text:
Methanol fuel cell anode - why not oxygen instead of water?
The methanol fuel cell:
\begin{align}
\text{Overall}&:& \ce{CH3OH + 3/2 O2 &-> 2H2O + CO2}\\
\text{Cathode}&:& \ce{O2 + 4H^+ + 4e^- &-> 2H2O}\\
\text{Anode}  &:& \ce{CH3OH + H2O &-> CO2 + 6H^+ + 6e^-}
\end{align}
All the equations balance and still work in theory if the methanol reacts with oxygen at the anode instead of water:
$$\text{Anode}: \ce{CH3OH + 1/2 O2 -> CO2 + 4H^+ + 4e^-}$$
So why is the first anode equation (with water) the correct one? How would a student know to use water at the anode and not oxygen?

In case of the last reaction at anode, what would get reduced at the cathode?

As the net electrode reaction, there is not ongoing oxidation AND reduction at the same time.

What will be the equation if pure methanol without water is used at the anode ?

How do you know if reaction X takes place instead of reaction Y a priori? You don't.

The answer is rather simple. A student is not supposed to know beforehand. The anode and cathode reaction will depend on what is being fed to the anode and cathode compartments in a fuel cell. In the anode section, the only components which are injected there are methanol + water along. Oxidation has to occur at the anode-by definition. If water and methanol are present, under those circumtances, methanol gets oxidized to carbon dioxide.

And what happens if pure methanol, without water, is used at the anode ?

Who will provide protons for the proton exchange membranes if water is not there? MeOH cannot provide protons.

So what happens if pure methanol, without water, is used at the anode ? Nothing ? No production of electricity ?

The success of a galvanic cell is dependent on the rate of electrode reactions. I think water helps there too.

Why don't you answer my question ? I'll repeat it for the fourth time. What happens if pure methanol, without water, is used at the anode ? Will it be a non-cell ? Will it be some electricity created ? Is water essential ?

The answer is right there in the comments. You need protons for the membranes to work and provide low resistance. High resistance means low voltage. I do not know what happens when pure MeOH is used.

Maurice  I have answered your question please read my answer above

The student is supposed to use the chemical principles they've learned and come up with reasonable answers.  A [redox, really any] reaction separated into its oxidation and and reduction moieties can possibly be used in a cell or fuel cell. The question was to separate the oxidation and reduction half reactions.
Overall: CH3OH + 3/2O2 = CO2 + 2H2O nicely balanced!
Reduction is easy: 3/2O2 + 6e- = 3O= This is the cathode and a source of H+ is needed to neutralize the oxide ion the logical source is H2O altho I don't know if a weak acid could be used.
The cathode reaction:  3/2O2 + 3H2O + 6e- = 6OH-
The oxidation reaction is a bit trickier; I will break it into simple steps!
CH3OH = C+4 + 4H+ + O= +6e-; what a mess! = CO+2 + 4H+ + 6e- The problem is now evident! There are just not enough O= ions to go around.  A source of O= ions is needed. H2O is a source of oxide ions; O2 is NOT a source of oxide ions. The anode reaction is: CH3OH + H2O = CO2 + 6H+ + 6e-. [The next project is to get the H+ and OH- ions together and possibly capture that energy.]
Please don't get all literal and object that O= and C+4 ions don't exist. The trick is to make the reactions as electrical as possible and keep the oxidation and reduction separate, then try to work out the chemistry; thinking about it outside the box might even elicit some fruitful ideas.

BTW It took me about 2 hours to come up with this. It can be a slow process. My point is to be patient and to work things out until they make sense.


text:
How does concentration of the electrolyte changing over time affect the rate of electrolysis in sodium sulphate solution?
During the electrolysis of sodium sulphate solution, hydrogen is produced at the cathode and oxygen (from the hydroxide ion) is produced at the anode. Then I believe that the rate of electrolysis is directly proportional to current and current is proportional to the concentration of sodium sulphate solution.
So would I be right in saying that the rate of electrolysis would increase as time passes since the volume of water depletes; increasing the concentration of the solution over time?

The rate of electrolysis can be measured by the amount of gasses evolved, or by the current passing thru the cell.
As electrolysis proceeds and conditions change, the rate might very well change. After all, you are correct in implying that the conductivity of a more concentrated solution of sodium sulfate will be greater than the conductivity of a less concentrated solution.
Conductivity is one factor in determining the rate of electrolysis. Another factor is the applied voltage. If the voltage remains the same and the conductivity decreases, the current will increase, and the rate of gas production will increase as well.
However, if the cell is connected to a constant current power source, decreasing the conductivity will decrease the voltage measured across the cell, but the current (and therefore the rate of electrolysis) will remain constant.

Most of that makes sense but I was just wondering why current would increase when conductivity decreased because I thought they were proportional.

To put it very briefly (too brief for an answer!): I=VxC. You must consider all 3 items in the equation.

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sort: votes


text:
Systematic and preferred names of acetamide
My book says to use carboxamide when it is a carboxylic acid derivative (acyl). Ethanamide $\ce{CH3CONH2}$ is a carboxylic acid derivative, so shouldn’t it be named ethan-1-carboxamide rather than ethanamide?

I am confused with this nomenclature and use of amide, carboxamide and imide. When should I use carboxamide and ethanamide?

Preferred IUPAC name:
Acetamide //
Systematic IUPAC name:
Ethanamide//
Other names
Acetic acid amide
Acetylamine
https://en.wikipedia.org/wiki/Acetamide

The first thing to do when constructing a systematic name is to find the parent structure - be it a cycle, or in this case an acyclic chain.
You choose the parent structure with the most skeletal carbon atoms - that is the longest chain.  Here the longest chain in 2 carbon atoms so we start with a parent chain of ethane.
Ethane been substituted with -$\ce{NH2}$ and $\ce{O}$ which describes an amide group.  So combining the amide suffix and the ethane parent, we get ethanamide.
NB: carboxyamide group
A carboxyamide group includes the carbon atom, so if you incorrectlydeclared the $\ce{CH3}$ group as the parent methane, you could say it is substituted with a carboxyamide group and would be called methanecarboxamide.
An example of the correct use of carboxyamide is cyclohexanecarboxyamide (pictured below).  Here the cyclohexane ring system is the parent and the can be described as carboyamide substituted.

The minutiae of systematic naming is described in the IUPAC Blue Book.  Here the relevant rule is P-44.3.2

This answer does not explain why the preferred IUPAC name is acetamide (not ethanamide) and how such names are formed. Also, the parent structure of an amide is not always the longest chain. The name of the second example is cyclohexanecarboxamide (not cyclohexanecarboxyamide). Rule P-44.3.2 is not relevant in this case; the relevant rules are in Subsection P-66.1.

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How to find the empirical formula of a compound containing C, H, O?
Problem: A 5.000 g sample of an organic compound containing C, H, and O atoms was combusted and produced 14.26 g CO2 and 3.335 g H2O. Its molar mass is 108.1 g/mol.
 Solution:
      Nc = 14.26 g/44.01 g/mol
         = 0.3240 mol C
      m  = Nc (12.01)
         = 3.8912 g
      Nh = 3.335 g/18.02 g/mol
         = 0.1851 mol H
      m  = Nh (1.008)
         = 0.1866 g
      Mo = 5.000 g - 3.891 g - 0.1866 g
         = 0.9224 g
      No = 0.9224 g/16.00 g/mol
         =0.05765 mol O

  C = 0.3240 mol C/0.05765
    = 5.6 (5)
    = 28
  H = 0.1851 mol H/0.05765
    = 3.2 (5)
    = 16
  O = 0.05765 mol O/0.5765
    = 1 (5)
    = 5

Comment: In finding the empirical formula of the compound, I keep getting C28H16O5 which has a molar mass of approx. 400+ that is not equal to the given molar mass above. I followed the process of finding the empirical formula, but I'm getting the same result. I don't know what else to do. Oh, and I also considered to get the mass percent composition first, but also the same result. I'm guessing that the formula could be either C7H8O (with 108.1 g/mol), I forgot the other one. What should I do?

1 mol water contains 2 mol H .Your approach looks otherwise correct.

Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. It helps in focusing on principles, spotting mistakes and improves orientation, Q/A reusability and the permanent value.
You may find useful  formatting mathematical/chemical expressions/formulas.

C7H8 and C28H16 are quite not matching ratios, considering their ratio is given by CO2/H2O ratio.

Your sample contains $0.324$ mol $\ce{C}$ (weighing $3.888$ g) and $0.371$ mole $\ce{H}$. Total mass : $4.259$ g. Out of $5.000$ g, the sample contains still $5.000 - 4.259 = 0.741$ g $\ce{O}$, or $0.741/ 16 = 0.0463$ mole $\ce{O}$. Compared to the $\ce{O}$ atoms, there is nearly $7$ $\ce{C}$ atoms and $8$ $\ce{H}$ atoms. This is $\ce{C7H8O}$.


text:
Acidic Strength of Benzoic Acid and Propanoic Acid
I want to know that why Benzoic Acid is more acidic than Propanoic acid. It should have to be less acidic, if we will observe the stability of conjugate bases. Because if you will look to the stability of Benzoate ion, then it is highly unstable : (See the image attached below)

Now, you can also see that Phenyl ring is increasing the Electron density and the Aromaticity of the ring is also hampered. So, it will now become highly unstable, and due to this the rate of formation of benzoate ion will be very less. And in case of Propanoic Acid, there is no +M effect.  So, there will be relatively high stability, only the +H effect of Ethyl group will increase the electron density.

Please check it, and tell me where I am neglecting any other factor. Because if we look at the pKa value then, we will observe that Benzoic Acid is stronger than Propanoic Acid. Can anyone explain me why ? Please !!

https://chemistry.stackexchange.com/questions/7591/comparing-acidities-of-substituted-and-aromatic-carboxylic-acids?rq=1

In the last canonical structure you show , the aromaticity is even though hampered , but it don't makes it highly unstable in any sense, as the ion is still non-aromatic instead of anti-aromatic.
By drawing all canonical forms of both acids, you will see that benzoate ion is more stable as it's a hybrid of more number of contributions than propionic acid.
Also , note that the inductive effect also destabilises the propanoate ion .


text:
Products formed on the electrolysis of wet HF
We know that, electroysis of water produces $\ce{O2}$ at anode and $\ce{H2}$ at cathode. In a similar way I thought that on electroysis of $\ce{HF}$ it would produce $\ce{F2}$ and $\ce{H2}$ at their respective electrodes. If this is true then fluorine could be prepared by electroysis of $\ce{HF}$. So to verify this I was searching on internet. Then I found an article on synthesis of $\ce{OF2}$.  It was interesting that on electroysis of wet $\ce{HF}$ it produces a mixture of $\ce{OF2,O2,O3}$ at the anode rather than simply producing $\ce{O2}$. Then I searched on internet for the mechanism but I couldn't find it.  So I need an appropriate mechanism for the electroysis of $\ce{HF}$. Thanks in advance.

You have not included into consideration respective electrode redox potentials nor the fact fluorine reacts with water.

Sorry . Now I have edited question again.


text:
Nitrogen dioxide vs Nitrogen peroxide: are they one and the same?
Are Nitogen Dioxide (NO2) & Nitrogen Peroxide (NO2) one & the same?

Welcome to the chemistry site of the StackExchange (SE) network.  Be aware that laconic, not elaborated questions are usually closed on the SE.
The site expects that you include the compact summary of your related current knowledge, involving your conclusion of searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself. Effort not shown may be considered as effort not done, possibly leading to the question closure. How do I ask a good question?.

No, they are not the same. A compound is called peroxide when they contains the peroxide bond ($\ce{-O-O -}$). However, in this context, nitrogen peroxide can be considered an equilibrium mixture of nitrogen dioxide and dinitrogen tetroxide in varying proportions. The first mention of this "compound" was in this 1944 paper, but not sure how accurate the information is in here considering an erratum has been published for this paper.

Yes I agree that a peroxide compund must have a peroxide bond (--O--O--) in it, but neither NO2 nor N2O4 has any. How it can still be called a Peroxide? Probably a dimer of N2O4 has --O--O-- bond in it.

@JagbirSinghKanet As I said, nitrogen peroxide is not an official term. It can be considered the mixture of NO2 and N2O4 but it can add to more confusion as nitrogen peroxide in itself doesn't exist. Also, N2O4 doesn't have a dimer.


text:
Connect glass tubes to a graham condenser
I am currently making a syphon coffee maker such as this one:

However, I find that all the silicon tubing used on this image between the condenser and the boiling flask / between the condenser and the Buchner funnel kind of ugly. I thought this could be solved by using a condenser (a Graham Condenser specifically for my use) with ground joints (24/40 to be precise). I found this one:

But the point of the syphon coffee maker is having a tube coming from the condenser to the bottom of the flask, the water being supposed to go up that glass tube. I therefore wonder if it is possible to attach a tube to the bottom of the Graham Condenser, going down the flask, to replicate the image.
Same thing for the top of the Graham Condenser, I would like a glass tube to be connected to the Buchner funnel.
Sorry if the information provided is not precise enough, please feel free to ask for some more details.

You're not going to get a siphon from this setup. Have a look here for a pretty cool example of what you are building, which is a drip-type rig.


text:
Electrochemistry: Conductivity Vs Molar Conductivity
Why does the molar conductivity (Λₘ) of an electrolyte increase while its conductivity (κ) decrease with a decrease in concentration (or with an increase in dilution)? It's really counterintuitive. Don't they both imply somthing similar? If yes, how can they show completely opposite behaviour in this case?
As per Wikipedia, Λₘ increases due to a decrease in solute–solute interaction. I would like to have an elaborative explanation for this.
A decrease in κ (in case of a strong electrolyte) upon dilution can be clearly understood by the fact that there is a decrease in number of ions per unit volume. But the fact that there is an increase in Λₘ upon dilution is just too counterintuitive for me and thus I can't shove this fact down my throat.
A lesser math ridden and a more mechanism based answer would be highly appreciated.

@Poutnik Yeah, I can get it, but kindly explain how conductivity and molar conductivity can manifest such contrasting results provided they both mean something very similar, i.e. ability of an electrolyte to conduct electricity.

So, as per your analogy x:sinx:[(sinx)/x]::c:κ:Λₘ , isn't it? well, do sinx and [(sinx)/x] mean something similar?

This analogy, no doubt, fits well mathematically but fails to address the idea that despite κ and Λₘ mean something similar, they show opposite variation upon dilution.

Yeah, I think this party baloon analogy/example is giving me a better idea. I just couldn't think that way. Thanks a lot :)

Comments deleted and converted to the answer.

Ion electromobility decreases with concentration due several effects slowing them down. Major ones are these:

The close ion neigbourhood has statistically slightly nonzero net charge of the opposite sign. When these opposite ions move in the opposite directions than "our" ion, there is nonzero net electrostatic force acting against the ion motion.
At higher concentrations ions has (weaker or stronger) tendency to form hydrated ion pairs, that are neutral and do not participate in conductivity.

As the consequence, conductivity curve, being initially proportional to concentration, bends to increase progressively slower with increasing concentration.
If concentration increases 10 times, conductivity does not increase 10 times, but (illustratively) just 8 times. As molar conductivity is conductivity divided by concentration, it is 0.8 times smaller than for the diluted solution.
As analogy, if you are compressing slowly leaking party balloon, the gas mass decreases, but the gas specific mass (density) increases, while both meaning something similar.
Another analogy at buying cables, imagine the price per meter decreases with a bought amount. So while the specific price = price/length decreases with the length, absolute price increases with length.

For low concentrations, by the Kohlrausch equation, the dependence of conductivity and molar conductivity  on molar concentration is approximately:
$$\kappa=\Lambda.c=\Lambda_0.c - a.c^{(\frac{3}{2})}$$
$$\Lambda=\Lambda_0 - a \sqrt{c}$$


text:
What effect would decreasing melt viscosity by increasing melt temperature have on filler disaggregation in an extruder/compounder?
PLA has a melting point of around 150-160 degrees Celcius (the grade I am working with). I have microcellulosic fillers with aspect ratios that I am introducing to the polymer for compounding in a Brabender single screw compounder with an L/D of 25:1 and two mixing zones in the screw. Due to a low affinity between the hydrophobic matrix and the hydrophilic filler phase, there's an aggregation of microcellulosic fillers. The extrudate is in the form of a 1.5-2 mm diameter filament. Microstructure analysis shows that some degree of aggregation persists even after I run the chipped composite filament a second time through the melt extruder/compounder.

I am wondering what would be the effect of reducing the melt viscosity by increasing the processing temperature to something like 230 degrees Celcius on the filler aggregation? Please do note that this temperature is within the safe range where neither the PLA nor microcellulosic filler I am using degrades.

Should I instead try to process the extrudate at a lower processing temperature which would not decrease the melt viscosity and hence subject the microcellulosic aggregates to higher shear stresses in the extruder, thereby facilitating filler disaggregation?


text:
Why is the value of $A$, in the expression for Boltzmann's distribution of speeds, not directly calculated?
I've read the proof for the expression for Boltzmann's distribution of speeds on this site. There, $f(v)$ is first factorized into $f(v_x)$, $f(v_y)$ and $f(v_z)$ and $A$ into $A_x$, $A_y$ and $A_z$. Then, to find the value of $A$,  $A_x$,  $A_y$ and $A_z$ are each calculated as:
$$A_x=A_y=A_z=({{\frac {m}{2\pi kT}}})^{1/2}$$ Thus the value of $A$ is found to be:
$$A=A_xA_yA_z=({\frac {m}{2\pi kT}})^{3/2}$$
My question is this: can A be not directly calculated from the equation $f(v)=Ae^{-{\epsilon}/{kT}}$, like $A_x$, $A_y$ and $A_z$ were?
I can't seem to find an answer to this online. But if what I just mentioned is in fact possible, please also mention the limits over which the term $e^{-{\frac {1}{2}m{v}^2}/{kT}}$ has to be integrated.

I've found that 'a normalizing constant is a constant by which an everywhere non-negative function must be multiplied so the area under its graph is 1.' This is what you mean when you say normalization constant right? Also, how do I find the normalization constant in the case of $f(v)$, if this is what you mean?

Thank you @Poutnik. But I don't think I can understand (I'm in 10th grade). So, could you please explain your latest comment a bit more? I understood what you said, until getting the symbolic value of the integral. But I couldn't understand much after that.

Doing it three dimensions the integral would instead be a triple integral dxdydz that you'd split up into three independent integrals identical to the one you solved for the 1D case, so it's the same math

After formulation of the answer, now redundant comments have been deleted.

It can be calculated directly, but the probability of a given speed must be integrated over the whole space of velocity vector components.
We integrate the Boltzmann distribution values
$$\exp{(-\frac {E}{kT})}=\exp{(-\frac {mv^2}{2kT})}$$
in the integral over the 3D space of velocity components $v_\mathrm{x}$, $v_\mathrm{y}$, $v_\mathrm{y}$, using the radial spherical coordinate $v \equiv r$, where $v=\sqrt{v_\mathrm{x}^2+v_\mathrm{y}^2+
v_\mathrm{z}^2}$:
$$C\int_0^{\infty}{4\pi v^2\exp{(\frac{-mv^2}{2kT})}\mathrm{d}v}=1$$
as for spherical coordinates, $\mathrm{d}V=4\pi r^2 \mathrm{d}r$
The value of the above finite integral:
$$\left(\dfrac{2\pi kT}{m}\right)^{3/2}$$
is the reciprocal value of the normalisation constant:
$$C=\left(\dfrac{m}{2\pi kT}\right)^{\frac 32}$$
which sets the probability that a molecule has some speed to certainty, as  $\int_0^{\infty}{p(v)\mathrm{d}v}=1$
We have now finally the complete normalized probability density function $p(v)$ for the speed $v$, known as the Maxwell-Boltzmann distribution:
$$p(v)=\left(\dfrac{m}{2\pi kT}\right)^{\frac 32}4\pi v^2\exp{(-\frac{mv^2}{2kT})}$$

The finite integral value can be conveniently obtained by putting the integrated function 4*pi*v^2*exp((-mv^2)/(2kT)) to the online symbolic integrator with limits 0 and $+\infty$.


text:
Dissolving NaOtBu in toluene
I am conducting a synthesis using Buchwald-Hartwig coupling, and many reaction conditions use NaOtBu and toluene. Also the internet says that the solubility of NaOtBu in toluene is 130 g/L at 20 ℃. I only use 42 mg of NaOtBu and 5 mL of toluene, but toluene doesn't readily dissolve NaOtBu. Due to this, NaOtBu powder sticks to the wall of reaction flask as I start stirring. And these powders that sticked to the wall does not dissolve in toluene again. As a trial, I weighed same amount of NaOtBu in vial, and added 5 mL of toluene. I stirred it for 2 hours, but still there are some powders that stick to the wall, and the toluene seems like suspension, not clear solution. Is this a natural phenomenon? In this situation, can I say that I added NaOtBu properly in the toluene? Currently, I weigh all the reagents in the solid powder form and put them into the reaction flask, then add anhydrous toluene by syringe.

How are you storing your NaOtBu? Is it kept in a desicattor? It will pick up water readily giving NaOH and tBuOH

@Waylander Yeah, I also blow Ar gas before sealing the NaOtBu. When I searched for it NaOtBu readily reacts with water, and in fact it released gas when I dropped water. So I think that the storage is not the problem.

How well does the reaction work using the NaOtBu as you have done?

Note the solubility of NaOtBu in toluene is 2.27g/100g which equates to about 19.5g/L so your 5ml of Toluene should still dissolve 42mg of NaOtBu

@Waylander I am still testing whether the reaction goes on varying the reaction conditions. As far as I know dissolution of NaOH is exothermic, but gas evolution is not observed prominently, therefore I concluded NaOtBu was not converted to NaOH during storage.. But undissolving NaOtBu is somewhat strange.

Well, that it's soluble doesn't mean it has to easily dissolve. Maybe heat it.

@Mithoron I am conducting my synthesis at 80 celsius degree, but refluxed toluene does not wash away the powders that sticked to the wall..

When running these reactions I never paid much attention to whether the NaOtBu went into solution, the reactions worked whatever.

Have you tried making your own reagent from sodium and HOtBu in refluxing toluene to assure that the sodium is liquid?

@Waylander Another person who have conducted this reaction said so. Maybe that really doesn't matter significantly. Thank you.

@user55119 I have not tried that, but dealing with metal sodium seems quite difficult.. I'll consider it.

Handling metallic sodium is not a problem. Cut it with a knife under dry, inert solvent such as pentane or hexane.. Handle pieces with tweezers. Weigh in an inert solvent.


text:
Ecell value for non spontaneous reaction
I've done my electroplating experiment but got stuck to calculate my E cell value.
So my equations are like these coz I used copper stripes and copper sulphate solution. Some website says E cell can't be 0 but it has to be -0.68V as it is -0.34-(+0.34)V which kinda make sense because it is non spontaneous reaction. But other websites are like it should be 0 because there is only copper reacting. Most of my classmates are using 0V as well so I don't know what to do. Thanks

This is kinda missing the point - only potential here is forced by outside source, so yeah, you could say it's zero - it's probably not completely true - that would be idealised situation, but it's not like you get much current if you just stick two ends of copper wire into water.

It is kind of concentration cell with potential difference $\Delta E = \frac {0.059}{2}\log{\left(\dfrac{c(\ce{Cu^2+},c)}{c(\ce{Cu^2+},\mathrm{a})}\right)}$

The redox potential of all electrodes depend on the concentration of the ions in solution. For Copper, it is $+0.34$ V when and only when the concentration of $\ce{Cu^{2+}}$ is equal to $1.000 M$. During the electrolysis, the concentration in contact with the electrodes is different from what it is in the bulk solution. This is why you should measure a voltage which is  $> 0.00$ V.

It doesn't matter. He's effectively not conducting any reaction. Delta G is zero and so is potential.

@Mithoron. It does matter. The author states that he was conducting a reaction. He says that I've done my electroplating experiment. So it may happen that the copper concentration against the anode is higher than its value in the bulk solution. And it is lower at the cathode. And Poutnik is right : it is similar to a concentration cell

Do you know how electroplating works? It's driven by external current source - a battery, or something else, is creating the actual potential difference. Some minuscule conc. issues can only cloud the picture here.

@Mithoron. Of course I know how electroplating  works, and I agree about your interpretation.


text:
Is there a recommended frequency when driving an alkaline water electrolyzer?
We are designing a small electrolyzer (about 1 kW input power) for HHO gas generation.
Sadly most of the literature online is bordering on the snake oil… However, it works fine with DC driving. Now from the electrical side the cell is a quite complex model (the current–voltage curve depends on anything under the sun) and we want to give a way to regulate the amount of gas produced. The idea would be some kind of PWM.
I found an article by Mazloomi et al. [1] (there are some thing inside I'm not very convinced of, first of all the circuit model) but the current transient matches what we see:

An initial peak (the electrolyzer plates charging as capacitor?)
An exponential rise (the actual electrochemistry starting up?)

My guess it that the minimum frequency for a PWM drive would be such that at least the current transient completes. But I also read that the gas formation processes are somewhat slow, due to bubble mechanics.
The question is: is there a recommended frequency for driving such an electrolyzer cell? I am thinking about three possible outcomes:

Ultra low frequency, like turning on 10 s the generator and let it idle 2s;
Low frequency, around 100 to 1000 Hz, where the transient completes (electrically) but I don't have a reasonably good idea of what's happening on the chemistry side;
High frequency (kHz range) where the electrical inductance in practice becomes a (lower) constant current and the generator is substantially at a DC steady state.

The first way is easy to do but the discontinuity in generation is problematic; the third way gives substantial thermal stress on the electronics due to switching losses. The middle way seems a good compromise to me. Are there some guidelines? Any kind of pointer can be useful.
Reference

Mazloomi, K.; Sulaiman, N.; Ahmad, S.; Md Yunus, N. Analysis of the Frequency Response of a Water Electrolysis Cell. Int. J. Electrochem. Sci 2013, 8. (PDF)

Why do you want to produce electrolysis with pulses ? is there any advantage ?

it's for controlling the amount of gas produced, the supply is not completely controlled and the application call for different flow rates

Sorry. I do not understand a word of your explanation.

It's brown gas, just electrolyze water and mix the outputs. Huge safety concerns for the obvious reasons. Anyway it seems there is not much around so I guess it's time for practical tests!

What ? A brown gas ? Which brown gas ? $\ce{NO2}$  ? What are you doing ? What is the composition of the solution to be electrolyzed ? I thought you wanted to produce hydrogen and oxygen !

@Maurice Not "brown gas", but "Brown's gas" aka Oxyhydrogen


text:
Is the formation of nitrogen dioxide spontaneous?
I was studying Thermodynamics from my Chemistry textbook (page 181-182, section 6.6 (a)).
I have a question regarding the following text from my book.

Now let us examine the following reactions:
$$\begin{align}
\ce{1/2 N2(g) + O2(g) &-> NO2(g)} & \Delta_\mathrm{r}H^⦵ &= \pu{+33.2 kJ mol-1} \\
\ce{C (graphite, s) + 2 S(l) &-> CS2(l)} & \Delta_\mathrm{r}H^⦵ &= \pu{+128.5 kJ mol-1}
\end{align}$$These reactions though endothermic, are
spontaneous.

I vistied the Wikipedia page on Standard Gibbs free energy of formation to check the values of $\Delta_\mathrm{f} G^⦵$ for $\ce{NO2}$ and $\ce{CS2}$ and verify if the above stated reactions were truly spontaneous. It turns out that $\Delta_\mathrm{f} G^⦵$ for $\ce{NO2}$ is $\pu{51.3 kJ mol-1}$; and $\Delta_\mathrm{f} G^⦵$ for $\ce{CS2}$ is $\pu{67.1 kJ mol-1}$.
I understand that as the change in entropy involved in the formation of $\ce{CS2}$ is positive, this reaction can be spontaneous at high temperatures even though it is not at $298 \pu K$.
This is not the case with the formation of $\ce{NO2}$ though. The change in entropy associated with its formation is negative and so $\Delta_\mathrm{f} G^⦵$ will always be positive. This means that this reaction will never be spontaneous.
Then why does my textbook state so?
Is that statement (bold text) wrongly stated or am I understanding it wrong?

If $\Delta G°$ is positive, the reaction is not spontaneous. If $\Delta G°$ is negative, the reaction is spontaneous, but it can be rather slow, so slow that it does not appear to happen even for years. Your inverse reactions, namely the decompositions of $\ce{NO2}$ or of $\ce{CS2}$ are good examples of this extremely slow behavior.

Hi, please note my edits, particularly the MathJax usage. \ce{...} is supposed to be used only for chemical formulae and equations. $\Delta G$ isn't a chemical formula or equation, you shouldn't use it there, as it leads to the $G$ being incorrectly written upright. Then, your physical quantities with units (like x kJ/mol) are much better typeset using $\pu{x kJ mol-1}$ rather than a mixture of \text and \ce. You may find https://chemistry.meta.stackexchange.com/q/86/16683 useful as a guide.

Note that $\Delta G_\mathrm{f}(T)=\Delta H_\mathrm{f}(T)-T\Delta S_\mathrm{f}(T)$ and that for  $\Delta  S_\mathrm{f}(T) \gt 0$ and high enough T, those reactions may become spontaneous in spite of being endothermic.

@Silica19 Compare the respective $\Delta G_\mathrm{f}^°$ and $\Delta H_\mathrm{f}^°$ values to get the sign of $\Delta S_\mathrm{f}^°$. Additionally, for high temperatures, the standard values are not that important, but  $\Delta H_\mathrm{f}(T)$ and $\Delta S_\mathrm{f}(T)$, using integration of $\int{\Delta C \mathrm{d}T}$ resp. $\int{\frac{\Delta C}{T} \mathrm{d}T}$

@Poutnik I've added details about sign of entropy in my question accordingly.

Note that the thermodynamic spontaneity of compound formation is based on the $\Delta G_\mathrm{f} = \Delta G_\mathrm{f}^° + RT \ln{Q}=RT \ln{(Q/K)}$, where Q is the reaction quotient and K is the equilibrium constant of the formation.


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Computing the overall charge of a compound with non-integer coefficients
I have a question about computing the overall charge of inorganic chemicals, where the composition is expressed using non-integer stoichiometric coefficients.
Let's say we have $\ce{Hg_{0.7}Cd_{0.3}Te}$, and let $\text{ON}(\ce{X})$ be the oxidation number of element $\ce{X}$.
Then the approach I would follow is, computing a weighted average of oxidation numbers of the elements in the composition, weighted by the fractional prevalence of the given element. I know that there is often more than a single oxidation state but I'm not aware of rules according to I should consider one instead of the other. So what I would do is
$$0.7 \cdot \text{ON}(\ce{Hg}) + 0.3 \cdot \text{ON}(\ce{Cd}) + 1\cdot \text{ON}(\ce{Te})$$
and I'd expect the above computation to return $0$ in the case of a charge-neutral composition.
Now, is this approach correct? Maybe I should avoid considering the fractional prevalence and only consider the oxidation numbers?

Here, $\ce{Hg}$ and $\ce{Cd}$ are both at oxidation number $+2$. $\ce{Te}$ is at oxidation number $-2$.

oh got it, actually doing the computation in this particular case it exactly returns 0... so the fractional prevalence of the element must indeed be considered...?

@maurice This might be true for this specific compound if you know the compound, but the result does not generalise. If the question is specifically about this compound and nothing else, then fine. But that isn't how it is worded and, as such, your answer is likely to be very misleading in general.

My reasoning is valid if the coefficients of $\ce{Hg}$ and $\ce{Cd}$ are such that their sum is equal to the coefficient of $\ce{Te}$. It is the case here.

@matt_black so the way I've depicted in my question does not hold for a general composition?

In many compounds a given element may occupy different lattice sites with different nominal charges. To further complicate things, often vacancies are charged (with different sites having different charges ).

@JamesArten Not really. SAE 304 stainless steel is (approximately) Fe77Ni10Cr20Mn2Si1 and thinking about the oxidation state of the components is entirely unhelpful as it would be for most metal alloys. The major place where net charge might be useful is describing many silicate minerals where specific structures are known to have specific charges. But the formulae for those are already well-known. It is impossible to generalise across multiple compound classes.

@maurice But you are importing specific knowledge of this compound that does not generalise. It might be true here, but it could easily be true by accident in some metal alloy without the implications you assume.

Rather the opposite direction: There is given formal redistribution of charges in context of the definition of the oxidation number, with the zero net charge. Respective elements have formally such effective oxidation numbers for the charge inventory to match the zero net charge. So the charge neutrality is primary, ONs secondary.

James Arten never speaks of alloys. He only speaks of compounds.

Compounds with fractional coefficients are in fact solid solutions, which is pretty much the same as alloys.

@IvanNeretin So diamond is just solid solution and alloy of carbon-12 and carbon-13.

@Poutnik In a way, yes.

@IvanNeretin Downscaled to the usual macromolecular scale, macromolecules are just solid solutions and alloys of respective atoms, that just happen to be in ratio of relatively small integers.  :-)

@Ivan Neretin. Solid solutions are not similar to alloys. An alloy can have any sort of coefficients. It is not the case for a solid solution. In a solid solution the total of negative charges must be equal to the total of positive charges. This is the case for James Arten's compound $\ce{Hg_{0.7}Cd_{0.3}Te}$. And I repeat : it is a compound, not an alloy.


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Standard Gibbs free energy when all the reactants and products are at standard condition
I have read in my textbook that for a reaction $\Delta G=\Delta G^o-RT\ln Q $, where $\Delta G^o$ is the Gibbs free energy change when the initial concentration of products and reactants are unity. But when I searched for the definition of $\Delta G^o$ and it is only mentioned that the reactants and products must be in their standard states, it doesn't mention that initial concentration of products and reactants should be  unity.
So if we carry out a reaction (with no gaseous reactants or product) at standard condition, then shouldn't $\Delta G=\Delta G^o $ instead of $\Delta G=\Delta G^o -RT\ln Q$  ?

See https://chemistry.stackexchange.com/a/139640

Standard states are: activities equal one for pure solids or liquids, fugacities equal one for gases and concentrations equal one Molar for solutions. So in the standard states Q equals one and lnQ is zero and DeltaG = DeltaG[0].  At equilibrium Delta G=0 so DeltaG[0] = RTlnKeq.  This is almost what you are stating, think it through. The standard free energy change is the free energy difference from reactants and products both in their standard states  to the equilibrium concentrations.

First note, I believe you mean this equation: $$\Delta G= \Delta G^\circ + RT \ln{Q}$$
The Gibbs free energy for reactions in their standard state is defined relative to their stoichiometric relations (what I believe you mean by "unity" here.) So for a sample reaction, $$2A + B \ce{<=>} 3C$$ we would define the standard Gibbs free energy as the free energy (non-expansion work, specifically) associated with one mole of this reaction happening. One mole of this reaction consumes two moles of $A$, one mole of $B$, and produces three moles of $C$. You can assume this reaction occurs in one liter of solvent in order to make the generalization to concentration easier.
The standard Gibbs free energy $\Delta G^\circ$ is associated to how this system comes to equilibrium; you can express the standard Gibbs free energy for a reversible reaction the following way: $$\Delta G^\circ = -RT \ln{K_{eq}} $$
This equation carries the implicit statement that our standard Gibbs free energy is only suitable when the system has accomplished equilibrium. We often run into systems that have not reached equilibrium yet—if we want to quantify Gibbs free energy for these systems too, then we will need to transform our equation a bit.
Here is where it is helpful to think of equilibrium as a sort of "home base." The equilibrium of a reaction is where a reaction will naturally tend if perturbed in some way. Adding more product can be a perturbation of the system, adding more reagent perturbs the reaction, et cetera. These perturbations take the system away from equilibrium, much like stretching or compressing a spring, and then over time they relax back to the equilibrium concentrations. We represent the perturbed system with $Q$. Any time you see $Q$, you know that it is a general statement about your chemical system not necessarily in equilibrium. If your system is already at equilibrium, then there is no change in the Gibbs free energy ($\Delta G=0$) which you can confirm for yourself by combining the previous equations listed with your $Q$ set equal to $K_{eq}$. If your system is a little bit away from equilibrium, then as it reestablishes equilibrium, it will do some work in some "direction" (chemically speaking.)


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For the same substance, which are stronger, the bonds in the solid or the bonds in the liquid?
As a teacher, I show students the particle model as an explanation of solids, liquids and gases, and changes of state. I did this today, explaining that the bonds in the solid break because the kinetic energy of the particles has increased so much that it overcomes the bonds between the particles. The bonds between the molecules in the liquid state are weaker. But... if this is true, then how come the level of kinetic energy that broke apart the solid bonds, is not also able to break apart the liquid bonds? Are the liquid bonds in fact stronger?

The situation is murky: at 30 °C, are the bonds in solid 1,4-dichlorobenzene mothballs, with high vapor pressure and boiling point of 174 °C, stronger than those of liquid gallium, with exceedingly low v. p. and b.p. of 2,400 °C?

@DrMoishePippik   ..... for the same substance ....

The bonds between the molecules in the liquid state are weaker.

The bonds in the solid are the same kind of bond as in a liquid. We know the liquid still has bonds between particles because liquids have a constant volume. It would be better to say that the solid turns into a liquid because there are fewer bonds, or bonds break continuously (with the current neighbor) and form continuously (with the new neighbor). In solid and liquid water, we have hydrogen bonds between molecules. In solid and liquid sodium chloride, we have ionic bonds between ions. In solid and liquid iron, we have metallic bonds between atoms.
Here is a simulation of ice melting. Notice the change in density as you pass zero degrees celsius, water molecules starting to rotate as the temperature increases, and rare switches in position (you would have to play this faster to see diffusion of water molecules in the liquid).

Source: https://phet.colorado.edu/sims/html/states-of-matter-basics/latest/states-of-matter-basics_en.html

[...] how come the level of kinetic energy that broke apart the solid bonds, is not also able to break apart the liquid bonds?

If I want to buy a house, and then buy furniture for it, the house will be more expensive than the furniture. However, if I have just enough money to buy the house, I can't buy the furniture until I get more money because I already spent it on the house.
So at the melting point, we put in sufficient energy to switch from continuous bonds between neighboring particles to bonds that break every now and then, allowing particles to move but not to escape (become a gas). To turn this liquid into a gas, we need to break all the intermolecular bonds, which has an additional energetic cost.
Often, the energy required to go from solid to liquid is lower than the energy required to go from liquid to gas. For example, in ice, each water molecule makes 4 hydrogen bonds. In liquid water, each water molecule makes between 3 and 4 hydrogen bonds. In steam, each water molecule makes about zero hydrogen bonds (except for the rare and fleeting instances where it collides with another water molecule).

Are the liquid bonds in fact stronger?

No, bond by bond they are of similar strength. If the liquid has a lower density thant the solid, the particles are a bit further apart from each other, so the bond will be slightly weaker. Also, averaged over time, there will be a lower number of bonds.

Ifthe energy required to go from solid to liquid is lower than the energy required to go from liquid to gas.
Can we conclude that the number of hydrogen bonds in ice  is less than in liquid water?

No, but the difference in number of bonds between solid and liquid is less than the difference in number of bonds between liquid and gas.

@alexigirl Couldn’t have responded to that comment better myself.

As a teacher, I tell my students that in a solid the atoms are vibrating around a central point. The higher the temperature, the stronger they vibrate. But, if the vibration energy is too high, it is like with a rubber band or an elastic, the bond is broken, and the system acquires a new degree of freedom : translation. The solid becomes a liquid. I know it is oversimplified, but for my high school students, it is sufficient

This makes it seem that once the bond is broken, it is like a snapped elastic band that can't reform. But Karsten Theis says that the bonds form and reform, and that they are of the same strength in the liquid as the solid.

I told you that it is a oversimplified message. Karsten Theis does not say that bonds have same strengths in solid and liquid. He says it is "slightly weaker" in the liquid. And there is a lower number of bonds

You could go for Velcro instead of rubber bands if you want to address the temporary nature of bonds in your analogy, and the capability of forming again when the substance freezes. Magnets are great for that purpose in a physical model. You did say the rubber bands are sufficient for the learning outcomes you are going for in your specific scenario.

I agree that simple analogies are what is needed at high school, but it would be great to  avoid teaching things that will need to be unlearnt later. My understanding of

... weaker bonds in the liquid is that the particles are. on average, slightly further apart, and electrostatic field strength decreases with distance. But would you therefore say that the bonds are weaker? At any time there will be some particles that are as close together in the liquid as they would be in the solid, would the bonding force then be the same as in the solid?


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Why does the amide get priority over the nitro group in the sulfonation of N-(4'-nitro[1,1'-biphenyl]-4-yl)acetamide?

An amide group is a moderate activating group and the nitro group is a strong deactivating group. Why did sulfonation occur in the ortho position of amide, when it could have added to the meta position of nitro (circled in red)?
I do understand the part that amide is a moderate activating group and that it favors ortho and para position.

What about this question? Is the ring on the right deactivated, causing the phenyl to act like a moderate activator for the left ring?

You're thinking only in terms of selectivity (i.e. whether it's ortho,para- or meta-directing), but the reactivity aspect (i.e. whether it's activating or deactivating) is equally, if not more important, here. Does it help if I put it this way: the nitro group deactivates every position on the ring it's attached to, just that the meta position is deactivated less than the others?

Oh!!!!! Thank you so much!!! So the ring on the right (top) is deactivated, while the left ring is activated.

Would it be okay if I asked you another question? I edited the post.


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Why is molar density of solids and liquids taken as 1?
In the expression for equilibrium constant, the molar density values of solids and liquids are taken as 1. Is the active mass taken as the molar density of reaction divided by molar density at standard states? Then why do we take molar concentration for gases instead of molar concentration / standard molar concentration?

Search the site as it was answered before.

What is "molar density"? The term in the equilibrium constant is usually called the chemical or thermodynamic activity AFAIK: https://en.wikipedia.org/wiki/Thermodynamic_activity


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What is the maximum osmotic pressure that can be exerted with water as the solvent at room temperature?
Lets say we have a membrane which separates a tank of pure water from a tank of water saturated with a solute(there are also undissolved chunks of solute which can immediately dissolve if water is added). Is there a theoretical maximum of osmotic pressure that can be achieved? What's the highest we can achieve using known solutes? I know there's an equation for estimating osmotic pressure for low molarity but I'm only interested in fully-saturated solutions. As such, the errors stemming from using that formula in extreme-molarity situations makes its usage here unreasonable. Is there an equation for the fully-saturated osmotic pressure? Does having multiple solutes effect anything? Any help is appreciated.

The value for any non-ideal concentration (including saturation) will have to be determined experimentally. There is no simple equation describing nonideality.

I updated my answer; you might want to take another look.

We can do an end-run around the van't Hoff equation, and its needed ideality approximations, by instead looking directly at experimentally-determined vapor pressure lowering in saturated aqueous solutions.  From each of these we can determine the activity of water, and from that we can calculate the osmotic pressure.
Conveniently, the online CRC Handbook provides a table of vapor pressures of saturated aqueous salt solutions $[1]$.  To find the one that would give the highest osmotic pressure, we want the want the one that yields the greatest vapor pressure lowering, i.e., the one with the lowest vapor pressure at a given temperature.  Of the 19 listed salts, the one whose saturated solution has the lowest vapor pressure at $\pu{25 ^{\circ}C}$ is, by a large margin, $\ce{LiCl(aq)}$.
Note that there may be solid substances whose aqueous saturated solutions yield an even lower water vapor pressure than does $\ce{LiCl(aq)}$*; but given how much lower $\ce{LiCl(aq)}$'s vapor pressure is than that of the 2nd place finisher on this list ($\pu{0.384 kPa}$ for $\ce{LiCl(aq)}$ vs. $\pu{1.536 kPa}$ for $\ce{K2CO3(aq)}$), I think we can safely say that $\ce{LiCl(aq)}$ is a strong contender (for even higher osmotic pressures, see my discussion of water-ethanol solutions at the end).
[*One candidate is saturated $\ce{NaOH(aq)}$, but I found it surprisingly hard to find data for its vapor pressure; this MSDS says 50% NaOH has a vapor pressure of $\pu{1.2 hPa}$ at $\pu{20 ^{\circ}C}$, which I assume means $\pu{0.12 kPa}$: http://www.labchem.com/tools/msds/msds/LC24150.pdf .  However, I don't know how much I can trust an MSDS for this type of information, particularly since the MSDS doesn't itself cite a source.]
With no approximations,
$$\int_{p'}^{p'+\Pi} \bar{V}(p) dp + R T\ln a =0,$$
where  $p'$ is the ambient pressure, $\Pi$ is the osmotic pressure, $\bar{V}(p)$ is the molar volume of water (which is a function of pressure), and $a$ is the activity.   This is the same equation that can be seen in the initial steps of the derivation of the van't Hoff equation in any physical chemistry textbook, modified to remove the two approximations typically made up to this point, namely that $\bar{V}$ is independent of pressure, and that $a = x$, where $x$ is the mole fraction. [Many more approximations are made subsequently in that derivation; we will avoid all of them, while introducing one of our own, to simplify the calculation of the pressure-dependence of the molar volume.]
The first step is to determine $a$ from the vapor pressure lowering.  To do this we use:
$$p_{vap} = a*p^{\circ}_{vap},$$
where $p^{\circ}_{vap}$ is the vapor pressure of pure water, and $p_{vap}$ is the vapor pressure of the solution. Note that, to account for non-ideality, we are using the activity instead of the mole fraction:
$$a = \gamma x,$$
where $\gamma$ is the activity coefficient ($\gamma =1$ in an ideal solution).
Using  $p^{\circ}_{vap} = \pu{3.169 kPa}$ (vapor pressure of pure $\ce{H2O}$ at $\pu{25 ^{\circ}C}$)  and $ p_{vap} = \pu{0.384 kPa}$ (vapor pressure of water over saturated  $\ce{LiCl(aq)}$ solution at $\pu{25 ^{\circ}C}$) gives:
$$\bf{a = 0.121174}$$
This is an extraordinarily low value. To achieve this in an ideal solution, you'd need to be (by mole fraction) $12\%$ water and $88\%$ $\ce{LiCl}$!
To account for the compression of water at high pressures, we use the compressibility of water: $k =\pu{45.8*10^-6 bar^-1}$ $[2]$.  And thus:
$$\bar{V}(p)=\bar{V}^o (1-k*p),$$
where $\bar{V}^o$ is the molar volume of water at $\pu{1 bar}$. Note that this is itself an approximation, since $k$ is a function of pressure.  A more sophisticated calculation would make use of an equation of state for water, but I think this is sufficient for our purposes.
Substituting:
$$\int_{p'}^{p'+\Pi} \bar{V}^o (1-k*p) dp + R T\ln a =0$$
Inserting $p' = \pu{1 bar}$, $\bar{V}^o =\pu{0.018068 L/mol}$, $k =\pu{45.8*10^-6 bar^-1}$, $T =\pu{298.15 K}$, and $a=0.121174$, and solving for $\Pi$, we obtain:
$$\bf{\Pi = \pu{3,000\bf{bar}}=\pu{2,900 \bf{atm }}=\pu{43,000 \bf{psi}} }$$
[I've rounded to 2 significant figures because $k$ has two sig figs, and, as explained above, assuming it's pressure-independent is itself an approximation.]
Note that this pressure is high enough to justify the extra work we did to account for the compressibility of water: Its molar volume at this pressure is 13% less than that at $\pu{1 bar}$.
To put this in context, this is approximately three times the pressure at the bottom of Challenger Deep, the deepest point in the world's oceans ($\pu{1127 bar}$ at $\pu{10925 m}$).  Or if you're a fan of U-shaped tubes, this would be enough to create a column of water $\pu{30 km}$ high! Indeed, we are beginning to approach the transition to Ice IV, which occurs at $\approx \pu{9500 bar}$ at $\pu{25 ^{\circ}C}$—see Martin Chaplin's phase diagram of water $[3]$.
Furthermore, I could calculate an even higher osmotic pressure by extrapolating the vapor pressure lowering of $\ce{LiCl}$ to a higher temperature.
Alas, such a high osmotic pressure is not achievable, because of the technological limits of currently-available semipermeable membranes, which would begin to leak or collapse at pressures an order of magnitude below this! According to a $2018$ paper by Davenport, Deshmukh, Werber, and Elimelech $[4]$, "At present, only a few studies have assessed RO above $\pu{100 bar}$, with maximum tested pressures of $\pu{200 bar}$." They add that $\pu{300 bar}$ remains "a challenging target from membrane materials and module design perspectives."  It appears the highest pressure rating currently available for commercial RO membranes is $\pu{1800 psi}$ ($\pu{124 bar}$) $[5,6]$.
Having said that, if technology were not a limit, we could achieve even higher osmotic pressures than those available from dissolving salts in water, by instead using a mixture of water with some liquid that is miscible with water in all proportions.  One example would be ethanol.  If you separated pure ethanol and pure water with a membrane permeable to water but not ethanol (and such membranes do exist—they are used by distilleries), the initial calculated osmotic pressure would be infinite, because the mole fraction of water on the ethanol side would be $0$, and thus the chemical potential of the water on the ethanol side would be $-\infty$.
Of course, this calculated osmotic pressure is not a true theoretical osmotic pressure, since the system needs to equilibrate. Thus the water would move from the water side to the ethanol side (leading to a finite mole fraction of water on the ethanol side) until the chemical potential of the water on both sides was the same. If the container is rigid, the theoretical hydrostatic pressure difference (equal to the theoretical osmotic pressure) one could generate would be enormous.
$[1]$ "Vapor Pressure of Saturated Salt Solutions," in CRC Handbook of Chemistry and Physics, 102nd Edition (Internet Version 2021), John R. Rumble, ed., CRC Press/Taylor & Francis, Boca Raton, FL.
https://hbcp.chemnetbase.com/faces/documents/06_42/06_42_0001.xhtml
$[2]$ Sears, Zemansky, Young, and Freedman, University Physics, 10th Ed., Section 11-6.
$[3]$ https://water.lsbu.ac.uk/water/water_phase_diagram.html
$[4]$ Douglas M. Davenport, Akshay Deshmukh, Jay R. Werber, and Menachem Elimelech.  High-Pressure Reverse Osmosis for Energy-Efficient Hypersaline Brine Desalination: Current Status, Design Considerations, and Research Needs.
Environmental Science & Technology Letters 2018 5 (8), 467-475.
DOI: 10.1021/acs.estlett.8b00274
https://pubs.acs.org/doi/10.1021/acs.estlett.8b00274
$[5]$ https://www.prnewswire.com/news-releases/first-commercial-order-of-a-next-generation-ultra-high-pressure-reverse-osmosis-uhp-ro-industrial-system-301058563.html
$[6]$ https://www.saltworkstech.com/articles/applying-ultra-high-pressure-reverse-osmosis-in-brine-management/

Wow! This is an amazing amount of scientific knowledge and effort! Thank you so much! I was originally wondering if one could use osmotic pressure to create a sort of cheap high energy density battery. Reading your initial comment made me realize that that'd basically be like compressing a virtually-incompressible fluid and storing energy in the minute change in volume at astronomical pressure. Something that'd work in theory but is nearly impossible in practice. I don't think any of the issues regarding that lie in the pressure differential across the membrane though. One could have a (1/3)

series of membranes separating sections of fluid of increasing concentrations. While the net osmotic pressure differential would be too much for a single membrane to withstand, a series of them could trivially withstand them. The real issue is the pressure on the container(especially if the container's volume would change, like it is in this case). In order to counteract the pressure, an opposing force needs to be generated. This could be supplied by a curved material with a lot of strain. With a spherical shell, the pressure differential that could be generated would be 2σln(b/a) where (2/3)

σ is the max stress the material can handle and b and a are the radii of the spherical shell. The other option is by constantly exchanging momentum with particles. An example of this would be a U-shaped pipe with fast fluid flowing through it. The fluid at the U-turn part would be accelerating towards the rest of the pipe and would create an opposing force. Enough of these and several layer of such pipes might allow for the creation of extreme pressure. Still impractical for energy storage. (3/3)

@Laff70: If you wanted to do energy storage on an industrial scale, you wouldn't need to go to the maximum possible osmotic pressure.  You could just use the energy to power an industrial reverse osmosis device, and then extract the energy by allowing it to run in the reverse direction. The question is: How robust, cost-efficient, and energy-efficient would this be compared with other industrial-scale energy-storage devices? You can do a search for 'reverse osmois for energy storage'.  Here's an example:  https://www.sciencedirect.com/science/article/abs/pii/S0011916421001594

I was hoping for something which could in theory compete with gasoline and lithium ion batteries in terms of energy density.

You don't need to store a the bulk liquid in high pressure tanks. The only high pressure system is the RO system itself. You can have fixed-size reservoirs of NaCl at intervals of 4% from 0-28% and each membrane only pumps across a 4% NaCl pressure-difference. This is far more energy dense than pumped storage, uses extremally low-cost ingredients, and can be situated on flat land. (I think) a downside is RO membranes are expensive and slow, so getting good round-trip and W/$ isn't trivial. The best bet would be a seasonal system that stores summer solar energy at high latitudes.


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Difference between DSF (disulfiram) and DSF-Cu
I though tetrathiomolybdate (TM) and disulfiram (DSF) were forms of copper chelators, so I don’t understand the difference between TM/TM-Cu and DSF/DSF-Cu in the below graph. Are they chemically different?

Chen D, Cui QC, Yang H, Dou QP (2006) Disulfiram, a clinically used anti-alcoholism drug and copper-binding agent, induces apoptotic cell death in breast cancer cultures and xenografts via inhibition of the proteasome activity. Cancer Res 66(21):10425–10433. https://doi.org/10.1158/0008-5472.CAN-06-2126

Welcome to [Chemistry.se]! I'm afraid, but this question is missing all context. It appears to be taken out of a publication; please cite it. I guess this publication should also explain the graphic, so I'm not sure how this site could be more helpful.

@Martin-マーチン the publication does explain that DSF-Cu is DSF laced with copper. But that’s all… I’m not sure why more context is needed as I know what DSF-Cu stands for. I’m asking about the chemical/relational differences between these compounds (DSF/DSF-Cu) and established chemical nomenclature. If you do not know the answer, that’s fine. I will wait for someone that can.

Context is always helpful. If you quote a passage of a text or an image you must also cite it. This is a standard scientific practice. I'm afraid that this is very likely not standard notation, but rather an abbreviation used in this publication. This is definitely not nomenclature though. || I'd like to recommend taking the [tour] to learn more about how this site works.

I’ve updated it. Then why is this notation practically used in most publications on pubmed/WoS. I’ll exemplify this one?: https://www.researchgate.net/figure/DSF-and-DSF-Cu-inhibited-myeloma-cell-viability-in-a-dose-dependent-and-time-dependent_fig1_339709780

It is right there: The DSF-copper complex inhibits the chymotrypsin-like activity of purified 20S proteasome and 26S proteasome in breast cancer cells. It has been shown that DSF is able to bind copper (31, 32). Indeed, when a solution of DSF was mixed with a solution of CuCl2 at 1:1 ratio, dramatic color change was observed, from light blue to dark green (data not shown). This indicates that a chemical reaction has occurred that involves formation of a DSF-copper complex.

These are all chemically different compounds—the paper is investigating whether or not a complex of disulfiram on copper has proteasome inhibiting properties. They test this by performing an MTT assay on breast cancer cells, and this data compares all of the groups they constructed for their experiment. The data is—in my opinion—poorly presented, so it makes it unclear what is going on. Let's dissect the groups they are testing:

In this plot, they have breast cancer cells for the assay tested against controls of the pure compounds investigated in the study. The cell line was cultured separately in presence of: DMSO (DM), copper (Cu) alone, disulfiram (DSF) alone, and tetrathiomolybdate (TM) alone. The figure represents the concentration in $\mu$M that the experiment used, and there is one concentration per compound.

This portion of the plot displays data from the MTT assay for the exact same experiment, but if the DSF is chelating on copper. They ran the individual DSF and Cu separately to ensure that this complex was the one performing proteasome inhibition and not one of its constituents. They tested three different concentrations, so there are three columns for this space. This chemical is chemically different than both DSF and Cu alone.

This portion of the plot acts as another control for the experiment. The authors want to show that it is the DSF-Cu complex that is doing the inhibition and that this does not generalize to just any copper complex they use. They chelated the copper with TM instead of DSF for the MTT assay to see if this would result in any proteasome inhibition. They tested two concentrations, so there are two columns. The information meant to be derived from this plot is that when DSF coordinates to copper, it has a unique ability to inhibit proteasomes that you don't find with its component parts alone or with similar cancer-relevant chelates.

I really don't think the data in the paper is presented poorly. If you take your time reading the paper then it is quite sufficiently explained what everything is. Apart from this, there is one sentence exactly explaining what DSF-Cu appears to be, at least to the extend necessary to understand the paper. I honestly think that the question itself is not a good fit for this site and explaining these obvious things is noise in my opinion.

@Martin-マーチン I mean, I disagree with you about what is a good fit for this site—I appreciate questions that are well-articulated and come from thoughtful people, but the purpose of this site is to provide insight about questions in chemistry and this fits the bill. In the sociological research regarding education, it's a well-known problem that teachers often overestimate just how "obvious" some material is, and thus I don't think people with such know-how are fair arbiters of what is obvious. The plot is OK—sufficient for a trained eye, but excellent data presentation is much more elegant imo

I appreciate you telling me about your opinion; I guess we really do view this site from different angles. However, this is not the place to dig deeper into this discussion, but I invite you to participate in these discussions on [meta]. I believe I have asked especially about this in the last year; it should be easy enough to find. Anyway, my opinion on the question is not an opinion on your answer, which I think explains the graphic well. I may be a bit late, but anyway: Welcome to [chemistry.se]!


text:
Origin of oxygen content among the products of burned carbohydrates
For simplicity, I am only considering complete combustion.
I am ignoring variants such as deoxyribose, just pure carbohydrates.
Balancing the equation is easy due to the definition of carbohydrate: we already have the correct amount of oxygen for the hydrogen.  We just need to supply the oxygen for the carbon.
$$\ce{C_mH_{2n}O_n + m O2 -> m CO2 + n H2O}$$
Here is the question: do we know whether the oxygen already in the carbohydrate is more likely to be in the water or the carbon dioxide?  I expect that this could be determined with isotopic labelling.
If it might make a difference, consider combustion in a laboratory.
Subsidiary questions.
Might the use of isotopes affect the reaction?  I know that biological processes can be sensitive to isotopes.
Might the answer be different in biological systems?

Not homework, just an old man's curiosity.

In biological systems, it's relatively easy to track the atoms, but the answer is different from combustion because the reactions are quite different even though the products are the same. The atoms from oxygen gas end up only in the water product, whereas the oxygen atoms on the carbohydrate can either be retained on carbon or end up in phosphate groups or water in the medium. Some of the oxygen atoms in the CO2 product also come from water in the medium.

@Andrew Of course, I should have thought of that, a very different route.


text:
Does hypochlorous acid oxidize hydrogen peroxide?
The following reaction is from https://ncert.nic.in/textbook/pdf/kech202.pdf, p. 11:

$$\ce{H2O2 + HOCl -> H3O+ + Cl- + O2}$$

I have to find if $\ce{H2O2}$ is acting like a reducing agent or an oxidising agent in this reaction.
If we see, the $\ce{Cl}$ atom in $\ce{HOCl}$ got reduced from $+1$ to $-1$ oxidation  state. That tells me that $\ce{H2O2}$ is acting as a reducing agent. But if I see the $\ce{O}$ atom, it is oxidised from $-2$ to $0$,  now this tells me that $\ce{H2O2}$ is acting as a oxidising agent.
But how is this possible? I know I'm doing something fundamentally wrong.

Evaluate formal oxidation numbers of chlorine and oxygen, the answer to the question should be clear from the result.

You are not doing much wrong except for assigning improper oxidation number to oxygen of the peroxo group.
If the redox reaction “feels” wrong, you can check yourself using a table of standard reduction potentials [1, p. 5-79]:
$$
\begin{align}
\ce{H2\overset{-1}{O}_2 &<=> \overset{0}{O}_2 + 2 H+ + 2 e-} &\quad E^\circ_1 &= \pu{-0.695 V} \tag{1}\\
\ce{H\overset{+1}{Cl}O + H+ + 2 e- &<=> \overset{-1}{Cl}^- + H2O} &\quad E^\circ_2 &= \pu{+1.482 V} \tag{2}\\
\hline
\ce{H2\overset{-1}{O}_2 + H\overset{+1}{Cl}O &-> H3O+ + \overset{-1}{Cl}^- + \overset{0}{O}_2} &\quad E^\circ &= +\pu{0.787 V} \tag{3}
\end{align}
$$
Since resulting $E^\circ = \pu{0.787 V} > 0,$ free Gibbs energy $Δ_\mathrm{r}G^\circ = -nFE^\circ < 0,$ and the redox reaction where hydrogen peroxide is oxidized by chlorate(I) can be considered a thermodynamically favorable process.
Alternatively, you can use Latimer diagram, which comes in handy for disproportionation and synproportionation reactions.
Reference

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.


text:
Green colored substances forming during electrolysis
I saw this but the OP didn't provide further information.
My setup is a glass container with the anode and cathode as slim metal pieces and using a 12v charger.
The solution was a saturated salt solution and tap water since I couldn't get distilled water.
My goal is to get sodium hydroxide, as explained in here, but white flakes started forming then it turned green and finally turned into orange.
The metal pieces didn't get affected they looked and felt the same didn't even get corroded oxidized.
Here is a picture of my setup; the wires of the charger are connected on top of that cap so they aren't touching the water:

Here is a picture from the internet of the material that formed; after some time that green turned into orange but not all of it

Note: while I was watching the positive side metal piece had some kind of greenish fumes coming out of its end. I found no way of filtering that green stuff out.

Please be more precise ! Explain the nature of your electrodes. In my opinion, this unwanted green stuff is produced by the electrodes which have been oxidized during the electrolysis. These electrode might be in copper, by judging the color of the insoluble stuff. Am I wrong ? And why do you use salt water ? You will never obtain sodium hydroxyde by this operation.

my electrodes are from a car space i dont know anything about it, also why i should not use salt? @Maurice

I really do not see how the following answer does not address this already: Green color substance forming during electrolysis. The photo of your setup is also not clear enough, these nails look rusty to me. So from the information given here the most likely candidate would still be some kind of iron hydroxide.

@Martin-マーチン they dosnt address how i can filter it off thats my question at the end of it i tried filtering them using mesh and many other things but non worked

I recommend you invest some time into your question and focus it towards what you want to know. Putting it below everything, especially these oversized images, makes it easy to be overlook. Don't add it as a note. And please use proper capitalisation and punctuation.

Does this answer your question? Green color substance forming during electrolysis

@Bredget. If you don't know the nature of your electrodes, we cannot help you, because the greenish stuff is clearly coming from the electrodes, and not from the solution.
Now if you electrolyze a salt solution, you will first produce some bubbles of $\ce{H2}$ and some ions $\ce{OH-}$ at the cathode (negative pole). Simultaneously, you will produce a mixture of oxagen $\ce{O2}$ plus chlorine bubbles $\ce{Cl2}$ at the anode (positive pole). But this chlorine is rather corrosive. It will soon attack and dissolve any metal at the electrode. And it may also react with the $\ce{OH-}$ ions coming from the cathode, and produce the reaction $\ce{Cl2 + 2 OH- ->ClO^- + Cl- + H2O}$. So the solution contains more and more $\ce{ClO-}$ ions, which are powerful oxidizing agents.

Thanks for your detailed answer, was really helpful. Do you suggest I use graphite as an anode?

Yes ! Graphite will not produce any corrosion like copper or iron. Graphite will not be chemically destroyed. But it may be victim of a physical corrosion. A graphite electrode will slowly be piulverized, with graphite powder falling down to the bottom of the solution. The best choice is platinum, which is not corroded at all.


text:
Why is methane not called carbon tetrahydride?
If aluminium hydride, $\ce{AlH3}$, is called "hydride" because it contains elemental hydrogen and sodium chloride, $\ce{NaCl}$, is called "chloride", and calcium carbide $\ce{CaC2}$ as another example, then why is methane not called carbon hydride or carbon tetrahydride.
Another question why hydrogen halides, e.g. $\ce{HCl}$, $\ce{HBr}$, $\ce{HF}$, are called like this and not halogen hydrides.

Aluminium hydride does not contain elemental hydrogen. From electronegativity considerations hydrides are closer to $\ce{H-}$ and therefore named this way. In methane the hydrogen is much more protic.

The suffix “-ide” indicates a negative charge, so “hydride” compounds are those which are polarized such that H is negative whereas “halide” compounds have a negative halogen and “oxide” a negative O

Methane is a carbon hydride - not even the simplest, technically. One could call it monocarbon tetrahydride, or tetrahydrogen monocarbide, and both would be as silly as calling water dihydrogen monoxide. You should read up about hydrocarbons. I think at the point of adventure with chemistry I suspect you are, the answer is "because nomenclature was made like that".

Because the halogens have a higher electronegativity than hydrogen. Ammonia is one of few exceptions, (Nitrogen has an electronegativity of 3.04, compared to hydrogen's 2.20) though usually, more electronegative elements have the -ide ending.

Closely related: https://chemistry.stackexchange.com/questions/49704/should-carbon-tetrachloride-really-be-considered-an-organic-compound

Also: https://chemistry.stackexchange.com/questions/16382/what-is-the-iupac-systematic-name-for-methane

The name methylene has existed in English (and French) for hundreds of years, and is based on the far older English mead and Greek root μέθυ (methy) for alcoholic beverage. In a similar fashion, ethyl has become part of the language from aether, or αἰθήρ.
So, $\ce{CH4}$ might be called carbon tetrahydride, as $\ce{SiH4}$, silane, is silicon tetrahydride, and $\ce{H2O}$ is dihydrogen oxide. More accurately, though, as Oscar Lanzi points out, it should be hydrogen carbide, because the hydrogen is more metallic, i.e, more generous with its electron, as in the case of $\ce{Be2C}$, beryllium carbide.
However, traditional names are more commonly used, as you can check on Google's Ngram Viewer for water. Personally, I'd rather that helium was helion, named as a metal by Pierre-Jules-Cesar Janssen, but c'est la vie

Changed calcium carbide to beryllium carbide. Calcium carbide is usually CaC2, an acetylide, whereas beryllium actually forms a methanide and so fits the argument better.

As DrMoishe Pippik describes in his answer, historical usage has rendered "methane" a sufficiently recognizable name to be promoted to technical usage. But beyond that, a look at electronegativities indicates carbon is more electronegative than hydrogen. Thus methane would be more accurately rendered as a hydrogen carbide than as a carbon hydride.
Metal carbides that give methane upon reaction with water or steam, such as $\ce{Be2C}$ or $\ce{Al4C3}$, are often called "methanides", which goes along with methane being a (covalent) carbide.


text:
What is the resulting solution after leaching hardwood ash with water?
After doing online research, I've noticed there are differing interpretations on what happens when ash is mixed with water, so I've come here seeking clarification on what indeed happens, and also clarification for why, since I don't know how to refute what I believe might be the misconceptions.
I understand a component of wood ash is potash (hardwood ash consisting greater portions of it than others). It seems wood-derived potash is primarily potassium carbonate ($\ce{K2CO3}$). I will assume this is the case for the rest of the question, but I'd also like to know what other potassium-containing compounds may be therein.
Another component of wood ash seems to be lime. Most of it seems to be calcium carbonate, but I also hear there is calcium oxide ($\ce{CaO}$). This is where things get tipsy-turvy for me. Taking into account what I've told so far, I don't know for sure what the solution is when ash is mixed with water.
One version is that potassium hydroxide ($\ce{KOH}$) is produced and remains in this solution. I don't know how to "refute" the following explanations supporting $\ce{KOH}$ formations if it's not the case or negligible:
(1) The $\ce{CaO}$ in the ash reacts with the water to form aqueous calcium hydroxide ($\ce{Ca(OH)2}$). The potash, primarily being $\ce{K2CO3}$, dissolves in the water too. Both of these then react with each other to produce $\ce{KOH}$ that becomes part of the solution per the equation: $\ce{Ca(OH)2 + K2CO3 -> CaCO3 + 2KOH}$.
(2) Due to the natural dissociation of water, there are some hydroxide ions present in solution. The potash still dissolves, and as part of that, potassium ions will also be present in solution. Both of these then react to produce $\ce{KOH}$.
The other version I've seen posted around is that no $\ce{KOH}$ is formed by mixing water and ash. this makes sense, especially when I consider the archaic job of an ash burner. Hardwood ash (being more potash rich) is leached with water. The potash dissolves into a solution. The mother liquor is then boiled to leave behind potash powder. If the potash had reacted to produce $\ce{KOH}$, I don't see how this could be the case especially since I think lye instead of potash powder would be left behind.
In short, is the solution water with primarily aqueous $\ce{K2CO3}$ or water with primarily aqueous $\ce{KOH}$? I'd also like an idea of what other impurities may be here in either case.

It's not really clear from your question, you are aware that in solution, there are only the ions? $\ce{K2CO3 + CaO + H2O -> 2K+ + Ca^{2+} + CO3^{2+} + 2 OH- }$, all in aqueous form.

I know there are ions. Sorry, I'm trying to say that I've heard some people say the solution is lye water, and others say there is no lye water and it's a myth. An explanation I've seen for lye water is that in the solution, the aqueous potassium ions interact with aqueous hydroxide ions to render the solution lye water (and that if the water is boiled out, potassium hydroxide solute will be left behind). Alternatively, when they say it's not lye water, and the water is boiled out, potash, being potassium carbonate mostly, is left behind.

Also, regarding your chemical equation, from what I understand, there would be no calcium ions as they should have precipitated out as calcium carbonate

You're right, calcium carbonate precipitates. That makes the answer to your question pretty obvious. You just need to know how the relative content of calcium and potassium in you wood.


text:
How to find the orbital a given wavefunction represents?
We were given the following question in chemistry class (under the topic “Atomic Structure”):

Which orbital does the following wavefunction represent? $$φ(r) = \frac{1}{81(6π)^{1/2}}\left(\frac{Z}{a}\right)^{7/2}r^2\exp\left(\frac{-Zr}{3a}\right)(3\cos^2θ - 1),$$
where $Z$ is the atomic mass, $a$ is the Bohr radius, and $r$ is the distance from the origin.

The given answer is a $3\mathrm d_{z^2}$ orbital.
This question was to be solved without having needed to plot diagrams, and the method was to apparently equate the power of $r$ to the azimuthal quantum number value and then use it to find the principal quantum number value. I felt a bit lost on how to do it here, and couldn't find this method anywhere else.
It would be great if you could help me on how to approach this question (on how we could specify that the orbital was a $3\mathrm d$ orbital and more specifically a $3\mathrm d_{z^2}$) since I feel I lack the understanding of how to relate the wavefunction and the orbitals, and would also be great if you could perhaps link any useful resources that I could refer to, to better my understanding of these concepts relating to wavefunctions.

An easier way is to search for nodes, i.e. regions where the wavefunction is zero.
Clearly, the $r^2$ term means that $\psi = 0$ at $r = 0$ (i.e. the nucleus). Also, the $\exp(-kr)$ term means that $\psi \to 0$ as $r \to +\infty$. Neither of those are particularly useful on their own; however, the absence of any other values of $r$ which yield $\psi = 0$ means that there are zero radial nodes.
This already narrows it down a lot: it's going to essentially either be a 1s, 2p, or 3d orbital. (Yes, in theory it could be 4f, 5g, ... but when was the last time you saw an illustration of one of those, let alone their mathematical form?)
Now, the angular part is:
$$3\cos^2\theta - 1$$
This part is zero when $\theta = \arccos(1/\sqrt{3})$, i.e., $\theta = 54.7^\circ$ or $125.3^\circ$. Assuming you know your spherical coordinates, then you can infer from this that the nodes are not planes (as you might be used to thinking) but rather cones*, which is a very special feature of the $\mathrm{d}_{z^2}$ orbital. See my answer here for an illustration: https://chemistry.stackexchange.com/a/59145/16683.
All in all, this lets us conclude that it's the $3\mathrm{d}_{z^2}$ orbital.
A useful further exercise is to inspect the mathematical form of the $\mathbf{4}\mathrm{d}_{z^2}$ orbital. It will look something like this:
$$\psi \propto r^2(c-r)(3\cos^2\theta - 1)$$
The angular form is the same, which allows us to tell that this is a $\mathrm{d}_{z^2}$ orbital. But crucially there is an extra term of $c - r$ (for some $c$), which implies a radial node at $r = c$, and hence a larger principal quantum number.

* If there's a technical / mathematical term for this, please let me know.

The general form of the radial wave function for hydrogen like atom is, $$ R(r) = - \sqrt{ \Bigl(\frac{2Z}{na_o}\Bigr)^3 \frac{(n-l-1)!}{2n(n+l)!}}e^{-\frac{Zr}{na_o}} \Bigl(\frac{2Zr}{na_o}\Bigr)^l L$$ where, $L$ is generalised Laguerre polynomial of degree $(n-l-1)$.  Here are some steps followed by me to identify the orbital which corresponds to the given wave function :

Let us first concentrate on the radial wave function. In this step we are interested to find the principal quantum number ($n$) of the orbital. Observe the  exponential part of the radial wave function. It is given by $ e^{-\frac{Zr}{na_o}}$. So clearly the value of n is $3$.
Observe the power of $r$. Radius (r) is always raised to the power of $l$ i.e., $r^l$. So the value of $l$ is $2$.
Now observe the angular part of the wave function. If it doesn't contain any imaginary part then it is corresponds to the orbital with zero magnetic quantum number ($m$).  Therefore, your orbital has quantum numbers $n=3$, $l=2$, and $m=0$ which corresponds to $3d_{z^2}$ orbital.

NOTE : $1.$ Generally orbitals with $m=0$ corresponds to the orbitals oriented along $Z-$axis.
$2.$ If the angular part of wave function contains imaginary part then exponential part of it is given by $e^{im\phi}$ where $i= \sqrt{-1}$. So from exponential part you can easily find the magnetic quantum number also.  And finally you can also deduce the values of $n$ or $l$ from constant part or from the power of atomic number ($Z$) i.e., $Z^{l+\frac{3}{2}}$.  Hope this helps.

"If it doesn't contain any imaginary part then it is corresponds to the orbital with zero magnetic quantum number" This is true if you're talking about the "true" orbitals / spherical harmonics $Y^m_\ell$, but not necessarily true if you're talking about the real-valued orbitals such as $\mathrm{d}{xy}$, which are linear combinations of those and don't have an imaginary part. It's, of course, not clear from the question which OP is expected to find, since the $\mathrm{d}{z^2}$ orbital is a member of both sets.

It's worth pointing out that in general, in chemistry, the real-valued orbitals are used, while physics tends to stick to the "true" harmonics and not worry so much about complex / imaginary wave functions.


text:
What happens when we react a sulfate salt solution with barium chloride and the formed precipitate with conc. hydrochloric acid?
When I performed this experiment in my school, I could not observe any sort of remarkable change (I must have performed it incorrectly or something).
I wanted to know what should have been the correct observation when the white precipitate (BaSO4 I think?) is reacted with conc. $\ce{HCl}$.
Also, if possible could you please tell me why this happens? And if possible if you could provide an image/video of the final observation, I would be grateful.

If I understand it correctly, you are not interested in the precipitation reaction itself,
$$\ce{Ba^2+(aq) + SO4^2-(aq) <=>> BaSO4(s) \tag{1}}$$
but in the eventual reaction
$$\ce{BaSO4(s) + H+(aq) ->[HCl conc] ??? \tag{2}}  $$
Concentrated $\ce{HCl}$ indirectly increases solubility of $\ce{BaSO4(s)}$ by decreasing concentration(more exactly activity) of sulfates by equilibrium
$$\ce{SO4^2-(aq) + H+(aq) <=>> HSO4-(aq)  \tag{3}}$$
what partially shifts the precipitation equilibrium (1) to the left in favor of dissolved barium and sulfate ions.
Depending on the ratio of the precipitate and acid, the (partial) dissolving of the precipitate $\ce{BaSO4(s)}$ in concentrated HCl
$$\ce{BaSO4(s) + H+(aq) <=>[HCl] Ba^2+(aq) + HSO4-(aq) \tag{4}}  $$
may or may not be noticeable.
That means if you put just a little amount of the precipitate to a large excess of the acid, it may dissolve completely or at least significantly. If, OTOH, you put large amount of precipitate to relatively small volume of acid, you may not notice at all, if e.g. 1% of it has dissolved.
Heating and/or stirring speeds up eventual dissolution, but heating is not advised due releasing gaseous hydrogen chloride. Aside of dissolution itself, not other noticeable visual, acoustic nor smell (aside of $\ce{HCl}$) effects are expected.


text:
Fractional Composition of Acid Question
I present to you a fractional composition question that my professor refuses to answer because he encourages me to use equations without understanding them. Any help you can provide would be very, very appreciated.
The fractional composition of $\ce{HA}$ for the disassociation of weak acid $\ce{HA}$ is presented as:
$$\text{fraction of HA in the form HA} \equiv\alpha_{\ce{HA}} = \frac{[\ce{HA}]}{\ce{[A-] + [HA]}}$$
The acid being $\ce{HA}$ should have a 1:1 ratio of $\ce{H+}$ and $\ce{A-}$ because they disassociated from the same acid $\ce{HA}$ where there is one mole of each. At least, this is what makes sense to me. The equation above only shows $\ce{[A-]}$ in the denominator and I figure I could replace that with $\ce{[H+]}$ if I wanted to. But it turns out I can't because it doesn't work. And I don't know why.
For example,
If there's an acid $\ce{HA}$ with a pH of 3, the concentration of $\ce{H+}$ in solution is $10^{-3}$. However, this concentration of $\ce{H+}$ does not mean that $\ce{[A-]} = 10^{-3}$. Why?
Here is a more in-depth example from my textbook:

Thank you for your time.

You can afford to replace [A-] by [H+] only in case if all H+ can be considered as becoming from acid dissociation, i.e. auto-disociation of water can be neglected. It is aceptable only if [H+] >> [OH-]. This condition is not met for just weakly acidic solutions with $\mathrm{pH} \approx 6+$, if the acid is too weak and/or its concentration is too low. The condition is not met either if other solution components affect ratio  of H+ and A- concentrations.

Please don't use screenshots of text. Don't forget to properly cite your sources. The last screenshot looks like it's from Harris' Exploring Chemical Analysis. "My textbook" is meaningless.

@Poutnik You're right. I completely overlooked this. Thank you.

In solution you have $\ce{HA}$, $\ce{A-}$, $\ce{H+}$ from dissociation of the acid and $\ce{H+}$ already in the solution. The last one conditions the dissociation of the weak acid.
\begin{array}{cccc}
\ce{HA &<=> &H+ &+ &A-}\\
x && 10^{-5.31}-x && -x
\end{array}
The equilibrium constant is
$$K_\mathrm{a} = \frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}.$$
You can see now, that an already presence of one of the species involved in this equation, tend to alter the concentration of the other since the $K_\mathrm{a}$ has to remain the same in every condition if $T$ remains constant.

Note that \ce{…} macro from mhchem is used strictly for chemical expressions, not math operators or variables. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Familiarize yourself with Which symbols are written in roman (upright) font and which are italicized?

I had a similar thought to this but HA being a weak acid does not completely disassociate and therefore will not have the same concentration as A-. Thus, they cannot both be denoted with X. Another issue is that in the example, [A-] is bigger than [H+]. Using your diagram, plugging x into 10^-5.31 - x  would yield a negative concentration.
I tried using the Henderson-Hasselbach equation to see if a base with the same A- as the acid was added to the solution. But it did not appear so. The other possibility is that a salt with the same A- was added. .....

... (continued) ... However, the math done to calculate the final [A-] uses only information about the acid.

@big_armpit try to calculate the "normal" dissociation and then set the $\ce{[H+]}=10^{-5.31}$ e calculate again the $\ce{[A-]}$


text:
Crystal packing prediction for simple organic molecules?
Is it yet possible to predict a simple organic molecule's crystal structure or how the molecules arrange in a crystal without diffraction measurements?
By crystal structure I mean, how would molecules arrange themselves? For example how an amide would pack, how an n-alkyl carboxylic acid would pack? Not partically interested in lattice systems or whether it's a monoclinic or orthorhombic, but more how the molecules would arrange.
Can anyone recommend softwares that can do this?
Thanks

https://mattermodeling.stackexchange.com/questions/157; https://mattermodeling.stackexchange.com/questions/3554; https://chemistry.stackexchange.com/questions/39070

Molecular modelling? All you need are good force fields. Organic molecules often have numerous solid forms, there obviously is no clear-cut single solution.

As of 2022, the crystal structure prediction even for simple organic molecules is far from simple.

This 2020 paper describes one of the largest efforts made so far. They built a software called Autopack based on human expert labels and structure data from Cambridge Structure Database. Autopack is free of cost and can be requested by filling a form and email to request online.


text:
How do I understand the IUPAC naming of 9-methyl-3-[(2-methylimidazol-1-yl)methyl]-2,3-dihydro-1H-carbazol-4-one?
I'm having trouble understanding the IUPAC naming convention of 9-methyl-3-[(2-methylimidazol-1-yl)methyl]-2,3-dihydro-1H-carbazol-4-one.
It would be great if you can help to breakdown the naming structure for me.

In order to name the compound that is given in the question, first we need to identify the parent structure. Fortunately, there is a principal characteristic group (the ketone), thus it is clear that the parent structure is the part that includes this group.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), retained names (also called trivial names) are used for various heterocycles with the maximum number of noncumulative double bonds. In Table 2.8 of the Blue Book, we find that the retained name for the parent structure of the given compound is carbazole.
Table 2.8 also shows the fixed numbering of carbazole. According to Rule P-14.4, this fixed numbering must be used. Therefore, the locant of the principal characteristic group is ‘4’. Since the principal characteristic group is expressed as a suffix, the given compound is a carbazol-4-one.
The original parent structure carbazole ($\ce{C12H9N}$) has the maximum number of noncumulative double bonds for this kind of structure. Compared to an ideal unsaturated compound, however, carbazole has one extra hydrogen somewhere. The position of this extra hydrogen must be indicated. A few different isomers are possible; for example, the original parent structure could be 9H-carbazole (i.e. with the extra hydrogen at the nitrogen).

In the given compound, two double bonds of the original carbazole are missing. Thus, the parent structure is a bit more saturated with the equivalent of four additional hydrogen atoms ($\ce{C12H13N}$). Such saturation is described using ‘hydro’ prefixes. The locants of the original indicated hydrogen and the four additional hydrogens are 1, 2, 3, 4, and 9. The indicated hydrogen gets the lowest locant, so the name of the unsubstituted parent structure is 2,3,4,9-tetrahydro-1H-carbazole.

In the substituted carbazol-4-one, however, the numbering is different because one of the double bonds was removed to make room for the ketone. According to Rule P-58.2.3, the indicated hydrogen is placed at the position that will accommodate this principal characteristic group. Therefore, the name becomes 1,2,3,9-tetrahydro-4H-carbazol-4-one.

Finally, we can add the other substituents. There is a simple methyl group at the nitrogen (locant 9) and the complex substituent group (2-methyl-1H-imidazol-1-yl)methyl at position 3. Therefore, the complete name of the given compound is 9-methyl-3-[(2-methyl-1H-imidazol-1-yl)methyl]-1,2,3,9-tetrahydro-4H-carbazol-4-one.


text:
Why is Pb3O4 called lead(II,IV) oxide?
We know that iron(II,III) oxide has the formula $\ce{Fe3O4}$ which is a combination of iron(II) oxide and iron(III) oxide with formulas $\ce{FeO}$ and $\ce{Fe2O3}$ respectively.
So why isn't this the case in $\ce{Pb3O4}$? Why can't we call it lead(II,III) oxide just like the iron example?
On the other hand, lead(II,IV) oxide should have the formula $\ce{Pb2O3}$ which is the same with lead(III) oxide.
Lead(II)oxide: $\ce{PbO}$ and lead(IV) oxide: $\ce{PbO2}$.

Because lead (III) is not a thing.

@IvanNeretin According to Wikipedia, the oxidation states of lead are $−4, −2, −1, +1, +2, +3$, and $+4$; with $+2$ and $+4$ being dominant.

These are not combinations of 2 oxides, but they are rather single oxides where metal atoms alternate their oxidation numbers ( II,III for Fe, II,IV for Pb ), often not exactly in stoichiometric ratios of small integers. This can be determined from X-ray crystallographic analysis. $\ce{Pb3O4}$ is AFAIK actually $\ce{Pb2[PbO4]}$

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lead(II,IV) oxide should have the formula Pb2O3 It would be true if Pb(II)/Pb(IV) ratio were 1:1, but it is 2:1. There is $\ce{Pb2^{II}Pb^{IV}O4}$ versus  $\ce{Fe^{II}Fe2^{III}O4}$

@Poutnik What would the iron oxide formula be if the Fe(II)/Fe(III) ratio were 1:1?

In the old valency notation (19th century), $\ce{Pb3O4}$ would have been  $\ce{O=Pb-O-Pb-O-Pb=O}$

@Maurice it can be considered as the salt lead(II) orthoplumbate(IV). // From wikipedia: Plumbate
"Lead tetroxide" $\ce{Pb3O4}$ may be thought of as lead(II) ortho-plumbate(IV), $\ce{Pb2^2+[PbO4]^4-}$. "Lead sesquioxide" $\ce{Pb2O3}$ is also known, and has the structure lead(II) meta-plumbate(IV), $\ce{Pb^2+[PbO3]^2-}$

@Poutnik Yes. It can be considered as the salt you mentioned. Bit in the 19th century ions were not yet known.

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@IvanNeretin technically, lead(III) is known in hexamethyldiplumbane. But yes, the oxide structure better matches with a combination of lead(II) ans lead (IV).

Both crystal structure and periodic characteristics favor the formulation $\ce{Pb^{II}_2Pb^{IV}O4}$.
Below is a picture of the crystal structure from Wikipedia.

Source
This structure has at least two features that would be consistent with a lead(II,IV) identification:

One third of the lead atoms have higher coordination number (six) to the electronegative oxygen and the rest have lower coordination number (two). The more highly coordinated lead atoms would naturally fit with a higher oxidation state, and $\ce{Pb^{II}_2Pb^{IV}O4}$ fits with the ratio of high to low coordinate lead atoms.

The six-coordinate lead atoms noted above have the same coordination as they do in the pure lead (IV) oxide $\ce{PbO2}$. Note the contrast here with iron, where six-coordination is typical for oxidation states +2 and +3.

In addition, $p$-block elements — unlike transition metals — strongly favor paired-electron structures, which means even-$Z$ elements favor even oxidation states and odd-$Z$ elements favor odd ones. Where the oxidation state has the "wrong" parity it's usually accompanied by homonuclear bonding which allows the electrons to be paired without being polarized (like peroxides with their oxygen-oxygen bond or lead(III) in $\ce{(CH3)3Pb–Pb(CH3)3}$). No such homonuclear bonding is evident in $\ce{Pb3O4}$, so the lead atoms ($Z=82$) would be assigned even oxidation states to fit the general trend.

@Poutnik gave us the Wikipedia link which basically contains the answer:

Lead tetroxide ("red lead"), a mixed oxide with formula $\ce{Pb3O4}$, may be thought of as lead(II) ortho-plumbate(IV), $\ce{[Pb^2+]2[PbO4]^4−}$. Lead sesquioxide, $\ce{Pb2O3}$, is also known, and has the structure lead(II) meta-plumbate(IV), $\ce{[Pb^{2+}][PbO3]^2−}$.

Since $\ce{Pb3O4}$ contains $\ce{Pb(II)}$ and $\ce{Pb(IV)}$, it is lead(II,IV) oxide. Magnetite is actually $\ce{[Fe^{2+}][(Fe^{3+})2O4]}$, so it is iron(II,III) oxide.


text:
Stability of Mercaptoundecanoic acid and thiol-PEG stored as solids
I am planning to use Mercaptoundecanoic acid (MUA, 98%) and thiol-PEG (O-[2-(3-Mercaptopropionylamino)ethyl]-O′-methylpolyethylene glycol) for functionalisation of gold nanoparticles.
The SDS recommends storage under inert gas conditions for both chemicals and either in the freezer or fridge as the chemicals are sensitive to air.
I am unsure on how sensitive the chemicals are and what part (for example the thiol group or the PEG part) are sensitive.
Storing solids under inert gas conditions is quite a hassle (always going into the glovebox takes forever to simply weigh out chemicals) so I was wondering if someone has input for:

what part of the molecule is 'sensitive' and will degrade if not stored/handled under inert gas?
Is there a smooth way to store solids under inertgas outside the glovebox (a potential product that I could order would help a lot! - like a gas-tight vial?)
Could these chemicals maybe be stable in some solvent (like ethanol?) so that the handling under inert gas gets easier?

Thanks!