from a vessel before conducting an air-sensitive reaction; sometimes it is to drive out remaining solvent from a completed reaction. And some of these actions need constant attention to prevent accidents. For example, when trying to remove solvents, it is important not to open the vessel to vacuum too quickly or the solvent can froth or boil, contaminating the vacuum line; the same is sometimes true when trying to evacuate gas from finely divided powders where the entrained gas can "blow" the fine powder into the vacuum line (I've done this preparing raney nickel which can be quite dangerous as the dry powder is spontaneously flammable in air!). It is unwise to attempt two such operations simultaneously as divided attention is dangerous when two actions both need fine control of the rate the vacuum is applied. But, once the dangerous stages are complete there is little reason two lines cannot be open at the same time.
When filling vessels with inert gas the major issue is that, if two vessels are both being flushed with inert gas, the overpressure can get too low to do a good job on one if the other is too open.
But the inert gas line is independent of the vacuum line so there is no reason not use both at the same time unless your lack of attention opens a valve the wrong way.
I suspect that the main reason for any advice recommending not to do two different actions at the same time is to prevent attention errors where one action needs to be carefully managed. But this is only a major issue when initiating the action as many reactions are quite safe when started so it is OK to leave them under vacuum or inert gas one the action is initiated (multiple vessels can be evacuated while others are flushed with inert gas as long as fine control is no longer needed).


text:
Physical significance of electrode potential for M^2+/M where M is a 3d series d-block metal
What is the actual meaning of electrode potential of $\ce{Cu^2+}$/$\ce{Cu}$ having a positive value? Does that mean that the reaction  (i.e. $\ce{Cu^2+ + 2e-  ->  Cu}$) will want to happen more vigorously than the elements having a negative value or is it the other way round?
From my understanding, since the configuration of $\ce{Cu^2+}$ is $[\ce{Ar}]\mathrm{3d^{9}}$ and the configuration of $\ce{Cu}$ is $[\ce{Ar}]\mathrm{3d^{10}4s^{1}}$, the reaction will happen more vigorously than others with a negative value becuase the end product has a fully filled d-subshell and should be more stable.

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text:
What is the purpose of the reflux ratio for solvent extraction?
Not distillation
If I understand it correctly, in my example, one has two components, A and B, starting at the "feed" in the aqueous phase. They get both extracted in the mixer settler, but component B is only partially extracted.
Thus B goes with the raffinate to the yellow circle where the reflux ratio comes into play. The reflux ratio tells one how much goes further to processing (down) and how much is sent back with a neutralization step to the mixer settler (up). What is the purpose of this step?
A that was extracted in the mixer settler and B, if extracted, goes into the organic phase (thus moves to the right) to the stripping stage, where it is stripped into the aqueous phase (thus moves to the left). Therefore, the mixer settler concentrates A.

Brown, Clifford G.; Sherrington, Leonard G. (1979): Solvent extraction used in industrial separation of rare earths. In: J. Chem. Technol. Biotechnol. 29 (4), S. 193–209. DOI: 10.1002/jctb.503290402. *

My thought is: If one imagines a reflux ratio of infinity or, in other words, nothing is going to further processing, then with time, the component in the raffinate gets more and more concentrated and pure. Thus, this would work as a batch process. If you want to have it continuous, you use a smaller reflux ratio. The trade-off is, of course, purity.

Reflux is a subpart of a distillation unit. You circled a recycle split that delivers a fraction of a product stream back upstream. The process diagram has another recycle stream up top to control solvent reuse. The recycle stream you circled is to control conversion to product. See https://en.wikibooks.org/wiki/Introduction_to_Chemical_Engineering_Processes/Why_use_recycle%3F and https://en.wikibooks.org/wiki/Introduction_to_Chemical_Engineering_Processes/How_to_Analyze_a_Recycle_System

In the context of liquid-liquid extraction, reflux may be understood as:

"back-recirculation of the scrubbing or stripping solutions"

Source: https://www.mdpi.com/2075-163X/13/6/714

So instead of using pure scrubbing solution in the process shown above, you add a certain fraction of the "Solution of B".
Another source describes the purpose of reflux:

To minimize the dilution of rare earth concentrations caused by the introduction of the barren aqueous scrub and the organic phase, reflux is sometimes used; some of the light rare earths from the aqueous raffinate are loaded back into the organic phase entering the first mixer settler, and some heavy rare earths are stripped from the organic product and added to the aqueous scrub.

So in general, the purpose of reflux seems to be to minimize solvent and acid usage, and to make steps more efficient.

[OP] What is the purpose of this step?

The process diagram posted by the OP is a bit puzzling because there is only one stream exiting the process. There should be another stream exiting (maybe from the stripping stages) for the solutes that are removed from the light rare earth solution. Different from the examples I show above, it seems that the reflux shown in this process concerns the organic phase, only. In the first example I show, the entirety of the organic solvent is kept in an internal loop, with aqueous solutions entering and exiting the process.
The state-of-the-art rare earth purification facilities are located in China. One report claims that the "hyperlink" technology is related to the reflux discussed here:

In common solvent extraction processes, the acidic solvation agents are usually saponified in the organic phase by addition of a base in order to achieve the extraction, whereupon the REE are stripped from that phase by a new acidic solution in the next step. This consumes a large amounts of chemicals and creates large amounts of acidic waste. Efforts to reduce this problem include the “hyperlink process”, which is proprietary but seems to involve reuse of the organic and aqueous phases and is supposed to decrease chemical consumption and the related waste production by 30% or more. (Yan et al., 2006)

Source: Rare Earth Elements - Purification, Separation and Recycling. Available from: https://www.researchgate.net/publication/316754287_Rare_Earth_Elements_-_Purification_Separation_and_Recycling [accessed Dec 28 2023].

Thanks for the explanation. If I understand it correctly, the reflux is there to have a certain rare earth element concentration in your organic phase. Thus, it is a bit loaded, which makes sense in your example for scrubbing, maybe.  But in my example, I do not think so.

To add a few points: In my example, we do not have a continuous process but a batch process. Thus, after 100 h, they empty the product stages. Thus, they get a loaded aqueous phase and a loaded organic phase. With stripping, they get the rare earth from the organic into the water phase.

If I understand hyperlink technology correctly, it is the use of a different flow sheet. Fuzzy-link separation (hyperlink) is a process where the extraction focuses on the most extractable element, e.g., from a La, Ce, Pr, Nd solution Nd, and the least extractable element, La. It thus separates the feed into two fractions, La/CePr, and CePr/Nd.


text:
Is NaCl precipitation possible in this mixture?
Consider the following mixing of solutions:
$$\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}$$
My chemistry book says that this equation is quite correct, especially in the sense that $\ce{NaCl}$ is always completely dissolved.
Naturally, I'm trying to prove it wrong. I read online that we can put at most 36g of salt into 100g water for it to dissolve completely. I also read on a different site that even by conventional means, one may produce a hydrochloric acid solution with 37 wt %. This sounds about as much as the previous number, but 36g per 100g solution would be a much larger proportion. Then, apparently we can dissolve 111g sodium hydroxide in 100g of water.
So the question is: Can't I then produce highly concentrated solutions of sodium hydroxide and hydrochloric acid, so that upon mixing the two, some salt solidifies, because the resulting amount of salt is larger than what would be admissible in a saturated salt solution?

There is no law forbidding that. As for the details, I encourage you to finish the calculations that you just started. Take as much of the first solution as to contain one mole of HCl...

You're not ready for this kind of heat…

@Martin-マーチン could you clarify?

Highly concentrated acid + highly concentrated base -> a lot of heat of neutralisation + (maybe what's left of a) neutral solution

I see, so you're saying it's going to explode.

It won't exactly explode, but boiling of something like that...

Correction of the textbook statement:  If HCl + NaOH neutralization is being performed by a chemist with common sense, NaCl always remains dissolved. // There are very few rules in chemistry that have status of non-breakable law. Very most of them have exceptions, proverbially confirming the rule. There is almost always possible to find a scenario where a particular rule is not valid.

Of course you can precipitate salt. Under pressure, the $\ce{HCl (aq)}$ could be far stronger, though it might not be completely ionized. However, that does not invalidate the equation, per se, other than that some of the product is $\ce{NaCl (aq)}$, and some $\ce{NaCl v}$.
This would be the same as the silver nitrate test for chlorine: $\ce{AgNO3 (aq) + NaCl(aq) -> NaNO3(aq) + AgCl v}$. $\ce{AgCl}$ is slightly soluble, 520 μg/100 g at 50 °C,  so the reaction actually can go backwards, but for convenience, it is written as if the precipitation were 100%.

But what you wrote in the first paragraph is, as I have indicated, precisely my point! Not everything stays ionised. Apart from that, up to boundary cases the silver nitrate test produces SOME AgCl if ANY NaCl is present, whereas if I mix my two solutions above, NaCl is only produced if their concentration was sufficiently high.

Assuming we have $\pu{100 g}$ of $\pu{37\%}$ $\ce{HCl}$, e.g.  $\pu{37 g HCl}$ + $\pu{63 g}$ $\ce{H2O}$
The equivalent mass of NaOH:
$$m_\ce{NaOH}=\frac{M_\ce{NaOH}}{M_\ce{HCl}} \cdot m_\ce{HCl}
\approx \frac{\pu{40 g mol-1}}{\pu{36.5 g mol-1}}\cdot \pu{37 g} \approx \pu{40.5 g } \ce{NaOH}$$
Solubility of $\ce{NaOH}$ is $$s= \dfrac{\pu{114 g } \ce{NaOH} }{ \pu{100 mL}} \approx  \dfrac{\pu{\pu{114 g} } \ce{NaOH} }{ \pu{100 g } \ce{H2O}}$$
The mass of water in $\ce{NaOH}$ solution is
$$m_\mathrm{\ce{H2O}\text{ in }\ce{NaOH}} = (\pu{40.5 g}) \frac{(\pu{100 g})}{(\pu{114 g})} \approx \pu{35.6 g } \ce{H2O}$$
We now sum water from HCl solution, from NaOH solkution and water from neutralization
$m_\mathrm{\ce{H2O},tot}=\pu{63 g} + \pu{35.5 g} + (\pu{40.5 g } \ce{NaOH}) \frac{M_\ce{H2O}}{M_\ce{NaOH}} = \\
= \pu{98.5 g} + \pu{40.5 g} \frac {\pu{18 g mol-1}}{\pu{40 g mol-1}}\approx 98.5 g + 18.2 g \approx \pu{116.7 g \ce{H2O}}$
There is formed $m_\ce{NaCl} = m_\ce{NaOH} \cdot \frac{M_\ce{NaCl}}{M_\ce{NaOH}} \approx \pu{40.5 g} \cdot \frac{\pu{58.5 g mol-1}}{\pu{40 g mol-1}} \approx \pu{59.2 g \ce{NaCl}}$
Now we have $\pu{59.2 g \pu{NaCl}}$ and $\pu{116.7 g \ce{H2O}}$, what gives approx. $\pu{50.8 g \ce{NaCl}}$ for $\pu{100 mL \ce{H2O}}$.
As $50.8 \gt 36$, about 1/3 of $\ce{NaCl}$ would precipitate .

I agree that your calculation is simpler. Kudos!

So, the molar concentrations are as follows:
Suppose we got twice 50g water, and dissolve HCl in the first portion and NaOH in the second.
We know that we can dissolve at most 36g of salt in 100g of water. Salt is sodium chloride, and sodium has an atomic mass of about 23, whereas chlorine has about 35. This gives us about .62 mole of salt in the 136g solution.
So we would need to dissolve strictly more than .62 moles of NaOH and HCl in 50g of water.
NaOH has a total molecular mass of about 40, and HCl has about 36. Multiplied by .62 this makes about 25 and 22 respectively. These are precisely the grams of each substance we need to dissolve in 50g of water.
I conclude (22/72 ~= .31) that even using conventional methods, we could get solutions that are sufficiently concentrated.
All of this depends on the reaction processing the largest part of the educts.
If any of this sounds wrong, please downvote!

Hmm, I think you didn't account for water created in neutralisation.

Right. I'll correct this, but not this evening.

RE:So the question is: Can't I then produce highly concentrated solutions of sodium hydroxide and hydrochloric acid, so that upon mixing the two, some salt solidifies, because the resulting amount of salt is larger than what would be admissible in a saturated salt solution?
Back of the envelope calculations say yes. However pragmatically don't try it experimentally without serious personal protection. Mixing highly concentrated $\ce{NaOH}$ solution with a highly concentrated $\ce{HCl}$ solution will create a lot of heat. Thus the solution is likely to "pop" and fling some of the concentrated solution around. Think of diluting concentrated sulfuric acid if you've ever done that.
Now for the calculations
For $\ce{NaCl}$
PubChem gives a solubility of 36.0 g/100 ml (25°C) and a density of 2.17 at 25°C
The weight percent of a saturated solution of $\ce{NaCl}$ is
$$\text{Wt% }\ce{NaCl} = \dfrac{36}{36+100} = 26.5\%$$
The molecular mass of $\ce{NaCl}$ is 58.4 grams/mole, so the solution is:
$$\dfrac{\ce{2170 x 0.265}}{58.4} = 9.84 \text{ molar}$$
In a liter of solution:
$$\text{g }\ce{NaCl} = \ce{2170 x 0.265} = 575$$
$$\text{g }\ce{H2O} = 2170 - 575 = 1595$$
For $\ce{NaOH}$
PubChem gives a solubility of 109 g per 100 ml water at 20°C: as does the Wikipedia Solubility Table. Wikipedia also gives a solubility of 119 g per 100 ml at 30°C. Interpolating between the two Wikipedia values gives 114 g /100 ml at 25°C. PubChem gives a density of 2.13 g/cu cm 25°C
The weight percent of a saturated solution of $\ce{NaOH}$ is
$$\text{Wt% }\ce{NaOH} = \dfrac{114}{114+100} = 53.3\%$$
The molecular mass of $\ce{NaCl}$ is 40.0 grams/mole, so the solution is:
$$\dfrac{\ce{2130 x 0.533}}{40.0} = 28.4 \text{ molar}$$
In a liter of solution:
$$\ce{\text{g }\ce{NaOH} = \ce{2130 x 0.533} = 1135}$$
$$\ce{\text{g }\ce{H2O} = 2130 - 1135 = 995}$$
For $\ce{HCl}$
Wikipedia states that a 37% solution is 12.4 molar and has a density of 1.189 kg/L. The MW of $\ce{HCl = 36.5}$.
In a liter of solution:
$$\text{g }\ce{HCl} = \ce{12.4 x 36.5 = 453}$$
$$\text{g }\ce{H2O} = \ce{1189 - 453 = 736}$$
Thus stoichiometrically about 1.00 Liter concentrated $\ce{NaOH}$ will be neutralized by 2.28 Liters of concentrated $\ce{HCl}$.
The neutralization creates roughly $28.4$ moles of water which is about 511 ml of water.
So 1.00 liters of concentrated $\ce{NaOH}$,  mixed with 2.28 Liters of concentrated $\ce{HCl}$ will have 28.4 moles $\ce{NaCl}$ and:
$$\text{g }\ce{H2O} = \ce{995 + 736 x 2.28 + 511 = 3184}$$
$$\text{g }\ce{NaCl} = \ce{28.4 x 58.4 =  1658}$$
Per 100 ml of water
$$\text{g }\ce{NaCl} = \dfrac{1658}{31.84} = 52.1$$
This exceeds the PubChem given solubility of 36.0 g/100 ml (25°C) so $\ce{52.1 - 36.0 = 16.1 g}$ per 100 ml of water must precipitate.

While diluting concentrated sulfuric acid in the sink I had a beaker get so hot that it cracked due to stress (scratch on glass, temperature differential- whatever). Fortunately I could just quickly flush the mess down the sink with lots and lots of water.


text:
Can Adding Alcohol Create Potable Water?
I've been exploring various methods for purifying water, and I'm curious about the potential effects of adding alcohol to make it potable. Could the addition of alcohol serve as a means of purifying water, or would it have adverse effects on its potability? Are there specific concentrations or types of alcohol that could potentially render water safe for consumption, or is this approach entirely ineffective or even harmful? Any insights or studies on this topic would be greatly appreciated. Thank you!

what are the contaminants making the water non-potable?

Please check your use of terminology. To purify means to remove contaminants, not to make potable. A pure substance is free of any others.

Ethanol is the only type of lower alcohols the human body can tolerate in amounts that are antibacterial.

Adding alcohol to water can make it safer to drink in some circumstances
There are many ways for a water supply to be contaminated with things that are bad for people. Alcohol can deal with some of them.
If the water supply is contaminated with inorganic poisons like lead or mercury, alcohol won't help. If it is contaminated with biological pathogens like cholera, E. coli or campylobacter, then a certain level of alcohol may render the water far safer for people.
Indeed the use of alcohol to make potable, safe drinks has been the dominant method of water "treatment" in human history possibly since before cities became common.
Only in the last 200 years have people had routine access to potable water free of pathogens (domestic water supplies use chlorination or ozonation to ensure bugs are killed but those are recent inventions). Before that the safe drink was an alcoholic beverage as the alcohol levels were mostly sufficient to kill the typical pathogens. Most spectacularly, when John Snow–no, not the Game of Thrones one, the father of modern epidemiology–was collecting the data about the London cholera epidemic in 1854 to understand its source, he observed that there were no cases in a brewery near the famous contaminated pump because the workers were all given generous supplies of beer.
More about the history of alcohol in "cleaning" water is given in this Skeptics.SE answer.
For most of human history alcohol has been a major way to make the water supply safer.

But microbiological safety of beer is result of many more processes than formation of alcohol itself, the most notably boiling.

@Poutnik That is also true. But many cultures added their wine to water to achieve the same effect. And the presence of alcohol in beer preserves the effect. And the production of some alcoholic drinks does not involve boiling.

But then it is rather diluted wine than water, while production of some drinks does not involve adding raw water.

@matt_black historical beer alcohol content seems to be reported around 1%. Is that really enough to preserve anything?

@infinitezero i suspect that report that historic beer was weak is nonsense. I've seen the claim before but the evidence seemed really speculative and weak.

@Poutnik The effect of adding wine to water works to reduce pathogens as does the production method of most alcoholic drinks. If the dominant drink is alcoholic rather than raw water, the effect is what matters. eg the John Snow example.

@matt_black But in the Q context, the primary criteria is if it is still possible to call the result potable/drinkable water. Not if boiling malt to make beer, or adding wine to water makes the result drinkable. E.g. according to local law, any drink with more than 0.5% of ethanol is considered as alcoholic drink. You would need to pour 1L of wine to about 25 L of water to call it "potable water". And I would hesitate to drink it if water source was questionable, even if just microbially.

I assume wine acids in high dilution could be more potent in killing bacteria than ethanol, especially in water of low bicarbonate content, boosted by gradual acetic fermentation.

@Poutnik The legal designation of the resulting drink is irrelevant. if creating a 5% alcohol solution kills bugs, that is a big contribution to creating a safe drink. Though we now have far better ways to achieve safe drinks, this was a big factor historically and probably saved many lives from the effects of dirty water.

Safe drink is not the same as water and it neclects chemical composition. You would not let 2 years old child to drink 5% ethanol as water unless as the last option if water itself was high risk of death or disease.

@Poutnik If the essence of the question was the distinction between a "safe drink" and "water" it should be moved to a linguistics site not a chemistry site. In other words, your objection is distracting from the essence of the question.

Well, you are distracting from the fact than alcohol does not make from general unsafe water safe drink.

@infinitezero 2.5, 3%, not 1%. Of course stronger beer was available, but more expensive, and you get drunk. 2.5% beer is pretty harmless in terms of direct intoxication.

It would not create potable water. It would create a disinfectant, with concentration higher than many/most spirits, if it should assure sufficient antimicrobial effect.
Low concentrations would additionally lead to formation of acetic acid, so there would be kind of vinegar after few days.
And, it would omit sensorical and chemical aspect to be potable water. If it was not good in this context before alcohol addition, it would not be good after it either.

Only a small selection of bacteria can grow while metabolizing ethanol of beer or wine into acetic acid and surviving the low pH environment, and none are human pathogens.  Further the conversion is very slow since it requires free oxygen.  We agree that modest ethanol levels in a dilution may not kill pathogens, but neither can they grow.

@stevea Well, they do not grow much in water either. They mostly stay and wait.

Additionally, producing beer involves boiling and wine is not made from water.

@Poutnik They grow, on less-than-clean surfaces, and from whatever else is in that water. With low amounts of alcohol, or vinegar (Roman "posca"), the germs have trouble staying alive, giving them much lower chances of overwhelming the defences in your stomach. I like living in modern central Europe, where the tap water is so clean they don't even chlorinate it at all. But in ancient Rome, or Lutetia, Londinium, that made quite a difference.

"Well, they do not grow much in water either."  If it was "just water" there wouldn't be bacteria & molds.


text:
Reason behind benzylamine being a primary amine
Benzylamine is a primary amine. Now a primary amine has the general formula R-NH₂ where R is an alkyl or aryl group. So here R will be equal to $\ce{C6H5CH2}$. But $\ce{C6H5CH2}$ is neither an alkyl group nor an aryl group. So why benzylamine is a primary amine?

Benzylamine falls in to the category of RCH2NH2 - in this case R=Ph but it is still a primary amine because  1 - it is attached to an alkyl carbon  2 - the N bears 2 hydrogens.

Does this mean NH(2) does not directly need to be attached to the R functional group to be called an anime? @Waylander

Also I think an alkyl carbon is a carbon atom that is part of an alkyl group. An alkyl group is a fragment of an alkane molecule that has one hydrogen atom removed.So I am unable to understand how the carbon attached to the nitrogen will be an alkyl carbon (I am unable to see how the carbon is the part of an alkyl group.As I think there is no alkyl group present) @Waylander

IF A is X and B is X then C as combination of A and B is usually X too. R need not to be even a hydrocarbon group and it would still be an amine, (or a compound with an amine group if nomenclature chooses different category name).

@Poutnik can you please briefly explain what you meant.I tried to understand what you are saying, but failed

Point out particularly what you do not understand there as I have already briefly explained it.

"A is X", "B is X".Now we are combining A and B to form C, which also will be X.I am unable to understand what C will be X try to imply.Also what is the relation between the R and the C here? @Poutnik

It was not reaction, it was an analogy. C(R) as combination of A(alkyl) and B(aryl) was meant being figuratively between A and B.  Like if you had 2 sets of fruits, one with fruits bigger than 10 cm(alkyls), the other smaller than 8 cm(aryls). and you told these are apples(amines).  Now you got one 9 cm big(aryl+alkyl combined). Is it possible it is an apple(amine) ?

Note that benzylamine as phenylmethylamine can have the phenyl substituent substituent replaced by any other and still being amine, or amine derivative of other compounds, depending on nomenclature priority.

The IUPAC gold book defines amines as:

Compounds formally derived from ammonia by replacing one, two or three hydrogen atoms by hydrocarbyl groups, and having the general structures RNH2 (primary amines), R2NH (secondary amines), R3N (tertiary amines).

https://doi.org/10.1351/goldbook.A00274
And it defines hydrocarbyl as:

Univalent groups formed by removing a hydrogen atom from a hydrocarbon, e.g. ethyl, phenyl.

https://doi.org/10.1351/goldbook.H02891
I would have said making a bond between ammonia and any organic molecule, replacing one hydrogen atom bound to nitrogen on ammonia, and one hydrogen of a C-H bond in the organic group, but that would also give rise to multifunctional molecules, e.g. with a hydroxyl and an amino group.

Thanks for your reply @Karsten.Your answer made perfect sense after all these wasted hours of searching. Every article I read was all about R can only be an alkyl or aryl group.

People posting on the internet often have beginning students as audience, and their definition is a bit quicker to write (and perhaps a bit easier to remember). Good for you discovering the inconsistencies and asking about them.

@RajeshPaul You can see benzylamine as a primary alkylamine  (methylamine) substituted by an aryl(phenyl).

Thanks! @Poutnik for pointing that out.I didn't know one can substitute, then check if the compound is a primary amine

The simplest amino acid glycine NH2-CH2-COOH is an acid in spite of the amine group and at the same time a primary amine, in spite of carboxy group. Due formal naming group priority, it is called aminoacid and not carboxyamine.


text:
what would be the mechanism for these reactions?
I believe the first one would have an intramolecular attack by the sulfur inverting the stereochem followed by an attack by OH- but I can’t get the second one.

Sodium nitrate/HCl is standard conditions for the diazotisation of primary amines. You may have seen this done with anilines, but it works generally for primary alkyl amines. The diazo group is then displaced by chloride. As reported here the reaction goes with retention of stereochemistry. If you use a less nucelophilic acid e.g. Aq. sulfuric acid you get the hydroxyacid again with retention

What about the first equation, the guessed mechanism is totally wrong if I'm not mistaken.


text:
Charge as a function of time in electrolysis
According to Faraday's laws of electrolysis,
$\frac{Q}{Z}$  = no. of moles of substance released (n), where Z is n-factor x 96500. Assuming the electrolyte is a strong electrolyte, specific conductivity (κ) =  $\frac{Λ(n_0-n)}{10 vol}$ and resistance = $\frac{l}{Aκ} = \frac{V}{i}$, where n0 is initial moles of substance, κ is specific conductance and Λ is molar conductance.
Combining all these equations together, we get $\frac{Λ(n_0-(Q/Z))}{10vol} = \frac{li}{AV}$, which after some working out gives us this, Q = Z$n_0$(1 - $e^{kt}$), charge passed as a function of time, where k = $\frac{ΛAV}{10Zlvol}$. Here I'm assuming that the electrolyte is strong and with the given constant potential difference the electrolysis is spontaneous, even with the change in concentration occuring. Is this correct? How would I change the equation to adjust it for weak electrolytes as well?

It is better to say amount of substance/substance amount than number of moles of substance, similarly as mass of substance/substance mass sounds better than number of kilograms of substance.

Specific conductance = conductivity.  Conductivity is just conductivity. Molar conductance should be called molar conductivity.  // Why "vol" and not V ?  What the coefficient 10 is about ?

Your equation says if the substance amount is unchanged ( n0 - n = 0)then the conductivity is zero. that is obviously false.

The title of the question mentions "a charge as a function of time". But the question itself after the title does not speak of time. Never. There are plenty of formula related to conductance, resistance,  number of moles, conductance, etc. but no time nowhere. It is about the same for "charge". There are no charge in the text of the question.

There is an important detail that at electrolysis, the current is not equal to product of voltage and conductance.

@Poutnik I used the symbol V for the constant potential difference applied, the coefficient 10 comes from a formula I changed to give κ the unit of S^-1 m^-1. As for the important detail you mention, I haven't seen or read that anywhere. Could you elaborate?

@Poutnik I really haven't seen U being used as potential difference in my books, that is a slightly irrelevant change considering it is clear what is what. As for ohmic resistance, I think it becomes clear that there is no linear relationship between current and potential difference since while solving the equations, R turns out to be a function of charge passed. However, I am rather interested in what how exactly the resistance we find is related to current and voltage if not in the usual way of V = iR. Could you clarify more on that topic?

Solution conductance, conductivity and molar conductivity are measured by small AC voltage of sufficient frequency, typically $\pu{1 kHz}$. In such a scenario, most of effects involved in electrolysis do not manifest.
In context of DC electrolysis:

The resulting effective resistance or conductance of the electrolytic cell, formally fitting the Ohm law, depends on many factors. The solution conductance or conductivity is just one of them.
Each electrode has for the given cell its voltammetric characteristic $I=f(E)$, which in large extent resembles a shape of a series of smoothed steps (one or several, often overlapping).
This characteristics dynamically changes under the current load and with change of composition of solution or electrode.
When an external voltage is applied, each of the electrodes establishes a momentary potential (being referred to SHE), where:

The electrode potential difference is equal to the external voltage.
The absolute value of currents passing the electrode are equal.

The external voltage is equal to the sum of potential differences
$$U_\text{ext}=U_\text{electrodes} + U_\text{charge transfer} \\
+ U_\text{local electrode concentration gradient} + U_\text{electromigration}$$
The particular terms are complicated, not exactly modelled functions of current and time.
As a huge simplification with limited accuracy, there is the approximate proportionality
$$I \propto (U - U_0) \cdot c,$$
where

$U_0$ is the voltage threshold when the current starts passing the cell.
$c$ is the concentration of limiting electroactive component, depending on the scenario.
The proportionality constant is the subject of dynamic changes or drifts, and the proportionality is just approximation.

If you want to control the charge passed the electrolytic cell, do not use the power source of constant voltage, but of constant current, with the max voltage limitation. Such sources are less common, but could be bought or assembled by skilled persons.
They internally keep constant  voltage difference on a small serial internal resistor, effectively keeping the passing current constant.
They can be formally modelled  as power sources of a huge voltage with a huge internal resistance. They, within limits, keep on the attached circuit such a voltage that is needed to keep the chosen current.


text:
What do the dots mean in DABCO's IUPAC name?
I came across this molecule with the following IUPAC name: 1,4-Diazabicyclo[2.2.2]octane
What do the dots inside the brackets mean, or where can I read about the IUPAC guidelines that require this?

Better question is what the numbers they separate mean...

https://chemistry.stackexchange.com/questions/37970/what-does-the-h-k-and-5-2-2-02-6-mean-in-dibenzoh-k-4-oxatricyclo5-2 https://chemistry.stackexchange.com/questions/27217/how-to-rationalise-the-iupac-name-for-tetrahedrane

The dots are used in bicyclic compounds and indicate the number of carbons in each bridge.  DABCO has three bridges, each with two carbons.


text:
Reasonable mechanism for conversion of 2-methylbut-3-en-2-ol to 1-Bromo-3-methylbut-2-ene

In the book "Designing Organic Syntheses - A programmed approach to the Synthon Approach" pg 23, the author proposes that one route to 1-bromo-3-methylbut-2-ene would be by reacting 2-methylbut-3-en-2-ol with hydrobromic acid. I am confused as to why these two reactants wouldn't produce 3-bromo-3-methylbut-1-ene instead. I am guessing the first step in a reasonable reaction mechanism would be protonation of the alcohol to form H30, which then leaves via SN1 forming a tertiary carbocation with a trigonal planar geometry. I would have thought that the nucleophilic bromide would have preferentially attacked the tertiary carbon since over the primary carbon since:

the tertiary carbocation is more stable than the primary carbocation
the tertiary carbocation has a trigonal planar geometry so sterics wouldn't prevent it from being approached by the bromide

Reaxy's link to the reaction described- href https://www.reaxys.com/reaxys/secured/hopinto.do?context=R&query=RX.ID%3D175529&database=RX&origin=ReaxysOutput&ln=
Background - I am not a chemist but I have been reading about Organic chemistry, on and off, since I took it in college 12 years ago.

Check out this logic: cation formation and then resonance. I am not able to expand further, but my thought was originally formed because the cation is allylic in nature and is resonance stabilised.

The sterics are still a large enough factor, bromine is a large atom

Also the higher substituted double bond is the most stable and the thermodynamic more stable product is preferred.

Hey, thanks for the responses so far and sorry for the delay - I was waiting for an email that someone replied to my post but I never got one. Anyway, @HarikrishnanM Allylic resonance is definitely at play, but what I thought was unusual in this case is that the electrons from the Pi bond are moving to form the less stable cation. I always read that primary cations basically dont form.

@Waylander - I will try to see if I can find some material that discuss the effects of sterics on species with triagonal planar geometries. I assumed it wasn't a factor but you might be correct.

@aventurin I will look into this as well. The author of the book with the reaction basically gave the reason that you had. I originally thought he was referring to zaitsev's rule, but I was confused by this reasoning as well because I thought Zaitsev's rule was only applied to elimination reactions.
Thanks all for the food for thought. Ill do some more reading and let you know if anything comes of it.

Alright! But resonance will lead to less stable cation - you're correct gregmagdits! Steric hindrance is also another factor a/c to the question a/c to @Waylander and Aventurin.

The conjugated cation is MORE stable than the tertiary cation. There are more electrons to cuddle to that tertiary cation.

Im a little hesitant to post an 'answer' to a question about the reaction mechanism for a particular reaction since there is no way to verify this information without performing specific tests or an ab initio calc, but someone posted an answer on Reddit that I thought made sense so I am repositing it here:

There are 2 possible react mechanisms here that give you this product:
SN1 where the driving force is greater stability of a tri-substituted alkene versus mono-substituted alkene. Both are allylic carbocations so the substitution of the alkene is more important.
SN2’ (read as SN2 prime) where HBr ptotonates the alcohol making it a good leaving group. Then Br- can nucleophilicly attack the terminal alkene that inturn displaces the water.
A key part for this reaction is temperature. The higher the heat, the more SN1 will be favored. I believe typical SN1 reactions need about 90C, but that is more a rule of thumb.

Since the reaction conditions on Reaxys are at 0 C I suspect that the mechanism for this reaction would be the SN2 route described above

Tertiary allylic carbonium ions form readily even at 0C. The protonating agent in hydrobromic acid is H3O+ not HBr making  SN2 less likely and the stability of the alkene important. Higher temperatures in SN1[and SN2] reactions usually promote elimination over substitution, not SN2. The bromide ion is simply attracted to the entire delocalized cation and the more stable alkene is formed.

First $HBr$ protonates the $-OH$ making it $-OH_2^+$, Now $-OH_2^+$ being a leaving group will leave the molecule creating a tertiary carbocation. But as you can observe there is p - $\pi$ Resonance Available. Hence we get 2 resonating structures.
The above clarifies the presence of two Resonating Structures.
But deciding on the major product needs more clarification on the part of Reaction conditions.
There are some interactions called ion pair interactions. The $Br^-$ present in the reaction medium mildly stabilizes the tertiary carbocation which dominate the stability from resonance. But we know that resonance doesn't change position of atoms rather changes only positions of bonds and electron pairs. But the caveat is that ion pair interactions are only dominant at around $-5^oC$. So if the reaction is done at around $-5^oC$ we can say the intermediate involving in the substitution will be the first i.e the tertiary carbocation and not primary one.
Now as we generally do organic reactions at room temperature or higher we can say that the significance of these ion pair interactions is gone and the stability of the product plays a bigger role in the formation of the final product and also if the reaction is performed at higher temperature, it tends to have a reversible nature in the product forming step, where by chance if any of the product is from the tertiary carbocation, it gets converted to the one with primary carbocation.
Hope this clarifies the doubt.

Ther are no primary and tertiary cations.  There is one conjugated allylic cation that reacts with bromide to give the more stable alkene.


text:
Acidic strength of alkanoic acids vs. benzoic acid vs. methanoic acid
I've heard some people say that benzoic acid has a $+\ce{M}$ effect due to the benzene ring which destabilizes the conjugate base. Also, this would be the dominating factor over the weak $-\ce{I}$ effect of the phenyl ring. This makes sense and also explains that benzoic acid is less acidic than methanoic acid (which has no significant effects). But this fails to explain why alkanoic acids (ethanoic and onwards) are less acidic than benzoic acid. The former has a $+\ce{I}$ effect, so is it that the overall electron donating effect of benzoic acid is less than that of alkanoic acids (i.e. the $-\ce{I}$ of phenyl reduces the EDG nature of? If this is the case, that surprises me because generally resonance is a strong enough factor to overcome a weak inductive effect.
Any help is appreciated!

There was quite a bit of questions on this topic. And you know what? In practice, you can either believe whatever lie for student you want, or better ignore all lame explanations and just remember facts. Some +M or +I certainly qualify as lame.

Fair enough, I guess. So you're saying this particular fact can't really be explained well theoretically?

I guess if you used some software, maybe you could get satisfying results. Otherwise I don't think so.

https://chemistry.stackexchange.com/questions/7591/comparing-acidities-of-substituted-and-aromatic-carboxylic-acids

But solvation likely isn't the explanation - I heard HCOOH is stronger even in gas phase.


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Determination of the composition of the mixture based on acid-base titration

The solution can contain $\ce{HCl}$, $\ce{H3PO4}$ or $\ce{NaH2PO4}$
individually or in all possible combinations. An aliquot of
$25.00\,\text{mL}$ of this solution consumes $14.56\,\text{mL}$ of
$0.1000\,\text{M}$ $\ce{NaOH}$ solution with methyl orange, while
another aliquot of $25.00\,\text{mL}$ of the same solution consumes
$47.50\,\text{mL}$ of base with phenolphthalein. Determine the
composition of the solution.

Much less base was used with methyl orange than with phenolphthalein. Since methyl orange changes color in the acidic region, then the stated volume of base is used for either $\ce{HCl}$ or $\ce{H3PO4}$ or both. How to determine this exactly?
In addition to phenolphthalein, a substance (or substances) was proven which was also proven with methyl orange and some additional substance. However, I don't understand how to determine the exact composition.

Hint: Evaluate the ratio of spent volumes.

Hmm, the ratio is around 0.3, but I don’t know how to use that further.

If V2/V1 = 2, then there is H3PO4. If V2/V1 < 2 then there is HCl + H3PO4. If V2/V1 > 2 There is H3PO4 + NaH2PO4.  If there was initially added HCl + NaH2PO4, than it depends what of these two gets the upper hand, forming H3PO4.

I’m sorry, but could you explain a bit more? How did you get to that?

It is difficult, but not impossible to solve this problem, namely find $3$ concentrations out of two measurements. Let's try.
The two titrations give two useful volumes of $\ce{NaOH}$ $0.1$M.:  $14.56$ mL and $47.50$ mL, which correspond to amount of $\ce{NaOH}$ equal to $1.456$ mmol and $4.75$ mmol. The acidic samples to be titrated contain the same total amounts of $\ce{H+}$ in $25$ mL. Their acidic concentrations $\ce{[H+]}$ are equal to $1.456$ mmol/$25$ mL = $0.0582$ mol/L in the first titration and $4.75$ mmol/$25$ mL = $0.19$ mol/L in the 2nd titration.
The first titration is done with methylorange, that is up to pH $4$, where $\ce{HCl}$ and $\ce{H3PO4}$ are titrated.
The second titration is done with phenolphtalein, that is up to pH $9$, where $\ce{HCl, H3PO4 twice}$ and $\ce{H2PO4^-}$ from $\ce{NaH2PO4}$ are titrated.
$$\ce{[HCl] + [H3PO4]}  =  0.0582 M   \tag{1} $$
$$\ce{[HCl] + 2[H3PO4] + [NaH2PO4]} = 0.19 M \tag{2}$$
If {$2$} is subtracted from ($1$) it gives : $$\ce{[H3PO4] + [NaH2PO4] = 0.1318} M \tag{3} $$
But the data says that the solution contains either $\ce{H3PO4}$ or $\ce{NaH2PO4}$. Both products never occur simultaneously. If $\ce{H3PO4}$ exist in solution, its concentration must be < $0.058$ M (from ($1$), and $\ce{[NaH2PO4]}$ must be higher than $0$, from ($3$), (must be between $0$ and about $0.1..$ M ). This is contrary to the initial text : only one of the two products $\ce{H3PO4}$ and $\ce{NaH2PO4}$ must be present. We just saw that if $\ce{H3PO4}$ is present, $\ce{NaH2PO4}$ must also be present in solution. This is excluded. So $\ce{H3PO4}$ is not in solution. As a consequence : $$\ce{[H3PO4]} = 0$$ and $\ce{NaH2PO4}$ is the only phosphorus containing compound in solution. The final composition of the acidic solution is : $$\ce{[HCl] = 0.0582 M}$$ $$\ce{[NaH2PO4] = 0.1318 M} $$
This is all the job to be done.

I think you misinterpreted the text. In my opinion, it does not mean HCl AND ( H3PO4 XOR NaH2PO4). Chemically, If there is no H3PO4, there is no HCl either.

@Poutnik. Try to solve the problem without my understanding. You will fail


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Water electrolysis with in-series electrodes
I'm trying to improve my efficiency when doing water electrolysis. For my specific application, I don't need to separate the hydrogen and oxygen gasses, in fact it's important that I don't. But it's also important that I don't crank up the voltage and create corrosion and by products, so I want to keep the voltage as low as possible.
So, to increase current I need to increase the surface area of my electrodes. So my idea is this: Use a 1x1" stainless steel strip for my cathode and anode each, about 2cm apart. Then insert an additional 1x1" stainless steel strip in the middle between them.
So like this (cathode on the left): |||
The current should flow from my cathode on the left, through the solution, to the left side of the middle strip, which now becomes an anode, out the right side of the middle strip, which now becomes a cathode, through the solution, and to the final anode. Thus I'm able to double my total surface area.
This makes sense to me, but can anyone confirm whether or not this is a sound solution or not? Since the middle strip becomes a cathode on one side, and anode on the other, would I get strange edge behavior where they meet?

It would make sense if the middle electrode galvanically separated the solutions on the left and the right. Otherwise it would not work.
If you need higher production using larger electrode areas, you have two options:

If the power source can provide high enough current, connect two or several electrodes in parallel, forming one electrode with large surface. If possible, you can create a combed structure of alternating anode and cathode, i.e. all odd electrodes mutually connected and the same for all even ones. That also leads to short distance between electrodes and overall high conductivity.
If the power source cannot provide high enough current or if the power source has too high voltage, you can create a series of cells, an anode of one cells connected by wire to cathode of other cell. This ways, at the same current, but with high voltage, you can multiple you gas output.

But beware of danger of explosion, as manipulation with electrodes under voltage may produce a spark and mighty explosion of hydrogen and oxygen mix.


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What liquid mixture has largest negative excess volume?
Math teacher here. I’m trying to teach about types of sets that when you add them together, their sizes don’t add (called non-measurable sets). I came across the chemistry experiment where you mix 50 ml water and 50 ml ethanol, and you get less than 100 ml total solution. Is there a more dramatic version of this experiment?

Does this answer your question? Is it possible to have Volume(solution) < Volume(solvent before adding solute)?

https://chemistry.stackexchange.com/questions/49913/what-solutes-will-give-the-largest-reduction-in-volume-when-dissolved-in-water

https://chemistry.stackexchange.com/questions/23534/how-does-volume-contraction-in-solvent-mixing-work

@Mithoron it is fine to flag similar questions as potentially relevant or duplicates. But none of those other questions comes close to answering this specific question. It isn't a question about the effect, it is a far more specific question about what mixtures have the largest effect which none of the other answers come close to addressing.

I agree with @matt_black in re: this is not a duplicate; a little Googling seems to indicate that a binary mixture of 1:1 water:ethanol is a good candidate for the answer if limited to liquid-liquid examples. However, I think the author as a non-chemist might be missing a potential opportunity if the scope is widened to solid-liquid examples where electrostriction is more pronounced.

There is a database that might contain a large number of data points on different mixtures here: https://www.ddbst.com/ddb-ve.html . It doesn't look easy to search but contains a lot of liquid-liquid mixture properties.

A quick glance at the graph $V^\mathrm{E}$ vs $x_1$ for the system water+ethanol typically show a minimum of around $\pu{-1.2 cm3 mol-1}$ at ambient temperature. This means that when we mix them, then the system contracts.
Ekezie et al. (2023) studied the excess volume and excess viscosity for methanol with several alkyl acetates at different temperatures:

where $\ce{M}$ is methanol, $\ce{MA}$ is methyl acetate, $\ce{BA}$ butyl acetate, and $\ce{PA}$ pentyl acetate. The graph labelled with (a) is at $\ce{25 °C}$ whilst the one with (d) is at $\ce{40 °C}$. We stick with the latter.
The system $\pu{M+MA}$ presents an expansion of about $\pu{8 cm3 mol-1}$ when the molar fraction of methanol is around $\pu{0.3}$. The system $\pu{M+PA}$ presents a contraction of around $\pu{-9 cm3 mol-1}$ when again the molar fraction of methanol is around $\ce{0.3}$. Both values are are approximately $\pu{6.5}$ and $\pu{7.5}$ times higher, in absolute value, with respect to the system water+methanol.
As an aside, the authors argument with respect to the positive excess volumes that:

On mixing, methyl acetate appears to disrupt the self association
of methanol hydrogen bonds leading to rupturing of the
bonds and hence, the observed positive deviation. Positive deviation
may also be attributed to weak dipole-dipole interaction dominant in
the mixed system.

Conversely, for the negative excess volume:

It is possible that on mixing, methanol formed new hydrogen bonds
with the dipolar oxygen of the esters. This behavior may be attributed to inductive effect which increased the electron density and thus, the proton accepting ability of the alkyl acetate. Increase in alkyl chain length of acetates increase the inductive effect.

References
Ekezie, Chukwuebuka M., Cookey, Grace A., & Maduelosi, Jane N.. (2023). Volumetric and viscometric studies of binary mixtures of methanol and some alkyl acetates at varying temperatures. South African Journal of Chemistry, 77, 133-137. https://dx.doi.org/10.17159/0379-4350/2023/v77a16

The presentation of the demonstration can make clear the volume change, though only a few percent reduction.
For example, pour 50 ml of water, with a bit of food coloring added, into a 100 ml cylinder. Carefully add 50 ml of ethanol, with a different color dye, on top, without mixing (float a drop of olive oil on the water, if need be, to separate the layers).
Put a one-hole rubber stopper with a 50 cm or so long glass tube on top, pushing the stopper in so that the alcohol rises in the tube, and mark the top of that meniscus in the tube.
As soon as the cylinder is shaken, the level in  the glass tube will drop greatly, because the apparent decrease in volume is magnified by the ratio of diameter of cylinder to glass tubing. Ah... a teachable moment -- have students prove, or disprove that statement!

Yes, this is a good way to demonstrate the effect. But are there liquids where the effect is larger?


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Model EC50/Kd values using ordinary differential equations
It is my first attempt at using ordinary differential equations to model Kd for antibody binding and help will be much appreciated.
To provide some context, I possess binding curve data for the EC50 of various antibodies, each targeting different epitopes on our glycoprotein. Our goal is to explore the feasibility of employing luminex technologies to multiplex these antibodies on a single bead, thereby significantly enhancing the throughput of our assays. However, determining the correct stoichiometric ratios for each antibody is crucial, considering potential cross-reactions or agonistic effects due to varying affinities. For instance, a high-affinity antibody might influence the affinity of another, depending on factors that require a more extensive background explanation.
We have a suite on 12 antibodies and planning experiments with all possible combinations in a multiplex of three would necessitate around 1726 different experiments. To streamline this process, we are considering the use of ordinary differential equations to model these binding experiments. This approach aims to elucidate how the different antibodies, with their distinct Kd values, will interact when multiplexed.
Could someone kindly provide a comprehensive walkthrough of how they would approach this? Specifically, how do I incorporate the EC50 values (which are linear, thus EC50 = Kd) into the equations, and what order of ODEs should I employ for these equations? Being new to this, any insights and patience in guiding me through the process would be immensely appreciated.

You should be able to write down the rate equations for binding in the simple case with no cross-reactions etc and solve numerically. But with cross -reactions you will need to know all these rate constants which I guess you don't know so you will have to guess them and keep on doing so until you can fit all the data simultaneously, by massive non-linear least squares fit. However, these rate constant values will probably not be unique. i.e. many values will fit the data, If this is important I can't tell.


text:
Co(NO3)2 forming colloid in H2O, and Methanol
I was just trying to make a stock solution of some Co(NO3)2 . One in H2O, another in Methanol. Both don't completely dissolve. They're almost dissolved and appear near translucent but when shining a laser pointer through, I can see the path of light (indicating it's a colloid).
I'm just making a 1M solution in 50 mL test tubes.
Anyone have any experience with this? Shouldn't this salt be completely soluble?
Perhaps i'm dissolving a lot within this 50 mL volume and it's forcing some particles to slightly aggregate?
Thank you!
P.S: The one in H2O I made was in early December. Every week or so a bit of precipitate collects at the top and when I shake it, it dissipates. Sort of like a sediment but on top instead of bottom. I assume this is just aggregated Co(NO3)2 and i'm just redissolving it. The one in Methanol, I just prepared today but it's exhibiting similar properties and I haven't observed this precipitate in the methanol yet given it's been less than a day.

A first thought would be impurities in $\ce{Co(NO3)2}$. There are varying grades of purity of laboratory chemicals, ranging from technical (just good enough for some crude uses) to American Chemical Society (ACS)... but even ACS grade reagents might have 5% impurities! 99.999% ("five nines") pure $\ce{Co(NO3)2}$ is available, if the use justifies the price.
Another possibility is some contaminant in the labware (or even in the lab air) is reacting with the $\ce{Co(NO3)2}$. For example, if a glass vessel were to be cleaned in an alkaline detergent, and not sufficiently washed (perhaps using an acid rinse), adherent alkali might precipitate an insoluble cobalt compound. You could test for this by rinsing all labware used to measure and to hold that stock solution with a small amount of nitric acid, and then rinsing again with highest purity water available. Use that highest quality water to make the stock solution.
Also consider the quality of glassware -- use good grade borosilicate, not soda-lime glassware.

i do use alkali to clean. i really hope it's not leaving some residue causing premature precipitation. wouldn't the pH need to be at a certain point though for the Co 2+ to precipitate as a Co(OH)2, or atleast have the process be kinetically/thermodynamically favourable? i feel like the tiny amount of residue left won't be enough to swing the pH that drastically, as i do rinse out the glassware thoroughly with water afterwards.

One uses the experimental method to verify a hypothesis. Make up some more solution, following the suggested procedure for acid rinse and water rinse.

For water solution, does the behaviour differ if kept acidic by addition of small amount of nitric acid? Co(II) salts should be stable, but if there are Fe salt traces, it can be hydrated Fe(III) oxide.


text:
Gibbs phase rule for azeotrope and Solid-Liquid phase diagram
Azeotrope
An azeotrope is a mixture of two or more components in fluidic states whose proportions cannot be altered or changed by simple distillation

Applying the phase rule to azeotrope. There are 2 phases: liquid and vapor, there are 2 components: X and Y. As the pressure is held constant, so F = 2 - 2 + 1 = 1.
Question:

What is the "1" degree of freedom represents? I cannot seem to find any proper explanation.

Thought: Is it because when we change the composition fraction, the mixture which includes both liquid and vapor still maintain?

Is there any exception to Gibbs phase rule?

Solid-liquid phase diagram

I wonder if the mixture is on the red line, what is the degree of freedom?
Thought: Since this is still a binary system, so F = 2 - 3 + 1 = 0. So, there is no intensive variable should be changed here. It is a bit strange here since the mixture is unchanged when increasing temperature (in the red line), so it should be 1?

Be aware that additional relations like reactions or the same phase co position for azeotropes decrease the number of degrees for freedom. So for the given pressure , there is no degree left and all is defined by the value of pressure.

So, you are saying that for azeotrope, the composition fraction is well defined, so another constraint like x(liquid) = ky(vapor) will decrease the degree of freedom?

The system has 1 degree of freedom,spent of choosing the pressure. It could be temperature too. But the 2nd parameter is determined by the chosen one. (It is not exactly correct as composition of azeotrope is slightly dependent on pressure. // see also  chem.libretexts.org - The Gibbs Phase Rule for Multicomponent Systems

What do you mean by choosing the pressure? The liquid-vapor phase diagram above has the pressure held constant

For ethanol/water azeotrope composition, the pressure is (pre)chosen the only degree of freedom. By setting the pressure, you cannot choose the temperature. By setting the temperature, you cannot choose the pressure.

"The pressure is (pre)chosen the only degree of freedom". This is rather new to me, my teacher did not say anything about it. So in other words, if I have the azeotropic mixture of ethanol/water, changing pressure would not result in any changes in the system. Is that right?

Let us continue this discussion in chat.

What is the "1" degree of freedom represents? I cannot seem to find any proper explanation.

The degrees of freedom of a fluid is the number of intensive variables that you must fix in order to have a complete thermodynamic description of that fluid. If you have $F = 2$, then you need two intensive variables (such as $p$ and $T$), so that $G$, $S$, $A$, $U$, $V$, etc. have a defined value.
For azeotropy, the degrees of freedom can be inferred by writing down the equilibrium problem. However, since we have an azeotrope, an additional mathematical restriction must be included (I will name the components just $\ce{A}$ and $\ce{B}$)
\begin{align}
  \mu_A^\mathrm{vapor} (p,T,y_A) &= \mu_A^\mathrm{liquid} (p,T,x_A) \tag{1} \\
  \mu_B^\mathrm{vapor} (p,T,y_B) &= \mu_B^\mathrm{liquid} (p,T,x_B) \tag{2} \\
  y_A + y_B &= 1 \tag{3} \\
  x_A + x_B &= 1 \tag{4} \\
  x_A &= y_A \hspace{3 cm} \color{blue}{\text{(Azeotropy condition)}} \tag{5} \\
\end{align}
I count six unknowns ($p,T,y_A,y_B,x_A,x_B$) and five equations. Hence, $F = 1$. All the equations are well known to us, the $\color{blue}{\text{blue}}$ one is new. This agrees with your diagram. Once you set the pressure $p$, you can point at the azeotrope directly with your finger without specifying anything more.
You may think that $x_B = y_B$ is another piece of information that we must add, but it is not. You can check that this equation can be derived combining Eqs. (3-5), so it is linearly dependent.

Is there any exception to Gibbs phase rule?

I think that one exception is the critical point, but I have not obtained a definite answer although I have asked many people. The critical point is zero-variant ($F = 0$). If you change the critical pressure $p_\mathrm{c}$ or the critical temperature $T_\mathrm{c}$, then the critical state is gone. If we apply Gibbs' phase rule, this means that
\begin{align}
  F &= 2 - \text{Num. of phases} + \text{Num. of components} \\
  \text{Num. of phases} & = 2  + \text{Num. of components} - F \\
  \text{Num. of phases} & = 2  + 1 - 0 \rightarrow
  \boxed{\text{Num. of phases = 3}} \tag{6} \\
\end{align}
This is indeed what happens at the triple point, there are three phases in equilibrium, and thus $F = 0$.
However, when you grab any Physical Chemistry book, the description of the critical point contradicts $P = 3$ ($P$ from now on is number of phases). Generally, it is spoken as an 'impossibility to distinguish the gas and liquid phases', in which both merge. You can go to YouTube for some videos on this, and if you are asked to point with your finger where the liquid and vapour phase are, you won't be able to do so.
If the liquid and vapour merge together, then we may just say that $P = 1$, but $P$ should be three. However, we cannot say that $P = 3$ either, because we cannot distinguish three phases.
The question for me is open, but some tend to state that the Gibbs's phase rule breaks down in the critical point.

I wonder if the mixture is on the red line, what is the degree of freedom?

For solid-liquid phase diagrams, that red line has the same interpretation as the simple ones in liquid/vapour equilibrium. It has $F = 1$.

Is the F = 1 in this case different from my understanding? In the liquid-vapor diagram that I browsed above, saying that we have a point in the liquid region, then F = 2 implies that I could change both T and composition fraction in two dimensions without breaking the phase. While the F = 1 in the azeotrope meaning that we should choose the pressure, and cannot be interpreted by the diagram?

@Shira Hi! I am very sorry, but I couldn't understand what you meant. Could you rephrase?

Lets say, I heard that we can calculate the degree of freedom of azeotrope like this: F = C - P + 1 (P = constant). There are 2 components X and Y, 2 phases (liquid and vapor) and there is another constraint x(A) = y(A) which decrease further by 1.  Therefore F = 2 - 2 + 1 - 1 = 0. What do you think about this idea?

@Shira I understood. The conclusion is the same, but you must have in mind that "pressure is fixed". That is fine, so your formula now becomes $F = C - P + \color{red}{1}$, instead of the most common one which is $F = C - P + \color{red}{2}$. I would suggest you to use this last one. Your conclusion in words would be all right ("all I need to fix the state is pressure"), but the statement "the degrees of freddom are zero" is wrong. Just use $F = C - P + 2$.

In case F = C - P + 2 is used, how could we come to the result F = 1? I am still a bit unclear

@Shira Sure. The equation must be extended to add the restriction I mentioned in my answer. The Gibbs' rule will become $F = C - P + 2 - r$ where $r$ is the number of restrictions. If $r = 1$ (in the case of azeotropy, by the relation $x_A = y_A$), we have that $F = 2 - 2 + 2 - 1 = 1$.

Well, in my university, we are taught to use F = C - P + 1 where P is constant. I wonder if we use the common equation F = C - P + 2 - r. And by your previous comment, the constant value of P is not a constraint, is it not?

Let us continue this discussion in chat.


text:
How does Tollen’s reagent reacts with hydroxylamine?
How does the Tollens' reagent react with hydroxyl amines? For example, let's consider phenylhydroxylamine $\ce{Ph-NH-OH}$ reaction with Tollens' reagent. First it yields $\ce{Ph-N=O}$ and on further oxidation, $\ce{Ph-NO2}$
There’s no carbonyl bond for O radical generation. How does it proceeds then? I could not find its mechanism online.

Why would you need some carbonyl bond? N-O radicals are created very easily. Hydrogen, like, barely holds onto oxygen.


text:
Effects when 100 mmol NaHCO3 are mixed with 50 mmol HCl
I read somewhere that when $\pu{100 mmol}\ \ce{NaHCO3}$  are mixed with $\pu{50 mmol}\ \ce{HCl}$, a buffer arises. I'm confused, since $\ce{HCl}$ is a strong acid, and I thought the classic buffer is weak acid, weak conjugate base in equal amounts.
Is it the fact that the strong acid is added in only half the amount as the weak base that creates the buffer?
Or is it that the $\pu{50 mmol}\ \ce{HCl}$ will turn half of the $\pu{100 mmol}\ \ce{NaHCO3}$ into $\pu{50 mmol}\ \ce{H2CO3}$ which then with the remaining $\pu{50 mmol}\ \ce{NaHCO3}$ comprises a classic buffer?

I'd very much like a reason for the down vote? This is my first question here, so honestly don't know this question constitutes a "bad question".

Formatting guides for texts and formulas/equations/expressions.  // Downvoting is anonymous and some users do not provide feedback. // Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

Your second idea is right, the acid partially neutralizes bicarbonate.
$$\ce{HCO3-(aq) + H3O+(aq) -> H2CO3(aq) + H2O}$$
It is effectively the buffer  $\ce{HCO3-(aq)/CO2(aq)}$, as $\ce{H2CO3(aq)}$ is about just one percent of total dissolved $\ce{CO2}$.
The problem is such a buffer is unstable because there is not usually established the equilibrium between aerial and dissolved $\ce{CO2}$ and it therefore gradually escapes from the solution, with pH drifting to higher values.
$$\ce{H2CO3(aq) <=> CO2(aq) + H2O <=> CO2(g) + H2O}$$


text:
Improving paraffin wax adhesion to steel - surfactants and adhesion modifiers
Context
I use paraffin wax as a lubricant for my bicycle chain. It is a personal / DIY project. Essentially, wax is melted in a slow cooker and a bicycle chain is immersed in the melted wax.
There are commercial products where there are proprietary additives which are better performing. From what I understand, there are friction modifiers (WS2, PTFE etc) surfactants and adhesion modifiers.
I am already adding some very small amounts of WS2 and PTFE so I am happy with friction modifiers part.
Question
The commercial blends have small amounts of surfactants and adhesion additives. Can any of you suggest a few surfactants and adhesion additives which are available for non-commercial and non-bulk quantities?
Thank you!

Drop some high-grade motor oil in the paraffin mix and get all the needed extra ingredients.


text:
Aromaticity of chlorobenzene
Hückel's rule predicts that; a planar ring molecule will have aromatic properties if it has 4n+2 π electrons in conjugation.
If 4n π electrons are in cyclic conjugation, then it is anti-aromatic.
So, I was looking at resonance structures of Chlorobenzene and;

C—Cl bond acquires a partial double bond character due to resonance. As a result, the bond cleavage in haloarene is difficult than haloalkane and therefore, they are less reactive towards nucleophilic substitution reaction. - My Textbook

But there are 8π electrons (6 from the ring and 2 from the chlorine) that are in conjugation, then this should be anti-aromatic. So, it seems like one of these should be wrong. Which is it?
Is chlorobenzene aromatic?
If yes, does that mean the lone pair of chlorine is not in resonance? If no, does that make it unstable, since it will be anti-aromatic?

I wish they would finally stop teaching Hückel's rules as a kind of dogma for aromaticity. The fail more often than that tey are useful.

To apply the $4n+2/4n$ rule at all is wrong unless the conjugated $\pi$ electrons are precisely in a single ring.  In the first structure you drew that is true, and you can confidently count six $\pi$ electrons and call that contributing structure aromatic. But when you draw the additional structures that bring in the chlorine atom, then with the pendant $\pi$ bond to the chlorine you don't have a single conjugated ring anymore. So no more $4n+2/4n$ rule.
We still may have aromaticity in all the contributing structures, not just the first. But we can't tell from the $4n+2/4n$ rule except for that first structure.

Does that mean the chlorine doesn't undergo resonance since it breaking the resonance of the benzene ring? I'm particularly interested in C-Cl bond length. I tried searching in Wiki but I got no info.

No. The chlorine could still conjugate, but we would not be able to evaluate  aromaticity with thr 4n+2/4n rule if it does. The rule is really applicable only under narrow circumstances.

For this question i will suggest you to read about the L.C.C.S rule, which says to take the largest conjugated cyclic system. Here the lone pair of chlorine is not in the single cyclic system. You can only apply Huckel's rule for a single closed ring.
Also i would like to add that anti-aromatic molecules do not show resonance even if they have conjugation due to the presence of unpaired electrons in ABMO(Anti Bonding Molecular Orbit). You can observe it by drawing the Frost Circle of the molecule.