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What makes carbon dioxide gas "sticky" in pipes?
There's some perceived wisdom where I work that when flowing $\ce{CO2}$ gas in pipes it acts somewhat "sticky", in the sense that it lingers around longer and takes more $\ce{N2}$ or air to purge out than, say, methane or $\ce{N2}$.
Unfortunately, no one here seems to know the physical or chemical basis for this observed result. I do have some quantitative data through various flow sensors and IR gas sensors that show it really does stay around longer than $\ce{N2}$ or $\ce{CH4}$, so I don't think it's due to bad observation.
As you can imagine, googling "why is $\ce{CO2}$ sticky" is not very helpful.  If anyone can explain to me why this happens, I'd be very grateful, or even just point me in the right direction to do my own research.

EDIT:
Thanks for your responses so far.
Here's some more info on our gas system: we buy our $\ce{CO2}$ in large bottles at ~400 psi. We then feed it from outside through a pressure regulator that regulates it down to 2 bar (~29 psi) absolute pressure. All the pipes until this point are stainless steel.
It's then fed through a proportional valve and the pressure is held at 1030 mBar over a set of flow and gas sensors, then exhausted to atmosphere. The pipes in this section are a mix of Polyurethane tubes, and two proprietary rubber-like materials called Tygon and Viton tubing. These are held together with nylon fittings. All of this is at ambient temperature.  In these pipes, only $\ce{N2}$, $\ce{CO2}$, and $\ce{CH4}$ flow.
Here are some graphs I collected of the system:

In this graph, the green & brown lines are each gas being flowed in the system. The two blueish lines are a rough analog of how much $\ce{CO2}$ is detected in the system. As you can see, even after 5 mins of methane and N2 purging, and continuous intermittent $\ce{N2}$ purges, $\ce{CO2}$ still creeps back into the system when no gas is flowing.
Here's a Graph of Gas Flow

The [Flow In] sensor is just after the 1030 Prop valve, and the [Flow Out] sensor is just before the exhaust. They are both identical Wheatstone bridge thermal flow sensors.
As you can see, there's a marked increase in flow loss (Flow In/Flow Out) from 2% to about 7% when CO2 is flowing, and some very strange transient spikes when I switch gasses to/from $\ce{CO2}$.
That's most of the data I've collected. Anecdotally, I've also noticed that the $\ce{CO2}$ valve is significantly harder to turn than the $\ce{N2}$ or $\ce{CH4}$ valves.
I hope this helps with the problem, and thank you again for your help so far.

There are some cross-country CO2 pipelines ( like 18" diameter). They pose unusual fracture concerns for the steel pipe. While following the technology , I never encountered "sticky" considerations. Operating pressures are above the critical point.

@Andrew I have added some more supporting information. Let me know what you think.

$\ce{CO2}$ solubility in water is about 100 times higher than other gasses. So if there is any moisture on the walls, $\ce{CO2}$ is much more likely to stay there.
$\ce{CO2}$ is also much more chemically active, forming lots of compounds through its carbonic acid. Some of these compounds are reversible, can absorb and emit $\ce{CO2}$ as its partial pressure or temperature changes.
There is likely water or mineral contamination in your system.

Indeed, unless you can pump out the lines to less than 1E-6 Torr, you have water adsorbed on the walls of your piping.

@JonCuster baking the lines helps too, but I suspect the OP's system isn't set up for that and it would be harder than getting a turbopump on there.  The materials may not be suitable for proper heating anyway.  Even with a pump it would be much quicker with some heat.

Also with a vacuum pump and voids separated by long runs, a process of pump out, N2 purge, pump out etc. can be more effective than simple pumping (He purge in my case as N2 is one thing I'm trying to get rid of to avoid it freezing).  And optimising changeover processes so ambient air never gets in

@ChrisH - Pump/purge cycles work wonders too... So many tricks of the trade out there, and so few places to learn them anymore... I just queried an editor at the Journal of Vacuum Science and Technology if they would be interested in a vacuum technique article based on some work we did on a project. They did say OK, but that there was no identified area editor for those anymore...

@JonCuster a varied engineering/science career has left me well placed to learn on the job - just as well as there's no other way.  But we need the time and space to fiddle about without causing big problems and that  opportunity seems hard to find these days

@JonCuster Thanks for the info, looking at my data again water adsorption does seem to fit. Do you have any idea how the water might have gotten into the system in the first place? Also, as you suspected I'm not confident my system is suitable for either vacuum pumping or heating. I have flexible pipes which might collapse in under a vacuum, and a lot of my pipework is intertwined with non heat-proof components. Do you have any advice on what techniques might work here, or where i might look for this kind of information?

@AlexanderJohnson circulate a gas over dessicant to remove the moisture if heating and vacuum are not an option. No need to purge the gas during this, if dessicant absorbs the water, circulation is enough. Once water is absorbed from the system, make sure feed gas is dry in the future. Also could separste CO2 and other gasses as much as possible, for example by using separate dessicant chambers for them. When water is removed CO2 will behave more similsrly to other gasses and less purging will be needed.

@AlexanderJohnson you might want to read a little about vacuum systems and what's required to get to ultrahigh vacuum. Any time a surface is exposed to atmosphere with water in it, the water will stick to the walls, and slowly outgas from them until it reaches equilibrium and the partial pressure of water in the gas is at the vapour pressure of water at the relevant temperature. Especially bad is if the system is below the dew point when this happens, then you get mass condensation of water, so there's oodles of water to outgas even if you start flushing the vapour out somehow

@AlexanderJohnson (I'm not saying you need to get to ultrahigh vacuum, I'm just saying the principles might help you.) Does the CO2 you're getting come with a purity specification for water and other contaminants? Simply flushing the system by flowing through a dry gas (nitrogen is usually cheapest I think) for a while after any time it's exposed to atmosphere might help, and this will work better at higher temperatures even if it's only by 10 or 20 K (look at a curve of H2O vapour pressure vs temperature)


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Are intermediate bulk container units suitable for storage of isopropyl alcohol?
I'm working in water treatment construction and my boss is going to be stocking isopropyl alcohol (IPA).
He wants to store the chemical in bulk, with amounts ranging from 5000 L to 10000 L in IBC units.
Is storage of IPA in plastic containers okay? Will the chemical react with plastic?

What type of plastic and storing conditions (most importantly, temperature, exposure to light and ambient polluting factors)? Caged IBC units are typically HDPE and should be suitable for IPA, but you might want to double-check with the supplier just in case they use something else or have additives/stabilizers in IPA.

You should have a discussion with your friendly local fire protection engineer. They may have issues with 10000 liters of isopropyl alcohol suddenly appearing on site.

Is there any legislation that can back this up?

@ImranMunir - legislation on what, exactly? Check with the IPA supplier for what containers they can ship in. Check with fire protection for how those containers need to be stored so your facility doesn't burn down too readily.

The volume makes this a non-trivial question like "what plastic resists IPA". As commented , there are fire protection concerns, maybe a berm will be required. Likely the answer is steel, possible with internal coating. There are strength considerations handled by steel; I think it will be difficult to find creep/long term strain, allowable stress values for a plastic vessel. When designing for low pressure plastic (FRP) water pipe, my company had to do extensive testing to determine allowable strength. The manufacturers had only estimated at strength levels; not adequate for storing significant amounts of flammable liquid.

If plastic is PE it will be fine. Most large plastic containers for chemicals are PE.
PP will also be fine, but it is usually for rigid boxes.
an article with a solvent compatibility table , because people here dont like images

I suppose you forgot to cite the source of the data, which is also a bit too excessive for what OP is about (single solvent).

Citing is not putting in an obfuscated link of where you found an image. It is giving credit to those who have produced this content. This is at least, but not limited to, including a clear text link of the source and authors where available. Apart from this, please make sure that you are allowed to use the image on [chemistry.se].

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


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Possible Plumbing Acid through Leaky Pipe in Floating Ceiling, How to Test/Clean?
So I had a plumber in the other day, and I don't know if he used acid drain cleaner or not. The pipe where the possible leak is, is over some drop ceiling (vinyl or expanded polystyrene).
Before I go probe around and clean it, how do I test if the liquid is indeed an acid? Is it safe to handle, given it may have been there for about a week? I know for sure:

There was no smell associated with the plumbing on that day.
There is still a single, small wet spot on drop ceiling tile.
There is no sign of a hole being corroded through the ceiling tile.

In short, how do I handle this clean up?

Do you mean there is a leak in the plumbing ? If so, I would not suspect material used to clean pipe. I would look to find the source of the leak.

Is the pipe metallic or made of some plastic ? A simple way of checking the acidity of liquid is using sodium bicarbonate $\ce{NaHCO3}$, this white powder will react and make bubbles, foam or moss if some acidity is present.

It's highly unlikely a chemical drain cleaner was acid; if one was used, it's far more likely to have been a strong base. That said, the best thing to do is:

Call the plumber and ask what was used and how to clean it up.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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When balancing equations, how do we handle chemicals in different states of matter?
Balancing chemical equations such as $$\mathrm{H_2 + O_2 \to H_2O}$$ or $$\mathrm{NaHCO_3 + H_3C_6H_5O_7 \to Na_3C_6H_5O_7 + H_2O + CO_2}$$ seems to be easy enough by finding the coefficients of each chemical. However, I have no idea when the states of the chemicals are introduced. For example, I can balance the equation $$\mathrm{Mg_3N_2 + H_2O \to NH_3 + Mg(OH)_2}$$ but I don't know how to balance it if the states are introduced such that it becomes $$\mathrm{Mg_3N_2 (s) + H_2O (l) \to NH_3 (g) + Mg(OH)_2 (s)}.$$

Is balancing equations without the states the same as balancing equations with states?

Ignore the states. They do not enter into balancing. Only mass balance counts i.e., the number of atoms on the left must equal the number of atoms on the right of the same element.

@M.Farooq Does this mean that the states are included for clarity, but does not really affect the balancing?

The matter state does not matter. It is the atom inventory that matters. Atoms do not transmute with change of matter state. There is observed no nuclear explosion when water freezes or ice melts.


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Water vapor permeation to the reaction flask
I am conducting a synthesis involving Buchwald-Hartwig coupling. The scale is $\approx10^{-4}$ mols of reagents. After running the reaction at 80 ℃ during 24 hours I found that sealing between the 3-neck reaction flask and the reflux condensor became loose, and the reaction flask is somewhat filled up relatively to the start of the reaction (5 mL toluene as solvent) although I did not add additional solvent during the reaction. As I add some toluene, the layer separation was observed and the toluene layer went to the upper layer. Initially I thought that this might be tert-butanol because it is the by-product of the reaction cycle. But I initially added only 0.0426 g of NaOtBu, so formed tBuOH would be very small amounts. In light of the fact that the sealing became loose, is there possibility that water vapor in the air went through the flask and condensed by the reflux condensor, and dropped to the flask? Is there any method to verify whether it is water or not? Ar gas was continuously flowed into the flask during the reaction and an oil bubbler was used.

------------ 2022/02/18 edited
My hypothesis is that some water came through the flask from exterior. So I tested some experiments. In this time I added only 10 mL of toluene in the flask and set up the instruments in the same manner. Considering the fact that sealing between the flask and reflux condensor became loose in the previous experiment, I didn't seal that portion intentionally. Below is ordinary instruments setting.

In the real experiment the portion between the three-necked flask and the reflux condensor is sealed, and the other portions are not sealed. Also the flask is wrapped with Al foils to prevent potential effects of the light. But these procedures were omitted in this time. The result is layer separation again:

Based on this result it seems obvious that some water came through the flask. But from where? Is the sealing between the flask and reflux condensor problem? Actually I had same problem with same instruments setting in the past, but the sealing was maintained during the reaction at that time. Then other unsealed portion, for example, the portion between reflux condensor and oil bubbler is the problem? Also I blocked the other open end of the bubbler with just pipette bulb, but can this be a problem too?

Seems unlikely, the Ar flow should have given a flow out through the gap stopping any air getting in. Is it possible your bubbler sucked back into the reaction mixture at some point?

@Waylander At a glance there seemed to be no change in the level of the oil in the bubbler. And I used silicon oil which is very viscous, but this unknown layer is not viscous. I added the image. I only used 5 mL of toluene at the beginning, but about 20 mL of unknown liquid is newly formed.

What does analysis of the toluene layer show? Has your reaction worked?

@Waylander The MS analysis request is waiting for approval. But from personal experience the color of toluene always turned brown after the reaction, so I presume that the product is formed. However if this unknown layer is formed again in the next synthesis, I think that this layer formation would be prevented. (Actually there was a similar experience in the past) I think the volume of the new layer is too large compared to my reaction scale, so it is hard to say that it is formed from the reaction cycle. So I assume that it may be from the exterior of the reaction.

On that small scale could you run the reaction in a sealed 20ml microwave reaction vial?

@Waylander There are no microwave reactor in the lab but I would consider to buy new equipment. Anyway the source of the layer is so mysterious...

Does not need to be in a microwave reactor, just stir and heat the sealed tube in an oil bath. I have done this many times.

@Waylander Oh, 'microwave reaction' doesn't mean that I need a instrument that irradiate microwave? Actually this synthesis was conducted via the similar method you said, the sealed three-necked flask heated with oil bath.

@Waylander I added some texts. I would appreciate if you could check it.

Is your gas supply dry? If the reaction is run in toluene at 80C then you do not need a water jacket condensor,so you can remove one potential source of water.

I found that inside the reflux condensor water drops are formed continuously and very slowly. Maybe there is very minute crack. Everyone be careful with apparatus...


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How do I calculate the maximum temperature of an exothermic NaOH reaction in water?
As the title suggests, let's say I want to make a 1 L 14M NaOH solution. For this I'd need:

(14 mol/ L) x 1 L x (39.99 g / mol ) = 559.86 g NaOH(s) {solute}
~1 L H2O (approx) {solvent}

The dissolution of NaOH is exothermic and the solution gets VERY hot (I physically measured over 70°C+). I want to predict the maximum and transient temperatures of the chemical system. Assuming the standard enthalpy of formation at 298.15 K / 25°C (ΔfH) for the following species:

ΔfH_NaOH(s) = -425.6 kJ/mol
ΔfH_Na+(aq) = -240.1 kJ/mol
ΔfH_OH-(aq) = -230 kJ/mol

The total enthalpy/heat of reaction via Hess' Law is:
-(-425.6 kJ/mol) + (-240.1 kJ/mol) + (-230 kJ/mol) = -44.5 kJ/mol
Using that number along with the following thermodynamic properties:

Molar Heat Capacity of NaOH(s) = 59.5 J/mol·K
(If necessary assume a standard 1L glass beaker with a thermal conductivity = 1 W/m·K)
Initial ambient temperature of 298 K.

How can I determine the max temperature that the solution will reach? I've tried doing calculations myself with these constants but I get answers with units [mol·K] via:
EDIT 1:
559.86 g NaOH x (1 mol NaOH / 39.99 g NaOH) x (-44,500 J / 1 mol NaOH) x (1 mol NaOH·K / 59.5 J)
= -10,471 mol NaOH·K = -10744.15 mol NaOH·°C
...which doesn't make sense as the answer should just be in temperature i.e. [K]. The number I calculated is also ridiculously high and obviously wrong. What am I missing?

How do you get a number in mol×K? Need to see the calculations.

@OscarLanzi refer to the newest edit cheers. I know the answer will likely involve some sort of integration for the transient/dynamic temperature but I'm not really sure how...

The heat capacity of water is different from the heat capacity of the NaOH solution. So the same amount of Joules will produce a different temperature increase in the beginning and at the end of the dissolution. So you would have to know the formula describing how the heat capacity of the solution changes with the concentration. This is not an easy task.

@Maurice thanks for the comment. Could you please elaborate on this, maybe even in a reply? I'm very determined to get to the bottom of this. I originally thought this sort of problem would be easily solvable especially in industry where I'd imagine scientists/engineers (etc.) would need to figure out the right thermo-resistant materials for building a reactor to avoid overheating (etc.).

Note that 14 M NaOH means 14 mol of NaOH (about 560 g) in 1 L of the solution, not to be added into 1 L of water. Just to be sure you are aware of it. // There is big problem of combination of 2 experimentally determined nonlinear quantities: ratio dependent dissolution enthalpy and ratio dependent specific heat.

@Hendrix 13. I am not familiar with this problem, and I don't know whether a formula  exists for describing how the specific heat of the solution changes with increasing $\ce{NaOH}$ concentration. Anyway you should also take into account the heat capacity of the container. if I were you, I would not make any difficult calculations. I would just try and measure the temperature in the container.

@Maurice I've measured it in the beaker and found it to go over 60°C+. I want to upscale from beaker to batch reactor which means I'll have a much higher NaOH input feedstock and higher max temperature. Are you suggesting that if there's no numerical model way to estimate the max temperature of the reactor with a given NaOH mass I'm better off just conservatively choosing thermo-resistant materials/metals for the reactor?

Yes. I am suggesting that no numerical model will estimate the max temperature of the reactor, and that the knowledge of the molar heat capacity of NaOH is useless. The heat capacity of NaOH ($1.5 J g^{-1}K^{-1}$) is nearly negligible with respect to the heat capacity of water ($4.18 J g^{-1} K^{-1}$} and aqueous solutions (unknown). You have not mentioned the mass of NaOH and water to be used. Is it $1 g$ ? $1 kg$? $100 kg$ ? Less ? More ?

don't forget the $\ce{HO- + H+ -> H2O}$ reaction, especially if your solution is buffered.

Yes for the reactor it'd be around say 50kg. Regardless it won't matter if there's no numerical solution or approximation. This is very surprising and very unexpected that there's no solution here.

I don't say there is no numerical solution. I only state the I don't know whether there is such a formula.

Preparation of 1 kg of solution is unusual. It is not possible to prepare 1 kg of 14 M NaOH without the knowledge of 14 M NaOH density, which could be found or interpolated in tabelated data. The easiest way is to prepare 1 L of 14 M solution and then to take 1 kg of it.

@Hendrix13 Chemical engineers may have some empirical formula or charts for that, but I have not found any. As the integral dissolution enthalpy is empirical data dependent on concentration. Eventually tabelated or calculated data is for infinite dilution.

The easiest way is to measure temperature by test dissolution, then to measure or estimate solution heat capacity and heat capacity of the vessel where is it would be prepared.

@Poutnik thanks for the great reply! That last comment seems to be the way to go I think at this point as I'm also asking some professors and industry contacts. Could you please expand upon your last comment in a reply so I can understand it easier? At this point I suspect it'd be the accepted answer.

Comments are not for extended discussion; this conversation has been moved to chat.

As someone with bachelors in both chemistry and chemical engineering, I can tell you that this sort calculation was not focused on in my chemistry courses, but is an important and standard exercise in multiple chemical engineering courses. If you search for "Adiabatic temperature rise", you can find calculation methodologies at varying levels of approximation.

This is very useful! If you did more work on the industrial 48-50% to 25% operation I'd like to discuss with you or read your final report. On the plant we use a fin fan to cool it after mixing, but winter can be challenging in terms of freezing of caustic since this is also an interesting non linear curve and should be considered for design of applications.

Hi @MorneBooysen, yes freezing caustic was a pain for us too. I'm unable to message you on here, but feel free to +1 the question if it helps and message me on linkedin at Ramon Skane (Perth, Australia) :)

After talking to a very experienced industrial chemist about this he helped me reach a very interesting answer which I believe is correct. I've provided good detail below to help anyone else with similar questions.
1. The Theory/Logic:
The first pointer he mentioned is that the molar heat of solution of NaOH in water does not hold true at NaOH concentrations approaching 14M NaOH, this is because as NaOH (s) dissolves it can form NaOH.nH2O as well as Na+ and OH- with (n = 1,2,,3.5,4...7 (etc.)) all depending on the concentration of NaOH. Refer below to page 11 of the Dow Caustic Manual:

This effect really kicks in when [NaOH] reaches > 20%wt and results in less temperature rise than expected from calculations when [NaOH] > 20wt%.
As an example experiment form literature validating this, refer to the graphic below from "Experimental Study of the Thermal Effect of the Dissolution Reaction for some Alkalis and Salts with Natural Mixing an Forced Stirring" below showing starting temperatures of Starting temperature 10°C and ending temperature close to 80°C thus a 70°C increase making 38%w/w NaOH. This is much closer to what I've seen in the lab when I've done it.

Combining all this theory, I made the following graph below which outlines the change in molar heat dissolution and hydrate formation with changing [NaOH]. It can be seen that the actual molar heat of dissolution I should be using is not -44.5 kJ/mol but really around -30 kJ/mol.

2. The Calculation/Application:
Along with the above assumptions, also assume that:

Specific Heat Capacity of Water (Cp,H2O(l)) = 4.184 kJ/kg°C
Solute Mass = (14 mol/ L) x 1 L x (39.99 g / mol ) = 559.86 g NaOH(s)
Assume 14M NaOH density is 1460 g/L
Solvent Mass = (1460 - 559.86)g = 900.14 g (i.e. NaOH is 38.3wt%)

Dissolving 559.86g NaOH (for 14 M) in 1L of water produces an enthalpy change (∆H) equivalent to -30kJ/mol x 14mol = - 420 kJ.
Taking the equation ∆H = mCp∆T and rearranging to solve for
∆T = ∆H/(m x Cp) = -420 kJ / (1.46 kg x 4.184 kJ/kg) = 68°C increase
Therefore, Tf = Ti + ∆T = (24.85 +68)°C = 93°C
This value is much closer to the value obtained in the experiment graphic above as well as what I originally measured in the lab.
I would suspect that if I made the 14M NaOH solution in a 1L pyrex beaker with open top, losses from the conduction of heat into beaker, radiation from beaker walls and conduction loses from liquid surface would keep the liquid closer to 70-80°C.
If the 14M was made in a larger quantity e.g. 100L in a 200L capacity drum, I would expect that the true 68°C increase in temperature over ambient to be realized as losses would be small relative to the amount of heat generated. Also, I acknowledge that a more accurate result would come about by measuring the actual density of 14M NaOH and using in calculations.
Many plastics are not capable of holding liquids at 93°C especially if their density is 1.5kg/L so, as another inferential answer to my question, using a metal container is certainly preferable.
EDIT1 & UPDATE:
Just thought I'd add the extrapolated curves I got based on these calculations by changing concentration.
The graph below includes a tail at the end after a smoothing of data series, I believe this is from the 6th degree polynomial I used to extrapolate the heat of dissolution trend where R^2 =/= 1. So for concentrations below, say 19, with this method seem more accurate. Assumes an overall 95% heat transfer efficiency.

Adding more practicality to this (which is what I'm aiming for with my research), is the graph below used with industrial "50% caustic" (my product's spec actually had a 49.9% alkalinity) which can be diluted with water to achieve various concentrations. Here instead I'm assuming 100% efficiency.

Hope this makes sense and is useful to others as it has been for me!

Why do you insist on dissolving in 1 L water when you just determined the water mass 900 g? Another mistake is considering the solution has the specific heat capacity of water while significant part of it is not water. //  Note that the molar ratio=f(conc %) can be formulated as the exact function. // See meta.stackexchange.com: new feature table support for easy syntax to format the text into a table, which is preferred to screen captures or scanned images.

I suggest using some inert object (e.g. from stainless steel or glass) and measure its heat capacity in a water bath with different initial water and object temperatures. Then measure the heat capacity and  calculate the specific heat capacity of the target 14 M NaOH  by doing the same with this solution. // Be aware the higher volumes mean higher peak temperature due slower heat dissipation due lower surface/mass ratio.

I am sorry your "answer" from March 14 is not easy to understand. For example, your first table gives values of $\Delta H$ for the dissolution of 1 mole NaOH in $n$ mol water. But this $\Delta H$ values is useless it the heat capacity of the solution is not given. And this heat capacity is changing during the dissolution process. Why not give the maximum  temperature change instead of the $\Delta H$ values ? Second you do not explain the difference between natural mixing and forced stirring. Natural mixing is probably done at the bottom of a beaker. But the result will not be homogeneous.

Note that the first table and figure are from the Dow Caustic manual. It's from their experiments I've just used some of their results to build correlations.


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In electrolysis, can conductivity be used to predict current?
Since conductivity is the reciprocal of resistivity, it can be used to calculate resistance in the following equation:
$$R=\frac{\rho l}{A}$$
$$R=\frac{l}{\kappa A}$$
Could this, along with Ohm's Law be used to predict the current in an electrolytic cell at a given voltage? My initial calculations with conductivities of around $20000\mu S/cm$ and a voltage of 10V yielded me an abnormally high resistance and current less than 1 amp. My actual experiment with 10V had a current of about 2.5A, which means the resistance of the solution is about 4, which doesn't seem right?
So I wanted to ask if this just doesn't work for electrolytic cells or am I doing something wrong?

An electrolytic cell  in DC scenario is in no means similar to a classical electronic resistor. While its conductivity does have influence of the passing current, it is far from a simple function. A model circuit simulating such a cell would be quite complicated. The most simple one is a power source in a serie with  a resistance. (Both are current and current history dependent.)

BTW, $20000\ \mu \pu{S}$ is not conductivity but conductance.

Sort of.
First it is important to mention that electrolysis uses up some voltage, regardless of current. So if you use 10 V and electrolysis of your reaction requires 2 V, then only 8 V will be available for the resistance.
Second is that your measuring device is affected by it too, and those usually use low voltage, where this effect is especially noticeable.
To measure the resistance of your electrolysis cell, you are better off using two measurments. One with a high voltage, low current, and a large value resistor. This way voltage drop in the electrolysis cell will be seen alone. And another one with high current, low voltage. This together with the previous result will help to get the resistance, when voltage drop from the previos result is subtracted from the available voltage in the second case.
And at higher current you may encounter gas bubbles that increase resistivity, and at higher still - plasma bubbles, that will decrease resistivity.

Thank you for the answer, that cleared up a lot, but just another question on how electrolysis uses up current, I just wanted to make sure whether or not I also have the subtract the voltage required for unwanted reactions during electrolysis from the applied voltage?

@AlanLiu amount of electrolysis reaction is directly proportional to current * time. All of the added voltage above the treshold is wasted on speeding up the reaction. If you want to predict how current will change as you change voltage, yes, subtract reaction voltage from the supply voltage.


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Sodium dithionite reduction of nitro group to amine
I've recently come across the synthesis of 3-aminopthalhydrazine from 3-nitropthalhydrazine where the reducing agent used was sodium dithionite ($\ce{Na2S2O4}$). I wanted to know what the mechanism of this reaction was but I'm unable find any mechanism online. Can anyone help me with that?


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How to find out amount of carbon in metal
I am an student of archaeology and on my final work I am working with mining hammers, on which metallography was used. But my supervisor want to use some kind of acid on the surface of the metal so it would be visible difference between parts witch higher amount of carbon in steel and in those that don’t have any or low amount. Is it even possible? Or if it would help, would it be visible just under microscope. Thans for any answers.

Steel and iron have different etch rates in different acids. Look up which ones. If you have a metallography lab around, they likely have several books with different etchant recipes.

@JonCuster Cool. I would've gone total apparative overkill and said you need to map the surface with XRD. ;) Crazy how used we get to just having equipment for 100s of k€.

@Karl - or EBSD with Auger mapping on top. Don’t mess around…

Have a look at Damascus steel (https://en.wikipedia.org/wiki/Damascus_steel) and procedures of etching.

By "metal" I assume you mean iron in some form. The most common economical method is combustion of a small sample; nearly all carbon in iron analysis is done with this method. There are other methods like non dispersive x-ray fluorescence in an election microscope, but it will be costly. Metallographic examination using, for example standard nital etch ( ethanol with little nitric acid), can show an experienced person a great deal about the steel/cast iron. But the general color would be relatively useless in determining the carbon content. Without some information regarding what "mining hammers" you are looking at, it would be difficult to guess what they are. For example, light and dark areas could indicate heat-treatment and have no connection with carbon content.

If you can spare a few grams of sample , commercial labs will offer a package deal on 6 or 7 standard element analysis on steel.  I don't know the price of a good deal today . It will include C,S,P, Si, Mn, Cr, Ni, Mo, maybe Cu. Results can be helpful for evaluating the steel. I am not sure about cast iron. If you can't spare material, a university should have a scanning electro microscope with EDAX , probably in physics department ; of course I have no clue about access. ( Every one should have an SEM with EDAX, they are impressive.)

Yes, I know, we have already use an election microscope, but I am trying to find other methods that would be visible be human eyes, cause we have already spent lot of money on election microscope and results are really weir and not really meaningful. The point of my work is to find out  the most preferable method medieval smiths, have taken care of the working parts of the hammers. So when it would be visible just by human eyes it would be best for me, cause I would be able to do this on more examples and it would not be so expensive as election microscope.

It sounds like you need a metallurgist.  The nature of inclusions ( silicates, oxides, sulfides) will tell a lot about the fabrication. The carbon can be completely changed by heat-treatment so it masks fabrication. Carbon could indicate if a "Damascus" type process was used.  A light microscope with modest magnifications such as 40 to 100X will be most informative. I am thinking of a traditionally polished sample , unetched , then etched with 2% nital., also that it is a steel rather than cast iron.


text:
Ellingham diagram/graph for tungsten
I've recently come across one of the methods to form tungsten carbide($\ce{WC}$) from wolframite ($\ce{(Fe,Mn)WO4}$) by using carbothermic reduction which was supposedly one of the first methods for extracting tungsten from its ores developed in 1783. But I wanted to know the particular temperature at which the reduction occurs, so I've searched online for the Ellingham diagram of wolframite but I've found no sources for that. There were Ellingham diagrams for many metals except for wolfram. I just want to know if anyone can cite a place where I can find it?

Wendel [(1)] studied the oxidation of tungsten and produced this diagram for the free energy of formation of various tungsten oxides:

Each of the lines represents free energy per mole of $\ce{O2}$, and $\ce{WO3}$ appears lower than any lower oxide (red, lowest line in the graph). Therefore $\ce{WO3}$ is the only thermodynamically stable oxide and the only one that would appear in an Ellingham diagram. Fitting that line onto the diagram we find that it crosses the $\ce{C/CO}$ line at about 1200°C.
This equilibrium termperature of 1200°C assumes the carbon monoxide is generated at one atmosphere partial pressure. The reaction could take place at somwhat lower temperature using a lower $\ce{CO}$ pressure. In practice, the reduction conditions are more likely driven by kinetics instead of thermodynamics. Wang et al.[2] report that carbothermic reduction at 1100°C gives a mixture of tungsten metal and carbides, which are then purified to ultrafine $\ce{WC}$ By reacting with more carbon at 1200°C:

In the current study, ultrafine and high-purity tungsten carbide (WC) powders are successfully prepared by a two-step process: carbothermic reduction of WO3 followed by carbonization reaction. The effects of the C/WO3 molar ratio, reaction temperature and reaction time on the phase transition and morphology evolution of the products are investigated. During the carbothermic reduction process, all the oxygen in yellow tungsten trioxide (WO3) is removed by carbon to generate a mixture of W, W2C and WC at 1100 °C; and then the as-prepared powder is mixed with an appropriate content of carbon black and carbonized at 1200 °C. The carbon content in the finally obtained WC powders is almost equal to the theoretical value. Furthermore, it is found that a high C/WO3 molar ratio at the first stage is beneficial for decreasing the particle size of the WC powders. When the C/WO3 molar ratio is 3.5, the single phase WC with a particle size of about 200 nm can be obtained. Therefore, this carbothermic reduction–carburization process may provide a simple, low-cost, and high efficiency route to prepare the WC powders in a large-scale.

References
1.
J. Wendel. Thermodynamics and Kinetics of TungstenbOxidation and Tungsten Oxide Sublimation in the Temperature Interval 200°–1100°C.  MS thesis submitted to Lund University (Sweden). (Lund, Sweden: European Spallation Source ESS AB, 2014), p. 14.
2.
Kai-Fei Wang, Guo-Dong Sun, Yue-Dong Wu, Guo-Hua Zhang (2019).
"Fabrication of ultrafine and high-purity tungsten carbide powders via a carbothermic reduction–carburization process".
Journal of Alloys and Compounds 784, 362-369,
ISSN 0925-8388,
https://doi.org/10.1016/j.jallcom.2019.01.055.


text:
I spilled some .1M Copper Sulfate on my sweater
I have accidentally spilled some .1M solution of Copper Sulfate (CuSO4) on my white cotton sweater. It left a light blue stain and I’m wondering how I might successfully remove the stain from the sweater. Any help would be useful. My immediate thought is bleach, and since the sweater is white, it wouldn’t do anything to the color; but I don’t know for sure if that will work.

Water will do. Bleach won't.

You could also try a weak acid, such as vinegar. Rinse,repeat. Bleach is alkaline, and might cause a precipitate.

The sweater took a long shower and the Copper Sulfate stain is gone. Did take bloody long though. Thank y’all for your help.


text:
Can a carbon that has 5 bonds be explained by saying that one of its bonds is a coordination bond?
In my organometallic chemistry class at the university, the doctor told us that compound μ³-CO-[Fe₄(CO)₁₂]²⁻ had the structure that appears in the image. As you can see, there is a pentavalent carbon.

When we asked her, the doctor told us that it was because one of the iron atoms donated electrons to carbon through a coordination bond and that carbon was still tetravalent.
I am not sure if this is the correct explanation, if I misunderstood the doctor or if my chemistry knowledge is worse than I thought and I haven't done enough research.
In any case, is it possible that this is the structure?

Useful links for text and formula formatting (not to be applied to titles):
Notation basics ,  Formatting of math/chem expressions  and
upright vs italic

$\ce{[Fe4(CO)13]^2-}$ is a canonical example among metal carbonyl clusters that has been synthesized by Hieber in 1930 and the structure and bonding have been determined and discussed in numerous publications later. https://doi.org/10.1021/ja00973a018 gives a good start.

Try to write chemical formulas ( or equations) like  μ³-CO-[Fe₄(CO)₁₂]²⁻ rather by MathJax as  $\ce{\mu^3-CO-[Fe4(CO)12]^2-}$ displayed inline as $\ce{\mu^3-CO-[Fe4(CO)12]^2-}$ or  with double dollars to  display it separately as $$\ce{\mu^3-CO-[Fe4(CO)12]^2-}$$

Heres a related discussion on hypervalent carbon: https://chemistry.stackexchange.com/questions/4027/carbon-with-5-bonds/4029

@Poutnik Ironically, μ³-CO-[Fe₄(CO)₁₂]²⁻ is better than $\ce{\mu^3-CO-[Fe4(CO)12]^2-}$ because the letter mu should be upright. Both formulas aren't IUPAC-complaint at least because the number of metal centers linked by the bridging atom should be denoted with right subscript (i.e. μ₃-, not μ³-). The superscript is used with eta notation for hapticity. And "CO-" part is equally misleading in both formulas.

@andselisk  I am aware of the issue of missing uprightness of MathJax Greek alphabet, but the comment was targeted on general formulas, focusing in the first step on rules, with exceptions involved in next steps. Nor was I addressing IUPAC recommendation, just MJ based formatting. Being away from academic life for many years, I admit my IUPAC related knowledge is rusty so I generally avoid any IUPAC related answers/comments unless I read such info.

If you have only ever heard of Lewis diagrams,* then you might misinterpret the shown representation.
Instead of bonding the focus of this diagram is more on structure, even though it compromises on that when showing a double bond in the carbonyl moiety, and to show that this particular carbon has four closest neighbours.
You can almost instantly spot this as lone pairs are omitted completely.
I do not want to dive deep into the bonding of this complex, it is far better to consult the literature for this, but one way of looking at is to look at the ligand first.
A Lewis diagram of carbon monoxide is $\ce{^\ominus|C#O|^\oplus}$, which emphasises on the triple bond nature of the compound. There are other resonance structures. This molecule is very versatile and has many bond modi.  One important feature is that it is a σ-donor through carbon.
Carbon monoxide bond in μ3-fashion to the iron cluster. This is a highly delocalised bond; a multi-centre-two-electron bond in first order approximation. The important take-away is that carbon still only utilises eight electrons for bonding.
In a more bonding complete diagram, there should be a lone pair and a positive formal charge at oxygen, a triple bond between oxygen and carbon, and a single line pointing to the middle of the vertex established by the iron cluster. Obviously you'll sacrifice some structural information. But diagrams do that anyway.
One more important point is, that you should not confuse valence with coordination. It is better to avoid valence all together, as its use has been and will remain inconsistent at best.
According to the IUPAC definition,† carbon is tetravalent. Always. It is a constant specific to this element.
Footnotes
* In a strict interpretation of a Lewis diagram, all electrons are shown and all lines represent two bonds. There is no difference whether these bonds covalent, ionic, or anything in between, there is no discrimination "from where the bond originates". There are no arrows in true Lewis diagrams.
† Valence in the Gold Book: https://doi.org/10.1351/goldbook.V06588


text:
I want to dechlorinate my tap water to use in an organic garden. I've purchased a product called Seachem Prime that is typically used for aquariums
I purchased a product called a seachem prime to dechlorinate my tap water so I can use it in my organic garden inside. I want to know if the product itself is safe to use on my plants and if there would be any adverse effects to the plant's ability to uptake nutrients. It's used to break the chemical bond of the chloramine and turn the nitrite and nitrate into "nontoxic" forms. Plants can utilize nitrate so I don't want to inhibit the health of the plant. I currently dose at 4 drops per 2 gallons of tap water. Is there any issue with using this method to dechlorinate and water/feed my plants?

See also a related question I asked a few weeks ago: reaction mechanism - Does hydrogen peroxide remove chloramine from water? - Chemistry Stack Exchange.

Are you sure your water is treated with chloramine? If the community just uses chlorine, then simply let the water stand a few days in a wide, open container.

For organic gardens, it makes more sense to let water dechlorinate itself spontaneously by chlorine escape than contaminate it by dechlorination products. They would accumulate in soil as water evaporates, while in aquarium, water is exchanged and chemicals are diluted.

Collect rainwater?

@Karl Rainwater is always the best option, if available. But it may be problematic for   cases of small gardens in an urban a/o low precipitation area.

Please try to find somewhere the composition of your product. It should be printed in small letters somewhere on the label. Without this information, we cannot help too much.

@Poutnik Well, small garden --> not much water needed. This dechlorination idea here only makes sense if you intend to do it for years. At that point, an invest into a reasonably sized cistern will pay off very soon. The dryer the region, the more expensive the water, I guess.

After quick search, it seems the Seachem prime is far from to be simply just a dechlorinator. Composion is not disclosed but I would be careful to use for your purpose. // It can be e.g. a hanging garden on am upper floor balcony, with difficult logistics to collect rain water. In a flat area like backyards of houses, I do agree to manage a rain collector would pay off, e.g. collecting rain from a house roof.

It is not just that a compound can remove chlorine/HOCl, it is the immediate products that can be powerful radicals leading to problematic side products. If  you water can is plastic, the photo induced radicals especially with chlorinated water can be a problem. Interestingly, your pure spring water bottle left in a car in sunlight, in time, gives an amazing odd taste! Do you believe that if that was tap water/air in that bottle, it would be safe to give to your dog?

@Poutnik It seems that stuff is simply a https://en.wikipedia.org/wiki/Sodium_dithionite solution.

@Karl I have seen such a reference, but it was  highly unofficial.

"Turn the nitrite and nitrate into nontoxic forms" It would be nice to see links or references that describe the product and ideally the suggested mode of action, and perhaps some information about the quality of the tap water. If the product is not toxic for fish and aquarium plants it probably will not harm house plants. Note the dithionate referred to in some comments may also function as a metal chelator, which is also an important function of such aquarium water additives.

Total waste of time and money. For many decades I have kept tropical fish and never used chlorine remover ( it used to be "hypo"). I let a gallon bottle sit overnight  and never had a problem with city tap water for fish, which are much more sensitive than plants.  I formerly raised all my own vegetable and flower sets using city water directly -no problem.

Note that "city water" depends upon which city one is in.  Chlorine might disappear overnight; chloramine might take a week.

I understand other compounds are used; however ,"overnight" has worked in the 6 municipalities I  have lived.


text:
Why is the anode in a galvanic cell negative, rather than positive?
I realise there are many questions and answers on this topic, but none has answered my question.
Why is the anode in a Galvanic cell negative? It makes more sense if it were positive because it is the side where electrons, having a negative charge, leave, and when electrons leave, it leaves behind a positive charge. Meanwhile on the other side, at the cathode side, incoming electrons should make the charge there increasingly negative as more and more electrons come over.
In the answers I have seen, people ask why is it defined this way/say that it is not a matter of definition. Please explain how it is not a mater of definition.
If you use money as an example, where electrons are debt, when debt moves away from you, you enter the black or become more positive. Given the same person and same situation, if you decided to define it differently, you could say that, since you had any debt to move to begin with, you are negative. Same situation, different definition.
Similarly in diagrams comparing galvanic and electrolytic cells, electrons move away from the anode to the cathode in the electrolytic cell, making the anode positive. This makes perfect sense according to my first reasoning. However, in a galvanic cell, even though electrons still move from the anode to the cathode, it is defined as negative. One answer I saw states that,
"the charge on the physical electrode does not change... simply, the actual chemical reaction at both electrodes is reversed so what was oxidation is now reduction and what was reduction is now oxidation, hence the labels anode and cathode are changed, even though the sign is not changed." (you can see this answer here)
This further confirms to me that positive and negative are merely definitions, the labels (man-bestowed name) are changed even though the signs (an intrinsic value that is determined by the movement of electrons) are not. Where is the logic behind calling the anode negative?

It's a definition. Blame Faraday if you want, but a decision had to be made about what to call the negative end and what to call the positive end.

And blame Ben Franklin for the conventional current flow from positive to negative, and the arrow placement in the diode symbol.

@ToddMinehardt Thank you! I finally have some sort of answer because I've been trying to figure out the logic behind the naming and it's driving me nuts.

@Ihavemanyquestions  Not the origin of terminology, but a mnemotechnic hint:  There is always oxidation at anodes (Greek anabasis = journey ( of electrons along the wire) upwards ), reduction at cathodes ( katabasis - journey  downwards). ( The terms have usage span since ancient Homer and Xenophon till current meteorology, like anabatic and katabatic air vertical convection.)

@Poutnik Thank you but like I already (hopefully) implied (forgot to mention explicitly), I already know that anodes is where oxidation occurs (electrons move away from it) and cathodes reduction (electrons go to it), no matter the cell type. My question is about why anodes are negative in a galvanic cell when electrons move away from the anode and logically this should make the anode more positive. For some bizarre reason my question has been shut down and downvoted even though no other answer specifically answers this part of my question.

Think longer about anode having only oxidation and it is clear why it is more positive at electrolytic mode and more negative at galvanic mode.

@Poutnik If anode only ever has oxidation and oxidation is loss of electrons, and losing electrons makes one less negative ie more positive, then should it not be always positive regardless of whether it is galvanic or electrolytic?

Think even longer. in context of electronics, without any electrochemistry. Imagine "black box " circuitry with electrons always leaving pin A and always coming to pin B. In the case 1, it is externally powered. In the case 2, it is active power source, powering other circuit. What is mutual relative polarity of pins A and B in cases 1 and 2?

Or, look at it from point of view of work done. For the electrode with higher potential, when the work is done on the cell or on the external circuit, the direction of the current must be the opposite. Therefore for the same direction, relative potentials must be switched.

It’s not the + and - which decides anode & cathode. Rather it’s the type of reaction that’s happening i.e. always oxidation occurs at anode & reduction occurs at cathode in cells ( galvanic and more) as well as electrolysis ( such as electrolysis of water)
Take an illustration:

In the galvanic cell, at anode(-), Zn oxidises to         Zn2+ And at cathode(+), Cu2+ reduces to Cu
In electrolysis of water, at anode(+) OH(-) oxidises to O2 & at cathode(-), H+ reduces to H2


text:
Can you use Michaelis–Menten kinetics to model the ATP production rate for glycolysis and OXPHOS with glucose as a substrate?
I am an applied mathematician researching cell metabolism, but I have little  experience in biochemistry. Currently, my PI and I are considering using Michaelis–Menten kinetics to study how varying glucose availability can impact a cell's rate of ATP production. As such we are considering or substrate to be glucose and our product to be our product:
\begin{align}
    v(\mathrm{gly})&=\frac{V_\max(\mathrm{gly})\cdot [\text{glucose}]}{K_\mathrm{M}(\mathrm{gly})+[\text{glucose}]}; \tag{1}\\
    v(\mathrm{ox})&=\frac{V_\max(\mathrm{ox})\cdot [\text{glucose}]}{K_\mathrm{M}(\mathrm{ox})+[\text{glucose}]}, \tag{2}
\end{align}
where $\mathrm{gly}$ and $\mathrm{ox}$ represent glycolysis and oxidative phosphorylation (OXPHOS), respectively. I have my doubts about using Michaelis–Menten kinetics given that OXPHOS and glycolysis use more then one step and more then one enzyme.
However, I was wondering if there is a method for averaging the $K_\mathrm{M}$'s for each step for an overall estimate for $K_\mathrm{M}$ for each metabolic pathway? I appreciate any advice you can offer and other possible modeling suggestions.

Course-grained metabolic model, e. g.  http://complex.upf.es/~andreea/2006/Bib/Browning.DevelopmentMinimalCell.p.pdf

Your Km should probably not be taken literally but rather serve as a saturation parameter. I'm not sure the the rate of glycolysis depends primarily on the amount of glucose present. For example, the blood sugar levels are high after a meal, but glycolysis is not necessarily at maximum capacity. Here is some further reading for mathematicians or physicists: https://www.pnas.org/content/pnas/103/32/11868.full.pdf

ATP production from glucose can generally be modeled pretty well by saturation kinetics, but the Ox-phos part is trickier. Can you share what organism you are working with?

Thank you these are excellent reads. We are working with humans cancer cells.


text:
Simple bistable and reversible electrochromic reaction in solution
I'm not a chemist and I'm looking for a simple bistable and reversible electrochemical reaction that changes color, specifically:

A liquid in a glass with two electrodes that when a voltage is applied, the liquid changes color.
When the power is removed, the new color remains (at least for a few minutes/hours).
When an opposite voltage is applied, the liquid changes to the original color and stays like that when power is removed.
It's possible to repeat this process many times without having to replace the electrodes.

It would be ideal if the chemical components were easy to get and non-toxic. The colors don't matter at all (transparent to red, blue to green, orange to white, anything really).

You open to liquid crystals and LCDs?

Hi @ToddMinehardt! Correct me if I'm wrong, but with liquid crystals and LCDs you need continuous power to keep them in a specific state, unless some kind of bistable liquid crystal exists?

You can carry out an electrolysis of a neutral solution containing sodium sulfate $\ce{Na2SO4}$ plus some ammonium acetate $\ce{CH3COONH4}$ as buffer, and some drops of bromthymol blue. The solution is initially green. A buffer is needed, as the "neutral" $\ce{Na2SO4}$ solution is always a bit acidic, probably due to the atmospheric $\ce{CO2}$. After some time, hydrogen and oxygen are produced on the electrodes. But the cathodic zone (around the minus pole) turns basic and the solution gets blue, because of the appearance of $\ce{OH-}$ ions in solution : $$\ce{2 H2O + 2 e- -> 2 OH- + H2}$$ Simultaneously the anodic region (around the positive pole) turns acidic as $\ce{H+}$ ions are produced and the solution turns yellow, according to the equation $$\ce{2H2O ->  4 H+ + O2 + 4 e-}$$ You may separate the two regions for example with a filter paper, allowing the ions to cross it. After some time, you obtain two separated solutions, one blue, and one yellow. If you remove the filter paper, the solution turns green again by stirring. You can inverse the polarity if you do want it, and start again.

Hi @Maurice, this is the kind of reaction that I'm looking for, thanks for that! The only thing I can think of is that H2 and O2 are being produced from the solution, so eventually the liquid would evaporate completely. Covering the glass would make the pressure to rise a lot inside, unless the H2 and O2 could be combined to form water again? But I don't think this is a spontaneous reaction.

@Marcos. The amount of water to be destroyed would take rather rather long. Look. The electrolysis of $18$ g water produces $24$ liters gaseous hydrogen and $12$ liters gaseous oxygen, at usual temperature and pressure. You will have to wait many days before being able to see the level of your solution decrease.


text:
How are names for atomic orbitals with high $l$ generated?
I'm looking at Table of Real Spherical Harmonics and trying to understand how the names for the real orbitals are generated. Note that I'm a physicist. In chemistry parlance, I'm trying to come up with an algorithm to generate the names like
$$
p_x, p_y, p_z, d_{yz}, d_{x^2-y^2}, \ldots, f_{y(3x^2-y^2)}, \ldots
$$
I know that the letters go like $s, p, d, f, g, h, \ldots$ but I don't know how to generate the subscripts. It is clear that the subscript is somehow extracted by looking at the polynomial expansion of the real spherical harmonic and dropping some terms for concision, but I can't figure out the exact rules.
For example, I'm not sure why
$$
Y_{3, -3} = f_{y(3x^2-y^2)} \propto (3x^2-y^2)y
$$
but
$$
Y_{4, -3} = g_{zy^3} \propto (3x^2-y^2)yz
$$
Direct questions:

Since the Cartesian expansions are so similar why aren't the subscripts more similar?
How is the order of terms in the subscript determined? Why do we have $Y_{3, -1} = f_{yz^2}$ but $Y_{4, -3}= g_{zy^3}$?
What is a reference that tabulates these things beyond $l=4$?

Side notes: Because of this trickiness of notation I tend to prefer the $Y_{l, m}$ notation because it is unambiguous and easily generalizable to high orbitals. Unfortunately it has the downside that it only makes sense if you're familiar with the complex $Y_l^m$ orbitals and how the real orbitals are the real and imaginary parts of those complex orbitals. You do also lose some of the nice visual intuition from $p_x, p_y, p_z$, but I argue this intuition is sort of a fool's dream because it basically breaks down by the $d$ orbitals when you see something like $d_{z^2}$ or $d_{x^2-y^2}$.
Some more context: I'm interested in this because I am trying to make large tables with visualizations for the complex and real orbitals going out to high $n$, for example, at least $n=6$. Currently I've made tables and labelled them using the $Y_{lm}$ convention described here, however, I'm getting pushback and being asked to label instead using the atomic orbital form. Unfortunately this is difficult because (1) I can't just put the formulation into code (2) I don't want to hardcode each name in and (3) even if I did want to hardcode in the atomic orbital names I wouldn't know how beyond what's listed on that Wikipedia page.

My suspicion is that it's entirely arbitrary and down to the whims of whoever is tabulating these... There's a possible argument for $d_{z^2}$ in that the "correct" term, $2z^2 - x^2 - y^2$, is equal to $3z^2 - r^2$ and we just drop the spherically symmetric part of this ($r^2$) to end up with something that's $\sim z^2$. (We're already dropping the powers of $r$ in the denominator, and that's probably because they are spherically symmetric.) I don't know how much this applies to the higher orbitals, maybe you could find a pattern.

@orthocresol That's my suspicion too. The tip about ignoring the spherically symmetric part does look like it might be helpful. I'll do some more looking with that in mind. I'll add a edit with some more context about why I'm curious about this.

Here's a stab at what is going on. A note upfront: The approach I lay out here only gives the sames answers up to the $f$ orbitals as shown at Wikipedia and The Orbitron. It is just the best I could come up with. I would still greatly appreciate any other answers that can further resolve the issue.
Following the quantum mechanics convention on wikipedia, the complex spherical harmonics are given by
\begin{align}
Y_l^m(\theta, \phi) =& (-1)^m \sqrt{\frac{(2l + 1)}{4\pi}\frac{(l-m)!}{(l+m)!}}P_l^m(\cos\theta) e^{im\phi}\\
=& N_{lm}P_l^m(\cos(\theta))e^{im\phi}
\end{align}
Where $N_{lm}$ is a scalar normalization factor and $P_l^m$ is an Associated Legendgre Polynomial.
These polynomials have a closed form
\begin{align}
P_l^m(x) =& (-1)^m 2^l (1-x^2)^{m/2} \sum_{k=m}^l \frac{k!}{(k-m)!} \binom{l}{k}\binom{\frac{l+k-1}{2}}{l} x^{k-m}\\
=& M_{lm} (1-x^2)^{m/2} G_{l-m}(x)\\
=& M_{lm}K_{lm} (1-x^2)^{m/2} x^{l-m} H_{l-m-1}
\end{align}
Where $M_{lm}$ is a normalization constant, $G_{l-m}$ is a polynomial of order $l-m$ and
$$
G_{l-m} = K_{lm}(x^{l-m} + H_{l-m-1}(x))
$$
Where $K_{lm}$ is a constant and $H_{l-m-1}$ is a polynomial of order $l-m-1$.
If $x = \cos(\theta)$ then $(1-x^2)^{m/2} = (\sin(\theta))^m$. Let
$$
C_{lm} = N_{lm}M_{lm}K_{lm}
$$
We then have
$$
Y_l^m(\theta, \phi) = C_{lm} e^{im\phi}(\sin(\theta))^m\left((\cos(\theta))^{l-m} + H_{l-m-1}(\cos\theta)\right)
$$
We now re-express in cartesian components. For that we have
\begin{align}
e^{i\phi} =& \frac{x+i y}{\sqrt{x^2 + y^2}}\\
\sin(\theta) =& \frac{\sqrt{x^2 + y^2}}{\sqrt{x^2+y^2+z^2}} = \frac{\sqrt{x^2 + y^2}}{r}\\
\cos(\theta) =& \frac{z}{\sqrt{x^2 + y^2 + z^2}} = \frac{z}{r}
\end{align}
Importantly, we can see that $e^{i\phi}\sin(\theta) = (x+iy)/r$.
With this we get
\begin{align}
Y_l^m(x, y, z) =& C_{lm} \frac{(x+iy)^m}{r^m}\left(\frac{z}{r^{l-m}} + H_{l-m-1}\left(\frac{z}{r}\right)\right)\\
=& \frac{C_{lm}}{r^l}(x+iy)^m\left(z^{l-m} + r^{l-m}H_{l-m-1}\left(z/r\right)\right)
\end{align}
From this complex spherical harmonic we can extract two real spherical harmonics. These will be related to (the signs of the actual harmonics may differ from the expressions below because of various conventions)
\begin{align}
&\frac{C_{lm}}{r^l}\text{Re}\left\{(x+iy)^m\right\}\left(z^{l-m} + r^{l-m}H_{l-m-1}\left(z/r\right)\right)\\
&\frac{C_{lm}}{r^l}\text{Im}\left\{(x+iy)^m\right\}\left(z^{l-m} + r^{l-m}H_{l-m-1}\left(z/r\right)\right)\\
\end{align}
These are the expressions from which extract the subscript for the atomic orbital name.
Focusing on $(x+iy)^m$. For every term in this polynomial the exponents for $x$ and $y$ add up to $m$.
$\text{Re/Im}\left\{(x+iy)^m\right\}$ will in general include terms with both $x$ and $y$ but there is one term (or zero terms) that has the biggest exponent for $x$ and one term (or zero terms) that has the biggest exponent for $y$.
The procedure for coming up with the subscript is as follows.

$z^{l-m}$ will appear in the subscript. This means we ignore all of $H_{l-m-1}(z/r)$.
Take $\text{Re/Im}\left\{(x+iy)^m\right\}$ depending on which orbital you are considering. Find the term with the biggest $x$ exponent and the term with the biggest $y$ exponent and add these two terms together. You will have something like $Ax^ay^b + B x^cy^d$. If one of these terms is missing then one of $A$ or $B$ is zero.
Now take $Ax^ay^b + B x^c y^d$ and factor out any possible scalars and factors of $x$ and $y$. You will now have something like $A x^a y^b(x^c + B y^c)$ or $A x^a y^b(Bx^c + y^c)$ (these are not the same $A, B, a, b, c$ as just before).
Strip off $A$ from your final expression so you now have $x^a y^b (B x^c + y^c)$. This expression will appear in the subscript. There are two parts here, the factored out part $x^a y^b$ and the remaining prat $(Bx^c + y^c)$ or $(x^c + By^c)$
We now put together the three parts.

First put in the factored out part like $x^a y^b$ with $x$ preceding $y$.
Then put the $z^{l-m}$ part.
Finally put the remaining part $(B x^c + y^c)$ or $(x^c + B y^c)$.

This should result in a subscript $x^a y^b z^{l-m} (B x^c + y^c)$ or $x^a y^b z^{l-m} (x^c + B y^c)$

A few notes:
First, and importantly, this scheme does not give the same answer as Wikipedia for two orbitals listed on that page.

Wikipedia says $Y_{4, -3} = g_{zy^3} \propto (3x^2-y^2) yz$. I would say this should be $g_{yz(3x^2-y^2)}$.
Similarly, Wikipedia says $Y_{4, 3} = g_{zx^3} \propto (x^2-3y^2)xz$. I would say this should be $g_{xz(x^2-3y^2)}$.
My ordering also differs from that on Wikipeda. But honestly, i don't think they were following a pattern for the order on Wikipedia.

Wikipedia uses the same notation as The Orbitron, and I wonder if Wikipedia got their notation from this webpage. The Orbitron goes out to $i$ orbitals and I see similar discrepencies between the scheme used on that page and my approach for $g$ and $h$ orbitals. The author stops quoting abbreviated polynomials for $i$ orbitals for the relevant $m$ quantum numbers so I can't compare past that. I've reached out to the author of the Orbitron for further comment and I'm also trying to reach out to Wikipedia folks.
One advantage of my approach over whatever is going on on Wikipedia and the Orbitron is that the $x, y$ part of the coefficient is the same for orbitals with the same $m$ quantum number. They only differ in the factor of $z$.
Note that $a + b + c + l-m = l$, so all the exponents always add up to $l$ which is a nice spherical harmonic like result.
Finally, the issue remains of determining the $a$, $b$, $c$ and $B$ factors arising from $\text{Re/Im}\left\{(x+iy)^m\right\}$ alone. A computer could work these out without too much trouble but it would be nice to have closed form expressions for them. It should be possible to work it out using the binomial theorem.
Finally an alternative approach would be to simply take the terms with the biggest exponents in $\text{Re\Im}\{(x+iy)^m\}$ and add them if there max $x$ and $y$ exponents are the same, but this also doesn't agree with Wikipedia as it would replace something like $f_{y(3x^2-y^2)}$ with $f_{y^3}$, but it does explain wikipedia terms like $g_{zy^3}\propto (3x^2-y^2)yz$.
More notes: I've done a little more research and I really can't find any authoritative reference that gives names for orbitals higher than $f$. If anyone is familiar with one please let me know. I'm going to personally consider orbitals $g$ and higher to NOT have official cartesian polynomial abbreviation names and pursue and advertise the $l, m$ notation as being the proper way to refer to these higher orbitals. More motivation for this below.
For orbitals $f$ and below the $xy$ part of "abbreviated polynomial" subscript DOES allow you to identify the azimuthal nodal planes, for example, $p_x$ is zero when $x=0$ and $d_{xy}$ is zero when $x=0$ or $y=0$. This even holds for the intimidating $f_{x(x^2-3y^2)}$ which is zero if $x=0, \pm \sqrt{3}y$. However, when you move to orbitals with $m>4$ (for example $g$ orbitals with $(l, m) = (4, \pm)$), there are now 3 nodal planes, and any subscript that identifies all of them is going to require multiple terms and it is really not convenient to have expressions like that. It is likely for this reason authors don't typically give subscripts for orbitals at this order.
Frankly, for anyone reading this, my opinion is that the $l, m$ notation is far superior above basically the $p$ orbitals. The number $|m|$ tells you how many azimuthal nodal planes there are and the number $l - |m|$ tells you how many polar nodla planes there are. If $m$ is positive then the orbital has a set of lobes aligned along the $x$ axis. if $m$ is negative then the orbital is just like the one with $+m$ but rotated about the $z$ axis by $\pm 360/(4|m|)$ degrees ($\pm$ depending on if $m$ is positive or negative by the Condon-Shortley phase). No need to abbreviate a polynomial to convey this geometric information when it is already communicated by the $l, m$ numbers when properly understood.

"No need to abbreviate a polynomial to convey this geometric information when it is already communicated by the l,m numbers when properly understood." By the same logic, there's no need to name elements when the atomic number is properly understood, but names are still useful.

@Andrew I see your point. I guess the point is that, just like we only have "proper" names for atomic species up to $Z\approx 120$, we only have "cartesian polynomial names" for atomic orbitals up to $\ell=3$. It's a little annoying because there seems to be some usage for larger $\ell$, but it is not well documented or reference in the literature. So it gives the impression of "mystical" high order orbitals that have wacky names where noone knows where they come from. This is the impression which I am trying to eliminate.


text:
How salt concentration can effect the rate of corrosion and how redox reaction plays a role in that
sub-questions:
What role do oxidation and reduction play in corrosion? Is this a complicated process?
Also, why does increase in salt concentration make it easier to rust?

IF you want "all the details" then read a few books about corrosion.

To be short, the reaction is not between the iron and salt but between the iron and the oxygen from the air. The salt and the moisture (water) just opens the electrochemical pathway for this, when certain fragments of iron surface will have slightly different potentials (because of mechanical roughness, impurities, damaged coatings, etc.), thus creating a local electrochemical cell (a battery in a way).
The oxygen which is dissolved in water readily reduces on the cathode while the iron oxidizes on the anode and transfers into the salty moisture (the solution) as $\ce{Fe^2+}$ ions, which migrate to the cathode via solution and, together with the oxygen gas being reduced, form the rust. And $\ce{Fe^2+}$ also gets oxidized to $\ce{Fe^3+}$ a lot; the ferric oxide $\ce{Fe2O3}$ and the mixed oxide $\ce{Fe3O4}$ actually slow the corrosion process by forming a protective layer. Also, the overpotential of the oxygen plays a role (connected with the phase change, if I have not mistaken).
The chloride anions also are forming a soluble complex with $\ce{Fe^2+}$ which partially removes the $\ce{Fe^2+}$ ions from the oxidation process, hampering the formation of the ferric oxide protective layer. So the chloride ions have two roles: providing the better conductivity in the electrolyte which favours the electrochemical pathway of corrosion and the concurring reaction which slows the formation of the protective layer. See, e.g., Ahmad, Z. Concrete Corrosion. Principles of Corrosion Engineering and Corrosion Control 2006, 609–646. DOI: 10.1016/B978-075065924-6/50013-1.


text:
E and Z nomeclature when we have a ring instead of double bond
I am confused as to whether the E/Z nomenclature can be extended to compounds of general structure like in the image. Clearly they are some sort of cis and trans but I cannot see a non-aromatic double bond for which the Cahn–Ingold–Prelog precedences do differ here and why... It intuitively seems that the description should go with respect to the quinone cycle but how? Phane nomenclature?
Thank You in advance!

I expect that the situation is exactly analogous to the one given in P-92.4.2.2 of the 2013 IUPAC Blue Book here: https://iupac.qmul.ac.uk/BlueBook/P9.html#92040202 Formally, following through with the CIP rules as stated in there (you need to construct something called a 'digraph') will allow you to assign the double bonds as either (E,E) or (Z,Z) (but not (E,Z) or (Z,E)). I'm rather lazy to do this myself, though, so will hold off on a proper answer.

Thank You very much! I will try to write a proper answer then myself.

How can I credit You properly in the answer, @orthocresol?

No need to. I didn't quite answer; I pointed you in the right direction, at most, and you did all the work :)

Still, thank You very much, as my problem was solved by Your comment, my answer is rather an excercise for me and maybe something useful  for others :)

This link maybe helpful.

So, @orthocresol actually answered the question in their comment, and I am just adding a summary of their reference. Later, @user55119 commented about this wonderful resource with CIP rules explanations, including my case:
http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/pinene.html#diethylidene
Nevertheless, I already spent some time inventing the bicycle, so here it is. To sum up, Cahn–Ingold–Prelog turns out to be more involved than usually mentioned in classes (at least ones I went through when I studied) :-)
First one needs to construct a digraph, or a tree-diagram, showing the connectivity (topology) of all the atoms. In complete digraph, lone pairs are included, but usually there is no need for this. Also, H atoms can be omitted. In either or both cases it is a simplified digraph. Also, You can draw the spheres around the stereocenter but these are not mandatory for either type of digraph. For example, digraph for this molecule:

the simplified digraph would be:

Numbering can be either systematic according to the general IUPAC rules or arbitrary, just to rank chains (as here).
Now, for double bonds we create duplicate atoms:

And for cycles, we open them as close to the stereocenter as possible, and also add the duplicate atoms:

And, what is usually not taught, for mancude cycles (those with MAximum Number of ConjUgated DoublE bonds) we calculate the mean atomic number for all possible Kekule structures we can draw for that atom. This number is then retained instead of duplicate atom in the digraph. This becomes interesting for ionized species:

We can see how incredibly many times the same sequence repeats. This looks really awful but allows to assign the stereochemical designation in a non-ambiguous way.
Now, we should recall the priority rules for Cahn–Ingold–Prelog:

higher atomic number precedes lower one;
higher isotope mass precedes lower one;
Z precedes E for double bonds.

For the molecules I asked the question about, we should select the in-cycle atom of one of the double bonds as a stereocenter. Then, while constructing the digraph, we will see that going one or the other way around the central ring formally makes the other quinone double bond either E or Z. Now let us look back on the first double bond. As we now have one substituent with Z some atoms along the digraph branch and one substituent with E along its digraph branch, our first double bond now has non-equal substituents at the end which belongs to the cycle and thus it, the first double bond can be now assigned configuration for real. Which will be Z for the left molecule and E for the right one.

Now in theory we should repeat that for the other double bond at the center but we always will get E,E or Z,Z, so it is basically redundant.
Also, E and Z are sometimes referred to as seqTrans and seqCis, by the way. Classsic trans and cis always follow the hydrocarbon path.


text:
Distillation of water and propanone
I distilled a solution consisting of propanone and water and recorded the boiling point of propanone depending on different concentrations. I thought that the boiling point elevation would be the appropriate theory here which means that the slope of the graph should be around 1.80 (ebullioscopic constant of propanone) but apparently, the slope is only around 0.15. Is there a reason for this and how can I explain the nonlinearity of the graph?

Convert the chart to quantities with expected linearity like $p_i=p_{i,0}x_i$. For ebulioscopic measurement the solute is expected nonvolatile.

Additionally, if staying at ebulioscopic scenario of addition of water to propanone, you would use the constant for propanone, not water.

Yes, I've made the changes but I still don't understand why the graph deviates so much. Is it because the solution is not ideal? why does it deviate to the right?

What is the quantity reported on the $Ox$ axis ? Is it the molality of water in the liquid to be distilled ? Is it the molality of the condensed vapor ?

Partly deviation (guessed) from the Raoult law, which is common, partly nonlinear dependence of b.p. on linear vapour pressure change. Linearity is just approximation for small changes.


text:
Reaction between concentrated sulfuric acid and sodium bromide

What are the products of the reaction between sodium bromide and concentrated sulfuric acid?

I did the question like this:
$$
\begin{align}
\ce{2Br- &-> Br2 + 2e-} \tag{R1.1} \\
\ce{SO4^2- 4H+ + 2e- &-> SO2 + 2H2O} \tag{R1.2} \\
\hline
\ce{SO4^2- + 2Br- + 4H+ &-> SO2 + Br2 + 2H2O} \tag{R1.3}
\end{align}
$$
Why is the answer
$$\ce{NaBr + H2SO4 -> NaHSO4(s) + HBr(g)}? \tag{R2}$$

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E.g. type $\ce{2e- +4H+ + SO4^2- -> SO2 + 2H2O}$ to get $\ce{2e- +4H+ + SO4^2- -> SO2 + 2H2O}$. That is for inline formula or equation within the text. To display it in dedicated space, use double dollars: $$\ce{2e- +4H+ + SO4^2- -> SO2 + 2H2O}$$ // Use 4 trailing spaces for forced newline, or double enter for an empty line.

Related: https://chemistry.stackexchange.com/questions/4750/why-do-sodium-halides-react-so-differently-with-sulfuric-acid

$\ce{H2SO4}$ reacts with $\ce{NaBr}$ in two steps. First $$\ce{H2SO4 + NaBr -> NaHSO4 + HBr \tag{1}}$$ This reaction occurs at all conditions of temperature and concentrations, even in dilute solutions, where the reagents and products are transformed into ions. Later on, if the sulfuric acid is hot and concentrated enough, $\ce{HBr}$ produced by the reaction ($1$) is partially oxidized by $\ce{H2SO4}$ according to a second equation $$\ce{2 HBr + H2SO4 <=> Br2 + SO2 + 2 H2O \tag{2}}$$ An excess of sulfuric acid helps getting rid of the $2\ce{H2O}$ molecules produced in ($2$) in a rection like $$\ce{H2SO4 + H2O -> HSO4^- + H3O+}$$ and so improves the yield in $\ce{Br2}$, as ($2$) in replaced by ($3$) $$\ce{2HBr + 2 H2SO4 -> Br2 + SO2 + H3O+ + HSO4- \tag{3}}$$

"This reaction occurs at all conditions of temperature and concentrations, even in dilute solutions" ||| What about chlorides and iodides? Why would the halides be protonated? Is this related to displacement of weak acid by strong acid? But aren't HCl, HBr, HI stronger than H2SO4? By the way, isn't NaHSO4 soluble? If we're in an aqueous environment, why would hydrogen halide be given off? Won't it just dissolve? Sorry ffor this barrage!

Why must H2O be removed?

@Cheng: I think it's better if you include this comment in your question.

Nobody can say that "$\ce{H2O}$ must be produced". I think that whatever the nature of the halogen $\ce{X}$, the reaction of $\ce{H2SO4}$ on $\ce{NaX}$ is the production of $\ce{HX}$. But this $\ce{HX}$ stays as it is, if $\ce{X}$ is fluorine and chlorine. But if $\ce{X}$ is bromine or iodine, the $\ce{HBr}$ or $\ce{HI}$ produced reacts with a new $\ce{H2SO4}$ molecule so that $\ce{Br}$ or $\ce{I}$ is oxidized into $\ce{Br2}$ and $\ce{I2}$ by this supplementary $\ce{H2SO4}$ molecule, which is reduced into $\ce{SO2}$ (and $\ce{H2O}$)

I mean how is removing H2O supposed to improve yield?

"This reaction occurs at all conditions of temperature and concentrations, even in dilute solutions, where the reagents and products are transformed into ions." do you have a reference for this?


text:
How does pH affect an extraction problem
I am not sure how to understand when pH will affect a given problem regarding extraction. For example, in a problem given that K$_D$ of the reaction is $1.86$ I am asked to calculate:

(a) How much propionic acid is left in octanol if you shake $100$ ml of octanol as
contains $1.0$ mM propionic acid with a $50$ ml aqueous solution buffered to pH $2$?
(b) How much propionic acid is left in octanol if you shake $100$ ml of octanol as
contains $1.0$ mM propionic acid with a $50$ ml aqueous solution buffered to pH $7$?

For these problems it is stated that "At pH $2$, only AH will be formed in water and we can ignore the proton equilibrium, so that n$_0$ = [AH]$_{aq}$ V$_{aq}$ + [AH]$_{org}$ V$_{org}$" but "
At pH $7$, A- is no longer negligible and n$_0$ = [AH]$_{aq}$ V$_{aq}$ + [AH]$_{org}$ V$_{org}$ +[A-]$_{aq}$ V$_{aq}$"
How does one know this?
The same is goes for the problem:

Propionic anhydride can be extracted from cyclohexane to water. The distribution equilibrium can be written as:
PRANH (org) <-> PRANH (H2O) K = $0.82$
In aqueous solution, propionic anhydride can be hydrolyzed to propionic acid (PRIC ACID) according to the following equilibrium:
PRANH + H2O <-> 2 PRIC ACID K = $0.0023$ M (at pH <$3.0$)
A cyclohexane solution contains 1$0$ mM propionic anhydride. At what volume ratio of cyclohexane to water will $80$% of the propionic anhydride be extracted over to the aqueous phase (pH $2$) with a shake?

Here I assume that I should either get if the pH doesn't matter:
$1-p = \frac{1}{1+ K_D V_{aq}/V_{org}}$
Or if pH matters then n$_0$ = [PRANH]$_{aq}$ V$_{aq}$ + [PRANH]$_{org}$ V$_{org}$ +[PRIC ACID]$_{aq}$ V$_{aq}$ and I would get that:
$1-p = \frac{1}{1+ K_D* V_{aq}/V_{org}+ [PRIC ACID]_{aq}/[PRANH]_{org}*V_{aq}/V_{org}}$
where $1$-p=$1$-$0.8$=$0.2$
I would assume that the second is correct (if either is correct) since it is given that PRANH hydrolyzes at pH<$3$ so if we're at pH=$2$ then that should occur. But am I correct in assuming that? But then what happens to the equilibrium constant given for the hydrolyzation, do I use that for something or should I only take into account the K$_D$ = $0.82$ for the extraction process?
Thankful for all help!

$\ce{HA(Oct)<=>HA(aq)<=>H+(aq) + A-(aq)}$. The former eq. is determined by KD, the latter by the acidity constant. What is unclear about that? Similarly for PRANH hydrolysis.

@Poutnik How do I from the way that the question is stated know when to take into account the ions and when not to as in the first question? For the PRANH hydrolysis I assume that I should take into account the hydrolysis but is the way that I am applying this correct? I have just "copied" what I understood from the first question to the hydrolysis question..

You use MathJax formatting in a weird way. // Useful links for text and formula formatting (not to be applied to titles):
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Generally, components being extracted can undergo in one phase a reaction to a form inactive in extraction. Typically salts do not like to be ectracted  to nonpolar solvents. The reactions often follow an equilibrium equation. From give equilibrium constants and eventually pH, there is calculated nonactive/total form ratio. If the ratio << 1, this equilibrium reaction can be neglected for the extraction evaluation. Otherwise it must be taken into consideration.

Or, you can define conditional distribution constant [A_org]/[A_water, total],  involving the mentioned ratio.


text:
Is sodium superoxide more abundant than sodium oxide?
Reference(NCERT book of India);
My professor has taught me that $\ce{Na2O}$ and $\ce{Na2O2}$ are the most common oxides of sodium, but in my reference book it is written that they are $\ce{NaO2}$ and $\ce{Na2O2}$. I have searched the net but couldn't find anything about their relative occurrence (percentage). Can anyone please clarify which of the above is correct?

What is your reference book? $\ce{NaO2}$ sounds farfetched... but exists: https://en.wikipedia.org/wiki/Sodium_superoxide Perhaps the question is, under which circumstances are they "most common"?

@BuckThorn I have edited my answer to include the screenshot.

Sone links for you: https://chemistry.stackexchange.com/questions/50796/why-do-the-alkali-metals-form-different-products-upon-combustion-in-air ... https://chemistry.stackexchange.com/questions/8387/stability-of-superoxides-of-alkali-metals/8813 ... https://chemistry.stackexchange.com/questions/24762/why-does-potassium-form-peroxides-but-sodium-does-not

Sodium superoxide $\ce{NaO2}$ is a rare substance that can only be made by burning sodium at high pressure. Sodium peroxide $\ce{Na2O2}$ is common: it is formed when metallic sodium is burning in oxygen or air at ordinary pressure. The usual (and "logical") monoxide $\ce{Na2O}$ is not common. It is the result of the pyrolysis of different salts, like sodium carbonate, at very high temperature ($\sim2000\ \mathrm{^\circ C}$). Both the Handbook of Chemistry and Physics and the Merck Index give the properties of $\ce{Na2O}$ and $\ce{Na2O2}$: density, melting point. The superoxide $\ce{NaO2}$ is not mentioned, at least in my edition.

@Maurice.Thanks for your answer,but can we comment which among $\ce{NaO2 and Na2O }$ is more common?

"The superoxide NaO2 is not mentioned" is not true. Sodium superoxide is entry 2650 in the CRC Handbook of Chemistry and Physics (96th edition) with mp 552 °C and density 2.2 g/ml. Besides, I'm not sure the missing database entry is a good argument that speaks anything about the abundance. Also, OP indeed asks for comparison between $\ce{NaO2}$ and $\ce{Na2O},$ not $\ce{Na2O2}.$


text:
Molar conductivity
Among $\ce{H+, Li+, K+, Rb+, Cs+}$ in an aqueous medium at infinite dilution at 25 °C, which ions have more ionic molar conductivities than that of $\ce{Na+}$?
I know that Molar conductivity is directly proportional to mobility, which is inversely proportional to hydration energy. By that logic, $\ce{K+, Rb+}$, and $\ce{Cs+}$ have molar conductivities that are greater than $\ce{Na+}$'s, but turns out, $\ce{H+}$'s molar conductivity is also higher in comparison with that of $\ce{Na+}$'s. Why is that so?

Do you know the ancient fable about a race run between a hare/rabbit and a turtle? Apply it.

@Poutnik Slow and steady wins the race? But here, the faster the ion moves, more is it's conductivity right?

No, try to fill the gaps. :-) The turtle won, because of cheating. The turtle ( or rather a land tortoise instead of a sea turtle)   that started the race was not the same as the one that finished it.

High mobility of hydronium and hydroxide ions is directly related to proton exchange between water molecules and respective ions. See also https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.05%3A_Hydrogen_and_Hydroxide_Ions

A similar question: https://chemistry.stackexchange.com/questions/64984/determining-the-order-of-molar-conductivity/65005#65005

The Grotthuss mechanism nicely illustrates the proton exchange on the embedded animated GIF image.


text:
Parabolic Arrhenius plot
I am currently doing a kinetic study on an enzyme catalysed reaction and plotting the Arrhenius graph ($\ln k$ vs $1/T$) and I got a parabola. My idea is to take the derivative at each point, meaning there would be different activation energies depending on the temperature, guessing it is due to enzymes dependency to heat. Does it make sense or would it be data forcing?
The parabola has the following shape:

so I am 99% sure it is a parabola. The only other possibility I see is to divide it in two straight lines but I don't know how to interpret it.
And the data is the following:

1/T
ln(k)

0.029412
3.35

0.028571
3.42

0.027778
4.18

0.027027
4.95

0.026316
5.67

0.025641
3.97

0.025
3.12

0.024390
1.41

0.023809
0.88

Temperature range is from 34 up to 42ºC (scale: 1ºC); 5 tries per temperature
The experiment was the following:
I prepared a 1% concentration solution of amylase and a 1% concentration solution of starch. I added lugol to starch in order to get a purple colored complex. I heated both amylase and starch to the desired temperature and then I let them react in a spectrophotometer, getting an absorbance vs time graph of this kind:

Using Beer-Lambert law I obtained the concentration vs time graph, and then I calculated the initial rate taking the derivative at $t=0$ of the best fit curve. Finally I computed $k$ and I plotted the Arrhenius graph

Much depends not just on the enzyme but on the T range, number of points. Is it a smooth curve? You can just fit a polynomial in that case.

I am going to modify the question in order to add this data, thank you for your answers !

I don't suppose you performed any of the measurements multiple times? It certainly looks like the rate peaks, but I wouldn't hazard a guess of the functional form from this data. It's too scattered for that, especially w/o uncertainty estimates on the data points (nice data all the same). Not sure it makes sense to think in terms of Arrhenius constants with data that peaks like this and with such a narrow T range.

What you are observing is a typical manifestation of an "Arrhenius break" (at the transition temperature $38;^\circ \rm C$) because of the conformational changes of enzyme due to denaturation. To calculate $E_a$, you would usually disregard the values above the transition temperature and use only values below it, which – as can be clearly seen from your plot – lie on a nice line as the Arrhenius equation predicts. Take a look at this exemplary article.

You may yet be able to find a more complex model: https://edepot.wur.nl/546299

I did not know about Arrhenius break but it seems you are completely right. Thank you very much to all for your help, have a nice day !


text:
Is boiling aqueous solution of copper(II) sulfate safe?
I want to create saturated coper(II) sulfate solution in order to grow crystals. Is boiling a solution of copper sulfate at home safe? Are there any risks? Is the vapour pure water?

It is safe. I have done it quite often. The vapor is pure water. It is difficult get nice crystals.

It will be hot, do not touch it.

One aspect of knowing what you're doing is knowing the properties relevant to your process, such as the compound you want may not be fully stable at the boiling point of water (or whatever solvent you're boiling off). Here, boiling is not best after all. See the answers below.

While the hazard associated with boiling copper sulfate solution is minimal (there can be fine aerosols of the solution carried by the bubbles during a vigorous boiling), the chemistry must be respected. To get beautiful blue crystals of the pentahydrate you may want to evaporate the solution at lower temperature (and thus more slowly) than boiling the solution. From Wikipedia:

Copper(II) sulfate pentahydrate decomposes before melting. It loses two water molecules upon heating at 63 °C (145 °F), followed by two more at 109 °C (228 °F) and the final water molecule at 200 °C (392 °F).[1][2]

Cited References
1.
Andrew Knox Galwey; Michael E. Green (1999). Thermal decomposition of ionic solids. Elsevier. pp. 228–229. ISBN 978-0-444-82437-0.
2.
Wiberg, Egon; Nils Wiberg; Arnold Frederick Holleman (2001). Inorganic chemistry. Academic Press. p. 1263. ISBN 978-0-12-352651-9.


text:
Ksp and Precipitation

A scientist slowly adds dissolved NaCl to form a solution of AgNO3. If a precipitate begins to form when the NaCl concentration is $5.5*10^{-4}$ M, what is the concentration of Ag$^+$ in the solution? K$_{sp}$ = $1.8*10^{-10}$.

A) $3.3*10^{-7}$ M
B) $9.39*10^{-14}$ M
C) $3.1*10^{6}$ M
D) $6.0*10^{-6}$ M
E) $6.0*10^{-8}$ M
I've been having trouble with this problem. I think it has something to do with the common ion effect, but I'm not sure.
Here's my current understanding:
The reaction is:  $\hspace 5mm$  NaCl(aq) + AgNO3 (aq) $\rightarrow$ NaNO3 (aq) + AgCl (s)
This makes the net ionic reaction:  $\hspace 5mm$ Cl$^-$ (aq) + Ag$^+$ (aq) $\rightarrow$ AgCl (s)
Whenever the concentration of NaCl, and therefore Cl$^-$ ions reaches $5.5*10^{-4}$ M, the Chlorine ions start reacting with the Silver ions. My best guess is that is when the maximum solubility is reached for AgCl, AgCl has to start precipitating out due to le Chatelier's principle. This means that [Ag$^{+}$]*$5.5*10^{-4}$ = $1.8*10^{-10}$, meaning that [Ag$^{+}$] = $3.27273*10^{-4}$ M, for an answer of A. Is this a valid line of reasoning?

Yes, that's pretty much correct.

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See also https://en.wikipedia.org/wiki/Solubility_equilibrium


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Distribution of molecules in a crystal
If I dissolve 0.999 mol NaCl and 0.001 mol citric acid in water and then evaporate the water, will the citric acid be uniformly distributed in the resulting crystals?
I found this post which suggest that it will result in a solid two-phase mixture but I am not 100% sure how this translates to my problem since citric acid is quite different from KCl.

See also https://en.wikipedia.org/wiki/Coprecipitation

If I dissolve 0.999 mol NaCl and 0.001 mol citric acid in water and then evaporate the water, will the citric acid be uniformly distributed in the resulting crystals?

No. First, sodium chloride crystals will form while the citric acid stays in solution (in the example, there is much more sodium chloride, and citric acid has a high solubility in water).
Once citric acid reaches its solubility and sufficient supersaturation to form citric acid nuclei, the citric acid will also form crystals, separately from the sodium chloride crystals.

I found this post which suggest that it will result in a solid two-phase mixture but I am not 100% sure how this translates to my problem since citric acid is quite different from KCl.

The post gives multiple answers; in combination they describe that while there might be some defects in the NaCl crystals showing an occasional potassium ion, there will be two distinct types of crystals.
However, if you have a mixture of very similar particles (such as different isotopes of a simple ion), you will get a single type of crystal with the two similar particles randomly distributed. There are also mixed or double salts where there is an ordered, stoichiometric mixture of cations or anions in a single crystal.

Thank you. Is there any way to grow crystals from so different molecules with a uniform distribution?


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Does Boltzmann distribution refers to microstates or particle's probability?
I am reading about the Boltzmann distribution and I can't grasp the following:
$$p_i = e^{-\frac{E_i}{k_B T}}$$
Is this the probability finding the system at a microstate with Energy $E_i$ or the probability finding a particle with energy $E_i$ (useful for transitions in spectroscopy etc).
This distribution is for canonical ensemble where system has constant $(T, V, N)$ (coupled to a heat bath). The probabilities for the microstates of the isolated system (system+heat bath) are equal whereas the probability for finding system A in a particular microstate in general will not (in accordance with the formula). So if we map the different microstates in distributions (i.e. the set of all microstates of system with a specific energy) the formula is interpreted as the fraction of microstates of A with the specific energy.
For example we can have $3$ distributions $D_1, D_2, D_3$ with energies $E_1, E_2, E_3$ respectively and each containing $Ω_1, Ω_2, Ω_3$ microstates. The probability therefore $p_i$ therefore, is simply the ratio:
$$\frac{Ω_i}{Ω_{system}}$$
This is how I understand it (maybe wrong). What I don't get is the connection between probabilities of microstates and the probabilities for particles (if any).


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Solubility of Silver Sulfate in Water
Q. What is the solubility of silver sulfate (K_sp=1.77x10^-5) in water, in g/L?
From my own working and other sources, we know that the molar solubility of silver sulfate is 0.0144mol/L. We also know that the molar mass of silver sulfate is 311.796g/mol, so I thought the solution would be m=nxMM=4.49g of silver sulfate per litre of water. However, whenever I google the solubility of silver sulfate at room temperature, it says that the solubility is 8.3 g/L? It seems that I'm out by a factor of 2 - where have I gone wrong?

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The solubility of sulfates is always bigger than what can be obtained from solubility products, because of the hydrolysis reaction between sulfate ions and water like $$\ce{SO4^{2-} + H2O <=> HSO4^- + OH-}$$ As a consequence, for silver sulfate : $$\ce{[Ag+] = 2 \frac{8.3g/L}{311.8 g/mol} = 0.0532 M = [HSO4^{-}] + \frac{[SO4^{2-}]}{2} > \frac{[SO4^{2-}]}{2} = \frac{0.0144 M}{2}  }$$ So, your calculations are perfect. If you do want it, you can even calculate the amount of the sulfate ions that have been hydrolyzed.

Thank you Poutnik for showing me my ridiculous mistake !

We all are unbelievably blind sometimes. :-)


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Oxidation of amino acids with N-bromosuccinimide
I've recently come across the reaction where an amino acid is oxidised to form an imine by using N-bromosuccinimide as reagent and then in hydrolysis it formed an aldehyde moiety.

The second step is well known (Stephan's aldehyde synthesis using $\ce{SnCl2}$ and $\ce{HCl}$) and I understand that. But I'm unable to understand the mechanism of the first step, how can NBS as a reagent do that?
My initial guess was that  as nucleophilicity and electronegativity are inversely proportional and oxygen is more electronegative than nitrogen from the order of E.N, we can say that nitrogen of $\ce{NH2}$ is more nucleophilic than carboxylic acid oxygen and hence nitrogen's free electrons forms a bond with the bromine of NBS but I'm unable to understand how to proceed next and how is $\ce{CO2}$ formed as products?
Can anyone help with the mechanism? I'm thinking it should be an elimination type mechanism in next step,But I'm not really sure!
Also the yield of product with this method of using NBS as reagent is considerable amount?

Is the aldehyde correctly rendered? Product seemingly should be $\ce{R - CHO}$ with no bend between.

The process is an oxidation, not a reduction of the amino acid.

True I forgot that should  be R-CHO, I've made a mistake

The first step is, as you correctly suggest, the N-bromination of the aminoacid as reported by Ramachandran et al here. An electrocyclic reaction then occurs to eliminate CO2 and HBr leaving the imine which on hydrolysis gives the aldehyde and NH3. Image from Wikipedia